https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Srim1027&feedformat=atom AoPS Wiki - User contributions [en] 2022-07-05T08:50:09Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_7&diff=157485 1986 AIME Problems/Problem 7 2021-07-07T14:17:31Z <p>Srim1027: /* Solutions */</p> <hr /> <div>== Problem ==<br /> The increasing [[sequence]] &lt;math&gt;1,3,4,9,10,12,13\cdots&lt;/math&gt; consists of all those positive [[integer]]s which are [[exponent|powers]] of 3 or sums of distinct powers of 3. Find the &lt;math&gt;100^{\mbox{th}}&lt;/math&gt; term of this sequence.<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> <br /> Rewrite all of the terms in base 3. Since the numbers are sums of ''distinct'' powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. &lt;math&gt;100&lt;/math&gt; is equal to &lt;math&gt;64 + 32 + 4&lt;/math&gt;, so in binary form we get &lt;math&gt;1100100&lt;/math&gt;. However, we must change it back to base 10 for the answer, which is &lt;math&gt;3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that the first term of the sequence is &lt;math&gt;1&lt;/math&gt;, the second is &lt;math&gt;3&lt;/math&gt;, the fourth is &lt;math&gt;9&lt;/math&gt;, and so on. Thus the &lt;math&gt;64th&lt;/math&gt; term of the sequence is &lt;math&gt;729&lt;/math&gt;. Now out of &lt;math&gt;64&lt;/math&gt; terms which are of the form &lt;math&gt;729&lt;/math&gt; + &lt;math&gt;'''S'''&lt;/math&gt;, &lt;math&gt;32&lt;/math&gt; of them include &lt;math&gt;243&lt;/math&gt; and &lt;math&gt;32&lt;/math&gt; do not. The smallest term that includes &lt;math&gt;243&lt;/math&gt;, i.e. &lt;math&gt;972&lt;/math&gt;, is greater than the largest term which does not, or &lt;math&gt;854&lt;/math&gt;. So the &lt;math&gt;96&lt;/math&gt;th term will be &lt;math&gt;972&lt;/math&gt;, then &lt;math&gt;973&lt;/math&gt;, then &lt;math&gt;975&lt;/math&gt;, then &lt;math&gt;976&lt;/math&gt;, and finally &lt;math&gt;\boxed{981}&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> After the &lt;math&gt;n&lt;/math&gt;th power of 3 in the sequence, the number of terms after that power but before the &lt;math&gt;(n+1)&lt;/math&gt;th power of 3 is equal to the number of terms before the &lt;math&gt;n&lt;/math&gt;th power, because those terms after the &lt;math&gt;n&lt;/math&gt;th power are just the &lt;math&gt;n&lt;/math&gt;th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of &lt;math&gt;3&lt;/math&gt; and the terms that come after them, we see that the &lt;math&gt;100&lt;/math&gt;th term is after &lt;math&gt;729&lt;/math&gt;, which is the &lt;math&gt;64&lt;/math&gt;th term. Also, note that the &lt;math&gt;k&lt;/math&gt;th term after the &lt;math&gt;n&lt;/math&gt;th power of 3 is equal to the power plus the &lt;math&gt;k&lt;/math&gt;th term in the entire sequence. Thus, the &lt;math&gt;100&lt;/math&gt;th term is &lt;math&gt;729&lt;/math&gt; plus the &lt;math&gt;36&lt;/math&gt;th term. Using the same logic, the &lt;math&gt;36&lt;/math&gt;th term is &lt;math&gt;243&lt;/math&gt; plus the &lt;math&gt;4&lt;/math&gt;th term, &lt;math&gt;9&lt;/math&gt;. We now have &lt;math&gt;729+243+9=\boxed{981}&lt;/math&gt;<br /> <br /> === Solution 4 ===<br /> Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the &lt;math&gt;2^{nth}&lt;/math&gt; term is equal to &lt;math&gt;3^n&lt;/math&gt;. From here, we can ballpark the range of the 100th term. The 64th term is &lt;math&gt;3^6&lt;/math&gt; = &lt;math&gt;729&lt;/math&gt; and the 128th term is &lt;math&gt;3^7&lt;/math&gt; = &lt;math&gt;2187&lt;/math&gt;. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (&lt;math&gt;2^n&lt;/math&gt;+&lt;math&gt;2^{n+1}&lt;/math&gt;)/2 term is equal to &lt;math&gt;3^n&lt;/math&gt; + &lt;math&gt;3^{n-1}&lt;/math&gt;. From here, we know that the 96th term is &lt;math&gt;3^6&lt;/math&gt; + &lt;math&gt;3^5&lt;/math&gt; = &lt;math&gt;972&lt;/math&gt;. From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is &lt;math&gt;972 + 1 = 973&lt;/math&gt;, the 98th term is &lt;math&gt;972 + 3 = 975&lt;/math&gt;, the 99th term is &lt;math&gt;972 + 3 + 1 = 976&lt;/math&gt;, and finally the 100th term is &lt;math&gt;972 + 9 = \boxed{981}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1986|num-b=6|num-a=8}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Srim1027 https://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_49&diff=132829 1950 AHSME Problems/Problem 49 2020-08-31T14:10:48Z <p>Srim1027: /* Problem */</p> <hr /> <div>==Problem==<br /> <br /> A triangle has a fixed base &lt;math&gt;AB&lt;/math&gt; that is &lt;math&gt;2&lt;/math&gt; inches long. The median from &lt;math&gt;A&lt;/math&gt; to side &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt; 1\frac{1}{2}&lt;/math&gt; inches long and can have any position emanating from &lt;math&gt;A&lt;/math&gt;. The locus of the vertex &lt;math&gt;C&lt;/math&gt; of the triangle is:<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\<br /> \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches} \qquad\\<br /> \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches} \qquad\\<br /> \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along } BA \qquad\\<br /> \textbf{(E)}\ \text{An ellipse with }A\text{ as focus}&lt;/math&gt;<br /> <br /> ==Solution==<br /> The locus of the median's endpoint on &lt;math&gt;BC&lt;/math&gt; is the circle about &lt;math&gt;A&lt;/math&gt; and of radius &lt;math&gt;1\frac{1}{2}&lt;/math&gt; inches. The locus of the vertex &lt;math&gt;C&lt;/math&gt; is then the circle twice as big and twice as far from &lt;math&gt;B&lt;/math&gt;, i.e. of radius &lt;math&gt;3&lt;/math&gt; inches and with center &lt;math&gt;4&lt;/math&gt; inches from &lt;math&gt;B&lt;/math&gt; along &lt;math&gt;BA&lt;/math&gt; which means that our answer is: &lt;math&gt;\textbf{(D)}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;A(a,y_A)&lt;/math&gt;, &lt;math&gt;B(b,y_B)&lt;/math&gt; and &lt;math&gt;C(x,y)&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;D\left(\frac{x+b}{2},\frac{y+y_B}{2}\right)&lt;/math&gt; is a midpoint of &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> Thus, the equation of needed locus is<br /> &lt;cmath&gt;\left(\frac{x+b}{2}-a\right)^2+\left(\frac{y+y_B}{2}-y_A\right)^2=\left(\frac{3}{2}\right)^2,&lt;/cmath&gt;<br /> which is equation of the circle:<br /> &lt;cmath&gt;(x-(2a-b))^2+(y-(2y_A-y_B))^2=3^2.&lt;/cmath&gt;<br /> <br /> Thus, D) is valid because &lt;cmath&gt;\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AHSME 50p box|year=1950|num-b=48|num-a=50}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> {{MAA Notice}}</div> Srim1027 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_17&diff=120130 2020 AMC 10B Problems/Problem 17 2020-03-26T15:30:09Z <p>Srim1027: /* Solution */</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #17]] and [[2020 AMC 12B Problems|2020 AMC 12B #15]]}}<br /> <br /> ==Problem==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let us use casework on the number of diagonals.<br /> <br /> Case 1: &lt;math&gt;0&lt;/math&gt; diagonals<br /> There are &lt;math&gt;2&lt;/math&gt; ways: either &lt;math&gt;1&lt;/math&gt; pairs with &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt; pairs with &lt;math&gt;4&lt;/math&gt;, and so on or &lt;math&gt;10&lt;/math&gt; pairs with &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; pairs with &lt;math&gt;3&lt;/math&gt;, etc.<br /> <br /> Case 2: &lt;math&gt;1&lt;/math&gt; diagonal<br /> There are &lt;math&gt;5&lt;/math&gt; possible diagonals to draw (everyone else pairs with the person next to them.<br /> <br /> Note that there cannot be 2 diagonals.<br /> <br /> Case 3: &lt;math&gt;3&lt;/math&gt; diagonals<br /> There are &lt;math&gt;5&lt;/math&gt; possible diagonals to draw. <br /> <br /> Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.<br /> <br /> Case 4: &lt;math&gt;5&lt;/math&gt; diagonals<br /> There is &lt;math&gt;1&lt;/math&gt; way to do this.<br /> <br /> Thus, in total there are &lt;math&gt;2+5+5+1=\boxed{13}&lt;/math&gt; possible ways.<br /> <br /> ==Video Solution==<br /> https://youtu.be/3BvJeZU3T-M (for AMC 10)<br /> https://youtu.be/0xgTR3UEqbQ (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2020|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Srim1027 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_17&diff=120129 2020 AMC 10B Problems/Problem 17 2020-03-26T15:29:50Z <p>Srim1027: /* Solution */</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #17]] and [[2020 AMC 12B Problems|2020 AMC 12B #15]]}}<br /> <br /> ==Problem==<br /> <br /> There are &lt;math&gt;10&lt;/math&gt; people standing equally spaced around a circle. Each person knows exactly &lt;math&gt;3&lt;/math&gt; of the other &lt;math&gt;9&lt;/math&gt; people: the &lt;math&gt;2&lt;/math&gt; people standing next to her or him, as well as the person directly across the circle. How many ways are there for the &lt;math&gt;10&lt;/math&gt; people to split up into &lt;math&gt;5&lt;/math&gt; pairs so that the members of each pair know each other?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> ==Solution==<br /> Let us use casework on the number of diagonals.<br /> <br /> Case 1: &lt;math&gt;0&lt;/math&gt; diagonals<br /> There are &lt;math&gt;2&lt;/math&gt; ways: either &lt;math&gt;1&lt;/math&gt; pairs with &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt; pairs with &lt;math&gt;4&lt;/math&gt;, and so on or &lt;math&gt;10&lt;/math&gt; pairs with &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt; pairs with &lt;math&gt;3&lt;/math&gt;, etc.<br /> <br /> Case 2: &lt;math&gt;1&lt;/math&gt; diagonal<br /> There are &lt;math&gt;5&lt;/math&gt; possible diagonals to draw (everyone else pairs with the person next to them.<br /> <br /> Note that there cannot be 2 diagonals.<br /> <br /> Case 3: &lt;math&gt;3&lt;/math&gt; diagonals<br /> There are &lt;math&gt;5% possible diagonals to draw. <br /> <br /> Note that there cannot be a case with 4 diagonals because then there would have to be 5 diagonals for the two remaining people, thus a contradiction.<br /> <br /> Case 4: &lt;/math&gt;5&lt;math&gt; diagonals<br /> There is &lt;/math&gt;1&lt;math&gt; way to do this.<br /> <br /> Thus, in total there are &lt;/math&gt;2+5+5+1=\boxed{13}$possible ways.<br /> <br /> ==Video Solution==<br /> https://youtu.be/3BvJeZU3T-M (for AMC 10)<br /> https://youtu.be/0xgTR3UEqbQ (for AMC 12)<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2020|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}</div> Srim1027 https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems&diff=98343 2003 AMC 8 Problems 2018-10-27T20:23:34Z <p>Srim1027: /* Problem 23 */</p> <hr /> <div>==Problem 1==<br /> Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum? <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Which of the following numbers has the smallest prime factor?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 55 \qquad\mathrm{(B)}\ 57 \qquad\mathrm{(C)}\ 58 \qquad\mathrm{(D)}\ 59 \qquad\mathrm{(E)}\ 61&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A burger at Ricky C's weighs 120 grams, of which 30 grams are filler. What percent of the burger is not filler?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 60\% \qquad\mathrm{(B)}\ 65\% \qquad\mathrm{(C)}\ 70\% \qquad\mathrm{(D)}\ 75\% \qquad\mathrm{(E)}\ 90\%&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted 7 children and 19 wheels. How many tricycles were there?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> If 20% of a number is 12, what is 30% of the same number?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 15\qquad\mathrm{(B)}\ 18 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 24 \qquad\mathrm{(E)}\ 30&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Given the areas of the three squares in the figure, what is the area of the interior triangle?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(-5,12)--(7,17)--(12,5)--(17,5)--(17,0)--(12,0)--(12,-12)--(0,-12)--(0,0)--(12,5)--(12,0)--cycle,linewidth(1));<br /> label(&quot;$25$&quot;,(14.5,1),N);<br /> label(&quot;$144$&quot;,(6,-7.5),N);<br /> label(&quot;$169$&quot;,(3.5,7),N);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 13 \qquad\mathrm{(B)}\ 30 \qquad\mathrm{(C)}\ 60 \qquad\mathrm{(D)}\ 300 \qquad\mathrm{(E)}\ 1800&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Blake and Jenny each took four 100-point tests. Blake averaged 78 on the four tests. Jenny scored 10 points higher than Blake on the first test, 10 points lower than him on the second test, and 20 points higher on both the third and fourth tests. What is the difference between Jenny's average and Blake's average on these four tests?<br /> <br /> &lt;math&gt; \mathrm{(A)}\ 10 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 25 \qquad\mathrm{(E)}\ 40 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 7|Solution]]<br /> <br /> ==Bake Sale==<br /> Problems 8, 9, and 10 use the data found in the accompanying paragraph and figures.<br /> <br /> Four friends, Art, Roger, Paul and Trisha, bake cookies, and all cookies have the same thickness. The shapes of the cookies differ, as shown.<br /> <br /> [[File:2003amc8bakesale.png]]<br /> <br /> Each friend uses the same amount of dough, and Art makes exactly 12 cookies. <br /> <br /> ===Problem 8===<br /> <br /> Who gets the fewest cookies from one batch of cookie dough?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{Art} \qquad\textbf{(B)}\ \text{Paul}\qquad\textbf{(C)}\ \text{Roger}\qquad\textbf{(D)}\ \text{Trisha}\qquad\textbf{(E)}\ \text{There is a tie for fewest.}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 8|Solution]]<br /> <br /> ===Problem 9===<br /> <br /> Art's cookies sell for 60 cents each. To earn the same amount from a single batch, how much should one of Roger's cookies cost in cents?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\textbf{(E)}\ 90&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 9|Solution]]<br /> <br /> ===Problem 10===<br /> <br /> How many cookies will be in one batch of Trisha's cookies?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Business is a little slow at Lou's Fine Shoes, so Lou decides to have a sale. On Friday, Lou increases all of Thursday's prices by 10%. Over the weekend, Lou advertises the sale: &quot;Ten percent off the listed price. Sale starts Monday.&quot; How much does a pair of shoes cost on Monday that cost 40 dollars on Thursday?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 36\qquad\textbf{(B)}\ 39.60\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 40.40\qquad\textbf{(E)}\ 44 &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> When a fair six-sided die is tossed on a table top, the bottom face cannot be seen. What is the probability that the product of the numbers on the five faces that can be seen is divisible by 6?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ 1&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Fourteen white cubes are put together to form the figure on the right. The complete surface of the figure, including the bottom, is painted red. The figure is then separated into individual cubes. How many of the individual cubes have exactly four red faces?<br /> <br /> &lt;asy&gt;<br /> import three;<br /> defaultpen(linewidth(0.8));<br /> real r=0.5;<br /> currentprojection=orthographic(3/4,8/15,7/15);<br /> draw(unitcube, white, thick(), nolight);<br /> draw(shift(1,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(2,0,1)*unitcube, white, thick(), nolight);<br /> draw(shift(0,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,1,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,2,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(0,3,1)*unitcube, white, thick(), nolight);<br /> draw(shift(1,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,0)*unitcube, white, thick(), nolight);<br /> draw(shift(2,3,1)*unitcube, white, thick(), nolight);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> In this addition problem, each letter stands for a different digit. <br /> <br /> &lt;math&gt; \setlength{\tabcolsep}{0.5mm}\begin{array}{cccc}&amp;T &amp; W &amp; O\\ + &amp;T &amp; W &amp; O\\ \hline F&amp; O &amp; U &amp; R\end{array} &lt;/math&gt;<br /> <br /> If &lt;math&gt;T = 7&lt;/math&gt; and the letter &lt;math&gt;O&lt;/math&gt; represents an even number, what is the only possible value for &lt;math&gt;W&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> A figure is constructed from unit cubes. Each cube shares at least one face with another cube. What is the minimum number of cubes needed to build a figure with the front and side views shown?<br /> <br /> &lt;asy&gt;<br /> defaultpen(linewidth(0.8));<br /> path p=unitsquare;<br /> draw(p^^shift(0,1)*p^^shift(1,0)*p);<br /> draw(shift(4,0)*p^^shift(5,0)*p^^shift(5,1)*p);<br /> label(&quot;FRONT&quot;, (1,0), S);<br /> label(&quot;SIDE&quot;, (5,0), S);&lt;/asy&gt;<br /> <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 /> <br /> [[2003 AMC 8 Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Ali, Bonnie, Carlo, and Dianna are going to drive together to a nearby theme park. The car they are using has 4 seats: 1 driver's seat, 1 front passenger seat, and 2 back passenger seats. Bonnie and Carlo are the only ones who know how to drive the car. How many possible seating arrangements are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> The six children listed below are from two families of three siblings each. Each child has blue or brown eyes and black or blond hair. Children from the same family have at least one of these characteristics in common. Which two children are Jim's siblings?<br /> <br /> &lt;cmath&gt; \begin{array}{c|c|c}\text{Child}&amp;\text{Eye Color}&amp;\text{Hair Color}\\ \hline\text{Benjamin}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Jim}&amp;\text{Brown}&amp;\text{Blonde}\\ \hline\text{Nadeen}&amp;\text{Brown}&amp;\text{Black}\\ \hline\text{Austin}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\text{Tevyn}&amp;\text{Blue}&amp;\text{Black}\\ \hline\text{Sue}&amp;\text{Blue}&amp;\text{Blonde}\\ \hline\end{array} &lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Nadeen and Austin}\qquad\textbf{(B)}\ \text{Benjamin and Sue}\qquad\textbf{(C)}\ \text{Benjamin and Austin}\qquad\textbf{(D)}\ \text{Nadeen and Tevyn}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Austin and Sue} &lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> Each of the twenty dots on the graph below represents one of Sarah's classmates. Classmates who are friends are connected with a line segment. For her birthday party, Sarah is inviting only the following: all of her friends and all of those classmates who are friends with at least one of her friends. How many classmates will not be invited to Sarah's party?<br /> &lt;asy&gt;/* AMC8 2003 #18 Problem */<br /> pair a=(102,256), b=(68,131), c=(162,101), d=(134,150);<br /> pair e=(269,105), f=(359,104), g=(303,12), h=(579,211);<br /> pair i=(534, 342), j=(442,432), k=(374,484), l=(278,501);<br /> pair m=(282,411), n=(147,451), o=(103,437), p=(31,373);<br /> pair q=(419,175), r=(462,209), s=(477,288), t=(443,358);<br /> pair oval=(282,303);<br /> draw(l--m--n--cycle);<br /> draw(p--oval);<br /> draw(o--oval);<br /> draw(b--d--oval);<br /> draw(c--d--e--oval);<br /> draw(e--f--g--h--i--j--oval);<br /> draw(k--oval);<br /> draw(q--oval);<br /> draw(s--oval);<br /> draw(r--s--t--oval);<br /> dot(a); dot(b); dot(c); dot(d); dot(e); dot(f); dot(g); dot(h);<br /> dot(i); dot(j); dot(k); dot(l); dot(m); dot(n); dot(o); dot(p);<br /> dot(q); dot(r); dot(s); dot(t);<br /> filldraw(yscale(.5)*Circle((282,606),80),white,black);<br /> label(scale(0.75)*&quot;Sarah&quot;, oval);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> How many integers between 1000 and 2000 have all three of the numbers 15, 20, and 25 as factors?<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 /> [[2003 AMC 8 Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> The area of trapezoid &lt;math&gt; ABCD&lt;/math&gt; is &lt;math&gt;164\text{ cm}^2&lt;/math&gt;. The altitude is 8 cm, &lt;math&gt;AB&lt;/math&gt; is 10 cm, and &lt;math&gt;CD&lt;/math&gt; is 17 cm. What is &lt;math&gt;BC&lt;/math&gt;, in centimeters?<br /> <br /> &lt;asy&gt;/* AMC8 2003 #21 Problem */<br /> size(4inch,2inch);<br /> draw((0,0)--(31,0)--(16,8)--(6,8)--cycle);<br /> draw((11,8)--(11,0), linetype(&quot;8 4&quot;));<br /> draw((11,1)--(12,1)--(12,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$D$&quot;, (31,0), SE);<br /> label(&quot;$B$&quot;, (6,8), NW);<br /> label(&quot;$C$&quot;, (16,8), NE);<br /> label(&quot;10&quot;, (3,5), W);<br /> label(&quot;8&quot;, (11,4), E);<br /> label(&quot;17&quot;, (22.5,5), E);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> The following figures are composed of squares and circles. Which figure has a shaded region with largest area?<br /> &lt;asy&gt;/* AMC8 2003 #22 Problem */<br /> size(3inch, 2inch);<br /> unitsize(1cm);<br /> pen outline = black+linewidth(1);<br /> filldraw((0,0)--(2,0)--(2,2)--(0,2)--cycle, mediumgrey, outline);<br /> filldraw(shift(3,0)*((0,0)--(2,0)--(2,2)--(0,2)--cycle), mediumgrey, outline);<br /> filldraw(Circle((7,1), 1), mediumgrey, black+linewidth(1));<br /> filldraw(Circle((1,1), 1), white, outline);<br /> filldraw(Circle((3.5,.5), .5), white, outline);<br /> filldraw(Circle((4.5,.5), .5), white, outline);<br /> filldraw(Circle((3.5,1.5), .5), white, outline);<br /> filldraw(Circle((4.5,1.5), .5), white, outline);<br /> filldraw((6.3,.3)--(7.7,.3)--(7.7,1.7)--(6.3,1.7)--cycle, white, outline);<br /> label(&quot;A&quot;, (1, 2), N);<br /> label(&quot;B&quot;, (4, 2), N);<br /> label(&quot;C&quot;, (7, 2), N);<br /> draw((0,-.5)--(.5,-.5), BeginArrow);<br /> draw((1.5, -.5)--(2, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (1, -.5));<br /> <br /> draw((3,-.5)--(3.5,-.5), BeginArrow);<br /> draw((4.5, -.5)--(5, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (4, -.5));<br /> <br /> draw((6,-.5)--(6.5,-.5), BeginArrow);<br /> draw((7.5, -.5)--(8, -.5), EndArrow);<br /> label(&quot;2 cm&quot;, (7, -.5));<br /> <br /> draw((6,1)--(6,-.5), linetype(&quot;4 4&quot;));<br /> draw((8,1)--(8,-.5), linetype(&quot;4 4&quot;));&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{A only}\qquad\textbf{(B)}\ \text{B only}\qquad\textbf{(C)}\ \text{C only}\qquad\textbf{(D)}\ \text{both A and B}\qquad\textbf{(E)}\ \text{all are equal}&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> In the pattern below, the cat moves clockwise through the four squares and the mouse moves counterclockwise through the eight exterior segments of the four squares.<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23a.png|800px]]<br /> &lt;/center&gt;<br /> <br /> If the pattern is continued, where would the cat and mouse be after the 247th move?<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob23b.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> A ship travels from point &lt;math&gt;A&lt;/math&gt; to point &lt;math&gt;B&lt;/math&gt; along a semicircular path, centered at Island &lt;math&gt;X&lt;/math&gt;. Then it travels along a straight path from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt;. Which of these graphs best shows the ship's distance from Island &lt;math&gt;X&lt;/math&gt; as it moves along its course?<br /> <br /> &lt;asy&gt;size(150);<br /> pair X=origin, A=(-5,0), B=(5,0), C=(0,5);<br /> draw(Arc(X, 5, 180, 360)^^B--C);<br /> dot(X);<br /> label(&quot;$X$&quot;, X, NE);<br /> label(&quot;$C$&quot;, C, N);<br /> label(&quot;$B$&quot;, B, E);<br /> label(&quot;$A$&quot;, A, W);<br /> &lt;/asy&gt;<br /> <br /> &lt;center&gt;<br /> [[Image:2003amc8prob24ans.png|800px]]<br /> &lt;/center&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> In the figure, the area of square &lt;math&gt;WXYZ&lt;/math&gt; is &lt;math&gt;25 \text{ cm}^2&lt;/math&gt;. The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = AC&lt;/math&gt;, and when &lt;math&gt;\triangle ABC&lt;/math&gt; is folded over side &lt;math&gt;\overline{BC}&lt;/math&gt;, point &lt;math&gt;A&lt;/math&gt; coincides with &lt;math&gt;O&lt;/math&gt;, the center of square &lt;math&gt;WXYZ&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;, in square centimeters?<br /> <br /> &lt;asy&gt;<br /> defaultpen(fontsize(8));<br /> size(225);<br /> pair Z=origin, W=(0,10), X=(10,10), Y=(10,0), O=(5,5), B=(-4,8), C=(-4,2), A=(-13,5);<br /> draw((-4,0)--Y--X--(-4,10)--cycle);<br /> draw((0,-2)--(0,12)--(-2,12)--(-2,8)--B--A--C--(-2,2)--(-2,-2)--cycle);<br /> dot(O);<br /> label(&quot;$A$&quot;, A, NW);<br /> label(&quot;$O$&quot;, O, NE);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, NW);<br /> label(&quot;$W$&quot;,W , NE);<br /> label(&quot;$X$&quot;, X, N);<br /> label(&quot;$Y$&quot;, Y, S);<br /> label(&quot;$Z\$&quot;, Z, SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{15}4\qquad\textbf{(B)}\ \frac{21}4\qquad\textbf{(C)}\ \frac{27}4\qquad\textbf{(D)}\ \frac{21}2\qquad\textbf{(E)}\ \frac{27}2&lt;/math&gt;<br /> <br /> [[2003 AMC 8 Problems/Problem 25|Solution]]<br /> <br /> ==See Also==<br /> {{AMC8 box|year=2003|before=[[2002 AMC 8 Problems|2002 AMC 8]]|after=[[2004 AMC 8 Problems|2004 AMC 8]]}}<br /> * [[AMC 8]]<br /> * [[AMC 8 Problems and Solutions]]<br /> * [[Mathematics competition resources]]<br /> <br /> <br /> {{MAA Notice}}</div> Srim1027