https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Hellopeople99&feedformat=atom AoPS Wiki - User contributions [en] 2022-09-26T11:49:09Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems&diff=140719 2017 AMC 10A Problems 2020-12-27T04:26:31Z <p>Hellopeople99: fixed page spacing</p> <hr /> <div>{{AMC10 Problems|year=2017|ab=A}}<br /> <br /> ==Problem 1==<br /> What is the value of &lt;math&gt;(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 70\qquad\textbf{(B)}\ 97\qquad\textbf{(C)}\ 127\qquad\textbf{(D)}\ 159\qquad\textbf{(E)}\ 729&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Pablo buys popsicles for his friends. The store sells single popsicles for &lt;math&gt;\$1&lt;/math&gt; each, &lt;math&gt;3&lt;/math&gt;-popsicle boxes for &lt;math&gt;\$2&lt;/math&gt; each, and &lt;math&gt;5&lt;/math&gt;-popsicle boxes for &lt;math&gt;\$3&lt;/math&gt;. What is the greatest number of popsicles that Pablo can buy with &lt;math&gt;\$8&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 8\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> Tamara has three rows of two &lt;math&gt;6&lt;/math&gt;-feet by &lt;math&gt;2&lt;/math&gt;-feet flower beds in her garden. The beds are separated and also surrounded by &lt;math&gt;1&lt;/math&gt;-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(0,10)--(15,10)--(15,0)--cycle);<br /> fill((0,0)--(0,10)--(15,10)--(15,0)--cycle, lightgray);<br /> draw((1,1)--(1,3)--(7,3)--(7,1)--cycle);<br /> fill((1,1)--(1,3)--(7,3)--(7,1)--cycle, white);<br /> draw((1,4)--(1,6)--(7,6)--(7,4)--cycle);<br /> fill((1,4)--(1,6)--(7,6)--(7,4)--cycle, white);<br /> draw((1,7)--(1,9)--(7,9)--(7,7)--cycle);<br /> fill((1,7)--(1,9)--(7,9)--(7,7)--cycle, white);<br /> <br /> draw((8,1)--(8,3)--(14,3)--(14,1)--cycle);<br /> fill((8,1)--(8,3)--(14,3)--(14,1)--cycle, white);<br /> draw((8,4)--(8,6)--(14,6)--(14,4)--cycle);<br /> fill((8,4)--(8,6)--(14,6)--(14,4)--cycle, white);<br /> draw((8,7)--(8,9)--(14,9)--(14,7)--cycle);<br /> fill((8,7)--(8,9)--(14,9)--(14,7)--cycle, white);<br /> <br /> defaultpen(fontsize(8, lineskip=1));<br /> label(&quot;2&quot;, (1.2, 2));<br /> label(&quot;6&quot;, (4, 1.2));<br /> defaultpen(linewidth(.2));<br /> draw((0,8)--(1,8), arrow=Arrows);<br /> draw((7,8)--(8,8), arrow=Arrows);<br /> draw((14,8)--(15,8), arrow=Arrows);<br /> draw((11,0)--(11,1), arrow=Arrows);<br /> draw((11,3)--(11,4), arrow=Arrows);<br /> draw((11,6)--(11,7), arrow=Arrows);<br /> label(&quot;1&quot;, (.5,7.8));<br /> label(&quot;1&quot;, (7.5,7.8));<br /> label(&quot;1&quot;, (14.5,7.8));<br /> label(&quot;1&quot;, (10.8,.5));<br /> label(&quot;1&quot;, (10.8,3.5));<br /> label(&quot;1&quot;, (10.8,6.5));<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt;\textbf{(A)}\ 72\qquad\textbf{(B)}\ 78\qquad\textbf{(C)}\ 90\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 150&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> Mia is “helping” her mom pick up &lt;math&gt;30&lt;/math&gt; toys that are strewn on the floor. Mia’s mom manages to put &lt;math&gt;3&lt;/math&gt; toys into the toy box every &lt;math&gt;30&lt;/math&gt; seconds, but each time immediately after those &lt;math&gt;30&lt;/math&gt; seconds have elapsed, Mia takes &lt;math&gt;2&lt;/math&gt; toys out of the box. How much time, in minutes, will it take Mia and her mom to put all &lt;math&gt;30&lt;/math&gt; toys into the box for the first time?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 14.5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 15.5&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> The sum of two nonzero real numbers is &lt;math&gt;4&lt;/math&gt; times their product. What is the sum of the reciprocals of the two numbers?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which of these statements necessarily follows logically?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{If Lewis did not receive an A, then he got all of the multiple choice questions wrong.}\\\textbf{(B)}\ \text{If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.}\\\textbf{(C)}\ \text{If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A. }\\\textbf{(D)}\ \text{If Lewis received an A, then he got all of the multiple choice questions right.}\\\textbf{(E)}\ \text{If Lewis received an A, then he got at least one of the multiple choice questions right.}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 40\%\qquad\textbf{(C)}\ 50\%\qquad\textbf{(D)}\ 60\%\qquad\textbf{(E)}\ 70\%&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> At a gathering of &lt;math&gt;30&lt;/math&gt; people, there are &lt;math&gt;20&lt;/math&gt; people who all know each other and &lt;math&gt;10&lt;/math&gt; people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 240\qquad\textbf{(B)}\ 245\qquad\textbf{(C)}\ 290\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 490&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> Minnie rides on a flat road at &lt;math&gt;20&lt;/math&gt; kilometers per hour (kph), downhill at &lt;math&gt;30&lt;/math&gt; kph, and uphill at &lt;math&gt;5&lt;/math&gt; kph. Penny rides on a flat road at &lt;math&gt;30&lt;/math&gt; kph, downhill at &lt;math&gt;40&lt;/math&gt; kph, and uphill at &lt;math&gt;10&lt;/math&gt; kph. Minnie goes from town &lt;math&gt;A&lt;/math&gt; to town &lt;math&gt;B&lt;/math&gt;, a distance of &lt;math&gt;10&lt;/math&gt; km all uphill, then from town &lt;math&gt;B&lt;/math&gt; to town &lt;math&gt;C&lt;/math&gt;, a distance of &lt;math&gt;15&lt;/math&gt; km all downhill, and then back to town &lt;math&gt;A&lt;/math&gt;, a distance of &lt;math&gt;20&lt;/math&gt; km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the &lt;math&gt;45&lt;/math&gt;-km ride than it takes Penny?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 45\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 65\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 95&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Joy has &lt;math&gt;30&lt;/math&gt; thin rods, one each of every integer length from &lt;math&gt;1&lt;/math&gt; cm through &lt;math&gt;30&lt;/math&gt; cm. She places the rods with lengths &lt;math&gt;3&lt;/math&gt; cm, &lt;math&gt;7&lt;/math&gt; cm, and &lt;math&gt;15&lt;/math&gt; cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> The region consisting of all points in three-dimensional space within &lt;math&gt;3&lt;/math&gt; units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume &lt;math&gt;216\pi&lt;/math&gt;. What is the length &lt;math&gt;\textit{AB}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> Let &lt;math&gt;S&lt;/math&gt; be a set of points &lt;math&gt;(x,y)&lt;/math&gt; in the coordinate plane such that two of the three quantities &lt;math&gt;3,~x+2,&lt;/math&gt; and &lt;math&gt;y-4&lt;/math&gt; are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{a single point} \qquad\textbf{(B)}\ \text{two intersecting lines} \\\qquad\textbf{(C)}\ \text{ three lines whose pairwise intersections are three distinct points} \\\qquad\textbf{(D)}\ \text{a triangle} \qquad\textbf{(E)}\ \text{three rays with a common endpoint}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> Define a sequence recursively by &lt;math&gt;F_{0}=0,~F_{1}=1,&lt;/math&gt; and &lt;math&gt;F_{n}=&lt;/math&gt; the remainder when &lt;math&gt;F_{n-1}+F_{n-2}&lt;/math&gt; is divided by &lt;math&gt;3,&lt;/math&gt; for all &lt;math&gt;n\geq 2.&lt;/math&gt; Thus the sequence starts &lt;math&gt;0,1,1,2,0,2,\ldots&lt;/math&gt; What is &lt;math&gt;F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was &lt;math&gt;A&lt;/math&gt; dollars. The cost of his movie ticket was &lt;math&gt;20\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his soda, while the cost of his soda was &lt;math&gt;5\%&lt;/math&gt; of the difference between &lt;math&gt;A&lt;/math&gt; and the cost of his movie ticket. To the nearest whole percent, what fraction of &lt;math&gt;A&lt;/math&gt; did Roger pay for his movie ticket and soda?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\%\qquad\textbf{(B)}\ 19\%\qquad\textbf{(C)}\ 22\%\qquad\textbf{(D)}\ 23\%\qquad\textbf{(E)}\ 25\%&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Chloé chooses a real number uniformly at random from the interval &lt;math&gt;[0, 2017]&lt;/math&gt;. Independently, Laurent chooses a real number uniformly at random from the interval &lt;math&gt;[0, 4034]&lt;/math&gt;. What is the probability that Laurent's number is greater than Chloé's number? (Assume they cannot be equal)<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{5}{6}\qquad\textbf{(E)}\ \frac{7}{8}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> There are 10 horses, named Horse 1, Horse 2, &lt;math&gt;\ldots&lt;/math&gt;, Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse &lt;math&gt;k&lt;/math&gt; runs one lap in exactly &lt;math&gt;k&lt;/math&gt; minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time &lt;math&gt;S&gt;0&lt;/math&gt;, in minutes, at which all 10 horses will again simultaneously be at the starting point is &lt;math&gt;S=2520&lt;/math&gt;. Let &lt;math&gt;T&gt;0&lt;/math&gt; be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of &lt;math&gt;T&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 /> [[2017 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Distinct points &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;Q&lt;/math&gt;, &lt;math&gt;R&lt;/math&gt;, &lt;math&gt;S&lt;/math&gt; lie on the circle &lt;math&gt;x^2+y^2=25&lt;/math&gt; and have integer coordinates. The distances &lt;math&gt;PQ&lt;/math&gt; and &lt;math&gt;RS&lt;/math&gt; are irrational numbers. What is the greatest possible value of the ratio &lt;math&gt;\frac{PQ}{RS}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 3\sqrt{5}\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 5\sqrt{2}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> Amelia has a coin that lands heads with probability &lt;math&gt;\tfrac{1}{3}&lt;/math&gt;, and Blaine has a coin that lands on heads with probability &lt;math&gt;\tfrac{2}{5}&lt;/math&gt;. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is &lt;math&gt;\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. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> Let &lt;math&gt;S(n)&lt;/math&gt; equal the sum of the digits of positive integer &lt;math&gt;n&lt;/math&gt;. For example, &lt;math&gt;S(1507) = 13&lt;/math&gt;. For a particular positive integer &lt;math&gt;n&lt;/math&gt;, &lt;math&gt;S(n) = 1274&lt;/math&gt;. Which of the following could be the value of &lt;math&gt;S(n+1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> A square with side length &lt;math&gt;x&lt;/math&gt; is inscribed in a right triangle with sides of length &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length &lt;math&gt;y&lt;/math&gt; is inscribed in another right triangle with sides of length &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, and &lt;math&gt;5&lt;/math&gt; so that one side of the square lies on the hypotenuse of the triangle. What is &lt;math&gt;\tfrac{x}{y}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Sides &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{AC}&lt;/math&gt; of equilateral triangle &lt;math&gt;ABC&lt;/math&gt; are tangent to a circle at points &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt; respectively. What fraction of the area of &lt;math&gt;\triangle ABC&lt;/math&gt; lies outside the circle?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad\textbf{(C)}\ \frac{1}{2}\qquad\textbf{(D)}\ \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad\textbf{(E)}\ \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> How many triangles with positive area have all their vertices at points &lt;math&gt;(i,j)&lt;/math&gt; in the coordinate plane, where &lt;math&gt;i&lt;/math&gt; and &lt;math&gt;j&lt;/math&gt; are integers between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;5&lt;/math&gt;, inclusive?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2128 \qquad\textbf{(B)}\ 2148 \qquad\textbf{(C)}\ 2160 \qquad\textbf{(D)}\ 2200 \qquad\textbf{(E)}\ 2300&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> For certain real numbers &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, the polynomial &lt;cmath&gt;g(x) = x^3 + ax^2 + x + 10&lt;/cmath&gt;has three distinct roots, and each root of &lt;math&gt;g(x)&lt;/math&gt; is also a root of the polynomial &lt;cmath&gt;f(x) = x^4 + x^3 + bx^2 + 100x + c.&lt;/cmath&gt;What is &lt;math&gt;f(1)&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ -9009\qquad\textbf{(B)}\ -8008\qquad\textbf{(C)}\ -7007\qquad\textbf{(D)}\ -6006\qquad\textbf{(E)}\ -5005&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> How many integers between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;, inclusive, have the property that some permutation of its digits is a multiple of &lt;math&gt;11&lt;/math&gt; between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999?&lt;/math&gt; For example, both &lt;math&gt;121&lt;/math&gt; and &lt;math&gt;211&lt;/math&gt; have this property.<br /> <br /> &lt;math&gt;\textbf{(A)}\ 226\qquad\textbf{(B)}\ 243\qquad\textbf{(C)}\ 270\qquad\textbf{(D)}\ 469\qquad\textbf{(E)}\ 486&lt;/math&gt;<br /> <br /> [[2017 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2017|ab=A|before=[[2016 AMC 10B Problems]]|after=[[2017 AMC 10B Problems]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[2017 AMC 10A]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Hellopeople99 https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems&diff=135702 2015 AMC 12A Problems 2020-10-23T23:34:24Z <p>Hellopeople99: /* Problem 25 */ fixed latex error with subscripts (copied problem statement from the solution page)</p> <hr /> <div>{{AMC12 Problems|year=2015|ab=A}}<br /> <br /> ==Problem 1==<br /> <br /> What is the value of &lt;math&gt;(2^0-1+5^2-0)^{-1}\times5?&lt;/math&gt;<br /> <br /> &lt;math&gt; \textbf{(A)}\ -125\qquad\textbf{(B)}\ -120\qquad\textbf{(C)}\ \frac{1}{5}\qquad\textbf{(D)}\ \frac{5}{24}\qquad\textbf{(E)}\ 25 &lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> <br /> Two of the three sides of a triangle are 20 and 15. Which of the following numbers is not a possible perimeter of the triangle?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 52\qquad\textbf{(B)}\ 57\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 67\qquad\textbf{(E)}\ 72 &lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> <br /> Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80. After he graded Payton's test, the class average became 81. What was Payton's score on the test?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95 &lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> <br /> The sum of two positive numbers is 5 times their difference. What is the ratio of the larger number to the smaller?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac54 \qquad\textbf{(B)}\ \frac32 \qquad\textbf{(C)}\ \frac95 \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac52 &lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> <br /> Amelia needs to estimate the quantity &lt;math&gt;\frac{a}{b} - c&lt;/math&gt;, where &lt;math&gt;a, b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of &lt;math&gt;\frac{a}{b} - c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{She rounds all three numbers up.}\\<br /> \qquad\textbf{(B)}\ \text{She rounds } a \text{ and } b \text{ up, and she rounds } c \text{ down.}\\<br /> \qquad\textbf{(C)}\ \text{She rounds } a \text{ and } c \text{ up, and she rounds } b \text{ down.} \\<br /> \qquad\textbf{(D)}\ \text{She rounds } a \text{ up, and she rounds } b \text{ and } c \text{ down.}\\<br /> \qquad\textbf{(E)}\ \text{She rounds } c \text{ up, and she rounds } a \text{ and } b \text{ down.} &lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> <br /> Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be &lt;math&gt;2 : 1&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> <br /> Two right circular cylinders have the same volume. The radius of the second cylinder is &lt;math&gt;10\%&lt;/math&gt; more than the radius of the first. What is the relationship between the heights of the two cylinders?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\ \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> <br /> The ratio of the length to the width of a rectangle is &lt;math&gt;4&lt;/math&gt; : &lt;math&gt;3&lt;/math&gt;. If the rectangle has diagonal of length &lt;math&gt;d&lt;/math&gt;, then the area may be expressed as &lt;math&gt;kd^2&lt;/math&gt; for some constant &lt;math&gt;k&lt;/math&gt;. What is &lt;math&gt;k&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac27 \qquad\textbf{(B)}\ \frac37 \qquad\textbf{(C)}\ \frac{12}{25} \qquad\textbf{(D)}\ \frac{16}{26} \qquad\textbf{(E)}\ \frac34&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> <br /> A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac16 \qquad\textbf{(C)}\ \frac15 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac12&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> <br /> Integers &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; with &lt;math&gt;x&gt;y&gt;0&lt;/math&gt; satisfy &lt;math&gt;x+y+xy=80&lt;/math&gt;. What is &lt;math&gt;x&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> <br /> On a sheet of paper, Isabella draws a circle of radius &lt;math&gt;2&lt;/math&gt;, a circle of radius &lt;math&gt;3&lt;/math&gt;, and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly &lt;math&gt;k \ge 0&lt;/math&gt; lines. How many different values of &lt;math&gt;k&lt;/math&gt; are possible?<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 /> [[2015 AMC 12A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> <br /> The parabolas &lt;math&gt;y=ax^2 - 2&lt;/math&gt; and &lt;math&gt;y=4 - bx^2&lt;/math&gt; intersect the coordinate axes in exactly four points, and these four points are the vertices of a kite of area &lt;math&gt;12&lt;/math&gt;. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 1\qquad\textbf{(B)}\ 1.5\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 2.5\qquad\textbf{(E)}\ 3&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> <br /> A league with 12 teams holds a round-robin tournament, with each team playing every other team exactly once. Games either end with one team victorious or else end in a draw. A team scores 2 points for every game it wins and 1 point for every game it draws. Which of the following is NOT a true statement about the list of 12 scores?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \text{There must be an even number of odd scores.}\\<br /> \qquad\textbf{(B)}\ \text{There must be an even number of even scores.}\\<br /> \qquad\textbf{(C)}\ \text{There cannot be two scores of }0\text{.}\\<br /> \qquad\textbf{(D)}\ \text{The sum of the scores must be at least }100\text{.}\\<br /> \qquad\textbf{(E)}\ \text{The highest score must be at least }12\text{.}&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> <br /> What is the value of &lt;math&gt;a&lt;/math&gt; for which &lt;math&gt;\frac{1}{\log_2 a} + \frac{1}{\log_3 a} + \frac{1}{\log_4 a} = 1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> <br /> What is the minimum number of digits to the right of the decimal point needed to express the fraction &lt;math&gt;\frac{123456789}{2^{26}\cdot 5^4}&lt;/math&gt; as a decimal?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 4\qquad\textbf{(B)}\ 22\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 104&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> <br /> Tetrahedron &lt;math&gt;ABCD&lt;/math&gt; has &lt;math&gt;AB=5,AC=3,BC=4,BD=4,AD=3,&lt;/math&gt; and &lt;math&gt;CD=\frac{12}{5}\sqrt{2}&lt;/math&gt;. What is the volume of the tetrahedron?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3\sqrt{2}\qquad\textbf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ \frac{24}{5}\qquad\textbf{(D)}\ 3\sqrt{3}\qquad\textbf{(E)}\ \frac{24}{5}\sqrt{2}&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> <br /> Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{47}{256} \qquad\textbf{(B)}\ \frac{3}{16} \qquad\textbf{(C)}\ \frac{49}{256} \qquad\textbf{(D)}\ \frac{25}{128} \qquad\textbf{(E)}\ \frac{51}{256}&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> <br /> The zeros of the function &lt;math&gt;f(x) = x^2-ax+2a&lt;/math&gt; are integers. What is the sum of the possible values of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> <br /> For some positive integers &lt;math&gt;p&lt;/math&gt;, there is a quadrilateral &lt;math&gt;ABCD&lt;/math&gt; with positive integer side lengths, perimeter &lt;math&gt;p&lt;/math&gt;, right angles at &lt;math&gt;B&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;AB=2&lt;/math&gt;, and &lt;math&gt;CD=AD&lt;/math&gt;. How many different values of &lt;math&gt;p&lt;2015&lt;/math&gt; are possible?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 30 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> <br /> Isosceles triangles &lt;math&gt;T&lt;/math&gt; and &lt;math&gt;T'&lt;/math&gt; are not congruent but have the same area and the same perimeter. The sides of &lt;math&gt;T&lt;/math&gt; have lengths of &lt;math&gt;5,5,&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt;, while those of &lt;math&gt;T'&lt;/math&gt; have lengths of &lt;math&gt;a,a,&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;. Which of the following numbers is closest to &lt;math&gt;b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 3 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> <br /> A circle of radius &lt;math&gt;r&lt;/math&gt; passes through both foci of, and exactly four points on, the ellipse with equation &lt;math&gt;x^2+16y^2=16&lt;/math&gt;. The set of all possible values of &lt;math&gt;r&lt;/math&gt; is an interval &lt;math&gt;[a,b)&lt;/math&gt;. What is &lt;math&gt;a+b&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 5\sqrt{2}+4 \qquad\textbf{(B)}\ \sqrt{17}+7 \qquad\textbf{(C)}\ 6\sqrt{2}+3 \qquad\textbf{(D)}\ \sqrt{15}+8 \qquad\textbf{(E)}\ 12&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> <br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;S(n)&lt;/math&gt; be the number of sequences of length &lt;math&gt;n&lt;/math&gt; consisting solely of the letters &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, with no more than three &lt;math&gt;A&lt;/math&gt;s in a row and no more than three &lt;math&gt;B&lt;/math&gt;s in a row. What is the remainder when &lt;math&gt;S(2015)&lt;/math&gt; is divided by 12?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 10&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> <br /> Let &lt;math&gt;S&lt;/math&gt; be a square of side length 1. Two points are chosen independently at random on the sides of &lt;math&gt;S&lt;/math&gt;. The probability that the straight-line distance between the points is at least &lt;math&gt;\frac12&lt;/math&gt; is &lt;math&gt;\frac{a-b\pi}{c}&lt;/math&gt;, where &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; are positive integers and &lt;math&gt;\text{gcd}(a,b,c) = 1&lt;/math&gt;. What is &lt;math&gt;a+b+c&lt;/math&gt;?<br /> <br /> &lt;math&gt; \textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> <br /> Rational numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are chosen at random among all rational numbers in the interval &lt;math&gt;[0,2)&lt;/math&gt; that can be written as fractions &lt;math&gt;\frac{n}{d}&lt;/math&gt; where &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; are integers with &lt;math&gt;1 \le d \le 5&lt;/math&gt;. What is the probability that<br /> &lt;cmath&gt;(\text{cos}(a\pi)+i\text{sin}(b\pi))^4&lt;/cmath&gt;<br /> is a real number?<br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{3}{50} \qquad\textbf{(B)}\ \frac{4}{25} \qquad\textbf{(C)}\ \frac{41}{200} \qquad\textbf{(D)}\ \frac{6}{25} \qquad\textbf{(E)}\ \frac{13}{50}&lt;/math&gt;<br /> <br /> [[2015 AMC 12A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> <br /> A collection of circles in the upper half-plane, all tangent to the &lt;math&gt;x&lt;/math&gt;-axis, is constructed in layers as follows. Layer &lt;math&gt;L_0&lt;/math&gt; consists of two circles of radii &lt;math&gt;70^2&lt;/math&gt; and &lt;math&gt;73^2&lt;/math&gt; that are externally tangent. For &lt;math&gt;k\ge1&lt;/math&gt;, the circles in &lt;math&gt;\bigcup_{j=0}^{k-1}L_j&lt;/math&gt; are ordered according to their points of tangency with the &lt;math&gt;x&lt;/math&gt;-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer &lt;math&gt;L_k&lt;/math&gt; consists of the &lt;math&gt;2^{k-1}&lt;/math&gt; circles constructed in this way. Let &lt;math&gt;S=\bigcup_{j=0}^{6}L_j&lt;/math&gt;, and for every circle &lt;math&gt;C&lt;/math&gt; denote by &lt;math&gt;r(C)&lt;/math&gt; its radius. What is<br /> &lt;cmath&gt;\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?&lt;/cmath&gt;<br /> <br /> &lt;asy&gt;<br /> import olympiad;<br /> size(350);<br /> defaultpen(linewidth(0.7));<br /> // define a bunch of arrays and starting points<br /> pair[] coord = new pair;<br /> int[] trav = {32,16,8,4,2,1};<br /> coord = (0,73^2); coord = (2*73*70,70^2);<br /> // draw the big circles and the bottom line<br /> path arc1 = arc(coord,coord.y,260,360);<br /> path arc2 = arc(coord,coord.y,175,280);<br /> fill((coord.x-910,coord.y)--arc1--cycle,gray(0.75));<br /> fill((coord.x+870,coord.y+425)--arc2--cycle,gray(0.75));<br /> draw(arc1^^arc2);<br /> draw((-930,0)--(70^2+73^2+850,0));<br /> // We now apply the findCenter function 63 times to get<br /> // the location of the centers of all 63 constructed circles.<br /> // The complicated array setup ensures that all the circles<br /> // will be taken in the right order<br /> for(int i = 0;i&lt;=5;i=i+1)<br /> {<br /> int skip = trav[i];<br /> for(int k=skip;k&lt;=64 - skip; k = k + 2*skip)<br /> {<br /> pair cent1 = coord[k-skip], cent2 = coord[k+skip];<br /> real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);<br /> real shiftx = cent1.x + sqrt(4*r1*rn);<br /> coord[k] = (shiftx,rn);<br /> }<br /> // Draw the remaining 63 circles<br /> }<br /> for(int i=1;i&lt;=63;i=i+1)<br /> {<br /> filldraw(circle(coord[i],coord[i].y),gray(0.75));<br /> }<br /> &lt;/asy&gt;<br /> <br /> <br /> &lt;math&gt; \textbf{(A)}\ \frac{286}{35} \qquad\textbf{(B)}\ \frac{583}{70} \qquad\textbf{(C)}\ \frac{715}{73}\qquad\textbf{(D)}\ \frac{143}{14} \qquad\textbf{(E)}\ \frac{1573}{146}&lt;/math&gt;<br /> <br /> <br /> [[2015 AMC 12A Problems/Problem 25|Solution]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2015|ab=A|before=[[2014 AMC 12B Problems]]|after=[[2015 AMC 12B Problems]]}}<br /> <br /> * [[AMC Problems and Solutions]]<br /> <br /> <br /> {{MAA Notice}}</div> Hellopeople99 https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_12&diff=119400 2020 AIME I Problems/Problem 12 2020-03-14T22:15:55Z <p>Hellopeople99: /* Solution 2 (Simpler, just basic mods and Fermat's theorem) */ fixed minor formatting error</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;n&lt;/math&gt; be the least positive integer for which &lt;math&gt;149^n-2^n&lt;/math&gt; is divisible by &lt;math&gt;3^3\cdot5^5\cdot7^7.&lt;/math&gt; Find the number of positive integer divisors of &lt;math&gt;n.&lt;/math&gt;<br /> <br /> == Solution 1==<br /> Lifting the Exponent shows that &lt;cmath&gt;v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1&lt;/cmath&gt; so thus, &lt;math&gt;3^2&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;. It also shows that &lt;cmath&gt;v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2&lt;/cmath&gt; so thus, &lt;math&gt;7^5&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;. <br /> <br /> The divisibility criterion implies &lt;math&gt;149^n\equiv 2^n \pmod{5}&lt;/math&gt;. This only occurs when &lt;math&gt;4 | n&lt;/math&gt;, so let &lt;math&gt;n=4m&lt;/math&gt;. We see &lt;cmath&gt;5 = v_5(149^{4m}-2^{4m}) = v_5((149^4)^{m}-(2^4)^{m}) = v_5(149^4-2^4)+v_5(m)&lt;/cmath&gt; Since &lt;math&gt;149^{4} \equiv 1 \pmod{25}&lt;/math&gt; and &lt;math&gt;16^1 \equiv 16 \pmod{25}&lt;/math&gt;, then &lt;math&gt;v_5(149^4-2^4)=1&lt;/math&gt;, so &lt;math&gt;v_5(m)=5&lt;/math&gt;, hence &lt;math&gt;4 \cdot 5^4&lt;/math&gt; divides &lt;math&gt;n&lt;/math&gt;. <br /> <br /> Since &lt;math&gt;3^2&lt;/math&gt;, &lt;math&gt;7^5&lt;/math&gt; and &lt;math&gt;4\cdot 5^4&lt;/math&gt; all divide &lt;math&gt;n&lt;/math&gt;, the smallest value of &lt;math&gt;n&lt;/math&gt; working is their LCM, also &lt;math&gt;3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5&lt;/math&gt;. Thus the number of divisors is &lt;math&gt;(2+1)(2+1)(4+1)(5+1) = \boxed{270}&lt;/math&gt;.<br /> <br /> ~kevinmathz + fireflame241<br /> <br /> == Solution 2 (Simpler, just basic mods and Fermat's theorem)==<br /> <br /> Note that for all n, &lt;math&gt;149^n - 2^n&lt;/math&gt; is divisible by &lt;math&gt;149-2 = 147&lt;/math&gt; because that is a factor. That is &lt;math&gt;3\cdot7^2&lt;/math&gt;, so now we can clearly see that the smallest &lt;math&gt;n&lt;/math&gt; to make the expression divisible by &lt;math&gt;3^3&lt;/math&gt; is just &lt;math&gt;3^2&lt;/math&gt;. Similarly, we can reason that the smallest n to make the expression divisible by &lt;math&gt;7^7&lt;/math&gt; is just &lt;math&gt;7^5&lt;/math&gt;. <br /> <br /> Finally, for &lt;math&gt;5^5&lt;/math&gt;, take mod &lt;math&gt;5&lt;/math&gt; and mod &lt;math&gt;25&lt;/math&gt; of each quantity (They happen to both be &lt;math&gt;-1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum &lt;math&gt;n&lt;/math&gt; for divisibility by &lt;math&gt;5&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, and other values are factors of &lt;math&gt;4&lt;/math&gt;. Testing all of them (just &lt;math&gt;1,2,4&lt;/math&gt; using mods-not too bad), &lt;math&gt;4&lt;/math&gt; is indeed the smallest value to make the expression divisible by &lt;math&gt;5&lt;/math&gt;, and this clearly is NOT divisible by &lt;math&gt;25&lt;/math&gt;.<br /> Therefore, the smallest &lt;math&gt;n&lt;/math&gt; to make this expression divisible by &lt;math&gt;5^5&lt;/math&gt; is &lt;math&gt;2^2 \cdot 5^4&lt;/math&gt;.<br /> <br /> Calculating the LCM of all these, one gets &lt;math&gt;2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5&lt;/math&gt;. Using the factor counting formula,<br /> the answer is &lt;math&gt;3\cdot 3\cdot 5\cdot 6&lt;/math&gt; = &lt;math&gt;\boxed{270}&lt;/math&gt;.<br /> <br /> -Solution by thanosaops<br /> <br /> ==See Also==<br /> <br /> {{AIME box|year=2020|n=I|num-b=11|num-a=13}}<br /> {{MAA Notice}}</div> Hellopeople99 https://artofproblemsolving.com/wiki/index.php?title=2015_AIME_II_Problems/Problem_9&diff=112578 2015 AIME II Problems/Problem 9 2019-12-05T06:30:32Z <p>Hellopeople99: /* Solution */ changed * to •</p> <hr /> <div>==Problem==<br /> A cylindrical barrel with radius &lt;math&gt;4&lt;/math&gt; feet and height &lt;math&gt;10&lt;/math&gt; feet is full of water. A solid cube with side length &lt;math&gt;8&lt;/math&gt; feet is set into the barrel so that the diagonal of the cube is vertical. The volume of water thus displaced is &lt;math&gt;v&lt;/math&gt; cubic feet. Find &lt;math&gt;v^2&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> import three; import solids;<br /> size(5cm);<br /> currentprojection=orthographic(1,-1/6,1/6);<br /> <br /> draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,360)),white,nolight);<br /> <br /> triple A =(8*sqrt(6)/3,0,8*sqrt(3)/3), B = (-4*sqrt(6)/3,4*sqrt(2),8*sqrt(3)/3), C = (-4*sqrt(6)/3,-4*sqrt(2),8*sqrt(3)/3), X = (0,0,-2*sqrt(2));<br /> <br /> draw(X--X+A--X+A+B--X+A+B+C);<br /> draw(X--X+B--X+A+B);<br /> draw(X--X+C--X+A+C--X+A+B+C);<br /> draw(X+A--X+A+C);<br /> draw(X+C--X+C+B--X+A+B+C,linetype(&quot;2 4&quot;));<br /> draw(X+B--X+C+B,linetype(&quot;2 4&quot;));<br /> <br /> draw(surface(revolution((0,0,0),(-2,-2*sqrt(3),0)--(-2,-2*sqrt(3),-10),Z,0,240)),white,nolight);<br /> draw((-2,-2*sqrt(3),0)..(4,0,0)..(-2,2*sqrt(3),0));<br /> draw((-4*cos(atan(5)),-4*sin(atan(5)),0)--(-4*cos(atan(5)),-4*sin(atan(5)),-10)..(4,0,-10)..(4*cos(atan(5)),4*sin(atan(5)),-10)--(4*cos(atan(5)),4*sin(atan(5)),0));<br /> draw((-2,-2*sqrt(3),0)..(-4,0,0)..(-2,2*sqrt(3),0),linetype(&quot;2 4&quot;)); &lt;/asy&gt;<br /> <br /> ==Solution==<br /> <br /> Our aim is to find the volume of the part of the cube submerged in the cylinder. <br /> In the problem, since three edges emanate from each vertex, the boundary of the cylinder touches the cube at three points. Because the space diagonal of the cube is vertical, by the symmetry of the cube, the three points form an equilateral triangle. Because the radius of the circle is &lt;math&gt;4&lt;/math&gt;, by the Law of Cosines, the side length s of the equilateral triangle is<br /> <br /> &lt;cmath&gt;s^2 = 2\cdot(4^2) - 2l\cdot(4^2)\cos(120^{\circ}) = 3(4^2)&lt;/cmath&gt;<br /> <br /> so &lt;math&gt;s = 4\sqrt{3}&lt;/math&gt;.* Again by the symmetry of the cube, the volume we want to find is the volume of a tetrahedron with right angles on all faces at the submerged vertex, so since the lengths of the legs of the tetrahedron are &lt;math&gt;\frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}&lt;/math&gt; (the three triangular faces touching the submerged vertex are all &lt;math&gt;45-45-90&lt;/math&gt; triangles) so <br /> <br /> &lt;cmath&gt;v = \frac{1}{3}(2\sqrt{6})\left(\frac{1}{2} \cdot (2\sqrt{6})^2\right) = \frac{1}{6} \cdot 48\sqrt{6} = 8\sqrt{6}&lt;/cmath&gt;<br /> <br /> so <br /> <br /> &lt;cmath&gt;v^2 = 64 \cdot 6 = \boxed{384}.&lt;/cmath&gt;<br /> <br /> In this case, our base was one of the isosceles triangles (not the larger equilateral one). To calculate volume using the latter, note that the height would be &lt;math&gt;2\sqrt{2}&lt;/math&gt;.<br /> <br /> *Note that in a 30-30-120 triangle, side length ratios are &lt;math&gt;1:1:\sqrt{3}&lt;/math&gt;.<br /> *Or, note that the altitude and the centroid of an equilateral triangle are the same point, so since the centroid is 4 units from the vertex (which is &lt;math&gt;\frac{2}{3}&lt;/math&gt; the length of the median), the altitude is 6, which gives a hypotenuse of &lt;math&gt;\frac{12}{\sqrt{3}}=4\sqrt{3}&lt;/math&gt; by &lt;math&gt;1:\frac{\sqrt{3}}{2}:\frac{1}{2}&lt;/math&gt; relationship for 30-60-90 triangles.<br /> <br /> ==See also==<br /> {{AIME box|year=2015|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]</div> Hellopeople99 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems&diff=98868 2018 AMC 10A Problems 2018-11-21T23:05:59Z <p>Hellopeople99: /* Problem 20 */ textified &quot;symmetric&quot;</p> <hr /> <div>==Problem 1==<br /> What is the value of &lt;cmath&gt;\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?&lt;/cmath&gt;&lt;math&gt;\textbf{(A) } \frac58 \qquad \textbf{(B) }\frac{11}7 \qquad \textbf{(C) } \frac85 \qquad \textbf{(D) } \frac{18}{11} \qquad \textbf{(E) } \frac{15}8&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 1|Solution]]<br /> <br /> ==Problem 2==<br /> Liliane has &lt;math&gt;50\%&lt;/math&gt; more soda than Jacqueline, and Alice has &lt;math&gt;25\%&lt;/math&gt; more soda than Jacqueline. What is the relationship between the amounts of soda that Liliane and Alice have?<br /> <br /> &lt;math&gt;\textbf{(A)}&lt;/math&gt; Liliane has &lt;math&gt;20\%&lt;/math&gt; more soda than Alice. <br /> <br /> &lt;math&gt;\textbf{(B)}&lt;/math&gt; Liliane has &lt;math&gt;25\%&lt;/math&gt; more soda than Alice. <br /> <br /> &lt;math&gt;\textbf{(C)}&lt;/math&gt; Liliane has &lt;math&gt;45\%&lt;/math&gt; more soda than Alice. <br /> <br /> &lt;math&gt;\textbf{(D)}&lt;/math&gt; Liliane has &lt;math&gt;75\%&lt;/math&gt; more soda than Alice.<br /> <br /> &lt;math&gt;\textbf{(E)}&lt;/math&gt; Liliane has &lt;math&gt;100\%&lt;/math&gt; more soda than Alice.<br /> <br /> [[2018 AMC 10A Problems/Problem 2|Solution]]<br /> <br /> ==Problem 3==<br /> A unit of blood expires after &lt;math&gt;10!=10\cdot 9 \cdot 8 \cdots 1&lt;/math&gt; seconds. Yasin donates a unit of blood at noon of January 1. On what day does his unit of blood expire?<br /> <br /> &lt;math&gt;\textbf{(A) }\text{January 2}\qquad\textbf{(B) }\text{January 12}\qquad\textbf{(C) }\text{January 22}\qquad\textbf{(D) }\text{February 11}\qquad\textbf{(E) }\text{February 12}&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 3|Solution]]<br /> <br /> ==Problem 4==<br /> How many ways can a student schedule 3 mathematics courses -- algebra, geometry, and number theory -- in a 6-period day if no two mathematics courses can be taken in consecutive periods? (What courses the student takes during the other 3 periods is of no concern here.)<br /> <br /> &lt;math&gt;\textbf{(A) }3\qquad\textbf{(B) }6\qquad\textbf{(C) }12\qquad\textbf{(D) }18\qquad\textbf{(E) }24&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 4|Solution]]<br /> <br /> ==Problem 5==<br /> Alice, Bob, and Charlie were on a hike and were wondering how far away the nearest town was. When Alice said, &quot;We are at least 6 miles away,&quot; Bob replied, &quot;We are at most 5 miles away.&quot; Charlie then remarked, &quot;Actually the nearest town is at most 4 miles away.&quot; It turned out that none of the three statements were true. Let &lt;math&gt;d&lt;/math&gt; be the distance in miles to the nearest town. Which of the following intervals is the set of all possible values of &lt;math&gt;d&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } (0,4) \qquad \textbf{(B) } (4,5) \qquad \textbf{(C) } (4,6) \qquad \textbf{(D) } (5,6) \qquad \textbf{(E) } (5,\infty) &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 5|Solution]]<br /> <br /> ==Problem 6==<br /> Sangho uploaded a video to a website where viewers can vote that they like or dislike a video. Each video begins with a score of 0, and the score increases by 1 for each like vote and decreases by 1 for each dislike vote. At one point Sangho saw that his video had a score of 90, and that &lt;math&gt;65\%&lt;/math&gt; of the votes cast on his video were like votes. How many votes had been cast on Sangho's video at that point?<br /> <br /> &lt;math&gt;\textbf{(A) } 200 \qquad \textbf{(B) } 300 \qquad \textbf{(C) } 400 \qquad \textbf{(D) } 500 \qquad \textbf{(E) } 600 &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 6|Solution]]<br /> <br /> ==Problem 7==<br /> For how many (not necessarily positive) integer values of &lt;math&gt;n&lt;/math&gt; is the value of &lt;math&gt;4000\cdot \left(\tfrac{2}{5}\right)^n&lt;/math&gt; an integer?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }3 \qquad<br /> \textbf{(B) }4 \qquad<br /> \textbf{(C) }6 \qquad<br /> \textbf{(D) }8 \qquad<br /> \textbf{(E) }9 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 7|Solution]]<br /> <br /> ==Problem 8==<br /> Joe has a collection of 23 coins, consisting of 5-cent coins, 10-cent coins, and 25-cent coins. He has 3 more 10-cent coins than 5-cent coins, and the total value of his collection is 320 cents. How many more 25-cent coins does Joe have than 5-cent coins?<br /> <br /> &lt;math&gt;\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4 &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 8|Solution]]<br /> <br /> ==Problem 9==<br /> All of the triangles in the diagram below are similar to iscoceles triangle &lt;math&gt;ABC&lt;/math&gt;, in which &lt;math&gt;AB=AC&lt;/math&gt;. Each of the 7 smallest triangles has area 1, and &lt;math&gt;\triangle ABC&lt;/math&gt; has area 40. What is the area of trapezoid &lt;math&gt;DBCE&lt;/math&gt;?<br /> <br /> &lt;asy&gt;<br /> unitsize(5);<br /> dot((0,0));<br /> dot((60,0));<br /> dot((50,10));<br /> dot((10,10));<br /> dot((30,30));<br /> draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0));<br /> draw((10,10)--(50,10));<br /> label(&quot;$B$&quot;,(0,0),SW);<br /> label(&quot;$C$&quot;,(60,0),SE);<br /> label(&quot;$E$&quot;,(50,10),E);<br /> label(&quot;$D$&quot;,(10,10),W);<br /> label(&quot;$A$&quot;,(30,30),N);<br /> draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10));<br /> draw((15,15)--(45,15));<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 9|Solution]]<br /> <br /> ==Problem 10==<br /> Suppose that real number &lt;math&gt;x&lt;/math&gt; satisfies &lt;cmath&gt;\sqrt{49-x^2}-\sqrt{25-x^2}=3&lt;/cmath&gt;. What is the value of &lt;math&gt;\sqrt{49-x^2}+\sqrt{25-x^2}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }8 \qquad<br /> \textbf{(B) }\sqrt{33}+8\qquad<br /> \textbf{(C) }9 \qquad<br /> \textbf{(D) }2\sqrt{10}+4 \qquad<br /> \textbf{(E) }12 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 10|Solution]]<br /> <br /> ==Problem 11==<br /> When &lt;math&gt;7&lt;/math&gt; fair standard &lt;math&gt;6&lt;/math&gt;-sided die are thrown, the probability that the sum of the numbers on the top faces is &lt;math&gt;10&lt;/math&gt; can be written as &lt;cmath&gt;\frac{n}{6^{7}}&lt;/cmath&gt;, where &lt;math&gt;n&lt;/math&gt; is a positive integer. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }42\qquad<br /> \textbf{(B) }49\qquad<br /> \textbf{(C) }56\qquad<br /> \textbf{(D) }63\qquad<br /> \textbf{(E) }84\qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 11|Solution]]<br /> <br /> ==Problem 12==<br /> How many ordered pairs of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfy the following system of equations?<br /> &lt;cmath&gt;x+3y=3&lt;/cmath&gt;<br /> &lt;cmath&gt;||x|-|y||=1&lt;/cmath&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }1\qquad<br /> \textbf{(B) }2\qquad<br /> \textbf{(C) }3\qquad<br /> \textbf{(D) }4\qquad<br /> \textbf{(E) }8\qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 12|Solution]]<br /> <br /> ==Problem 13==<br /> A paper triangle with sides of lengths 3, 4, and 5 inches, as shown, is folded so that point &lt;math&gt;A&lt;/math&gt; falls on point &lt;math&gt;B&lt;/math&gt;. What is the length in inches of the crease?<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,3), NE);<br /> label(&quot;$C$&quot;, (4,0), SE);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 13|Solution]]<br /> <br /> ==Problem 14==<br /> What is the greatest integer less than or equal to &lt;cmath&gt;\frac{3^{100}+2^{100}}{3^{96}+2^{96}}?&lt;/cmath&gt;<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }80\qquad<br /> \textbf{(B) }81 \qquad<br /> \textbf{(C) }96 \qquad<br /> \textbf{(D) }97 \qquad<br /> \textbf{(E) }625\qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 14|Solution]]<br /> <br /> ==Problem 15==<br /> Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points &lt;math&gt;A&lt;/math&gt; and &lt;math&gt;B&lt;/math&gt;, as shown in the diagram. The distance &lt;math&gt;AB&lt;/math&gt; can be written in the form &lt;math&gt;\tfrac{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;asy&gt;<br /> draw(circle((0,0),13));<br /> draw(circle((5,-6.2),5));<br /> draw(circle((-5,-6.2),5));<br /> label(&quot;$B$&quot;, (9.5,-9.5), S);<br /> label(&quot;$A$&quot;, (-9.5,-9.5), S);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93 &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 15|Solution]]<br /> <br /> ==Problem 16==<br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 16|Solution]]<br /> <br /> ==Problem 17==<br /> Let &lt;math&gt;S&lt;/math&gt; be a set of 6 integers taken from &lt;math&gt;\{1,2,\dots,12\}&lt;/math&gt; with the property that if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are elements of &lt;math&gt;S&lt;/math&gt; with &lt;math&gt;a&lt;b&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; is not a multiple of &lt;math&gt;a&lt;/math&gt;. What is the least possible value of an element in &lt;math&gt;S?&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)}\ 7&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 17|Solution]]<br /> <br /> ==Problem 18==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 18|Solution]]<br /> <br /> ==Problem 19==<br /> A number &lt;math&gt;m&lt;/math&gt; is randomly selected from the set &lt;math&gt;\{11,13,15,17,19\}&lt;/math&gt;, and a number &lt;math&gt;n&lt;/math&gt; is randomly selected from &lt;math&gt;\{1999,2000,2001,\ldots,2018\}&lt;/math&gt;. What is the probability that &lt;math&gt;m^n&lt;/math&gt; has a units digit of &lt;math&gt;1&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{5} \qquad \textbf{(B) } \frac{1}{4} \qquad \textbf{(C) } \frac{3}{10} \qquad \textbf{(D) } \frac{7}{20} \qquad \textbf{(E) } \frac{2}{5} &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 19|Solution]]<br /> <br /> ==Problem 20==<br /> A scanning code consists of a &lt;math&gt;7 \times 7&lt;/math&gt; grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of &lt;math&gt;49&lt;/math&gt; squares. A scanning code is called &lt;math&gt;\text{symmetric}&lt;/math&gt; if its look does not change when the entire square is rotated by a multiple of &lt;math&gt;90 ^{\circ}&lt;/math&gt; counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 20|Solution]]<br /> <br /> ==Problem 21==<br /> Which of the following describes the set of values of &lt;math&gt;a&lt;/math&gt; for which the curves &lt;math&gt;x^2+y^2=a^2&lt;/math&gt; and &lt;math&gt;y=x^2-a&lt;/math&gt; in the real &lt;math&gt;xy&lt;/math&gt;-plane intersect at exactly &lt;math&gt;3&lt;/math&gt; points?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }a=\frac14 \qquad<br /> \textbf{(B) }\frac14 &lt; a &lt; \frac12 \qquad<br /> \textbf{(C) }a&gt;\frac14 \qquad<br /> \textbf{(D) }a=\frac12 \qquad<br /> \textbf{(E) }a&gt;\frac12 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 21|Solution]]<br /> <br /> ==Problem 22==<br /> Let &lt;math&gt;a, b, c,&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; be positive integers such that &lt;math&gt;\gcd(a, b)=24&lt;/math&gt;, &lt;math&gt;\gcd(b, c)=36&lt;/math&gt;, &lt;math&gt;\gcd(c, d)=54&lt;/math&gt;, and &lt;math&gt;70&lt;\gcd(d, a)&lt;100&lt;/math&gt;. Which of the following must be a divisor of &lt;math&gt;a&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 22|Solution]]<br /> <br /> ==Problem 23==<br /> Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square &lt;math&gt;S&lt;/math&gt; so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from &lt;math&gt;S&lt;/math&gt; to the hypotenuse is 2 units. What fraction of the field is planted?<br /> <br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(0,3)--(0,0));<br /> draw((0,0)--(0.3,0)--(0.3,0.3)--(0,0.3)--(0,0));<br /> fill(origin--(0.3,0)--(0.3,0.3)--(0,0.3)--cycle, gray);<br /> label(&quot;$4$&quot;, (2,0), N);<br /> label(&quot;$3$&quot;, (0,1.5), E);<br /> label(&quot;$2$&quot;, (.8,1), E);<br /> label(&quot;$S$&quot;, (0,0), NE);<br /> draw((0.3,0.3)--(1.4,1.9), dashed);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75} &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 23|Solution]]<br /> <br /> ==Problem 24==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; with &lt;math&gt;AB=50&lt;/math&gt; and &lt;math&gt;AC=10&lt;/math&gt; has area &lt;math&gt;120&lt;/math&gt;. Let &lt;math&gt;D&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AB}&lt;/math&gt;, and let &lt;math&gt;E&lt;/math&gt; be the midpoint of &lt;math&gt;\overline{AC}&lt;/math&gt;. The angle bisector of &lt;math&gt;\angle BAC&lt;/math&gt; intersects &lt;math&gt;\overline{DE}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt; at &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt;, respectively. What is the area of quadrilateral &lt;math&gt;FDBG&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }60 \qquad<br /> \textbf{(B) }65 \qquad<br /> \textbf{(C) }70 \qquad<br /> \textbf{(D) }75 \qquad<br /> \textbf{(E) }80 \qquad<br /> &lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 24|Solution]]<br /> <br /> ==Problem 25==<br /> For a positive integer &lt;math&gt;n&lt;/math&gt; and nonzero digits &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt;, let &lt;math&gt;A_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;-digit integer each of whose digits is equal to &lt;math&gt;a&lt;/math&gt;; let &lt;math&gt;B_n&lt;/math&gt; be the &lt;math&gt;n&lt;/math&gt;-digit integer each of whose digits is equal to &lt;math&gt;b&lt;/math&gt;, and let &lt;math&gt;C_n&lt;/math&gt; be the &lt;math&gt;2n&lt;/math&gt;-digit (not &lt;math&gt;n&lt;/math&gt;-digit) integer each of whose digits is equal to &lt;math&gt;c&lt;/math&gt;. What is the greatest possible value of &lt;math&gt;a + b + c&lt;/math&gt; for which there are at least two values of &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;C_n - B_n = A_n^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 12} \qquad \textbf{(B)} \text{ 14} \qquad \textbf{(C)} \text{ 16} \qquad \textbf{(D)} \text{ 18} \qquad \textbf{(E)} \text{ 20}&lt;/math&gt;<br /> <br /> [[2018 AMC 10A Problems/Problem 25|Solution]]<br /> <br /> ==See also==<br /> {{AMC10 box|year=2018|ab=A|before=[[2017 AMC 10B]]|after=[[2018 AMC 10B]]}}<br /> * [[AMC 10]]<br /> * [[AMC 10 Problems and Solutions]]<br /> * [[Mathematics competitions]]<br /> * [[Mathematics competition resources]]<br /> {{MAA Notice}}</div> Hellopeople99 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_11&diff=98465 2018 AMC 10A Problems/Problem 11 2018-11-02T03:53:39Z <p>Hellopeople99: /* Solution 1 */ changed &quot;*&quot; to &quot;\cdot&quot; for easier reading</p> <hr /> <div>When 7 fair standard 6-sided dice are thrown, the probability that the sum of the numbers on the top faces is 10 can be written as &lt;cmath&gt;\frac{n}{6^7},&lt;/cmath&gt;where &lt;math&gt;n&lt;/math&gt; is a positive integer. What is &lt;math&gt;n&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 42 \qquad \textbf{(B) } 49 \qquad \textbf{(C) } 56 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 84 &lt;/math&gt;<br /> <br /> == Solutions ==<br /> ===Solution 1===<br /> The minimum number that can be shown on the face of a die is 1, so the least possible sum of the top faces of the 7 dies is 7. <br /> <br /> In order for the sum to be exactly 10, 1 to 3 dices' number on the top face must be increased by a total of 3. <br /> <br /> There are 3 ways to do so:<br /> 3, 2+1, and 1+1+1<br /> <br /> There are &lt;math&gt;7&lt;/math&gt; for Case 1, &lt;math&gt;7\cdot 6 = 42&lt;/math&gt; for Case 2, and &lt;math&gt;\frac{7\cdot 6\cdot 5}{3!} = 35&lt;/math&gt; for Case 3.<br /> <br /> Therefore, the answer is &lt;math&gt;7+42+35 = \boxed {\textbf{(E) } 84}&lt;/math&gt;<br /> <br /> Solution by PancakeMonster2004<br /> <br /> ===Solution 2===<br /> Rolling a sum of 10 with 7 dice can be represented with stars and bars, with 10 stars and 6 bars. Each star represents one of the dots on the die's faces and the bars represent separation between different dice. However, we must note that each die must have at least one dot on a face, so there must already be 7 stars predetermined. We are left with 3 stars and 6 bars, which we can rearrange in &lt;math&gt;\dbinom{9}{3}=\boxed{\textbf{(E) } 84}&lt;/math&gt; ways. (RegularHexagon)<br /> <br /> ===Solution 3===<br /> Add possibilities. There are &lt;math&gt;3&lt;/math&gt; ways to sum to &lt;math&gt;10&lt;/math&gt;, listed below.<br /> <br /> &lt;cmath&gt;4,1,1,1,1,1,1: 7&lt;/cmath&gt;<br /> &lt;cmath&gt;3,2,1,1,1,1,1: 42&lt;/cmath&gt;<br /> &lt;cmath&gt;2,2,2,1,1,1,1: 35.&lt;/cmath&gt;<br /> <br /> Add up the possibilities: &lt;math&gt;35+42+7=\boxed{\textbf{(E) } 84}&lt;/math&gt;.<br /> <br /> Thus we have repeated Solution 1 exactly, but with less explanation.<br /> <br /> ~kevinmathz<br /> <br /> ===Solution 4 (overkill)===<br /> We can use generating functions, where &lt;math&gt;(x+x^2+...+x^6)&lt;/math&gt; is the function for each die. We want to find the coefficient of &lt;math&gt;x^{10}&lt;/math&gt; in &lt;math&gt;(x+x^2+...+x^6)^7&lt;/math&gt;, which is the coefficient of &lt;math&gt;x^3&lt;/math&gt; in &lt;math&gt;\left(\frac{1-x^7}{1-x}\right)^7&lt;/math&gt;. This evaluates to &lt;math&gt;\dbinom{-7}{3} \cdot (-1)^3=\boxed{\textbf{(E) } 84}&lt;/math&gt;<br /> <br /> -wannabecharmander<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Probability Problems]]</div> Hellopeople99