https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Wlm7&feedformat=atom AoPS Wiki - User contributions [en] 2021-12-05T03:06:14Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_13&diff=137894 2020 AMC 8 Problems/Problem 13 2020-11-20T00:04:11Z <p>Wlm7: /* Solution 1 */</p> <hr /> <div>Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> After Jamal adds &lt;math&gt;x&lt;/math&gt; purple socks, he has &lt;math&gt;18+x&lt;/math&gt; purple socks and &lt;math&gt;6+18+12+x=x+36&lt;/math&gt; total socks, for a probability of drawing a purple sock of &lt;cmath&gt;\dfrac{18+x}{36+x}=\dfrac{3}{5}.&lt;/cmath&gt; Since &lt;math&gt;\dfrac{18+9}{36+9}=\dfrac{27}{45}=\dfrac35&lt;/math&gt;, the answer is &lt;math&gt;\boxed{\textbf{(B) }9}&lt;/math&gt;. ~icematrix<br /> <br /> ==Solution 2==<br /> The total number of socks that Jamal has is &lt;math&gt;\, 6+18+12=36&lt;/math&gt; socks. We are trying to determine how many purple socks he added to his drawer. Let's say he adds &lt;math&gt;x&lt;/math&gt; purple socks. This means that the total number of purple socks in his drawer will be &lt;math&gt;(18+x)&lt;/math&gt; and the new total number of socks in his drawer will be &lt;math&gt;(36+x)&lt;/math&gt;. The ratio of purple socks to total socks in his drawer is now &lt;math&gt;\frac{60}{100}=\frac{3}{5}&lt;/math&gt;. This leads to the equation &lt;math&gt;\frac{18+x}{36+x}=\frac{3}{5}&lt;/math&gt;. Cross multiplying this equation gives us &lt;math&gt;90+5x=108+3x \implies 2x=18 \implies x=9&lt;/math&gt;. Thus, Jamal added 9 purple socks &lt;math&gt;\implies\boxed{\textbf{(B) }9}&lt;/math&gt;.&lt;br&gt;<br /> ~[http://artofproblemsolving.com/community/user/jmansuri junaidmansuri]<br /> <br /> ==Solution 3==<br /> <br /> Let Jamal add &lt;math&gt;x&lt;/math&gt; more purple socks. Then we are told that &lt;math&gt;\frac{18+x}{6+18+12+x}=\frac35&lt;/math&gt;. Cross multiplying and simplifying tells us that &lt;math&gt;x=\textbf{(B)}9&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Video Solution==<br /> https://youtu.be/x9Di0yxUqeU<br /> <br /> ~savannahsolver<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137893 2020 AMC 8 Problems 2020-11-20T00:00:29Z <p>Wlm7: /* Problem 22 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137892 2020 AMC 8 Problems 2020-11-20T00:00:19Z <p>Wlm7: /* Problem 22 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137891 2020 AMC 8 Problems 2020-11-19T23:59:59Z <p>Wlm7: /* Problem 22 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\dfrac{N}{2}$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137890 2020 AMC 8 Problems 2020-11-19T23:59:38Z <p>Wlm7: /* Problem 22 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\dfrac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137889 2020 AMC 8 Problems 2020-11-19T23:59:16Z <p>Wlm7: /* Problem 21 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35),grey);<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137888 2020 AMC 8 Problems 2020-11-19T23:57:34Z <p>Wlm7: /* Problem 20 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.5cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.5cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.5cm}{0.15mm}\text{ .}2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35));<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems&diff=137887 2020 AMC 8 Problems 2020-11-19T23:56:32Z <p>Wlm7: /* Problem 20 */</p> <hr /> <div>==Problem 1==<br /> Luka is making lemonade to sell at a school fundraiser. His recipe requires &lt;math&gt;4&lt;/math&gt; times as much water as sugar and twice as much sugar as lemon juice. He uses &lt;math&gt;3&lt;/math&gt; cups of lemon juice. How many cups of water does he need?<br /> <br /> &lt;math&gt;\textbf{(A) } 6\qquad\textbf{(B) } 8\qquad\textbf{(C) } 12\qquad\textbf{(D) } 18\qquad\textbf{(E) } 24\qquad&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Four friends do yardwork for their neighbors over the weekend, earning &lt;math&gt;\$15, \$20, \$25,&lt;/math&gt; and &lt;math&gt;\$40,&lt;/math&gt; respectively. They decide to split their earnings equally among themselves. In total how much will the friend who earned &lt;math&gt;\$40&lt;/math&gt; give to the others?<br /> <br /> &lt;math&gt;\textbf{(A) }\$5 \qquad \textbf{(B) }\$10 \qquad \textbf{(C) }\$15 \qquad \textbf{(D) }\$20 \qquad \textbf{(E) }\$25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Carrie has a rectangular garden that measures &lt;math&gt;6&lt;/math&gt; feet by &lt;math&gt;8&lt;/math&gt; feet. She plants the entire garden with strawberry plants. Carrie is able to plant &lt;math&gt;4&lt;/math&gt; strawberry plants per square foot, and she harvests an average of &lt;math&gt;10&lt;/math&gt; strawberries per plant. How many strawberries can she expect to harvest?<br /> <br /> &lt;math&gt;\textbf{(A) }560 \qquad \textbf{(B) }960 \qquad \textbf{(C) }1120 \qquad \textbf{(D) }1920 \qquad \textbf{(E) }3840&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?<br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(250);<br /> real side1 = 1.5;<br /> real side2 = 4.0;<br /> real side3 = 6.5;<br /> real pos = 2.5;<br /> pair s1 = (-10,-2.19);<br /> pair s2 = (15,2.19);<br /> pen grey1 = rgb(100/256, 100/256, 100/256);<br /> pen grey2 = rgb(183/256, 183/256, 183/256);<br /> fill(circle(origin + s1, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> draw(side1*dir(60*i)+s1--side1*dir(60*i-60)+s1,linewidth(1.25));<br /> }<br /> fill(circle(origin, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i),1), grey2);<br /> draw(side2*dir(60*i)--side2*dir(60*i-60),linewidth(1.25));<br /> }<br /> fill(circle(origin+s2, 1), grey1);<br /> for (int i = 0; i &lt; 6; ++i) {<br /> fill(circle(pos*dir(60*i)+s2,1), grey2);<br /> fill(circle(2*pos*dir(60*i)+s2,1), grey1);<br /> fill(circle(sqrt(3)*pos*dir(60*i+30)+s2,1), grey1);<br /> draw(side3*dir(60*i)+s2--side3*dir(60*i-60)+s2,linewidth(1.25));<br /> }<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of &lt;math&gt;5&lt;/math&gt; cups. What percent of the total capacity of the pitcher did each cup receive?<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> How many integers between &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;2400&lt;/math&gt; have four distinct digits arranged in increasing order? (For example, &lt;math&gt;2347&lt;/math&gt; is one integer.)<br /> <br /> &lt;math&gt;\textbf{(A) }\text{9} \qquad \textbf{(B) }\text{10} \qquad \textbf{(C) }\text{15} \qquad \textbf{(D) }\text{21}\qquad \textbf{(E) }\text{28}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Ricardo has &lt;math&gt;2020&lt;/math&gt; coins, some of which are pennies (&lt;math&gt;1&lt;/math&gt;-cent coins) and the rest of which are nickels (&lt;math&gt;5&lt;/math&gt;-cent coins). He has at least one penny and at least one nickel. What is the difference in cents between the greatest possible and least possible amounts of money that Ricardo can have?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{8062} \qquad \textbf{(B) }\text{8068} \qquad \textbf{(C) }\text{8072} \qquad \textbf{(D) }\text{8076}\qquad \textbf{(E) }\text{8082}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Akash's birthday cake is in the form of a &lt;math&gt;4 \times 4 \times 4&lt;/math&gt; inch cube. The cake has icing on the top and the four side faces, and no icing on the bottom. Suppose the cake is cut into &lt;math&gt;64&lt;/math&gt; smaller cubes, each measuring &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; inch, as shown below. How many small pieces will have icing on exactly two sides?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> currentprojection=orthographic(1.75,7,2);<br /> <br /> //++++ edit colors, names are self-explainatory ++++<br /> //pen top=rgb(27/255, 135/255, 212/255);<br /> //pen right=rgb(254/255,245/255,182/255);<br /> //pen left=rgb(153/255,200/255,99/255);<br /> pen top = rgb(170/255, 170/255, 170/255);<br /> pen left = rgb(81/255, 81/255, 81/255);<br /> pen right = rgb(165/255, 165/255, 165/255);<br /> pen edges=black;<br /> int max_side = 4;<br /> //+++++++++++++++++++++++++++++++++++++++<br /> <br /> path3 leftface=(1,0,0)--(1,1,0)--(1,1,1)--(1,0,1)--cycle;<br /> path3 rightface=(0,1,0)--(1,1,0)--(1,1,1)--(0,1,1)--cycle;<br /> path3 topface=(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle;<br /> <br /> for(int i=0; i&lt;max_side; ++i){<br /> for(int j=0; j&lt;max_side; ++j){<br /> <br /> draw(shift(i,j,-1)*surface(topface),top);<br /> draw(shift(i,j,-1)*topface,edges);<br /> <br /> draw(shift(i,-1,j)*surface(rightface),right);<br /> draw(shift(i,-1,j)*rightface,edges);<br /> <br /> draw(shift(-1,j,i)*surface(leftface),left);<br /> draw(shift(-1,j,i)*leftface,edges);<br /> <br /> }<br /> }<br /> <br /> picture CUBE;<br /> draw(CUBE,surface(leftface),left,nolight);<br /> draw(CUBE,surface(rightface),right,nolight);<br /> draw(CUBE,surface(topface),top,nolight);<br /> draw(CUBE,topface,edges);<br /> draw(CUBE,leftface,edges);<br /> draw(CUBE,rightface,edges);<br /> <br /> int[][] heights = {{4,4,4,4},{4,4,4,4},{4,4,4,4},{4,4,4,4}};<br /> <br /> for (int i = 0; i &lt; max_side; ++i) {<br /> for (int j = 0; j &lt; max_side; ++j) {<br /> for (int k = 0; k &lt; min(heights[i][j], max_side); ++k) {<br /> add(shift(i,j,k)*CUBE);<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }\text{12} \qquad \textbf{(B) }\text{16} \qquad \textbf{(C) }\text{18} \qquad \textbf{(D) }\text{20}\qquad \textbf{(E) }\text{24}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Zara has a collection of &lt;math&gt;4&lt;/math&gt; marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> After school, Maya and Naomi headed to the beach, &lt;math&gt;6&lt;/math&gt; miles away. Maya decided to bike while Naomi took a bus. The graph below shows their journeys, indicating the time and distance traveled. What was the difference, in miles per hour, between Naomi's and Maya's average speeds?<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> unitsize(1.25cm);<br /> dotfactor = 10;<br /> pen shortdashed=linetype(new real[] {2.7,2.7});<br /> <br /> for (int i = 0; i &lt; 6; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> draw((i,0)--(i,6), grey);<br /> draw((0,j)--(6,j), grey);<br /> }<br /> }<br /> <br /> for (int i = 1; i &lt;= 6; ++i) {<br /> draw((-0.1,i)--(0.1,i),linewidth(1.25));<br /> draw((i,-0.1)--(i,0.1),linewidth(1.25));<br /> label(string(5*i), (i,0), 2*S);<br /> label(string(i), (0, i), 2*W); <br /> }<br /> <br /> draw((0,0)--(0,6)--(6,6)--(6,0)--(0,0)--cycle,linewidth(1.25));<br /> <br /> label(rotate(90) * &quot;Distance (miles)&quot;, (-0.5,3), W);<br /> label(&quot;Time (minutes)&quot;, (3,-0.5), S);<br /> <br /> dot(&quot;Naomi&quot;, (2,6), 3*dir(305));<br /> dot((6,6));<br /> <br /> label(&quot;Maya&quot;, (4.45,3.5));<br /> <br /> draw((0,0)--(1.15,1.3)--(1.55,1.3)--(3.15,3.2)--(3.65,3.2)--(5.2,5.2)--(5.4,5.2)--(6,6),linewidth(1.35));<br /> draw((0,0)--(0.4,0.1)--(1.15,3.7)--(1.6,3.7)--(2,6),linewidth(1.35)+shortdashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }12 \qquad \textbf{(C) }18 \qquad \textbf{(D) }20 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> For a positive integer &lt;math&gt;n,&lt;/math&gt; the factorial notation &lt;math&gt;n!&lt;/math&gt; represents the product of the integers<br /> from &lt;math&gt;n&lt;/math&gt; to &lt;math&gt;1.&lt;/math&gt; (For example, &lt;math&gt;6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1.&lt;/math&gt;) What value of &lt;math&gt;N&lt;/math&gt; satisfies the following equation?<br /> &lt;cmath&gt;5! \cdot 9! = 12 \cdot N!&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }10 \qquad \textbf{(B) }11 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Jamal has a drawer containing &lt;math&gt;6&lt;/math&gt; green socks, &lt;math&gt;18&lt;/math&gt; purple socks, and &lt;math&gt;12&lt;/math&gt; orange socks. After adding more purple socks, Jamal noticed that there is now a &lt;math&gt;60\%&lt;/math&gt; chance that a sock randomly selected from the drawer is purple. How many purple socks did Jamal add?<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \qquad \textbf{(D) }18 \qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> There are &lt;math&gt;20&lt;/math&gt; cities in the County of Newton. Their populations are shown in the bar chart below. The average population of all the cities is indicated by the horizontal dashed line. Which of the following is closest to the total population of all &lt;math&gt;20&lt;/math&gt; cities?<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(300);<br /> <br /> pen shortdashed=linetype(new real[] {6,6});<br /> <br /> // axis<br /> draw((0,0)--(0,9300), linewidth(1.25));<br /> draw((0,0)--(11550,0), linewidth(1.25));<br /> <br /> for (int i = 2000; i &lt; 9000; i = i + 2000) {<br /> draw((0,i)--(11550,i), linewidth(0.5)+1.5*grey);<br /> label(string(i), (0,i), W);<br /> }<br /> <br /> <br /> for (int i = 500; i &lt; 9300; i=i+500) {<br /> draw((0,i)--(150,i),linewidth(1.25));<br /> if (i % 2000 == 0) {<br /> draw((0,i)--(250,i),linewidth(1.25));<br /> }<br /> }<br /> <br /> int[] data = {8750, 3800, 5000, 2900, 6400, 7500, 4100, 1400, 2600, 1470, 2600, 7100, 4070, 7500, 7000, 8100, 1900, 1600, 5850, 5750};<br /> int data_length = 20;<br /> <br /> int r = 550;<br /> for (int i = 0; i &lt; data_length; ++i) {<br /> fill(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0)--cycle, 1.5*grey);<br /> draw(((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+1)*r,0)--((i+1)*r, data[i])--((i+2)*r-100, data[i])--((i+2)*r-100,0));<br /> }<br /> <br /> draw((0,4750)--(11450,4750),shortdashed);<br /> <br /> label(&quot;Cities&quot;, (11450*0.5,0), S);<br /> label(rotate(90)*&quot;Population&quot;, (0,9000*0.5), 10*W);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }65{,}000 \qquad \textbf{(B) }75{,}000 \qquad \textbf{(C) }85{,}000 \qquad \textbf{(D) }95{,}000 \qquad \textbf{(E) }105{,}000&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Suppose &lt;math&gt;15\%&lt;/math&gt; of &lt;math&gt;x&lt;/math&gt; equals &lt;math&gt;20\%&lt;/math&gt; of &lt;math&gt;y.&lt;/math&gt; What percentage of &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;y?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Each of the points &lt;math&gt;A,B,C,D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; in the figure below represents a different digit from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;6.&lt;/math&gt; Each of the five lines shown passes through some of these points. The digits along each line are added to produce five sums, one for each line. The total of the five sums is &lt;math&gt;47.&lt;/math&gt; What is the digit represented by &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // made by SirCalcsALot<br /> <br /> size(200);<br /> dotfactor = 10;<br /> <br /> pair p1 = (-28,0);<br /> pair p2 = (-111,213);<br /> draw(p1--p2,linewidth(1));<br /> <br /> pair p3 = (-160,0);<br /> pair p4 = (-244,213);<br /> draw(p3--p4,linewidth(1));<br /> <br /> pair p5 = (-316,0);<br /> pair p6 = (-67,213);<br /> draw(p5--p6,linewidth(1));<br /> <br /> pair p7 = (0, 68);<br /> pair p8 = (-350,10);<br /> draw(p7--p8,linewidth(1));<br /> <br /> pair p9 = (0, 150);<br /> pair p10 = (-350, 62);<br /> draw(p9--p10,linewidth(1));<br /> <br /> pair A = intersectionpoint(p1--p2, p5--p6);<br /> dot(&quot;$A$&quot;, A, 2*W);<br /> <br /> pair B = intersectionpoint(p5--p6, p3--p4);<br /> dot(&quot;$B$&quot;, B, 2*WNW);<br /> <br /> pair C = intersectionpoint(p7--p8, p5--p6);<br /> dot(&quot;$C$&quot;, C, 1.5*NW);<br /> <br /> pair D = intersectionpoint(p3--p4, p7--p8);<br /> dot(&quot;$D$&quot;, D, 2*NNE);<br /> <br /> pair EE = intersectionpoint(p1--p2, p7--p8);<br /> dot(&quot;$E$&quot;, EE, 2*NNE);<br /> <br /> pair F = intersectionpoint(p1--p2, p9--p10);<br /> dot(&quot;$F$&quot;, F, 2*NNE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }1 \qquad \textbf{(B) }2 \qquad \textbf{(C) }3 \qquad \textbf{(D) }4 \qquad \textbf{(E) }5&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> How many factors of &lt;math&gt;2020&lt;/math&gt; have more than &lt;math&gt;3&lt;/math&gt; factors? (As an example, &lt;math&gt;12&lt;/math&gt; has &lt;math&gt;6&lt;/math&gt; factors, namely &lt;math&gt;1, 2, 3, 4, 6,&lt;/math&gt; and &lt;math&gt;12.&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{(A) }6 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8 \qquad \textbf{(D) }9 \qquad \textbf{(E) }10&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; is inscribed in a semicircle with diameter &lt;math&gt;\overline{FE},&lt;/math&gt; as shown in the figure. Let &lt;math&gt;DA=16,&lt;/math&gt; and let &lt;math&gt;FD=AE=9.&lt;/math&gt; What is the area of &lt;math&gt;ABCD?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> draw(arc((0,0),17,180,0));<br /> draw((-17,0)--(17,0));<br /> fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey);<br /> draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle);<br /> dot(&quot;$A$&quot;,(8,0), 1.25*S);<br /> dot(&quot;$B$&quot;,(8,15), 1.25*N);<br /> dot(&quot;$C$&quot;,(-8,15), 1.25*N);<br /> dot(&quot;$D$&quot;,(-8,0), 1.25*S);<br /> dot(&quot;$E$&quot;,(17,0), 1.25*S);<br /> dot(&quot;$F$&quot;,(-17,0), 1.25*S);<br /> label(&quot;$16$&quot;,(0,0),N);<br /> label(&quot;$9$&quot;,(12.5,0),N);<br /> label(&quot;$9$&quot;,(-12.5,0),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }240 \qquad \textbf{(B) }248 \qquad \textbf{(C) }256 \qquad \textbf{(D) }264 \qquad \textbf{(E) }272&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number is called flippy if its digits alternate between two distinct digits. For example, &lt;math&gt;2020&lt;/math&gt; and &lt;math&gt;37373&lt;/math&gt; are flippy, but &lt;math&gt;3883&lt;/math&gt; and &lt;math&gt;123123&lt;/math&gt; are not. How many five-digit flippy numbers are divisible by &lt;math&gt;15?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scientist walking through a forest recorded as integers the heights of &lt;math&gt;5&lt;/math&gt; trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?<br /> <br /> &lt;cmath&gt;<br /> \begingroup<br /> \setlength{\tabcolsep}{10pt}<br /> \renewcommand{\arraystretch}{1.5}<br /> \begin{tabular}{|c|c|}<br /> \hline Tree 1 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 2 &amp; 11 meters \\<br /> Tree 3 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 4 &amp; \rule{0.4cm}{0.15mm} meters \\<br /> Tree 5 &amp; \rule{0.4cm}{0.15mm} meters \\ \hline<br /> Average height &amp; \rule{0.4cm}{0.15mm}.2 meters \\<br /> \hline<br /> \end{tabular}<br /> \endgroup&lt;/cmath&gt;<br /> &lt;math&gt;\newline \textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A game board consists of &lt;math&gt;64&lt;/math&gt; squares that alternate in color between black and white. The figure below shows square &lt;math&gt;P&lt;/math&gt; in the bottom row and square &lt;math&gt;Q&lt;/math&gt; in the top row. A marker is placed at &lt;math&gt;P.&lt;/math&gt; A step consists of moving the marker onto one of the adjoining white squares in the row above. How many &lt;math&gt;7&lt;/math&gt;-step paths are there from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q?&lt;/math&gt; (The figure shows a sample path.)<br /> <br /> &lt;asy&gt;<br /> // diagram by SirCalcsALot<br /> size(200);<br /> int[] x = {6, 5, 4, 5, 6, 5, 6};<br /> int[] y = {1, 2, 3, 4, 5, 6, 7};<br /> int N = 7;<br /> for (int i = 0; i &lt; 8; ++i) {<br /> for (int j = 0; j &lt; 8; ++j) {<br /> draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j));<br /> if ((i+j) % 2 == 0) {<br /> filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black);<br /> }<br /> }<br /> }<br /> for (int i = 0; i &lt; N; ++i) {<br /> draw(circle((x[i],y[i])+(0.5,0.5),0.35));<br /> }<br /> label(&quot;$P$&quot;, (5.5, 0.5));<br /> label(&quot;$Q$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> When a positive integer &lt;math&gt;N&lt;/math&gt; is fed into a machine, the output is a number calculated according to the rule shown below.<br /> <br /> &lt;asy&gt;<br /> size(300);<br /> defaultpen(linewidth(0.8)+fontsize(13));<br /> real r = 0.05;<br /> draw((0.9,0)--(3.5,0),EndArrow(size=7));<br /> filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65));<br /> fill(circle((5.5,1.25),0.8),white);<br /> fill(circle((5.5,1.25),0.5),gray(0.65));<br /> fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white);<br /> fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white);<br /> fill((4.6,-0.25-r)--(6.4,-0.25-r)--(6.4,-0.75-r)--(4.6,-0.75-r)--cycle,gray(0.65));<br /> fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65));<br /> label(&quot;$N$&quot;,(0.45,0));<br /> draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7));<br /> draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7));<br /> label(&quot;if $N$ is even&quot;,(9.25,1.25),N);<br /> label(&quot;if $N$ is odd&quot;,(9.25,-1.25),N);<br /> label(&quot;$\frac N2$&quot;,(12,1.25));<br /> label(&quot;$3N+1$&quot;,(12.6,-1.25));<br /> &lt;/asy&gt;<br /> <br /> For example, starting with an input of &lt;math&gt;N=7,&lt;/math&gt; the machine will output &lt;math&gt;3 \cdot 7 +1 = 22.&lt;/math&gt; Then if the output is repeatedly inserted into the machine five more times, the final output is &lt;math&gt;26.&lt;/math&gt;<br /> &lt;cmath&gt;7 \to 22 \to 11 \to 34 \to 17 \to 52 \to 26&lt;/cmath&gt;When the same &lt;math&gt;6&lt;/math&gt;-step process is applied to a different starting value of &lt;math&gt;N,&lt;/math&gt; the final output is &lt;math&gt;1.&lt;/math&gt; What is the sum of all such integers &lt;math&gt;N?&lt;/math&gt;<br /> &lt;cmath&gt;N \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to \rule{0.5cm}{0.15mm} \to 1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }73 \qquad \textbf{(B) }74 \qquad \textbf{(C) }75 \qquad \textbf{(D) }82 \qquad \textbf{(E) }83&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Five different awards are to be given to three students. Each student will receive at least one<br /> award. In how many different ways can the awards be distributed?<br /> <br /> &lt;math&gt;\textbf{(A) }120 \qquad \textbf{(B) }150 \qquad \textbf{(C) }180 \qquad \textbf{(D) }210 \qquad \textbf{(E) }240&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A large square region is paved with &lt;math&gt;n^2&lt;/math&gt; gray square tiles, each measuring &lt;math&gt;s&lt;/math&gt; inches on a side. A border &lt;math&gt;d&lt;/math&gt; inches wide surrounds each tile. The figure below shows the case for &lt;math&gt;n=3&lt;/math&gt;. When &lt;math&gt;n=24&lt;/math&gt;, the &lt;math&gt;576&lt;/math&gt; gray tiles cover &lt;math&gt;64\%&lt;/math&gt; of the area of the large square region. What is the ratio &lt;math&gt;\frac{d}{s}&lt;/math&gt; for this larger value of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br /> filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br /> filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br /> filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br /> filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br /> filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br /> filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br /> filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br /> filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br /> filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }\frac6{25} \qquad \textbf{(B) }\frac14 \qquad \textbf{(C) }\frac9{25} \qquad \textbf{(D) }\frac7{16} \qquad \textbf{(E) }\frac9{16}&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> [[2020 AMC 8 Problems/Problem 25|Solution]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_25&diff=137734 2020 AMC 8 Problems/Problem 25 2020-11-19T02:00:40Z <p>Wlm7: /* Solution 1 */</p> <hr /> <div>Rectangles &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;R_2,&lt;/math&gt; and squares &lt;math&gt;S_1,\,S_2,\,&lt;/math&gt; and &lt;math&gt;S_3,&lt;/math&gt; shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of &lt;math&gt;S_2&lt;/math&gt; in units?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0));<br /> draw((3,0)--(3,1)--(0,1));<br /> draw((3,1)--(3,2)--(5,2));<br /> draw((3,2)--(2,2)--(2,1)--(2,3));<br /> label(&quot;$R_1$&quot;,(3/2,1/2));<br /> label(&quot;$S_3$&quot;,(4,1));<br /> label(&quot;$S_2$&quot;,(5/2,3/2));<br /> label(&quot;$S_1$&quot;,(1,2));<br /> label(&quot;$R_2$&quot;,(7/2,5/2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> For each square &lt;math&gt;S_{k}&lt;/math&gt;, let the sidelength of this square be denoted by &lt;math&gt;s_{k}&lt;/math&gt;.<br /> <br /> As the diagram shows, &lt;math&gt;s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.&lt;/math&gt; We subtract the second equation from the first, getting &lt;math&gt;2s_{2}=1302&lt;/math&gt;, and thus &lt;math&gt;s_{2}=651&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(A)}\text{ }651}&lt;/math&gt; ~icematrix, edits by starrynight7210<br /> <br /> ==Solution 2==<br /> <br /> WLOG, assume that &lt;math&gt;S_1=S_3&lt;/math&gt; and &lt;math&gt;R_1=R_2&lt;/math&gt;. Let the sum of the lengths of &lt;math&gt;S_1&lt;/math&gt; and &lt;math&gt;S_2&lt;/math&gt; be &lt;math&gt;x&lt;/math&gt; and let the length of &lt;math&gt;S_2&lt;/math&gt; be &lt;math&gt;y&lt;/math&gt;. We have the system &lt;cmath&gt;x+y =3322&lt;/cmath&gt; &lt;cmath&gt;x-y=2020&lt;/cmath&gt;<br /> <br /> which we solve to find that &lt;math&gt;y=\textbf{(A) }651&lt;/math&gt;.<br /> <br /> -franzliszt<br /> <br /> ==Solution 3==<br /> Since each pair of boxes has a sum of &lt;math&gt;3322&lt;/math&gt; or &lt;math&gt;2000&lt;/math&gt; and a difference of &lt;math&gt;S_2&lt;/math&gt;, we see that the answer is &lt;math&gt;\dfrac{3322 - 2000}{2} = \dfrac{1322}{2} = \boxed{(\text{A}) 651}.&lt;/math&gt;<br /> <br /> -A_MatheMagician<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=122648 2019 AIME I Problems/Problem 14 2020-05-19T15:38:55Z <p>Wlm7: /* Note to solution */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(k)&lt;/math&gt; is the Euler Totient Function of integer &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;\phi(k)&lt;/math&gt; is the number of positive integers less than &lt;math&gt;k&lt;/math&gt; relatively prime to &lt;math&gt;k&lt;/math&gt;. Define the numbers &lt;math&gt;k_1,k_2,k_3,\cdots,k_n&lt;/math&gt; to be the prime factors of &lt;math&gt;k&lt;/math&gt;. Then, we have&lt;cmath&gt;\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).&lt;/cmath&gt;A property of the Totient function is that, for any prime &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;\phi(p)=p-1&lt;/math&gt;.<br /> <br /> Euler's Totient Theorem states that&lt;cmath&gt;a^{\phi(k)} \equiv 1\pmod k&lt;/cmath&gt;if &lt;math&gt;\gcd(a,k)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=122613 2019 AIME I Problems/Problem 14 2020-05-19T00:27:09Z <p>Wlm7: /* Note to solution */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(k)&lt;/math&gt; is the [[Euler Totient Function]] of integer &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;\phi(k)&lt;/math&gt; is the number of positive integers less than &lt;math&gt;k&lt;/math&gt; relatively prime to &lt;math&gt;k&lt;/math&gt;. Define the numbers &lt;math&gt;k_1,k_2,k_3,\cdots,k_n&lt;/math&gt; to be the prime factors of &lt;math&gt;k&lt;/math&gt;. Then, we have &lt;cmath&gt;\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).&lt;/cmath&gt; A property of the Totient function is that, for any prime &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;\phi(p)=p-1&lt;/math&gt;.<br /> <br /> [[Euler's Totient Theorem]] states that &lt;cmath&gt;a^{\phi(k)} \equiv 1\pmod k&lt;/cmath&gt; if &lt;math&gt;\gcd(a,k)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=122611 2019 AIME I Problems/Problem 14 2020-05-19T00:22:42Z <p>Wlm7: /* Note to solution */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(k)&lt;/math&gt; is the [[Euler Totient Function]] of integer &lt;math&gt;k&lt;/math&gt;. &lt;math&gt;\phi(k)&lt;/math&gt; is the number of positive integers less than &lt;math&gt;k&lt;/math&gt; relatively prime to &lt;math&gt;k&lt;/math&gt;. Define the numbers &lt;math&gt;k_1,k_2,k_3,\cdots,k_n&lt;/math&gt; to be the prime factors of &lt;math&gt;k&lt;/math&gt;. Then, we have &lt;math&gt;\phi(k)=k\cdot \prod^n_{i=1}\left(1-\dfrac{1}{k_i}\right).&lt;/math&gt;&lt;math&gt; A property of the Totient function is that, for any prime &lt;/math&gt;p&lt;math&gt;, &lt;/math&gt;\phi(p)=p-1&lt;math&gt;.<br /> <br /> [[Euler's Totient Theorem]] states that &lt;cmath&gt;a^{\phi(k)} \equiv 1\pmod k&lt;/cmath&gt; if &lt;/math&gt;\gcd(a,k)=1&lt;math&gt;.<br /> <br /> Furthermore, the order &lt;/math&gt;a&lt;math&gt; modulo &lt;/math&gt;n&lt;math&gt; for an integer &lt;/math&gt;a&lt;math&gt; relatively prime to &lt;/math&gt;n&lt;math&gt; is defined as the smallest positive integer &lt;/math&gt;d&lt;math&gt; such that &lt;/math&gt;a^{d} \equiv 1\pmod n&lt;math&gt;. An important property of the order &lt;/math&gt;d&lt;math&gt; is that &lt;/math&gt;d|\phi(n)$.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=122608 2019 AIME I Problems/Problem 14 2020-05-19T00:16:10Z <p>Wlm7: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(p)&lt;/math&gt; is the [[Euler Totient Function]] of integer &lt;math&gt;p&lt;/math&gt;.<br /> [[Euler's Totient Theorem]]: define &lt;math&gt;\phi(p)&lt;/math&gt; as the number of positive integers less than &lt;math&gt;p&lt;/math&gt; but relatively prime to &lt;math&gt;p&lt;/math&gt;. We have &lt;cmath&gt;\phi(p)=p\cdot \prod^n_{i=1}\left(1-\dfrac{1}{p_i}\right)&lt;/cmath&gt; where &lt;math&gt;p_1,p_2,...,p_n&lt;/math&gt; are the prime factors of &lt;math&gt;p&lt;/math&gt;. Then, we have &lt;cmath&gt;a^{\phi(p)} \equiv 1\pmod p&lt;/cmath&gt; if &lt;math&gt;\gcd(a,p)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=122607 2019 AIME I Problems/Problem 14 2020-05-19T00:15:35Z <p>Wlm7: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is a positive divisor of &lt;math&gt;16&lt;/math&gt;.<br /> <br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of &lt;math&gt;16&lt;/math&gt;. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p\left(1 - \dfrac{1}{p}\right) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv 1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, the smallest possible &lt;math&gt;p&lt;/math&gt; is thus &lt;math&gt;\boxed{097}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(p)&lt;/math&gt; is the [[Euler Totient Function]] of integer &lt;math&gt;p&lt;/math&gt;.<br /> [[Euler's Totient Theorem]]: define &lt;math&gt;\phi(p)&lt;/math&gt; as the number of positive integers less than &lt;math&gt;p&lt;/math&gt; but relatively prime to &lt;math&gt;p&lt;/math&gt;. We have &lt;cmath&gt;\phi(p)=p\cdot \prod^n_{i=1}\left(1-\dfrac{1}{p_i}\right)&lt;/cmath&gt; where &lt;math&gt;p_1,p_2,...,p_n&lt;/math&gt; are the prime factors of &lt;math&gt;p&lt;/math&gt;. Then, we have &lt;cmath&gt;a^{\phi(p)} \equiv 1\pmod p&lt;/cmath&gt; if &lt;math&gt;\gcd(a,p)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_I_Problems/Problem_14&diff=122603 2019 AIME I Problems/Problem 14 2020-05-19T00:09:32Z <p>Wlm7: /* Solution */</p> <hr /> <div>==Problem 14==<br /> Find the least odd prime factor of &lt;math&gt;2019^8+1&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> We know that &lt;math&gt;2019^8 \equiv -1 \pmod{p}&lt;/math&gt; for some prime &lt;math&gt;p&lt;/math&gt;. We want to find the smallest odd possible value of &lt;math&gt;p&lt;/math&gt;. By squaring both sides of the congruence, we find &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;2019^{16} \equiv 1 \pmod{p}&lt;/math&gt;, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4, 8,&lt;/math&gt; or &lt;math&gt;16&lt;/math&gt;.<br /> <br /> However, if the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;1, 2, 4,&lt;/math&gt; or &lt;math&gt;8,&lt;/math&gt; then &lt;math&gt;2019^8&lt;/math&gt; will be equivalent to &lt;math&gt;1 \pmod{p},&lt;/math&gt; which contradicts the given requirement that &lt;math&gt;2019^8\equiv -1\pmod{p}&lt;/math&gt;.<br /> <br /> Therefore, the order of &lt;math&gt;2019&lt;/math&gt; modulo &lt;math&gt;p&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;. Because all orders modulo &lt;math&gt;p&lt;/math&gt; divide &lt;math&gt;\phi(p)&lt;/math&gt;, we see that &lt;math&gt;\phi(p)&lt;/math&gt; is a multiple of 16. As &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;\phi(p) = p(1 - \frac{1}{p}) = p - 1&lt;/math&gt;. Therefore, &lt;math&gt;p\equiv1 \pmod{16}&lt;/math&gt;. The two smallest primes equivalent to &lt;math&gt;1 \pmod{16}&lt;/math&gt; are &lt;math&gt;17&lt;/math&gt; and &lt;math&gt;97&lt;/math&gt;. As &lt;math&gt;2019^8 \not\equiv -1 \pmod{17}&lt;/math&gt; and &lt;math&gt;2019^8 \equiv -1 \pmod{97}&lt;/math&gt;, our answer is &lt;math&gt;\boxed{97}&lt;/math&gt;.<br /> <br /> ===Note to solution===<br /> &lt;math&gt;\phi(k)&lt;/math&gt; is the [[Euler Totient Function]] of integer &lt;math&gt;k&lt;/math&gt;.<br /> [[Euler's Totient Theorem]]: define &lt;math&gt;\phi(p)&lt;/math&gt; as the number of positive integers less than &lt;math&gt;p&lt;/math&gt; but relatively prime to &lt;math&gt;p&lt;/math&gt;, then we have &lt;cmath&gt;\phi(p)=p\cdot \prod^n_{i=1}(1-\frac{1}{p_i})&lt;/cmath&gt; where &lt;math&gt;p_1,p_2,...,p_n&lt;/math&gt; are the prime factors of &lt;math&gt;p&lt;/math&gt;. Then, we have &lt;cmath&gt;a^{\phi(p)} \equiv 1\ (\mathrm{mod}\ p)&lt;/cmath&gt; if &lt;math&gt;\gcd(a,p)=1&lt;/math&gt;.<br /> <br /> Furthermore, the order &lt;math&gt;a&lt;/math&gt; modulo &lt;math&gt;n&lt;/math&gt; for an integer &lt;math&gt;a&lt;/math&gt; relatively prime to &lt;math&gt;n&lt;/math&gt; is defined as the smallest positive integer &lt;math&gt;d&lt;/math&gt; such that &lt;math&gt;a^{d} \equiv 1\pmod n&lt;/math&gt;. An important property of the order &lt;math&gt;d&lt;/math&gt; is that &lt;math&gt;d|\phi(n)&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> On The Spot STEM:<br /> <br /> https://youtu.be/_vHq5_5qCd8<br /> <br /> <br /> <br /> https://youtu.be/IF88iO5keFo<br /> <br /> ==See Also==<br /> {{AIME box|year=2019|n=I|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10B_Problems/Problem_22&diff=110339 2011 AMC 10B Problems/Problem 22 2019-10-15T00:58:45Z <p>Wlm7: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> A pyramid has a square base with sides of length &lt;math&gt;1&lt;/math&gt; and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5\sqrt{2} - 7 \qquad\textbf{(B)}\ 7 - 4\sqrt{3} \qquad\textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad\textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad\textbf{(E)}\ \frac{\sqrt{3}}{9}&lt;/math&gt;<br /> [[Category: Introductory Geometry Problems]]<br /> <br /> == Solution ==<br /> It is often easier to first draw a diagram for such a problem.<br /> <br /> [[Image:2011AMC10B22.png|center|260px]]<br /> <br /> Sometimes, it may also be easier to think of the problem in 2D. Take a cross section of the pyramid through the apex and two points from the base that are opposite to each other. Place it in two dimensions.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(35mm);<br /> defaultpen(linewidth(2pt)+fontsize(10pt));<br /> pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));<br /> pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);<br /> draw(A--B--C--cycle);<br /> draw(W--X--Y--Z--cycle,red);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> The dimensions of this triangle are &lt;math&gt;1, 1,&lt;/math&gt; and &lt;math&gt;\sqrt{2}&lt;/math&gt; because the sidelengths of the pyramid are &lt;math&gt;1&lt;/math&gt; and the base of the triangle is the diagonal of the pyramid's base. This is a &lt;math&gt;45-45-90&lt;/math&gt; triangle. Also, we can let the dimensions of the rectangle be &lt;math&gt;s&lt;/math&gt; and &lt;math&gt;s\sqrt{2}&lt;/math&gt; because the longer side was the diagonal of the cube's base and the shorter cube was a side of the cube.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(35mm);<br /> defaultpen(linewidth(2pt)+fontsize(12pt));<br /> pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));<br /> pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);<br /> draw(A--B--C--cycle);<br /> draw(W--X--Y--Z--cycle,red);<br /> label(&quot;$1$&quot;,(A--C),NW); label(&quot;$1$&quot;,(B--C),NE); label(&quot;$\sqrt{2}$&quot;,(A--B),S);<br /> label(&quot;$s$&quot;,(W--Z),E,red); label(&quot;$s$&quot;,(X--Y),W,red); label(&quot;$s\sqrt{2}$&quot;,(W--X),N,red);<br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> The two triangles on the right and left of the rectangle are also &lt;math&gt;45-45-90&lt;/math&gt; triangles because the rectangle is perpendicular to the base, and they share a &lt;math&gt;45^\circ&lt;/math&gt; angle with the larger triangle. Therefore, the legs of the right triangles can be expressed as &lt;math&gt;s.&lt;/math&gt;<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(35mm);<br /> defaultpen(linewidth(2pt)+fontsize(12pt));<br /> pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));<br /> pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);<br /> draw(A--B--C--cycle);<br /> draw(W--X--Y--Z--cycle,red);<br /> label(&quot;$1$&quot;,(A--C),NW); label(&quot;$1$&quot;,(B--C),NE); label(&quot;$\sqrt{2}$&quot;,(A--B),S);<br /> label(&quot;$s$&quot;,(W--Z),E,red); label(&quot;$s$&quot;,(X--Y),W,red); label(&quot;$s\sqrt{2}$&quot;,(W--X),N,red);<br /> label(&quot;$s$&quot;,(A--W),N); label(&quot;$s&quot;,(X--B),N); <br /> &lt;/asy&gt;<br /> &lt;/center&gt;<br /> <br /> Now we can just use segment addition to find the value of &lt;math&gt;s.&lt;/math&gt;<br /> &lt;cmath&gt;\sqrt{2}=s+s\sqrt{2}+s=2s+s\sqrt{2}=(2+\sqrt{2})s&lt;/cmath&gt;<br /> &lt;cmath&gt;s=\frac{\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1&lt;/cmath&gt;<br /> <br /> The volume of the cube is &lt;math&gt;s^3 = (\sqrt{2}-1)^3 = (\sqrt{2}-1)(3-2\sqrt{2}) = 3\sqrt{2}-3-4+2\sqrt{2} = \boxed{\textbf{(A)} 5\sqrt{2}-7}&lt;/math&gt;<br /> <br /> <br /> == Solution 2 ==<br /> <br /> Let &lt;math&gt;s&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt; be the slant height and the distance from one side of the base to the point right under the apex, respectively. Then using the Pythagorean Theorem, the altitude from the apex to the base is &lt;cmath&gt;a=\sqrt{s^2-d^2}=\sqrt{\left(\dfrac{\sqrt{3}}{2}\right)^2+\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{2}}{2}&lt;/cmath&gt;<br /> Notice that the smaller pyramid on top of the cube is similar to the larger pyramid. Thus, letting &lt;math&gt;x&lt;/math&gt; be the edge length of the cube, then the altitude of the smaller pyramid is &lt;math&gt;\dfrac{\sqrt{2}}{2}x&lt;/math&gt;.<br /> <br /> Since the altitude of the larger pyramid is the sum of the edge length of the cube and the altitude of the smaller pyramid, we have &lt;cmath&gt;\dfrac{\sqrt{2}}{2}=x+\dfrac{\sqrt{2}}{2}x&lt;/cmath&gt; &lt;cmath&gt;x=\sqrt{2}-1&lt;/cmath&gt;<br /> <br /> Finally, &lt;math&gt;V_{cube}=x^3=(\sqrt{2}-1)^3=\boxed{\textbf{(A)} 5\sqrt{2}-7}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2011|ab=B|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&diff=106870 2008 AMC 12A Problems/Problem 22 2019-06-23T15:58:13Z <p>Wlm7: /* Solution 2 (without trigonometry) */</p> <hr /> <div>{{duplicate|[[2008 AMC 12A Problems|2008 AMC 12A #22]] and [[2008 AMC 10A Problems/Problem 25|2008 AMC 10A #25]]}}<br /> ==Problem==<br /> A round table has radius &lt;math&gt;4&lt;/math&gt;. Six rectangular place mats are placed on the table. Each place mat has width &lt;math&gt;1&lt;/math&gt; and length &lt;math&gt;x&lt;/math&gt; as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length &lt;math&gt;x&lt;/math&gt;. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;asy&gt;unitsize(4mm);<br /> defaultpen(linewidth(.8)+fontsize(8));<br /> draw(Circle((0,0),4));<br /> path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br /> draw(mat);<br /> draw(rotate(60)*mat);<br /> draw(rotate(120)*mat);<br /> draw(rotate(180)*mat);<br /> draw(rotate(240)*mat);<br /> draw(rotate(300)*mat);<br /> label(&quot;$$x$$&quot;,(-1.55,2.1),E);<br /> label(&quot;$$1$$&quot;,(-0.5,3.8),S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> ==Solution==<br /> === Solution 1 (trigonometry) ===<br /> Let one of the mats be &lt;math&gt;ABCD&lt;/math&gt;, and the center be &lt;math&gt;O&lt;/math&gt; as shown: <br /> <br /> &lt;asy&gt;unitsize(8mm);<br /> defaultpen(linewidth(.8)+fontsize(8));<br /> draw(Circle((0,0),4));<br /> path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br /> draw(mat);<br /> draw(rotate(60)*mat);<br /> draw(rotate(120)*mat);<br /> draw(rotate(180)*mat);<br /> draw(rotate(240)*mat);<br /> draw(rotate(300)*mat);<br /> label(&quot;$$x$$&quot;,(-1.55,2.1),E);<br /> label(&quot;$$x$$&quot;,(0.03,1.5),E);<br /> label(&quot;$$A$$&quot;,(-3.6,2.5513),E);<br /> label(&quot;$$B$$&quot;,(-3.15,1.35),E);<br /> label(&quot;$$C$$&quot;,(0.05,3.20),E);<br /> label(&quot;$$D$$&quot;,(-0.75,4.15),E);<br /> label(&quot;$$O$$&quot;,(0.00,-0.10),E);<br /> label(&quot;$$1$$&quot;,(-0.1,3.8),S);<br /> label(&quot;$$4$$&quot;,(-0.4,2.2),S);<br /> draw((0,0)--(0,3.103));<br /> draw((0,0)--(-2.687,1.5513));<br /> draw((0,0)--(-0.5,3.9686));&lt;/asy&gt;<br /> <br /> Since there are &lt;math&gt;6&lt;/math&gt; mats, &lt;math&gt;\Delta BOC&lt;/math&gt; is [[equilateral]]. So, &lt;math&gt;BC=CO=x&lt;/math&gt;. Also, &lt;math&gt;\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ&lt;/math&gt;. <br /> <br /> By the [[Law of Cosines]]: &lt;math&gt;4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;x&lt;/math&gt; must be positive, &lt;math&gt;x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow C&lt;/math&gt;.<br /> <br /> === Solution 2 (without trigonometry) ===<br /> Draw &lt;math&gt;OD&lt;/math&gt; and &lt;math&gt;OC&lt;/math&gt; as in the diagram. Draw the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;DC&lt;/math&gt; and call the intersection &lt;math&gt;E&lt;/math&gt;<br /> <br /> <br /> &lt;asy&gt;unitsize(8mm);<br /> defaultpen(linewidth(.8)+fontsize(8));<br /> draw(Circle((0,0),4));<br /> path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle);<br /> draw(mat);<br /> draw(rotate(60)*mat);<br /> draw(rotate(120)*mat);<br /> draw(rotate(180)*mat);<br /> draw(rotate(240)*mat);<br /> draw(rotate(300)*mat);<br /> pair D = rotate(300)*(-3.687,1.5513);<br /> pair C = rotate(300)*(-2.687,1.5513);<br /> pair EE = foot((0.00,0.00),D,C);<br /> draw(D--EE--(0,0));<br /> label(&quot;$$x$$&quot;,(-1.55,2.1),E);<br /> label(&quot;$$x$$&quot;,(0.03,1.5),E);<br /> label(&quot;$$A$$&quot;,(-3.6,2.5513),E);<br /> label(&quot;$$B$$&quot;,(-3.15,1.35),E);<br /> label(&quot;$$C$$&quot;,(0.05,3.20),E);<br /> label(&quot;$$D$$&quot;,(-0.75,4.15),E);<br /> label(&quot;$$O$$&quot;,(0.00,-0.10),E);<br /> label(&quot;$$1$$&quot;,(-0.1,3.8),S);<br /> label(&quot;$$4$$&quot;,(-0.4,2.2),S);<br /> draw((0,0)--(0,3.103));<br /> draw((0,0)--(-2.687,1.5513));<br /> draw((0,0)--(-0.5,3.9686));<br /> label(&quot;$$E$$&quot;, EE,SE);<br /> &lt;/asy&gt;<br /> <br /> As proved in the first solution, &lt;math&gt; \angle OCD = 150^\circ&lt;/math&gt;. <br /> That makes &lt;math&gt;\Delta OCE&lt;/math&gt; a &lt;math&gt;30-60-90&lt;/math&gt; triangle, so &lt;math&gt;OE = \frac{x}{2}&lt;/math&gt; and &lt;math&gt;CE= \frac{x\sqrt 3}{2}&lt;/math&gt;<br /> <br /> Since &lt;math&gt; \Delta OEC&lt;/math&gt; is a right triangle, <br /> &lt;math&gt;\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;x&lt;/math&gt; gives &lt;math&gt; x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow C &lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> &lt;asy&gt;unitsize(8mm);<br /> defaultpen(linewidth(.8)+fontsize(8));<br /> draw(Circle((0,0),4));<br /> path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br /> draw(mat);<br /> draw(rotate(60)*mat);<br /> draw(rotate(120)*mat);<br /> draw(rotate(180)*mat);<br /> draw(rotate(240)*mat);<br /> draw(rotate(300)*mat);<br /> label(&quot;$$x$$&quot;,(-1.95,3),E);<br /> label(&quot;$$A$$&quot;,(-3.6,2.5513),E);<br /> label(&quot;$$C$$&quot;,(0.05,3.20),E);<br /> label(&quot;$$E$$&quot;,(0.40,-3.60),E);<br /> label(&quot;$$B$$&quot;,(-0.75,4.15),E);<br /> label(&quot;$$D$$&quot;,(-2.62,1.5),E);<br /> label(&quot;$$F$$&quot;,(-2.64,-1.43),E);<br /> label(&quot;$$G$$&quot;,(-0.2,-2.8),E);<br /> label(&quot;$$\sqrt{3}x$$&quot;,(-1.5,-0.5),E);<br /> label(&quot;$$M$$&quot;,(-2,-0.9),E);<br /> label(&quot;$$O$$&quot;,(0.00,-0.10),E);<br /> label(&quot;$$1$$&quot;,(-2.7,2.3),S);<br /> label(&quot;$$1$$&quot;,(0.1,-3.4),S);<br /> label(&quot;$$8$$&quot;,(-0.3,0),S);<br /> draw((0,-3.103)--(-2.687,1.5513));<br /> draw((0.5,-3.9686)--(-0.5,3.9686));&lt;/asy&gt;<br /> <br /> Looking at the diagram above, we know that &lt;math&gt;BE&lt;/math&gt; is a diameter of circle &lt;math&gt;O&lt;/math&gt; due to symmetry. Due to Thales' theorem, triangle &lt;math&gt;ABE&lt;/math&gt; is a right triangle with &lt;math&gt;A = 90 ^\circ&lt;/math&gt;. &lt;math&gt;AE&lt;/math&gt; lies on &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;GE&lt;/math&gt; because &lt;math&gt;BAD&lt;/math&gt; is also a right angle. To find the length of &lt;math&gt;DG&lt;/math&gt;, notice that if we draw a line from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;M&lt;/math&gt;, the midpoint of line &lt;math&gt;DG&lt;/math&gt;, it creates two &lt;math&gt;30&lt;/math&gt; - &lt;math&gt;60&lt;/math&gt; - &lt;math&gt;90&lt;/math&gt; triangles. Therefore, &lt;math&gt;MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x&lt;/math&gt;. &lt;math&gt;AE = 2 + \sqrt{3}x&lt;/math&gt;<br /> <br /> Use the Pythagorean theorem on triangle &lt;math&gt;ABE&lt;/math&gt;, we get &lt;cmath&gt;(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0&lt;/cmath&gt; Using the pythagorean theorem to solve, we get &lt;cmath&gt;x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}&lt;/cmath&gt; &lt;math&gt;x&lt;/math&gt; must be positive, therefore &lt;cmath&gt;x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow C&lt;/cmath&gt;<br /> <br /> ~Zeric Hang<br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=A|num-b=21|num-a=23}}<br /> {{AMC10 box|year=2008|ab=A|num-b=24|after=Last Question}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Introductory Trigonometry Problems]]<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Answer_Key&diff=105994 2003 AMC 10B Answer Key 2019-05-28T17:54:12Z <p>Wlm7: </p> <hr /> <div>#C<br /> #D<br /> #B<br /> #A<br /> #C<br /> #D<br /> #B<br /> #B<br /> #B<br /> #C<br /> #A<br /> #C<br /> #E<br /> #D<br /> #E<br /> #E<br /> #B<br /> #D<br /> #E<br /> #D<br /> #C<br /> #B<br /> #D<br /> #E<br /> #B</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_8_Problems/Problem_11&diff=104737 2006 AMC 8 Problems/Problem 11 2019-03-22T19:20:35Z <p>Wlm7: </p> <hr /> <div>== Problem ==<br /> How many two-digit numbers have digits whose sum is a perfect square? <br /> <br /> &lt;math&gt; \textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19 &lt;/math&gt;<br /> <br /> == Solution ==<br /> There is &lt;math&gt; 1 &lt;/math&gt; integer whose digits sum to &lt;math&gt; 1 &lt;/math&gt;: &lt;math&gt;10&lt;/math&gt;.<br /> <br /> There are &lt;math&gt; 4 &lt;/math&gt; integers whose digits sum to &lt;math&gt; 4 &lt;/math&gt;: &lt;math&gt;13, 22, 31, \text{and } 40&lt;/math&gt;.<br /> <br /> There are &lt;math&gt; 9 &lt;/math&gt; integers whose digits sum to &lt;math&gt; 9 &lt;/math&gt;: &lt;math&gt;18, 27, 36, 45, 54, 63, 72, 81, \text{and } 90&lt;/math&gt;.<br /> <br /> There are &lt;math&gt; 3&lt;/math&gt; integers whose digits sum to &lt;math&gt; 16 &lt;/math&gt;: &lt;math&gt;79, 88, \text{and } 97&lt;/math&gt;.<br /> <br /> Two digits cannot sum to &lt;math&gt;25&lt;/math&gt; or any greater square since the greatest sum of digits of a two-digit number is &lt;math&gt; 9 + 9 = 18 &lt;/math&gt;.<br /> <br /> Thus, the answer is &lt;math&gt; 1 + 4 + 9 + 3 = \boxed{\textbf{(C)} 17} &lt;/math&gt;.<br /> ==See Also==<br /> {{AMC8 box|year=2006|num-b=10|num-a=12}}<br /> <br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_7&diff=100089 2004 AMC 12A Problems/Problem 7 2019-01-04T22:29:46Z <p>Wlm7: /* Solution */</p> <hr /> <div>{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #7]] and [[2004 AMC 10A Problems/Problem 8|2004 AMC 10A #8]]}}<br /> ==Problem==<br /> A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; start with &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;14&lt;/math&gt;, and &lt;math&gt;13&lt;/math&gt; tokens, respectively. How many rounds will there be in the game?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40 &lt;/math&gt;<br /> <br /> ==Solution==<br /> We look at a set of three rounds, where the players begin with &lt;math&gt;x+1&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt;, and &lt;math&gt;x-1&lt;/math&gt; tokens.<br /> After three rounds, there will be a net loss of &lt;math&gt;1&lt;/math&gt; token per player (they receive two tokens and lose three). Therefore, after &lt;math&gt;36&lt;/math&gt; rounds -- or &lt;math&gt;12&lt;/math&gt; three-round sets, &lt;math&gt;A,B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; will have &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, and &lt;math&gt;1&lt;/math&gt; tokens, respectively. After &lt;math&gt;1&lt;/math&gt; more round, player &lt;math&gt;A&lt;/math&gt; will give away &lt;math&gt;3&lt;/math&gt; tokens, leaving it empty-handed, and thus the game will end. We then have there are &lt;math&gt;36+1=\boxed{\mathrm{(B)}\ 37}&lt;/math&gt; rounds until the game ends.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2004|ab=A|num-b=6|num-a=8}}<br /> {{AMC10 box|year=2004|ab=A|num-b=7|num-a=9}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99836 2009 AMC 10A Problems/Problem 5 2018-12-30T16:33:02Z <p>Wlm7: /* Solution 2 -- Find And Harness a Pattern */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2 -- Find And Harness a Pattern==<br /> We note that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;,<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;,<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;,<br /> <br /> and &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;.<br /> <br /> We can clearly see the pattern: If &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;111\cdots111&lt;/math&gt;, with &lt;math&gt;n&lt;/math&gt; ones (and for the sake of simplicity, assume that &lt;math&gt;n&lt;10&lt;/math&gt;), then the sum of the digits of &lt;math&gt;X^2&lt;/math&gt; is <br /> <br /> &lt;math&gt;1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1&lt;/math&gt;<br /> <br /> &lt;math&gt;=(1+2+3\cdots n)+(1+2+3+\cdots n-1)&lt;/math&gt;<br /> <br /> &lt;math&gt;=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}&lt;/math&gt;<br /> <br /> &lt;math&gt;=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2.&lt;/math&gt;<br /> <br /> Aha! We know that &lt;math&gt;111,111,111&lt;/math&gt; has &lt;math&gt;9&lt;/math&gt; digits, so its digit sum is &lt;math&gt;9^2=\boxed{81(E)}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> <br /> &lt;math&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;=11111111100000000+1111111110000000+\cdots+111111111&lt;/math&gt;<br /> <br /> &lt;math&gt;=12,345,678,987,654,321&lt;/math&gt; <br /> <br /> The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99835 2009 AMC 10A Problems/Problem 5 2018-12-30T16:32:11Z <p>Wlm7: /* Solution 2 -- Find and Harness a Pattern*/</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2 -- Find And Harness a Pattern==<br /> We note that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;,<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;,<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;,<br /> <br /> and &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;.<br /> <br /> We can clearly see the pattern: If &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;111\cdots111&lt;/math&gt;, with &lt;math&gt;n&lt;/math&gt; ones (and for the sake of simplicity, assume that &lt;math&gt;n&lt;10&lt;/math&gt;), then the sum of the digits of &lt;math&gt;X^2&lt;/math&gt; is &lt;math&gt;1+2+3+4+5\cdots n+(n-1)+(n-2)\cdots+1=(1+2+3\cdots n)+(1+2+3+\cdots n-1)=\dfrac{n(n+1)}{2}+\dfrac{(n-1)n}{2}=\dfrac{n(n+1+n-1)}{2}=\dfrac{2n^2}{2}=n^2&lt;/math&gt;. Aha! We know that &lt;math&gt;111,111,111&lt;/math&gt; has &lt;math&gt;9&lt;/math&gt; digits, so its digit sum is &lt;math&gt;9^2=\boxed{81(E)}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> <br /> &lt;math&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;=11111111100000000+1111111110000000+\cdots+111111111&lt;/math&gt;<br /> <br /> &lt;math&gt;=12,345,678,987,654,321&lt;/math&gt; <br /> <br /> The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99834 2009 AMC 10A Problems/Problem 5 2018-12-30T16:25:22Z <p>Wlm7: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> <br /> &lt;math&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;=11111111100000000+1111111110000000+\cdots+111111111&lt;/math&gt;<br /> <br /> &lt;math&gt;=12,345,678,987,654,321&lt;/math&gt; <br /> <br /> The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99833 2009 AMC 10A Problems/Problem 5 2018-12-30T16:24:05Z <p>Wlm7: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> <br /> &lt;math&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)&lt;/math&gt;<br /> <br /> &lt;math&gt;=11111111100000000+1111111110000000+\cdots+111111111&lt;/math&gt;<br /> <br /> &lt;math&gt;=12,345,678,987,654,321&lt;/math&gt; <br /> <br /> The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99832 2009 AMC 10A Problems/Problem 5 2018-12-30T16:23:15Z <p>Wlm7: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> &lt;math&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)&lt;/math&gt;<br /> &lt;math&gt;=11111111100000000+1111111110000000+\cdots+111111111&lt;/math&gt;<br /> &lt;math&gt;=12,345,678,987,654,321&lt;/math&gt; The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99831 2009 AMC 10A Problems/Problem 5 2018-12-30T16:22:42Z <p>Wlm7: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> &lt;math&gt;111111111^2\=111111111(100000000+10000000\cdots+10+1)\=11111111100000000+1111111110000000+\cdots+111111111\=<br /> 12,345,678,987,654,321&lt;/math&gt; The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99830 2009 AMC 10A Problems/Problem 5 2018-12-30T16:21:43Z <p>Wlm7: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> \begin{align*}<br /> 111111111^2&amp;=111111111(100000000+10000000\cdots+10+1)\\<br /> &amp;=11111111100000000+1111111110000000+\cdots+111111111\\<br /> &amp;= 12,345,678,987,654,321\end{align*} The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99829 2009 AMC 10A Problems/Problem 5 2018-12-30T16:21:28Z <p>Wlm7: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> \begin{align*}<br /> 111111111^2&amp;=111111111(100000000+10000000\cdots+10+1)\\<br /> &amp;=11111111100000000+1111111110000000+\cdots+111111111\\<br /> &amp;= 12,345,678,987,654,321\end{align*} The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99828 2009 AMC 10A Problems/Problem 5 2018-12-30T16:21:02Z <p>Wlm7: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> \begin{align*}<br /> 111111111^2&amp;=111111111(100000000+10000000\cdots+10+1)\\<br /> &amp;=11111111100000000+1111111110000000+\cdots+111111111\\<br /> &amp;= 12,345,678,987,654,321\end{align*} The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99827 2009 AMC 10A Problems/Problem 5 2018-12-30T16:18:59Z <p>Wlm7: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> &lt;cmath&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)=11111111100000000+1111111110000000+\cdots+111111111 = 12,345,678,987,654,321&lt;/cmath&gt; The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99826 2009 AMC 10A Problems/Problem 5 2018-12-30T16:18:39Z <p>Wlm7: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> We see that &lt;math&gt;111^2&lt;/math&gt; can be written as &lt;math&gt;111(100+10+1)=11100+1110+111=12321&lt;/math&gt;.<br /> <br /> We can apply this strategy to find &lt;math&gt;111,111,111^2&lt;/math&gt;, as seen below.<br /> &lt;cmath&gt;111111111^2=111111111(100000000+10000000\cdots+10+1)=11111111100000000+1111111110000000+\cdots+1111111110+111111111 = 12,345,678,987,654,321&lt;/cmath&gt; The digit sum is thus &lt;math&gt;1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99825 2009 AMC 10A Problems/Problem 5 2018-12-30T16:14:11Z <p>Wlm7: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe a pattern with the squares of number comprised of only &lt;math&gt;1&lt;/math&gt;'s and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> You can see that<br /> &lt;math&gt;111*111&lt;/math&gt; can be written as<br /> <br /> &lt;math&gt;111+1110+11100&lt;/math&gt;, which is &lt;math&gt;12321&lt;/math&gt;.<br /> We can apply the same fact into 111,111,111, receiving<br /> &lt;math&gt;111111111+1111111110+11111111100... = 12,345,678,987,654,321&lt;/math&gt; whose digits sum up to &lt;math&gt;81\longrightarrow \fbox{E}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99824 2009 AMC 10A Problems/Problem 5 2018-12-30T16:13:38Z <p>Wlm7: /* Solution 2*/</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2==<br /> Note that <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.&lt;/math&gt;<br /> <br /> We observe the pattern and use it to find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> You can see that<br /> &lt;math&gt;111*111&lt;/math&gt; can be written as<br /> <br /> &lt;math&gt;111+1110+11100&lt;/math&gt;, which is &lt;math&gt;12321&lt;/math&gt;.<br /> We can apply the same fact into 111,111,111, receiving<br /> &lt;math&gt;111111111+1111111110+11111111100... = 12,345,678,987,654,321&lt;/math&gt; whose digits sum up to &lt;math&gt;81\longrightarrow \fbox{E}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_10A_Problems/Problem_5&diff=99823 2009 AMC 10A Problems/Problem 5 2018-12-30T16:11:32Z <p>Wlm7: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> &lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;onlyinclude&gt;What is the sum of the digits of the square of &lt;math&gt;\text 111,111,111&lt;/math&gt;?&lt;!-- don't remove the following tag, for PoTW on the Wiki front page--&gt;&lt;/onlyinclude&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Using the standard multiplication algorithm, &lt;math&gt;111,111,111^2=12,345,678,987,654,321,&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{(E)}.&lt;/math&gt;<br /> (I hope you didn't seriously multiply it outright...)<br /> <br /> ==Solution 2(Pattern)==<br /> Note that: <br /> <br /> &lt;math&gt;11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321&lt;/math&gt;<br /> <br /> We see a pattern and find that &lt;math&gt;111,111,111^2=12,345,678,987,654,321&lt;/math&gt; whose digit sum is &lt;math&gt;81\longrightarrow \fbox{E}.&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> You can see that<br /> &lt;math&gt;111*111&lt;/math&gt; can be written as<br /> <br /> &lt;math&gt;111+1110+11100&lt;/math&gt;, which is &lt;math&gt;12321&lt;/math&gt;.<br /> We can apply the same fact into 111,111,111, receiving<br /> &lt;math&gt;111111111+1111111110+11111111100... = 12,345,678,987,654,321&lt;/math&gt; whose digits sum up to &lt;math&gt;81\longrightarrow \fbox{E}.&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> We can also do something a little bit more clever rather than just adding up the digits though. Realize too that<br /> <br /> &lt;math&gt;1^2 = 1&lt;/math&gt;<br /> <br /> &lt;math&gt;11^2 = 121&lt;/math&gt;<br /> <br /> &lt;math&gt;111^2 = 12321&lt;/math&gt;<br /> <br /> &lt;math&gt;1,111^2 = 1234321&lt;/math&gt;<br /> <br /> We clearly see the pattern, the number of digits determines how high the number goes, as with &lt;math&gt;111^2&lt;/math&gt;, it has &lt;math&gt;3&lt;/math&gt; digits so it goes up to &lt;math&gt;1,2,3&lt;/math&gt; then decreases back down. If we start adding up the digits, we see that the first one is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;2 + 1 + 1 = 4&lt;/math&gt;, the third one is &lt;math&gt;1 + 2 + 3 + 2 + 1 = 9&lt;/math&gt;, and the fourth one is &lt;math&gt;16&lt;/math&gt;. We instantly see a pattern and find that these are all square numbers. If the number you square has &lt;math&gt;4&lt;/math&gt; digits, you do &lt;math&gt;4^2&lt;/math&gt; to see what the added digits of that particular square will be. In this case, we are dealing with &lt;math&gt;111,111,111&lt;/math&gt; which has &lt;math&gt;9&lt;/math&gt; digits so &lt;math&gt;9^2&lt;/math&gt; equals &lt;math&gt;81\longrightarrow \fbox{E}&lt;/math&gt;<br /> <br /> ==See also==<br /> {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems&diff=96059 1983 AHSME Problems 2018-07-09T14:59:13Z <p>Wlm7: /* Problem 30 */</p> <hr /> <div>== Problem 1 ==<br /> <br /> If &lt;math&gt;x \neq 0, \frac x{2} = y^2&lt;/math&gt; and &lt;math&gt;\frac{x}{4} = 4y&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad<br /> \textbf{(B)}\ 16\qquad<br /> \textbf{(C)}\ 32\qquad<br /> \textbf{(D)}\ 64\qquad<br /> \textbf{(E)}\ 128 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Point &lt;math&gt;P&lt;/math&gt; is outside circle &lt;math&gt;C&lt;/math&gt; on the plane. At most how many points on &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;3 \text{cm}&lt;/math&gt; from P? <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Three primes &lt;math&gt;p,q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; satisfy &lt;math&gt;p+q = r&lt;/math&gt; and &lt;math&gt;1 &lt; p &lt; q&lt;/math&gt;. Then &lt;math&gt;p&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 7\qquad<br /> \textbf{(D)}\ 13\qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Position &lt;math&gt;A,B,C,D,E,F&lt;/math&gt; such that &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; are parallel, as are sides &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;, <br /> and sides &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt;. Each side has length of &lt;math&gt;1&lt;/math&gt; and it is given that &lt;math&gt;\measuredangle FAB = \measuredangle BCD = 60^\circ&lt;/math&gt;. <br /> The area of the figure is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ \frac{3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has a right angle at &lt;math&gt;C&lt;/math&gt;. If &lt;math&gt;\sin A = \frac{2}{3}&lt;/math&gt;, then &lt;math&gt;\tan B&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{3}{5}\qquad<br /> \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad<br /> \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad<br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ 2&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> When &lt;math&gt;x^5, x+\frac{1}{x}&lt;/math&gt; and &lt;math&gt;1+\frac{2}{x} + \frac{3}{x^2}&lt;/math&gt; are multiplied, the product is a polynomial of degree.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> Alice sells an item at &amp;#036; &lt;math&gt;10&lt;/math&gt; less than the list price and receives &lt;math&gt;10\%&lt;/math&gt; of her selling price as her commission. <br /> Bob sells the same item at &amp;#036; &lt;math&gt;20&lt;/math&gt; less than the list price and receives &lt;math&gt;20\%&lt;/math&gt; of his selling price as his commission. <br /> If they both get the same commission, then the list price in dollars is<br /> <br /> &lt;math&gt;\textbf{(A) } 20\qquad<br /> \textbf{(B) } 30\qquad<br /> \textbf{(C) } 50\qquad<br /> \textbf{(D) } 70\qquad<br /> \textbf{(E) } 100 &lt;/math&gt;<br /> <br /> <br /> [[1983 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Let &lt;math&gt;f(x) = \frac{x+1}{x-1}&lt;/math&gt;. Then for &lt;math&gt;x^2 \neq 1, f(-x)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{f(x)}\qquad<br /> \textbf{(B)}\ -f(x)\qquad<br /> \textbf{(C)}\ \frac{1}{f(-x)}\qquad<br /> \textbf{(D)}\ -f(-x)\qquad<br /> \textbf{(E)}\ f(x)&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In a certain population the ratio of the number of women to the number of men is &lt;math&gt;11&lt;/math&gt; to &lt;math&gt;10&lt;/math&gt;. <br /> If the average (arithmetic mean) age of the women is &lt;math&gt;34&lt;/math&gt; and the average age of the men is &lt;math&gt;32&lt;/math&gt;, <br /> then the average age of the population is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\frac{9}{10}\qquad<br /> \textbf{(B)}\ 32\frac{20}{21}\qquad<br /> \textbf{(C)}\ 33\qquad<br /> \textbf{(D)}\ 33\frac{1}{21}\qquad<br /> \textbf{(E)}\ 33\frac{1}{10} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Segment &lt;math&gt;AB&lt;/math&gt; is both a diameter of a circle of radius &lt;math&gt;1&lt;/math&gt; and a side of an equilateral triangle &lt;math&gt;ABC&lt;/math&gt;. <br /> The circle also intersects &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; at points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively. The length of &lt;math&gt;AE&lt;/math&gt; is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{3}{2} \qquad <br /> \text{(B)} \ \frac{5}{3} \qquad <br /> \text{(C)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ \frac{2+\sqrt 3}{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Simplify &lt;math&gt;\sin (x-y) \cos y + \cos (x-y) \sin y&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ \sin x\qquad<br /> \textbf{(C)}\ \cos x\qquad<br /> \textbf{(D)}\ \sin x \cos 2y\qquad<br /> \textbf{(E)}\ \cos x\cos 2y &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> If &lt;math&gt;\log_2 \Big(\log_3 (\log_2 x) \Big) = 0&lt;/math&gt;, then &lt;math&gt;x^{-1/2}&lt;/math&gt; equals <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{3} \qquad <br /> \text{(B)} \ \frac{1}{2 \sqrt 3} \qquad <br /> \text{(C)}\ \frac{1}{3\sqrt 3}\qquad<br /> \text{(D)}\ \frac{1}{\sqrt{42}}\qquad<br /> \text{(E)}\ \text{none of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> If &lt;math&gt;xy = a, xz =b,&lt;/math&gt; and &lt;math&gt;yz = c&lt;/math&gt;, and none of these quantities is zero, then &lt;math&gt;x^2+y^2+z^2&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad<br /> \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad<br /> \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad<br /> \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad<br /> \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The units digit of &lt;math&gt;3^{1001}* 7^{1002}* 13^{1003}&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 9 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Three balls marked &lt;math&gt;1,2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;, are placed in an urn. One ball is drawn, its number is recorded, <br /> then the ball is returned to the urn. This process is repeated and then repeated once more, <br /> and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is &lt;math&gt;6&lt;/math&gt;, <br /> what is the probability that the ball numbered &lt;math&gt;2&lt;/math&gt; was drawn all three times? <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{27} \qquad <br /> \text{(B)} \ \frac{1}{8} \qquad <br /> \text{(C)} \ \frac{1}{7} \qquad <br /> \text{(D)}\ \frac{1}{6}\qquad<br /> \text{(E)}\ \frac{1}{3} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Let &lt;math&gt;x = .123456789101112....998999&lt;/math&gt;, where the digits are obtained by writing the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;999&lt;/math&gt; in order. <br /> The &lt;math&gt;1983&lt;/math&gt;rd digit to the right of the decimal point is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> The diagram to the right shows several numbers in the complex plane. The circle is the unit circle centered at the origin. <br /> One of these numbers is the reciprocal of &lt;math&gt;F&lt;/math&gt;. Which one? <br /> <br /> &lt;math&gt;\text{(A)} \ A \qquad <br /> \text{(B)} \ B \qquad <br /> \text{(C)} \ C \qquad <br /> \text{(D)} \ D \qquad <br /> \text{(E)} \ E &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Let &lt;math&gt;f&lt;/math&gt; be a polynomial function such that, for all real &lt;math&gt;x&lt;/math&gt;,<br /> &lt;math&gt;f(x^2 + 1) = x^4 + 5x^2 + 3&lt;/math&gt;.<br /> For all real &lt;math&gt;x, f(x^2-1)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ x^4+5x^2+1\qquad<br /> \textbf{(B)}\ x^4+x^2-3\qquad<br /> \textbf{(C)}\ x^4-5x^2+1\qquad<br /> \textbf{(D)}\ x^4+x^2+3\qquad\\<br /> \textbf{(E)}\ \text{None of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Point &lt;math&gt;D&lt;/math&gt; is on side &lt;math&gt;CB&lt;/math&gt; of triangle &lt;math&gt;ABC&lt;/math&gt;. If <br /> &lt;math&gt;\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6&lt;/math&gt;, <br /> then the length of &lt;math&gt;AD&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 2 \qquad <br /> \text{(B)} \ 2.5 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 3.5 \qquad <br /> \text{(E)} \ 4 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> If &lt;math&gt;\tan{\alpha}&lt;/math&gt; and &lt;math&gt;\tan{\beta}&lt;/math&gt; are the roots of &lt;math&gt;x^2 - px + q = 0&lt;/math&gt;, and &lt;math&gt;\cot{\alpha}&lt;/math&gt; and &lt;math&gt;\cot{\beta}&lt;/math&gt;<br /> are the roots of &lt;math&gt;x^2 - rx + s = 0&lt;/math&gt;, then &lt;math&gt;rs&lt;/math&gt; is necessarily <br /> <br /> &lt;math&gt;\text{(A)} \ pq \qquad <br /> \text{(B)} \ \frac{1}{pq} \qquad <br /> \text{(C)} \ \frac{p}{q^2} \qquad <br /> \text{(D)}\ \frac{q}{p^2}\qquad<br /> \text{(E)}\ \frac{p}{q}&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Find the smallest positive number from the numbers below <br /> <br /> &lt;math&gt;\text{(A)} \ 10-3\sqrt{11} \qquad <br /> \text{(B)} \ 3\sqrt{11}-10 \qquad <br /> \text{(C)}\ 18-5\sqrt{13}\qquad\\<br /> \text{(D)}\ 51-10\sqrt{26}\qquad<br /> \text{(E)}\ 10\sqrt{26}-51 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> Consider the two functions &lt;math&gt;f(x) = x^2+2bx+1&lt;/math&gt; and &lt;math&gt;g(x) = 2a(x+b)&lt;/math&gt;, where the variable &lt;math&gt;x&lt;/math&gt; and the constants &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers. <br /> Each such pair of the constants a and b may be considered as a point &lt;math&gt;(a,b)&lt;/math&gt; in an &lt;math&gt;ab&lt;/math&gt;-plane. <br /> Let S be the set of such points &lt;math&gt;(a,b)&lt;/math&gt; for which the graphs of &lt;math&gt;y = f(x)&lt;/math&gt; and &lt;math&gt;y = g(x)&lt;/math&gt; do NOT intersect (in the &lt;math&gt;xy&lt;/math&gt;- plane.). The area of &lt;math&gt;S&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ \pi \qquad <br /> \text{(C)} \ 4 \qquad <br /> \text{(D)} \ 4 \pi \qquad <br /> \text{(E)} \ \infty &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> In the adjoining figure the five circles are tangent to one another consecutively and to the lines <br /> &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; (&lt;math&gt;L_1&lt;/math&gt; is the line that is above the circles and &lt;math&gt;L_2&lt;/math&gt; is the line that goes under the circles). <br /> If the radius of the largest circle is &lt;math&gt;18&lt;/math&gt; and that of the smallest one is &lt;math&gt;8&lt;/math&gt;, then the radius of the middle circle is<br /> <br /> &lt;asy&gt;<br /> size(250);defaultpen(linewidth(0.7));<br /> real alpha=5.797939254, x=71.191836;<br /> int i;<br /> for(i=0; i&lt;5; i=i+1) {<br /> real r=8*(sqrt(6)/2)^i;<br /> draw(Circle((x+r)*dir(alpha), r));<br /> x=x+2r;<br /> }<br /> real x=71.191836+40+20*sqrt(6), r=18;<br /> pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2);<br /> pair A1=300*dir(origin--A), B1=300*dir(origin--B);<br /> draw(B1--origin--A1);<br /> pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X,<br /> Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y,<br /> Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z;<br /> clip(X--Y--Y1--X1--cycle);<br /> label(&quot;L_2$&quot;, Z, S);<br /> label(&quot;$L_1$&quot;, Z1, dir(2*alpha)*dir(90));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ 12 \qquad <br /> \text{(B)} \ 12.5 \qquad <br /> \text{(C)} \ 13 \qquad <br /> \text{(D)} \ 13.5 \qquad <br /> \text{(E)} \ 14 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> How many non-congruent right triangles are there such that the perimeter in &lt;math&gt;\text{cm}&lt;/math&gt; and the area in &lt;math&gt;\text{cm}^2&lt;/math&gt; are numerically equal? <br /> <br /> &lt;math&gt;\text{(A)} \ \text{none} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ 2 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ \infty&lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> If &lt;math&gt;60^a = 3&lt;/math&gt; and &lt;math&gt;60^b = 5&lt;/math&gt;, then &lt;math&gt;12^{[(1-a-b)/2(1-b)]} &lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ \sqrt{3} \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ \sqrt{5} \qquad <br /> \text{(D)} \ 3 \qquad <br /> \text{(E)} \ \sqrt{12} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 25|Solution]]<br /> <br /> == Problem 26 ==<br /> <br /> The probability that event &lt;math&gt;A&lt;/math&gt; occurs is &lt;math&gt;\frac{3}{4}&lt;/math&gt;; the probability that event B occurs is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. <br /> Let &lt;math&gt;p&lt;/math&gt; be the probability that both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; occur. The smallest interval necessarily containing &lt;math&gt;p&lt;/math&gt; is the interval<br /> <br /> &lt;math&gt;\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad<br /> \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad<br /> \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big] &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27 ==<br /> <br /> A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance <br /> of &lt;math&gt;10&lt;/math&gt; m from the point where the sphere touches the ground. At the same instant a meter stick <br /> (held vertically with one end on the ground) casts a shadow of length &lt;math&gt;2&lt;/math&gt; m. What is the radius of the sphere in meters? <br /> (Assume the sun's rays are parallel and the meter stick is a line segment.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{5}{2}\qquad<br /> \textbf{(B)}\ 9 - 4\sqrt{5}\qquad<br /> \textbf{(C)}\ 8\sqrt{10} - 23\qquad<br /> \textbf{(D)}\ 6-\sqrt{15}\qquad<br /> \textbf{(E)}\ 10\sqrt{5}-20 &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28 ==<br /> <br /> Triangle &lt;math&gt;\triangle ABC&lt;/math&gt; in the figure has area &lt;math&gt;10&lt;/math&gt;. Points &lt;math&gt;D, E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, all distinct from &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, <br /> are on sides &lt;math&gt;AB, BC&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; respectively, and &lt;math&gt;AD = 2, DB = 3&lt;/math&gt;. If triangle &lt;math&gt;\triangle ABE&lt;/math&gt; and quadrilateral &lt;math&gt;DBEF&lt;/math&gt; <br /> have equal areas, then that area is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B);<br /> draw(A--B--C--A--E--F--D);<br /> pair point=incenter(A,B,C);<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, dir(point--C));<br /> label(&quot;$D$&quot;, D, dir(point--D));<br /> label(&quot;$E$&quot;, E, dir(point--E));<br /> label(&quot;$F$&quot;, F, dir(point--F));<br /> label(&quot;$2$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (7,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad<br /> \textbf{(B)}\ 5\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad<br /> \textbf{(E)}\ \text{not uniquely determined} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29 ==<br /> <br /> A point &lt;math&gt;P&lt;/math&gt; lies in the same plane as a given square of side &lt;math&gt;1&lt;/math&gt;. Let the vertices of the square, <br /> taken counterclockwise, be &lt;math&gt;A, B, C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Also, let the distances from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, be &lt;math&gt;u, v&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;. <br /> What is the greatest distance that &lt;math&gt;P&lt;/math&gt; can be from &lt;math&gt;D&lt;/math&gt; if &lt;math&gt;u^2 + v^2 = w^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2}\qquad<br /> \textbf{(B)}\ 2\sqrt{2}\qquad<br /> \textbf{(C)}\ 2 + \sqrt{2}\qquad<br /> \textbf{(D)}\ 3\sqrt{2}\qquad<br /> \textbf{(E)}\ 3+\sqrt{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30 ==<br /> <br /> Distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are on a semicircle with diameter &lt;math&gt;MN&lt;/math&gt; and center &lt;math&gt;C&lt;/math&gt;. <br /> The point &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;CN&lt;/math&gt; and &lt;math&gt;\angle CAP = \angle CBP = 10^{\circ}&lt;/math&gt;. If &lt;math&gt;\stackrel{\frown}{MA} = 40^{\circ}&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{BN}&lt;/math&gt; equals<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair C=origin, N=dir(0), B=dir(20), A=dir(135), M=dir(180), P=(3/7)*dir(C--N);<br /> draw(M--N^^C--A--P--B--C^^Arc(origin,1,0,180));<br /> markscalefactor=0.03;<br /> draw(anglemark(C,A,P));<br /> draw(anglemark(C,B,P));<br /> pair point=C;<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, S);<br /> label(&quot;$M$&quot;, M, dir(point--M));<br /> label(&quot;$N$&quot;, N, dir(point--N));<br /> label(&quot;$P$&quot;, P, S);<br /> label(&quot;$40^\circ$&quot;, C, NW);<br /> label(&quot;$10^\circ$&quot;, B, SE);<br /> label(&quot;$10^\circ$&quot;, A+(0.05,0.02), SW);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10^{\circ}\qquad<br /> \textbf{(B)}\ 15^{\circ}\qquad<br /> \textbf{(C)}\ 20^{\circ}\qquad<br /> \textbf{(D)}\ 25^{\circ}\qquad<br /> \textbf{(E)}\ 30^{\circ} &lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME box|year=1983|before=[[1982 AHSME]]|after=[[1984 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems&diff=96058 1983 AHSME Problems 2018-07-09T14:58:18Z <p>Wlm7: /* Problem 30 */</p> <hr /> <div>== Problem 1 ==<br /> <br /> If &lt;math&gt;x \neq 0, \frac x{2} = y^2&lt;/math&gt; and &lt;math&gt;\frac{x}{4} = 4y&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad<br /> \textbf{(B)}\ 16\qquad<br /> \textbf{(C)}\ 32\qquad<br /> \textbf{(D)}\ 64\qquad<br /> \textbf{(E)}\ 128 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Point &lt;math&gt;P&lt;/math&gt; is outside circle &lt;math&gt;C&lt;/math&gt; on the plane. At most how many points on &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;3 \text{cm}&lt;/math&gt; from P? <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Three primes &lt;math&gt;p,q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; satisfy &lt;math&gt;p+q = r&lt;/math&gt; and &lt;math&gt;1 &lt; p &lt; q&lt;/math&gt;. Then &lt;math&gt;p&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 7\qquad<br /> \textbf{(D)}\ 13\qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Position &lt;math&gt;A,B,C,D,E,F&lt;/math&gt; such that &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; are parallel, as are sides &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;, <br /> and sides &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt;. Each side has length of &lt;math&gt;1&lt;/math&gt; and it is given that &lt;math&gt;\measuredangle FAB = \measuredangle BCD = 60^\circ&lt;/math&gt;. <br /> The area of the figure is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ \frac{3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has a right angle at &lt;math&gt;C&lt;/math&gt;. If &lt;math&gt;\sin A = \frac{2}{3}&lt;/math&gt;, then &lt;math&gt;\tan B&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{3}{5}\qquad<br /> \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad<br /> \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad<br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ 2&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> When &lt;math&gt;x^5, x+\frac{1}{x}&lt;/math&gt; and &lt;math&gt;1+\frac{2}{x} + \frac{3}{x^2}&lt;/math&gt; are multiplied, the product is a polynomial of degree.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> Alice sells an item at &amp;#036; &lt;math&gt;10&lt;/math&gt; less than the list price and receives &lt;math&gt;10\%&lt;/math&gt; of her selling price as her commission. <br /> Bob sells the same item at &amp;#036; &lt;math&gt;20&lt;/math&gt; less than the list price and receives &lt;math&gt;20\%&lt;/math&gt; of his selling price as his commission. <br /> If they both get the same commission, then the list price in dollars is<br /> <br /> &lt;math&gt;\textbf{(A) } 20\qquad<br /> \textbf{(B) } 30\qquad<br /> \textbf{(C) } 50\qquad<br /> \textbf{(D) } 70\qquad<br /> \textbf{(E) } 100 &lt;/math&gt;<br /> <br /> <br /> [[1983 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Let &lt;math&gt;f(x) = \frac{x+1}{x-1}&lt;/math&gt;. Then for &lt;math&gt;x^2 \neq 1, f(-x)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{f(x)}\qquad<br /> \textbf{(B)}\ -f(x)\qquad<br /> \textbf{(C)}\ \frac{1}{f(-x)}\qquad<br /> \textbf{(D)}\ -f(-x)\qquad<br /> \textbf{(E)}\ f(x)&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In a certain population the ratio of the number of women to the number of men is &lt;math&gt;11&lt;/math&gt; to &lt;math&gt;10&lt;/math&gt;. <br /> If the average (arithmetic mean) age of the women is &lt;math&gt;34&lt;/math&gt; and the average age of the men is &lt;math&gt;32&lt;/math&gt;, <br /> then the average age of the population is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\frac{9}{10}\qquad<br /> \textbf{(B)}\ 32\frac{20}{21}\qquad<br /> \textbf{(C)}\ 33\qquad<br /> \textbf{(D)}\ 33\frac{1}{21}\qquad<br /> \textbf{(E)}\ 33\frac{1}{10} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Segment &lt;math&gt;AB&lt;/math&gt; is both a diameter of a circle of radius &lt;math&gt;1&lt;/math&gt; and a side of an equilateral triangle &lt;math&gt;ABC&lt;/math&gt;. <br /> The circle also intersects &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; at points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively. The length of &lt;math&gt;AE&lt;/math&gt; is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{3}{2} \qquad <br /> \text{(B)} \ \frac{5}{3} \qquad <br /> \text{(C)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ \frac{2+\sqrt 3}{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Simplify &lt;math&gt;\sin (x-y) \cos y + \cos (x-y) \sin y&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ \sin x\qquad<br /> \textbf{(C)}\ \cos x\qquad<br /> \textbf{(D)}\ \sin x \cos 2y\qquad<br /> \textbf{(E)}\ \cos x\cos 2y &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> If &lt;math&gt;\log_2 \Big(\log_3 (\log_2 x) \Big) = 0&lt;/math&gt;, then &lt;math&gt;x^{-1/2}&lt;/math&gt; equals <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{3} \qquad <br /> \text{(B)} \ \frac{1}{2 \sqrt 3} \qquad <br /> \text{(C)}\ \frac{1}{3\sqrt 3}\qquad<br /> \text{(D)}\ \frac{1}{\sqrt{42}}\qquad<br /> \text{(E)}\ \text{none of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> If &lt;math&gt;xy = a, xz =b,&lt;/math&gt; and &lt;math&gt;yz = c&lt;/math&gt;, and none of these quantities is zero, then &lt;math&gt;x^2+y^2+z^2&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad<br /> \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad<br /> \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad<br /> \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad<br /> \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The units digit of &lt;math&gt;3^{1001}* 7^{1002}* 13^{1003}&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 9 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Three balls marked &lt;math&gt;1,2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;, are placed in an urn. One ball is drawn, its number is recorded, <br /> then the ball is returned to the urn. This process is repeated and then repeated once more, <br /> and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is &lt;math&gt;6&lt;/math&gt;, <br /> what is the probability that the ball numbered &lt;math&gt;2&lt;/math&gt; was drawn all three times? <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{27} \qquad <br /> \text{(B)} \ \frac{1}{8} \qquad <br /> \text{(C)} \ \frac{1}{7} \qquad <br /> \text{(D)}\ \frac{1}{6}\qquad<br /> \text{(E)}\ \frac{1}{3} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Let &lt;math&gt;x = .123456789101112....998999&lt;/math&gt;, where the digits are obtained by writing the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;999&lt;/math&gt; in order. <br /> The &lt;math&gt;1983&lt;/math&gt;rd digit to the right of the decimal point is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> The diagram to the right shows several numbers in the complex plane. The circle is the unit circle centered at the origin. <br /> One of these numbers is the reciprocal of &lt;math&gt;F&lt;/math&gt;. Which one? <br /> <br /> &lt;math&gt;\text{(A)} \ A \qquad <br /> \text{(B)} \ B \qquad <br /> \text{(C)} \ C \qquad <br /> \text{(D)} \ D \qquad <br /> \text{(E)} \ E &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Let &lt;math&gt;f&lt;/math&gt; be a polynomial function such that, for all real &lt;math&gt;x&lt;/math&gt;,<br /> &lt;math&gt;f(x^2 + 1) = x^4 + 5x^2 + 3&lt;/math&gt;.<br /> For all real &lt;math&gt;x, f(x^2-1)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ x^4+5x^2+1\qquad<br /> \textbf{(B)}\ x^4+x^2-3\qquad<br /> \textbf{(C)}\ x^4-5x^2+1\qquad<br /> \textbf{(D)}\ x^4+x^2+3\qquad\\<br /> \textbf{(E)}\ \text{None of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Point &lt;math&gt;D&lt;/math&gt; is on side &lt;math&gt;CB&lt;/math&gt; of triangle &lt;math&gt;ABC&lt;/math&gt;. If <br /> &lt;math&gt;\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6&lt;/math&gt;, <br /> then the length of &lt;math&gt;AD&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 2 \qquad <br /> \text{(B)} \ 2.5 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 3.5 \qquad <br /> \text{(E)} \ 4 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> If &lt;math&gt;\tan{\alpha}&lt;/math&gt; and &lt;math&gt;\tan{\beta}&lt;/math&gt; are the roots of &lt;math&gt;x^2 - px + q = 0&lt;/math&gt;, and &lt;math&gt;\cot{\alpha}&lt;/math&gt; and &lt;math&gt;\cot{\beta}&lt;/math&gt;<br /> are the roots of &lt;math&gt;x^2 - rx + s = 0&lt;/math&gt;, then &lt;math&gt;rs&lt;/math&gt; is necessarily <br /> <br /> &lt;math&gt;\text{(A)} \ pq \qquad <br /> \text{(B)} \ \frac{1}{pq} \qquad <br /> \text{(C)} \ \frac{p}{q^2} \qquad <br /> \text{(D)}\ \frac{q}{p^2}\qquad<br /> \text{(E)}\ \frac{p}{q}&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Find the smallest positive number from the numbers below <br /> <br /> &lt;math&gt;\text{(A)} \ 10-3\sqrt{11} \qquad <br /> \text{(B)} \ 3\sqrt{11}-10 \qquad <br /> \text{(C)}\ 18-5\sqrt{13}\qquad\\<br /> \text{(D)}\ 51-10\sqrt{26}\qquad<br /> \text{(E)}\ 10\sqrt{26}-51 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> Consider the two functions &lt;math&gt;f(x) = x^2+2bx+1&lt;/math&gt; and &lt;math&gt;g(x) = 2a(x+b)&lt;/math&gt;, where the variable &lt;math&gt;x&lt;/math&gt; and the constants &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers. <br /> Each such pair of the constants a and b may be considered as a point &lt;math&gt;(a,b)&lt;/math&gt; in an &lt;math&gt;ab&lt;/math&gt;-plane. <br /> Let S be the set of such points &lt;math&gt;(a,b)&lt;/math&gt; for which the graphs of &lt;math&gt;y = f(x)&lt;/math&gt; and &lt;math&gt;y = g(x)&lt;/math&gt; do NOT intersect (in the &lt;math&gt;xy&lt;/math&gt;- plane.). The area of &lt;math&gt;S&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ \pi \qquad <br /> \text{(C)} \ 4 \qquad <br /> \text{(D)} \ 4 \pi \qquad <br /> \text{(E)} \ \infty &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> In the adjoining figure the five circles are tangent to one another consecutively and to the lines <br /> &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; (&lt;math&gt;L_1&lt;/math&gt; is the line that is above the circles and &lt;math&gt;L_2&lt;/math&gt; is the line that goes under the circles). <br /> If the radius of the largest circle is &lt;math&gt;18&lt;/math&gt; and that of the smallest one is &lt;math&gt;8&lt;/math&gt;, then the radius of the middle circle is<br /> <br /> &lt;asy&gt;<br /> size(250);defaultpen(linewidth(0.7));<br /> real alpha=5.797939254, x=71.191836;<br /> int i;<br /> for(i=0; i&lt;5; i=i+1) {<br /> real r=8*(sqrt(6)/2)^i;<br /> draw(Circle((x+r)*dir(alpha), r));<br /> x=x+2r;<br /> }<br /> real x=71.191836+40+20*sqrt(6), r=18;<br /> pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2);<br /> pair A1=300*dir(origin--A), B1=300*dir(origin--B);<br /> draw(B1--origin--A1);<br /> pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X,<br /> Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y,<br /> Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z;<br /> clip(X--Y--Y1--X1--cycle);<br /> label(&quot;$L_2$&quot;, Z, S);<br /> label(&quot;$L_1$&quot;, Z1, dir(2*alpha)*dir(90));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ 12 \qquad <br /> \text{(B)} \ 12.5 \qquad <br /> \text{(C)} \ 13 \qquad <br /> \text{(D)} \ 13.5 \qquad <br /> \text{(E)} \ 14 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> How many non-congruent right triangles are there such that the perimeter in &lt;math&gt;\text{cm}&lt;/math&gt; and the area in &lt;math&gt;\text{cm}^2&lt;/math&gt; are numerically equal? <br /> <br /> &lt;math&gt;\text{(A)} \ \text{none} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ 2 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ \infty&lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> If &lt;math&gt;60^a = 3&lt;/math&gt; and &lt;math&gt;60^b = 5&lt;/math&gt;, then &lt;math&gt;12^{[(1-a-b)/2(1-b)]} &lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ \sqrt{3} \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ \sqrt{5} \qquad <br /> \text{(D)} \ 3 \qquad <br /> \text{(E)} \ \sqrt{12} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 25|Solution]]<br /> <br /> == Problem 26 ==<br /> <br /> The probability that event &lt;math&gt;A&lt;/math&gt; occurs is &lt;math&gt;\frac{3}{4}&lt;/math&gt;; the probability that event B occurs is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. <br /> Let &lt;math&gt;p&lt;/math&gt; be the probability that both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; occur. The smallest interval necessarily containing &lt;math&gt;p&lt;/math&gt; is the interval<br /> <br /> &lt;math&gt;\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad<br /> \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad<br /> \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big] &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27 ==<br /> <br /> A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance <br /> of &lt;math&gt;10&lt;/math&gt; m from the point where the sphere touches the ground. At the same instant a meter stick <br /> (held vertically with one end on the ground) casts a shadow of length &lt;math&gt;2&lt;/math&gt; m. What is the radius of the sphere in meters? <br /> (Assume the sun's rays are parallel and the meter stick is a line segment.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{5}{2}\qquad<br /> \textbf{(B)}\ 9 - 4\sqrt{5}\qquad<br /> \textbf{(C)}\ 8\sqrt{10} - 23\qquad<br /> \textbf{(D)}\ 6-\sqrt{15}\qquad<br /> \textbf{(E)}\ 10\sqrt{5}-20 &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28 ==<br /> <br /> Triangle &lt;math&gt;\triangle ABC&lt;/math&gt; in the figure has area &lt;math&gt;10&lt;/math&gt;. Points &lt;math&gt;D, E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, all distinct from &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, <br /> are on sides &lt;math&gt;AB, BC&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; respectively, and &lt;math&gt;AD = 2, DB = 3&lt;/math&gt;. If triangle &lt;math&gt;\triangle ABE&lt;/math&gt; and quadrilateral &lt;math&gt;DBEF&lt;/math&gt; <br /> have equal areas, then that area is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B);<br /> draw(A--B--C--A--E--F--D);<br /> pair point=incenter(A,B,C);<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, dir(point--C));<br /> label(&quot;$D$&quot;, D, dir(point--D));<br /> label(&quot;$E$&quot;, E, dir(point--E));<br /> label(&quot;$F$&quot;, F, dir(point--F));<br /> label(&quot;$2$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (7,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad<br /> \textbf{(B)}\ 5\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad<br /> \textbf{(E)}\ \text{not uniquely determined} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29 ==<br /> <br /> A point &lt;math&gt;P&lt;/math&gt; lies in the same plane as a given square of side &lt;math&gt;1&lt;/math&gt;. Let the vertices of the square, <br /> taken counterclockwise, be &lt;math&gt;A, B, C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Also, let the distances from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, be &lt;math&gt;u, v&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;. <br /> What is the greatest distance that &lt;math&gt;P&lt;/math&gt; can be from &lt;math&gt;D&lt;/math&gt; if &lt;math&gt;u^2 + v^2 = w^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2}\qquad<br /> \textbf{(B)}\ 2\sqrt{2}\qquad<br /> \textbf{(C)}\ 2 + \sqrt{2}\qquad<br /> \textbf{(D)}\ 3\sqrt{2}\qquad<br /> \textbf{(E)}\ 3+\sqrt{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30 ==<br /> <br /> Distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are on a semicircle with diameter &lt;math&gt;MN&lt;/math&gt; and center &lt;math&gt;C&lt;/math&gt;. <br /> The point &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;CN&lt;/math&gt; and &lt;math&gt;\angle CAP = \angle CBP = 10^{\circ}&lt;/math&gt;. If &lt;math&gt;\stackrel{\frown}{MA} = 40^{\circ}&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{BN}&lt;/math&gt; equals<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair C=origin, N=dir(0), B=dir(20), A=dir(135), M=dir(180), P=(3/7)*dir(C--N);<br /> draw(M--N^^C--A--P--B--C^^Arc(origin,1,0,180));<br /> markscalefactor=0.03;<br /> draw(anglemark(C,A,P));<br /> draw(anglemark(C,B,P));<br /> pair point=C;<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, S);<br /> label(&quot;$M$&quot;, M, dir(point--M));<br /> label(&quot;$N$&quot;, N, dir(point--N));<br /> label(&quot;$P$&quot;, P, S);<br /> label(&quot;$40^\circ$&quot;, C+(-0.15,0), NW);<br /> label(&quot;$10^\circ$&quot;, B+(0,0.05), SE);<br /> label(&quot;$10^\circ$&quot;, A+(0.05,0.02), SW);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10^{\circ}\qquad<br /> \textbf{(B)}\ 15^{\circ}\qquad<br /> \textbf{(C)}\ 20^{\circ}\qquad<br /> \textbf{(D)}\ 25^{\circ}\qquad<br /> \textbf{(E)}\ 30^{\circ} &lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME box|year=1983|before=[[1982 AHSME]]|after=[[1984 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems&diff=96057 1983 AHSME Problems 2018-07-09T14:57:33Z <p>Wlm7: /* Problem 30 */</p> <hr /> <div>== Problem 1 ==<br /> <br /> If &lt;math&gt;x \neq 0, \frac x{2} = y^2&lt;/math&gt; and &lt;math&gt;\frac{x}{4} = 4y&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad<br /> \textbf{(B)}\ 16\qquad<br /> \textbf{(C)}\ 32\qquad<br /> \textbf{(D)}\ 64\qquad<br /> \textbf{(E)}\ 128 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Point &lt;math&gt;P&lt;/math&gt; is outside circle &lt;math&gt;C&lt;/math&gt; on the plane. At most how many points on &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;3 \text{cm}&lt;/math&gt; from P? <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Three primes &lt;math&gt;p,q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; satisfy &lt;math&gt;p+q = r&lt;/math&gt; and &lt;math&gt;1 &lt; p &lt; q&lt;/math&gt;. Then &lt;math&gt;p&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 7\qquad<br /> \textbf{(D)}\ 13\qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Position &lt;math&gt;A,B,C,D,E,F&lt;/math&gt; such that &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; are parallel, as are sides &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;, <br /> and sides &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt;. Each side has length of &lt;math&gt;1&lt;/math&gt; and it is given that &lt;math&gt;\measuredangle FAB = \measuredangle BCD = 60^\circ&lt;/math&gt;. <br /> The area of the figure is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ \frac{3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has a right angle at &lt;math&gt;C&lt;/math&gt;. If &lt;math&gt;\sin A = \frac{2}{3}&lt;/math&gt;, then &lt;math&gt;\tan B&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{3}{5}\qquad<br /> \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad<br /> \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad<br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ 2&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> When &lt;math&gt;x^5, x+\frac{1}{x}&lt;/math&gt; and &lt;math&gt;1+\frac{2}{x} + \frac{3}{x^2}&lt;/math&gt; are multiplied, the product is a polynomial of degree.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> Alice sells an item at &amp;#036; &lt;math&gt;10&lt;/math&gt; less than the list price and receives &lt;math&gt;10\%&lt;/math&gt; of her selling price as her commission. <br /> Bob sells the same item at &amp;#036; &lt;math&gt;20&lt;/math&gt; less than the list price and receives &lt;math&gt;20\%&lt;/math&gt; of his selling price as his commission. <br /> If they both get the same commission, then the list price in dollars is<br /> <br /> &lt;math&gt;\textbf{(A) } 20\qquad<br /> \textbf{(B) } 30\qquad<br /> \textbf{(C) } 50\qquad<br /> \textbf{(D) } 70\qquad<br /> \textbf{(E) } 100 &lt;/math&gt;<br /> <br /> <br /> [[1983 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Let &lt;math&gt;f(x) = \frac{x+1}{x-1}&lt;/math&gt;. Then for &lt;math&gt;x^2 \neq 1, f(-x)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{f(x)}\qquad<br /> \textbf{(B)}\ -f(x)\qquad<br /> \textbf{(C)}\ \frac{1}{f(-x)}\qquad<br /> \textbf{(D)}\ -f(-x)\qquad<br /> \textbf{(E)}\ f(x)&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In a certain population the ratio of the number of women to the number of men is &lt;math&gt;11&lt;/math&gt; to &lt;math&gt;10&lt;/math&gt;. <br /> If the average (arithmetic mean) age of the women is &lt;math&gt;34&lt;/math&gt; and the average age of the men is &lt;math&gt;32&lt;/math&gt;, <br /> then the average age of the population is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\frac{9}{10}\qquad<br /> \textbf{(B)}\ 32\frac{20}{21}\qquad<br /> \textbf{(C)}\ 33\qquad<br /> \textbf{(D)}\ 33\frac{1}{21}\qquad<br /> \textbf{(E)}\ 33\frac{1}{10} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Segment &lt;math&gt;AB&lt;/math&gt; is both a diameter of a circle of radius &lt;math&gt;1&lt;/math&gt; and a side of an equilateral triangle &lt;math&gt;ABC&lt;/math&gt;. <br /> The circle also intersects &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; at points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively. The length of &lt;math&gt;AE&lt;/math&gt; is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{3}{2} \qquad <br /> \text{(B)} \ \frac{5}{3} \qquad <br /> \text{(C)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ \frac{2+\sqrt 3}{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Simplify &lt;math&gt;\sin (x-y) \cos y + \cos (x-y) \sin y&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ \sin x\qquad<br /> \textbf{(C)}\ \cos x\qquad<br /> \textbf{(D)}\ \sin x \cos 2y\qquad<br /> \textbf{(E)}\ \cos x\cos 2y &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> If &lt;math&gt;\log_2 \Big(\log_3 (\log_2 x) \Big) = 0&lt;/math&gt;, then &lt;math&gt;x^{-1/2}&lt;/math&gt; equals <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{3} \qquad <br /> \text{(B)} \ \frac{1}{2 \sqrt 3} \qquad <br /> \text{(C)}\ \frac{1}{3\sqrt 3}\qquad<br /> \text{(D)}\ \frac{1}{\sqrt{42}}\qquad<br /> \text{(E)}\ \text{none of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> If &lt;math&gt;xy = a, xz =b,&lt;/math&gt; and &lt;math&gt;yz = c&lt;/math&gt;, and none of these quantities is zero, then &lt;math&gt;x^2+y^2+z^2&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad<br /> \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad<br /> \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad<br /> \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad<br /> \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The units digit of &lt;math&gt;3^{1001}* 7^{1002}* 13^{1003}&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 9 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Three balls marked &lt;math&gt;1,2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;, are placed in an urn. One ball is drawn, its number is recorded, <br /> then the ball is returned to the urn. This process is repeated and then repeated once more, <br /> and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is &lt;math&gt;6&lt;/math&gt;, <br /> what is the probability that the ball numbered &lt;math&gt;2&lt;/math&gt; was drawn all three times? <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{27} \qquad <br /> \text{(B)} \ \frac{1}{8} \qquad <br /> \text{(C)} \ \frac{1}{7} \qquad <br /> \text{(D)}\ \frac{1}{6}\qquad<br /> \text{(E)}\ \frac{1}{3} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Let &lt;math&gt;x = .123456789101112....998999&lt;/math&gt;, where the digits are obtained by writing the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;999&lt;/math&gt; in order. <br /> The &lt;math&gt;1983&lt;/math&gt;rd digit to the right of the decimal point is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> The diagram to the right shows several numbers in the complex plane. The circle is the unit circle centered at the origin. <br /> One of these numbers is the reciprocal of &lt;math&gt;F&lt;/math&gt;. Which one? <br /> <br /> &lt;math&gt;\text{(A)} \ A \qquad <br /> \text{(B)} \ B \qquad <br /> \text{(C)} \ C \qquad <br /> \text{(D)} \ D \qquad <br /> \text{(E)} \ E &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Let &lt;math&gt;f&lt;/math&gt; be a polynomial function such that, for all real &lt;math&gt;x&lt;/math&gt;,<br /> &lt;math&gt;f(x^2 + 1) = x^4 + 5x^2 + 3&lt;/math&gt;.<br /> For all real &lt;math&gt;x, f(x^2-1)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ x^4+5x^2+1\qquad<br /> \textbf{(B)}\ x^4+x^2-3\qquad<br /> \textbf{(C)}\ x^4-5x^2+1\qquad<br /> \textbf{(D)}\ x^4+x^2+3\qquad\\<br /> \textbf{(E)}\ \text{None of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Point &lt;math&gt;D&lt;/math&gt; is on side &lt;math&gt;CB&lt;/math&gt; of triangle &lt;math&gt;ABC&lt;/math&gt;. If <br /> &lt;math&gt;\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6&lt;/math&gt;, <br /> then the length of &lt;math&gt;AD&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 2 \qquad <br /> \text{(B)} \ 2.5 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 3.5 \qquad <br /> \text{(E)} \ 4 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> If &lt;math&gt;\tan{\alpha}&lt;/math&gt; and &lt;math&gt;\tan{\beta}&lt;/math&gt; are the roots of &lt;math&gt;x^2 - px + q = 0&lt;/math&gt;, and &lt;math&gt;\cot{\alpha}&lt;/math&gt; and &lt;math&gt;\cot{\beta}&lt;/math&gt;<br /> are the roots of &lt;math&gt;x^2 - rx + s = 0&lt;/math&gt;, then &lt;math&gt;rs&lt;/math&gt; is necessarily <br /> <br /> &lt;math&gt;\text{(A)} \ pq \qquad <br /> \text{(B)} \ \frac{1}{pq} \qquad <br /> \text{(C)} \ \frac{p}{q^2} \qquad <br /> \text{(D)}\ \frac{q}{p^2}\qquad<br /> \text{(E)}\ \frac{p}{q}&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Find the smallest positive number from the numbers below <br /> <br /> &lt;math&gt;\text{(A)} \ 10-3\sqrt{11} \qquad <br /> \text{(B)} \ 3\sqrt{11}-10 \qquad <br /> \text{(C)}\ 18-5\sqrt{13}\qquad\\<br /> \text{(D)}\ 51-10\sqrt{26}\qquad<br /> \text{(E)}\ 10\sqrt{26}-51 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> Consider the two functions &lt;math&gt;f(x) = x^2+2bx+1&lt;/math&gt; and &lt;math&gt;g(x) = 2a(x+b)&lt;/math&gt;, where the variable &lt;math&gt;x&lt;/math&gt; and the constants &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers. <br /> Each such pair of the constants a and b may be considered as a point &lt;math&gt;(a,b)&lt;/math&gt; in an &lt;math&gt;ab&lt;/math&gt;-plane. <br /> Let S be the set of such points &lt;math&gt;(a,b)&lt;/math&gt; for which the graphs of &lt;math&gt;y = f(x)&lt;/math&gt; and &lt;math&gt;y = g(x)&lt;/math&gt; do NOT intersect (in the &lt;math&gt;xy&lt;/math&gt;- plane.). The area of &lt;math&gt;S&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ \pi \qquad <br /> \text{(C)} \ 4 \qquad <br /> \text{(D)} \ 4 \pi \qquad <br /> \text{(E)} \ \infty &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> In the adjoining figure the five circles are tangent to one another consecutively and to the lines <br /> &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; (&lt;math&gt;L_1&lt;/math&gt; is the line that is above the circles and &lt;math&gt;L_2&lt;/math&gt; is the line that goes under the circles). <br /> If the radius of the largest circle is &lt;math&gt;18&lt;/math&gt; and that of the smallest one is &lt;math&gt;8&lt;/math&gt;, then the radius of the middle circle is<br /> <br /> &lt;asy&gt;<br /> size(250);defaultpen(linewidth(0.7));<br /> real alpha=5.797939254, x=71.191836;<br /> int i;<br /> for(i=0; i&lt;5; i=i+1) {<br /> real r=8*(sqrt(6)/2)^i;<br /> draw(Circle((x+r)*dir(alpha), r));<br /> x=x+2r;<br /> }<br /> real x=71.191836+40+20*sqrt(6), r=18;<br /> pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2);<br /> pair A1=300*dir(origin--A), B1=300*dir(origin--B);<br /> draw(B1--origin--A1);<br /> pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X,<br /> Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y,<br /> Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z;<br /> clip(X--Y--Y1--X1--cycle);<br /> label(&quot;$L_2$&quot;, Z, S);<br /> label(&quot;$L_1$&quot;, Z1, dir(2*alpha)*dir(90));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ 12 \qquad <br /> \text{(B)} \ 12.5 \qquad <br /> \text{(C)} \ 13 \qquad <br /> \text{(D)} \ 13.5 \qquad <br /> \text{(E)} \ 14 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> How many non-congruent right triangles are there such that the perimeter in &lt;math&gt;\text{cm}&lt;/math&gt; and the area in &lt;math&gt;\text{cm}^2&lt;/math&gt; are numerically equal? <br /> <br /> &lt;math&gt;\text{(A)} \ \text{none} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ 2 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ \infty&lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> If &lt;math&gt;60^a = 3&lt;/math&gt; and &lt;math&gt;60^b = 5&lt;/math&gt;, then &lt;math&gt;12^{[(1-a-b)/2(1-b)]} &lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ \sqrt{3} \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ \sqrt{5} \qquad <br /> \text{(D)} \ 3 \qquad <br /> \text{(E)} \ \sqrt{12} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 25|Solution]]<br /> <br /> == Problem 26 ==<br /> <br /> The probability that event &lt;math&gt;A&lt;/math&gt; occurs is &lt;math&gt;\frac{3}{4}&lt;/math&gt;; the probability that event B occurs is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. <br /> Let &lt;math&gt;p&lt;/math&gt; be the probability that both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; occur. The smallest interval necessarily containing &lt;math&gt;p&lt;/math&gt; is the interval<br /> <br /> &lt;math&gt;\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad<br /> \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad<br /> \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big] &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27 ==<br /> <br /> A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance <br /> of &lt;math&gt;10&lt;/math&gt; m from the point where the sphere touches the ground. At the same instant a meter stick <br /> (held vertically with one end on the ground) casts a shadow of length &lt;math&gt;2&lt;/math&gt; m. What is the radius of the sphere in meters? <br /> (Assume the sun's rays are parallel and the meter stick is a line segment.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{5}{2}\qquad<br /> \textbf{(B)}\ 9 - 4\sqrt{5}\qquad<br /> \textbf{(C)}\ 8\sqrt{10} - 23\qquad<br /> \textbf{(D)}\ 6-\sqrt{15}\qquad<br /> \textbf{(E)}\ 10\sqrt{5}-20 &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28 ==<br /> <br /> Triangle &lt;math&gt;\triangle ABC&lt;/math&gt; in the figure has area &lt;math&gt;10&lt;/math&gt;. Points &lt;math&gt;D, E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, all distinct from &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, <br /> are on sides &lt;math&gt;AB, BC&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; respectively, and &lt;math&gt;AD = 2, DB = 3&lt;/math&gt;. If triangle &lt;math&gt;\triangle ABE&lt;/math&gt; and quadrilateral &lt;math&gt;DBEF&lt;/math&gt; <br /> have equal areas, then that area is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B);<br /> draw(A--B--C--A--E--F--D);<br /> pair point=incenter(A,B,C);<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, dir(point--C));<br /> label(&quot;$D$&quot;, D, dir(point--D));<br /> label(&quot;$E$&quot;, E, dir(point--E));<br /> label(&quot;$F$&quot;, F, dir(point--F));<br /> label(&quot;$2$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (7,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad<br /> \textbf{(B)}\ 5\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad<br /> \textbf{(E)}\ \text{not uniquely determined} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29 ==<br /> <br /> A point &lt;math&gt;P&lt;/math&gt; lies in the same plane as a given square of side &lt;math&gt;1&lt;/math&gt;. Let the vertices of the square, <br /> taken counterclockwise, be &lt;math&gt;A, B, C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Also, let the distances from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, be &lt;math&gt;u, v&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;. <br /> What is the greatest distance that &lt;math&gt;P&lt;/math&gt; can be from &lt;math&gt;D&lt;/math&gt; if &lt;math&gt;u^2 + v^2 = w^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2}\qquad<br /> \textbf{(B)}\ 2\sqrt{2}\qquad<br /> \textbf{(C)}\ 2 + \sqrt{2}\qquad<br /> \textbf{(D)}\ 3\sqrt{2}\qquad<br /> \textbf{(E)}\ 3+\sqrt{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30 ==<br /> <br /> Distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are on a semicircle with diameter &lt;math&gt;MN&lt;/math&gt; and center &lt;math&gt;C&lt;/math&gt;. <br /> The point &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;CN&lt;/math&gt; and &lt;math&gt;\angle CAP = \angle CBP = 10^{\circ}&lt;/math&gt;. If &lt;math&gt;\stackrel{\frown}{MA} = 40^{\circ}&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{BN}&lt;/math&gt; equals<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair C=origin, N=dir(0), B=dir(20), A=dir(135), M=dir(180), P=(3/7)*dir(C--N);<br /> draw(M--N^^C--A--P--B--C^^Arc(origin,1,0,180));<br /> markscalefactor=0.03;<br /> draw(anglemark(C,A,P));<br /> draw(anglemark(C,B,P));<br /> pair point=C;<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, S);<br /> label(&quot;$M$&quot;, M, dir(point--M));<br /> label(&quot;$N$&quot;, N, dir(point--N));<br /> label(&quot;$P$&quot;, P, S);<br /> label(&quot;$40^\circ$&quot;, C+(-0.15,0), NW);<br /> label(&quot;$10^\circ$&quot;, B+(0,0.05), dir(200));<br /> label(&quot;$10^\circ$&quot;, A+(0.05,0.02), dir(330));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10^{\circ}\qquad<br /> \textbf{(B)}\ 15^{\circ}\qquad<br /> \textbf{(C)}\ 20^{\circ}\qquad<br /> \textbf{(D)}\ 25^{\circ}\qquad<br /> \textbf{(E)}\ 30^{\circ} &lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME box|year=1983|before=[[1982 AHSME]]|after=[[1984 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems&diff=96056 1983 AHSME Problems 2018-07-09T14:57:09Z <p>Wlm7: /* Problem 30 */</p> <hr /> <div>== Problem 1 ==<br /> <br /> If &lt;math&gt;x \neq 0, \frac x{2} = y^2&lt;/math&gt; and &lt;math&gt;\frac{x}{4} = 4y&lt;/math&gt;, then &lt;math&gt;x&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad<br /> \textbf{(B)}\ 16\qquad<br /> \textbf{(C)}\ 32\qquad<br /> \textbf{(D)}\ 64\qquad<br /> \textbf{(E)}\ 128 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> Point &lt;math&gt;P&lt;/math&gt; is outside circle &lt;math&gt;C&lt;/math&gt; on the plane. At most how many points on &lt;math&gt;C&lt;/math&gt; are &lt;math&gt;3 \text{cm}&lt;/math&gt; from P? <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> Three primes &lt;math&gt;p,q&lt;/math&gt;, and &lt;math&gt;r&lt;/math&gt; satisfy &lt;math&gt;p+q = r&lt;/math&gt; and &lt;math&gt;1 &lt; p &lt; q&lt;/math&gt;. Then &lt;math&gt;p&lt;/math&gt; equals<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 7\qquad<br /> \textbf{(D)}\ 13\qquad<br /> \textbf{(E)}\ 17 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Position &lt;math&gt;A,B,C,D,E,F&lt;/math&gt; such that &lt;math&gt;AF&lt;/math&gt; and &lt;math&gt;CD&lt;/math&gt; are parallel, as are sides &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;EF&lt;/math&gt;, <br /> and sides &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;ED&lt;/math&gt;. Each side has length of &lt;math&gt;1&lt;/math&gt; and it is given that &lt;math&gt;\measuredangle FAB = \measuredangle BCD = 60^\circ&lt;/math&gt;. <br /> The area of the figure is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ \frac{3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has a right angle at &lt;math&gt;C&lt;/math&gt;. If &lt;math&gt;\sin A = \frac{2}{3}&lt;/math&gt;, then &lt;math&gt;\tan B&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{3}{5}\qquad<br /> \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad<br /> \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad<br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ 2&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> When &lt;math&gt;x^5, x+\frac{1}{x}&lt;/math&gt; and &lt;math&gt;1+\frac{2}{x} + \frac{3}{x^2}&lt;/math&gt; are multiplied, the product is a polynomial of degree.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> Alice sells an item at &amp;#036; &lt;math&gt;10&lt;/math&gt; less than the list price and receives &lt;math&gt;10\%&lt;/math&gt; of her selling price as her commission. <br /> Bob sells the same item at &amp;#036; &lt;math&gt;20&lt;/math&gt; less than the list price and receives &lt;math&gt;20\%&lt;/math&gt; of his selling price as his commission. <br /> If they both get the same commission, then the list price in dollars is<br /> <br /> &lt;math&gt;\textbf{(A) } 20\qquad<br /> \textbf{(B) } 30\qquad<br /> \textbf{(C) } 50\qquad<br /> \textbf{(D) } 70\qquad<br /> \textbf{(E) } 100 &lt;/math&gt;<br /> <br /> <br /> [[1983 AHSME Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Let &lt;math&gt;f(x) = \frac{x+1}{x-1}&lt;/math&gt;. Then for &lt;math&gt;x^2 \neq 1, f(-x)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{f(x)}\qquad<br /> \textbf{(B)}\ -f(x)\qquad<br /> \textbf{(C)}\ \frac{1}{f(-x)}\qquad<br /> \textbf{(D)}\ -f(-x)\qquad<br /> \textbf{(E)}\ f(x)&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> In a certain population the ratio of the number of women to the number of men is &lt;math&gt;11&lt;/math&gt; to &lt;math&gt;10&lt;/math&gt;. <br /> If the average (arithmetic mean) age of the women is &lt;math&gt;34&lt;/math&gt; and the average age of the men is &lt;math&gt;32&lt;/math&gt;, <br /> then the average age of the population is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\frac{9}{10}\qquad<br /> \textbf{(B)}\ 32\frac{20}{21}\qquad<br /> \textbf{(C)}\ 33\qquad<br /> \textbf{(D)}\ 33\frac{1}{21}\qquad<br /> \textbf{(E)}\ 33\frac{1}{10} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Segment &lt;math&gt;AB&lt;/math&gt; is both a diameter of a circle of radius &lt;math&gt;1&lt;/math&gt; and a side of an equilateral triangle &lt;math&gt;ABC&lt;/math&gt;. <br /> The circle also intersects &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;BD&lt;/math&gt; at points &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt;, respectively. The length of &lt;math&gt;AE&lt;/math&gt; is <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{3}{2} \qquad <br /> \text{(B)} \ \frac{5}{3} \qquad <br /> \text{(C)} \ \frac{\sqrt 3}{2} \qquad <br /> \text{(D)}\ \sqrt{3}\qquad<br /> \text{(E)}\ \frac{2+\sqrt 3}{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> Simplify &lt;math&gt;\sin (x-y) \cos y + \cos (x-y) \sin y&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ \sin x\qquad<br /> \textbf{(C)}\ \cos x\qquad<br /> \textbf{(D)}\ \sin x \cos 2y\qquad<br /> \textbf{(E)}\ \cos x\cos 2y &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> If &lt;math&gt;\log_2 \Big(\log_3 (\log_2 x) \Big) = 0&lt;/math&gt;, then &lt;math&gt;x^{-1/2}&lt;/math&gt; equals <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{3} \qquad <br /> \text{(B)} \ \frac{1}{2 \sqrt 3} \qquad <br /> \text{(C)}\ \frac{1}{3\sqrt 3}\qquad<br /> \text{(D)}\ \frac{1}{\sqrt{42}}\qquad<br /> \text{(E)}\ \text{none of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> If &lt;math&gt;xy = a, xz =b,&lt;/math&gt; and &lt;math&gt;yz = c&lt;/math&gt;, and none of these quantities is zero, then &lt;math&gt;x^2+y^2+z^2&lt;/math&gt; equals:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad<br /> \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad<br /> \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad<br /> \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad<br /> \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The units digit of &lt;math&gt;3^{1001}* 7^{1002}* 13^{1003}&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 9 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Three balls marked &lt;math&gt;1,2&lt;/math&gt;, and &lt;math&gt;3&lt;/math&gt;, are placed in an urn. One ball is drawn, its number is recorded, <br /> then the ball is returned to the urn. This process is repeated and then repeated once more, <br /> and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is &lt;math&gt;6&lt;/math&gt;, <br /> what is the probability that the ball numbered &lt;math&gt;2&lt;/math&gt; was drawn all three times? <br /> <br /> &lt;math&gt;<br /> \text{(A)} \ \frac{1}{27} \qquad <br /> \text{(B)} \ \frac{1}{8} \qquad <br /> \text{(C)} \ \frac{1}{7} \qquad <br /> \text{(D)}\ \frac{1}{6}\qquad<br /> \text{(E)}\ \frac{1}{3} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Let &lt;math&gt;x = .123456789101112....998999&lt;/math&gt;, where the digits are obtained by writing the integers &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;999&lt;/math&gt; in order. <br /> The &lt;math&gt;1983&lt;/math&gt;rd digit to the right of the decimal point is<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad<br /> \textbf{(B)}\ 3\qquad<br /> \textbf{(C)}\ 5\qquad<br /> \textbf{(D)}\ 7\qquad<br /> \textbf{(E)}\ 8 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> The diagram to the right shows several numbers in the complex plane. The circle is the unit circle centered at the origin. <br /> One of these numbers is the reciprocal of &lt;math&gt;F&lt;/math&gt;. Which one? <br /> <br /> &lt;math&gt;\text{(A)} \ A \qquad <br /> \text{(B)} \ B \qquad <br /> \text{(C)} \ C \qquad <br /> \text{(D)} \ D \qquad <br /> \text{(E)} \ E &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Let &lt;math&gt;f&lt;/math&gt; be a polynomial function such that, for all real &lt;math&gt;x&lt;/math&gt;,<br /> &lt;math&gt;f(x^2 + 1) = x^4 + 5x^2 + 3&lt;/math&gt;.<br /> For all real &lt;math&gt;x, f(x^2-1)&lt;/math&gt; is<br /> <br /> &lt;math&gt;\textbf{(A)}\ x^4+5x^2+1\qquad<br /> \textbf{(B)}\ x^4+x^2-3\qquad<br /> \textbf{(C)}\ x^4-5x^2+1\qquad<br /> \textbf{(D)}\ x^4+x^2+3\qquad\\<br /> \textbf{(E)}\ \text{None of these} &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Point &lt;math&gt;D&lt;/math&gt; is on side &lt;math&gt;CB&lt;/math&gt; of triangle &lt;math&gt;ABC&lt;/math&gt;. If <br /> &lt;math&gt;\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6&lt;/math&gt;, <br /> then the length of &lt;math&gt;AD&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 2 \qquad <br /> \text{(B)} \ 2.5 \qquad <br /> \text{(C)} \ 3 \qquad <br /> \text{(D)} \ 3.5 \qquad <br /> \text{(E)} \ 4 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> If &lt;math&gt;\tan{\alpha}&lt;/math&gt; and &lt;math&gt;\tan{\beta}&lt;/math&gt; are the roots of &lt;math&gt;x^2 - px + q = 0&lt;/math&gt;, and &lt;math&gt;\cot{\alpha}&lt;/math&gt; and &lt;math&gt;\cot{\beta}&lt;/math&gt;<br /> are the roots of &lt;math&gt;x^2 - rx + s = 0&lt;/math&gt;, then &lt;math&gt;rs&lt;/math&gt; is necessarily <br /> <br /> &lt;math&gt;\text{(A)} \ pq \qquad <br /> \text{(B)} \ \frac{1}{pq} \qquad <br /> \text{(C)} \ \frac{p}{q^2} \qquad <br /> \text{(D)}\ \frac{q}{p^2}\qquad<br /> \text{(E)}\ \frac{p}{q}&lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Find the smallest positive number from the numbers below <br /> <br /> &lt;math&gt;\text{(A)} \ 10-3\sqrt{11} \qquad <br /> \text{(B)} \ 3\sqrt{11}-10 \qquad <br /> \text{(C)}\ 18-5\sqrt{13}\qquad\\<br /> \text{(D)}\ 51-10\sqrt{26}\qquad<br /> \text{(E)}\ 10\sqrt{26}-51 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> Consider the two functions &lt;math&gt;f(x) = x^2+2bx+1&lt;/math&gt; and &lt;math&gt;g(x) = 2a(x+b)&lt;/math&gt;, where the variable &lt;math&gt;x&lt;/math&gt; and the constants &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are real numbers. <br /> Each such pair of the constants a and b may be considered as a point &lt;math&gt;(a,b)&lt;/math&gt; in an &lt;math&gt;ab&lt;/math&gt;-plane. <br /> Let S be the set of such points &lt;math&gt;(a,b)&lt;/math&gt; for which the graphs of &lt;math&gt;y = f(x)&lt;/math&gt; and &lt;math&gt;y = g(x)&lt;/math&gt; do NOT intersect (in the &lt;math&gt;xy&lt;/math&gt;- plane.). The area of &lt;math&gt;S&lt;/math&gt; is <br /> <br /> &lt;math&gt;\text{(A)} \ 1 \qquad <br /> \text{(B)} \ \pi \qquad <br /> \text{(C)} \ 4 \qquad <br /> \text{(D)} \ 4 \pi \qquad <br /> \text{(E)} \ \infty &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> In the adjoining figure the five circles are tangent to one another consecutively and to the lines <br /> &lt;math&gt;L_1&lt;/math&gt; and &lt;math&gt;L_2&lt;/math&gt; (&lt;math&gt;L_1&lt;/math&gt; is the line that is above the circles and &lt;math&gt;L_2&lt;/math&gt; is the line that goes under the circles). <br /> If the radius of the largest circle is &lt;math&gt;18&lt;/math&gt; and that of the smallest one is &lt;math&gt;8&lt;/math&gt;, then the radius of the middle circle is<br /> <br /> &lt;asy&gt;<br /> size(250);defaultpen(linewidth(0.7));<br /> real alpha=5.797939254, x=71.191836;<br /> int i;<br /> for(i=0; i&lt;5; i=i+1) {<br /> real r=8*(sqrt(6)/2)^i;<br /> draw(Circle((x+r)*dir(alpha), r));<br /> x=x+2r;<br /> }<br /> real x=71.191836+40+20*sqrt(6), r=18;<br /> pair A=tangent(origin, (x+r)*dir(alpha), r, 1), B=tangent(origin, (x+r)*dir(alpha), r, 2);<br /> pair A1=300*dir(origin--A), B1=300*dir(origin--B);<br /> draw(B1--origin--A1);<br /> pair X=(69,-5), X1=reflect(origin, (x+r)*dir(alpha))*X,<br /> Y=(200,-5), Y1=reflect(origin, (x+r)*dir(alpha))*Y,<br /> Z=(130,0), Z1=reflect(origin, (x+r)*dir(alpha))*Z;<br /> clip(X--Y--Y1--X1--cycle);<br /> label(&quot;$L_2$&quot;, Z, S);<br /> label(&quot;$L_1$&quot;, Z1, dir(2*alpha)*dir(90));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ 12 \qquad <br /> \text{(B)} \ 12.5 \qquad <br /> \text{(C)} \ 13 \qquad <br /> \text{(D)} \ 13.5 \qquad <br /> \text{(E)} \ 14 &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 22|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> How many non-congruent right triangles are there such that the perimeter in &lt;math&gt;\text{cm}&lt;/math&gt; and the area in &lt;math&gt;\text{cm}^2&lt;/math&gt; are numerically equal? <br /> <br /> &lt;math&gt;\text{(A)} \ \text{none} \qquad <br /> \text{(B)} \ 1 \qquad <br /> \text{(C)} \ 2 \qquad <br /> \text{(D)} \ 4 \qquad <br /> \text{(E)} \ \infty&lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> If &lt;math&gt;60^a = 3&lt;/math&gt; and &lt;math&gt;60^b = 5&lt;/math&gt;, then &lt;math&gt;12^{[(1-a-b)/2(1-b)]} &lt;/math&gt;<br /> <br /> &lt;math&gt;\text{(A)} \ \sqrt{3} \qquad <br /> \text{(B)} \ 2 \qquad <br /> \text{(C)} \ \sqrt{5} \qquad <br /> \text{(D)} \ 3 \qquad <br /> \text{(E)} \ \sqrt{12} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 25|Solution]]<br /> <br /> == Problem 26 ==<br /> <br /> The probability that event &lt;math&gt;A&lt;/math&gt; occurs is &lt;math&gt;\frac{3}{4}&lt;/math&gt;; the probability that event B occurs is &lt;math&gt;\frac{2}{3}&lt;/math&gt;. <br /> Let &lt;math&gt;p&lt;/math&gt; be the probability that both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; occur. The smallest interval necessarily containing &lt;math&gt;p&lt;/math&gt; is the interval<br /> <br /> &lt;math&gt;\textbf{(A)}\ \Big[\frac{1}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(B)}\ \Big[\frac{5}{12},\frac{1}{2}\Big]\qquad<br /> \textbf{(C)}\ \Big[\frac{1}{2},\frac{2}{3}\Big]\qquad<br /> \textbf{(D)}\ \Big[\frac{5}{12},\frac{2}{3}\Big]\qquad<br /> \textbf{(E)}\ \Big[\frac{1}{12},\frac{2}{3}\Big] &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 26|Solution]]<br /> <br /> == Problem 27 ==<br /> <br /> A large sphere is on a horizontal field on a sunny day. At a certain time the shadow of the sphere reaches out a distance <br /> of &lt;math&gt;10&lt;/math&gt; m from the point where the sphere touches the ground. At the same instant a meter stick <br /> (held vertically with one end on the ground) casts a shadow of length &lt;math&gt;2&lt;/math&gt; m. What is the radius of the sphere in meters? <br /> (Assume the sun's rays are parallel and the meter stick is a line segment.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{5}{2}\qquad<br /> \textbf{(B)}\ 9 - 4\sqrt{5}\qquad<br /> \textbf{(C)}\ 8\sqrt{10} - 23\qquad<br /> \textbf{(D)}\ 6-\sqrt{15}\qquad<br /> \textbf{(E)}\ 10\sqrt{5}-20 &lt;/math&gt;<br /> <br /> [[1983 AHSME Problems/Problem 27|Solution]]<br /> <br /> == Problem 28 ==<br /> <br /> Triangle &lt;math&gt;\triangle ABC&lt;/math&gt; in the figure has area &lt;math&gt;10&lt;/math&gt;. Points &lt;math&gt;D, E&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt;, all distinct from &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, <br /> are on sides &lt;math&gt;AB, BC&lt;/math&gt; and &lt;math&gt;CA&lt;/math&gt; respectively, and &lt;math&gt;AD = 2, DB = 3&lt;/math&gt;. If triangle &lt;math&gt;\triangle ABE&lt;/math&gt; and quadrilateral &lt;math&gt;DBEF&lt;/math&gt; <br /> have equal areas, then that area is<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B);<br /> draw(A--B--C--A--E--F--D);<br /> pair point=incenter(A,B,C);<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, dir(point--C));<br /> label(&quot;$D$&quot;, D, dir(point--D));<br /> label(&quot;$E$&quot;, E, dir(point--E));<br /> label(&quot;$F$&quot;, F, dir(point--F));<br /> label(&quot;$2$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (7,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad<br /> \textbf{(B)}\ 5\qquad<br /> \textbf{(C)}\ 6\qquad<br /> \textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad<br /> \textbf{(E)}\ \text{not uniquely determined} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 28|Solution]]<br /> <br /> == Problem 29 ==<br /> <br /> A point &lt;math&gt;P&lt;/math&gt; lies in the same plane as a given square of side &lt;math&gt;1&lt;/math&gt;. Let the vertices of the square, <br /> taken counterclockwise, be &lt;math&gt;A, B, C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. Also, let the distances from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;A, B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, respectively, be &lt;math&gt;u, v&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;. <br /> What is the greatest distance that &lt;math&gt;P&lt;/math&gt; can be from &lt;math&gt;D&lt;/math&gt; if &lt;math&gt;u^2 + v^2 = w^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 + \sqrt{2}\qquad<br /> \textbf{(B)}\ 2\sqrt{2}\qquad<br /> \textbf{(C)}\ 2 + \sqrt{2}\qquad<br /> \textbf{(D)}\ 3\sqrt{2}\qquad<br /> \textbf{(E)}\ 3+\sqrt{2} &lt;/math&gt; <br /> <br /> [[1983 AHSME Problems/Problem 29|Solution]]<br /> <br /> == Problem 30 ==<br /> <br /> Distinct points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are on a semicircle with diameter &lt;math&gt;MN&lt;/math&gt; and center &lt;math&gt;C&lt;/math&gt;. <br /> The point &lt;math&gt;P&lt;/math&gt; is on &lt;math&gt;CN&lt;/math&gt; and &lt;math&gt;\angle CAP = \angle CBP = 10^{\circ}&lt;/math&gt;. If &lt;math&gt;\stackrel{\frown}{MA} = 40^{\circ}&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{BN}&lt;/math&gt; equals<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.7)+fontsize(10));<br /> pair C=origin, N=dir(0), B=dir(20), A=dir(135), M=dir(180), P=(3/7)*dir(C--N);<br /> draw(M--N^^C--A--P--B--C^^Arc(origin,1,0,180));<br /> markscalefactor=0.03;<br /> draw(anglemark(C,A,P));<br /> draw(anglemark(C,B,P));<br /> pair point=C;<br /> label(&quot;$A$&quot;, A, dir(point--A));<br /> label(&quot;$B$&quot;, B, dir(point--B));<br /> label(&quot;$C$&quot;, C, S);<br /> label(&quot;$M$&quot;, M, dir(point--M));<br /> label(&quot;$N$&quot;, N, dir(point--N));<br /> label(&quot;$P$&quot;, P, S);<br /> label(&quot;$40^\circ$&quot;, C+(-0.15,0), NW);<br /> label(&quot;$10^\circ$&quot;, B+(0,0.05), dir(200);<br /> label(&quot;$10^\circ$&quot;, A+(0.05,0.02), dir(330));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10^{\circ}\qquad<br /> \textbf{(B)}\ 15^{\circ}\qquad<br /> \textbf{(C)}\ 20^{\circ}\qquad<br /> \textbf{(D)}\ 25^{\circ}\qquad<br /> \textbf{(E)}\ 30^{\circ} &lt;/math&gt;<br /> <br /> == See also ==<br /> <br /> * [[AMC 12 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> {{AHSME box|year=1983|before=[[1982 AHSME]]|after=[[1984 AHSME]]}} <br /> <br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems&diff=95920 2016 AMC 10B Problems 2018-07-06T17:23:56Z <p>Wlm7: </p> <hr /> <div>==Problem 1== <br /> <br /> What is the value of &lt;math&gt;\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}&lt;/math&gt; when &lt;math&gt;a= \frac{1}{2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> If &lt;math&gt;n\heartsuit m=n^3m^2&lt;/math&gt;, what is &lt;math&gt;\frac{2\heartsuit 4}{4\heartsuit 2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Let &lt;math&gt;x=-2016&lt;/math&gt;. What is the value of &lt;math&gt;\bigg|&lt;/math&gt; &lt;math&gt;||x|-x|-|x|&lt;/math&gt; &lt;math&gt;\bigg|&lt;/math&gt; &lt;math&gt;-x&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Zoey read &lt;math&gt;15&lt;/math&gt; books, one at a time. The first book took her &lt;math&gt;1&lt;/math&gt; day to read, the second book took her &lt;math&gt;2&lt;/math&gt; days to read, the third book took her &lt;math&gt;3&lt;/math&gt; days to read, and so on, with each book taking her &lt;math&gt;1&lt;/math&gt; more day to read than the previous book. Zoey finished the first book on a Monday, and the second on a Wednesday. On what day the week did she finish her &lt;math&gt;15&lt;/math&gt;th book?<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Sunday}\qquad\textbf{(B)}\ \text{Monday}\qquad\textbf{(C)}\ \text{Wednesday}\qquad\textbf{(D)}\ \text{Friday}\qquad\textbf{(E)}\ \text{Saturday}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The mean age of Amanda's &lt;math&gt;4&lt;/math&gt; cousins is &lt;math&gt;8&lt;/math&gt;, and their median age is &lt;math&gt;5&lt;/math&gt;. What is the sum of the ages of Amanda's youngest and oldest cousins?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 19\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 25&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number &lt;math&gt;S&lt;/math&gt;. What is the smallest possible value for the sum of the digits of &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> The ratio of the measures of two acute angles is &lt;math&gt;5:4&lt;/math&gt;, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> What is the tens digit of &lt;math&gt;2015^{2016}-2017?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> All three vertices of &lt;math&gt;\bigtriangleup ABC&lt;/math&gt; are lying on the parabola defined by &lt;math&gt;y=x^2&lt;/math&gt;, with &lt;math&gt;A&lt;/math&gt; at the origin and &lt;math&gt;\overline{BC}&lt;/math&gt; parallel to the &lt;math&gt;x&lt;/math&gt;-axis. The area of the triangle is &lt;math&gt;64&lt;/math&gt;. What is the length of &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> A thin piece of wood of uniform density in the shape of an equilateral triangle with side length &lt;math&gt;3&lt;/math&gt; inches weighs &lt;math&gt;12&lt;/math&gt; ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of &lt;math&gt;5&lt;/math&gt; inches. Which of the following is closest to the weight, in ounces, of the second piece?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> Carl decided to fence in his rectangular garden. He bought &lt;math&gt;20&lt;/math&gt; fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly &lt;math&gt;4&lt;/math&gt; yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Two different numbers are selected at random from &lt;math&gt;( 1, 2, 3, 4, 5)&lt;/math&gt; and multiplied together. What is the probability that the product is even?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0.2\qquad\textbf{(B)}\ 0.4\qquad\textbf{(C)}\ 0.5\qquad\textbf{(D)}\ 0.7\qquad\textbf{(E)}\ 0.8&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> At Megapolis Hospital one year, multiple-birth statistics were as follows: Sets of twins, triplets, and quadruplets accounted for &lt;math&gt;1000&lt;/math&gt; of the babies born. There were four times as many sets of triplets as sets of quadruplets, and there was three times as many sets of twins as sets of triplets. How many of these &lt;math&gt;1000&lt;/math&gt; babies were in sets of quadruplets?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 25\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 100\qquad\textbf{(E)}\ 160&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> How many squares whose sides are parallel to the axis and whose vertices have coordinates that are integers lie entirely within the region bounded by the line &lt;math&gt;y=\pi x&lt;/math&gt;, the line &lt;math&gt;y=-0.1&lt;/math&gt; and the line &lt;math&gt;x=5.1?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> All the numbers &lt;math&gt;1, 2, 3, 4, 5, 6, 7, 8, 9&lt;/math&gt; are written in a &lt;math&gt;3\times3&lt;/math&gt; array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to &lt;math&gt;18&lt;/math&gt;. What is the number in the center?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> The sum of an infinite geometric series is a positive number &lt;math&gt;S&lt;/math&gt;, and the second term in the series is &lt;math&gt;1&lt;/math&gt;. What is the smallest possible value of &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1+\sqrt{5}}{2} \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ \sqrt{5} \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> All the numbers &lt;math&gt;2, 3, 4, 5, 6, 7&lt;/math&gt; are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 312 \qquad \textbf{(B)}\ 343 \qquad \textbf{(C)}\ 625 \qquad \textbf{(D)}\ 729 \qquad \textbf{(E)}\ 1680&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> In how many ways can &lt;math&gt;345&lt;/math&gt; be written as the sum of an increasing sequence of two or more consecutive positive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Rectangle &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5&lt;/math&gt; and &lt;math&gt;BC=4&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; lies on &lt;math&gt;\overline{AB}&lt;/math&gt; so that &lt;math&gt;EB=1&lt;/math&gt;, point &lt;math&gt;G&lt;/math&gt; lies on &lt;math&gt;\overline{BC}&lt;/math&gt; so that &lt;math&gt;CG=1&lt;/math&gt;. and point &lt;math&gt;F&lt;/math&gt; lies on &lt;math&gt;\overline{CD}&lt;/math&gt; so that &lt;math&gt;DF=2&lt;/math&gt;. Segments &lt;math&gt;\overline{AG}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; intersect &lt;math&gt;\overline{EF}&lt;/math&gt; at &lt;math&gt;Q&lt;/math&gt; and &lt;math&gt;P&lt;/math&gt;, respectively. What is the value of &lt;math&gt;\frac{PQ}{EF}&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt; pair A1=(2,0),A2=(4,4);<br /> pair B1=(0,4),B2=(5,1);<br /> pair C1=(5,0),C2=(0,4); <br /> draw(A1--A2);<br /> draw(B1--B2);<br /> draw(C1--C2);<br /> draw((0,0)--B1--(5,4)--C1--cycle);<br /> dot((20/7,12/7));<br /> dot((3.07692307692,2.15384615384));<br /> label(&quot;$Q$&quot;,(3.07692307692,2.15384615384),N);<br /> label(&quot;$P$&quot;,(20/7,12/7),W);<br /> label(&quot;$A$&quot;,(0,4), NW);<br /> label(&quot;$B$&quot;,(5,4), NE);<br /> label(&quot;$C$&quot;,(5,0),SE);<br /> label(&quot;$D$&quot;,(0,0),SW);<br /> label(&quot;$F$&quot;,(2,0),S); label(&quot;$G$&quot;,(5,1),E);<br /> label(&quot;$E$&quot;,(4,4),N);<br /> <br /> dot(A1); dot(A2);<br /> dot(B1); dot(B2);<br /> dot(C1); dot(C2);<br /> dot((0,0)); dot((5,4));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad<br /> \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad<br /> \textbf{(C)}~\frac{9}{82} \qquad<br /> \textbf{(D)}~\frac{10}{91}\qquad<br /> \textbf{(E)}~\frac19&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius &lt;math&gt;2&lt;/math&gt; centered at &lt;math&gt;A(2,2)&lt;/math&gt; to the circle of radius &lt;math&gt;3&lt;/math&gt; centered at &lt;math&gt;A’(5,6)&lt;/math&gt;. What distance does the origin &lt;math&gt;O(0,0)&lt;/math&gt;, move under this transformation?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> What is the area of the region enclosed by the graph of the equation &lt;math&gt;x^2+y^2=|x|+|y|?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won &lt;math&gt;10&lt;/math&gt; games and lost &lt;math&gt;10&lt;/math&gt; games; there were no ties. How many sets of three teams &lt;math&gt;\{A, B, C\}&lt;/math&gt; were there in which &lt;math&gt;A&lt;/math&gt; beat &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt; beat &lt;math&gt;C&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; beat &lt;math&gt;A?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In regular hexagon &lt;math&gt;ABCDEF&lt;/math&gt;, points &lt;math&gt;W&lt;/math&gt;, &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; are chosen on sides &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;\overline{CD}&lt;/math&gt;, &lt;math&gt;\overline{EF}&lt;/math&gt;, and &lt;math&gt;\overline{FA}&lt;/math&gt; respectively, so lines &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;ZW&lt;/math&gt;, &lt;math&gt;YX&lt;/math&gt;, and &lt;math&gt;ED&lt;/math&gt; are parallel and equally spaced. What is the ratio of the area of hexagon &lt;math&gt;WCXYFZ&lt;/math&gt; to the area of hexagon &lt;math&gt;ABCDEF&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> How many four-digit integers &lt;math&gt;abcd&lt;/math&gt;, with &lt;math&gt;a \neq 0&lt;/math&gt;, have the property that the three two-digit integers &lt;math&gt;ab&lt;bc&lt;cd&lt;/math&gt; form an increasing arithmetic sequence? One such number is &lt;math&gt;4692&lt;/math&gt;, where &lt;math&gt;a=4&lt;/math&gt;, &lt;math&gt;b=6&lt;/math&gt;, &lt;math&gt;c=9&lt;/math&gt;, and &lt;math&gt;d=2&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> Let &lt;math&gt;f(x)=\sum_{k=2}^{10}(\lfloor kx \rfloor -k \lfloor x \rfloor)&lt;/math&gt;, where &lt;math&gt;\lfloor r \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;r&lt;/math&gt;. How many distinct values does &lt;math&gt;f(x)&lt;/math&gt; assume for &lt;math&gt;x \ge 0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 32\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 46\qquad\textbf{(E)}\ \text{infinitely many}&lt;/math&gt;<br /> <br /> [[2016 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> <br /> ==See also==<br /> {{AMC10 box|year=2016|ab=B|before=[[2016 AMC 10A Problems]]|after=[[2017 AMC 10A Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2016 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_8_Problems/Problem_25&diff=95633 2017 AMC 8 Problems/Problem 25 2018-06-26T14:23:23Z <p>Wlm7: /* Solution */</p> <hr /> <div>==Problem 25==<br /> <br /> In the figure shown, &lt;math&gt;\overline{US}&lt;/math&gt; and &lt;math&gt;\overline{UT}&lt;/math&gt; are line segments each of length 2, and &lt;math&gt;m\angle TUS = 60^\circ&lt;/math&gt;. Arcs &lt;math&gt;\overarc{TR}&lt;/math&gt; and &lt;math&gt;\overarc{SR}&lt;/math&gt; are each one-sixth of a circle with radius 2. What is the area of the region shown? <br /> <br /> &lt;asy&gt;draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label(&quot;$U$&quot;, (2,3.464), N); label(&quot;$S$&quot;, (1,1.732), W); label(&quot;$T$&quot;, (3,1.732), E); label(&quot;$R$&quot;, (2,0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the centers of the circles containing arcs &lt;math&gt;\overarc{SR}&lt;/math&gt; and &lt;math&gt;\overarc{TR}&lt;/math&gt; be &lt;math&gt;S'&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt;, respectively. Extend &lt;math&gt;\overline{US}&lt;/math&gt; and &lt;math&gt;\overline{UT}&lt;/math&gt; to &lt;math&gt;S'&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt;, and connect point &lt;math&gt;S'&lt;/math&gt; with point &lt;math&gt;T'&lt;/math&gt;. <br /> &lt;asy&gt;draw((1,1.732)--(2,3.464)--(3,1.732));<br /> draw((1,1.732)--(0,0)--(4,0)--(3,1.732));<br /> draw(arc((0,0),(2,0),(1,1.732)));<br /> draw(arc((4,0),(3,1.732),(2,0)));<br /> label(&quot;$U$&quot;, (2,3.464), N);<br /> label(&quot;$S$&quot;, (1,1.732), W);<br /> label(&quot;$T$&quot;, (3,1.732), E);<br /> label(&quot;$R$&quot;, (2,0), S);<br /> label(&quot;$S'$&quot;, (0,0), W);<br /> label(&quot;$T'$&quot;, (4,0), E);&lt;/asy&gt;<br /> We can clearly see that &lt;math&gt;\triangle US'T'&lt;/math&gt; is an equilateral triangle, because two of its angles are &lt;math&gt;60^\circ&lt;/math&gt;, which is the degree measure of &lt;math&gt;\frac{1}{6}&lt;/math&gt; a circle. The area of the figure is equal to &lt;math&gt;[\triangle US'T']&lt;/math&gt; minus the combined area of the &lt;math&gt;2&lt;/math&gt; sectors of the circles. Using the area formula for an equilateral triangle, &lt;math&gt;\frac{a^2\sqrt{3}}{4},&lt;/math&gt; where &lt;math&gt;a&lt;/math&gt; is the side length of the equilateral triangle, &lt;math&gt;[\triangle US'T']&lt;/math&gt; is &lt;math&gt;\frac{\sqrt 3}{4} \cdot 4^2 = 4\sqrt 3.&lt;/math&gt; The combined area of the &lt;math&gt;2&lt;/math&gt; sectors is &lt;math&gt;2\cdot\frac16\cdot\pi r^2&lt;/math&gt;, which is &lt;math&gt;\frac 13\pi \cdot 2^2 = \frac{4\pi}{3}.&lt;/math&gt; Thus, our final answer is &lt;math&gt;\boxed{\textbf{(B)}\ 4\sqrt{3}-\frac{4\pi}{3}}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2017|num-b=24|after=Last Problem}}<br /> <br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Dodecagon&diff=95173 Dodecagon 2018-06-15T17:08:58Z <p>Wlm7: </p> <hr /> <div>A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is &lt;math&gt;1800^{\circ}&lt;/math&gt;. Each of its exterior angles has measure &lt;math&gt;30^{\circ}&lt;/math&gt;.<br /> <br /> A regular dodecagon can be seen below:<br /> <br /> &lt;asy&gt;<br /> for(int i = 0; i &lt;= 11; ++i) {<br /> draw(dir(360/12*i)--dir(360/12*(i + 1)));<br /> }<br /> pair A,B,C,D,E,F,G,H,I,J,K,L;<br /> A=dir(360/12*0);<br /> B=dir(360/12*1);<br /> C=dir(360/12*2);<br /> D=dir(360/12*3);<br /> E=dir(360/12*4);<br /> F=dir(360/12*5);<br /> G=dir(360/12*6);<br /> H=dir(360/12*7);<br /> I=dir(360/12*8);<br /> J=dir(360/12*9);<br /> K=dir(360/12*10);<br /> L=dir(360/12*11);<br /> label(&quot;A&quot;,A,dir(0));<br /> label(&quot;B&quot;,B,dir(30));<br /> label(&quot;C&quot;,C,dir(60));<br /> label(&quot;D&quot;,D,dir(90));<br /> label(&quot;E&quot;,E,dir(120));<br /> label(&quot;F&quot;,F,dir(150));<br /> label(&quot;G&quot;,G,dir(180));<br /> label(&quot;H&quot;,H,dir(210));<br /> label(&quot;I&quot;,I,dir(240));<br /> label(&quot;J&quot;,J,dir(270));<br /> label(&quot;K&quot;,K,dir(300));<br /> label(&quot;L&quot;,L,dir(330));<br /> draw(dir(360/12*0)--dir(360/12*6));<br /> dot((dir(360/12*0)+dir(360/12*6))/2);<br /> pair O = (dir(360/12*0)+dir(360/12*6))/2;<br /> label(&quot;O&quot;,O,S);<br /> draw(A--O);<br /> draw(Circle(O,1));<br /> &lt;/asy&gt;<br /> The area of a regular dodecagon can be calculated by the formula &lt;math&gt;3R^2&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the circumradius of the dodecagon. In this case, &lt;math&gt;R&lt;/math&gt; would be &lt;math&gt;OA&lt;/math&gt;. Also, each small triangle (&lt;math&gt;AOB&lt;/math&gt;, &lt;math&gt;BOC&lt;/math&gt;, etc.) is congruent, so &lt;math&gt;\angle AOB=\angle BOC=\angle COD&lt;/math&gt; (etc) &lt;math&gt;=30^{\circ}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> * [[Polygon]]<br /> * [[Geometry]]<br /> <br /> [[Category:Geometry]]<br /> <br /> {{stub}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Dodecagon&diff=95172 Dodecagon 2018-06-15T17:07:57Z <p>Wlm7: </p> <hr /> <div>A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is &lt;math&gt;1800^{\circ}&lt;/math&gt;.<br /> <br /> A regular dodecagon can be seen below:<br /> <br /> &lt;asy&gt;<br /> for(int i = 0; i &lt;= 11; ++i) {<br /> draw(dir(360/12*i)--dir(360/12*(i + 1)));<br /> }<br /> pair A,B,C,D,E,F,G,H,I,J,K,L;<br /> A=dir(360/12*0);<br /> B=dir(360/12*1);<br /> C=dir(360/12*2);<br /> D=dir(360/12*3);<br /> E=dir(360/12*4);<br /> F=dir(360/12*5);<br /> G=dir(360/12*6);<br /> H=dir(360/12*7);<br /> I=dir(360/12*8);<br /> J=dir(360/12*9);<br /> K=dir(360/12*10);<br /> L=dir(360/12*11);<br /> label(&quot;A&quot;,A,dir(0));<br /> label(&quot;B&quot;,B,dir(30));<br /> label(&quot;C&quot;,C,dir(60));<br /> label(&quot;D&quot;,D,dir(90));<br /> label(&quot;E&quot;,E,dir(120));<br /> label(&quot;F&quot;,F,dir(150));<br /> label(&quot;G&quot;,G,dir(180));<br /> label(&quot;H&quot;,H,dir(210));<br /> label(&quot;I&quot;,I,dir(240));<br /> label(&quot;J&quot;,J,dir(270));<br /> label(&quot;K&quot;,K,dir(300));<br /> label(&quot;L&quot;,L,dir(330));<br /> draw(dir(360/12*0)--dir(360/12*6));<br /> dot((dir(360/12*0)+dir(360/12*6))/2);<br /> pair O = (dir(360/12*0)+dir(360/12*6))/2;<br /> label(&quot;O&quot;,O,S);<br /> draw(A--O);<br /> draw(Circle(O,1));<br /> &lt;/asy&gt;<br /> The area of a regular dodecagon can be calculated by the formula &lt;math&gt;3R^2&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the circumradius of the dodecagon. In this case, &lt;math&gt;R&lt;/math&gt; would be &lt;math&gt;OA&lt;/math&gt;. Also, each small triangle (&lt;math&gt;AOB&lt;/math&gt;, &lt;math&gt;BOC&lt;/math&gt;, etc.) is congruent, so &lt;math&gt;\angle AOB=\angle BOC=\angle COD&lt;/math&gt; (etc) &lt;math&gt;=30^{\circ}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> * [[Polygon]]<br /> * [[Geometry]]<br /> <br /> [[Category:Geometry]]<br /> <br /> {{stub}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Dodecagon&diff=95171 Dodecagon 2018-06-15T17:05:54Z <p>Wlm7: </p> <hr /> <div>A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is &lt;math&gt;1800^{\circ}&lt;/math&gt;.<br /> <br /> A regular dodecagon can be seen below:<br /> <br /> &lt;asy&gt;<br /> for(int i = 0; i &lt;= 11; ++i) {<br /> draw(dir(360/12*i)--dir(360/12*(i + 1)));<br /> }<br /> pair A,B,C,D,E,F,G,H,I,J,K,L;<br /> A=dir(360/12*0);<br /> B=dir(360/12*1);<br /> C=dir(360/12*2);<br /> D=dir(360/12*3);<br /> E=dir(360/12*4);<br /> F=dir(360/12*5);<br /> G=dir(360/12*6);<br /> H=dir(360/12*7);<br /> I=dir(360/12*8);<br /> J=dir(360/12*9);<br /> K=dir(360/12*10);<br /> L=dir(360/12*11);<br /> label(&quot;A&quot;,A,dir(0));<br /> label(&quot;B&quot;,B,dir(30));<br /> label(&quot;C&quot;,C,dir(60));<br /> label(&quot;D&quot;,D,dir(90));<br /> label(&quot;E&quot;,E,dir(120));<br /> label(&quot;F&quot;,F,dir(150));<br /> label(&quot;G&quot;,G,dir(180));<br /> label(&quot;H&quot;,H,dir(210));<br /> label(&quot;I&quot;,I,dir(240));<br /> label(&quot;J&quot;,J,dir(270));<br /> label(&quot;K&quot;,K,dir(300));<br /> label(&quot;L&quot;,L,dir(330));<br /> draw(dir(360/12*0)--dir(360/12*6));<br /> dot((dir(360/12*0)+dir(360/12*6))/2);<br /> pair O = (dir(360/12*0)+dir(360/12*6))/2;<br /> label(&quot;O&quot;,O,S);<br /> draw(A--G);<br /> draw(Circle(O,1));<br /> &lt;/asy&gt;<br /> The area of a regular dodecagon can be calculated by the formula &lt;math&gt;3R^2&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the circumradius of the dodecagon. In this case, &lt;math&gt;R&lt;/math&gt; would be &lt;math&gt;OA&lt;/math&gt;.<br /> <br /> ==See Also==<br /> * [[Polygon]]<br /> * [[Geometry]]<br /> <br /> [[Category:Geometry]]<br /> <br /> {{stub}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Dodecagon&diff=95170 Dodecagon 2018-06-15T17:05:19Z <p>Wlm7: </p> <hr /> <div>A '''dodecagon''' is a 12-sided [[polygon]]. The sum of its internal [[angle]]s is &lt;math&gt;1800^{\circ}&lt;/math&gt;.<br /> <br /> A regular dodecagon can be seen below:<br /> <br /> [asy]<br /> for(int i = 0; i &lt;= 11; ++i) {<br /> draw(dir(360/12*i)--dir(360/12*(i + 1)));<br /> }<br /> pair A,B,C,D,E,F,G,H,I,J,K,L;<br /> A=dir(360/12*0);<br /> B=dir(360/12*1);<br /> C=dir(360/12*2);<br /> D=dir(360/12*3);<br /> E=dir(360/12*4);<br /> F=dir(360/12*5);<br /> G=dir(360/12*6);<br /> H=dir(360/12*7);<br /> I=dir(360/12*8);<br /> J=dir(360/12*9);<br /> K=dir(360/12*10);<br /> L=dir(360/12*11);<br /> label(&quot;A&quot;,A,dir(0));<br /> label(&quot;B&quot;,B,dir(30));<br /> label(&quot;C&quot;,C,dir(60));<br /> label(&quot;D&quot;,D,dir(90));<br /> label(&quot;E&quot;,E,dir(120));<br /> label(&quot;F&quot;,F,dir(150));<br /> label(&quot;G&quot;,G,dir(180));<br /> label(&quot;H&quot;,H,dir(210));<br /> label(&quot;I&quot;,I,dir(240));<br /> label(&quot;J&quot;,J,dir(270));<br /> label(&quot;K&quot;,K,dir(300));<br /> label(&quot;L&quot;,L,dir(330));<br /> draw(dir(360/12*0)--dir(360/12*6));<br /> dot((dir(360/12*0)+dir(360/12*6))/2);<br /> pair O = (dir(360/12*0)+dir(360/12*6))/2;<br /> label(&quot;O&quot;,O,S);<br /> draw(A--G);<br /> draw(Circle(O,1));<br /> [/asy]<br /> The area of a regular dodecagon can be calculated by the formula &lt;math&gt;3R^2&lt;/math&gt;, where &lt;math&gt;R&lt;/math&gt; is the circumradius of the dodecagon. In this case, &lt;math&gt;R&lt;/math&gt; would be &lt;math&gt;OA&lt;/math&gt;.<br /> <br /> ==See Also==<br /> * [[Polygon]]<br /> * [[Geometry]]<br /> <br /> [[Category:Geometry]]<br /> <br /> {{stub}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=94941 Gmaas 2018-06-06T18:22:05Z <p>Wlm7: /* Gmaas Facts */</p> <hr /> <div>=== Gmaas Facts ===<br /> <br /> - Gmass often uses his less-known account, wlm7.<br /> <br /> -There is an ongoing debate over whether it is &quot;Gmaas&quot; or &quot;GMAAS&quot;, and there is a side group that says it is &quot;Gmass&quot;.<br /> <br /> -Gmaas hates the AM-GM inequality because he knows that it is really the AM-GMaas inequality. He does not like being considered less than anything.<br /> <br /> - Gmaas eats pusheen for breakfast.<br /> <br /> - Gmaas's favorite roblox game is Phantom Forces, and he is currently rank 169.<br /> <br /> - None of these facts are true, except the ones that are...<br /> <br /> - ...None of the trues are facts, except the ares that one<br /> <br /> - Gmaas has made friends with the one and only Spob from Introduction to Counting and Probability (1673)<br /> <br /> - Gmaas once lived on Lorien before the Mogadorians decided to waltz in and kill.<br /> <br /> -Gmaas was secretly one of the Garde, Number 11. None of the other Garde knew of him. His Cepan is sseraj.<br /> <br /> - gm lion isn't even his final form.<br /> <br /> - piphi posts GMAAS pictures in his blog<br /> https://artofproblemsolving.com/community/c580181<br /> <br /> -Gmaas once farted. The result was the Big Bang.<br /> <br /> -Gmaas' burps created all gaseous planets.<br /> <br /> - Gmaas knows these digits of pi from memory: 3.1415926535897932384626433333333333333333333333333333333333333333333333333333333333333333333333333333 repeating, thereby disproving the assumption that pi is irrational.<br /> <br /> -Gmass beat Luke Robatille in an epic showdown of catnip consumption.<br /> <br /> -since Gmaass invented pi, he knows that the digits are really 9.587979087879877897087r09780970987098790870987609870987908067897578786. digit-ese for &quot;Gmaass is the ruler of the earth.&quot;<br /> <br /> - Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br /> <br /> - Gmaas has a summer house on Mars.<br /> <br /> -Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time, using a time-turner.<br /> <br /> -Gmaas also attended Hogwarts and was a prefect.<br /> <br /> -Mrs.Norris is Gmaas's archenemy.<br /> <br /> -Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin. Because cats are demigod counselors too.<br /> <br /> -Gmaas has completed over 2,000 quests, and is very popular throughout Camp Half-Blood. He has also been to Camp Jupiter.<br /> <br /> -Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br /> <br /> -Gmass knows that the real names are Gmassa Lisa, The Last Domestic Meal, and Far-away Light.<br /> <br /> -Gmaas actually attended all the Ivy Leagues.<br /> <br /> - I am Gmaas<br /> <br /> -I too am Gmaas<br /> <br /> -But it is I who is Gmaas<br /> <br /> -Gmaas is us all<br /> <br /> -Gmaas is all of us yet none of us<br /> <br /> - Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br /> <br /> -Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br /> <br /> -Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br /> <br /> -Gmaas crossed the Delaware River with Washington.<br /> <br /> -Gmaas also crossed the Atlantic with the pilgrims.<br /> <br /> -if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br /> <br /> -Chuck norris makes Gmaas jokes.<br /> <br /> -Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br /> <br /> - Gmaas killed Big Brother by farting on him. Though he was caught by the Ministry of Love, he escaped easily.<br /> <br /> -Everyone thinks that Gmaas is a god<br /> <br /> -Gmaas also owns Animal Farm. Napoleon was his servant.<br /> <br /> -Gmaas is the only one who knows where Amelia Earhart is.<br /> <br /> -Gmaas is the only cat that has been proven transcendental.<br /> <br /> - Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br /> <br /> -Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmass in't very happy about that, either.<br /> <br /> -The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br /> <br /> -Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br /> <br /> -Gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br /> <br /> -Gmaas does merely not use USD. He owns it.<br /> <br /> -Gmaas really knows that roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> -In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br /> <br /> -&quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> - Gmaas is the smartest living being in the universe.<br /> <br /> -It was Gmaas who helped Sun Wukong on the Journey to the West.<br /> <br /> -Gmaas is the real creator of Wikipedia.<br /> <br /> -It is said Gmaas could hack any website he desires.<br /> <br /> -Gmaas is the basis of Greek mythology.<br /> <br /> -Gmaas once sold Google to a man for around &lt;math&gt;12&lt;/math&gt; dollars!<br /> <br /> -Gmaas uses a HP printer.<br /> <br /> -Gmaas owns all AoPS staff including Richard Rusczyk.<br /> <br /> -Gmaas was there when Yoda was born.<br /> <br /> - Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br /> <br /> - sseraj once spelled Gmaas as gmASS on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br /> <br /> -Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br /> <br /> -Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmaas in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also &lt;math&gt;\boxed{\text{loves}}&lt;/math&gt; &lt;b&gt;Cat&lt;/b&gt;ch that fish.<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;math&gt;289547987693&lt;/math&gt; robux and &lt;math&gt;190348&lt;/math&gt; in CPR.<br /> <br /> -This is all hypothetical.<br /> <br /> - EDIT: This is all factual <br /> <br /> -Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> -The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999999999999999999999999999999999999th dimension.<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise.<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, cyumi, srej, squareman, Electro3.0, sturdywill, g1zq, alleycat, batta, beastgert, and lakecomo224. His best account, though, is wlm7.<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> -Gmaas knows how to hack into top secret aops community pages.<br /> <br /> -Gmaas was a river clant cat who crossed the event horizon of a black hole and came out the other end!<br /> <br /> -Gmaas is king of the first men, the anduls.<br /> <br /> -Gmaas is a well known professor at MEOWston Academy.<br /> <br /> -Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout.<br /> <br /> -Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat. (He also out-grumped grumpy cat!!!)<br /> <br /> -Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM.<br /> <br /> - owner of sseraj, not pet.<br /> <br /> - embodiment of life and universe and beyond .<br /> <br /> - Watches memes.<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls.<br /> <br /> -Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmaas is a certified Slytherin.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;/math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is everyone's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> ~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> -EDIT. The above fact is slightly irrelevant.<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99999.\overline{9}}{100000}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> -EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;math&gt;57843504&lt;/math&gt; regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his smelly butt.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - Gmass is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once.<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;math&gt;-GMAAS&lt;/math&gt; seconds.<br /> <br /> -Gmaas beat Superman in a fight with ease<br /> <br /> -Gmaas was an admin of Roblox<br /> <br /> -Gmaas traveled around the world, paying so much &lt;math&gt;MONEY&lt;/math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;math&gt;0.\overline{99}&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt;, the difference is greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br /> <br /> -Picture of Gmaas http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmaas in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot;. anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br /> <br /> - gmaas has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmaas is the ruler of the universe and has been known to be the creator of the species &quot;Gmaasians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> - Gmaas burped and caused an earthquake.<br /> <br /> - Gmaas once drank from your pretty teacup.<br /> <br /> -GMAAS IS HERE.... PURRRRRRRRRRRRRRRRRRRR<br /> <br /> -Gmass made, and currently owns the Matrix.<br /> <br /> -The above fact is true. Therefore, this is an illusion.<br /> <br /> -Gmaas is the reason Salah will become better than Ronaldo.<br /> <br /> -Who is Gmaas?<br /> -Gmass is a heavenly being<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - &lt;s&gt;Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.&lt;/s&gt;Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br /> <br /> -Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;math&gt;Another&lt;/math&gt; sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> - Yet Another Gmaas sighting ? [https://supportforums.cisco.com/t5/user/viewprofilepage/user-id/45046]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: &lt;img&gt; http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg &lt;/img&gt;<br /> <br /> - When Gmaas was mad, he started world wars 1 &amp; 2. It is only because of Gmaas that we have not had World War 3.<br /> <br /> - Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br /> <br /> - Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br /> <br /> -Gmaas likes to whiz on the wilzo<br /> <br /> -Gmaas has been spotted in AMC 8 Basics<br /> <br /> -Gmaas is cool<br /> <br /> -Gmaas hemoon card that does over 9000000 dmg<br /> <br /> -Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br /> <br /> -Kirby once swallowed Gmaas. Gmaas had to spit him out.</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_6&diff=94018 1962 AHSME Problems/Problem 6 2018-04-18T19:55:23Z <p>Wlm7: /* Solution */</p> <hr /> <div>==Problem==<br /> A square and an equilateral triangle have equal perimeters. The area of the triangle is &lt;math&gt;9 \sqrt{3}&lt;/math&gt; square inches. Expressed in inches the diagonal of the square is: <br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{9}{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{none of these} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> To solve for the perimeter of the triangle we plug in the formula for the area of an equilateral triangle which is &lt;math&gt;\dfrac{x^2\sqrt{3}}{4}&lt;/math&gt;. This has to be equal to &lt;math&gt;9 \sqrt{3}&lt;/math&gt;, which means that &lt;math&gt;x=6&lt;/math&gt;, or the side length of the triangle is &lt;math&gt;6&lt;/math&gt;. Thus, the triangle (and the square) have a perimeter of &lt;math&gt;18&lt;/math&gt;. It follows that each side of the square is &lt;math&gt;\dfrac{18}{4}=4.5&lt;/math&gt;. If we draw the diagonal, we create a 45-45-90 triangle, whose hypotenuse (also the diagonal of the square) is &lt;math&gt;\boxed{\text{(D)} \dfrac{9\sqrt{2}}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AHSME 40p box|year=1962|before=Problem 5|num-a=7}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=93991 Gmaas 2018-04-16T23:16:56Z <p>Wlm7: /* Gmaas Facts */</p> <hr /> <div>=== Gmaas Facts ===<br /> - gm lion isn't even his final form.<br /> <br /> - piphi posts GMAAS pictures in his blog<br /> https://artofproblemsolving.com/community/c580181<br /> <br /> -Gmaas once farted. The result was the Big Bang.<br /> <br /> - Gmaas knows these digits of pi from memory: 3.1415926535897932384626433. Edit: He invented pi and pie.<br /> <br /> -since Gmaass invented pi, he knows that the digits are really 9.587979087879877897087r09780970987098790870987609870987908067897578786. digit-ese for &quot;Gmaass is the ruler of the earth.&quot;<br /> <br /> -<br /> <br /> -Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time.<br /> <br /> -Gmaas also attended Hogwarts and was a prefect.<br /> <br /> -Mrs.Norris is Gmaas's archenemy.<br /> <br /> -Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin.<br /> <br /> -Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br /> <br /> -Gmaas actually attended all the Ivy Leagues.<br /> <br /> - I am Gmaas<br /> <br /> -I too am Gmaas<br /> <br /> -But it is I who is Gmaas<br /> <br /> -Gmass is us all<br /> <br /> - Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br /> <br /> -Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br /> <br /> -Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br /> <br /> -Gmaas crossed the Delaware River with Washington.<br /> <br /> -Gmaas also crossed the Atlantic with the pilgrims.<br /> <br /> -if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br /> <br /> -Chuck norris makes Gmaas jokes.<br /> <br /> -Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br /> <br /> -Gmaas also owns Animal Farm. Napoleon was his servant.<br /> <br /> -Gmaas is the only one who knows where Amelia Earhart is.<br /> <br /> -Gmaas is the only cat that has been proven transcendental.<br /> <br /> - Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br /> <br /> -Grumpy cat reads Gmaas memes<br /> <br /> - The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br /> <br /> -Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br /> <br /> -gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br /> <br /> - Gmaas does merely not use USD. He owns it.<br /> <br /> -Gmaas really knows that roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> -In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br /> <br /> -&quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> - Gmaas is the smartest living being in the universe.<br /> <br /> -It was Gmaas who helped Monkey King on the Journey to the West.<br /> <br /> -Gmaas is the real creator of Wikipedia.<br /> <br /> -It is said Gmaas could hack any website he desires.<br /> <br /> -Gmaas is the basis of Greek mythology<br /> <br /> -Gmaas once sold Google to a man for around &lt;math&gt;12.<br /> <br /> -Gmaas uses a HP printer.<br /> <br /> -Gmaas owns all AoPS staff including Richard Rusczyk<br /> <br /> - Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br /> <br /> - sseraj once spelled gmaas as gmaas on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br /> <br /> -Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br /> <br /> -Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmaas in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;/math&gt;289547987693&lt;math&gt; robux and &lt;/math&gt;190348&lt;math&gt; in CPR.<br /> <br /> -This is all hypothetical EDIT: This is all factual <br /> <br /> -Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999th dimension<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, Lord_Baltimore, cyumi, squareman, Electro3.0, and lakecomo224<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> - knows how to hack into top secret ops community pages<br /> <br /> -Gmaass was a river clant cat who crossed the event horizon of a black hole. and Gmaass came out the other end!<br /> <br /> - Gmaas is king of the first men, the anduls<br /> <br /> -Gmaas is a well known professor at MEOWston Academy<br /> <br /> -Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout<br /> <br /> - Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat (He also out grumpied grumpy cat!!!)<br /> <br /> -Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM<br /> <br /> - owner of sseraj, not pet<br /> <br /> - embodiment of life and universe and beyond <br /> <br /> - Watches memes<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br /> <br /> -Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmaas is a certified Slytherin and probably the cutest cat ever.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;/math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is 5space's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;/math&gt;\frac{3972}{7891}&lt;math&gt;), or malnourished (with probability &lt;/math&gt;\frac{3919}{7891}&lt;math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> ~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> -EDIT. The above fact is slightly irrelevant.<br /> <br /> - Gmaas might have been viewing (with a &lt;/math&gt;\frac{99999.\overline{9}}{100000}&lt;math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;/math&gt;\tan(90)&lt;math&gt; ritual. Draw a pentagram and write the numerical value of &lt;/math&gt;\tan(90)&lt;math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> -EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;/math&gt;57843504&lt;math&gt; regenerations and has used &lt;/math&gt;3&lt;math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;/math&gt;\frac{13}{32}&lt;math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows. (Except him.)<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his smelly butt.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - He (or she) is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;/math&gt;-GMAAS&lt;math&gt; seconds.<br /> <br /> -Gmaas beat Superman in a fight with ease<br /> <br /> -Gmaas was an admin of Roblox<br /> <br /> -Gmaas traveled around the world, paying so much &lt;/math&gt;MONEY&lt;math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;/math&gt;0.\overline{99}&lt;math&gt; from &lt;/math&gt;1&lt;math&gt;, the difference is greater than &lt;/math&gt;0&lt;math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Ender is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will<br /> <br /> -Picture of Gmaas http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmaas in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot; anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br /> <br /> - gmaas has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmaas is the ruler of the universe and has been known to be the creator of the species &quot;Gmaasians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> - Gmaas burped and caused an earthquake.<br /> <br /> - Gmaas once drank from your pretty teacup.<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.[/s]Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br /> <br /> -Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;/math&gt;Another$ sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: &lt;img&gt; http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg &lt;/img&gt;<br /> <br /> - When Gmaas was mad, he started world wars 1 &amp; 2. It is only because of Gmaas that we have not had World War 3.<br /> <br /> - Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br /> <br /> - Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br /> <br /> -Gmaas likes to whiz on the wilzo<br /> <br /> -Gmaas is ded<br /> <br /> -Gmaas has been spotted in AMC 8 Basics<br /> <br /> -Gmaas is cool</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=92308 AoPS Wiki:FAQ 2018-02-24T15:37:26Z <p>Wlm7: /* AoPS Acronyms */</p> <hr /> <div>{{shortcut|[[A:FAQ]]}}<br /> <br /> This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&amp;t=414129 here].<br /> <br /> == General==<br /> <br /> <br /> ==== Can I change my user name? ====<br /> <br /> :No, unless it has personal information.<br /> <br /> ==== Why can't I change my avatar? ====<br /> <br /> : You must be a user for two weeks before being able to change your avatar.<br /> : It is almost impossible to get a small enough photo to fit the 24 KB restrictions.<br /> : Just be happy with what you have.<br /> <br /> ==== Something looks weird (e.g., blurry, missing line, etc.) ====<br /> <br /> : This is likely due to your browser zoom level. Please make sure your zoom level is at 100% for correct rendering of the web page.<br /> <br /> ==== Can I make more than one account?==== <br /> <br /> :Short answer: No.<br /> <br /> :Long answer: Multiple accounts (multis) are banned on AoPS. Having more than one account leads to issues of not remembering on what account you did what. Using multiple accounts to &quot;game&quot; the system, (e.g. increase rating for posts or in online games) will lead to bans on all accounts associated with you. If you have already made additional accounts, please choose one account and stop using the others.<br /> <br /> ====What software does Art of Problem Solving use to run the website?====<br /> <br /> :* Search: Solr<br /> :* Wiki: MediaWiki<br /> :* Asymptote and &lt;math&gt;\text{\LaTeX}&lt;/math&gt; are generated through their respective binary packages<br /> :* Videos: YouTube<br /> <br /> :All other parts of the website are custom built.<br /> <br /> ====Can you make an AoPS App?====<br /> <br /> :No. The logistics and costs involved are too great and there would be very little, if any benefit to the app over the website page itself.<br /> <br /> == Forums ==<br /> <br /> ====How do I create a forum?====<br /> <br /> :To create an AoPS forum, a user must be on the AoPS community for at least 2 weeks. To create a forum, hover over the community tab, then click &quot;My AoPS.&quot; You should now see your avatar, and a list of your friends. Now, click on the &quot;My Forums&quot; tab. There, you would be able to see which forum you moderate or administrate, as well as the private forums you can access. Click on the &quot;+&quot; button at the top right. This should lead you to a forum creating page. From there, you may ruin everybody's life with your terrible writings.<br /> <br /> ====Why do some posts say they were posted in the future?====<br /> :The AoPS clock is based on the official Naval Observatory time. This is considered the most accurate time. If your computer's system clock is behind the correct time, recent posts may indicate they were posted in the future. Please correct your computer's clock or enable clock synchronization so that your clock is always correct. Mac users may wish to check [http://www.macinstruct.com/node/92 Synchronize your Mac's Clock With A Time Server]. You can also check [http://www.time.gov the US offical time.]<br /> <br /> ====I've lost admin access to a forum or blog I created, how do I get it back?====<br /> :Please send an email to sheriff@aops.com. They will research your request and restore admin access to your forum if appropriate.<br /> <br /> ====What should I do if I find a glitch in the community?====<br /> :First, search the Site Support forum to check if your issue has already been reported. You can search here.<br /> <br /> :If your issue isn't reported, try refreshing your browser page. Most issues go away after a refresh and there is no need to report the issue unless it continues after you refresh your browser. You can refresh on most browsers in Windows with Ctrl + Shift + R and on Mac with Cmd + Shift + R.<br /> <br /> :Some commonly known glitches that should not be reported are avatars appearing twice in the topic list or private message, friends appearing more than once in the friend list, the edit icon showing next to a message when it shouldn't, and not being able to search for a forum after going back a page in the community.<br /> <br /> :Some glitches I have found that should be reported is a little space like thing that counts as &lt;math&gt;8&lt;/math&gt; characters, and capitol quote and hide tags to nest more than &lt;math&gt;3&lt;/math&gt;.<br /> <br /> ==== How do I format my post, e.g. bold text, add URLs, etc.? ====<br /> <br /> :AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found [[BBCode:Tutorial|here]].<br /> <br /> ==== How do I hide content in the forums? ====<br /> :Wrap the content you want to hide in [hide] tags.<br /> [hide]Content[/hide]<br /> :If you want to customize the label, instead of saying &quot;Click here to reveal hidden text&quot;, you can do something like:<br /> [hide=Label to display]Content[/hide]<br /> <br /> ====I got the message &quot;Too many messages.&quot; when trying to send a private message, why?====<br /> :To prevent PM spam abuse<br /> like &quot;hvbowibvibviorybvirusbshbiuovvisuvib&quot;, users with less than five forum posts are limited to four private messages within a forty-eight hour period.<br /> <br /> ==== If I make more posts, it means I'm a better user, right? ====<br /> :Post quality is far more important than post quantity.<br /> <br /> ==== I have made some posts but my post count did not increase. Why? ====<br /> :When you post in some of the forums, such as the Test Forum, Mafia Forum, Fun Factory, and most user created forums, the post does not count towards your overall post count. And still. Quality over quantity.<br /> <br /> ==== I believe a post needs corrective action. What should I do? ====<br /> :If you believe a post needs moderative action, you may report it by clicking the &quot;!&quot; icon on the upper-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better. Usually, you shouldn't publicly post such things on a thread itself, which is called &quot;backseat moderation&quot; and is considered rude and unproductive.<br /> <br /> ==== How do I post images? ====<br /> :While AoPS forums have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restorations, etc. We also have limited disk space which causes us to remove attachments from time to time. Therefore, we recommend using a third party image hosting solution. There are many options, but we recommend imgur.com.<br /> <br /> :* Go to imgur.com<br /> :* Click upload images<br /> :* Follow the on screen instructions to upload the image to imgur.<br /> :* After uploading, view your image<br /> :* Right click (cmd-click on mac) and select something similar to &quot;Copy image address&quot;<br /> :* In your post, paste the image address and surround it by [img]..[/img] tags<br /> <br /> ====How do I become a moderator of a forum?====<br /> :The creators of user-created forums select moderators in several different ways. Contact the creator(s) of the forum for more information.<br /> <br /> :The AoPS staff will select moderators for the major forums on an as-needed basis. If the AoPS staff determines that a forum needs additional moderators, then the staff will reach out to productive users and invite them to become moderators. These are some features that the AoPS staff may consider when selecting new moderators.<br /> <br /> :* An established history of productive posting. No unproductive, spammy, or troll posts in the past several years. (For example, a high school student might have made some poor choices while in elementary school -- but then learned from those mistakes.)<br /> :* No use of \8char or other strategies to get around rules or restrictions on the forums.<br /> :* When the user reports posts, the report is useful and helpful, and the reported post (or thread, or user) really did reflect a serious issue worthy of administrators' attention. The user filled in the &quot;Further details&quot; field with valuable information. The user does not report the use of phrases such as &quot;gosh, golly gee&quot; as profanity.<br /> :* No multis.<br /> :* No backseat moderating, no posting of &quot;\requestlock&quot;.<br /> :* When the user revives an old thread, the new post is valuable and relevant.<br /> :* The user does not use the forums to cheat on Alcumus, homework, contests, or similar activities.<br /> <br /> == Blogs ==<br /> ==== How come I can't create a blog? ====<br /> :One needs to have at least 5 posts in order to make a blog.<br /> <br /> ==== How do I make my blog look nice? ====<br /> :Many AoPSers make their blogs look awesome by applying [[CSS|CSS]], which is a high-level stylesheet language. This can be done by typing CSS code into the CSS box in the Blog Control Panel.<br /> <br /> == Alcumus ==<br /> <br /> ==== How is rating computed? ====<br /> :The rating is more of a prediction of what percentage of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. However, if you are predicted to get most correct, and you miss one or two problems, the rating, or prediction of percentage correct, will go down.<br /> <br /> ==== I am stuck on a problem, changing the topic does not change the problem. ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before moving on.<br /> <br /> ==== Why can't I change topics? ====<br /> :Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before the topic changes to the currently selected topic.<br /> <br /> == Contests ==<br /> ==== Where can I find past contest questions and solutions? ====<br /> :In the [http://www.artofproblemsolving.com/Forum/resources.php Contests] section.<br /> <br /> ==== How do I get problems onto the contest page? ====<br /> <br /> :Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.<br /> <br /> ==== What are the guidelines for posting problems to be added to the contests section? ====<br /> :Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&amp;t=195579 guidelines in this post].<br /> <br /> ==== Why is the wiki missing many contest questions? ====<br /> :Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations.<br /> <br /> ==== What if I find an error on a problem? ====<br /> :Please post an accurate description of the problem in [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&amp;t=426693 this thread]. If the problem is on the wiki, you can edit it yourself.<br /> <br /> == LaTeX and Asymptote ==<br /> ==== What is LaTeX, and how do I use it? ====<br /> <br /> :&lt;math&gt;\text{\LaTeX}&lt;/math&gt; is a typesetting markup language that is useful to produce properly formatted mathematical and scientific expressions.<br /> <br /> ==== How can I download LaTeX to use on the forums? ====<br /> <br /> :There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br /> <br /> ==== How can I download LaTeX for personal use? ====<br /> :You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br /> <br /> ==== Where can I find a list of LaTeX commands? ====<br /> :See [[LaTeX:Symbols|here]].<br /> <br /> ==== Where can I test LaTeX commands? ====<br /> <br /> :[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. You can also use our [http://artofproblemsolving.com/community/c67_test_forum Test Forum].<br /> <br /> ==== Where can I find examples of Asymptote diagrams and code? ====<br /> <br /> :Search this wiki for the &lt;tt&gt;&lt;nowiki&gt;&lt;asy&gt;&lt;/nowiki&gt;&lt;/tt&gt; tag or the Forums for the &lt;tt&gt;&lt;nowiki&gt;[asy]&lt;/nowiki&gt;&lt;/tt&gt; tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br /> <br /> ==== How can I draw 3D diagrams? ====<br /> <br /> :See [[Asymptote: 3D graphics]].<br /> <br /> ==== What is the cse5 package? ==== <br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example &lt;tt&gt;D()&lt;/tt&gt; for &lt;tt&gt;draw()&lt;/tt&gt; and &lt;tt&gt;dot()&lt;/tt&gt;, and so forth.<br /> <br /> ==== What is the olympiad package? ====<br /> <br /> :See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&amp;t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br /> <br /> == AoPSWiki ==<br /> ==== Is there a guide for wiki syntax? ====<br /> <br /> :See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br /> <br /> ==== What do I do if I see a mistake in the wiki? ====<br /> <br /> :Click &quot;edit&quot; and correct the error! (But like, don't touch the acronyms. Plz.)<br /> <br /> ==== Why can't I edit the wiki? ====<br /> <br /> :Ha ha ha...<br /> <br /> :Peasant.<br /> <br /> == Miscellaneous ==<br /> ==== Is it possible to join the AoPS Staff? ====<br /> <br /> :Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns, graders, and teaching assistants. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br /> <br /> ==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br /> <br /> :You must have graduated from high school, or at least be in the second term of your senior year.<br /> <br /> ==AoPS Acronyms==<br /> *'''AFK'''- Away from keyboard<br /> *'''AoPS'''- Art of Problem Solving, the website you're not on right now, because we're inside a simulation!<br /> *'''AIME'''- American Invitational Mathematics Examination<br /> *'''AMC'''- American Math Competitions<br /> *'''ATM'''- At the Moment/Automated teller machine (probably less relevant)<br /> *'''brb'''- Be right Back<br /> *'''BTW'''- By the way<br /> *'''CEMC''' - Centre for Mathematics and Computing<br /> *'''C&amp;P or C+P or CP''' - Counting and Probability or Contests and Programs<br /> *'''EBWOP'''- Editing by way of post<br /> *'''FTW'''- For the Win, a game on AoPS<br /> *'''gg'''- Good Game<br /> *'''gj'''- Good Job<br /> *'''glhf'''-Good Luck Have Fun<br /> *'''gtg''' - Got to go<br /> *'''HSM''' - High School Math Forum<br /> *'''ID(R)K'''-I Don't (Really) Know<br /> *'''iff'''-If and only if<br /> *'''IIRC'''- If I recall correctly<br /> *'''IKR'''- I know, right?<br /> *'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br /> *'''JMO'''- United States of America Junior Mathematical Olympiad<br /> *'''lol'''- Laugh Out Loud<br /> *'''MC'''- Mathcounts, a popular math contest for Middle School students.<br /> *'''NFL'''- Not for long/No friends left/Nation Football League<br /> *'''mod(s)'''- Moderator(s)<br /> *'''MOEMS'''- Math Olympiads for Elementary and Middle Schools<br /> *'''MO(S)P'''- Mathematical Olympiad (Summer) Program<br /> *'''MSM'''- Middle School Math Forum<br /> *'''NIMO'''-National Internet Math Olympiad<br /> *'''NT'''- Number Theory<br /> *'''OBC'''- Online by computer<br /> *'''OMG'''- Oh My Gosh<br /> *'''OMO'''-Online Math Olympiad<br /> *'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br /> *'''OSPD'''- Oatsquire Pedigree<br /> *'''PIG'''- Poorly Intellectual Guy/Girl<br /> *'''QED'''- Quod erat demonstrandum, Latin for &quot;Which was to be proven&quot;; some English mathematicians use it as an acronym for Quite Elegantly Done<br /> *'''QS&amp;A'''- Questions, Suggestions, and Announcements Forum<br /> *'''ro(t)fl''' - Rolling on the floor laughing<br /> *'''smh''' - Shaking my head<br /> *'''sqrt''' - Square root<br /> *'''Sticky'''- A post pinned to the top of a forum - a thing no one reads but really should read<br /> *'''ToS'''- Terms of Service - a thing no one reads but really should read<br /> *'''USA(J)MO'''- USA (Junior) Mathematical Olympiad<br /> *'''V/LA'''- Vacation or Long Absence/Limited Access<br /> *'''WLOG'''- Without loss of generality<br /> *'''wrt'''- With respect to<br /> *'''wtg''' - Way to go<br /> *'''tytia'''- Thank you, that is all<br /> *'''xD'''- Bursting Laugh<br /> <br /> == FTW! ==<br /> <br /> Please see the [//artofproblemsolving.com/ftw/faq For the Win! FAQ] for many common questions.<br /> <br /> == School ==<br /> <br /> ==== What if I miss a class? ====<br /> :There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board. Don't worry, though, classroom participation usually isn't weighted heavily.<br /> <br /> ==== Is there audio or video in class? ====<br /> :There is generally no audio or video in the class. The classes are generally text and image based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity and being less pedagogically effective, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video.<br /> <br /> ==== What if I want to drop out of a class? ====<br /> :For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br /> <br /> ==== For my homework, there is supposed to be a green bar but it's orange, why? ====<br /> <br /> :For the bar to turn green, the writing problem must be attempted. Once it is attempted the bar will turn green.<br /> <br /> ==== I need more time for my homework, what should I do? ====<br /> <br /> :There is a &quot;Request Extension&quot; button in the homework tab of your class. This will automatically extend the due date to 2 days after the normal deadline. If you want more time you need to ask for it in the little comment box, stating the reason why you want the extension, and how much time you want. This request will be looked at by the teachers and they will decide if you get the extension or not. Note that you can only use this button 3 times.<br /> :Otherwise, you can send an email to extensions@aops.com with your username, class name and ID (the number in the class page URL after https://artofproblemsolving.com/class/) and reason for extension. Someone should get back to you within a couple days.</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Constant_function&diff=80193 Constant function 2016-09-03T23:34:47Z <p>Wlm7: </p> <hr /> <div>A '''constant function''' is a [[function]] which has a [[constant]] output: the value of the function does not depend on the value of its input. Equivalently, a constant function is a function whose [[range]] has only a single value. For example, &lt;math&gt;f(x)=4&lt;/math&gt; is a constant function because &lt;math&gt;4&lt;/math&gt; always equals &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;g(x)=\log_{34} e^{\pi-\frac{1}{2}}&lt;/math&gt; is a constant function since &lt;math&gt;\log_{34} e^{\pi-\frac{1}{2}}&lt;/math&gt; is always equal to &lt;math&gt;\log_{34} e^{\pi-\frac{1}{2}}&lt;/math&gt;. So, basically, a constant function is a function that for whatever input you put in, it always returns the same value, or a constant.<br /> <br /> {{stub}}<br /> [[Category:Functions]]<br /> [[Category:Definition]]</div> Wlm7 https://artofproblemsolving.com/wiki/index.php?title=Constant_function&diff=80192 Constant function 2016-09-03T23:32:29Z <p>Wlm7: </p> <hr /> <div>A '''constant function''' is a [[function]] which has a [[constant]] output: the value of the function does not depend on the value of its input. Equivalently, a constant function is a function whose [[range]] has only a single value. For example, &lt;math&gt;f(x)=4&lt;/math&gt; is a constant function, and so is &lt;math&gt;g(x)=\log_{34} e^{\pi-\frac{1}{2}}&lt;/math&gt;. So basically, a constant function that whatever input you put in, it always returns the same value.<br /> <br /> {{stub}}<br /> [[Category:Functions]]<br /> [[Category:Definition]]</div> Wlm7