https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Siddaops2013&feedformat=atom AoPS Wiki - User contributions [en] 2022-01-28T12:27:07Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_1&diff=92785 2004 AIME I Problems/Problem 1 2018-03-07T14:06:56Z <p>Siddaops2013: /* Solution */</p> <hr /> <div>== Problem ==<br /> The digits of a positive integer &lt;math&gt; n &lt;/math&gt; are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when &lt;math&gt; n &lt;/math&gt; is divided by &lt;math&gt;37&lt;/math&gt;?<br /> <br /> == Solution ==<br /> A brute-force solution to this question is fairly quick, but we'll try something slightly more clever: our numbers have the form &lt;math&gt;{\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}} &lt;/math&gt;&lt;math&gt;= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n&lt;/math&gt;, for &lt;math&gt;n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace&lt;/math&gt;.<br /> <br /> Now, note that &lt;math&gt;3\cdot 37 = 111&lt;/math&gt; so &lt;math&gt;30 \cdot 37 = 1110&lt;/math&gt;, and &lt;math&gt;90 \cdot 37 = 3330&lt;/math&gt; so &lt;math&gt;87 \cdot 37 = 3219&lt;/math&gt;. So the [[remainder]]s are all congruent to &lt;math&gt;n - 9 \pmod{37}&lt;/math&gt;. However, these numbers are negative for our choices of &lt;math&gt;n&lt;/math&gt;, so in fact the remainders must equal &lt;math&gt;n + 28&lt;/math&gt;.<br /> <br /> Adding these numbers up, we get &lt;math&gt;(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7)\cdot28 = \boxed{217}&lt;/math&gt;, our answer.<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Siddaops2013 https://artofproblemsolving.com/wiki/index.php?title=Greedy_algorithm&diff=81593 Greedy algorithm 2016-11-24T21:44:59Z <p>Siddaops2013: </p> <hr /> <div>In [[mathematics]] and [[computer science]], a &lt;b&gt;greedy algorithm&lt;/b&gt; is one that selects for the maximal immediate benefit, without regard for how this selection affects future choices.<br /> <br /> =Introduction=<br /> As with all [[algorithm|algorithms]], greedy algorithms seek to maximize the overall utility of some process. They operate by making the immediately optimal choice at each sub-stage of the process, hoping that this will maximize the utility of the entire process. More formally, when we reframe the problem in terms of forming a [[set]] with a desired property, at each step a greedy algorithm will add the element into the [[set]] if and only it does not cause the [[set]] to lose the desired property.<br /> <br /> Greedy algorithms are among the simplest types of [[algorithm|algorithms]]; as such, they are among the first examples taught when demonstrating the subject. They have the advantage of being ruthlessly efficient, when correct, and they are usually among the most natural approaches to a problem. For this reason, they are often referred to as &quot;naïve methods&quot;. In many cases, more complicated [[algorithm|algorithms]] are formed by adjusting the greedy process to be correct, often through the use of clever [[sorting]]. <br /> <br /> =Examples of greedy algorithms=<br /> Many real-life scenarios are good examples of greedy algorithms. For example, consider the problem of converting an arbitrary number of cents into standard coins; in other words, consider the problem of making change. The process you almost certainly follow, without consciously considering it, is first using the largest number of quarters you can, then the largest number of dimes, then nickels, then pennies. This is an example of working &lt;b&gt;greedily&lt;/b&gt;: at each step, we chose the maximal immediate benefit (number of coins we could give).<br /> <br /> Some problems are not so obviously algorithmic processes. For example, the [[Rearrangement inequality]] states that if &lt;math&gt;(a_1, a_2, \hdots, a_n)&lt;/math&gt; and &lt;math&gt;(b_1, b_2, \hdots, b_n)&lt;/math&gt; are [[increasing sequences]], we have<br /> &lt;cmath&gt;a_1b_n+a_2b_{n-1}+\hdots+a_nb_1\leq a_{\sigma(1)}b_{\sigma(1)}+a_{\sigma(2)}b_{\sigma(2)}+\hdots+a_{\sigma(n)}b_{\sigma(n)} \leq a_1b_1+a_2b_2+\hdots+a_nb_n&lt;/cmath&gt;<br /> where &lt;math&gt;\sigma&lt;/math&gt; denotes any [[permutation]] of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; (so &lt;math&gt;\sigma(1), \sigma(2), \hdots, \sigma(n)&lt;/math&gt; are the numbers &lt;math&gt;1, 2, 3, \hdots, n&lt;/math&gt; in any order). This may have an intimidating looking statement, but it is equivalent to saying that, in the particular scenario it deals with, the greedy algorithm is the best! This is because we can view the situation algorithmically: for each of the &lt;math&gt;a_i&lt;/math&gt;, we are selecting a &lt;math&gt;b_j&lt;/math&gt; to pair it with - with the caveat that each &lt;math&gt;b_j&lt;/math&gt; must be used exactly once. The [[Rearrangement inequality]] states that the largest &lt;math&gt;a_i&lt;/math&gt; should be paired with the largest &lt;math&gt;b_j&lt;/math&gt; to achieve the maximal [[dot product]]. <br /> <br /> This idea of transforming problems into algorithmic process is a very important one. Take the following example:<br /> {| class=&quot;wikitable&quot;<br /> |-<br /> | Given a 10 by 10 grid of unit squares, what is the maximum number of squares that can be shaded such that no row and no column is completely shaded?<br /> |}<br /> The first step is to transform this into an algorithmic process, which we can do as follows: for each row, in order, shade in some (but not all) of the 10 squares. <br /> <br /> =Proving correctness=<br /> Of course, greedy algorithms are not generally very interesting unless they're &lt;i&gt;correct&lt;/i&gt;; in other words, they always produce the maximal overall benefit. In order to prove the correctness of a greedy algorithm, we must show that it is never beneficial to take less than the maximal benefit at any step of the process. This is most often done by focusing on two specific steps, and showing that if the benefit of one decreases and the benefit of the other increases, the overall benefit is decreased, and so taking the maximal benefit at the first step is most beneficial.<br /> <br /> For example, let's prove the correctness of the [[rearrangement inequality]] mentioned above. Suppose we have two sequences, &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. We now want to prove that &lt;math&gt;a_1b_1+a_2b_2+\hdots+a_nb_n&lt;/math&gt; is maximized when &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are similarly sorted; i.e. they are either both [[increasing sequence|increasing]] or both [[decreasing sequence|decreasing]]. More formally, we wish to show that the [[dot product]] &lt;math&gt;a \cdot b&lt;/math&gt; is maximized when the sequences are similarly sorted. We first reformat the problem algorithmically: we consider the &lt;math&gt;a_i&lt;/math&gt; in order, at each step choosing a &lt;math&gt;b_j&lt;/math&gt; to pair it with.<br /> <br /> We claim that the greedy algorithm produces the best result; i.e. if &lt;math&gt;a_1 \geq a_2 \geq \hdots a_n&lt;/math&gt; and &lt;math&gt;b_1 \geq b_2 \geq \hdots \geq b_n&lt;/math&gt;, then &lt;math&gt;a_1b_1+a_2b_2+\hdots+a_nb_n&lt;/math&gt; is the maximal possible answer. Suppose that there exists a better [[algorithm]]. Consider the &lt;b&gt;first&lt;/b&gt; step &lt;math&gt;i&lt;/math&gt; in which we pair &lt;math&gt;a_i&lt;/math&gt; with &lt;math&gt;b_j&lt;/math&gt; such that &lt;math&gt;i&lt;j&lt;/math&gt; (in other words, &lt;math&gt;a_i&lt;/math&gt; is in a &quot;higher position&quot; than &lt;math&gt;b_j&lt;/math&gt; is) - if this step didn't exist, we'd always be pairing &lt;math&gt;a_i&lt;/math&gt; with &lt;math&gt;b_i&lt;/math&gt;, and be done immediately. Then, at some step &lt;math&gt;k&lt;/math&gt;, we pair &lt;math&gt;a_k&lt;/math&gt; with &lt;math&gt;b_i&lt;/math&gt; - we also know that &lt;math&gt;k&gt;i&lt;/math&gt;, since we've already done the steps where &lt;math&gt;k \leq i&lt;/math&gt;.<br /> <br /> We want to show that the result of this process is no better than if we simply paired &lt;math&gt;a_i&lt;/math&gt; and &lt;math&gt;b_i&lt;/math&gt; (note that it is not necessary to prove that it is &lt;i&gt;worse&lt;/i&gt;, only that it cannot be &lt;i&gt;better&lt;/i&gt;). In other words, we need to show that &lt;math&gt;a_ib_j+a_kb_i\leq a_ib_i+a_kb_j&lt;/math&gt;. But this [[factoring|factors]] nicely: we want to show &lt;math&gt;(a_i-a_k)(b_j-b_i)\leq 0&lt;/math&gt;. Fortunately, this is quite easy: we know that &lt;math&gt;i&lt;j&lt;/math&gt;, so &lt;math&gt;b_i \geq b_j&lt;/math&gt;, but we also know that &lt;math&gt;i&lt;k&lt;/math&gt;, so &lt;math&gt;a_i \geq a_k&lt;/math&gt;. Thus &lt;math&gt;a_i-a_k&lt;/math&gt; is [[nonnegative]], but &lt;math&gt;b_j-b_i&lt;/math&gt; is [[nonpositive]], proving the statement.<br /> <br /> ==Adjusting greedy algorithms==<br /> Of course, the immediate application of greedy algorithms does not always produce the optimal result. For example, if asked what the maximum number of elements in the set &lt;math&gt;\{0.6, 0.5, 0.4, 0.3, 0.2, 0.1\}&lt;/math&gt; can be chosen with sum at most 1, a particularly naive greedy algorithm will conclude the answer is two, as it will put the &lt;math&gt;0.6&lt;/math&gt; term into the &quot;greedy [[set]]&quot;, not put the &lt;math&gt;0.5&lt;/math&gt; term in, put the &lt;math&gt;0.4&lt;/math&gt; term in, and put none of the remaining terms in. Obviously, this is not the correct result, as the [[set]] &lt;math&gt;\{0.4, 0.3, 0.2, 0.1\}&lt;/math&gt; is much bigger and still has sum at most 1.<br /> <br /> What went wrong here is the ordering of the elements - we certainly don't want to be considering the largest elements first! Instead we should process the elements in the order of smallest to largest, and this will indeed give us the correct [[set]] (the proof is straightforward: why would we ever take a larger element when a smaller one is available?). This concept, of sorting the elements in a convenient way prior to processing, is <br /> key to all but the most basic of greedy algorithms, and much more complicated problems can be taken down using this strategy.<br /> <br /> ==When greedy algorithms fail==<br /> Of course, greedy algorithms are not always the optimal process, even after adjusting the order of their processing. For example, there is no way to salvage a greedy algorithm to do the following classic problem: given the following triangle of numbers, at each step we will move either left or right, and add the number we reach to a running total. Our goal is to minimize the final total.<br /> &lt;asy&gt;<br /> defaultpen(fontsize(10pt));<br /> unitsize(.7inch);<br /> real x=.7;//Scales figure in x-direction<br /> label(&quot;Start&quot;,(0,2));<br /> draw((.2x,1.8)--(.8x,1.2));<br /> label(&quot;1&quot;,(1x,1));<br /> draw((-.2x,1.8)--(-.8x,1.2));<br /> label(&quot;2&quot;,(-1x,1));<br /> draw((-1.2x,.8)--(-1.8x,.2));<br /> label(&quot;1&quot;,(-2x,0));<br /> draw((-.8x,.8)--(-.2x,.2));<br /> draw((.8x,.8)--(.2x,.2));<br /> label(&quot;over 9000&quot;,(0,0));<br /> draw((1.2x,.8)--(1.8x,.2));<br /> label(&quot;over 9000&quot;,(2x,0));<br /> &lt;/asy&gt;<br /> <br /> Obviously, the optimal path is to go left twice - but a greedy algorithm will begin by moving to the right! This is an example of when &lt;i&gt;all&lt;/i&gt; paths must be considered, and taking a shortcut by using a greedy algorithm is insufficient. As an aside, it may appear that, in the general version of this problem with &lt;math&gt;n&lt;/math&gt; layers, we have to consider all &lt;math&gt;2^n&lt;/math&gt; possible paths - but there is a much more clever approach to this problem, which - as a conclusion to this article - we offer as an exercise to the reader.</div> Siddaops2013 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=81008 2013 AMC 10A Problems 2016-11-08T21:10:10Z <p>Siddaops2013: /* Problem 10 */</p> <hr /> <div>==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid \$105, Dorothy paid \$125, and Sammy paid \$175. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Siddaops2013 https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems&diff=81007 2013 AMC 10A Problems 2016-11-08T21:09:59Z <p>Siddaops2013: /* Problem 10 */</p> <hr /> <div>==Problem 1==<br /> <br /> A taxi ride costs \$1.50 plus \$0.25 per mile traveled. How much does a 5-mile taxi ride cost?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50 \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Alice is making a batch of cookies and needs &lt;math&gt;2\frac{1}{2}&lt;/math&gt; cups of sugar. Unfortunately, her measuring cup holds only &lt;math&gt;\frac{1}{4}&lt;/math&gt; cup of sugar. How many times must she fill that cup to get the correct amount of sugar?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\textbf{(E)}\ 20 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Square &lt;math&gt; ABCD &lt;/math&gt; has side length &lt;math&gt;10&lt;/math&gt;. Point &lt;math&gt;E&lt;/math&gt; is on &lt;math&gt;\overline{BC}&lt;/math&gt;, and the area of &lt;math&gt; \triangle ABE &lt;/math&gt;<br /> is &lt;math&gt;40&lt;/math&gt;. What is &lt;math&gt; BE &lt;/math&gt;?<br /> &lt;asy&gt;<br /> pair A,B,C,D,E;<br /> A=(0,0);<br /> B=(0,50);<br /> C=(50,50);<br /> D=(50,0);<br /> E = (30,50);<br /> draw(A--B);<br /> draw(B--E);<br /> draw(E--C);<br /> draw(C--D);<br /> draw(D--A);<br /> draw(A--E);<br /> dot(A);<br /> dot(B);<br /> dot(C);<br /> dot(D);<br /> dot(E);<br /> label(&quot;A&quot;,A,SW);<br /> label(&quot;B&quot;,B,NW);<br /> label(&quot;C&quot;,C,NE);<br /> label(&quot;D&quot;,D,SE);<br /> label(&quot;E&quot;,E,N);<br /> <br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> A softball team played ten games, scoring 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 runs. They lost by one run in exactly five games. In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Tom, Dorothy, and Sammy went on a vacation and agreed to split the costs evenly. During their trip Tom paid \$105, Dorothy paid \$125, and Sammy paid \$175. In order to share costs equally, Tom gave Sammy &lt;math&gt;t&lt;/math&gt; dollars, and Dorothy gave Sammy &lt;math&gt;d&lt;/math&gt; dollars. What is &lt;math&gt;t-d&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 13 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> What is the value of &lt;math&gt;\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}} ?&lt;/math&gt; <br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ \frac{5}{3} \qquad\textbf{(D)}\ 2013 \qquad\textbf{(E)}\ 2^{4024} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on &lt;math&gt;20\%&lt;/math&gt; of her three-point shots and &lt;math&gt;30\%&lt;/math&gt; of her two-point shots. Shenille attempted &lt;math&gt;30&lt;/math&gt; shots. How many points did she score?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> AA flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly 10 ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB=AC=28&lt;/math&gt; and &lt;math&gt;BC=20&lt;/math&gt;. Points &lt;math&gt;D,E,&lt;/math&gt; and &lt;math&gt;F&lt;/math&gt; are on sides &lt;math&gt;\overline{AB}&lt;/math&gt;, &lt;math&gt;\overline{BC}&lt;/math&gt;, and &lt;math&gt;\overline{AC}&lt;/math&gt;, respectively, such that &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{EF}&lt;/math&gt; are parallel to &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. What is the perimeter of parallelogram &lt;math&gt;ADEF&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> size(180);<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br /> real r=5/7;<br /> pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);<br /> pair bottom=(C.x+(D.x-A.x),C.y+(D.y-A.y));<br /> pair E=extension(D,bottom,B,C);<br /> pair top=(E.x+D.x,E.y+D.y);<br /> pair F=extension(E,top,A,C);<br /> draw(A--B--C--cycle^^D--E--F);<br /> dot(A^^B^^C^^D^^E^^F);<br /> label(&quot;$A$&quot;,A,NW);<br /> label(&quot;$B$&quot;,B,SW);<br /> label(&quot;$C$&quot;,C,SE);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E,S);<br /> label(&quot;$F$&quot;,F,dir(0));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }48\qquad<br /> \textbf{(B) }52\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }60\qquad<br /> \textbf{(E) }72\qquad&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> How many three-digit numbers are not divisible by &lt;math&gt;5&lt;/math&gt;, have digits that sum to less than &lt;math&gt;20&lt;/math&gt;, and have the first digit equal to the third digit?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> A solid cube of side length &lt;math&gt;1&lt;/math&gt; is removed from each corner of a solid cube of side length &lt;math&gt;3&lt;/math&gt;. How many edges does the remaining solid have?<br /> <br /> <br /> &lt;math&gt; \textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> Two sides of a triangle have lengths &lt;math&gt;10&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> A triangle with vertices &lt;math&gt;(6, 5)&lt;/math&gt;, &lt;math&gt;(8, -3)&lt;/math&gt;, and &lt;math&gt;(9, 1)&lt;/math&gt; is reflected about the line &lt;math&gt;x=8&lt;/math&gt; to create a second triangle. What is the area of the union of the two triangles?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Let points &lt;math&gt;A = (0, 0)&lt;/math&gt;, &lt;math&gt;B = (1, 2)&lt;/math&gt;, &lt;math&gt;C=(3, 3)&lt;/math&gt;, and &lt;math&gt;D = (4, 0)&lt;/math&gt;. Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; is cut into equal area pieces by a line passing through &lt;math&gt;A&lt;/math&gt;. This line intersects &lt;math&gt;\overline{CD}&lt;/math&gt; at point &lt;math&gt;\bigg(\frac{p}{q}, \frac{r}{s}\bigg)&lt;/math&gt;, where these fractions are in lowest terms. What is &lt;math&gt;p+q+r+s&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> In base &lt;math&gt;10&lt;/math&gt;, the number &lt;math&gt;2013&lt;/math&gt; ends in the digit &lt;math&gt;3&lt;/math&gt;. In base &lt;math&gt;9&lt;/math&gt;, on the other hand, the same number is written as &lt;math&gt;(2676)_{9}&lt;/math&gt; and ends in the digit &lt;math&gt;6&lt;/math&gt;. For how many positive integers &lt;math&gt;b&lt;/math&gt; does the base-&lt;math&gt;b&lt;/math&gt;-representation of &lt;math&gt;2013&lt;/math&gt; end in the digit &lt;math&gt;3&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A unit square is rotated &lt;math&gt;45^\circ&lt;/math&gt; about its center. What is the area of the region swept out by the interior of the square? <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8} &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A group of &lt;math&gt;12&lt;/math&gt; pirates agree to divide a treasure chest of gold coins among themselves as follows. The &lt;math&gt;k^{\text{th}}&lt;/math&gt; pirate to take a share takes &lt;math&gt;\frac{k}{12}&lt;/math&gt; of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the &lt;math&gt;12^{\text{th}}&lt;/math&gt; pirate receive?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 21|Solution]]<br /> ==Problem 22==<br /> <br /> Six spheres of radius &lt;math&gt;1&lt;/math&gt; are positioned so that their centers are at the vertices of a regular hexagon of side length &lt;math&gt;2&lt;/math&gt;. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 22|Solution]]<br /> ==Problem 23==<br /> <br /> In &lt;math&gt;\triangle ABC&lt;/math&gt;, &lt;math&gt;AB = 86&lt;/math&gt;, and &lt;math&gt;AC=97&lt;/math&gt;. A circle with center &lt;math&gt;A&lt;/math&gt; and radius &lt;math&gt;AB&lt;/math&gt; intersects &lt;math&gt;\overline{BC}&lt;/math&gt; at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;X&lt;/math&gt;. Moreover &lt;math&gt;\overline{BX}&lt;/math&gt; and &lt;math&gt;\overline{CX}&lt;/math&gt; have integer lengths. What is &lt;math&gt;BC&lt;/math&gt;?<br /> <br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 23|Solution]]<br /> ==Problem 24==<br /> Central High School is competing against Northern High School in a backgammon match. Each school has three players, and the contest rules require that each player play two games against each of the other school's players. The match takes place in six rounds, with three games played simultaneously in each round. In how many different ways can the match be scheduled?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 540\qquad\textbf{(B)}\ 600\qquad\textbf{(C)}\ 720\qquad\textbf{(D)}\ 810\qquad\textbf{(E)}\ 900&lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 24|Solution]]<br /> ==Problem 25==<br /> <br /> All 20 diagonals are drawn in a regular octagon. At how many distinct points in the interior<br /> of the octagon (not on the boundary) do two or more diagonals intersect?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128 &lt;/math&gt;<br /> <br /> <br /> [[2013 AMC 10A Problems/Problem 25|Solution]]<br /> ==See also==<br /> {{AMC10 box|year=2013|ab=A|before=[[2012 AMC 10B Problems]]|after=[[2013 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2013 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Siddaops2013