https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Notmyopic667&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-03T18:29:14Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems&diff=153528 2016 AMC 8 Problems 2021-05-11T15:21:39Z <p>Notmyopic667: /* Problem 2 */</p> <hr /> <div><br /> <br /> ==Problem 1==<br /> <br /> The longest professional tennis match lasted a total of 11 hours and 5 minutes. How many minutes is that?<br /> <br /> &lt;math&gt;\textbf{(A)} 605 \qquad\textbf{(B)} 655\qquad\textbf{(C)} 665\qquad\textbf{(D)} 1005\qquad \textbf{(E)} 1105&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 1|Solution<br /> ]]<br /> <br /> ==Problem 2==<br /> <br /> In rectangle &lt;math&gt;ABCD&lt;/math&gt;, &lt;math&gt;AB=6&lt;/math&gt; and &lt;math&gt;AD=8&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{AD}&lt;/math&gt;. What is the area of &lt;math&gt;\triangle AMC&lt;/math&gt;?<br /> <br /> &lt;asy&gt;draw((0,4)--(0,0)--(6,0)--(6,8)--(0,8)--(0,4)--(6,8)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (6, 0), SE);<br /> label(&quot;$C$&quot;, (6,8), NE);<br /> label(&quot;$D$&quot;, (0, 8), NW);<br /> label(&quot;$M$&quot;, (0, 4), W);<br /> label(&quot;$4$&quot;, (0, 2), W);<br /> label(&quot;$6$&quot;, (3, 0), S);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 2|Solution<br /> ]]<br /> <br /> ==Problem 3==<br /> <br /> Four students take an exam. Three of their scores are &lt;math&gt;70, 80,&lt;/math&gt; and &lt;math&gt;90&lt;/math&gt;. If the average of their four scores is &lt;math&gt;70&lt;/math&gt;, then what is the remaining score?<br /> <br /> &lt;math&gt;\textbf{(A) }40\qquad\textbf{(B) }50\qquad\textbf{(C) }55\qquad\textbf{(D) }60\qquad \textbf{(E) }70&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 3|Solution<br /> ]]<br /> <br /> ==Problem 4==<br /> <br /> When Cheenu was a boy he could run &lt;math&gt;15&lt;/math&gt; miles in &lt;math&gt;3&lt;/math&gt; hours and &lt;math&gt;30&lt;/math&gt; minutes. As an old man he can now walk &lt;math&gt;10&lt;/math&gt; miles in &lt;math&gt;4&lt;/math&gt; hours. How many minutes longer does it take for him to travel a mile now compared to when he was a boy?<br /> <br /> &lt;math&gt;\textbf{(A) }6\qquad\textbf{(B) }10\qquad\textbf{(C) }15\qquad\textbf{(D) }18\qquad \textbf{(E) }30&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 4|Solution<br /> ]]<br /> <br /> ==Problem 5==<br /> <br /> The number &lt;math&gt;N&lt;/math&gt; is a two-digit number.<br /> <br /> • When &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;9&lt;/math&gt;, the remainder is &lt;math&gt;1&lt;/math&gt;.<br /> <br /> • When &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;10&lt;/math&gt;, the remainder is &lt;math&gt;3&lt;/math&gt;.<br /> <br /> What is the remainder when &lt;math&gt;N&lt;/math&gt; is divided by &lt;math&gt;11&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }7&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 5|Solution<br /> ]]<br /> <br /> ==Problem 6==<br /> The following bar graph represents the length (in letters) of the names of 19 people. What is the median length of these names?<br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }7&lt;/math&gt;<br /> &lt;asy&gt;<br /> unitsize(0.9cm);<br /> draw((-0.5,0)--(10,0), linewidth(1.5));<br /> draw((-0.5,1)--(10,1));<br /> draw((-0.5,2)--(10,2));<br /> draw((-0.5,3)--(10,3));<br /> draw((-0.5,4)--(10,4));<br /> draw((-0.5,5)--(10,5));<br /> draw((-0.5,6)--(10,6));<br /> draw((-0.5,7)--(10,7));<br /> label(&quot;frequency&quot;,(-0.5,8));<br /> label(&quot;0&quot;, (-1, 0));<br /> label(&quot;1&quot;, (-1, 1));<br /> label(&quot;2&quot;, (-1, 2));<br /> label(&quot;3&quot;, (-1, 3));<br /> label(&quot;4&quot;, (-1, 4));<br /> label(&quot;5&quot;, (-1, 5));<br /> label(&quot;6&quot;, (-1, 6));<br /> label(&quot;7&quot;, (-1, 7));<br /> filldraw((0,0)--(0,7)--(1,7)--(1,0)--cycle, black);<br /> filldraw((2,0)--(2,3)--(3,3)--(3,0)--cycle, black);<br /> filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, black);<br /> filldraw((6,0)--(6,4)--(7,4)--(7,0)--cycle, black);<br /> filldraw((8,0)--(8,4)--(9,4)--(9,0)--cycle, black);<br /> label(&quot;3&quot;, (0.5, -0.5));<br /> label(&quot;4&quot;, (2.5, -0.5));<br /> label(&quot;5&quot;, (4.5, -0.5));<br /> label(&quot;6&quot;, (6.5, -0.5));<br /> label(&quot;7&quot;, (8.5, -0.5));<br /> label(&quot;name length&quot;, (4.5, -1));<br /> &lt;/asy&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 6|Solution<br /> ]]<br /> <br /> <br /> ==Problem 7==<br /> <br /> Which of the following numbers is not a perfect square?<br /> <br /> &lt;math&gt;\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 7|Solution<br /> ]]<br /> <br /> ==Problem 8==<br /> <br /> Find the value of the expression<br /> &lt;cmath&gt;100-98+96-94+92-90+\cdots+8-6+4-2.&lt;/cmath&gt;&lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 8|Solution<br /> ]]<br /> <br /> ==Problem 9==<br /> <br /> What is the sum of the distinct prime integer divisors of &lt;math&gt;2016&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }9\qquad\textbf{(B) }12\qquad\textbf{(C) }16\qquad\textbf{(D) }49\qquad \textbf{(E) }63&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 9|Solution<br /> ]]<br /> <br /> ==Problem 10==<br /> <br /> Suppose that &lt;math&gt;a * b&lt;/math&gt; means &lt;math&gt;3a-b.&lt;/math&gt; What is the value of &lt;math&gt;x&lt;/math&gt; if<br /> &lt;cmath&gt;2 * (5 * x)=1&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac{1}{10} \qquad\textbf{(B) }2\qquad\textbf{(C) }\frac{10}{3} \qquad\textbf{(D) }10\qquad \textbf{(E) }14&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 10|Solution<br /> ]]<br /> <br /> ==Problem 11==<br /> <br /> Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is &lt;math&gt;132.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 11|Solution<br /> ]]<br /> <br /> ==Problem 12==<br /> <br /> Jefferson Middle School has the same number of boys and girls. &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the girls and &lt;math&gt;\frac{2}{3}&lt;/math&gt;<br /> of the boys went on a field trip. What fraction of the students on the field trip were girls?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 12|Solution<br /> ]]<br /> <br /> ==Problem 13==<br /> <br /> Two different numbers are randomly selected from the set &lt;math&gt;{ - 2, -1, 0, 3, 4, 5}&lt;/math&gt; and multiplied together. What is the probability that the product is &lt;math&gt;0&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 13|Solution<br /> ]]<br /> <br /> ==Problem 14==<br /> <br /> Karl's car uses a gallon of gas every &lt;math&gt;35&lt;/math&gt; miles, and his gas tank holds &lt;math&gt;14&lt;/math&gt; gallons when it is full. One day, Karl started with a full tank of gas, <br /> drove &lt;math&gt;350&lt;/math&gt; miles, bought &lt;math&gt;8&lt;/math&gt; gallons of gas, and continued driving to his destination. When he arrived, his gas tank was half full. How many miles did Karl drive that day? <br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }525\qquad\textbf{(B)}\mbox{ }560\qquad\textbf{(C)}\mbox{ }595\qquad\textbf{(D)}\mbox{ }665\qquad\textbf{(E)}\mbox{ }735&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 14|Solution<br /> ]]<br /> <br /> ==Problem 15==<br /> <br /> What is the largest power of &lt;math&gt;2&lt;/math&gt; that is a divisor of &lt;math&gt;13^4 - 11^4&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }8\qquad \textbf{(B)}\mbox{ }16\qquad \textbf{(C)}\mbox{ }32\qquad \textbf{(D)}\mbox{ }64\qquad \textbf{(E)}\mbox{ }128&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 15|Solution<br /> ]]<br /> <br /> ==Problem 16==<br /> <br /> Annie and Bonnie are running laps around a &lt;math&gt;400&lt;/math&gt;-meter oval track. They started together, but Annie has pulled ahead because she runs &lt;math&gt;25\%&lt;/math&gt; faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?<br /> <br /> &lt;math&gt;\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 16|Solution<br /> ]]<br /> <br /> ==Problem 17==<br /> <br /> An ATM password at Fred's Bank is composed of four digits from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt;, with repeated digits allowable. If no password may begin with the sequence &lt;math&gt;9,1,1,&lt;/math&gt; then how many passwords are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 17|Solution<br /> ]]<br /> <br /> ==Problem 18==<br /> <br /> In an All-Area track meet, &lt;math&gt;216&lt;/math&gt; sprinters enter a &lt;math&gt;100-&lt;/math&gt;meter dash competition. The track has &lt;math&gt;6&lt;/math&gt; lanes, so only &lt;math&gt;6&lt;/math&gt; sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 18|Solution<br /> ]]<br /> <br /> ==Problem 19==<br /> <br /> The sum of &lt;math&gt;25&lt;/math&gt; consecutive even integers is &lt;math&gt;10,000&lt;/math&gt;. What is the largest of these &lt;math&gt;25&lt;/math&gt; consecutive integers?<br /> <br /> &lt;math&gt;\textbf{(A)}\mbox{ }360\qquad\textbf{(B)}\mbox{ }388\qquad\textbf{(C)}\mbox{ }412\qquad\textbf{(D)}\mbox{ }416\qquad\textbf{(E)}\mbox{ }424&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 19|Solution<br /> ]]<br /> <br /> ==Problem 20==<br /> <br /> The least common multiple of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; is &lt;math&gt;12&lt;/math&gt;, and the least common multiple of &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. What is the least possible value of the least common multiple of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }20\qquad\textbf{(B) }30\qquad\textbf{(C) }60\qquad\textbf{(D) }120\qquad \textbf{(E) }180&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 20|Solution<br /> ]]<br /> <br /> ==Problem 21==<br /> <br /> A top hat contains 3 red chips and 2 green chips. Chips are drawn randomly, one at a time without replacement, until all 3 of the reds are drawn or until both green chips are drawn. What is the probability that the 3 reds are drawn?<br /> <br /> &lt;math&gt;\textbf{(A) }\dfrac{3}{10}\qquad\textbf{(B) }\dfrac{2}{5}\qquad\textbf{(C) }\dfrac{1}{2}\qquad\textbf{(D) }\dfrac{3}{5}\qquad \textbf{(E) }\dfrac{7}{10}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 21|Solution<br /> ]]<br /> <br /> ==Problem 22==<br /> Rectangle &lt;math&gt;DEFA&lt;/math&gt; below is a &lt;math&gt;3 \times 4&lt;/math&gt; rectangle with &lt;math&gt;DC=CB=BA=1&lt;/math&gt;. The area of the &quot;bat wings&quot; (shaded area) is<br /> &lt;asy&gt;<br /> draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0));<br /> draw((3,0)--(1,4)--(0,0));<br /> fill((0,0)--(1,4)--(1.5,3)--cycle, black);<br /> fill((3,0)--(2,4)--(1.5,3)--cycle, black);<br /> label(&quot;$A$&quot;,(3.05,4.2));<br /> label(&quot;$B$&quot;,(2,4.2));<br /> label(&quot;$C$&quot;,(1,4.2));<br /> label(&quot;$D$&quot;,(0,4.2));<br /> label(&quot;$E$&quot;, (0,-0.2));<br /> label(&quot;$F$&quot;, (3,-0.2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }2 \frac{1}{2}\qquad\textbf{(C) }3\qquad\textbf{(D) }3 \frac{1}{2}\qquad \textbf{(E) }4&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 22|Solution<br /> ]]<br /> <br /> ==Problem 23==<br /> <br /> Two congruent circles centered at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; each pass through the other circle's center. The line containing both &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; is extended to intersect the circles at points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt;. The circles intersect at two points, one of which is &lt;math&gt;E&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle CED&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 23|Solution<br /> ]]<br /> <br /> ==Problem 24==<br /> <br /> The digits &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; are each used once to write a five-digit number &lt;math&gt;PQRST&lt;/math&gt;. The three-digit number &lt;math&gt;PQR&lt;/math&gt; is divisible by &lt;math&gt;4&lt;/math&gt;, the three-digit number &lt;math&gt;QRS&lt;/math&gt; is divisible by &lt;math&gt;5&lt;/math&gt;, and the three-digit number &lt;math&gt;RST&lt;/math&gt; is divisible by &lt;math&gt;3&lt;/math&gt;. What is &lt;math&gt;P&lt;/math&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 /> [[2016 AMC 8 Problems/Problem 24|Solution<br /> ]]<br /> <br /> ==Problem 25== <br /> <br /> A semicircle is inscribed in an isosceles triangle with base &lt;math&gt;16&lt;/math&gt; and height &lt;math&gt;15&lt;/math&gt; so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?<br /> <br /> &lt;asy&gt;draw((0,0)--(8,15)--(16,0)--(0,0));<br /> draw(arc((8,0),7.0588,0,180));&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}&lt;/math&gt;<br /> <br /> [[2016 AMC 8 Problems/Problem 25|Solution<br /> ]]<br /> <br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_12&diff=149863 2016 AMC 8 Problems/Problem 12 2021-03-23T01:30:58Z <p>Notmyopic667: /* Problem */</p> <hr /> <div>==Problem 12==<br /> <br /> Jefferson Middle School has the same number of boys and girls. &lt;math&gt;\frac{3}{4}&lt;/math&gt; of the girls and &lt;math&gt;\frac{2}{3}&lt;/math&gt;<br /> of the boys went on a field trip. What fraction of the students on the field trip were girls?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{9}{17}\qquad\textbf{(C) }\frac{7}{13}\qquad\textbf{(D) }\frac{2}{3}\qquad \textbf{(E) }\frac{14}{15}&lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> <br /> Set the number of children to a number that is divisible by two, four, and three. In this question, the number of children in the school is not a specific number because there are no actual numbers in the question, only ratios.This way, we can calculate the answer without dealing with decimals.<br /> &lt;math&gt;120&lt;/math&gt; is a number that works. There will be &lt;math&gt;60&lt;/math&gt; girls and &lt;math&gt;60&lt;/math&gt; boys. So, there will be <br /> &lt;math&gt;60\cdot\frac{3}{4}&lt;/math&gt; = &lt;math&gt;45&lt;/math&gt; girls on the trip and &lt;math&gt;60\cdot\frac{2}{3}&lt;/math&gt; = &lt;math&gt;40&lt;/math&gt; boys on the trip. <br /> The total number of children on the trip is &lt;math&gt;85&lt;/math&gt;, so the fraction of girls on the trip is &lt;math&gt;\frac{45}{85}&lt;/math&gt; or &lt;math&gt;\boxed{\textbf{(B)} \frac{9}{17}}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> <br /> Let there be &lt;math&gt;b&lt;/math&gt; boys and &lt;math&gt;g&lt;/math&gt; girls in the school. We see &lt;math&gt;g=b&lt;/math&gt;, which means &lt;math&gt;\frac{3}{4}b+\frac{2}{3}b=\frac{17}{12}b&lt;/math&gt; kids went on the trip and &lt;math&gt;\frac{3}{4}b&lt;/math&gt; kids are girls. So, the answer is &lt;math&gt;\frac{\frac{3}{4}b}{\frac{17}{12}b}=\frac{9}{17}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(B)} \frac{9}{17}}&lt;/math&gt;<br /> {{AMC8 box|year=2016|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_3&diff=149307 2013 AMC 8 Problems/Problem 3 2021-03-12T23:41:12Z <p>Notmyopic667: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;math&gt;4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000&lt;/math&gt;<br /> <br /> ==Solution==<br /> We see that if we group two numbers at a time, we see that it would make 1 on each. There are 500 pairs so the answer inside the parentheses is 1 times 500 equals 500. Now we just multiply that by 4 to get 2000. So the answer is E.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_3&diff=149306 2013 AMC 8 Problems/Problem 3 2021-03-12T23:40:57Z <p>Notmyopic667: /* Solution */</p> <hr /> <div>==Problem==<br /> What is the value of &lt;math&gt;4 \cdot (-1+2-3+4-5+6-7+\cdots+1000)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -10 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 500 \qquad \textbf{(E)}\ 2000&lt;/math&gt;<br /> <br /> ==Solution==<br /> ==We see that if we group two numbers at a time, we see that it would make 1 on each. There are 500 pairs so the answer inside the parentheses is 1 times 500 equals 500. Now we just multiply that by 4 to get 2000. So the answer is E.==<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2013|num-b=2|num-a=4}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_22&diff=148644 2015 AMC 8 Problems/Problem 22 2021-03-06T02:17:41Z <p>Notmyopic667: /* Problem */</p> <hr /> <div>== Problem 22==<br /> <br /> On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?<br /> <br /> &lt;math&gt;\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080 &lt;/math&gt;<br /> <br /> ==Solution ==<br /> As we read through this text, we find that the given information means that the number of students in the group has &lt;math&gt;12&lt;/math&gt; factors, since each arrangement is a factor. The smallest integer with &lt;math&gt;12&lt;/math&gt; factors is &lt;math&gt;2^2\cdot3\cdot5=\boxed{\textbf{(C) }60}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/E2om3xIvtYM<br /> <br /> ~savannahsolver<br /> <br /> == Video Solution ==<br /> https://youtu.be/HISL2-N5NVg?t=5241<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC8 box|year=2015|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_22&diff=141359 2019 AMC 8 Problems/Problem 22 2021-01-02T23:25:54Z <p>Notmyopic667: /* Problem 22 */</p> <hr /> <div>==Problem 22==<br /> A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was &lt;math&gt;84\%&lt;/math&gt; of the original price, by what percent was the price increased and decreased&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Suppose the fraction of discount is &lt;math&gt;x&lt;/math&gt;. That means &lt;math&gt;(1-x)(1+x)=0.84&lt;/math&gt;; so &lt;math&gt;1-x^{2}=0.84&lt;/math&gt;, and &lt;math&gt;(x^{2})=0.16&lt;/math&gt;, obtaining &lt;math&gt;x=0.4&lt;/math&gt;. Therefore, the price was increased and decreased by &lt;math&gt;40&lt;/math&gt;%, or &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ==Solution 2 (Answer options)==<br /> We can try out every option and see which one works out. By this method, we get &lt;math&gt;\boxed{\textbf{(E)}\ 40}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Let x be the discount. We can also work in reverse such as (&lt;math&gt;84&lt;/math&gt;)&lt;math&gt;(\frac{100}{100-x})&lt;/math&gt;&lt;math&gt;(\frac{100}{100+x})&lt;/math&gt; = &lt;math&gt;100&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;8400&lt;/math&gt; = &lt;math&gt;(100+x)(100-x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; gives us &lt;math&gt;x = 40, -40&lt;/math&gt;. But &lt;math&gt;x&lt;/math&gt; has to be positive. Thus &lt;math&gt;x&lt;/math&gt; = &lt;math&gt;40&lt;/math&gt;.<br /> <br /> ==Solution 4 ~ using the answer choices==<br /> <br /> Let our original cost be &lt;math&gt;\$100.&lt;/math&gt; We are looking for a result of &lt;math&gt;\$ 84,&lt;/math&gt; then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try &lt;math&gt;\boxed{40\%}&lt;/math&gt;, and we have the answer; it worked.<br /> <br /> ==Video explaining solution== <br /> <br /> Associated video - https://www.youtube.com/watch?v=aJX27Cxvwlc<br /> <br /> https://youtu.be/gX_l0PGsQao<br /> <br /> https://www.youtube.com/watch?v=_TheVi-6LWE<br /> <br /> https://www.youtube.com/watch?v=RcBDdB35Whk&amp;list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&amp;index=4 ~ MathEx<br /> <br /> https://youtu.be/h2GlK7itc2g<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2019|num-b=21|num-a=23}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_13&diff=141342 2019 AMC 8 Problems/Problem 13 2021-01-02T20:50:50Z <p>Notmyopic667: /* Solution 2 */</p> <hr /> <div>==Problem 13==<br /> A ''palindrome'' is a number that has the same value when read from left to right or from right to left. (For example, 12321 is a palindrome.) Let &lt;math&gt;N&lt;/math&gt; be the least three-digit integer which is not a palindrome but which is the sum of three distinct two-digit palindromes. What is the sum of the digits of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Note that the only positive 2-digit palindromes are multiples of 11, namely &lt;math&gt;11, 22, \ldots, 99&lt;/math&gt;. Since &lt;math&gt;N&lt;/math&gt; is the sum of 2-digit palindromes, &lt;math&gt;N&lt;/math&gt; is necessarily a multiple of 11. The smallest 3-digit multiple of 11 which is not a palindrome is 110, so &lt;math&gt;N=110&lt;/math&gt; is a candidate solution. We must check that 110 can be written as the sum of three distinct 2-digit palindromes; this suffices as &lt;math&gt;110=77+22+11&lt;/math&gt;. Then &lt;math&gt;N = 110&lt;/math&gt;, and the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;1+1+0 = \boxed{\textbf{(A) }2}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=bOnNFeZs7S8<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|num-b=12|num-a=14}}<br /> <br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_1&diff=141332 2019 AMC 8 Problems/Problem 1 2021-01-02T20:04:34Z <p>Notmyopic667: /* See also */</p> <hr /> <div>== Problem 1 ==<br /> Ike and Mike go into a sandwich shop with a total of &lt;math&gt;\$30.00&lt;/math&gt; to spend. Sandwiches cost &lt;math&gt;\$4.50&lt;/math&gt; each and soft drinks cost &lt;math&gt;\$1.00&lt;/math&gt; each. Ike and Mike plan to buy as many sandwiches as they can,<br /> and use any remaining money to buy soft drinks. Counting both sandwiches and soft drinks, how<br /> many items will they buy?<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 /> == Solution 1 ==<br /> We know that the sandwiches cost &lt;math&gt;4.50&lt;/math&gt; dollars. Guessing will bring us to multiplying &lt;math&gt;4.50&lt;/math&gt; by 6, which gives us &lt;math&gt;27.00&lt;/math&gt;. Since they can spend &lt;math&gt;30.00&lt;/math&gt; they have &lt;math&gt;3&lt;/math&gt; dollars left. Since sodas cost &lt;math&gt;1.00&lt;/math&gt; dollar each, they can buy 3 sodas, which makes them spend &lt;math&gt;30.00&lt;/math&gt; Since they bought 6 sandwiches and 3 sodas, they bought a total of &lt;math&gt;9&lt;/math&gt; items. Therefore, the answer is &lt;math&gt;\boxed{D = 9 }&lt;/math&gt;<br /> <br /> - SBose<br /> <br /> == Solution 2 (Using Algebra) ==<br /> Let &lt;math&gt;s&lt;/math&gt; be the number of sandwiches and &lt;math&gt;d&lt;/math&gt; be the number of sodas. We have to satisfy the equation of<br /> &lt;cmath&gt;4.50s+d=30&lt;/cmath&gt;<br /> In the question, it states that Ike and Mike buys as many sandwiches as possible. <br /> So, we drop the number of sodas for a while.<br /> We have: <br /> &lt;cmath&gt;4.50s=30&lt;/cmath&gt;<br /> &lt;cmath&gt;s=\frac{30}{4.5}&lt;/cmath&gt;<br /> &lt;cmath&gt;s=6R30&lt;/cmath&gt;<br /> We don't want a remainder so the maximum number of sandwiches is &lt;math&gt;6&lt;/math&gt;.<br /> The total money spent is &lt;math&gt;6\cdot 4.50=27&lt;/math&gt;.<br /> The number of dollar left to spent on sodas is &lt;math&gt;30-27=3&lt;/math&gt; dollars.<br /> &lt;math&gt;3&lt;/math&gt; dollars can buy &lt;math&gt;3&lt;/math&gt; sodas leading us to a total of <br /> &lt;math&gt;6+3=9&lt;/math&gt; items. <br /> Hence, the answer is &lt;math&gt;\boxed{(D) = 9}&lt;/math&gt;<br /> <br /> - by interactivemath<br /> <br /> ==See also==<br /> {{AMC8 box|year=2019|before = First Problem|num-a=2}}<br /> <br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_16&diff=141270 2018 AMC 8 Problems/Problem 16 2021-01-01T19:56:40Z <p>Notmyopic667: /* Solution 1 */</p> <hr /> <div>==Problem 16==<br /> Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?<br /> <br /> &lt;math&gt;\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just &lt;math&gt;5&lt;/math&gt; factorial. Now we multiply this product by &lt;math&gt;2!&lt;/math&gt; because there are &lt;math&gt;2&lt;/math&gt; factorial ways to arrange the Arabic books within themselves, and &lt;math&gt;4!&lt;/math&gt; ways to arrange the Spanish books within themselves. Multiplying all these together, we have &lt;math&gt;2!\cdot 4!\cdot 5!=\boxed{\textbf{(C) }5760}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_16&diff=141269 2018 AMC 8 Problems/Problem 16 2021-01-01T19:56:22Z <p>Notmyopic667: /* Solution 1 */</p> <hr /> <div>==Problem 16==<br /> Professor Chang has nine different language books lined up on a bookshelf: two Arabic, three German, and four Spanish. How many ways are there to arrange the nine books on the shelf keeping the Arabic books together and keeping the Spanish books together?<br /> <br /> &lt;math&gt;\textbf{(A) }1440\qquad\textbf{(B) }2880\qquad\textbf{(C) }5760\qquad\textbf{(D) }182,440\qquad \textbf{(E) }362,880&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Since the Arabic books and Spanish books have to be kept together, we can treat them both as just one book. That means we're trying to find the number of ways you can arrange one Arabic book, one Spanish book, and three German books, which is just &lt;math&gt;5&lt;/math&gt; factorial. Now we multiply this product by &lt;math&gt;2!&lt;/math&gt; because there are &lt;math&gt;2&lt;/math&gt; factorial ways to arrange the Arabic books within themselves, and &lt;math&gt;4!&lt;/math&gt; ways to arrange the Spanish books within themselves. Multiplying all these together, we have &lt;math&gt;2!\cdot 4!\cdot 5!=\boxed{\textbf{(C) }5760}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=15|num-a=17}}<br /> <br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_22&diff=140958 2020 AMC 8 Problems/Problem 22 2020-12-29T23:28:30Z <p>Notmyopic667: /* Problem */</p> <hr /> <div>==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; size(300); defaultpen(linewidth(0.8)+fontsize(13)); real r = 0.05; draw((0.9,0)--(3.5,0),EndArrow(size=7)); filldraw((4,2.5)--(7,2.5)--(7,-2.5)--(4,-2.5)--cycle,gray(0.65)); fill(circle((5.5,1.25),0.8),white); fill(circle((5.5,1.25),0.5),gray(0.65)); fill((4.3,-r)--(6.7,-r)--(6.7,-1-r)--(4.3,-1-r)--cycle,white); fill((4.3,-1.25+r)--(6.7,-1.25+r)--(6.7,-2.25+r)--(4.3,-2.25+r)--cycle,white); 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)); fill((4.6,-1.5+r)--(6.4,-1.5+r)--(6.4,-2+r)--(4.6,-2+r)--cycle,gray(0.65)); label(&quot;$N$&quot;,(0.45,0)); draw((7.5,1.25)--(11.25,1.25),EndArrow(size=7)); draw((7.5,-1.25)--(11.25,-1.25),EndArrow(size=7)); label(&quot;if$N$is even&quot;,(9.25,1.25),N); label(&quot;if$N$is odd&quot;,(9.25,-1.25),N); label(&quot;$\frac N2$&quot;,(12,1.25)); label(&quot;$3N+1$&quot;,(12.6,-1.25)); &lt;/asy&gt;<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;&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;&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 /> &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 /> ==Solution 1==<br /> We start with the final output of &lt;math&gt;1&lt;/math&gt; and work backwards, taking care to consider all possible inputs that could have resulted in any particular output. This produces the following set of possibilities at each stage:<br /> &lt;cmath&gt;\{1\}\rightarrow\{2\}\rightarrow\{4\}\rightarrow\{1,8\}\rightarrow\{2,16\}\rightarrow\{4,5,32\}\rightarrow\{1,8,10,64\}&lt;/cmath&gt;<br /> where, for example, &lt;math&gt;2&lt;/math&gt; must come from &lt;math&gt;4&lt;/math&gt; (as there is no integer &lt;math&gt;n&lt;/math&gt; satisfying &lt;math&gt;3n+1=2&lt;/math&gt;), but &lt;math&gt;16&lt;/math&gt; could come from &lt;math&gt;32&lt;/math&gt; or &lt;math&gt;5&lt;/math&gt; (as &lt;math&gt;\frac{32}{2} = 3 \cdot 5 + 1 = 16&lt;/math&gt;, and &lt;math&gt;32&lt;/math&gt; is even while &lt;math&gt;5&lt;/math&gt; is odd). By construction, the last set in this sequence contains all the numbers which will lead to the number &lt;math&gt;1&lt;/math&gt; at the end of the &lt;math&gt;6&lt;/math&gt;-step process, and their sum is &lt;math&gt;1+8+10+64=\boxed{\textbf{(E) }83}&lt;/math&gt;.<br /> <br /> ==Solution 2 (variant of Solution 1)==<br /> As in Solution 1, we work backwards from &lt;math&gt;1&lt;/math&gt;, this time showing the possible cases in a tree diagram:<br /> [[File:Prob22-diagram.png|middle|center]]<br /> The possible numbers are those at the &quot;leaves&quot; of the tree (the ends of the various branches), which are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;. Thus the answer is &lt;math&gt;1+8+64+10=\boxed{\textbf{(E) }83}&lt;/math&gt;.<br /> <br /> ==Solution 3 (algebraic)==<br /> We begin by finding the inverse of the function that the machine uses. Call the input &lt;math&gt;I&lt;/math&gt; and the output &lt;math&gt;O&lt;/math&gt;. If &lt;math&gt;I&lt;/math&gt; is even, &lt;math&gt;O=\frac{I}{2}&lt;/math&gt;, and if &lt;math&gt;I&lt;/math&gt; is odd, &lt;math&gt;O=3I+1&lt;/math&gt;. We can therefore see that &lt;math&gt;I=2O&lt;/math&gt; when &lt;math&gt;I&lt;/math&gt; is even and &lt;math&gt;I=\frac{O-1}{3}&lt;/math&gt; when &lt;math&gt;I&lt;/math&gt; is odd. Therefore, starting with &lt;math&gt;1&lt;/math&gt;, if &lt;math&gt;I&lt;/math&gt; is even, &lt;math&gt;I=2&lt;/math&gt;, and if &lt;math&gt;I&lt;/math&gt; is odd, &lt;math&gt;I=0&lt;/math&gt;, but the latter is not valid since &lt;math&gt;0&lt;/math&gt; is not actually odd. This means that the 2nd-to-last number in the sequence has to be &lt;math&gt;2&lt;/math&gt;. Now, substituting &lt;math&gt;2&lt;/math&gt; into the inverse formulae, if &lt;math&gt;I&lt;/math&gt; is even, &lt;math&gt;I=4&lt;/math&gt; (which is indeed even), and if &lt;math&gt;I&lt;/math&gt; is odd, &lt;math&gt;I=\frac{1}{3}&lt;/math&gt;, which is not an integer. This means the 3rd-to-last number in the sequence has to be &lt;math&gt;4&lt;/math&gt;. Substituting in &lt;math&gt;4&lt;/math&gt;, if &lt;math&gt;I&lt;/math&gt; is even, &lt;math&gt;I=8&lt;/math&gt;, but if &lt;math&gt;I&lt;/math&gt; is odd, &lt;math&gt;I=1&lt;/math&gt;. Both of these are valid solutions, so the 4th-to-last number can be either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. If it is &lt;math&gt;1&lt;/math&gt;, then by the argument we have just made, the 5th-to-last number has to be &lt;math&gt;2&lt;/math&gt;, the 6th-to-last number has to be &lt;math&gt;4&lt;/math&gt;, and the 7th-to-last number, which is the first number, must be either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. In this way, we have ultimately found two solutions: &lt;math&gt;N=1&lt;/math&gt; and &lt;math&gt;N=8&lt;/math&gt;.<br /> <br /> On the other hand, if the 4th-to-last number is &lt;math&gt;8&lt;/math&gt;, substituting &lt;math&gt;8&lt;/math&gt; into the inverse formulae shows that the 5th-to-last number is either &lt;math&gt;16&lt;/math&gt; or &lt;math&gt;\frac{7}{3}&lt;/math&gt;, but the latter is not an integer. Substituting &lt;math&gt;16&lt;/math&gt; shows that if &lt;math&gt;I&lt;/math&gt; is even, &lt;math&gt;I=32&lt;/math&gt;, but if I is odd, &lt;math&gt;I=5&lt;/math&gt;, and both of these are valid. If the 6th-to-last number is &lt;math&gt;32&lt;/math&gt;, then the first number must be &lt;math&gt;64&lt;/math&gt;, since &lt;math&gt;\frac{31}{3}&lt;/math&gt; is not an integer; if the 6th-to-last number is &lt;math&gt;5,&lt;/math&gt; then the first number has to be &lt;math&gt;10&lt;/math&gt;, as &lt;math&gt;\frac{4}{3}&lt;/math&gt; is not an integer. This means that, in total, there are &lt;math&gt;4&lt;/math&gt; solutions for &lt;math&gt;N&lt;/math&gt;, specifically, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;10&lt;/math&gt;, and &lt;math&gt;64&lt;/math&gt;, which sum to &lt;math&gt;\boxed{\textbf{(E) }83}&lt;/math&gt;.<br /> <br /> ==Video Solutions==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/lhDFmiKNPBg<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=21|num-a=23}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_21&diff=140953 2020 AMC 8 Problems/Problem 21 2020-12-29T23:08:34Z <p>Notmyopic667: /* See also */</p> <hr /> <div>==Problem==<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;//diagram by SirCalcsALot<br /> size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i &lt; 8; ++i) { for (int j = 0; j &lt; 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i &lt; N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(&quot;$P$&quot;, (5.5, 0.5)); label(&quot;$Q$&quot;, (6.5, 7.5)); &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 /> ==Solution 1==<br /> Notice that, in order to step onto any particular white square, the marker must have come from one of the &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt; white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from &lt;math&gt;P&lt;/math&gt; to that square is the sum of the numbers of ways to move from &lt;math&gt;P&lt;/math&gt; to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from &lt;math&gt;P&lt;/math&gt; to that square, as already stated).<br /> &lt;asy&gt;<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 /> label(&quot;$1$&quot;, (5.5, .5));<br /> label(&quot;$1$&quot;, (4.5, 1.5));<br /> label(&quot;$1$&quot;, (6.5, 1.5));<br /> label(&quot;$1$&quot;, (3.5, 2.5));<br /> label(&quot;$1$&quot;, (7.5, 2.5));<br /> label(&quot;$2$&quot;, (5.5, 2.5));<br /> label(&quot;$1$&quot;, (2.5, 3.5));<br /> label(&quot;$3$&quot;, (6.5, 3.5));<br /> label(&quot;$3$&quot;, (4.5, 3.5));<br /> label(&quot;$4$&quot;, (3.5, 4.5));<br /> label(&quot;$3$&quot;, (7.5, 4.5));<br /> label(&quot;$6$&quot;, (5.5, 4.5));<br /> label(&quot;$10$&quot;, (4.5, 5.5));<br /> label(&quot;$9$&quot;, (6.5, 5.5));<br /> label(&quot;$19$&quot;, (5.5, 6.5));<br /> label(&quot;$9$&quot;, (7.5, 6.5));<br /> label(&quot;$28$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> The answer is therefore &lt;math&gt;\boxed{\textbf{(A) }28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Suppose we &quot;extend&quot; the chessboard infinitely with &lt;math&gt;2&lt;/math&gt; additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.<br /> <br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 10; ++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 /> draw((8,0) -- (8,8),red);<br /> label(&quot;$P$&quot;, (5.5,.5));<br /> label(&quot;$Q$&quot;, (6.5,7.5));<br /> label(&quot;$X$&quot;, (8.5,3.5));<br /> label(&quot;$Y$&quot;, (8.5,5.5));<br /> &lt;/asy&gt;<br /> The total number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt;, including invalid paths which cross over the red line, is then the number of paths which make &lt;math&gt;4&lt;/math&gt; steps up-and-right and &lt;math&gt;3&lt;/math&gt; steps up-and-left, which is &lt;math&gt;\binom{4+3}{3} = \binom{7}{3} = 35&lt;/math&gt;. We need to subtract the number of invalid paths, i.e. the number of paths that pass through &lt;math&gt;X&lt;/math&gt; or &lt;math&gt;Y&lt;/math&gt;. To get to &lt;math&gt;X&lt;/math&gt;, the marker has to make &lt;math&gt;3&lt;/math&gt; up-and-right steps, after which it can proceed to &lt;math&gt;Q&lt;/math&gt; with &lt;math&gt;3&lt;/math&gt; steps up-and-left and &lt;math&gt;1&lt;/math&gt; step up-and-right. Thus, the number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; that pass through &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;1 \cdot \binom{3+1}{3} = 4&lt;/math&gt;. Similarly, the number of paths that pass through &lt;math&gt;Y&lt;/math&gt; is &lt;math&gt;\binom{4+1}{1}\cdot 1 = 5&lt;/math&gt;. However, we have now double-counted the invalid paths which pass through both &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;; from the diagram, it is clear that there are only &lt;math&gt;2&lt;/math&gt; of these (as the marker can get from &lt;math&gt;X&lt;/math&gt; to &lt;math&gt;Y&lt;/math&gt; by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is &lt;math&gt;4+5-2=7&lt;/math&gt;, and the number of valid paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;35-7 = \boxed{\textbf{(A) }28}&lt;/math&gt;.<br /> <br /> ==Video Solutions==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/hGCxt8G9g-s<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=20|num-a=22}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_21&diff=140952 2020 AMC 8 Problems/Problem 21 2020-12-29T23:08:23Z <p>Notmyopic667: /* See also */</p> <hr /> <div>==Problem==<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;//diagram by SirCalcsALot<br /> size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i &lt; 8; ++i) { for (int j = 0; j &lt; 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i &lt; N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label(&quot;$P$&quot;, (5.5, 0.5)); label(&quot;$Q$&quot;, (6.5, 7.5)); &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 /> ==Solution 1==<br /> Notice that, in order to step onto any particular white square, the marker must have come from one of the &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;2&lt;/math&gt; white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from &lt;math&gt;P&lt;/math&gt; to that square is the sum of the numbers of ways to move from &lt;math&gt;P&lt;/math&gt; to each of the white squares immediately beneath it. To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from &lt;math&gt;P&lt;/math&gt; to that square, as already stated).<br /> &lt;asy&gt;<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 /> label(&quot;$1$&quot;, (5.5, .5));<br /> label(&quot;$1$&quot;, (4.5, 1.5));<br /> label(&quot;$1$&quot;, (6.5, 1.5));<br /> label(&quot;$1$&quot;, (3.5, 2.5));<br /> label(&quot;$1$&quot;, (7.5, 2.5));<br /> label(&quot;$2$&quot;, (5.5, 2.5));<br /> label(&quot;$1$&quot;, (2.5, 3.5));<br /> label(&quot;$3$&quot;, (6.5, 3.5));<br /> label(&quot;$3$&quot;, (4.5, 3.5));<br /> label(&quot;$4$&quot;, (3.5, 4.5));<br /> label(&quot;$3$&quot;, (7.5, 4.5));<br /> label(&quot;$6$&quot;, (5.5, 4.5));<br /> label(&quot;$10$&quot;, (4.5, 5.5));<br /> label(&quot;$9$&quot;, (6.5, 5.5));<br /> label(&quot;$19$&quot;, (5.5, 6.5));<br /> label(&quot;$9$&quot;, (7.5, 6.5));<br /> label(&quot;$28$&quot;, (6.5, 7.5));<br /> &lt;/asy&gt;<br /> The answer is therefore &lt;math&gt;\boxed{\textbf{(A) }28}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Suppose we &quot;extend&quot; the chessboard infinitely with &lt;math&gt;2&lt;/math&gt; additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.<br /> <br /> &lt;asy&gt;<br /> int N = 7;<br /> for (int i = 0; i &lt; 10; ++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 /> draw((8,0) -- (8,8),red);<br /> label(&quot;$P$&quot;, (5.5,.5));<br /> label(&quot;$Q$&quot;, (6.5,7.5));<br /> label(&quot;$X$&quot;, (8.5,3.5));<br /> label(&quot;$Y$&quot;, (8.5,5.5));<br /> &lt;/asy&gt;<br /> The total number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt;, including invalid paths which cross over the red line, is then the number of paths which make &lt;math&gt;4&lt;/math&gt; steps up-and-right and &lt;math&gt;3&lt;/math&gt; steps up-and-left, which is &lt;math&gt;\binom{4+3}{3} = \binom{7}{3} = 35&lt;/math&gt;. We need to subtract the number of invalid paths, i.e. the number of paths that pass through &lt;math&gt;X&lt;/math&gt; or &lt;math&gt;Y&lt;/math&gt;. To get to &lt;math&gt;X&lt;/math&gt;, the marker has to make &lt;math&gt;3&lt;/math&gt; up-and-right steps, after which it can proceed to &lt;math&gt;Q&lt;/math&gt; with &lt;math&gt;3&lt;/math&gt; steps up-and-left and &lt;math&gt;1&lt;/math&gt; step up-and-right. Thus, the number of paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; that pass through &lt;math&gt;X&lt;/math&gt; is &lt;math&gt;1 \cdot \binom{3+1}{3} = 4&lt;/math&gt;. Similarly, the number of paths that pass through &lt;math&gt;Y&lt;/math&gt; is &lt;math&gt;\binom{4+1}{1}\cdot 1 = 5&lt;/math&gt;. However, we have now double-counted the invalid paths which pass through both &lt;math&gt;X&lt;/math&gt; and &lt;math&gt;Y&lt;/math&gt;; from the diagram, it is clear that there are only &lt;math&gt;2&lt;/math&gt; of these (as the marker can get from &lt;math&gt;X&lt;/math&gt; to &lt;math&gt;Y&lt;/math&gt; by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is &lt;math&gt;4+5-2=7&lt;/math&gt;, and the number of valid paths from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;Q&lt;/math&gt; is &lt;math&gt;35-7 = \boxed{\textbf{(A) }28}&lt;/math&gt;.<br /> <br /> ==Video Solutions==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/hGCxt8G9g-s<br /> <br /> ==See also==</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_19&diff=140951 2020 AMC 8 Problems/Problem 19 2020-12-29T22:50:38Z <p>Notmyopic667: /* Problem */</p> <hr /> <div>==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 /> ==Solution 1==<br /> A number is divisible by &lt;math&gt;15&lt;/math&gt; precisely if it is divisible by &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;. The latter means the last digit must be either &lt;math&gt;5&lt;/math&gt; or &lt;math&gt;0&lt;/math&gt;, and the former means the sum of the digits must be divisible by &lt;math&gt;3&lt;/math&gt;. If the last digit is &lt;math&gt;0&lt;/math&gt;, the first digit would be &lt;math&gt;0&lt;/math&gt; (because the digits alternate), which is not possible. Hence the last digit must be &lt;math&gt;5&lt;/math&gt;, and the number is of the form &lt;math&gt;5\square 5\square 5&lt;/math&gt;. If the unknown digit is &lt;math&gt;x&lt;/math&gt;, we deduce &lt;math&gt;5+x+5+x+5 \equiv 0 \pmod{3} \Rightarrow 2x \equiv 0 \pmod{3}&lt;/math&gt;. We know &lt;math&gt;2^{-1}&lt;/math&gt; exists modulo &lt;math&gt;3&lt;/math&gt; because 2 is relatively prime to 3, so we conclude that &lt;math&gt;x&lt;/math&gt; (i.e. the second and fourth digit of the number) must be a multiple of &lt;math&gt;3&lt;/math&gt;. It can be &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, or &lt;math&gt;9&lt;/math&gt;, so there are &lt;math&gt;\boxed{\textbf{(B) }4}&lt;/math&gt; options: &lt;math&gt;50505&lt;/math&gt;, &lt;math&gt;53535&lt;/math&gt;, &lt;math&gt;56565&lt;/math&gt;, and &lt;math&gt;59595&lt;/math&gt;.<br /> <br /> ==Solution 2 (variant of Solution 1)==<br /> As in Solution 1, we find that such numbers must start with &lt;math&gt;5&lt;/math&gt; and alternate with &lt;math&gt;5&lt;/math&gt; (i.e. must be of the form &lt;math&gt;5\square 5\square 5&lt;/math&gt;), where the two digits between the &lt;math&gt;5&lt;/math&gt;s need to be the same. Call that digit &lt;math&gt;x&lt;/math&gt;. For the number to be divisible by &lt;math&gt;3&lt;/math&gt;, the sum of the digits must be divisible by &lt;math&gt;3&lt;/math&gt;; since the sum of the three &lt;math&gt;5&lt;/math&gt;s is &lt;math&gt;15&lt;/math&gt;, which is already a multiple of &lt;math&gt;3&lt;/math&gt;, it must also be the case that &lt;math&gt;x+x=2x&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Thus, the problem reduces to finding the number of digits from &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;9&lt;/math&gt; for which &lt;math&gt;2x&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. This leads to &lt;math&gt;x=0&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, or &lt;math&gt;9&lt;/math&gt;, so there are &lt;math&gt;\boxed{\textbf{(B) }4}&lt;/math&gt; possible numbers (namely &lt;math&gt;50505&lt;/math&gt;, &lt;math&gt;53535&lt;/math&gt;, &lt;math&gt;56565&lt;/math&gt;, and &lt;math&gt;59595&lt;/math&gt;).<br /> <br /> ==Video Solution==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/VnOecUiP-SA<br /> <br /> ==See also== <br /> {{AMC8 box|year=2020|num-b=18|num-a=20}}<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_18&diff=140937 2020 AMC 8 Problems/Problem 18 2020-12-29T19:27:37Z <p>Notmyopic667: /* Problem */</p> <hr /> <div>==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 /> draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); <br /> &lt;/asy&gt;<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 /> ==Solution 1==<br /> &lt;asy&gt; draw(arc((0,0),17,180,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*S); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*S); dot(&quot;$E$&quot;,(17,0), 1.25*S); dot(&quot;$F$&quot;,(-17,0), 1.25*S); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); dot(&quot;$O$&quot;, (0,0), 1.25*S); draw((0,0)--(-8,15));&lt;/asy&gt;<br /> <br /> Let &lt;math&gt;O&lt;/math&gt; be the center of the semicircle. The diameter of the semicircle is &lt;math&gt;9+16+9=34&lt;/math&gt;, so &lt;math&gt;OC = 17&lt;/math&gt;. By symmetry, &lt;math&gt;O&lt;/math&gt; is in fact the midpoint of &lt;math&gt;DA&lt;/math&gt;, so &lt;math&gt;OD=OA=\frac{16}{2}= 8&lt;/math&gt;. By the Pythagorean theorem in right-angled triangle &lt;math&gt;ODC&lt;/math&gt; (or &lt;math&gt;OBA&lt;/math&gt;), we have that &lt;math&gt;CD&lt;/math&gt; (or &lt;math&gt;AB&lt;/math&gt;) is &lt;math&gt;\sqrt{17^2-8^2}=15&lt;/math&gt;. Accordingly, the area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;16\cdot 15=\boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> ==Solution 2 (coordinate geometry)==<br /> Let the midpoint of segment &lt;math&gt;FE&lt;/math&gt; be the origin. Evidently, point &lt;math&gt;D=(-8,0)&lt;/math&gt; and &lt;math&gt;A=(8,0)&lt;/math&gt;. Since points &lt;math&gt;C&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; share &lt;math&gt;x&lt;/math&gt;-coordinates with &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt; respectively, it suffices to find the &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;B&lt;/math&gt; (which will be the height of the rectangle) and multiply this by &lt;math&gt;DA&lt;/math&gt; (which we know is &lt;math&gt;16&lt;/math&gt;). The radius of the semicircle is &lt;math&gt;\frac{9+16+9}{2} = 17&lt;/math&gt;, so the whole circle has equation &lt;math&gt;x^2+y^2=289&lt;/math&gt;; as already stated, &lt;math&gt;B&lt;/math&gt; has the same &lt;math&gt;x&lt;/math&gt;-coordinate as &lt;math&gt;A&lt;/math&gt;, i.e. &lt;math&gt;8&lt;/math&gt;, so substituting this into the equation shows that &lt;math&gt;y=\pm15&lt;/math&gt;. Since &lt;math&gt;y&gt;0&lt;/math&gt; at &lt;math&gt;B&lt;/math&gt;, the y-coordinate of &lt;math&gt;B&lt;/math&gt; is &lt;math&gt;15&lt;/math&gt;. Therefore, the answer is &lt;math&gt;16\cdot 15 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> <br /> (Note that the synthetic solution (Solution 1) is definitely faster and more elegant. However, this is the solution that you should use if you can't see any other easier strategy.)<br /> <br /> ==Solution 3==<br /> We can use a result from the Art of Problem Solving &lt;i&gt;Introduction to Algebra&lt;/i&gt; book Sidenote: for a semicircle with diameter &lt;math&gt;(1+n)&lt;/math&gt;, such that the &lt;math&gt;1&lt;/math&gt; part is on one side and the &lt;math&gt;n&lt;/math&gt; part is on the other side, the height from the end of the &lt;math&gt;1&lt;/math&gt; side (or the start of the &lt;math&gt;n&lt;/math&gt; side) is &lt;math&gt;\sqrt{n}&lt;/math&gt;. To use this formula, we scale the figure down by &lt;math&gt;9&lt;/math&gt;; this will give the height a length of &lt;math&gt;\sqrt{\frac{16+9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}&lt;/math&gt;. Now, scaling back up by &lt;math&gt;9&lt;/math&gt;, the height &lt;math&gt;DC&lt;/math&gt; is &lt;math&gt;9 \cdot \frac{5}{3} = 15&lt;/math&gt;. The answer is then &lt;math&gt;15 \cdot 16 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> -[[User:Sweetmango77|SweetMango77]]<br /> <br /> ==Solution 4 (Power Of A Point)==<br /> Draw the other half of the circle as follows:<br /> &lt;asy&gt; <br /> draw(arc((0,0),17,360,0)); draw((-17,0)--(17,0)); fill((-8,0)--(-8,15)--(8,15)--(8,0)--cycle, 1.5*grey); draw((-8,0)--(-8,15)--(8,15)--(8,0)--cycle); dot(&quot;$A$&quot;,(8,0), 1.25*SE); dot(&quot;$B$&quot;,(8,15), 1.25*N); dot(&quot;$C$&quot;,(-8,15), 1.25*N); dot(&quot;$D$&quot;,(-8,0), 1.25*SW); dot(&quot;$E$&quot;,(17,0), 1.25*E); dot(&quot;$F$&quot;,(-17,0), 1.25*W); label(&quot;$16$&quot;,(0,0),N); label(&quot;$9$&quot;,(12.5,0),N); label(&quot;$9$&quot;,(-12.5,0),N); draw((-8,-15)--(-8,0)--(8,0)--(8,-15)--cycle); dot(&quot;$B'$&quot;,(8,-15), 1.25*S); dot(&quot;$C'\$&quot;,(-8,-15), 1.25*S);<br /> &lt;/asy&gt;<br /> By Power of a Point, &lt;math&gt;FD\cdot DE = CD\cdot C'D&lt;/math&gt;. By symmetry, &lt;math&gt;CD = C'D&lt;/math&gt;. We see that &lt;math&gt;FD = 9&lt;/math&gt; and &lt;math&gt;DE = 25&lt;/math&gt;. Substituting in these values, &lt;math&gt;9\cdot 25 = CD^2&lt;/math&gt;, giving &lt;math&gt;CD^2 = 225&lt;/math&gt; and &lt;math&gt;CD = 15&lt;/math&gt;.<br /> The area of the rectangle is therefore &lt;math&gt;15\cdot 16 = \boxed{\textbf{(A) }240}&lt;/math&gt;.<br /> ==Video Solution==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/VnOecUiP-SA<br /> <br /> {{AMC8 box|year=2020|num-b=17|num-a=19}}<br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_17&diff=140936 2020 AMC 8 Problems/Problem 17 2020-12-29T19:20:57Z <p>Notmyopic667: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> How many positive integer 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 /> ==Solution 1==<br /> Since &lt;math&gt;2020 = 2^2 \cdot 5 \cdot 101&lt;/math&gt;, we can simply list its factors: &lt;cmath&gt;1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.&lt;/cmath&gt; There are &lt;math&gt;12&lt;/math&gt; of these; only &lt;math&gt;1, 2, 4, 5, 101&lt;/math&gt; (i.e. &lt;math&gt;5&lt;/math&gt; of them) don't have &lt;math&gt;3&lt;/math&gt; factors, so the remaining &lt;math&gt;12-5 = \boxed{\textbf{(B) }7}&lt;/math&gt; factors have more than &lt;math&gt;3&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> As in Solution 1, we prime factorize &lt;math&gt;2020&lt;/math&gt; as &lt;math&gt;2^2\cdot 5\cdot 101&lt;/math&gt;, and we recall the standard formula that the number of positive factors of an integer is found by adding &lt;math&gt;1&lt;/math&gt; to each exponent in its prime factorization, and then multiplying these. Thus &lt;math&gt;2020&lt;/math&gt; has &lt;math&gt;(2+1)(1+1)(1+1) = 12&lt;/math&gt; factors. The only number which has one factor is &lt;math&gt;1&lt;/math&gt;. For a number to have exactly two factors, it must be prime, and the only prime factors of &lt;math&gt;2020&lt;/math&gt; are &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;101&lt;/math&gt;. For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Thus, there are &lt;math&gt;5&lt;/math&gt; factors of &lt;math&gt;2020&lt;/math&gt; which themselves have &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, or &lt;math&gt;3&lt;/math&gt; factors (namely &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;101&lt;/math&gt;), so the number of factors of &lt;math&gt;2020&lt;/math&gt; that have more than &lt;math&gt;3&lt;/math&gt; factors is &lt;math&gt;12-5=\boxed{\textbf{(B) }7}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/VnOecUiP-SA<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_17&diff=140934 2020 AMC 8 Problems/Problem 17 2020-12-29T19:19:34Z <p>Notmyopic667: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> How many positive integer 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 /> ==Solution 1==<br /> Since &lt;math&gt;2020 = 2^2 \cdot 5 \cdot 101&lt;/math&gt;, we can simply list its factors: &lt;cmath&gt;1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.&lt;/cmath&gt; There are &lt;math&gt;12&lt;/math&gt; of these; only &lt;math&gt;1, 2, 4, 5, 101&lt;/math&gt; (i.e. &lt;math&gt;5&lt;/math&gt; of them) have at least &lt;math&gt;3&lt;/math&gt; factors, so the remaining &lt;math&gt;12-5 = \boxed{\textbf{(B) }7}&lt;/math&gt; factors have more than &lt;math&gt;3&lt;/math&gt; factors.<br /> <br /> ==Solution 2==<br /> As in Solution 1, we prime factorize &lt;math&gt;2020&lt;/math&gt; as &lt;math&gt;2^2\cdot 5\cdot 101&lt;/math&gt;, and we recall the standard formula that the number of positive factors of an integer is found by adding &lt;math&gt;1&lt;/math&gt; to each exponent in its prime factorization, and then multiplying these. Thus &lt;math&gt;2020&lt;/math&gt; has &lt;math&gt;(2+1)(1+1)(1+1) = 12&lt;/math&gt; factors. The only number which has one factor is &lt;math&gt;1&lt;/math&gt;. For a number to have exactly two factors, it must be prime, and the only prime factors of &lt;math&gt;2020&lt;/math&gt; are &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;101&lt;/math&gt;. For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of &lt;math&gt;2020&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Thus, there are &lt;math&gt;5&lt;/math&gt; factors of &lt;math&gt;2020&lt;/math&gt; which themselves have &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, or &lt;math&gt;3&lt;/math&gt; factors (namely &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, and &lt;math&gt;101&lt;/math&gt;), so the number of factors of &lt;math&gt;2020&lt;/math&gt; that have more than &lt;math&gt;3&lt;/math&gt; factors is &lt;math&gt;12-5=\boxed{\textbf{(B) }7}&lt;/math&gt;.<br /> <br /> ==Video Solution==<br /> https://youtu.be/SPNobOd4t1c (Includes all the problems and has a free class update)<br /> <br /> <br /> <br /> https://youtu.be/VnOecUiP-SA<br /> <br /> ==See also==<br /> {{AMC8 box|year=2020|num-b=16|num-a=18}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Notmyopic667 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_23&diff=140721 2018 AMC 8 Problems/Problem 23 2020-12-27T04:39:00Z <p>Notmyopic667: /* Problem */</p> <hr /> <div>==Problem 23==<br /> From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon?<br /> <br /> &lt;asy&gt;<br /> size(3cm);<br /> pair A[];<br /> for (int i=0; i&lt;9; ++i) {<br /> A[i] = rotate(22.5+45*i)*(1,0);<br /> }<br /> filldraw(A--A--A--A--A--A--A--A--cycle,gray,black);<br /> for (int i=0; i&lt;8; ++i) { dot(A[i]); }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A) } \frac{2}{7} \qquad \textbf{(B) } \frac{5}{42} \qquad \textbf{(C) } \frac{11}{14} \qquad \textbf{(D) } \frac{5}{7} \qquad \textbf{(E) } \frac{6}{7}&lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> We will use constructive counting to solve this. There are &lt;math&gt;2&lt;/math&gt; cases: Either all &lt;math&gt;3&lt;/math&gt; points are adjacent, or exactly &lt;math&gt;2&lt;/math&gt; points are adjacent.<br /> <br /> If all &lt;math&gt;3&lt;/math&gt; points are adjacent, then we have &lt;math&gt;8&lt;/math&gt; choices. If we have exactly &lt;math&gt;2&lt;/math&gt; adjacent points, then we will have &lt;math&gt;8&lt;/math&gt; places to put the adjacent points and also &lt;math&gt;4&lt;/math&gt; places to put the remaining point, so we have &lt;math&gt;8\cdot4&lt;/math&gt; choices. The total amount of choices is &lt;math&gt;{8 \choose 3} = 8\cdot7&lt;/math&gt;.<br /> Thus our answer is &lt;math&gt;\frac{8+8\cdot4}{8\cdot7}= \frac{1+4}{7}=\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ===Solution 2 ===<br /> We can decide &lt;math&gt;2&lt;/math&gt; adjacent points with &lt;math&gt;8&lt;/math&gt; choices. The remaining point will have &lt;math&gt;6&lt;/math&gt; choices. However, we have counted the case with &lt;math&gt;3&lt;/math&gt; adjacent points twice, so we need to subtract this case once. The case with the &lt;math&gt;3&lt;/math&gt; adjacent points has &lt;math&gt;8&lt;/math&gt; arrangements, so our answer is &lt;math&gt;\frac{8\cdot6-8}{{8 \choose 3 }}&lt;/math&gt;&lt;math&gt;=\frac{8\cdot6-8}{8 \cdot 7 \cdot 6 \div 6}\Longrightarrow\boxed{\textbf{(D) } \frac 57}&lt;/math&gt;<br /> <br /> ===Solution 3 (Stars and Bars)===<br /> Let &lt;math&gt;1&lt;/math&gt; point of the triangle be fixed at the top. Then, there are &lt;math&gt;{7 \choose 2} = 21&lt;/math&gt; ways to chose the other 2 points. There must be &lt;math&gt;3&lt;/math&gt; spaces in the points and &lt;math&gt;3&lt;/math&gt; points themselves. This leaves 2 extra points to be placed anywhere. By stars and bars, there are 3 triangle points (n) and &lt;math&gt;2&lt;/math&gt; extra points (k-1) distributed so by the stars and bars formula, &lt;math&gt;{n+k-1 \choose k-1}&lt;/math&gt;, there are &lt;math&gt;{4 \choose 2} = 6&lt;/math&gt; ways to arrange the bars and stars. Thus, the probability is &lt;math&gt;\frac{(21 - 6)}{21} = \boxed{\frac{5}{7}}&lt;/math&gt;.<br /> The stars and bars formula might be inaccurate<br /> <br /> ==Video Solution==<br /> https://www.youtube.com/watch?v=VNflxl7VpL0 - Happytwin<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2018|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Notmyopic667