https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Memories&feedformat=atom AoPS Wiki - User contributions [en] 2021-11-27T06:28:36Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_II_Problems/Problem_7&diff=148871 2020 AIME II Problems/Problem 7 2021-03-08T01:23:16Z <p>Memories: /* Solution (Official MAA) */</p> <hr /> <div>==Problem==<br /> Two congruent right circular cones each with base radius &lt;math&gt;3&lt;/math&gt; and height &lt;math&gt;8&lt;/math&gt; have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance &lt;math&gt;3&lt;/math&gt; from the base of each cone. A sphere with radius &lt;math&gt;r&lt;/math&gt; lies within both cones. The maximum possible value of &lt;math&gt;r^2&lt;/math&gt; is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;m+n&lt;/math&gt;.<br /> <br /> ==Solution (Official MAA)==<br /> Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the intersection of their axes of symmetry. Let &lt;math&gt;A&lt;/math&gt; be the point in the cross section where the bases of the cones meet, and let &lt;math&gt;C&lt;/math&gt; be the center of the sphere. Let the axis of symmetry of one of the cones extend from its vertex, &lt;math&gt;B&lt;/math&gt;, to the center of its base, &lt;math&gt;D&lt;/math&gt;. Let the sphere be tangent to &lt;math&gt;\overline{AB}&lt;/math&gt; at &lt;math&gt;E&lt;/math&gt;. The right triangles &lt;math&gt;\triangle ABD&lt;/math&gt; and &lt;math&gt;\triangle CBE&lt;/math&gt; are similar, implying that the radius of the sphere is&lt;cmath&gt;CE = AD \cdot\frac{BC}{AB} = AD \cdot\frac{BD-CD}{AB} =3\cdot\frac5{\sqrt{8^2+3^2}} = \frac{15}{\sqrt{73}}=\sqrt{\frac{225}{73}}.&lt;/cmath&gt;The requested sum is &lt;math&gt;225+73=298&lt;/math&gt;.<br /> &lt;asy&gt;<br /> unitsize(0.6cm);<br /> pair A = (0,0); <br /> pair TriangleOneLeft = (-6,0); <br /> pair TriangleOneDown = (-3,-8); <br /> pair TriangleOneMid = (-3,0);<br /> <br /> pair D = (0,-3); <br /> pair TriangleTwoDown = (0,-6); <br /> pair B = (-8,-3);<br /> <br /> pair C = IP(TriangleOneMid -- TriangleOneDown, B--D);<br /> pair EE = foot(C, A, B); <br /> real radius = arclength(C--EE); <br /> path circ = Circle(C, radius);<br /> <br /> <br /> <br /> draw(A--B--TriangleTwoDown--cycle);<br /> draw(B--D); <br /> draw(A--TriangleOneLeft--TriangleOneDown--cycle); <br /> draw(circ); <br /> draw(C--EE); <br /> draw(TriangleOneMid -- TriangleOneDown, gray);<br /> <br /> dot(&quot;$B$&quot;, B, W); <br /> dot(&quot;$E$&quot;, EE, NW); <br /> dot(&quot;$A$&quot;, A, NE); <br /> dot(&quot;$D$&quot;, D, E); <br /> dot(&quot;$C$&quot;, C, SE);<br /> &lt;/asy&gt;<br /> <br /> Not part of MAA's solution, but this: https://www.geogebra.org/calculator/xv4nm97a is a good visual of the cones in GeoGebra.<br /> <br /> ==Video Solution==<br /> https://youtu.be/bz5N-jI2e0U?t=44<br /> <br /> ==Video Solution 2==<br /> https://www.youtube.com/watch?v=0XJddG43pIk ~ MathEx<br /> <br /> ==Video Solution 3==<br /> https://youtu.be/dHGXtB0FxXs<br /> <br /> ~IceMatrix<br /> <br /> ==See Also==<br /> {{AIME box|year=2020|n=II|num-b=6|num-a=8}}<br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_11&diff=147198 2008 AIME I Problems/Problem 11 2021-02-16T21:51:31Z <p>Memories: Undo revision 147197 by Memories (talk)</p> <hr /> <div>== Problem ==<br /> Consider sequences that consist entirely of &lt;math&gt;A&lt;/math&gt;'s and &lt;math&gt;B&lt;/math&gt;'s and that have the property that every run of consecutive &lt;math&gt;A&lt;/math&gt;'s has even length, and every run of consecutive &lt;math&gt;B&lt;/math&gt;'s has odd length. Examples of such sequences are &lt;math&gt;AA&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;AABAA&lt;/math&gt;, while &lt;math&gt;BBAB&lt;/math&gt; is not such a sequence. How many such sequences have length 14?<br /> <br /> __TOC__<br /> <br /> == Solutions ==<br /> === Solution 1a ===<br /> Let &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; denote, respectively, the number of sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. If a sequence ends in an &lt;math&gt;A&lt;/math&gt;, then it must have been formed by appending two &lt;math&gt;A&lt;/math&gt;s to the end of a string of length &lt;math&gt;n-2&lt;/math&gt;. If a sequence ends in a &lt;math&gt;B,&lt;/math&gt; it must have either been formed by appending one &lt;math&gt;B&lt;/math&gt; to a string of length &lt;math&gt;n-1&lt;/math&gt; ending in an &lt;math&gt;A&lt;/math&gt;, or by appending two &lt;math&gt;B&lt;/math&gt;s to a string of length &lt;math&gt;n-2&lt;/math&gt; ending in a &lt;math&gt;B&lt;/math&gt;. Thus, we have the [[recursion]]s<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a_n &amp;= a_{n-2} + b_{n-2}\\<br /> b_n &amp;= a_{n-1} + b_{n-2} <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> By counting, we find that &lt;math&gt;a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0&lt;/math&gt;. <br /> &lt;cmath&gt;\begin{array}{|r||r|r|||r||r|r|}<br /> \hline<br /> n &amp; a_n &amp; b_n &amp; n &amp; a_n &amp; b_n\\<br /> \hline<br /> 1&amp;0&amp;1&amp;<br /> 8&amp;6&amp;10\\<br /> 2&amp;1&amp;0&amp;<br /> 9&amp;11&amp;11\\<br /> 3&amp;1&amp;2&amp;<br /> 10&amp;16&amp;21\\<br /> 4&amp;1&amp;1&amp;<br /> 11&amp;22&amp;27\\<br /> 5&amp;3&amp;3&amp;<br /> 12&amp;37&amp;43\\<br /> 6&amp;2&amp;4&amp;<br /> 13&amp;49&amp;64\\<br /> 7&amp;6&amp;5&amp;<br /> 14&amp;80&amp;92\\<br /> \hline<br /> \end{array}<br /> &lt;/cmath&gt;<br /> Therefore, the number of such strings of length &lt;math&gt;14&lt;/math&gt; is &lt;math&gt;a_{14} + b_{14} = \boxed{172}&lt;/math&gt;.<br /> <br /> === Solution 1b ===<br /> Let &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; denote, respectively, the number of sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. <br /> <br /> Additionally, let &lt;math&gt;t_n&lt;/math&gt; denote the total number of sequences of length &lt;math&gt;n&lt;/math&gt;. Then, &lt;math&gt;t_n=a_n+b_n&lt;/math&gt;, as the total amount of sequences of length &lt;math&gt;n&lt;/math&gt; consists of the sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;A&lt;/math&gt; and the sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;B&lt;/math&gt;.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a_n &amp;= a_{n-2} + b_{n-2}\\<br /> b_n &amp;= a_{n-1} + b_{n-2} <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> The recursion for &lt;math&gt;a_n&lt;/math&gt; tells us that &lt;math&gt;a_n=a_{n-2}+b_{n-2}&lt;/math&gt;. However, this is also the definition for &lt;math&gt;t_{n-2}&lt;/math&gt;. Therefore, &lt;math&gt;a_n=t_{n-2}&lt;/math&gt;.<br /> <br /> We also know from our recursion for &lt;math&gt;b_n&lt;/math&gt; that &lt;math&gt;b_n=a_{n-1}+b_{n-2}&lt;/math&gt;. Substituting for &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; into our recursion for &lt;math&gt;t_n&lt;/math&gt; gives us &lt;math&gt;t_n=t_{n-2}+a_{n-1}+b_{n-2}&lt;/math&gt;.<br /> <br /> Furthermore, note that since &lt;math&gt;a_n=t_{n-2}&lt;/math&gt;, &lt;math&gt;a_{n-1}=t_{n-3}&lt;/math&gt;. Furthermore, using our definition for &lt;math&gt;t_{n-2}&lt;/math&gt;, we can rewrite &lt;math&gt;b_{n-2}&lt;/math&gt; as &lt;math&gt;t_{n-2}-a_{n-2}&lt;/math&gt;. Substituting for &lt;math&gt;a_{n-1}&lt;/math&gt; and &lt;math&gt;b_{n-2}&lt;/math&gt; into our recursion for &lt;math&gt;t_n&lt;/math&gt; gives us &lt;math&gt;t_n=t_{n-2}+t_{n-3}+t_{n-2}-a_{n-2}&lt;/math&gt;.<br /> <br /> Finally, note that since &lt;math&gt;a_n=t_{n-2}&lt;/math&gt;, &lt;math&gt;a_{n-2}=t_{n-4}&lt;/math&gt;. Substituting for &lt;math&gt;a_{n-2}&lt;/math&gt; into our recursion for &lt;math&gt;t_n&lt;/math&gt; gives us &lt;math&gt;t_n=2t_{n-2}+t_{n-3}-t_{n-4}&lt;/math&gt;. We now have a recursion only in terms of &lt;math&gt;t&lt;/math&gt;.<br /> <br /> By counting, we find that &lt;math&gt;t_1=1&lt;/math&gt;, &lt;math&gt;t_2=1&lt;/math&gt;, &lt;math&gt;t_3=3&lt;/math&gt;, and &lt;math&gt;t_4=2&lt;/math&gt;.<br /> &lt;cmath&gt;\begin{array}{|r|r||r|r|} <br /> \hline <br /> n &amp; t_n &amp; n &amp; t_n\\ <br /> \hline <br /> 1&amp;1&amp;8&amp;16\\ <br /> 2&amp;1&amp;9&amp;22\\<br /> 3&amp;3&amp;10&amp;37\\ <br /> 4&amp;2&amp;11&amp;49\\ <br /> 5&amp;6&amp;12&amp;80\\ <br /> 6&amp;6&amp;13&amp;113\\ <br /> 7&amp;11&amp;14&amp;172\\ <br /> \hline<br /> \end{array}<br /> &lt;/cmath&gt;<br /> <br /> Therefore, the number of such sequences of length 14 is &lt;math&gt;\boxed{172}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> We replace &quot;14&quot; with &quot;&lt;math&gt;2n&lt;/math&gt;&quot;. We first note that we must have an even number of chunks of &lt;math&gt;B&lt;/math&gt;'s, because of parity issues. We then note that every chunk of &lt;math&gt;B&lt;/math&gt;'s except the last one must end in the sequence &lt;math&gt;BAA&lt;/math&gt;, since we need at least two &lt;math&gt;A&lt;/math&gt;'s to separate it from the next chunk of &lt;math&gt;B&lt;/math&gt;'s. The last chunk of &lt;math&gt;B&lt;/math&gt;'s must, of course, end with a &lt;math&gt;B&lt;/math&gt;. Thus our sequence must look like this :<br /> &lt;cmath&gt;<br /> \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} BAA \boxed{\quad A\text{'s} \quad} \cdots \boxed{\quad B\text{'s} \quad}BAA \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} B \boxed{\quad A\text{'s} \quad} ,<br /> &lt;/cmath&gt;<br /> where each box holds an even number of letters (possibly zero).<br /> <br /> If we want a sequence with &lt;math&gt;2k&lt;/math&gt; chunks of &lt;math&gt;B&lt;/math&gt;'s, then we have &lt;math&gt;(2n - 6k + 2)&lt;/math&gt; letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have &lt;math&gt;(n - 3k + 1)&lt;/math&gt; pairs of letters to put into &lt;math&gt;4k + 1&lt;/math&gt; boxes. By a classic balls-and-bins argument, the number of ways to do this is<br /> &lt;cmath&gt;<br /> \binom{n + k + 1}{4k} .<br /> &lt;/cmath&gt;<br /> It follows that the total number of desirable sequences is<br /> &lt;cmath&gt;<br /> \sum_k \binom{n + k + 1}{4k} .<br /> &lt;/cmath&gt;<br /> For &lt;math&gt;n = 7&lt;/math&gt;, this evaluates as &lt;math&gt;\binom{8}{0} + \binom{9}{4} + \binom{10}{8} = 1 + 126 + 45 = \boxed{172}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> There must be an even amount of runs of consecutive &lt;math&gt;B&lt;/math&gt;s due to parity. Thus, we can split this sequence into the following cases: &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;BAAB&lt;/math&gt;, &lt;math&gt;AABAAB&lt;/math&gt;, &lt;math&gt;BAABAA&lt;/math&gt;, &lt;math&gt;AABAABAA&lt;/math&gt;, &lt;math&gt;BAABAABAAB&lt;/math&gt;, &lt;math&gt;AABAABAABAAB&lt;/math&gt;, &lt;math&gt;BAABAABAABAA&lt;/math&gt;, and &lt;math&gt;AABAABAABAABAA&lt;/math&gt;, in which the amount of letters in one run does not necessarily represent the amount of letters there can be.<br /> <br /> For the first case and the last case, there is only one possible sequence of letters.<br /> <br /> For all other cases, we can insert two of the same letter at a time into a run that has the exact same letter. For example, for the second case, we can insert two &lt;math&gt;A&lt;/math&gt;s and make the sequence &lt;math&gt;BAAAAB&lt;/math&gt;. There are three &quot;slots&quot; in which we can insert two additional letters in, and we must insert five groups of new letters. By stars and bars, the number of ways for the second case is &lt;math&gt;\binom{7}{2}=21&lt;/math&gt;.<br /> <br /> Applying this logic to all of the other cases gives us &lt;math&gt;\binom{7}{3}&lt;/math&gt;, &lt;math&gt;\binom{7}{3}&lt;/math&gt;, &lt;math&gt;\binom{7}{4}&lt;/math&gt;, &lt;math&gt;\binom{8}{6}&lt;/math&gt;, &lt;math&gt;\binom{8}{1}&lt;/math&gt;, and &lt;math&gt;\binom{8}{1}&lt;/math&gt;. Adding 1+&lt;math&gt;\binom{7}{2}&lt;/math&gt;+&lt;math&gt;\binom{7}{3}&lt;/math&gt;+&lt;math&gt;\binom{7}{3}&lt;/math&gt;+&lt;math&gt;\binom{7}{4}&lt;/math&gt;+&lt;math&gt;\binom{8}{6}&lt;/math&gt;+&lt;math&gt;\binom{8}{1}&lt;/math&gt;+&lt;math&gt;\binom{8}{1}&lt;/math&gt; gives us the answer &lt;math&gt;\boxed{172}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_11&diff=147197 2008 AIME I Problems/Problem 11 2021-02-16T21:50:49Z <p>Memories: /* Solution 1b */</p> <hr /> <div>== Problem ==<br /> Consider sequences that consist entirely of &lt;math&gt;A&lt;/math&gt;'s and &lt;math&gt;B&lt;/math&gt;'s and that have the property that every run of consecutive &lt;math&gt;A&lt;/math&gt;'s has even length, and every run of consecutive &lt;math&gt;B&lt;/math&gt;'s has odd length. Examples of such sequences are &lt;math&gt;AA&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, and &lt;math&gt;AABAA&lt;/math&gt;, while &lt;math&gt;BBAB&lt;/math&gt; is not such a sequence. How many such sequences have length 14?<br /> <br /> __TOC__<br /> <br /> == Solutions ==<br /> === Solution 1a ===<br /> Let &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; denote, respectively, the number of sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. If a sequence ends in an &lt;math&gt;A&lt;/math&gt;, then it must have been formed by appending two &lt;math&gt;A&lt;/math&gt;s to the end of a string of length &lt;math&gt;n-2&lt;/math&gt;. If a sequence ends in a &lt;math&gt;B,&lt;/math&gt; it must have either been formed by appending one &lt;math&gt;B&lt;/math&gt; to a string of length &lt;math&gt;n-1&lt;/math&gt; ending in an &lt;math&gt;A&lt;/math&gt;, or by appending two &lt;math&gt;B&lt;/math&gt;s to a string of length &lt;math&gt;n-2&lt;/math&gt; ending in a &lt;math&gt;B&lt;/math&gt;. Thus, we have the [[recursion]]s<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a_n &amp;= a_{n-2} + b_{n-2}\\<br /> b_n &amp;= a_{n-1} + b_{n-2} <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> By counting, we find that &lt;math&gt;a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0&lt;/math&gt;. <br /> &lt;cmath&gt;\begin{array}{|r||r|r|||r||r|r|}<br /> \hline<br /> n &amp; a_n &amp; b_n &amp; n &amp; a_n &amp; b_n\\<br /> \hline<br /> 1&amp;0&amp;1&amp;<br /> 8&amp;6&amp;10\\<br /> 2&amp;1&amp;0&amp;<br /> 9&amp;11&amp;11\\<br /> 3&amp;1&amp;2&amp;<br /> 10&amp;16&amp;21\\<br /> 4&amp;1&amp;1&amp;<br /> 11&amp;22&amp;27\\<br /> 5&amp;3&amp;3&amp;<br /> 12&amp;37&amp;43\\<br /> 6&amp;2&amp;4&amp;<br /> 13&amp;49&amp;64\\<br /> 7&amp;6&amp;5&amp;<br /> 14&amp;80&amp;92\\<br /> \hline<br /> \end{array}<br /> &lt;/cmath&gt;<br /> Therefore, the number of such strings of length &lt;math&gt;14&lt;/math&gt; is &lt;math&gt;a_{14} + b_{14} = \boxed{172}&lt;/math&gt;.<br /> <br /> === Solution 1b ===<br /> Let &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; denote, respectively, the number of sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;. <br /> <br /> Additionally, let &lt;math&gt;t_n&lt;/math&gt; denote the total number of sequences of length &lt;math&gt;n&lt;/math&gt;. Then, &lt;math&gt;t_n=a_n+b_n&lt;/math&gt;, as the total amount of sequences of length &lt;math&gt;n&lt;/math&gt; consists of the sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;A&lt;/math&gt; and the sequences of length &lt;math&gt;n&lt;/math&gt; ending in &lt;math&gt;B&lt;/math&gt;.<br /> &lt;cmath&gt;<br /> \begin{align*}<br /> a_n &amp;= a_{n-2} + b_{n-2}\\<br /> b_n &amp;= a_{n-1} + b_{n-2} <br /> \end{align*}<br /> &lt;/cmath&gt;<br /> The recursion for &lt;math&gt;a_n&lt;/math&gt; tells us that &lt;math&gt;a_n=a_{n-2}+b_{n-2}&lt;/math&gt;. However, this is also the definition for &lt;math&gt;t_{n-2}&lt;/math&gt;. Therefore, &lt;math&gt;a_n=t_{n-2}&lt;/math&gt;.<br /> <br /> We also know from our recursion for &lt;math&gt;b_n&lt;/math&gt; that &lt;math&gt;b_n=a_{n-1}+b_{n-2}&lt;/math&gt;. Substituting for &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; into our recursion for &lt;math&gt;t_n&lt;/math&gt; gives us &lt;math&gt;t_n=t_{n-2}+a_{n-1}+b_{n-2}&lt;/math&gt;.<br /> <br /> Furthermore, note that since &lt;math&gt;a_n=t_{n-2}&lt;/math&gt;, &lt;math&gt;a_{n-1}=t_{n-3}&lt;/math&gt;. Furthermore, using our definition for &lt;math&gt;t_{n-2}&lt;/math&gt;, we can rewrite &lt;math&gt;b_{n-2}&lt;/math&gt; as &lt;math&gt;t_{n-2}-a_{n-2}&lt;/math&gt;. Substituting for &lt;math&gt;a_{n-1}&lt;/math&gt; and &lt;math&gt;b_{n-2}&lt;/math&gt; into our recursion for &lt;math&gt;t_n&lt;/math&gt; gives us &lt;math&gt;t_n=t_{n-2}+t_{n-3}+t_{n-2}-a_{n-2}&lt;/math&gt;.<br /> <br /> Finally, note that since &lt;math&gt;a_n=t_{n-2}&lt;/math&gt;, &lt;math&gt;a_{n-2}=t_{n-4}&lt;/math&gt;. Substituting for &lt;math&gt;a_{n-2}&lt;/math&gt; into our recursion for &lt;math&gt;t_n&lt;/math&gt; gives us &lt;math&gt;t_n=2t_{n-2}+t_{n-3}-t_{n-4}&lt;/math&gt;. We now have a recursion only in terms of &lt;math&gt;t&lt;/math&gt;.<br /> <br /> By counting, we find that &lt;math&gt;t_1=1&lt;/math&gt;, &lt;math&gt;t_2=1&lt;/math&gt;, &lt;math&gt;t_3=3&lt;/math&gt;, and &lt;math&gt;t_4=2&lt;/math&gt;.<br /> &lt;cmath&gt;\begin{array}{|r||r|r|r|r|} <br /> \hline <br /> || &amp; 0 &amp; 1\\ <br /> \hline <br /> 0&amp;(0,0)&amp;(0,1)\\ <br /> 1&amp;(1,0)&amp;(1,1)\\<br /> 2&amp;10&amp;37\\ <br /> 3&amp;11&amp;49\\ <br /> 4&amp;12&amp;80\\ <br /> 5&amp;13&amp;113\\ <br /> 6&amp;14&amp;172\\<br /> 7&amp;(0,7)&amp;(1,7)\\<br /> <br /> \hline<br /> \end{array}<br /> &lt;/cmath&gt;<br /> <br /> Therefore, the number of such sequences of length 14 is &lt;math&gt;\boxed{172}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> We replace &quot;14&quot; with &quot;&lt;math&gt;2n&lt;/math&gt;&quot;. We first note that we must have an even number of chunks of &lt;math&gt;B&lt;/math&gt;'s, because of parity issues. We then note that every chunk of &lt;math&gt;B&lt;/math&gt;'s except the last one must end in the sequence &lt;math&gt;BAA&lt;/math&gt;, since we need at least two &lt;math&gt;A&lt;/math&gt;'s to separate it from the next chunk of &lt;math&gt;B&lt;/math&gt;'s. The last chunk of &lt;math&gt;B&lt;/math&gt;'s must, of course, end with a &lt;math&gt;B&lt;/math&gt;. Thus our sequence must look like this :<br /> &lt;cmath&gt;<br /> \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} BAA \boxed{\quad A\text{'s} \quad} \cdots \boxed{\quad B\text{'s} \quad}BAA \boxed{\quad A\text{'s} \quad} \boxed{\quad B\text{'s} \quad} B \boxed{\quad A\text{'s} \quad} ,<br /> &lt;/cmath&gt;<br /> where each box holds an even number of letters (possibly zero).<br /> <br /> If we want a sequence with &lt;math&gt;2k&lt;/math&gt; chunks of &lt;math&gt;B&lt;/math&gt;'s, then we have &lt;math&gt;(2n - 6k + 2)&lt;/math&gt; letters with which to fill the boxes. Since each box must have an even number of letters, we may put the letters in the boxes in pairs. Then we have &lt;math&gt;(n - 3k + 1)&lt;/math&gt; pairs of letters to put into &lt;math&gt;4k + 1&lt;/math&gt; boxes. By a classic balls-and-bins argument, the number of ways to do this is<br /> &lt;cmath&gt;<br /> \binom{n + k + 1}{4k} .<br /> &lt;/cmath&gt;<br /> It follows that the total number of desirable sequences is<br /> &lt;cmath&gt;<br /> \sum_k \binom{n + k + 1}{4k} .<br /> &lt;/cmath&gt;<br /> For &lt;math&gt;n = 7&lt;/math&gt;, this evaluates as &lt;math&gt;\binom{8}{0} + \binom{9}{4} + \binom{10}{8} = 1 + 126 + 45 = \boxed{172}&lt;/math&gt;.<br /> <br /> ===Solution 3===<br /> There must be an even amount of runs of consecutive &lt;math&gt;B&lt;/math&gt;s due to parity. Thus, we can split this sequence into the following cases: &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;BAAB&lt;/math&gt;, &lt;math&gt;AABAAB&lt;/math&gt;, &lt;math&gt;BAABAA&lt;/math&gt;, &lt;math&gt;AABAABAA&lt;/math&gt;, &lt;math&gt;BAABAABAAB&lt;/math&gt;, &lt;math&gt;AABAABAABAAB&lt;/math&gt;, &lt;math&gt;BAABAABAABAA&lt;/math&gt;, and &lt;math&gt;AABAABAABAABAA&lt;/math&gt;, in which the amount of letters in one run does not necessarily represent the amount of letters there can be.<br /> <br /> For the first case and the last case, there is only one possible sequence of letters.<br /> <br /> For all other cases, we can insert two of the same letter at a time into a run that has the exact same letter. For example, for the second case, we can insert two &lt;math&gt;A&lt;/math&gt;s and make the sequence &lt;math&gt;BAAAAB&lt;/math&gt;. There are three &quot;slots&quot; in which we can insert two additional letters in, and we must insert five groups of new letters. By stars and bars, the number of ways for the second case is &lt;math&gt;\binom{7}{2}=21&lt;/math&gt;.<br /> <br /> Applying this logic to all of the other cases gives us &lt;math&gt;\binom{7}{3}&lt;/math&gt;, &lt;math&gt;\binom{7}{3}&lt;/math&gt;, &lt;math&gt;\binom{7}{4}&lt;/math&gt;, &lt;math&gt;\binom{8}{6}&lt;/math&gt;, &lt;math&gt;\binom{8}{1}&lt;/math&gt;, and &lt;math&gt;\binom{8}{1}&lt;/math&gt;. Adding 1+&lt;math&gt;\binom{7}{2}&lt;/math&gt;+&lt;math&gt;\binom{7}{3}&lt;/math&gt;+&lt;math&gt;\binom{7}{3}&lt;/math&gt;+&lt;math&gt;\binom{7}{4}&lt;/math&gt;+&lt;math&gt;\binom{8}{6}&lt;/math&gt;+&lt;math&gt;\binom{8}{1}&lt;/math&gt;+&lt;math&gt;\binom{8}{1}&lt;/math&gt; gives us the answer &lt;math&gt;\boxed{172}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=2008|n=I|num-b=10|num-a=12}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems&diff=147177 2021 AMC 10B Problems 2021-02-16T16:40:35Z <p>Memories: /* Problem 18 */</p> <hr /> <div>{{AMC10 Problems|year=2021|ab=B}}<br /> <br /> ==Problem 1==<br /> How many integer values of &lt;math&gt;x&lt;/math&gt; satisfy &lt;math&gt;|x| &lt; 3\pi&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> What is the value of &lt;math&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3} + 6&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the &lt;math&gt;28&lt;/math&gt; students in the program, &lt;math&gt;25\%&lt;/math&gt; of the juniors and &lt;math&gt;10\%&lt;/math&gt; of the seniors are on the debate team. How many juniors are in the program?<br /> <br /> &lt;math&gt;\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> At a math contest, &lt;math&gt;57&lt;/math&gt; students are wearing blue shirts, and another &lt;math&gt;75&lt;/math&gt; students are wearing yellow shirts. The &lt;math&gt;132&lt;/math&gt; students are assigned into &lt;math&gt;66&lt;/math&gt; pairs. In exactly &lt;math&gt;23&lt;/math&gt; of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?<br /> <br /> &lt;math&gt;\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give &lt;math&gt;24&lt;/math&gt;, while the other two multiply to &lt;math&gt;30&lt;/math&gt;. What is the sum of the ages of Jonie's four cousins?<br /> <br /> &lt;math&gt;\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is &lt;math&gt;84&lt;/math&gt;, and the afternoon class's mean score is &lt;math&gt;70&lt;/math&gt;. The ratio of the number of students in the morning class to the number of students in the afternoon class is &lt;math&gt;\frac{3}{4}&lt;/math&gt;. What is the mean of the scores of all the students?<br /> <br /> &lt;math&gt;\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> In a plane, four circles with radii &lt;math&gt;1,3,5,&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are tangent to line &lt;math&gt;l&lt;/math&gt; at the same point &lt;math&gt;A,&lt;/math&gt; but they may be on either side of &lt;math&gt;l&lt;/math&gt;. Region &lt;math&gt;S&lt;/math&gt; consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Mr. Zhou places all the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;225&lt;/math&gt; into a &lt;math&gt;15&lt;/math&gt; by &lt;math&gt;15&lt;/math&gt; grid. He places &lt;math&gt;1&lt;/math&gt; in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> add(grid(7,7));<br /> label(&quot;$\dots$&quot;, (0.5,0.5));<br /> label(&quot;$\dots$&quot;, (1.5,0.5));<br /> label(&quot;$\dots$&quot;, (2.5,0.5));<br /> label(&quot;$\dots$&quot;, (3.5,0.5));<br /> label(&quot;$\dots$&quot;, (4.5,0.5));<br /> label(&quot;$\dots$&quot;, (5.5,0.5));<br /> label(&quot;$\dots$&quot;, (6.5,0.5));<br /> label(&quot;$\dots$&quot;, (1.5,0.5));<br /> label(&quot;$\dots$&quot;, (0.5,1.5));<br /> label(&quot;$\dots$&quot;, (0.5,2.5));<br /> label(&quot;$\dots$&quot;, (0.5,3.5));<br /> label(&quot;$\dots$&quot;, (0.5,4.5));<br /> label(&quot;$\dots$&quot;, (0.5,5.5));<br /> label(&quot;$\dots$&quot;, (0.5,6.5));<br /> label(&quot;$\dots$&quot;, (6.5,0.5));<br /> label(&quot;$\dots$&quot;, (6.5,1.5));<br /> label(&quot;$\dots$&quot;, (6.5,2.5));<br /> label(&quot;$\dots$&quot;, (6.5,3.5));<br /> label(&quot;$\dots$&quot;, (6.5,4.5));<br /> label(&quot;$\dots$&quot;, (6.5,5.5));<br /> label(&quot;$\dots$&quot;, (0.5,6.5));<br /> label(&quot;$\dots$&quot;, (1.5,6.5));<br /> label(&quot;$\dots$&quot;, (2.5,6.5));<br /> label(&quot;$\dots$&quot;, (3.5,6.5));<br /> label(&quot;$\dots$&quot;, (4.5,6.5));<br /> label(&quot;$\dots$&quot;, (5.5,6.5));<br /> label(&quot;$\dots$&quot;, (6.5,6.5));<br /> label(&quot;$17$&quot;, (1.5,1.5));<br /> label(&quot;$18$&quot;, (1.5,2.5));<br /> label(&quot;$19$&quot;, (1.5,3.5));<br /> label(&quot;$20$&quot;, (1.5,4.5));<br /> label(&quot;$21$&quot;, (1.5,5.5));<br /> label(&quot;$16$&quot;, (2.5,1.5));<br /> label(&quot;$5$&quot;, (2.5,2.5));<br /> label(&quot;$6$&quot;, (2.5,3.5));<br /> label(&quot;$7$&quot;, (2.5,4.5));<br /> label(&quot;$22$&quot;, (2.5,5.5));<br /> label(&quot;$15$&quot;, (3.5,1.5));<br /> label(&quot;$4$&quot;, (3.5,2.5));<br /> label(&quot;$1$&quot;, (3.5,3.5));<br /> label(&quot;$8$&quot;, (3.5,4.5));<br /> label(&quot;$23$&quot;, (3.5,5.5));<br /> label(&quot;$14$&quot;, (4.5,1.5));<br /> label(&quot;$3$&quot;, (4.5,2.5));<br /> label(&quot;$2$&quot;, (4.5,3.5));<br /> label(&quot;$9$&quot;, (4.5,4.5));<br /> label(&quot;$24$&quot;, (4.5,5.5));<br /> label(&quot;$13$&quot;, (5.5,1.5));<br /> label(&quot;$12$&quot;, (5.5,2.5));<br /> label(&quot;$11$&quot;, (5.5,3.5));<br /> label(&quot;$10$&quot;, (5.5,4.5));<br /> label(&quot;$25$&quot;, (5.5,5.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The point &lt;math&gt;P(a,b)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane is first rotated counterclockwise by &lt;math&gt;90^\circ&lt;/math&gt; around the point &lt;math&gt;(1,5)&lt;/math&gt; and then reflected about the line &lt;math&gt;y = -x&lt;/math&gt;. The image of &lt;math&gt;P&lt;/math&gt; after these two transformations is at &lt;math&gt;(-6,3)&lt;/math&gt;. What is &lt;math&gt;b - a ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> An inverted cone with base radius &lt;math&gt;12 \text{cm}&lt;/math&gt; and height &lt;math&gt;18\text{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of &lt;math&gt;24\text{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A) }1.5 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }4.5 \qquad \textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?<br /> <br /> &lt;math&gt;\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Let &lt;math&gt;N = 34 \cdot 34 \cdot 63 \cdot 270&lt;/math&gt;. What is the ratio of the sum of the odd divisors of &lt;math&gt;N&lt;/math&gt; to the sum of the even divisors of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer and &lt;math&gt;d&lt;/math&gt; be a digit such that the value of the numeral &lt;math&gt;\underline{32d}&lt;/math&gt; in base &lt;math&gt;n&lt;/math&gt; equals &lt;math&gt;263&lt;/math&gt;, and the value of the numeral &lt;math&gt;\underline{324}&lt;/math&gt; in base &lt;math&gt;n&lt;/math&gt; equals the value of the numeral &lt;math&gt;\underline{11d1}&lt;/math&gt; in base six. What is &lt;math&gt;n + d ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38, 38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A)} ~5\frac{1}{2} \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~6\frac{1}{2} \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~7\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> The real number &lt;math&gt;x&lt;/math&gt; satisfies the equation &lt;math&gt;x+\frac{1}{x} = \sqrt{5}&lt;/math&gt;. What is the value of &lt;math&gt;x^{11}-7x^{7}+x^3?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, &lt;math&gt;1357&lt;/math&gt;, &lt;math&gt;89&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; are all uphill integers, but &lt;math&gt;32&lt;/math&gt;, &lt;math&gt;1240&lt;/math&gt;, and &lt;math&gt;466&lt;/math&gt; are not. How many uphill integers are divisible by &lt;math&gt;15&lt;/math&gt;?<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 /> [[2021 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given &lt;math&gt;2&lt;/math&gt; cards out of a set of &lt;math&gt;10&lt;/math&gt; cards numbered &lt;math&gt;1, 2, 3,\cdots , 10&lt;/math&gt;. The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon &lt;math&gt;11&lt;/math&gt;, Oscar &lt;math&gt;4&lt;/math&gt;, Aditi &lt;math&gt;7&lt;/math&gt;, Tyrone &lt;math&gt;16&lt;/math&gt;, Kim &lt;math&gt;17&lt;/math&gt;. Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; ~Ravon was given card &lt;math&gt;3&lt;/math&gt; .<br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; ~Aditi was given card &lt;math&gt;3&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; ~Ravon was given card &lt;math&gt;4&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; ~Aditi was given card &lt;math&gt;4&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; ~Tyrone was given card &lt;math&gt;7&lt;/math&gt;.<br /> <br /> [[2021 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Suppose that &lt;math&gt;S&lt;/math&gt; is a finite set of positive integers. If the greatest integer in &lt;math&gt;S&lt;/math&gt; is removed from &lt;math&gt;S&lt;/math&gt;, then the average value (arithmetic mean) of the integers remaining is &lt;math&gt;32&lt;/math&gt;. If the least integer is &lt;math&gt;S&lt;/math&gt; is also removed, then the average value of the integers remaining is &lt;math&gt;35&lt;/math&gt;. If the greatest integer is then returned to the set, the average value of the integers rises of &lt;math&gt;40&lt;/math&gt;. The greatest integer in the original set &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;72&lt;/math&gt; greater than the least integer in &lt;math&gt;S&lt;/math&gt;. What is the average value of all the integers in the set &lt;math&gt;S ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~36.2 \qquad\textbf{(B)} ~36.4 \qquad\textbf{(C)} ~36.6 \qquad\textbf{(D)} ~36.8 \qquad\textbf{(E)} ~37&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> The figure is constructed from &lt;math&gt;11&lt;/math&gt; line segments, each of which has length &lt;math&gt;2&lt;/math&gt;. The area of pentagon &lt;math&gt;ABCDE&lt;/math&gt; can be written is &lt;math&gt;\sqrt{m} + \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. What is &lt;math&gt;m + n ?&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(-2.4638,4.10658);<br /> pair B=(-4,2.6567453480756127);<br /> pair C=(-3.47132,0.6335248637894945);<br /> pair D=(-1.464483379039766,0.6335248637894945);<br /> pair E=(-0.956630463955801,2.6567453480756127);<br /> pair F=(-2,2);<br /> pair G=(-3,2);<br /> draw(A--B--C--D--E--A);<br /> draw(A--F--A--G);<br /> draw(B--F--C);<br /> draw(E--G--D);<br /> label(&quot;A&quot;,A,N);<br /> label(&quot;B&quot;,B,W);<br /> label(&quot;C&quot;,C,W);<br /> label(&quot;D&quot;,D,dir(0));<br /> label(&quot;E&quot;,E,dir(0));<br /> dot(A^^B^^C^^D^^E^^F^^G);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<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,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Ang, Ben, and Jasmin each have &lt;math&gt;5&lt;/math&gt; blocks, colored red, blue, yellow, white, and green; and there are &lt;math&gt;5&lt;/math&gt; empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives &lt;math&gt;3&lt;/math&gt; blocks all of the same color is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m + n ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square, The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }64 \qquad \textbf{(B) }66 \qquad \textbf{(C) }68 \qquad \textbf{(D) }70 \qquad \textbf{(E) }72&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one &quot;wall&quot; among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; can be changed into any of the following by one move: &lt;math&gt;(3,2),(2, 1, 2),(4),(4, 1),(2, 2),&lt;/math&gt; or &lt;math&gt;(1, 1, 2)&lt;/math&gt;.<br /> &lt;asy&gt;<br /> /* CREDITS */<br /> /* Made by samrocksnature */<br /> /* Modified commas an periods by forester2015 */<br /> <br /> /* Import and Set variables */<br /> import graph; size(10cm);<br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */<br /> pen dotstyle = black; /* point style */<br /> real xmin = -20.98617651190462, xmax = 71.97119514540032, ymin = -24.802885633158763, ymax = 28.83570218998556; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw((14,4)--(13.010050506338834,3.0100505063388336), linewidth(1));<br /> draw((14,4)--(13.010050506338834,4.989949493661166), linewidth(1));<br /> draw((10,4)--(14,4), linewidth(1));<br /> draw((4,6)--(8,6), linewidth(1));<br /> draw((4,2)--(8,2), linewidth(1));<br /> draw((8,2)--(8,6), linewidth(1));<br /> draw((4,6)--(4,2), linewidth(1));<br /> draw((6,6)--(6,2), linewidth(1));<br /> draw((-6,6)--(-6,2), linewidth(1));<br /> draw((-6,6)--(2,6), linewidth(1));<br /> draw((2,6)--(2,2), linewidth(1));<br /> draw((2,2)--(-6,2), linewidth(1));<br /> draw((-4,2)--(-4,6), linewidth(1));<br /> draw((-2,6)--(-2,2), linewidth(1));<br /> draw((0,2)--(0,6), linewidth(1));<br /> draw((50,6)--(50,2), linewidth(1));<br /> draw((50,2)--(58,2), linewidth(1));<br /> draw((58,2)--(58,6), linewidth(1));<br /> draw((58,6)--(50,6), linewidth(1));<br /> draw((52,6)--(52,2), linewidth(1));<br /> draw((54,6)--(54,2), linewidth(1));<br /> draw((56,6)--(56,2), linewidth(1));<br /> draw((32,6)--(32,2), linewidth(1));<br /> draw((46,2)--(46,6), linewidth(1));<br /> draw((34,6)--(34,2), linewidth(1));<br /> draw((36,2)--(36,6), linewidth(1));<br /> draw((38,6)--(38,2), linewidth(1));<br /> draw((40,2)--(40,6), linewidth(1));<br /> draw((42,6)--(42,2), linewidth(1));<br /> draw((44,2)--(44,6), linewidth(1));<br /> draw((16,6)--(16,2), linewidth(1));<br /> draw((28,2)--(28,6), linewidth(1));<br /> draw((18,6)--(18,2), linewidth(1));<br /> draw((20,6)--(20,2), linewidth(1));<br /> draw((22,6)--(22,2), linewidth(1));<br /> draw((24,6)--(24,2), linewidth(1));<br /> draw((26,6)--(26,2), linewidth(1));<br /> draw((16,6)--(22,6), linewidth(1));<br /> draw((24,6)--(28,6), linewidth(1));<br /> draw((16,2)--(22,2), linewidth(1));<br /> draw((24,2)--(28,2), linewidth(1));<br /> draw((32,6)--(36,6), linewidth(1));<br /> draw((32,2)--(36,2), linewidth(1));<br /> draw((38,6)--(40,6), linewidth(1));<br /> draw((38,2)--(40,2), linewidth(1));<br /> draw((42,6)--(46,6), linewidth(1));<br /> draw((42,2)--(46,2), linewidth(1));<br /> <br /> /* dots and labels */<br /> label(&quot;,&quot;,(59,2));<br /> label(&quot;.&quot;,(60,2));<br /> label(&quot;.&quot;,(61,2));<br /> label(&quot;.&quot;,(62,2));<br /> label(&quot;,&quot;,(29,2));<br /> label(&quot;,&quot;,(47,2));<br /> &lt;/asy&gt;<br /> Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?<br /> <br /> &lt;math&gt;\textbf{(A)} ~(6, 1, 1) \qquad\textbf{(B)} ~(6, 2, 1) \qquad\textbf{(C)} ~(6, 2, 2) \qquad\textbf{(D)} ~(6, 3, 1) \qquad\textbf{(E)} ~(6, 3, 2)&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of lattice points in the coordinate plane, both of whose coordinates are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt;, inclusive. Exactly &lt;math&gt;300&lt;/math&gt; points in &lt;math&gt;S&lt;/math&gt; lie on or below a line with equation &lt;math&gt;y = mx&lt;/math&gt;. The possible values of &lt;math&gt;m&lt;/math&gt; lie in an interval of length &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~31 \qquad\textbf{(B)} ~47 \qquad\textbf{(C)} ~62 \qquad\textbf{(D)} ~72 \qquad\textbf{(E)} ~85&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=B|before=[[2021 AMC 10A]]|after=[[2022 AMC 10A]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=User:Sugar_rush&diff=146991 User:Sugar rush 2021-02-15T23:29:37Z <p>Memories: /* User Count */</p> <hr /> <div>==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;User Count&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;If this is your first time visiting this page, edit it by incrementing the user count below by one.&lt;/font&gt;&lt;/div&gt;<br /> &lt;center&gt;&lt;font size=&quot;120px”&gt;16<br /> &lt;/font&gt;&lt;/center&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#ccccff;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;About Me&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-bottom:10px&quot;&gt;&lt;font color=&quot;black&quot;&gt;sugar_rush is 14 years old.&lt;br&gt;<br /> sugar_rush was born on January 12, 2007.<br /> <br /> sugar_rush is in 8th grade.<br /> <br /> sugar_rush is a competitive runner and pro at math.<br /> <br /> &lt;/font&gt;&lt;/div&gt;<br /> &lt;/div&gt;<br /> &lt;div style=&quot;border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center&quot;&gt;<br /> <br /> ==&lt;font color=&quot;black&quot; style=&quot;font-family: ITC Avant Garde Gothic Std, Verdana&quot;&gt;&lt;div style=&quot;margin-left:10px&quot;&gt;Goals&lt;/div&gt;&lt;/font&gt;==<br /> &lt;div style=&quot;margin-left: 10px; margin-right: 10px; margin-bottom:10px&quot;&gt;Qualify for AIME and get at least 6 on 2021 AIME I (never qualified for AIME before)<br /> <br /> Get at least 120 on AMC 12A 2021 and 150 on AMC 10B 2021<br /> <br /> &lt;/div&gt;<br /> &lt;/div&gt;<br /> Also: here is this guy's youtube channel: https://www.youtube.com/c/PunxsutawnyPhil</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems&diff=146990 2021 AMC 10B Problems 2021-02-15T23:27:25Z <p>Memories: /* Problem 18 */</p> <hr /> <div>{{AMC10 Problems|year=2021|ab=B}}<br /> <br /> ==Problem 1==<br /> How many integer values of &lt;math&gt;x&lt;/math&gt; satisfy &lt;math&gt;|x| &lt; 3\pi&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> What is the value of &lt;math&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3} + 6&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the &lt;math&gt;28&lt;/math&gt; students in the program, &lt;math&gt;25\%&lt;/math&gt; of the juniors and &lt;math&gt;10\%&lt;/math&gt; of the seniors are on the debate team. How many juniors are in the program?<br /> <br /> &lt;math&gt;\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> At a math contest, &lt;math&gt;57&lt;/math&gt; students are wearing blue shirts, and another &lt;math&gt;75&lt;/math&gt; students are wearing yellow shirts. The &lt;math&gt;132&lt;/math&gt; students are assigned into &lt;math&gt;66&lt;/math&gt; pairs. In exactly &lt;math&gt;23&lt;/math&gt; of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?<br /> <br /> &lt;math&gt;\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give &lt;math&gt;24&lt;/math&gt;, while the other two multiply to &lt;math&gt;30&lt;/math&gt;. What is the sum of the ages of Jonie's four cousins?<br /> <br /> &lt;math&gt;\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is &lt;math&gt;84&lt;/math&gt;, and the afternoon class's mean score is &lt;math&gt;70&lt;/math&gt;. The ratio of the number of students in the morning class to the number of students in the afternoon class is &lt;math&gt;\frac{3}{4}&lt;/math&gt;. What is the mean of the scores of all the students?<br /> <br /> &lt;math&gt;\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> In a plane, four circles with radii &lt;math&gt;1,3,5,&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are tangent to line &lt;math&gt;l&lt;/math&gt; at the same point &lt;math&gt;A,&lt;/math&gt; but they may be on either side of &lt;math&gt;l&lt;/math&gt;. Region &lt;math&gt;S&lt;/math&gt; consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Mr. Zhou places all the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;225&lt;/math&gt; into a &lt;math&gt;15&lt;/math&gt; by &lt;math&gt;15&lt;/math&gt; grid. He places &lt;math&gt;1&lt;/math&gt; in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> add(grid(7,7));<br /> label(&quot;$\dots$&quot;, (0.5,0.5));<br /> label(&quot;$\dots$&quot;, (1.5,0.5));<br /> label(&quot;$\dots$&quot;, (2.5,0.5));<br /> label(&quot;$\dots$&quot;, (3.5,0.5));<br /> label(&quot;$\dots$&quot;, (4.5,0.5));<br /> label(&quot;$\dots$&quot;, (5.5,0.5));<br /> label(&quot;$\dots$&quot;, (6.5,0.5));<br /> label(&quot;$\dots$&quot;, (1.5,0.5));<br /> label(&quot;$\dots$&quot;, (0.5,1.5));<br /> label(&quot;$\dots$&quot;, (0.5,2.5));<br /> label(&quot;$\dots$&quot;, (0.5,3.5));<br /> label(&quot;$\dots$&quot;, (0.5,4.5));<br /> label(&quot;$\dots$&quot;, (0.5,5.5));<br /> label(&quot;$\dots$&quot;, (0.5,6.5));<br /> label(&quot;$\dots$&quot;, (6.5,0.5));<br /> label(&quot;$\dots$&quot;, (6.5,1.5));<br /> label(&quot;$\dots$&quot;, (6.5,2.5));<br /> label(&quot;$\dots$&quot;, (6.5,3.5));<br /> label(&quot;$\dots$&quot;, (6.5,4.5));<br /> label(&quot;$\dots$&quot;, (6.5,5.5));<br /> label(&quot;$\dots$&quot;, (0.5,6.5));<br /> label(&quot;$\dots$&quot;, (1.5,6.5));<br /> label(&quot;$\dots$&quot;, (2.5,6.5));<br /> label(&quot;$\dots$&quot;, (3.5,6.5));<br /> label(&quot;$\dots$&quot;, (4.5,6.5));<br /> label(&quot;$\dots$&quot;, (5.5,6.5));<br /> label(&quot;$\dots$&quot;, (6.5,6.5));<br /> label(&quot;$17$&quot;, (1.5,1.5));<br /> label(&quot;$18$&quot;, (1.5,2.5));<br /> label(&quot;$19$&quot;, (1.5,3.5));<br /> label(&quot;$20$&quot;, (1.5,4.5));<br /> label(&quot;$21$&quot;, (1.5,5.5));<br /> label(&quot;$16$&quot;, (2.5,1.5));<br /> label(&quot;$5$&quot;, (2.5,2.5));<br /> label(&quot;$6$&quot;, (2.5,3.5));<br /> label(&quot;$7$&quot;, (2.5,4.5));<br /> label(&quot;$22$&quot;, (2.5,5.5));<br /> label(&quot;$15$&quot;, (3.5,1.5));<br /> label(&quot;$4$&quot;, (3.5,2.5));<br /> label(&quot;$1$&quot;, (3.5,3.5));<br /> label(&quot;$8$&quot;, (3.5,4.5));<br /> label(&quot;$23$&quot;, (3.5,5.5));<br /> label(&quot;$14$&quot;, (4.5,1.5));<br /> label(&quot;$3$&quot;, (4.5,2.5));<br /> label(&quot;$2$&quot;, (4.5,3.5));<br /> label(&quot;$9$&quot;, (4.5,4.5));<br /> label(&quot;$24$&quot;, (4.5,5.5));<br /> label(&quot;$13$&quot;, (5.5,1.5));<br /> label(&quot;$12$&quot;, (5.5,2.5));<br /> label(&quot;$11$&quot;, (5.5,3.5));<br /> label(&quot;$10$&quot;, (5.5,4.5));<br /> label(&quot;$25$&quot;, (5.5,5.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The point &lt;math&gt;P(a,b)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane is first rotated counterclockwise by &lt;math&gt;90^\circ&lt;/math&gt; around the point &lt;math&gt;(1,5)&lt;/math&gt; and then reflected about the line &lt;math&gt;y = -x&lt;/math&gt;. The image of &lt;math&gt;P&lt;/math&gt; after these two transformations is at &lt;math&gt;(-6,3)&lt;/math&gt;. What is &lt;math&gt;b - a ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> An inverted cone with base radius &lt;math&gt;12 \text{cm}&lt;/math&gt; and height &lt;math&gt;18\text{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of &lt;math&gt;24\text{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A) }1.5 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }4.5 \qquad \textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?<br /> <br /> &lt;math&gt;\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Let &lt;math&gt;N = 34 \cdot 34 \cdot 63 \cdot 270&lt;/math&gt;. What is the ratio of the sum of the odd divisors of &lt;math&gt;N&lt;/math&gt; to the sum of the even divisors of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer and &lt;math&gt;d&lt;/math&gt; be a digit such that the value of the numeral &lt;math&gt;\underline{32d}&lt;/math&gt; in base &lt;math&gt;n&lt;/math&gt; equals &lt;math&gt;263&lt;/math&gt;, and the value of the numeral &lt;math&gt;\underline{324}&lt;/math&gt; in base &lt;math&gt;n&lt;/math&gt; equals the value of the numeral &lt;math&gt;\underline{11d1}&lt;/math&gt; in base six. What is &lt;math&gt;n + d ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38, 38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A)} ~5\frac{1}{2} \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~6\frac{1}{2} \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~7\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> The real number &lt;math&gt;x&lt;/math&gt; satisfies the equation &lt;math&gt;x+\frac{1}{x} = \sqrt{5}&lt;/math&gt;. What is the value of &lt;math&gt;x^{11}-7x^{7}+x^3?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, &lt;math&gt;1357&lt;/math&gt;, &lt;math&gt;89&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; are all uphill integers, but &lt;math&gt;32&lt;/math&gt;, &lt;math&gt;1240&lt;/math&gt;, and &lt;math&gt;466&lt;/math&gt; are not. How many uphill integers are divisible by &lt;math&gt;15&lt;/math&gt;?<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 /> [[2021 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given &lt;math&gt;2&lt;/math&gt; cards out of a set of &lt;math&gt;10&lt;/math&gt; cards numbered &lt;math&gt;1, 2, 3,\cdots , 10&lt;/math&gt;. The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon &lt;math&gt;11&lt;/math&gt;, Oscar &lt;math&gt;4&lt;/math&gt;, Aditi &lt;math&gt;7&lt;/math&gt;, Tyrone &lt;math&gt;16&lt;/math&gt;, Kim &lt;math&gt;17&lt;/math&gt;. Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; ~Ravon was given card &lt;math&gt;3&lt;/math&gt; .<br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; ~Aditi was given card &lt;math&gt;3&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; ~Ravon was given card &lt;math&gt;4&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; ~Aditi was given card &lt;math&gt;4&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; ~Tyrone was given card &lt;math&gt;7&lt;/math&gt;.<br /> <br /> [[2021 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Suppose that &lt;math&gt;S&lt;/math&gt; is a finite set of positive integers. If the greatest integer in &lt;math&gt;S&lt;/math&gt; is removed from &lt;math&gt;S&lt;/math&gt;, then the average value (arithmetic mean) of the integers remaining is &lt;math&gt;32&lt;/math&gt;. If the least integer is &lt;math&gt;S&lt;/math&gt; is also removed, then the average value of the integers remaining is &lt;math&gt;35&lt;/math&gt;. If the greatest integer is then returned to the set, the average value of the integers rises of &lt;math&gt;40&lt;/math&gt;. The greatest integer in the original set &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;72&lt;/math&gt; greater than the least integer in &lt;math&gt;S&lt;/math&gt;. What is the average value of all the integers in the set &lt;math&gt;S ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~36.2 \qquad\textbf{(B)} ~36.4 \qquad\textbf{(C)} ~36.6 \qquad\textbf{(D)} ~36.8 \qquad\textbf{(E)} ~37&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> The figure is constructed from &lt;math&gt;11&lt;/math&gt; line segments, each of which has length &lt;math&gt;2&lt;/math&gt;. The area of pentagon &lt;math&gt;ABCDE&lt;/math&gt; can be written is &lt;math&gt;\sqrt{m} + \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. What is &lt;math&gt;m + n ?&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(-2.4638,4.10658);<br /> pair B=(-4,2.6567453480756127);<br /> pair C=(-3.47132,0.6335248637894945);<br /> pair D=(-1.464483379039766,0.6335248637894945);<br /> pair E=(-0.956630463955801,2.6567453480756127);<br /> pair F=(-2,2);<br /> pair G=(-3,2);<br /> draw(A--B--C--D--E--A);<br /> draw(A--F--A--G);<br /> draw(B--F--C);<br /> draw(E--G--D);<br /> label(&quot;A&quot;,A,N);<br /> label(&quot;B&quot;,B,W);<br /> label(&quot;C&quot;,C,W);<br /> label(&quot;D&quot;,D,E);<br /> label(&quot;E&quot;,E,dir(0));<br /> dot(A^^B^^C^^D^^E^^F^^G);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<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,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Ang, Ben, and Jasmin each have &lt;math&gt;5&lt;/math&gt; blocks, colored red, blue, yellow, white, and green; and there are &lt;math&gt;5&lt;/math&gt; empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives &lt;math&gt;3&lt;/math&gt; blocks all of the same color is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m + n ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square, The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }64 \qquad \textbf{(B) }66 \qquad \textbf{(C) }68 \qquad \textbf{(D) }70 \qquad \textbf{(E) }72&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one &quot;wall&quot; among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; can be changed into any of the following by one move: &lt;math&gt;(3,2),(2, 1, 2),(4),(4, 1),(2, 2),&lt;/math&gt; or &lt;math&gt;(1, 1, 2)&lt;/math&gt;.<br /> &lt;asy&gt;<br /> /* CREDITS */<br /> /* Made by samrocksnature */<br /> /* Modified commas an periods by forester2015 */<br /> <br /> /* Import and Set variables */<br /> import graph; size(10cm);<br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */<br /> pen dotstyle = black; /* point style */<br /> real xmin = -20.98617651190462, xmax = 71.97119514540032, ymin = -24.802885633158763, ymax = 28.83570218998556; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw((14,4)--(13.010050506338834,3.0100505063388336), linewidth(1));<br /> draw((14,4)--(13.010050506338834,4.989949493661166), linewidth(1));<br /> draw((10,4)--(14,4), linewidth(1));<br /> draw((4,6)--(8,6), linewidth(1));<br /> draw((4,2)--(8,2), linewidth(1));<br /> draw((8,2)--(8,6), linewidth(1));<br /> draw((4,6)--(4,2), linewidth(1));<br /> draw((6,6)--(6,2), linewidth(1));<br /> draw((-6,6)--(-6,2), linewidth(1));<br /> draw((-6,6)--(2,6), linewidth(1));<br /> draw((2,6)--(2,2), linewidth(1));<br /> draw((2,2)--(-6,2), linewidth(1));<br /> draw((-4,2)--(-4,6), linewidth(1));<br /> draw((-2,6)--(-2,2), linewidth(1));<br /> draw((0,2)--(0,6), linewidth(1));<br /> draw((50,6)--(50,2), linewidth(1));<br /> draw((50,2)--(58,2), linewidth(1));<br /> draw((58,2)--(58,6), linewidth(1));<br /> draw((58,6)--(50,6), linewidth(1));<br /> draw((52,6)--(52,2), linewidth(1));<br /> draw((54,6)--(54,2), linewidth(1));<br /> draw((56,6)--(56,2), linewidth(1));<br /> draw((32,6)--(32,2), linewidth(1));<br /> draw((46,2)--(46,6), linewidth(1));<br /> draw((34,6)--(34,2), linewidth(1));<br /> draw((36,2)--(36,6), linewidth(1));<br /> draw((38,6)--(38,2), linewidth(1));<br /> draw((40,2)--(40,6), linewidth(1));<br /> draw((42,6)--(42,2), linewidth(1));<br /> draw((44,2)--(44,6), linewidth(1));<br /> draw((16,6)--(16,2), linewidth(1));<br /> draw((28,2)--(28,6), linewidth(1));<br /> draw((18,6)--(18,2), linewidth(1));<br /> draw((20,6)--(20,2), linewidth(1));<br /> draw((22,6)--(22,2), linewidth(1));<br /> draw((24,6)--(24,2), linewidth(1));<br /> draw((26,6)--(26,2), linewidth(1));<br /> draw((16,6)--(22,6), linewidth(1));<br /> draw((24,6)--(28,6), linewidth(1));<br /> draw((16,2)--(22,2), linewidth(1));<br /> draw((24,2)--(28,2), linewidth(1));<br /> draw((32,6)--(36,6), linewidth(1));<br /> draw((32,2)--(36,2), linewidth(1));<br /> draw((38,6)--(40,6), linewidth(1));<br /> draw((38,2)--(40,2), linewidth(1));<br /> draw((42,6)--(46,6), linewidth(1));<br /> draw((42,2)--(46,2), linewidth(1));<br /> <br /> /* dots and labels */<br /> label(&quot;,&quot;,(59,2));<br /> label(&quot;.&quot;,(60,2));<br /> label(&quot;.&quot;,(61,2));<br /> label(&quot;.&quot;,(62,2));<br /> label(&quot;,&quot;,(29,2));<br /> label(&quot;,&quot;,(47,2));<br /> &lt;/asy&gt;<br /> Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?<br /> <br /> &lt;math&gt;\textbf{(A)} ~(6, 1, 1) \qquad\textbf{(B)} ~(6, 2, 1) \qquad\textbf{(C)} ~(6, 2, 2) \qquad\textbf{(D)} ~(6, 3, 1) \qquad\textbf{(E)} ~(6, 3, 2)&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of lattice points in the coordinate plane, both of whose coordinates are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt;, inclusive. Exactly &lt;math&gt;300&lt;/math&gt; points in &lt;math&gt;S&lt;/math&gt; lie on or below a line with equation &lt;math&gt;y = mx&lt;/math&gt;. The possible values of &lt;math&gt;m&lt;/math&gt; lie in an interval of length &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~31 \qquad\textbf{(B)} ~47 \qquad\textbf{(C)} ~62 \qquad\textbf{(D)} ~72 \qquad\textbf{(E)} ~85&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=B|before=[[2021 AMC 10A]]|after=[[2022 AMC 10A]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems&diff=146988 2021 AMC 10B Problems 2021-02-15T23:27:08Z <p>Memories: /* Problem 18 */</p> <hr /> <div>{{AMC10 Problems|year=2021|ab=B}}<br /> <br /> ==Problem 1==<br /> How many integer values of &lt;math&gt;x&lt;/math&gt; satisfy &lt;math&gt;|x| &lt; 3\pi&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~9 \qquad\textbf{(B)} ~10 \qquad\textbf{(C)} ~18 \qquad\textbf{(D)} ~19 \qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> What is the value of &lt;math&gt;\sqrt{(3-2\sqrt{3})^2}+\sqrt{(3+2\sqrt{3})^2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~0 \qquad\textbf{(B)} ~4\sqrt{3}-6 \qquad\textbf{(C)} ~6 \qquad\textbf{(D)} ~4\sqrt{3} \qquad\textbf{(E)} ~4\sqrt{3} + 6&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the &lt;math&gt;28&lt;/math&gt; students in the program, &lt;math&gt;25\%&lt;/math&gt; of the juniors and &lt;math&gt;10\%&lt;/math&gt; of the seniors are on the debate team. How many juniors are in the program?<br /> <br /> &lt;math&gt;\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> At a math contest, &lt;math&gt;57&lt;/math&gt; students are wearing blue shirts, and another &lt;math&gt;75&lt;/math&gt; students are wearing yellow shirts. The &lt;math&gt;132&lt;/math&gt; students are assigned into &lt;math&gt;66&lt;/math&gt; pairs. In exactly &lt;math&gt;23&lt;/math&gt; of these pairs, both students are wearing blue shirts. In how many pairs are both students wearing yellow shirts?<br /> <br /> &lt;math&gt;\textbf{(A)} ~23 \qquad\textbf{(B)} ~32 \qquad\textbf{(C)} ~37 \qquad\textbf{(D)} ~41 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The ages of Jonie's four cousins are distinct single-digit positive integers. Two of the cousins' ages multiplied together give &lt;math&gt;24&lt;/math&gt;, while the other two multiply to &lt;math&gt;30&lt;/math&gt;. What is the sum of the ages of Jonie's four cousins?<br /> <br /> &lt;math&gt;\textbf{(A)} ~21 \qquad\textbf{(B)} ~22 \qquad\textbf{(C)} ~23 \qquad\textbf{(D)} ~24 \qquad\textbf{(E)} ~25&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is &lt;math&gt;84&lt;/math&gt;, and the afternoon class's mean score is &lt;math&gt;70&lt;/math&gt;. The ratio of the number of students in the morning class to the number of students in the afternoon class is &lt;math&gt;\frac{3}{4}&lt;/math&gt;. What is the mean of the scores of all the students?<br /> <br /> &lt;math&gt;\textbf{(A)} ~74 \qquad\textbf{(B)} ~75 \qquad\textbf{(C)} ~76 \qquad\textbf{(D)} ~77 \qquad\textbf{(E)} ~78&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> In a plane, four circles with radii &lt;math&gt;1,3,5,&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are tangent to line &lt;math&gt;l&lt;/math&gt; at the same point &lt;math&gt;A,&lt;/math&gt; but they may be on either side of &lt;math&gt;l&lt;/math&gt;. Region &lt;math&gt;S&lt;/math&gt; consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region &lt;math&gt;S&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Mr. Zhou places all the integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;225&lt;/math&gt; into a &lt;math&gt;15&lt;/math&gt; by &lt;math&gt;15&lt;/math&gt; grid. He places &lt;math&gt;1&lt;/math&gt; in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> add(grid(7,7));<br /> label(&quot;$\dots$&quot;, (0.5,0.5));<br /> label(&quot;$\dots$&quot;, (1.5,0.5));<br /> label(&quot;$\dots$&quot;, (2.5,0.5));<br /> label(&quot;$\dots$&quot;, (3.5,0.5));<br /> label(&quot;$\dots$&quot;, (4.5,0.5));<br /> label(&quot;$\dots$&quot;, (5.5,0.5));<br /> label(&quot;$\dots$&quot;, (6.5,0.5));<br /> label(&quot;$\dots$&quot;, (1.5,0.5));<br /> label(&quot;$\dots$&quot;, (0.5,1.5));<br /> label(&quot;$\dots$&quot;, (0.5,2.5));<br /> label(&quot;$\dots$&quot;, (0.5,3.5));<br /> label(&quot;$\dots$&quot;, (0.5,4.5));<br /> label(&quot;$\dots$&quot;, (0.5,5.5));<br /> label(&quot;$\dots$&quot;, (0.5,6.5));<br /> label(&quot;$\dots$&quot;, (6.5,0.5));<br /> label(&quot;$\dots$&quot;, (6.5,1.5));<br /> label(&quot;$\dots$&quot;, (6.5,2.5));<br /> label(&quot;$\dots$&quot;, (6.5,3.5));<br /> label(&quot;$\dots$&quot;, (6.5,4.5));<br /> label(&quot;$\dots$&quot;, (6.5,5.5));<br /> label(&quot;$\dots$&quot;, (0.5,6.5));<br /> label(&quot;$\dots$&quot;, (1.5,6.5));<br /> label(&quot;$\dots$&quot;, (2.5,6.5));<br /> label(&quot;$\dots$&quot;, (3.5,6.5));<br /> label(&quot;$\dots$&quot;, (4.5,6.5));<br /> label(&quot;$\dots$&quot;, (5.5,6.5));<br /> label(&quot;$\dots$&quot;, (6.5,6.5));<br /> label(&quot;$17$&quot;, (1.5,1.5));<br /> label(&quot;$18$&quot;, (1.5,2.5));<br /> label(&quot;$19$&quot;, (1.5,3.5));<br /> label(&quot;$20$&quot;, (1.5,4.5));<br /> label(&quot;$21$&quot;, (1.5,5.5));<br /> label(&quot;$16$&quot;, (2.5,1.5));<br /> label(&quot;$5$&quot;, (2.5,2.5));<br /> label(&quot;$6$&quot;, (2.5,3.5));<br /> label(&quot;$7$&quot;, (2.5,4.5));<br /> label(&quot;$22$&quot;, (2.5,5.5));<br /> label(&quot;$15$&quot;, (3.5,1.5));<br /> label(&quot;$4$&quot;, (3.5,2.5));<br /> label(&quot;$1$&quot;, (3.5,3.5));<br /> label(&quot;$8$&quot;, (3.5,4.5));<br /> label(&quot;$23$&quot;, (3.5,5.5));<br /> label(&quot;$14$&quot;, (4.5,1.5));<br /> label(&quot;$3$&quot;, (4.5,2.5));<br /> label(&quot;$2$&quot;, (4.5,3.5));<br /> label(&quot;$9$&quot;, (4.5,4.5));<br /> label(&quot;$24$&quot;, (4.5,5.5));<br /> label(&quot;$13$&quot;, (5.5,1.5));<br /> label(&quot;$12$&quot;, (5.5,2.5));<br /> label(&quot;$11$&quot;, (5.5,3.5));<br /> label(&quot;$10$&quot;, (5.5,4.5));<br /> label(&quot;$25$&quot;, (5.5,5.5));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> The point &lt;math&gt;P(a,b)&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt;-plane is first rotated counterclockwise by &lt;math&gt;90^\circ&lt;/math&gt; around the point &lt;math&gt;(1,5)&lt;/math&gt; and then reflected about the line &lt;math&gt;y = -x&lt;/math&gt;. The image of &lt;math&gt;P&lt;/math&gt; after these two transformations is at &lt;math&gt;(-6,3)&lt;/math&gt;. What is &lt;math&gt;b - a ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> An inverted cone with base radius &lt;math&gt;12 \text{cm}&lt;/math&gt; and height &lt;math&gt;18\text{cm}&lt;/math&gt; is full of water. The water is poured into a tall cylinder whose horizontal base has a radius of &lt;math&gt;24\text{cm}&lt;/math&gt;. What is the height in centimeters of the water in the cylinder?<br /> <br /> &lt;math&gt;\textbf{(A) }1.5 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }4.5 \qquad \textbf{(E) }6&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?<br /> <br /> &lt;math&gt;\textbf{(A)} ~24 \qquad\textbf{(B)} ~30 \qquad\textbf{(C)} ~48 \qquad\textbf{(D)} ~60 \qquad\textbf{(E)} ~64&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> Let &lt;math&gt;N = 34 \cdot 34 \cdot 63 \cdot 270&lt;/math&gt;. What is the ratio of the sum of the odd divisors of &lt;math&gt;N&lt;/math&gt; to the sum of the even divisors of &lt;math&gt;N&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer and &lt;math&gt;d&lt;/math&gt; be a digit such that the value of the numeral &lt;math&gt;\underline{32d}&lt;/math&gt; in base &lt;math&gt;n&lt;/math&gt; equals &lt;math&gt;263&lt;/math&gt;, and the value of the numeral &lt;math&gt;\underline{324}&lt;/math&gt; in base &lt;math&gt;n&lt;/math&gt; equals the value of the numeral &lt;math&gt;\underline{11d1}&lt;/math&gt; in base six. What is &lt;math&gt;n + d ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> Three equally spaced parallel lines intersect a circle, creating three chords of lengths &lt;math&gt;38, 38,&lt;/math&gt; and &lt;math&gt;34&lt;/math&gt;. What is the distance between two adjacent parallel lines?<br /> <br /> &lt;math&gt;\textbf{(A)} ~5\frac{1}{2} \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~6\frac{1}{2} \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~7\frac{1}{2}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> The real number &lt;math&gt;x&lt;/math&gt; satisfies the equation &lt;math&gt;x+\frac{1}{x} = \sqrt{5}&lt;/math&gt;. What is the value of &lt;math&gt;x^{11}-7x^{7}+x^3?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, &lt;math&gt;1357&lt;/math&gt;, &lt;math&gt;89&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; are all uphill integers, but &lt;math&gt;32&lt;/math&gt;, &lt;math&gt;1240&lt;/math&gt;, and &lt;math&gt;466&lt;/math&gt; are not. How many uphill integers are divisible by &lt;math&gt;15&lt;/math&gt;?<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 /> [[2021 AMC 10B Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Ravon, Oscar, Aditi, Tyrone, and Kim play a card game. Each person is given &lt;math&gt;2&lt;/math&gt; cards out of a set of &lt;math&gt;10&lt;/math&gt; cards numbered &lt;math&gt;1, 2, 3,\cdots , 10&lt;/math&gt;. The score of a player is the sum of the numbers of their cards. The scores of the players are as follows: Ravon &lt;math&gt;11&lt;/math&gt;, Oscar &lt;math&gt;4&lt;/math&gt;, Aditi &lt;math&gt;7&lt;/math&gt;, Tyrone &lt;math&gt;16&lt;/math&gt;, Kim &lt;math&gt;17&lt;/math&gt;. Which of the following statements is true?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; ~Ravon was given card &lt;math&gt;3&lt;/math&gt; .<br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; ~Aditi was given card &lt;math&gt;3&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; ~Ravon was given card &lt;math&gt;4&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; ~Aditi was given card &lt;math&gt;4&lt;/math&gt;.<br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; ~Tyrone was given card &lt;math&gt;7&lt;/math&gt;.<br /> <br /> [[2021 AMC 10B Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> A fair &lt;math&gt;6&lt;/math&gt;-sided die is repeatedly rolled until an odd number appears. What is the probability that every even number appears at least once before the first occurrence of an odd number?<br /> <br /> &lt;math&gt;\textbf{(A)} ~\frac{1}{120} \qquad\textbf{(B)} ~\frac{1}{32} \qquad\textbf{(C)} ~\frac{1}{20} \qquad\textbf{(D)} ~\frac{3}{20} \qquad\textbf{(E)} ~\frac{1}{6}&lt;/math&gt;<br /> <br /> <br /> [[2021 AMC 10B Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> Suppose that &lt;math&gt;S&lt;/math&gt; is a finite set of positive integers. If the greatest integer in &lt;math&gt;S&lt;/math&gt; is removed from &lt;math&gt;S&lt;/math&gt;, then the average value (arithmetic mean) of the integers remaining is &lt;math&gt;32&lt;/math&gt;. If the least integer is &lt;math&gt;S&lt;/math&gt; is also removed, then the average value of the integers remaining is &lt;math&gt;35&lt;/math&gt;. If the greatest integer is then returned to the set, the average value of the integers rises of &lt;math&gt;40&lt;/math&gt;. The greatest integer in the original set &lt;math&gt;S&lt;/math&gt; is &lt;math&gt;72&lt;/math&gt; greater than the least integer in &lt;math&gt;S&lt;/math&gt;. What is the average value of all the integers in the set &lt;math&gt;S ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~36.2 \qquad\textbf{(B)} ~36.4 \qquad\textbf{(C)} ~36.6 \qquad\textbf{(D)} ~36.8 \qquad\textbf{(E)} ~37&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> The figure is constructed from &lt;math&gt;11&lt;/math&gt; line segments, each of which has length &lt;math&gt;2&lt;/math&gt;. The area of pentagon &lt;math&gt;ABCDE&lt;/math&gt; can be written is &lt;math&gt;\sqrt{m} + \sqrt{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are positive integers. What is &lt;math&gt;m + n ?&lt;/math&gt;<br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(-2.4638,4.10658);<br /> pair B=(-4,2.6567453480756127);<br /> pair C=(-3.47132,0.6335248637894945);<br /> pair D=(-1.464483379039766,0.6335248637894945);<br /> pair E=(-0.956630463955801,2.6567453480756127);<br /> pair F=(-2,2);<br /> pair G=(-3,2);<br /> draw(A--B--C--D--E--A);<br /> draw(A--F--A--G);<br /> draw(B--F--C);<br /> draw(E--G--D);<br /> label(&quot;A&quot;,A,N);<br /> label(&quot;B&quot;,B,W);<br /> label(&quot;C&quot;,C,W);<br /> label(&quot;D&quot;,D,E);<br /> label(&quot;E&quot;,E,dir(0));<br /> dot(A^^B^^C^^D^^E^^F^^G);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A square piece of paper has side length &lt;math&gt;1&lt;/math&gt; and vertices &lt;math&gt;A,B,C,&lt;/math&gt; and &lt;math&gt;D&lt;/math&gt; in that order. As shown in the figure, the paper is folded so that vertex &lt;math&gt;C&lt;/math&gt; meets edge &lt;math&gt;\overline{AD}&lt;/math&gt; at point &lt;math&gt;C'&lt;/math&gt;, and edge &lt;math&gt;\overline{AB}&lt;/math&gt; at point &lt;math&gt;E&lt;/math&gt;. Suppose that &lt;math&gt;C'D = \frac{1}{3}&lt;/math&gt;. What is the perimeter of triangle &lt;math&gt;\bigtriangleup AEC' ?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> pair A=(0,1);<br /> pair CC=(0.666666666666,1);<br /> pair D=(1,1);<br /> pair F=(1,0.62);<br /> pair C=(1,0);<br /> pair B=(0,0);<br /> pair G=(0,0.25);<br /> pair H=(-0.13,0.41);<br /> pair E=(0,0.5);<br /> dot(A^^CC^^D^^C^^B^^E);<br /> draw(E--A--D--F);<br /> draw(G--B--C--F, dashed);<br /> fill(E--CC--F--G--H--E--CC--cycle, gray);<br /> draw(E--CC--F--G--H--E--CC);<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,NE);<br /> label(&quot;E&quot;,E,NW);<br /> label(&quot;C'&quot;,CC,N);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Ang, Ben, and Jasmin each have &lt;math&gt;5&lt;/math&gt; blocks, colored red, blue, yellow, white, and green; and there are &lt;math&gt;5&lt;/math&gt; empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives &lt;math&gt;3&lt;/math&gt; blocks all of the same color is &lt;math&gt;\frac{m}{n}&lt;/math&gt;, where &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;m + n ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~47 \qquad\textbf{(B)} ~94 \qquad\textbf{(C)} ~227 \qquad\textbf{(D)} ~471 \qquad\textbf{(E)} ~542&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> A square with side length &lt;math&gt;8&lt;/math&gt; is colored white except for &lt;math&gt;4&lt;/math&gt; black isosceles right triangular regions with legs of length &lt;math&gt;2&lt;/math&gt; in each corner of the square and a black diamond with side length &lt;math&gt;2\sqrt{2}&lt;/math&gt; in the center of the square, as shown in the diagram. A circular coin with diameter &lt;math&gt;1&lt;/math&gt; is dropped onto the square and lands in a random location where the coin is completely contained within the square, The probability that the coin will cover part of the black region of the square can be written as &lt;math&gt;\frac{1}{196}(a+b\sqrt{2}+\pi)&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are positive integers. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> /* Made by samrocksnature */<br /> draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0));<br /> fill((2,0)--(0,2)--(0,0)--cycle, black);<br /> fill((6,0)--(8,0)--(8,2)--cycle, black);<br /> fill((8,6)--(8,8)--(6,8)--cycle, black);<br /> fill((0,6)--(2,8)--(0,8)--cycle, black);<br /> fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black);<br /> filldraw(circle((2.6,3.31),0.47),gray);<br /> <br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) }64 \qquad \textbf{(B) }66 \qquad \textbf{(C) }68 \qquad \textbf{(D) }70 \qquad \textbf{(E) }72&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one &quot;wall&quot; among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; can be changed into any of the following by one move: &lt;math&gt;(3,2),(2, 1, 2),(4),(4, 1),(2, 2),&lt;/math&gt; or &lt;math&gt;(1, 1, 2)&lt;/math&gt;.<br /> &lt;asy&gt;<br /> /* CREDITS */<br /> /* Made by samrocksnature */<br /> /* Modified commas an periods by forester2015 */<br /> <br /> /* Import and Set variables */<br /> import graph; size(10cm);<br /> real labelscalefactor = 0.5; /* changes label-to-point distance */<br /> pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */<br /> pen dotstyle = black; /* point style */<br /> real xmin = -20.98617651190462, xmax = 71.97119514540032, ymin = -24.802885633158763, ymax = 28.83570218998556; /* image dimensions */<br /> <br /> /* draw figures */<br /> draw((14,4)--(13.010050506338834,3.0100505063388336), linewidth(1));<br /> draw((14,4)--(13.010050506338834,4.989949493661166), linewidth(1));<br /> draw((10,4)--(14,4), linewidth(1));<br /> draw((4,6)--(8,6), linewidth(1));<br /> draw((4,2)--(8,2), linewidth(1));<br /> draw((8,2)--(8,6), linewidth(1));<br /> draw((4,6)--(4,2), linewidth(1));<br /> draw((6,6)--(6,2), linewidth(1));<br /> draw((-6,6)--(-6,2), linewidth(1));<br /> draw((-6,6)--(2,6), linewidth(1));<br /> draw((2,6)--(2,2), linewidth(1));<br /> draw((2,2)--(-6,2), linewidth(1));<br /> draw((-4,2)--(-4,6), linewidth(1));<br /> draw((-2,6)--(-2,2), linewidth(1));<br /> draw((0,2)--(0,6), linewidth(1));<br /> draw((50,6)--(50,2), linewidth(1));<br /> draw((50,2)--(58,2), linewidth(1));<br /> draw((58,2)--(58,6), linewidth(1));<br /> draw((58,6)--(50,6), linewidth(1));<br /> draw((52,6)--(52,2), linewidth(1));<br /> draw((54,6)--(54,2), linewidth(1));<br /> draw((56,6)--(56,2), linewidth(1));<br /> draw((32,6)--(32,2), linewidth(1));<br /> draw((46,2)--(46,6), linewidth(1));<br /> draw((34,6)--(34,2), linewidth(1));<br /> draw((36,2)--(36,6), linewidth(1));<br /> draw((38,6)--(38,2), linewidth(1));<br /> draw((40,2)--(40,6), linewidth(1));<br /> draw((42,6)--(42,2), linewidth(1));<br /> draw((44,2)--(44,6), linewidth(1));<br /> draw((16,6)--(16,2), linewidth(1));<br /> draw((28,2)--(28,6), linewidth(1));<br /> draw((18,6)--(18,2), linewidth(1));<br /> draw((20,6)--(20,2), linewidth(1));<br /> draw((22,6)--(22,2), linewidth(1));<br /> draw((24,6)--(24,2), linewidth(1));<br /> draw((26,6)--(26,2), linewidth(1));<br /> draw((16,6)--(22,6), linewidth(1));<br /> draw((24,6)--(28,6), linewidth(1));<br /> draw((16,2)--(22,2), linewidth(1));<br /> draw((24,2)--(28,2), linewidth(1));<br /> draw((32,6)--(36,6), linewidth(1));<br /> draw((32,2)--(36,2), linewidth(1));<br /> draw((38,6)--(40,6), linewidth(1));<br /> draw((38,2)--(40,2), linewidth(1));<br /> draw((42,6)--(46,6), linewidth(1));<br /> draw((42,2)--(46,2), linewidth(1));<br /> <br /> /* dots and labels */<br /> label(&quot;,&quot;,(59,2));<br /> label(&quot;.&quot;,(60,2));<br /> label(&quot;.&quot;,(61,2));<br /> label(&quot;.&quot;,(62,2));<br /> label(&quot;,&quot;,(29,2));<br /> label(&quot;,&quot;,(47,2));<br /> &lt;/asy&gt;<br /> Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?<br /> <br /> &lt;math&gt;\textbf{(A)} ~(6, 1, 1) \qquad\textbf{(B)} ~(6, 2, 1) \qquad\textbf{(C)} ~(6, 2, 2) \qquad\textbf{(D)} ~(6, 3, 1) \qquad\textbf{(E)} ~(6, 3, 2)&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be the set of lattice points in the coordinate plane, both of whose coordinates are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;30&lt;/math&gt;, inclusive. Exactly &lt;math&gt;300&lt;/math&gt; points in &lt;math&gt;S&lt;/math&gt; lie on or below a line with equation &lt;math&gt;y = mx&lt;/math&gt;. The possible values of &lt;math&gt;m&lt;/math&gt; lie in an interval of length &lt;math&gt;\frac{a}{b}&lt;/math&gt;, where &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;a + b ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)} ~31 \qquad\textbf{(B)} ~47 \qquad\textbf{(C)} ~62 \qquad\textbf{(D)} ~72 \qquad\textbf{(E)} ~85&lt;/math&gt;<br /> <br /> [[2021 AMC 10B Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2021|ab=B|before=[[2021 AMC 10A]]|after=[[2022 AMC 10A]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_23&diff=134881 2016 AMC 12A Problems/Problem 23 2020-10-10T21:09:18Z <p>Memories: /* Solution 8: 3D geometry */</p> <hr /> <div>==Problem==<br /> <br /> Three numbers in the interval &lt;math&gt;\left[0,1\right]&lt;/math&gt; are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?<br /> <br /> &lt;math&gt;\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56&lt;/math&gt;<br /> <br /> ==Solution==<br /> <br /> ===Solution 1: Super WLOG===<br /> <br /> WLOG assume &lt;math&gt;a&lt;/math&gt; is the largest. Scale the triangle to &lt;math&gt;1,{b}/{a},{c}/{a}&lt;/math&gt; or &lt;math&gt;1,x,y&lt;/math&gt; with &lt;math&gt;x,y \in [0,1]&lt;/math&gt; and equally likely to be any pair of numbers within the interval. Then, &lt;math&gt;x+y&gt;1&lt;/math&gt;, meaning the solution is &lt;math&gt;\boxed{\textbf{(C)}\;1/2}&lt;/math&gt;, as shown in the graph below.<br /> &lt;asy&gt;<br /> pair A = (0,0);<br /> pair B = (1,0);<br /> pair C = (1,1);<br /> pair D = (0,1);<br /> pair E = (0,0);<br /> <br /> draw(A--B--C--D--cycle);<br /> draw(B--D,dashed);<br /> fill(B--D--C--cycle,gray);<br /> <br /> label(&quot;$0$&quot;,A,SW);<br /> label(&quot;$1$&quot;,B,S);<br /> label(&quot;$1$&quot;,D,W);<br /> label(&quot;$y$&quot;,(0,.5),W);<br /> label(&quot;$x$&quot;,(.5,0),S);<br /> label(&quot;$x+y&gt;1$&quot;,(5/7,5/7));<br /> &lt;/asy&gt;<br /> <br /> ===Solution 2: Conditional Probability===<br /> <br /> WLOG, let the largest of the three numbers drawn be &lt;math&gt;a&gt;0&lt;/math&gt;. Then the other two numbers are drawn uniformly and independently from the interval &lt;math&gt;[0,a]&lt;/math&gt;. The probability that their sum is greater than &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\;1/2.}&lt;/math&gt;<br /> <br /> ===Solution 3: Calculus===<br /> <br /> When &lt;math&gt;a&gt;b&lt;/math&gt;, consider two cases:<br /> <br /> 1) &lt;math&gt;0&lt;a&lt;\frac{1}{2}&lt;/math&gt;, then <br /> &lt;math&gt;\int_{0}^{\frac{1}{2}} \int_{0}^{a}2b \,\text{d}b\,\text{d}a=\frac{1}{24}&lt;/math&gt;<br /> <br /> 2)&lt;math&gt;\frac{1}{2}&lt;a&lt;1&lt;/math&gt;, then <br /> &lt;math&gt;\int_{\frac{1}{2}}^{1} \left(\int_{0}^{1-a}2b \,\text{d}b + \int_{1-a}^{a}1+b-a \,db\right)\text{d}a=\frac{5}{24}&lt;/math&gt;<br /> <br /> &lt;math&gt;a&lt;b&lt;/math&gt; is the same. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> ===Solution 4: Geometry===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. The region where, WLOG, side &lt;math&gt;z&lt;/math&gt; is too long, &lt;math&gt;z\geq x+y&lt;/math&gt;, is a pyramid with a base of area &lt;math&gt;\frac{1}{2}&lt;/math&gt; and height &lt;math&gt;1&lt;/math&gt;, so its volume is &lt;math&gt;\frac{\frac{1}{2}\cdot 1}{3}=\frac{1}{6}&lt;/math&gt;. Accounting for the corresponding cases in &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; multiplies our answer by &lt;math&gt;3&lt;/math&gt;, so we have excluded a total volume of &lt;math&gt;\frac{1}{2}&lt;/math&gt; from the space of possible probabilities. Subtracting this from &lt;math&gt;1&lt;/math&gt; leaves us with a final answer of &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> === Solution 5: More Calculus ===<br /> <br /> The probability of this occurring is the volume of the corresponding region within a &lt;math&gt;1 \times 1 \times 1&lt;/math&gt; cube, where each point &lt;math&gt;(x,y,z)&lt;/math&gt; corresponds to a choice of values for each of &lt;math&gt;x, y,&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. We take a horizontal cross section of the cube, essentially picking a value for z. The area where the triangle inequality will not hold is when &lt;math&gt;x + y &lt; z&lt;/math&gt;, which has area &lt;math&gt;\frac{z^2}{2}&lt;/math&gt; or when &lt;math&gt;x+z&lt;y&lt;/math&gt; or &lt;math&gt;y+z&lt;x&lt;/math&gt;, which have an area of &lt;math&gt;\frac{(1-z)^2}{2}+\frac{(1-z)^2}{2} = (1-z)^2.&lt;/math&gt; Integrating this expression from 0 to 1 in the form<br /> <br /> &lt;math&gt;\int_0^1 \left(\frac{z^2}{2} + (1-z)^2\right) dz = \bigg[\frac{z^3}{2} - z^2 + z \biggr |_0^1 = \frac{1}{2} -1 + 1 = \frac{1}{2}&lt;/math&gt;<br /> <br /> === Solution 6: Geometry in 2-D ===<br /> WLOG assume that &lt;math&gt;z&lt;/math&gt; is the largest number and hence the largest side. Then &lt;math&gt;x,y \leq z&lt;/math&gt;. We can set up a square that is &lt;math&gt;z&lt;/math&gt; by &lt;math&gt;z&lt;/math&gt; in the &lt;math&gt;xy&lt;/math&gt; plane. We are wanting all the points within this square that satisfy &lt;math&gt;x+y &gt; z&lt;/math&gt;. This happens to be a line dividing the square into 2 equal regions. Thus the answer is &lt;math&gt;\frac{1}{2}&lt;/math&gt;.<br /> <br /> <br /> [][] diagram for this problem goes here (z by z square)<br /> <br /> === Solution 7: More WLOG, Complementary Probability ===<br /> The triangle inequality simplifies to considering only one case: &lt;math&gt;\text{the smallest side+ the second smallest side} &gt; \text{the largest side}&lt;/math&gt;. Consider the complement (the same statement, except with a less than or equal to). Assume (WLOG) &lt;math&gt;a&lt;/math&gt; is the largest, so on average &lt;math&gt;a=1/2&lt;/math&gt; (now equal to becomes a degenerate case with probability &lt;math&gt;0&lt;/math&gt;, so we no longer need to consider it). We now want &lt;math&gt;b+c&lt;1/2&lt;/math&gt;, so imagine choosing &lt;math&gt;b+c&lt;/math&gt; at once rather than independently. But we know that &lt;math&gt;b+c&lt;/math&gt; is between &lt;math&gt;0&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. The complement is thus: &lt;math&gt;(1/2-0)/2=1/4&lt;/math&gt;. But keep in mind that we choose each &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; randomly and independently, so if there are &lt;math&gt;k&lt;/math&gt; ways to choose &lt;math&gt;b+c&lt;/math&gt; together, there are &lt;math&gt;2k&lt;/math&gt; ways to choose them separately, and therefore the complement actually doubles to match each case (a good example of this is to restrict b and c to integers such that if &lt;math&gt;b+c=3&lt;/math&gt;, then we only count this once, but in reality: we have two cases &lt;math&gt;1+2&lt;/math&gt;, and &lt;math&gt;2+1&lt;/math&gt;; similar reasoning also generalizes to non-integral values). The complement is then actually &lt;math&gt;2(1/4)=1/2&lt;/math&gt;. Therefore, our desired probability is given by &lt;math&gt;1-\text{complement}=1/2, C&lt;/math&gt;<br /> <br /> === Solution 8: 3D geometry ===<br /> <br /> We can draw a 3D space where each coordinate is in the range [0,1]. Drawing the lines &lt;math&gt;x+y&gt;z, x+z&gt;y,&lt;/math&gt; and &lt;math&gt;y+z&gt;x,&lt;/math&gt; We have a 3D space that consists of two tetrahedrons. One is a regular tetrahedron with side length &lt;math&gt;\sqrt2&lt;/math&gt; and the other has 3 sides of length &lt;math&gt;\sqrt2&lt;/math&gt; and 3 sides of length &lt;math&gt;1.&lt;/math&gt; The volume of this region is &lt;math&gt;\frac 1 2&lt;/math&gt;. Hence our solution is &lt;math&gt;C.&lt;/math&gt;<br /> <br /> === Solution 9: Cheap Solution ===<br /> <br /> Pretend that the values of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; are integers ranging from &lt;math&gt;[1,n]&lt;/math&gt;. Test out the probability of the first few values of &lt;math&gt;n&lt;/math&gt; (for example, &lt;math&gt;P(2) = \frac{5}{8}, P(3) = \frac{15}{27}, P(4) = \frac{34}{64}, P(5) = \frac{65}{125}&lt;/math&gt;). Since real numbers contain infinite increments, the answer is the limit as &lt;math&gt;n&lt;/math&gt; approaches infinity of &lt;math&gt;P(n)&lt;/math&gt;, which is easily hypothesized as &lt;math&gt;\frac{1}{2}&lt;/math&gt;, or &lt;math&gt;C.&lt;/math&gt;<br /> -solution by fidgetboss_4000 get rect<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2016|ab=A|num-b=22|num-a=24}}<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems&diff=126328 2020 AMC 12A Problems 2020-06-24T13:52:41Z <p>Memories: /* Problem 1 */</p> <hr /> <div>{{AMC12 Problems|year=2020|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> Carlos took &lt;math&gt;70\%&lt;/math&gt; of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> The acronym AMC is shown in the rectangular grid below with grid lines spaced &lt;math&gt;1&lt;/math&gt; unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> import olympiad;<br /> unitsize(25);<br /> for (int i = 0; i &lt; 3; ++i) {<br /> for (int j = 0; j &lt; 9; ++j) {<br /> pair A = (j,i);<br /> <br /> }<br /> }<br /> for (int i = 0; i &lt; 3; ++i) {<br /> for (int j = 0; j &lt; 9; ++j) {<br /> if (j != 8) {<br /> draw((j,i)--(j+1,i), dashed);<br /> }<br /> if (i != 2) {<br /> draw((j,i)--(j,i+1), dashed);<br /> }<br /> }<br /> }<br /> draw((0,0)--(2,2),linewidth(2));<br /> draw((2,0)--(2,2),linewidth(2));<br /> draw((1,1)--(2,1),linewidth(2));<br /> draw((3,0)--(3,2),linewidth(2));<br /> draw((5,0)--(5,2),linewidth(2));<br /> draw((4,1)--(3,2),linewidth(2));<br /> draw((4,1)--(5,2),linewidth(2));<br /> draw((6,0)--(8,0),linewidth(2));<br /> draw((6,2)--(8,2),linewidth(2));<br /> draw((6,0)--(6,2),linewidth(2));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> A driver travels for &lt;math&gt;2&lt;/math&gt; hours at &lt;math&gt;60&lt;/math&gt; miles per hour, during which her car gets &lt;math&gt;30&lt;/math&gt; miles per gallon of gasoline. She is paid &lt;math&gt;\$0.50&lt;/math&gt; per mile, and her only expense is gasoline at &lt;math&gt;\$2.00&lt;/math&gt; per gallon. What is her net rate of pay, in dollars per hour, after this expense?<br /> <br /> &lt;math&gt;\textbf{(A) }20 \qquad\textbf{(B) }22 \qquad\textbf{(C) }24 \qquad\textbf{(D) } 25\qquad\textbf{(E) } 26&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> <br /> ==Problem 4==<br /> <br /> How many &lt;math&gt;4&lt;/math&gt;-digit positive integers (that is, integers between &lt;math&gt;1000&lt;/math&gt; and &lt;math&gt;9999&lt;/math&gt;, inclusive) having only even digits are divisible by &lt;math&gt;5?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> <br /> ==Problem 5==<br /> <br /> The &lt;math&gt;25&lt;/math&gt; integers from &lt;math&gt;-10&lt;/math&gt; to &lt;math&gt;14,&lt;/math&gt; inclusive, can be arranged to form a &lt;math&gt;5&lt;/math&gt;-by-&lt;math&gt;5&lt;/math&gt; square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br /> <br /> &lt;math&gt;\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> <br /> ==Problem 6==<br /> <br /> In the plane figure shown below, &lt;math&gt;3&lt;/math&gt; of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;asy&gt;<br /> import olympiad;<br /> unitsize(25);<br /> filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7));<br /> filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7));<br /> filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7));<br /> for (int i = 0; i &lt; 5; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> pair A = (j,i);<br /> <br /> }<br /> }<br /> for (int i = 0; i &lt; 5; ++i) {<br /> for (int j = 0; j &lt; 6; ++j) {<br /> if (j != 5) {<br /> draw((j,i)--(j+1,i));<br /> }<br /> if (i != 4) {<br /> draw((j,i)--(j,i+1));<br /> }<br /> }<br /> }<br /> &lt;/asy&gt;<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 /> [[2020 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> <br /> ==Problem 7==<br /> <br /> Seven cubes, whose volumes are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;27&lt;/math&gt;, &lt;math&gt;64&lt;/math&gt;, &lt;math&gt;125&lt;/math&gt;, &lt;math&gt;216&lt;/math&gt;, and &lt;math&gt;343&lt;/math&gt; cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?<br /> <br /> &lt;math&gt;\textbf{(A) } 644 \qquad \textbf{(B) } 658 \qquad \textbf{(C) } 664 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 749&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> <br /> ==Problem 8==<br /> <br /> What is the median of the following list of &lt;math&gt;4040&lt;/math&gt; numbers&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;cmath&gt;1, 2, 3, ..., 2020, 1^2, 2^2, 3^2, ..., 2020^2&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) } 1974.5 \qquad \textbf{(B) } 1975.5 \qquad \textbf{(C) } 1976.5 \qquad \textbf{(D) } 1977.5 \qquad \textbf{(E) } 1978.5&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> <br /> ==Problem 9==<br /> <br /> How many solutions does the equation &lt;math&gt;\tan{(2x)} = \cos{(\tfrac{x}{2})}&lt;/math&gt; have on the interval &lt;math&gt;[0, 2\pi]?&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 /> [[2020 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> <br /> ==Problem 10==<br /> <br /> There is a unique positive integer &lt;math&gt;n&lt;/math&gt; such that&lt;cmath&gt;\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.&lt;/cmath&gt;What is the sum of the digits of &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> <br /> ==Problem 11==<br /> <br /> A frog sitting at the point &lt;math&gt;(1, 2)&lt;/math&gt; begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length &lt;math&gt;1&lt;/math&gt;, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices &lt;math&gt;(0,0), (0,4), (4,4),&lt;/math&gt; and &lt;math&gt;(4,0)&lt;/math&gt;. What is the probability that the sequence of jumps ends on a vertical side of the square&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{2} \qquad \textbf{(B) } \frac{5}{8} \qquad \textbf{(C) } \frac{2}{3} \qquad \textbf{(D) } \frac{3}{4} \qquad \textbf{(E) } \frac{7}{8}&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> <br /> ==Problem 12==<br /> <br /> Line &lt;math&gt;\ell&lt;/math&gt; in the coordinate plane has the equation &lt;math&gt;3x - 5y + 40 = 0&lt;/math&gt;. This line is rotated &lt;math&gt;45^{\circ}&lt;/math&gt; counterclockwise about the point &lt;math&gt;(20, 20)&lt;/math&gt; to obtain line &lt;math&gt;k&lt;/math&gt;. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of the &lt;math&gt;x&lt;/math&gt;-intercept of line &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> <br /> ==Problem 13==<br /> <br /> There are integers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, each greater than 1, such that&lt;cmath&gt;\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}} = \sqrt{N^{25}}&lt;/cmath&gt;for all &lt;math&gt;N &gt; 1&lt;/math&gt;. What is &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> <br /> ==Problem 14==<br /> <br /> Regular octagon &lt;math&gt;ABCDEFGH&lt;/math&gt; has area &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;m&lt;/math&gt; be the area of quadrilateral &lt;math&gt;ACEG&lt;/math&gt;. What is &lt;math&gt;\tfrac{m}{n}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> In the complex plane, let &lt;math&gt;A&lt;/math&gt; be the set of solutions to &lt;math&gt;z^3 - 8 = 0&lt;/math&gt; and let &lt;math&gt;B&lt;/math&gt; be the set of solutions to &lt;math&gt;z^3 - 8z^2 - 8z + 64 = 0&lt;/math&gt;. What is the greatest distance between a point of &lt;math&gt;A&lt;/math&gt; and a point of &lt;math&gt;B?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9 + \sqrt{3}&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> <br /> ==Problem 16==<br /> <br /> A point is chosen at random within the square in the coordinate plane whose vertices are &lt;math&gt;(0, 0), (2020, 0), (2020, 2020),&lt;/math&gt; and &lt;math&gt;(0, 2020)&lt;/math&gt;. The probability that the point is within &lt;math&gt;d&lt;/math&gt; units of a lattice point is &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;. (A point &lt;math&gt;(x, y)&lt;/math&gt; is a lattice point if &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are both integers.) What is &lt;math&gt;d&lt;/math&gt; to the nearest tenth&lt;math&gt;?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> <br /> ==Problem 17==<br /> <br /> The vertices of a quadrilateral lie on the graph of &lt;math&gt;y = \ln x&lt;/math&gt;, and the &lt;math&gt;x&lt;/math&gt;-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is &lt;math&gt;\ln \frac{91}{90}&lt;/math&gt;. What is the &lt;math&gt;x&lt;/math&gt;-coordinate of the leftmost vertex?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 13&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> <br /> ==Problem 18==<br /> <br /> Quadrilateral &lt;math&gt;ABCD&lt;/math&gt; satisfies &lt;math&gt;\angle ABC = \angle ACD = 90^{\circ}, AC = 20&lt;/math&gt;, and &lt;math&gt;CD = 30&lt;/math&gt;. Diagonals &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{BD}&lt;/math&gt; intersect at point &lt;math&gt;E&lt;/math&gt;, and &lt;math&gt;AE = 5&lt;/math&gt;. What is the area of quadrilateral &lt;math&gt;ABCD&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 330 \qquad\textbf{(B) } 340 \qquad\textbf{(C) } 350 \qquad\textbf{(D) } 360 \qquad\textbf{(E) } 370&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> <br /> ==Problem 19==<br /> <br /> There exists a unique strictly increasing sequence of nonnegative integers &lt;math&gt;a_1 &lt; a_2 &lt; … &lt; a_k&lt;/math&gt; such that&lt;cmath&gt;\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.&lt;/cmath&gt;What is &lt;math&gt;k?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> <br /> ==Problem 20==<br /> <br /> Let &lt;math&gt;T&lt;/math&gt; be the triangle in the coordinate plane with vertices &lt;math&gt;\left(0,0\right)&lt;/math&gt;, &lt;math&gt;\left(4,0\right)&lt;/math&gt;, and &lt;math&gt;\left(0,3\right)&lt;/math&gt;. Consider the following five isometries (rigid transformations) of the plane: rotations of &lt;math&gt;90^{\circ}&lt;/math&gt;, &lt;math&gt;180^{\circ}&lt;/math&gt;, and &lt;math&gt;270^{\circ}&lt;/math&gt; counterclockwise around the origin, reflection across the &lt;math&gt;x&lt;/math&gt;-axis, and reflection across the &lt;math&gt;y&lt;/math&gt;-axis. How many of the &lt;math&gt;125&lt;/math&gt; sequences of three of these transformations (not necessarily distinct) will return &lt;math&gt;T&lt;/math&gt; to its original position? (For example, a &lt;math&gt;180^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by a reflection across the &lt;math&gt;y&lt;/math&gt;-axis will return &lt;math&gt;T&lt;/math&gt; to its original position, but a &lt;math&gt;90^{\circ}&lt;/math&gt; rotation, followed by a reflection across the &lt;math&gt;x&lt;/math&gt;-axis, followed by another reflection across the &lt;math&gt;x&lt;/math&gt;-axis will not return &lt;math&gt;T&lt;/math&gt; to its original position.)<br /> <br /> &lt;math&gt;\textbf{(A) } 12\qquad\textbf{(B) } 15\qquad\textbf{(C) }17 \qquad\textbf{(D) }20 \qquad\textbf{(E) }25&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> <br /> ==Problem 21==<br /> <br /> How many positive integers &lt;math&gt;n&lt;/math&gt; are there such that &lt;math&gt;n&lt;/math&gt; is a multiple of &lt;math&gt;5&lt;/math&gt;, and the least common multiple of &lt;math&gt;5!&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt; equals &lt;math&gt;5&lt;/math&gt; times the greatest common divisor of &lt;math&gt;10!&lt;/math&gt; and &lt;math&gt;n?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> <br /> ==Problem 22==<br /> <br /> Let &lt;math&gt;(a_n)&lt;/math&gt; and &lt;math&gt;(b_n)&lt;/math&gt; be the sequences of real numbers such that<br /> &lt;cmath&gt;$<br /> (2 + i)^n = a_n + b_ni<br />$&lt;/cmath&gt;for all integers &lt;math&gt;n\geq 0&lt;/math&gt;, where &lt;math&gt;i = \sqrt{-1}&lt;/math&gt;. What is&lt;cmath&gt;\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) }\frac 38\qquad\textbf{(B) }\frac7{16}\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac9{16}\qquad\textbf{(E) }\frac47&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly &lt;math&gt;7&lt;/math&gt;. Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{7}{36} \qquad\textbf{(B) } \frac{5}{24} \qquad\textbf{(C) } \frac{2}{9} \qquad\textbf{(D) } \frac{17}{72} \qquad\textbf{(E) } \frac{1}{4}&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> <br /> ==Problem 24==<br /> <br /> Suppose that &lt;math&gt;\triangle ABC&lt;/math&gt; is an equilateral triangle of side length &lt;math&gt;s&lt;/math&gt;, with the property that there is a unique point &lt;math&gt;P&lt;/math&gt; inside the triangle such that &lt;math&gt;AP = 1&lt;/math&gt;, &lt;math&gt;BP = \sqrt{3}&lt;/math&gt;, and &lt;math&gt;CP = 2&lt;/math&gt;. What is &lt;math&gt;s?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 + \sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5 + \sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> <br /> ==Problem 25==<br /> <br /> The number &lt;math&gt;a = \tfrac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers, has the property that the sum of all real numbers &lt;math&gt;x&lt;/math&gt; satisfying&lt;cmath&gt;\lfloor x \rfloor \cdot \{x\} = a \cdot x^2&lt;/cmath&gt;is &lt;math&gt;420&lt;/math&gt;, where &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; denotes the greatest integer less than or equal to &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;\{x\} = x - \lfloor x \rfloor&lt;/math&gt; denotes the fractional part of &lt;math&gt;x&lt;/math&gt;. What is &lt;math&gt;p + q?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332&lt;/math&gt;<br /> <br /> [[2020 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> <br /> ==See also==<br /> {{AMC12 box|year=2020|ab=A|before=[[2019 AMC 12B Problems]]|after=[[2020 AMC 12B Problems]]}}<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_18&diff=118322 2016 AMC 10A Problems/Problem 18 2020-02-23T15:40:02Z <p>Memories: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Note that the sum of the numbers on each face must be 18, because &lt;math&gt;\frac{1+2+\cdots+8}{2}=18&lt;/math&gt;.<br /> So now consider the opposite edges (two edges which are parallel but not on same face of the cube);<br /> they must have the same sum value too.<br /> Now think about the points &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. If they are not on the same edge, they must be endpoints of opposite edges, and we should have &lt;math&gt;1+X=8+Y&lt;/math&gt;. However, this scenario would yield no solution for &lt;math&gt;[2,7]&lt;/math&gt;, which is a contradiction. <br /> <br /> The points &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are therefore on the same side and all edges parallel must also sum to &lt;math&gt;9&lt;/math&gt;.<br /> <br /> Now we have &lt;math&gt;4&lt;/math&gt; parallel sides &lt;math&gt;1-8, 2-7, 3-6, 4-5&lt;/math&gt;.<br /> Thinking about &lt;math&gt;4&lt;/math&gt; endpoints, we realize they need to sum to &lt;math&gt;18&lt;/math&gt;.<br /> It is easy to notice only &lt;math&gt;1-7-6-4&lt;/math&gt; and &lt;math&gt;8-2-3-5&lt;/math&gt; would work.<br /> <br /> So if we fix one direction &lt;math&gt;1-8 (&lt;/math&gt;or &lt;math&gt;8-1)&lt;/math&gt; all other &lt;math&gt;3&lt;/math&gt; parallel sides must lay in one particular direction. &lt;math&gt;(1-8,7-2,6-3,4-5)&lt;/math&gt; or &lt;math&gt;(8-1,2-7,3-6,5-4)&lt;/math&gt;<br /> <br /> Now, the problem is the same as arranging &lt;math&gt;4&lt;/math&gt; points in a two-dimensional square, which is &lt;math&gt;\frac{4!}{4}=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Again, all faces sum to &lt;math&gt;18.&lt;/math&gt; If &lt;math&gt;x,y,z&lt;/math&gt; are the vertices next to &lt;math&gt;1&lt;/math&gt;, then the remaining vertices are &lt;math&gt;17-x-y, 17-y-z, 17-x-z, x+y+z-16.&lt;/math&gt; Now it remains to test possibilities. Note that we must have &lt;math&gt;x+y+z&gt;17.&lt;/math&gt; Without loss of generality, let &lt;math&gt;x&lt;y&lt;z.&lt;/math&gt;<br /> &lt;cmath&gt;3,7,8\text{: Does not work.}&lt;/cmath&gt; <br /> &lt;cmath&gt;4,6,8\text{: Works}&lt;/cmath&gt;<br /> &lt;cmath&gt;4,7,8\text{: Works.}&lt;/cmath&gt; <br /> &lt;cmath&gt;5,6,7\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;5,6,8\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;5,7,8\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;6,7,8\text{: Works.}&lt;/cmath&gt;<br /> <br /> Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with &lt;math&gt;[x, y, z]&lt;/math&gt; and one with &lt;math&gt;[z, y, x]&lt;/math&gt;) for each combination of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; from above. So, our answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> <br /> We know the sum of each face is &lt;math&gt;18.&lt;/math&gt; If we look at an edge of the cube whose numbers sum to &lt;math&gt;x&lt;/math&gt;, it must be possible to achieve the sum &lt;math&gt;18-x&lt;/math&gt; in two distinct ways, looking at the two faces which contain the edge. If &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; were on the same edge, it is possible to achieve the desired sum only with the numbers &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; since the values must be distinct. Similarly, if &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; were on the same edge, the only way to get the sum is with &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. This means that &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are not on the same edge as &lt;math&gt;8&lt;/math&gt;, or in other words they are diagonally across from it on the same face, or on the other end of the cube.<br /> <br /> Now we look at three cases, each yielding two solutions which are reflections of each other:<br /> <br /> 1) &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are diagonally opposite &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 2) &lt;math&gt;6&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;7&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 3) &lt;math&gt;7&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;6&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> <br /> This means the answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2016|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_18&diff=118321 2016 AMC 10A Problems/Problem 18 2020-02-23T15:38:15Z <p>Memories: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Note that the sum of the numbers on each face must be 18, because &lt;math&gt;\frac{1+2+\cdots+8}{2}=18&lt;/math&gt;.<br /> So now consider the opposite edges (two edges which are parallel but not on same face of the cube);<br /> they must have the same sum value too.<br /> Now think about the points &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. If they are not on the same edge, they must be endpoints of opposite edges, and we should have &lt;math&gt;1+X=8+Y&lt;/math&gt;, but no solution for &lt;math&gt;[2,7]&lt;/math&gt;, which is a contradiction. <br /> <br /> The points &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are therefore on the same side and all edges parallel must also sum to &lt;math&gt;9&lt;/math&gt;.<br /> <br /> Now we have &lt;math&gt;4&lt;/math&gt; parallel sides &lt;math&gt;1-8, 2-7, 3-6, 4-5&lt;/math&gt;.<br /> thinking about &lt;math&gt;4&lt;/math&gt; endpoints number need to have a sum of &lt;math&gt;18&lt;/math&gt;.<br /> It is easy to notice only &lt;math&gt;1-7-6-4&lt;/math&gt; and &lt;math&gt;8-2-3-5&lt;/math&gt; would work.<br /> <br /> So if we fix one direction &lt;math&gt;1-8 (&lt;/math&gt;or &lt;math&gt;8-1)&lt;/math&gt; all other &lt;math&gt;3&lt;/math&gt; parallel sides must lay in one particular direction. &lt;math&gt;(1-8,7-2,6-3,4-5)&lt;/math&gt; or &lt;math&gt;(8-1,2-7,3-6,5-4)&lt;/math&gt;<br /> <br /> Now, the problem is the same as arranging &lt;math&gt;4&lt;/math&gt; points in a two-dimensional square, which is &lt;math&gt;\frac{4!}{4}=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Again, all faces sum to &lt;math&gt;18.&lt;/math&gt; If &lt;math&gt;x,y,z&lt;/math&gt; are the vertices next to &lt;math&gt;1&lt;/math&gt;, then the remaining vertices are &lt;math&gt;17-x-y, 17-y-z, 17-x-z, x+y+z-16.&lt;/math&gt; Now it remains to test possibilities. Note that we must have &lt;math&gt;x+y+z&gt;17.&lt;/math&gt; Without loss of generality, let &lt;math&gt;x&lt;y&lt;z.&lt;/math&gt;<br /> &lt;cmath&gt;3,7,8\text{: Does not work.}&lt;/cmath&gt; <br /> &lt;cmath&gt;4,6,8\text{: Works}&lt;/cmath&gt;<br /> &lt;cmath&gt;4,7,8\text{: Works.}&lt;/cmath&gt; <br /> &lt;cmath&gt;5,6,7\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;5,6,8\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;5,7,8\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;6,7,8\text{: Works.}&lt;/cmath&gt;<br /> <br /> Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with &lt;math&gt;[x, y, z]&lt;/math&gt; and one with &lt;math&gt;[z, y, x]&lt;/math&gt;) for each combination of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; from above. So, our answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> <br /> We know the sum of each face is &lt;math&gt;18.&lt;/math&gt; If we look at an edge of the cube whose numbers sum to &lt;math&gt;x&lt;/math&gt;, it must be possible to achieve the sum &lt;math&gt;18-x&lt;/math&gt; in two distinct ways, looking at the two faces which contain the edge. If &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; were on the same edge, it is possible to achieve the desired sum only with the numbers &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; since the values must be distinct. Similarly, if &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; were on the same edge, the only way to get the sum is with &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. This means that &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are not on the same edge as &lt;math&gt;8&lt;/math&gt;, or in other words they are diagonally across from it on the same face, or on the other end of the cube.<br /> <br /> Now we look at three cases, each yielding two solutions which are reflections of each other:<br /> <br /> 1) &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are diagonally opposite &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 2) &lt;math&gt;6&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;7&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 3) &lt;math&gt;7&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;6&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> <br /> This means the answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2016|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Memories https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_18&diff=118320 2016 AMC 10A Problems/Problem 18 2020-02-23T15:37:46Z <p>Memories: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Each vertex of a cube is to be labeled with an integer &lt;math&gt;1&lt;/math&gt; through &lt;math&gt;8&lt;/math&gt;, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br /> <br /> &lt;math&gt;\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Note that the sum of the numbers on each face must be 18, because &lt;math&gt;\frac{1+2+\cdots+8}{2}=18&lt;/math&gt;.<br /> So now consider the opposite edges (two edges which are parallel but not on same face of the cube);<br /> they must have the same sum value too.<br /> Now think about the points &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;. If they are not on the same edge, they must be endpoints of opposite edges, and we should have &lt;math&gt;1+X=8+Y&lt;/math&gt;, but no solution for &lt;math&gt;[2,7]&lt;/math&gt;, which is a contradiction. <br /> <br /> The points &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; are therefore on the same side and all edges parallel must also sum to &lt;math&gt;9&lt;/math&gt;.<br /> <br /> Now we have &lt;math&gt;4&lt;/math&gt; parallel sides &lt;math&gt;1-8, 2-7, 3-6, 4-5&lt;/math&gt;.<br /> thinking about &lt;math&gt;4&lt;/math&gt; endpoints number need to have a sum of &lt;math&gt;18&lt;/math&gt;.<br /> It is easy to notice only &lt;math&gt;1-7-6-4&lt;/math&gt; and &lt;math&gt;8-2-3-5&lt;/math&gt; would work.<br /> <br /> So if we fix one direction &lt;math&gt;1-8 (&lt;/math&gt;or &lt;math&gt;8-1)&lt;/math&gt; all other &lt;math&gt;3&lt;/math&gt; parallel sides must lay in one particular direction. &lt;math&gt;(1-8,7-2,6-3,4-5)&lt;/math&gt; or &lt;math&gt;(8-1,2-7,3-6,5-4)&lt;/math&gt;<br /> <br /> Now, the problem is same as arranging &lt;math&gt;4&lt;/math&gt; points in a two-dimensional square, which is &lt;math&gt;\frac{4!}{4}=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> <br /> Again, all faces sum to &lt;math&gt;18.&lt;/math&gt; If &lt;math&gt;x,y,z&lt;/math&gt; are the vertices next to &lt;math&gt;1&lt;/math&gt;, then the remaining vertices are &lt;math&gt;17-x-y, 17-y-z, 17-x-z, x+y+z-16.&lt;/math&gt; Now it remains to test possibilities. Note that we must have &lt;math&gt;x+y+z&gt;17.&lt;/math&gt; Without loss of generality, let &lt;math&gt;x&lt;y&lt;z.&lt;/math&gt;<br /> &lt;cmath&gt;3,7,8\text{: Does not work.}&lt;/cmath&gt; <br /> &lt;cmath&gt;4,6,8\text{: Works}&lt;/cmath&gt;<br /> &lt;cmath&gt;4,7,8\text{: Works.}&lt;/cmath&gt; <br /> &lt;cmath&gt;5,6,7\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;5,6,8\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;5,7,8\text{: Does not work.}&lt;/cmath&gt;<br /> &lt;cmath&gt;6,7,8\text{: Works.}&lt;/cmath&gt;<br /> <br /> Keeping in mind that a) solutions that can be obtained by rotating each other count as one solution and b) a cyclic sequence is always asymmetrical, we can see that there are two solutions (one with &lt;math&gt;[x, y, z]&lt;/math&gt; and one with &lt;math&gt;[z, y, x]&lt;/math&gt;) for each combination of &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt; from above. So, our answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> == Solution 3 ==<br /> <br /> We know the sum of each face is &lt;math&gt;18.&lt;/math&gt; If we look at an edge of the cube whose numbers sum to &lt;math&gt;x&lt;/math&gt;, it must be possible to achieve the sum &lt;math&gt;18-x&lt;/math&gt; in two distinct ways, looking at the two faces which contain the edge. If &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt; were on the same edge, it is possible to achieve the desired sum only with the numbers &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; since the values must be distinct. Similarly, if &lt;math&gt;8&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; were on the same edge, the only way to get the sum is with &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt;. This means that &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are not on the same edge as &lt;math&gt;8&lt;/math&gt;, or in other words they are diagonally across from it on the same face, or on the other end of the cube.<br /> <br /> Now we look at three cases, each yielding two solutions which are reflections of each other:<br /> <br /> 1) &lt;math&gt;6&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; are diagonally opposite &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 2) &lt;math&gt;6&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;7&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> 3) &lt;math&gt;7&lt;/math&gt; is diagonally across the cube from &lt;math&gt;8&lt;/math&gt;, while &lt;math&gt;6&lt;/math&gt; is diagonally across from &lt;math&gt;8&lt;/math&gt; on the same face.<br /> <br /> This means the answer is &lt;math&gt;3\cdot 2=\boxed{\textbf{(C) }6.}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2016|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Memories