https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mariekitty&feedformat=atom AoPS Wiki - User contributions [en] 2021-05-11T07:12:52Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_12B_Problems&diff=51119 2012 AMC 12B Problems 2013-02-18T15:50:06Z <p>Mariekitty: /* Problem 9 */</p> <hr /> <div>{{AMC12 Problems|year=2012|ab=B}}<br /> <br /> == Problem 1 ==<br /> <br /> Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there in all 4 of the third-grade classrooms?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> <br /> A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?<br /> &lt;asy&gt;<br /> draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); <br /> draw(circle((10,5),5));&lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> <br /> For a science project, Sammy observed a chipmunk and squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide? <br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> Suppose that the euro is worth 1.30 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Two integers have a sum of 26. when two more integers are added to the first two, the sum is 41. Finally, when two more integers are added to the sum of the previous 4 integers, the sum is 57. What is the minimum number of even integers among the 6 integers? <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 /> [[2012 AMC 12B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> In order to estimate the value of &lt;math&gt;x-y&lt;/math&gt; where &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are real numbers with &lt;math&gt;x &gt; y &gt; 0&lt;/math&gt;, Xiaoli rounded &lt;math&gt;x&lt;/math&gt; up by a small amount, rounded &lt;math&gt;y&lt;/math&gt; down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Her estimate is larger than }x-y&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(B)}\ \text{Her estimate is smaller than }x-y&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(C)}\ \text{Her estimate equals }x-y&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \text{Her estimate equals }y - x&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(E)}\ \text{Her estimate is 0}&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> Small lights are hung on a string 6 inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of 2 red lights followed by 3 green lights. How many feet separate the 3rd red light and the 21st red light?<br /> <br /> '''Note:''' 1 foot is equal to 12 inches.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> What is the area of the polygon whose vertices are the points of intersection of the curves &lt;math&gt;x^2 + y^2 =25&lt;/math&gt; and &lt;math&gt;(x-4)^2 + 9y^2 = 81&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 24\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 42&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> In the equation below, &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt; are consecutive positive integers, and &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;A+B&lt;/math&gt; represent number bases: &lt;cmath&gt;132_A+43_B=69_{A+B}.&lt;/cmath&gt;<br /> What is &lt;math&gt;A+B&lt;/math&gt;?<br /> <br /> [[2012 AMC 12B Problems/Problem 11|Solution]]<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 17 &lt;/math&gt;<br /> <br /> == Problem 12 ==<br /> <br /> How many sequences of zeros and ones of length 20 have all the zeros consecutive, or all the ones consecutive, or both?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 190\qquad\textbf{(B)}\ 192\qquad\textbf{(C)}\ 211\qquad\textbf{(D)}\ 380\qquad\textbf{(E)}\ 382 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13 ==<br /> <br /> Two parabolas have equations &lt;math&gt;y= x^2 + ax +b&lt;/math&gt; and &lt;math&gt;y= x^2 + cx +d&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, and &lt;math&gt;d&lt;/math&gt; are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have a least one point in common?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{25}{36}\qquad\textbf{(C)}\ \frac{5}{6}\qquad\textbf{(D)}\ \frac{31}{36}\qquad\textbf{(E)}\ 1 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> Bernardo and Silvia play the following game. An integer between 0 and 999 inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she addes 50 to it and passes the result to Bernardo. The winner is the last person who produces a number less than 1000. Let ''N'' be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of ''N''?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> Jesse cuts a circular paper disk of radius 12 along two radii to form two sectors, the smaller having a central angle of 120 degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{\sqrt{10}}{10}\qquad\textbf{(D)}\ \frac{\sqrt{5}}{6}\qquad\textbf{(E)}\ \frac{\sqrt{5}}{5}&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those girls but disliked by the third. In how many different ways is this possible?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> Square &lt;math&gt;PQRS&lt;/math&gt; lies in the first quadrant. Points &lt;math&gt;(3,0), (5,0), (7,0),&lt;/math&gt; and &lt;math&gt;(13,0)&lt;/math&gt; lie on lines &lt;math&gt;SP, RQ, PQ,&lt;/math&gt; and &lt;math&gt;SR&lt;/math&gt;, respectively. What is the sum of the coordinates of the center of the square &lt;math&gt;PQRS&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 6.2\qquad\textbf{(C)}\ 6.4\qquad\textbf{(D)}\ 6.6\qquad\textbf{(E)}\ 6.8 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Let &lt;math&gt;(a_1,a_2, \dots ,a_{10})&lt;/math&gt; be a list of the first 10 positive integers such that for each &lt;math&gt;2 \le i \le 10&lt;/math&gt; either &lt;math&gt;a_i+1&lt;/math&gt; or &lt;math&gt;a_i-1&lt;/math&gt; or both appear somewhere before &lt;math&gt;a_i&lt;/math&gt; in the list. How many such lists are there?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> A unit cube has vertices &lt;math&gt;P_1,P_2,P_3,P_4,P_1',P_2',P_3',&lt;/math&gt; and &lt;math&gt;P_4'&lt;/math&gt;. Vertices &lt;math&gt;P_2&lt;/math&gt;, &lt;math&gt;P_3&lt;/math&gt;, and &lt;math&gt;P_4&lt;/math&gt; are adjacent to &lt;math&gt;P_1&lt;/math&gt;, and for &lt;math&gt;1\le i\le 4,&lt;/math&gt; vertices &lt;math&gt;P_i&lt;/math&gt; and &lt;math&gt;P_i'&lt;/math&gt; are opposite to each other. A regular octahedron has one vertex in each of the segments &lt;math&gt;P_1P_2&lt;/math&gt;, &lt;math&gt;P_1P_3&lt;/math&gt;, &lt;math&gt;P_1P_4&lt;/math&gt;, &lt;math&gt;P_1'P_2'&lt;/math&gt;, &lt;math&gt;P_1'P_3'&lt;/math&gt;, and &lt;math&gt;P_1'P_4'&lt;/math&gt;. What is the octahedron's side length?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{3\sqrt{2}}{4}\qquad\textbf{(B)}\ \frac{7\sqrt{6}}{16}\qquad\textbf{(C)}\ \frac{\sqrt{5}}{2}\qquad\textbf{(D)}\ \frac{2\sqrt{3}}{3}\qquad\textbf{(E)}\ \frac{\sqrt{6}}{2} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> A trapezoid has side lengths 3, 5, 7, and 11. The sums of all the possible areas of the trapezoid can be written in the form of &lt;math&gt;r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3&lt;/math&gt;, where &lt;math&gt;r_1&lt;/math&gt;, &lt;math&gt;r_2&lt;/math&gt;, and &lt;math&gt;r_3&lt;/math&gt; are rational numbers and &lt;math&gt;n_1&lt;/math&gt; and &lt;math&gt;n_2&lt;/math&gt; are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to &lt;math&gt;r_1+r_2+r_3+n_1+n_2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Square &lt;math&gt;AXYZ&lt;/math&gt; is inscribed in equiangular hexagon &lt;math&gt;ABCDEF&lt;/math&gt; with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;\overline{BC}&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt; on &lt;math&gt;\overline{DE}&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; on &lt;math&gt;\overline{EF}&lt;/math&gt;. Suppose that &lt;math&gt;AB=40&lt;/math&gt;, and &lt;math&gt;EF=41(\sqrt{3}-1)&lt;/math&gt;. What is the side-length of the square?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}\qquad\textbf{(C)}\ 20\sqrt{3}+16&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ 20\sqrt{2}+13\sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> A bug travels from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;B&lt;/math&gt; along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?<br /> <br /> &lt;asy&gt;<br /> size(10cm);<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));<br /> draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));<br /> draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));<br /> draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));<br /> draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));<br /> draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));<br /> draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));<br /> draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));<br /> draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));<br /> draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));<br /> dot((0,0));<br /> dot((22,0));<br /> label(&quot;$A$&quot;,(0,0),WNW);<br /> label(&quot;$B$&quot;,(22,0),E);<br /> filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);<br /> filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);<br /> filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);<br /> filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);<br /> filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);<br /> filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);<br /> filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);<br /> filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);<br /> filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);<br /> filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);<br /> filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);<br /> filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);<br /> filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);<br /> filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);<br /> filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);<br /> filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);<br /> filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);<br /> filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);<br /> filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);<br /> filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);<br /> filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);<br /> filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);<br /> filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);&lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400&lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> Consider all polynomials of a complex variable, &lt;math&gt;P(z)=4z^4+az^3+bz^2+cz+d&lt;/math&gt;, where &lt;math&gt;a,b,c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers, &lt;math&gt;0\le d\le c\le b\le a\le 4&lt;/math&gt;, and the polynomial has a zero &lt;math&gt;z_0&lt;/math&gt; with &lt;math&gt;|z_0|=1.&lt;/math&gt; What is the sum of all values &lt;math&gt;P(1)&lt;/math&gt; over all the polynomials with these properties?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 84\qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 108\qquad\textbf{(E)}\ 120 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Define the function &lt;math&gt;f_1&lt;/math&gt; on the positive integers by setting &lt;math&gt;f_1(1)=1&lt;/math&gt; and if &lt;math&gt;n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}&lt;/math&gt; is the prime factorization of &lt;math&gt;n&gt;1&lt;/math&gt;, then &lt;cmath&gt;f_1(n)=(p_1+1)^{e_1-1}(p_2+1)^{e_2-1}\cdots (p_k+1)^{e_k-1}.&lt;/cmath&gt;<br /> For every &lt;math&gt;m\ge 2&lt;/math&gt;, let &lt;math&gt;f_m(n)=f_1(f_{m-1}(n))&lt;/math&gt;. For how many &lt;math&gt;N&lt;/math&gt; in the range &lt;math&gt;1\le N\le 400&lt;/math&gt; is the sequence &lt;math&gt;(f_1(N),f_2(N),f_3(N),\dots )&lt;/math&gt; unbounded?<br /> <br /> '''Note:''' A sequence of positive numbers is unbounded if for every integer &lt;math&gt;B&lt;/math&gt;, there is a member of the sequence greater than &lt;math&gt;B&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19 &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> Let &lt;math&gt;S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}&lt;/math&gt;. <br /> Let &lt;math&gt;T&lt;/math&gt; be the set of all right triangles whose vertices are in &lt;math&gt;S&lt;/math&gt;. For every right triangle &lt;math&gt;t=\triangle{ABC}&lt;/math&gt; with vertices &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt; in counter-clockwise order and right angle at &lt;math&gt;A&lt;/math&gt;, let &lt;math&gt;f(t)=\tan(\angle{CBA})&lt;/math&gt;. What is &lt;cmath&gt;\prod_{t\in T} f(t)?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24} &lt;/math&gt;<br /> <br /> [[2012 AMC 12B Problems/Problem 25|Solution]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B/Answer_Key&diff=51088 2006 AMC 12B/Answer Key 2013-02-17T15:30:26Z <p>Mariekitty: Answer Key to 2006 AMC 12B</p> <hr /> <div>==Answer Key to 2006 AMC 12B==<br /> <br /> 1. C<br /> <br /> 2. A<br /> <br /> 3. A<br /> <br /> 4. A<br /> <br /> 5. A<br /> <br /> 6 .B<br /> <br /> 7. B<br /> <br /> 8. E<br /> <br /> 9. B<br /> <br /> 10. A<br /> <br /> 11. E<br /> <br /> 12. D<br /> <br /> 13. C<br /> <br /> 14. D<br /> <br /> 15. B<br /> <br /> 16. C<br /> <br /> 17. C<br /> <br /> 18. B<br /> <br /> 19. B<br /> <br /> 20. C<br /> <br /> 21. C<br /> <br /> 22. B<br /> <br /> 23. E<br /> <br /> 24. C<br /> <br /> 25. B</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=Template:AMC12_Problems&diff=51032 Template:AMC12 Problems 2013-02-11T19:43:22Z <p>Mariekitty: </p> <hr /> <div>{| class=&quot;wikitable&quot; style=&quot;margin:0.5em auto; font-size:95%; border:1px solid black; width:50%;&quot;<br /> | style=&quot;background:#ccf;text-align:center;&quot; colspan=&quot;3&quot; | '''[[{{{year}}} AMC 12{{{ab|}}}]]''' ('''[[{{{year}}} AMC 12{{{ab|}}} Answer Key|Answer Key]]''')&lt;br /&gt;Printable version: &lt;span class=&quot;noprint plainlinks&quot;&gt;[{{fullurl:{{FULLPAGENAME}}|printable=yes}} Wiki]&lt;/span&gt; | [http://www.artofproblemsolving.com/Forum/resources.php?c=182&amp;cid=44&amp;year={{{year}}} AoPS Resources] • [http://www.artofproblemsolving.com/Forum/resources/files/usa/USA-AMC_12-AHSME-{{{year}}}-44 PDF] <br /> |-<br /> | style=&quot;font-size:80%;&quot;|<br /> &lt;font style=&quot;text-align:center;&quot;&gt;Instructions&lt;/font&gt;<br /> # This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. <br /> # You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.<br /> # No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will ''require'' the use of a calculator). <br /> # Figures are not necessarily drawn to scale.<br /> # You will have '''75 minutes''' working time to complete the test. <br /> |-<br /> | style=&quot;text-align:center;&quot;| [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 1|1]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 2|2]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 3|3]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 4|4]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 5|5]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 6|6]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 7|7]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 8|8]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 9|9]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 10|10]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 11|11]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 12|12]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 13|13]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 14|14]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 15|15]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 16|16]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 17|17]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 18|18]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 19|19]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 20|20]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 21|21]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 22|22]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 23|23]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 24|24]] '''•''' [[{{{year}}} AMC 12{{{ab|}}} Problems/Problem 25|25]]<br /> |}&lt;noinclude&gt;<br /> <br /> To use, put at the top of an AMC 12 Problems page (like [[2010 AMC 12A Problems]]) and follow the syntax &lt;pre&gt;<br /> {{AMC12 Problems|year=2010|ab=A}}<br /> &lt;/pre&gt;<br /> <br /> &lt;/noinclude&gt;[[Category:AMC 12 Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=User:Mariekitty&diff=44109 User:Mariekitty 2012-01-02T14:14:43Z <p>Mariekitty: </p> <hr /> <div>= Mariekitty's AoPSWiki :) =<br /> <br /> ==Bio==<br /> *Apparently i have an AoPSwiki, i just found out randomly......<br /> Umm, i'll start by telling you more about myself. I am a student that lives in MA, <br /> I'm athletic, funny and outgoing when you really get to know me.<br /> For my spare time, I like to do math, swim and read. <br /> <br /> ==My Blog==<br /> *This is my blog on AoPS, [http://www.artofproblemsolving.com/Forum/blog.php?u=83394 feel free to visit]<br /> <br /> ==Links==<br /> * [[AoPSWiki:Words of the Week]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_15&diff=44108 2010 AMC 10A Problems/Problem 15 2012-01-02T14:02:27Z <p>Mariekitty: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br /> <br /> Brian: &quot;Mike and I are different species.&quot;<br /> <br /> Chris: &quot;LeRoy is a frog.&quot;<br /> <br /> LeRoy: &quot;Chris is a frog.&quot;<br /> <br /> Mike: &quot;Of the four of us, at least two are toads.&quot;<br /> <br /> How many of these amphibians are frogs?<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 /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> We can begin by first looking at Chris and LeRoy.<br /> <br /> Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.<br /> <br /> Clearly, Chris and LeRoy are different species, and so we have at least &lt;math&gt;1&lt;/math&gt; frog out of the two of them.<br /> <br /> Now suppose Mike is a toad. Then what he says is true because we already have &lt;math&gt;2&lt;/math&gt; toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.<br /> <br /> Therefore, Mike must be a frog. His statement must be false, which means that there is at most &lt;math&gt;1&lt;/math&gt; toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.<br /> <br /> Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have &lt;math&gt;3&lt;/math&gt; frogs total. &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.<br /> <br /> As Mike is a frog, his statement is false, hence there is at most one toad.<br /> <br /> As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.<br /> <br /> Hence we must have one toad and three frogs. &lt;math&gt; \boxed{\textbf{(D)}} &lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=14|num-a=16|ab=A}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_15&diff=44107 2010 AMC 10A Problems/Problem 15 2012-01-02T14:02:03Z <p>Mariekitty: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.<br /> <br /> Brian: &quot;Mike and I are different species.&quot;<br /> <br /> Chris: &quot;LeRoy is a frog.&quot;<br /> <br /> LeRoy: &quot;Chris is a frog.&quot;<br /> <br /> Mike: &quot;Of the four of us, at least two are toads.&quot;<br /> <br /> How many of these amphibians are frogs?<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 /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> We can begin by first looking at Chris and LeRoy.<br /> <br /> Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog.<br /> <br /> Clearly, Chris and LeRoy are different species, and so we have at least &lt;math&gt;1&lt;/math&gt; frog out of the two of them.<br /> <br /> Now suppose Mike is a toad. Then what he says is true because we already have &lt;math&gt;2&lt;/math&gt; toads. However, if Brian is a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction.<br /> <br /> Therefore, Mike must be a frog. His statement must be false, which means that there is at most &lt;math&gt;1&lt;/math&gt; toad. Since either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false.<br /> <br /> Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have &lt;math&gt;3&lt;/math&gt; frogs total. &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;<br /> <br /> === Solution 2 ===<br /> <br /> Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.<br /> <br /> As Mike is a frog, his statement is false, hence there is at most one toad.<br /> <br /> As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.<br /> <br /> Hence we must have one toad and three frogs. &lt;math&gt;\boxed{(D)}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|num-b=14|num-a=16|ab=A}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_13&diff=44106 2010 AMC 10A Problems/Problem 13 2012-01-02T13:58:45Z <p>Mariekitty: /* Problem */</p> <hr /> <div>==Problem==<br /> Angelina drove at an average rate of &lt;math&gt;80&lt;/math&gt; kph and then stopped &lt;math&gt;20&lt;/math&gt; minutes for gas. After the stop, she drove at an average rate of &lt;math&gt;100&lt;/math&gt; kph. Altogether she drove &lt;math&gt;250&lt;/math&gt; km in a total trip time of &lt;math&gt;3&lt;/math&gt; hours including the stop. Which equation could be used to solve for the time &lt;math&gt;t&lt;/math&gt; in hours that she drove before her stop?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 80t+100(\frac{8}{3}-t)=250 \qquad \textbf{(B)}\ 80t=250 \qquad \textbf{(C)}\ 100t=250 \qquad \textbf{(D)}\ 90t=250 \qquad \textbf{(E)}\ 80(\frac{8}{3}-t)+100t=250&lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> The answer is &lt;math&gt;A&lt;/math&gt; because she drove at &lt;math&gt;80&lt;/math&gt; kmh for &lt;math&gt;t&lt;/math&gt; hours (the amount of time before the stop), and 100 kmh for &lt;math&gt;\frac{8}{3}-t&lt;/math&gt; because she wasn't driving for &lt;math&gt;20&lt;/math&gt; minutes, or &lt;math&gt;\frac{1}{3}&lt;/math&gt; hours. Multiplying by &lt;math&gt;t&lt;/math&gt; gives the total distance, which is &lt;math&gt;250&lt;/math&gt; kms. Therefore, the answer is &lt;math&gt;80t+100(\frac{8}{3}-t)=250&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{(A)}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=12|num-a=14}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_11&diff=44105 2010 AMC 10A Problems/Problem 11 2012-01-02T13:54:23Z <p>Mariekitty: </p> <hr /> <div>== Problem 11 ==<br /> The length of the interval of solutions of the inequality &lt;math&gt;a \le 2x + 3 \le b&lt;/math&gt; is &lt;math&gt;10&lt;/math&gt;. What is &lt;math&gt;b - a&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 6<br /> \qquad<br /> \mathrm{(B)}\ 10<br /> \qquad<br /> \mathrm{(C)}\ 15<br /> \qquad<br /> \mathrm{(D)}\ 20<br /> \qquad<br /> \mathrm{(E)}\ 30<br /> &lt;/math&gt;<br /> <br /> <br /> ==Solution==<br /> <br /> Since we are given the range of the solutions, we must re-write the inequalities so that we have &lt;math&gt; x &lt;/math&gt; in terms of &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt;. <br /> <br /> &lt;math&gt; a\le 2x+3\le b &lt;/math&gt;<br /> <br /> Subtract &lt;math&gt; 3 &lt;/math&gt; from all of the quantities:<br /> <br /> &lt;math&gt; a-3\le 2x\le b-3 &lt;/math&gt;<br /> <br /> Divide all of the quantities by &lt;math&gt; 2 &lt;/math&gt;.<br /> <br /> &lt;math&gt; \frac{a-3}{2}\le x\le \frac{b-3}{2} &lt;/math&gt;<br /> <br /> Since we have the range of the solutions, we can make them equal to &lt;math&gt; 10 &lt;/math&gt;.<br /> <br /> &lt;math&gt; \frac{b-3}{2}-\frac{a-3}{2} = 10 &lt;/math&gt;<br /> <br /> Multiply both sides by 2.<br /> <br /> &lt;math&gt; (b-3) - (a-3) = 20 &lt;/math&gt;<br /> <br /> Re-write without using parentheses.<br /> <br /> &lt;math&gt; b-3-a+3 = 20 &lt;/math&gt;<br /> <br /> Simplify.<br /> <br /> &lt;math&gt; b-a = 20 &lt;/math&gt;<br /> <br /> We need to find &lt;math&gt; b - a &lt;/math&gt; for the problem, so the answer is &lt;math&gt; \boxed{20\ \textbf{(D)}} &lt;/math&gt;<br /> <br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=A|num-b=10|num-a=12}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_10&diff=44081 2010 AMC 10A Problems/Problem 10 2012-01-02T00:09:23Z <p>Mariekitty: </p> <hr /> <div>== Problem 10 ==<br /> Marvin had a birthday on Tuesday, May 27 in the leap year &lt;math&gt;2008&lt;/math&gt;. In what year will his birthday next fall on a Saturday?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 2011<br /> \qquad<br /> \mathrm{(B)}\ 2012<br /> \qquad<br /> \mathrm{(C)}\ 2013<br /> \qquad<br /> \mathrm{(D)}\ 2015<br /> \qquad<br /> \mathrm{(E)}\ 2017<br /> &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;\boxed{(E)}&lt;/math&gt; &lt;math&gt; 2017 &lt;/math&gt;<br /> <br /> There are &lt;math&gt;365&lt;/math&gt; days in a non-leap year. There are &lt;math&gt;7&lt;/math&gt; days in a week. Since &lt;math&gt;365 = 52 \cdot 7 + 1&lt;/math&gt; (or &lt;math&gt;365&lt;/math&gt; is congruent to &lt;math&gt;1 \mod{ 7}&lt;/math&gt;), the same date (after February) moves &quot;forward&quot; one day in the subsequent year, if that year is not a leap year.<br /> <br /> For example:<br /> <br /> &lt;math&gt;5/27/08&lt;/math&gt; Tue<br /> <br /> &lt;math&gt;5/27/09&lt;/math&gt; Wed<br /> <br /> However, a leap year has &lt;math&gt;366&lt;/math&gt; days, and &lt;math&gt;366 = 52 \cdot 7 + 2&lt;/math&gt; . So the same date (after February) moves &quot;forward&quot; '''two''' days in the subsequent year, if that year is a leap year.<br /> <br /> For example:<br /> &lt;math&gt;5/27/11&lt;/math&gt; Fri<br /> <br /> &lt;math&gt;5/27/12&lt;/math&gt; Sun<br /> <br /> You can keep count forward to find that the first time this date falls on a Saturday is in &lt;math&gt; 2017&lt;/math&gt;:<br /> <br /> &lt;math&gt;5/27/13&lt;/math&gt; Mon<br /> <br /> &lt;math&gt;5/27/14&lt;/math&gt; Tue<br /> <br /> &lt;math&gt;5/27/15&lt;/math&gt; Wed<br /> <br /> &lt;math&gt;5/27/16&lt;/math&gt; Fri<br /> <br /> &lt;math&gt;5/27/17&lt;/math&gt; Sat<br /> <br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=A|num-b=9|num-a=11}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_10&diff=44080 2010 AMC 10A Problems/Problem 10 2012-01-02T00:07:20Z <p>Mariekitty: </p> <hr /> <div>== Problem 9 ==<br /> A &lt;i&gt;palindrome&lt;/i&gt;, such as &lt;math&gt;83438&lt;/math&gt;, is a number that remains the same when its digits are reversed. The numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;x+32&lt;/math&gt; are three-digit and four-digit palindromes, respectively. What is the sum of the digits of &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 20<br /> \qquad<br /> \mathrm{(B)}\ 21<br /> \qquad<br /> \mathrm{(C)}\ 22<br /> \qquad<br /> \mathrm{(D)}\ 23<br /> \qquad<br /> \mathrm{(E)}\ 24<br /> &lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution==<br /> <br /> &lt;math&gt;\boxed{(E)}&lt;/math&gt; &lt;math&gt; 2017 &lt;/math&gt;<br /> <br /> There are &lt;math&gt;365&lt;/math&gt; days in a non-leap year. There are &lt;math&gt;7&lt;/math&gt; days in a week. Since &lt;math&gt;365 = 52 \cdot 7 + 1&lt;/math&gt; (or &lt;math&gt;365&lt;/math&gt; is congruent to &lt;math&gt;1 \mod{ 7}&lt;/math&gt;), the same date (after February) moves &quot;forward&quot; one day in the subsequent year, if that year is not a leap year.<br /> <br /> For example:<br /> <br /> &lt;math&gt;5/27/08&lt;/math&gt; Tue<br /> <br /> &lt;math&gt;5/27/09&lt;/math&gt; Wed<br /> <br /> However, a leap year has &lt;math&gt;366&lt;/math&gt; days, and &lt;math&gt;366 = 52 \cdot 7 + 2&lt;/math&gt; . So the same date (after February) moves &quot;forward&quot; '''two''' days in the subsequent year, if that year is a leap year.<br /> <br /> For example:<br /> &lt;math&gt;5/27/11&lt;/math&gt; Fri<br /> <br /> &lt;math&gt;5/27/12&lt;/math&gt; Sun<br /> <br /> You can keep count forward to find that the first time this date falls on a Saturday is in &lt;math&gt; 2017&lt;/math&gt;:<br /> <br /> &lt;math&gt;5/27/13&lt;/math&gt; Mon<br /> <br /> &lt;math&gt;5/27/14&lt;/math&gt; Tue<br /> <br /> &lt;math&gt;5/27/15&lt;/math&gt; Wed<br /> <br /> &lt;math&gt;5/27/16&lt;/math&gt; Fri<br /> <br /> &lt;math&gt;5/27/17&lt;/math&gt; Sat<br /> <br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=A|num-b=19|num-a=11}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_8&diff=44079 2010 AMC 10A Problems/Problem 8 2012-01-02T00:00:16Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 8 ==<br /> Tony works &lt;math&gt;2&lt;/math&gt; hours a day and is paid &amp;#36;&lt;math&gt;0.50&lt;/math&gt; per hour for each full year of his age. During a six month period Tony worked &lt;math&gt;50&lt;/math&gt; days and earned &amp;#36;&lt;math&gt;630&lt;/math&gt;. How old was Tony at the end of the six month period?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 9<br /> \qquad<br /> \mathrm{(B)}\ 11<br /> \qquad<br /> \mathrm{(C)}\ 12<br /> \qquad<br /> \mathrm{(D)}\ 13<br /> \qquad<br /> \mathrm{(E)}\ 14<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Tony worked &lt;math&gt;2&lt;/math&gt; hours a day and is paid &lt;math&gt;0.50&lt;/math&gt; dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is &lt;math&gt;12&lt;/math&gt; years old, he gets &lt;math&gt;12&lt;/math&gt; dollars a day. We also know that he worked &lt;math&gt;50&lt;/math&gt; days and earned &lt;math&gt;630&lt;/math&gt; dollars. If he was &lt;math&gt;12&lt;/math&gt; years old at the beginning of his working period, he would have earned &lt;math&gt;12 * 50 = 600&lt;/math&gt; dollars. If he was &lt;math&gt;13&lt;/math&gt; years old at the beginning of his working period, he would have earned &lt;math&gt;13 * 50 = 650&lt;/math&gt; dollars. Because he earned &lt;math&gt;630&lt;/math&gt; dollars, we know that he was &lt;math&gt;13&lt;/math&gt; for some period of time, but not the whole time, because then the money earned would be greater than or equal to &lt;math&gt;650&lt;/math&gt;. This is why he was &lt;math&gt;12&lt;/math&gt; when he began, but turned &lt;math&gt;13&lt;/math&gt; sometime in the middle and earned &lt;math&gt;630&lt;/math&gt; dollars in total. So the answer is &lt;math&gt;13&lt;/math&gt;.The answer is &lt;math&gt;\boxed{D}&lt;/math&gt;. We could find out for how long he was &lt;math&gt;12&lt;/math&gt; and &lt;math&gt;13&lt;/math&gt;. &lt;math&gt;12 \cdot x + 13 \cdot (50-x) = 630&lt;/math&gt;. Then &lt;math&gt;x&lt;/math&gt; is &lt;math&gt;20&lt;/math&gt; and we know that he was &lt;math&gt;12&lt;/math&gt; for &lt;math&gt;20&lt;/math&gt; days, and &lt;math&gt;13&lt;/math&gt; for &lt;math&gt;30&lt;/math&gt; days. Thus, the answer is &lt;math&gt;13&lt;/math&gt;.<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=7|num-a=9}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_7&diff=44078 2010 AMC 10A Problems/Problem 7 2012-01-01T23:59:57Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 7 ==<br /> Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 1<br /> \qquad<br /> \mathrm{(B)}\ \sqrt{2}<br /> \qquad<br /> \mathrm{(C)}\ \sqrt{3}<br /> \qquad<br /> \mathrm{(D)}\ 2<br /> \qquad<br /> \mathrm{(E)}\ 2\sqrt{2}<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> Crystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is &lt;math&gt;\sqrt{2}&lt;/math&gt; miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to &lt;math&gt;\sqrt{((\sqrt{2})^2+1^2)}&lt;/math&gt;, which is equal to &lt;math&gt;\sqrt{3}&lt;/math&gt;. The answer is &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=6|num-a=8}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_6&diff=44073 2010 AMC 10A Problems/Problem 6 2012-01-01T23:38:03Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 6 ==<br /> For positive numbers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; the operation &lt;math&gt;\spadesuit(x, y)&lt;/math&gt; is defined as <br /> <br /> &lt;cmath&gt;\spadesuit(x, y) = x -\dfrac{1}{y}&lt;/cmath&gt;<br /> <br /> What is &lt;math&gt;\spadesuit(2,\spadesuit(2, 2))&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \dfrac{2}{3}<br /> \qquad<br /> \mathrm{(B)}\ 1<br /> \qquad<br /> \mathrm{(C)}\ \dfrac{4}{3}<br /> \qquad<br /> \mathrm{(D)}\ \dfrac{5}{3}<br /> \qquad<br /> \mathrm{(E)}\ 2<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;math&gt;\spadesuit(2, 2) = 2 - \frac{1}{2} = \frac{3}{2}&lt;/math&gt;. Then, &lt;math&gt;\spadesuit(2, \frac{3}{2})&lt;/math&gt; is &lt;math&gt;2-\frac{1}{\frac{3}{2}} = 2- \frac{2}{3} = \frac{4}{3}&lt;/math&gt;<br /> The answer is &lt;math&gt;\boxed{C}&lt;/math&gt;<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=5|num-a=7}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_5&diff=44072 2010 AMC 10A Problems/Problem 5 2012-01-01T23:36:04Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 5 ==<br /> The area of a circle whose circumference is &lt;math&gt;24\pi&lt;/math&gt; is &lt;math&gt;k\pi&lt;/math&gt;. What is the value of &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 6<br /> \qquad<br /> \mathrm{(B)}\ 12<br /> \qquad<br /> \mathrm{(C)}\ 24<br /> \qquad<br /> \mathrm{(D)}\ 36<br /> \qquad<br /> \mathrm{(E)}\ 144<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> If the circumference of a circle is &lt;math&gt;24\pi&lt;/math&gt;, the radius would be &lt;math&gt;12&lt;/math&gt;. Since the area of a circle is &lt;math&gt;\pi r^2&lt;/math&gt;, the area is &lt;math&gt;144\pi&lt;/math&gt;. The answer is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=4|num-a=6}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_4&diff=44071 2010 AMC 10A Problems/Problem 4 2012-01-01T23:35:27Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 4 ==<br /> A book that is to be recorded onto compact discs takes &lt;math&gt;412&lt;/math&gt; minutes to read aloud. Each disc can hold up to &lt;math&gt;56&lt;/math&gt; minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 50.2<br /> \qquad<br /> \mathrm{(B)}\ 51.5<br /> \qquad<br /> \mathrm{(C)}\ 52.4<br /> \qquad<br /> \mathrm{(D)}\ 53.8<br /> \qquad<br /> \mathrm{(E)}\ 55.2<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Assuming that there were fractions of compact discs, it would take &lt;math&gt;412/56 ~= 7.357&lt;/math&gt; CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have &lt;math&gt;412/8 = 51.5&lt;/math&gt; minutes on each of the 8 discs. The answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=3|num-a=5}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_3&diff=44070 2010 AMC 10A Problems/Problem 3 2012-01-01T23:35:00Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 3 ==<br /> Tyrone had &lt;math&gt;97&lt;/math&gt; marbles and Eric had &lt;math&gt;11&lt;/math&gt; marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 3<br /> \qquad<br /> \mathrm{(B)}\ 13<br /> \qquad<br /> \mathrm{(C)}\ 18<br /> \qquad<br /> \mathrm{(D)}\ 25<br /> \qquad<br /> \mathrm{(E)}\ 29<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let &lt;math&gt;x&lt;/math&gt; be the number of marbles Tyrone gave to Eric. Then, &lt;math&gt;97-x = 2\cdot(11+x)&lt;/math&gt;. Solving for &lt;math&gt;x&lt;/math&gt; yields &lt;math&gt;75=3x&lt;/math&gt; and &lt;math&gt;x = 25&lt;/math&gt;. The answer is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|num-b=2|num-a=4}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_2&diff=44069 2010 AMC 10A Problems/Problem 2 2012-01-01T23:34:21Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 2 ==<br /> <br /> Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width? <br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(8mm);<br /> defaultpen(linewidth(.8pt));<br /> <br /> draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);<br /> draw((0,3)--(0,4)--(1,4)--(1,3)--cycle);<br /> draw((1,3)--(1,4)--(2,4)--(2,3)--cycle);<br /> draw((2,3)--(2,4)--(3,4)--(3,3)--cycle);<br /> draw((3,3)--(3,4)--(4,4)--(4,3)--cycle);<br /> <br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ \dfrac{5}{4}<br /> \qquad<br /> \mathrm{(B)}\ \dfrac{4}{3}<br /> \qquad<br /> \mathrm{(C)}\ \dfrac{3}{2}<br /> \qquad<br /> \mathrm{(D)}\ 2<br /> \qquad<br /> \mathrm{(E)}\ 3<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Let the length of the small square be &lt;math&gt;x&lt;/math&gt;, intuitively, the length of the big square is &lt;math&gt;4x&lt;/math&gt;. It can be seen that the width of the rectangle is &lt;math&gt;3x&lt;/math&gt;. Thus, the length of the rectangle is &lt;math&gt;4x/3x = 4/3&lt;/math&gt; times large as the width. The answer is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> <br /> == See also ==<br /> {{AMC10 box|year=2010|ab=A|num-b=1|num-a=3}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10A_Problems/Problem_1&diff=44068 2010 AMC 10A Problems/Problem 1 2012-01-01T23:33:04Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem 1 ==<br /> Mary’s top book shelf holds five books with the following widths, in centimeters: &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;\dfrac{1}{2}&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2.5&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;. What is the average book width, in centimeters?<br /> <br /> &lt;math&gt;<br /> \mathrm{(A)}\ 1<br /> \qquad<br /> \mathrm{(B)}\ 2<br /> \qquad<br /> \mathrm{(C)}\ 3<br /> \qquad<br /> \mathrm{(D)}\ 4<br /> \qquad<br /> \mathrm{(E)}\ 5<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> <br /> <br /> To find the average, we add up the widths &lt;math&gt;6&lt;/math&gt;, &lt;math&gt;\dfrac{1}{2}&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2.5&lt;/math&gt;, and &lt;math&gt;10&lt;/math&gt;, to get a total sum of &lt;math&gt;20&lt;/math&gt;. Since there are &lt;math&gt;5&lt;/math&gt; books, the average book width is &lt;math&gt;\frac{20}{5}=4&lt;/math&gt; The answer is &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2010|ab=A|before=First Question|num-a=2}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12A_Problems/Problem_4&diff=44057 2009 AMC 12A Problems/Problem 4 2012-01-01T19:31:55Z <p>Mariekitty: /* Solution */</p> <hr /> <div>{{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #4]] and [[2009 AMC 10A Problems|2009 AMC 10A #2]]}}<br /> <br /> == Problem ==<br /> Four coins are picked out of a piggy bank that contains a collection of pennies, nickels, dimes, and quarters. Which of the following could &lt;em&gt;not&lt;/em&gt; be the total value of the four coins, in cents?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 35 \qquad \textbf{(D)}\ 45 \qquad \textbf{(E)}\ 55&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> As all five options are divisible by &lt;math&gt;5&lt;/math&gt;, we may not use any pennies. (This is because a penny is the only coin that is not divisible by &lt;math&gt;5&lt;/math&gt;, and if we used between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;4&lt;/math&gt; pennies, the sum would not be divisible by &lt;math&gt;5&lt;/math&gt;.)<br /> <br /> Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is &lt;math&gt;4\cdot 5 = 20&lt;/math&gt;. Therefore the option that is not reachable is &lt;math&gt;\boxed{15}&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;(A)&lt;/math&gt;.<br /> <br /> We can verify that we can indeed get the other ones:<br /> * &lt;math&gt;25 = 10+5+5+5&lt;/math&gt;<br /> * &lt;math&gt;35 = 10+10+10+5&lt;/math&gt;<br /> * &lt;math&gt;45 = 25+10+5+5&lt;/math&gt;<br /> * &lt;math&gt;55 = 25+10+10+10&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2009|ab=A|num-b=3|num-a=5}}<br /> {{AMC10 box|year=2009|ab=A|num-b=1|num-a=3}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_15&diff=44035 2007 AMC 10A Problems/Problem 15 2011-12-31T16:33:16Z <p>Mariekitty: /* Alternate Solution */</p> <hr /> <div>==Problem==<br /> Four circles of radius &lt;math&gt;1&lt;/math&gt; are each tangent to two sides of a square and externally tangent to a circle of radius &lt;math&gt;2&lt;/math&gt;, as shown. What is the area of the square?<br /> <br /> [[Image:2007 AMC 10A -15 for wiki.png]]<br /> <br /> &lt;math&gt;\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> ----<br /> <br /> The diagonal has length &lt;math&gt;\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}&lt;/math&gt;. Therefore the sides have length &lt;math&gt;2+3\sqrt{2}&lt;/math&gt;, and the area is<br /> <br /> &lt;cmath&gt;A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> === Solution 2 === <br /> <br /> ----<br /> <br /> <br /> Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides &lt;math&gt;3, 3, 3\sqrt{2}&lt;/math&gt;. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then &lt;cmath&gt; A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2007|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_15&diff=44034 2007 AMC 10A Problems/Problem 15 2011-12-31T16:32:45Z <p>Mariekitty: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> Four circles of radius &lt;math&gt;1&lt;/math&gt; are each tangent to two sides of a square and externally tangent to a circle of radius &lt;math&gt;2&lt;/math&gt;, as shown. What is the area of the square?<br /> <br /> [[Image:2007 AMC 10A -15 for wiki.png]]<br /> <br /> &lt;math&gt;\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> <br /> ----<br /> <br /> The diagonal has length &lt;math&gt;\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}&lt;/math&gt;. Therefore the sides have length &lt;math&gt;2+3\sqrt{2}&lt;/math&gt;, and the area is<br /> <br /> &lt;cmath&gt;A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> == Alternate Solution == <br /> <br /> Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides &lt;math&gt;3, 3, 3\sqrt{2}&lt;/math&gt;. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then &lt;cmath&gt; A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2007|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_10A_Problems/Problem_15&diff=44033 2007 AMC 10A Problems/Problem 15 2011-12-31T16:32:20Z <p>Mariekitty: /* Solution */</p> <hr /> <div>==Problem==<br /> Four circles of radius &lt;math&gt;1&lt;/math&gt; are each tangent to two sides of a square and externally tangent to a circle of radius &lt;math&gt;2&lt;/math&gt;, as shown. What is the area of the square?<br /> <br /> [[Image:2007 AMC 10A -15 for wiki.png]]<br /> <br /> &lt;math&gt;\text{(A)}\ 32 \qquad \text{(B)}\ 22 + 12\sqrt {2}\qquad \text{(C)}\ 16 + 16\sqrt {3}\qquad \text{(D)}\ 48 \qquad \text{(E)}\ 36 + 16\sqrt {2}&lt;/math&gt;<br /> <br /> ===Solution 1===<br /> The diagonal has length &lt;math&gt;\sqrt{2}+1+2+2+1+\sqrt{2}=6+2\sqrt{2}&lt;/math&gt;. Therefore the sides have length &lt;math&gt;2+3\sqrt{2}&lt;/math&gt;, and the area is<br /> <br /> &lt;cmath&gt;A=(2+3\sqrt{2})^2=4+6\sqrt{2}+6\sqrt{2}+18=22+12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> == Alternate Solution == <br /> <br /> Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides &lt;math&gt;3, 3, 3\sqrt{2}&lt;/math&gt;. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then &lt;cmath&gt; A = (2+3\sqrt{2})^2 = 22 + 12\sqrt{2} \Rightarrow \text{(B)}&lt;/cmath&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2007|ab=A|num-b=14|num-a=16}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_2&diff=44011 2004 AMC 10B Problems/Problem 2 2011-12-29T23:41:15Z <p>Mariekitty: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> How many two-digit positive integers have at least one 7 as a digit?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30 &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Ten numbers (&lt;math&gt;70,71,\dots,79&lt;/math&gt;) have &lt;math&gt;7&lt;/math&gt; as the tens digit. Nine numbers (&lt;math&gt;17,27,\dots,97&lt;/math&gt;) have it as the ones digit. Number &lt;math&gt;77&lt;/math&gt; is in both sets.<br /> <br /> Thus the result is &lt;math&gt;10+9-1=18 \Rightarrow&lt;/math&gt; &lt;math&gt; \boxed{(B)}&lt;/math&gt;.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=1|num-a=3}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_1&diff=44010 2004 AMC 10B Problems/Problem 1 2011-12-29T23:37:57Z <p>Mariekitty: /* Solution */</p> <hr /> <div>==Problem==<br /> Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 297 \qquad \mathrm{(B) \ } 330\qquad \mathrm{(C) \ } 363\qquad \mathrm{(D) \ } 396\qquad \mathrm{(E) \ } 726 &lt;/math&gt;<br /> <br /> ==Solution==<br /> There are &lt;math&gt;22-12+1=11&lt;/math&gt; rows of &lt;math&gt;33&lt;/math&gt; seats, giving &lt;math&gt;11\times 33=\boxed{(C) 363}&lt;/math&gt; seats.<br /> <br /> == See also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|before=First Question|num-a=2}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12A_Problems/Problem_8&diff=44008 2004 AMC 12A Problems/Problem 8 2011-12-29T20:43:08Z <p>Mariekitty: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2004 AMC 12A Problems|2004 AMC 12A #8]] and [[2004 AMC 10A Problems/Problem 9|2004 AMC 10A #9]]}}<br /> == Problem ==<br /> In the overlapping [[triangle]]s &lt;math&gt;\triangle{ABC}&lt;/math&gt; and &lt;math&gt;\triangle{ABE}&lt;/math&gt; sharing common [[edge | side]] &lt;math&gt;AB&lt;/math&gt;, &lt;math&gt;\angle{EAB}&lt;/math&gt; and &lt;math&gt;\angle{ABC}&lt;/math&gt; are [[right angle]]s, &lt;math&gt;AB=4&lt;/math&gt;, &lt;math&gt;BC=6&lt;/math&gt;, &lt;math&gt;AE=8&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BE}&lt;/math&gt; intersect at &lt;math&gt;D&lt;/math&gt;. What is the difference between the areas of &lt;math&gt;\triangle{ADE}&lt;/math&gt; and &lt;math&gt;\triangle{BDC}&lt;/math&gt;? <br /> <br /> [[Image:AMC10_2004A_9.gif|center]]<br /> <br /> &lt;math&gt;\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad&lt;/math&gt;<br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> If we let &lt;math&gt;[\ldots]&lt;/math&gt; denote area, &lt;math&gt;[ABE] - [ABC] = [ADE] + [ABD] - [ABD] - [BDC] = [ADE] - [BDC]&lt;/math&gt;. Using the given, &lt;math&gt;[ABE] = \frac 12 \cdot 8 \cdot 4&lt;/math&gt; and &lt;math&gt;[ABC] = \frac 12 \cdot 6 \cdot 4&lt;/math&gt;, and their difference is &lt;math&gt;16 - 12 = 4\ \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Since &lt;math&gt;AE \perp AB&lt;/math&gt; and &lt;math&gt;BC \perp AB&lt;/math&gt;, &lt;math&gt;AE \parallel BC&lt;/math&gt;. By [[alternate interior angles]] and AA~, we find that &lt;math&gt;\triangle ADE \sim \triangle CDB&lt;/math&gt;, with side length ratio &lt;math&gt;\frac{4}{3}&lt;/math&gt;. Their heights also have the same ratio, and since the two heights add up to &lt;math&gt;4&lt;/math&gt;, we have that &lt;math&gt;h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}&lt;/math&gt; and &lt;math&gt;h_{CDB} = 3 \cdot \frac 47 = \frac {12}7&lt;/math&gt;. Subtracting the areas, &lt;math&gt;\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{(B)}&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131320 AoPS topic]<br /> {{AMC12 box|year=2004|ab=A|num-b=7|num-a=9}}<br /> {{AMC10 box|year=2004|ab=A|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_23&diff=44006 2003 AMC 10B Problems/Problem 23 2011-12-29T18:12:48Z <p>Mariekitty: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A regular octagon &lt;math&gt; ABCDEFGH &lt;/math&gt; has an area of one square unit. What is the area of the rectangle &lt;math&gt; ABEF &lt;/math&gt;?<br /> <br /> &lt;asy&gt; unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label(&quot;$A$&quot;,A,NNW); label(&quot;$B$&quot;,B,NNE); label(&quot;$C$&quot;,C,ENE); label(&quot;$D$&quot;,D,ESE); label(&quot;$E$&quot;,E,SSE); label(&quot;$F$&quot;,F,SSW); label(&quot;$G$&quot;,G,WSW); label(&quot;$H$&quot;,H,WNW);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> Here is an easy way to look at this, where &lt;math&gt;p&lt;/math&gt; is the perimeter, and &lt;math&gt;a&lt;/math&gt; is the [[apothem]]:<br /> <br /> Area of Octagon: &lt;math&gt; \frac{ap}{2}=1 &lt;/math&gt;.<br /> <br /> Area of Rectangle: &lt;math&gt; \frac{p}{8}\times 2a=\frac{ap}{4} &lt;/math&gt;.<br /> <br /> You can see from this that the octagon's area is twice as large as the rectangle's area is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{1}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_23&diff=44005 2003 AMC 10B Problems/Problem 23 2011-12-29T18:06:57Z <p>Mariekitty: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> A regular octagon &lt;math&gt; ABCDEFGH &lt;/math&gt; has an area of one square unit. What is the area of the rectangle &lt;math&gt; ABEF &lt;/math&gt;?<br /> <br /> &lt;asy&gt; unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label(&quot;$A$&quot;,A,NNW); label(&quot;$B$&quot;,B,NNE); label(&quot;$C$&quot;,C,ENE); label(&quot;$D$&quot;,D,ESE); label(&quot;$E$&quot;,E,SSE); label(&quot;$F$&quot;,F,SSW); label(&quot;$G$&quot;,G,WSW); label(&quot;$H$&quot;,H,WNW);&lt;/asy&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4} &lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> An easy way to look at this:<br /> <br /> Area of Octagon: &lt;math&gt; \frac{ap}{2}=1 &lt;/math&gt;.<br /> <br /> Area of Rectangle: &lt;math&gt; \frac{p}{8}\times 2a=\frac{ap}{4} &lt;/math&gt;.<br /> <br /> You can see from this that the octagon's area is twice as large as the rectangle's area is &lt;math&gt;\boxed{\textbf{(D)}\ \frac{1}{2}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=22|num-a=24}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_15&diff=44004 2003 AMC 10B Problems/Problem 15 2011-12-29T17:54:30Z <p>Mariekitty: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> There are &lt;math&gt;100&lt;/math&gt; players in a single tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest &lt;math&gt;28&lt;/math&gt; players are given a bye, and the remaining &lt;math&gt;72&lt;/math&gt; players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is<br /> <br /> &lt;math&gt;\textbf{(A) } \text{a prime number} <br /> <br /> \qquad\textbf{(B) } \text{divisible by 2} <br /> <br /> \qquad\textbf{(C) } \text{divisible by 5} <br /> <br /> \qquad\textbf{(D) } \text{divisible by 7} <br /> <br /> \qquad\textbf{(E) } \text{divisible by 11}&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> 28 people receive byes, so in the first round there are &lt;math&gt;36&lt;/math&gt; matches played. In the second round there are &lt;math&gt;28 + 36 = 64&lt;/math&gt; people, so there are 32 matches. In the subsequent rounds, there are &lt;math&gt;16, 8, 4, 2, 1&lt;/math&gt; matches played, for a total of &lt;math&gt;36 + 32 + 16 + 8 + 4 + 2 + 1 = 99&lt;/math&gt; matches. Divisible by 11 &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{(E)} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2003|ab=B|num-b=14|num-a=16}}<br /> [[Category:Introductory Number Theory Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_12A_Problems/Problem_1&diff=43973 2003 AMC 12A Problems/Problem 1 2011-12-28T21:04:41Z <p>Mariekitty: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2003 AMC 12A Problems|2003 AMC 12A #1]] and [[2003 AMC 10A Problems|2003 AMC 10A #1]]}}<br /> == Problem ==<br /> What is the difference between the sum of the first &lt;math&gt;2003&lt;/math&gt; even counting numbers and the sum of the first &lt;math&gt;2003&lt;/math&gt; odd counting numbers? <br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4006 &lt;/math&gt;<br /> <br /> == Solution ==<br /> ===Solution 1===<br /> <br /> The first &lt;math&gt;2003&lt;/math&gt; even counting numbers are &lt;math&gt;2,4,6,...,4006&lt;/math&gt;. <br /> <br /> The first &lt;math&gt;2003&lt;/math&gt; odd counting numbers are &lt;math&gt;1,3,5,...,4005&lt;/math&gt;. <br /> <br /> Thus, the problem is asking for the value of &lt;math&gt;(2+4+6+...+4006)-(1+3+5+...+4005)&lt;/math&gt;. <br /> <br /> &lt;math&gt;(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005) &lt;/math&gt; <br /> <br /> &lt;math&gt;= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;<br /> <br /> ===Solution 2===<br /> Using the sum of an [[arithmetic progression]] formula, we can write this as &lt;math&gt;\frac{2003}{2}(2 + 4006) - \frac{2003}{2}(1 + 4005) = \frac{2003}{2} \cdot 2 = \boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;.<br /> <br /> <br /> <br /> ===Solution 3===<br /> The formula for the sum of the first &lt;math&gt;n&lt;/math&gt; even numbers, is &lt;math&gt;S_E=n^{2}+n&lt;/math&gt;, (E standing for even).<br /> <br /> Sum of first &lt;math&gt;n&lt;/math&gt; odd numbers, is &lt;math&gt;S_O=n^{2}&lt;/math&gt;, (O standing for odd).<br /> <br /> Knowing this, plug &lt;math&gt;2003&lt;/math&gt; for &lt;math&gt;n&lt;/math&gt;, <br /> <br /> &lt;math&gt;S_E-S_O= (2003^{2}+2003)-(2003^{2})=2003 \Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\mathrm{(D)}\ 2003}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC10 box|year=2003|ab=A|before=First Question|num-a=2}}<br /> {{AMC12 box|year=2003|ab=A|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_19&diff=43972 2002 AMC 10B Problems/Problem 19 2011-12-28T20:02:53Z <p>Mariekitty: /* See Also */</p> <hr /> <div>== Problem ==<br /> <br /> Suppose that &lt;math&gt;\{a_n\}&lt;/math&gt; is an arithmetic sequence with<br /> &lt;cmath&gt; a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.&lt;/cmath&gt;<br /> What is the value of &lt;math&gt;a_2 - a_1 ?&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1 &lt;/math&gt;<br /> <br /> == Solution ==<br /> Adding the two given equations together gives <br /> <br /> &lt;math&gt; a_1+a_2+...+a_{200}=300 &lt;/math&gt;.<br /> <br /> Now, let the common difference be &lt;math&gt; d &lt;/math&gt;. Notice that &lt;math&gt; a_2-a_1=d &lt;/math&gt;, so we merely need to find &lt;math&gt; d &lt;/math&gt; to get the answer. The formula for an arithmetic sum is <br /> <br /> &lt;math&gt; \frac{n}{2}(2a_1+d(n-1)) &lt;/math&gt;,<br /> <br /> where &lt;math&gt; a_1 &lt;/math&gt; is the first term, &lt;math&gt; n &lt;/math&gt; is the number of terms, and &lt;math&gt; d &lt;/math&gt; is the common difference. Now we use this formula to find a closed form for the first given equation and the sum of the given equations. For the first equation, we have &lt;math&gt; n=100 &lt;/math&gt;. Therefore, we have <br /> <br /> &lt;math&gt; 50(2a_1+99d)=100 &lt;/math&gt;,<br /> <br /> or <br /> <br /> &lt;math&gt; 2a_1+99d=2 &lt;/math&gt;. *'''(1)'''<br /> <br /> For the sum of the equations (shown at the beginning of the solution) we have &lt;math&gt; n=200 &lt;/math&gt;, so<br /> <br /> &lt;math&gt; 100(2a_1+199d)=300 &lt;/math&gt;<br /> <br /> or <br /> <br /> &lt;math&gt; 2a_1+199d=3 &lt;/math&gt; *'''(2)'''<br /> <br /> Now we have a system of equations in terms of &lt;math&gt; a_1 &lt;/math&gt; and &lt;math&gt; d &lt;/math&gt;.<br /> Subtracting '''(1)''' from '''(2)''' eliminates &lt;math&gt; a_1 &lt;/math&gt;, yielding &lt;math&gt; 100d=1 &lt;/math&gt;, and &lt;math&gt; d=a_2-a_1=\frac{1}{100}=.01, \boxed{\text{C}} &lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2002|ab=B|num-b=18|num-a=20}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_17&diff=43971 2002 AMC 10B Problems/Problem 17 2011-12-28T19:54:16Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> A regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; has sides of length two. Find the area of &lt;math&gt;\triangle ADG&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> &lt;asy&gt;<br /> unitsize(1cm);<br /> defaultpen(0.8);<br /> pair[] A = new pair;<br /> A=(0,0);<br /> for (int i=1; i&lt;8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1));<br /> draw( A--A--A--A--A--A--A--A--cycle );<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,A,SE);<br /> label(&quot;$C$&quot;,A,SE);<br /> label(&quot;$D$&quot;,A,NE);<br /> label(&quot;$E$&quot;,A,NE);<br /> label(&quot;$F$&quot;,A,NW);<br /> label(&quot;$G$&quot;,A,NW);<br /> label(&quot;$H$&quot;,A,SW);<br /> filldraw( A--A--A--cycle, lightgray, black );<br /> pair P = intersectionpoint( A--A, A--A );<br /> draw( A--P );<br /> draw( P -- A, dashed );<br /> label(&quot;$P$&quot;,P,NE);<br /> draw( A--A, dashed );<br /> pair Q = intersectionpoint( A--A, A--A );<br /> label(&quot;$Q$&quot;,Q,NW);<br /> &lt;/asy&gt;<br /> <br /> The area of the triangle &lt;math&gt;ADG&lt;/math&gt; can be computed as &lt;math&gt;\frac{DG \cdot AP}2&lt;/math&gt;. We will now find &lt;math&gt;DG&lt;/math&gt; and &lt;math&gt;AP&lt;/math&gt;.<br /> <br /> Clearly, &lt;math&gt;PFG&lt;/math&gt; is a right isosceles triangle with hypotenuse of lenght &lt;math&gt;2&lt;/math&gt;, hence &lt;math&gt;PG=\sqrt 2&lt;/math&gt;. <br /> The same holds for triangle &lt;math&gt;QED&lt;/math&gt; and its leg &lt;math&gt;QD&lt;/math&gt;. The length of &lt;math&gt;PQ&lt;/math&gt; is equal to &lt;math&gt;FE=2&lt;/math&gt;.<br /> Hence &lt;math&gt;GD = 2 + 2\sqrt 2&lt;/math&gt;, and &lt;math&gt;AP = PD = 2 + \sqrt 2&lt;/math&gt;.<br /> <br /> Then the area of &lt;math&gt;ADG&lt;/math&gt; equals &lt;math&gt;\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = 4+3\sqrt 2 \Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{(C)}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}}</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_14&diff=43970 2002 AMC 10B Problems/Problem 14 2011-12-28T19:46:54Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The number &lt;math&gt;25^{64}\cdot 64^{25}&lt;/math&gt; is the square of a positive integer &lt;math&gt;N&lt;/math&gt;. In decimal representation, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Since, &lt;math&gt;N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}&lt;/math&gt;.<br /> <br /> Combing the &lt;math&gt;2&lt;/math&gt;'s and &lt;math&gt;5&lt;/math&gt;'s gives us, &lt;math&gt;(2\cdot 5)^{64}\cdot 2^{(75-64)}=(2\cdot 5)^{64}\cdot 2^{11}=10^{64}\cdot 2^{11}&lt;/math&gt;. <br /> <br /> This is &lt;math&gt;2048&lt;/math&gt; with sixty-four, &lt;math&gt;0&lt;/math&gt;'s on the end. So, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;2+4+8=14\Longrightarrow\mathrm{ (B) \ }&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC10 box|year=2002|ab=B|num-b=13|num-a=15}}<br /> <br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=User:Mariekitty&diff=43969 User:Mariekitty 2011-12-28T19:37:11Z <p>Mariekitty: /* Mariekitty's AoPSWiki :) */</p> <hr /> <div>= Mariekitty's AoPSWiki :) =<br /> <br /> ==Bio==<br /> *Apparently i have an AoPSwiki, i just found out randomly......<br /> Umm, i'll start by telling you more about myself. I am a student that lives in MA, <br /> I'm athletic, funny and outgoing when you really get to know me.<br /> For my spare time, I like to do math, swim and read. <br /> <br /> ==My Blog==<br /> *This is my blog on AoPS, [http://www.artofproblemsolving.com/Forum/blog.php?u=83394 feel free to visit]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems/Problem_14&diff=43965 2002 AMC 10B Problems/Problem 14 2011-12-28T19:33:45Z <p>Mariekitty: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The number &lt;math&gt;25^{64}\cdot 64^{25}&lt;/math&gt; is the square of a positive integer &lt;math&gt;N&lt;/math&gt;. In decimal representation, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35 &lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Since, &lt;math&gt;N=5^{64}\cdot 8^{25}=5^{64}\cdot (2^{3})^{25}=5^{64}\cdot 2^{75}&lt;/math&gt;.<br /> <br /> Combing the &lt;math&gt;2&lt;/math&gt;'s and &lt;math&gt;5&lt;/math&gt;'s gives us, &lt;math&gt;(2\cdot 5)^{64}\cdot 2^{(75-64)}=(2\cdot 5)^{64}\cdot 2^{11}=10^{64}\cdot 2^{11}&lt;/math&gt;. <br /> <br /> This is &lt;math&gt;2048&lt;/math&gt; with sixty-four, &lt;math&gt;0&lt;/math&gt;'s on the end. So, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is &lt;math&gt;2+4+8=14\Longrightarrow\mathrm{ (B) \ }&lt;/math&gt;</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems&diff=43954 2002 AMC 10B Problems 2011-12-28T17:15:24Z <p>Mariekitty: /* Problem 8 */</p> <hr /> <div>==Problem 1==<br /> The ratio &lt;math&gt;\frac{2^{2001}\cdot3^{2003}}{6^{2002}}&lt;/math&gt; is:<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{3}\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } \frac{2}{3}\qquad \mathrm{(E) \ } \frac{3}{2} &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> == Problem 2 ==<br /> For the nonzero numbers &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c,&lt;/math&gt; define<br /> &lt;cmath&gt;(a,b,c)=\frac{abc}{a+b+c}&lt;/cmath&gt;<br /> Find &lt;math&gt;(2,4,6)&lt;/math&gt;.<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> == Problem 3 ==<br /> The arithmetic mean of the nine numbers in the set &lt;math&gt;\{9,99,999,9999,\ldots,999999999\}&lt;/math&gt; is a &lt;math&gt;9&lt;/math&gt;-digit number &lt;math&gt;M&lt;/math&gt;, all of whose digits are distinct. The number &lt;math&gt;M&lt;/math&gt; does not contain the digit<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> == Problem 4 ==<br /> <br /> What is the value of<br /> <br /> &lt;cmath&gt;(3x-2)(4x+1)-(3x-2)4x+1&lt;/cmath&gt;<br /> <br /> when &lt;math&gt;x=4&lt;/math&gt;?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> == Problem 5 ==<br /> <br /> Circles of radius &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.<br /> <br /> &lt;center&gt;&lt;asy&gt;<br /> unitsize(5mm);<br /> defaultpen(linewidth(.8pt)+fontsize(10pt));<br /> dotfactor=4;<br /> <br /> real r1=3; real r2=2; real r3=5;<br /> pair A=(-2,0), B=(3,0), C=(0,0);<br /> pair X=(1,0), Y=(5,0);<br /> path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r3);<br /> fill(circleC,gray);<br /> fill(circleA,white);<br /> fill(circleB,white);<br /> draw(circleA); draw(circleB); draw(circleC);<br /> draw(A--X); draw(B--Y);<br /> <br /> pair[] ps={A,B}; dot(ps);<br /> <br /> label(&quot;$3$&quot;,midpoint(A--X),N);<br /> label(&quot;$2$&quot;,midpoint(B--Y),N);<br /> &lt;/asy&gt;&lt;/center&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 3\pi\qquad \mathrm{(B) \ } 4\pi\qquad \mathrm{(C) \ } 6\pi\qquad \mathrm{(D) \ } 9\pi\qquad \mathrm{(E) \ } 12\pi &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> == Problem 6 ==<br /> <br /> For how many positive integers &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n^2-3n+2&lt;/math&gt; a prime number?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \text{none}\qquad \mathrm{(B) \ } \text{one}\qquad \mathrm{(C) \ } \text{two}\qquad \mathrm{(D) \ } \text{more than two, but finitely many}\qquad \mathrm{(E) \ } \text{infinitely many} &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> == Problem 7 ==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer such that &lt;math&gt;\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}&lt;/math&gt; is an integer. Which of the following statements is '''not''' true?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 2\text{ divides }n\qquad \mathrm{(B) \ } 3\text{ divides }n\qquad \mathrm{(C) \ } 6\text{ divides }n\qquad \mathrm{(D) \ } 7\text{ divides }n\qquad \mathrm{(E) \ } n&gt;84 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> == Problem 8 ==<br /> <br /> Suppose July of year &lt;math&gt;N&lt;/math&gt; has five Mondays. Which of the following must occur five times in the August of year &lt;math&gt;N&lt;/math&gt;? (Note: Both months have &lt;math&gt;31&lt;/math&gt; days.)<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{Monday} \qquad \textbf{(B)}\ \text{Tuesday} \qquad \textbf{(C)}\ \text{Wednesday} \qquad \textbf{(D)}\ \text{Thursday} \qquad \textbf{(E)}\ \text{Friday}&lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> == Problem 9 ==<br /> <br /> Using the letters &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;M&lt;/math&gt;, &lt;math&gt;O&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt;, and &lt;math&gt;U&lt;/math&gt;, we can form five-letter &quot;words&quot;. If these &quot;words&quot; are arranged in alphabetical order, then the &quot;word&quot; &lt;math&gt;USAMO&lt;/math&gt; occupies position<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> == Problem 10 ==<br /> <br /> Suppose that &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are nonzero real numbers, and that the equation &lt;math&gt;x^2+ax+b=0&lt;/math&gt; has positive solutions &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Then the pair &lt;math&gt;(a,b)&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } (-2,1)\qquad \mathrm{(B) \ } (-1,2)\qquad \mathrm{(C) \ } (1,-2)\qquad \mathrm{(D) \ } (2,-1)\qquad \mathrm{(E) \ } (4,4) &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> == Problem 11 ==<br /> <br /> The product of three consecutive positive integers is &lt;math&gt;8&lt;/math&gt; times their sum. What is the sum of the squares?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 50\qquad \mathrm{(B) \ } 77\qquad \mathrm{(C) \ } 110\qquad \mathrm{(D) \ } 149\qquad \mathrm{(E) \ } 194 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> == Problem 12 ==<br /> <br /> For which of the following values of &lt;math&gt;k&lt;/math&gt; does the equation &lt;math&gt;\frac{x-1}{x-2} = \frac{x-k}{x-6}&lt;/math&gt; have no solution for &lt;math&gt;x&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 /> [[2002 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> == Problem 13==<br /> <br /> Find the value(s) of &lt;math&gt;x&lt;/math&gt; such that &lt;math&gt;8xy - 12y + 2x - 3 = 0&lt;/math&gt; is true for all values of &lt;math&gt;y&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac32 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14&lt;/math&gt;<br /> <br /> <br /> [[2002 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> == Problem 14 ==<br /> <br /> The number &lt;math&gt;25^{64}\cdot 64^{25}&lt;/math&gt; is the square of a positive integer &lt;math&gt;N&lt;/math&gt;. In decimal representation, the sum of the digits of &lt;math&gt;N&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> == Problem 15 ==<br /> <br /> The positive integers &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;A-B&lt;/math&gt;, and &lt;math&gt;A+B&lt;/math&gt; are all prime numbers. The sum of these four primes is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \text{even}\qquad \mathrm{(B) \ } \text{divisible by }3\qquad \mathrm{(C) \ } \text{divisible by }5\qquad \mathrm{(D) \ } \text{divisible by }7\qquad \mathrm{(E) \ } \text{prime}&lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> == Problem 16 ==<br /> <br /> For how many integers &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;\frac{n}{20-n}&lt;/math&gt; the square of an integer?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 10&lt;/math&gt;<br /> <br /> <br /> [[2002 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> == Problem 17 ==<br /> <br /> A regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; has sides of length two. Find the area of &lt;math&gt;\triangle ADG&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2&lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> == Problem 18 ==<br /> <br /> Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?<br /> <br /> &lt;math&gt;\textbf{(A) } 8\qquad \textbf{(B) } 9\qquad \textbf{(C) } 10\qquad \textbf{(D) } 12\qquad \textbf{(E) } 16&lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> == Problem 19 ==<br /> <br /> Suppose that &lt;math&gt;\{a_n\}&lt;/math&gt; is an arithmetic sequence with<br /> &lt;cmath&gt; a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.&lt;/cmath&gt;<br /> What is the value of &lt;math&gt;a_2 - a_1 ?&lt;/math&gt;<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> == Problem 20 ==<br /> <br /> Let &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; be real numbers such that &lt;math&gt;a-7b+8c=4&lt;/math&gt; and &lt;math&gt;8a+4b-c=7.&lt;/math&gt; Then &lt;math&gt;a^2-b^2+c^2&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> == Problem 21 ==<br /> <br /> Andy's lawn has twice as much area as Beth's lawn and three times as much as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } \text{Andy}\qquad \mathrm{(B) \ } \text{Beth}\qquad \mathrm{(C) \ } \text{Carlos}\qquad \mathrm{(D) \ } \text{Andy and Carlos tie for first.}\qquad \mathrm{(E) \ } \text{All three tie.} &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> == Problem 22 ==<br /> <br /> Let &lt;math&gt;\triangle{XOY}&lt;/math&gt; be a right-angled triangle with &lt;math&gt;m\angle{XOY}=90^\circ&lt;/math&gt;. Let &lt;math&gt;M&lt;/math&gt; and &lt;math&gt;N&lt;/math&gt; be the midpoints of the legs &lt;math&gt;OX&lt;/math&gt; and &lt;math&gt;OY&lt;/math&gt;, respectively. Given &lt;math&gt;XN=19&lt;/math&gt; and &lt;math&gt;YM=22&lt;/math&gt;, find &lt;math&gt;XY&lt;/math&gt;.<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> == Problem 23 ==<br /> <br /> Let &lt;math&gt;\{a_k\}&lt;/math&gt; be a sequence of integers such that &lt;math&gt;a_1=1&lt;/math&gt; and &lt;math&gt;a_{m+n}=a_m+a_n+mn,&lt;/math&gt; for all positive integers &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n.&lt;/math&gt; Then &lt;math&gt;a_{12}&lt;/math&gt; is<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> == Problem 24 ==<br /> <br /> Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius &lt;math&gt;20&lt;/math&gt; feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point &lt;math&gt;10&lt;/math&gt; vertical feet above the bottom?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> == Problem 25 ==<br /> <br /> When &lt;math&gt;15&lt;/math&gt; is appended to a list of integers, the mean is increased by &lt;math&gt;2&lt;/math&gt;. When &lt;math&gt;1&lt;/math&gt; is appended to the enlarged list, the mean of the enlarged list is decreased by &lt;math&gt;1&lt;/math&gt;. How many integers were in the original list?<br /> <br /> &lt;math&gt; \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 &lt;/math&gt;<br /> <br /> [[2002 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> * [[AMC Problems and Solutions]]</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=User:Mariekitty&diff=43758 User:Mariekitty 2011-12-18T01:20:56Z <p>Mariekitty: /* Mariekitty's AoPSWiki :) */</p> <hr /> <div>== Mariekitty's AoPSWiki :) ==<br /> <br /> &lt;nowiki&gt;Apparently i have an AoPSwiki, i just found out randomly......<br /> <br /> Umm, i'll start by telling you more about myself. I am a student that lives in MA, <br /> I'm athletic, funny and outgoing when you really get to know me.<br /> For my spare time, I like to do math, swim and read. &lt;/nowiki&gt;<br /> <br /> This is my blog on AoPS, feel free to visit: [http://www.artofproblemsolving.com/Forum/blog.php?u=83394]<br /> --[[User:Mariekitty|Mariekitty]] 18:14, 18 August 2010 (UTC)</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=User:Mariekitty&diff=35681 User:Mariekitty 2010-08-18T18:15:46Z <p>Mariekitty: /* Mariekitty's AoPSWiki :) */</p> <hr /> <div>== Mariekitty's AoPSWiki :) ==<br /> <br /> &lt;nowiki&gt;Apparently i have an AoPSwiki, i just found out randomly......<br /> <br /> Umm, i'll start by telling you more about myself. I am a student that lives in MA, <br /> I'm athletic, funny and outgoing when you really get to know me.<br /> For my spare time, I like to do math, swim and read. &lt;/nowiki&gt;<br /> <br /> This is my blog on AoPS, feel free to visit: [http://www.artofproblemsolving.com/Forum/blog.php?u=83394]<br /> --[[User:Mariekitty|Mariekitty]] 18:14, 18 August 2010 (UTC)<br /> :)</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=User:Mariekitty&diff=35680 User:Mariekitty 2010-08-18T18:14:41Z <p>Mariekitty: /* Mariekitty's AoPSWiki :) */</p> <hr /> <div>== Mariekitty's AoPSWiki :) ==<br /> <br /> Apparently i have an AoPSwiki, i just found out randomly......<br /> <br /> Umm, i'll start by telling you more about myself. I am a student that lives in MA, <br /> I'm athletic, funny and outgoing when you really get to know me.<br /> For my spare time, I like to do math, swim and read<br /> <br /> This is my blog on AoPS, feel free to visit: [http://www.artofproblemsolving.com/Forum/blog.php?u=83394]<br /> --[[User:Mariekitty|Mariekitty]] 18:14, 18 August 2010 (UTC)</div> Mariekitty https://artofproblemsolving.com/wiki/index.php?title=User:Mariekitty&diff=35678 User:Mariekitty 2010-08-18T18:08:35Z <p>Mariekitty: Created page with '== Mariekitty's AoPSWiki :) == Apparently i have an AoPSwiki, i just found out randomly...... what exactly are we supposed to do here??? hmhmhmmmmmm, i wander if we're allowed…'</p> <hr /> <div>== Mariekitty's AoPSWiki :) ==<br /> <br /> Apparently i have an AoPSwiki, i just found out randomly......<br /> what exactly are we supposed to do here???<br /> <br /> <br /> hmhmhmmmmmm, i wander if we're allowed to visit other people's wiki's, huh.....</div> Mariekitty