Difference between revisions of "2020 AMC 10A Problems/Problem 15"
Advancedjus (talk | contribs) |
(→Video Solution) |
||
(28 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
− | ==Problem | + | ==Problem== |
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? | A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>? | ||
Line 5: | Line 5: | ||
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math> | <math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math> | ||
− | == Solution == | + | ==Solution== |
+ | The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. | ||
+ | This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math> | ||
+ | In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>10</math> <math>2</math>s, etc.) | ||
+ | The probability that the divisor chosen is a perfect square is <cmath>\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }</cmath> | ||
+ | |||
+ | ~mshell214, edited by Rzhpamath | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | Education, The Study of Everything | ||
+ | |||
+ | https://youtu.be/ipZV6QfN3iU | ||
+ | |||
+ | https://youtu.be/ZGwAasE32Y4 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 11:51, 7 November 2020
Contents
Problem
A positive integer divisor of is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as , where and are relatively prime positive integers. What is ?
Solution
The prime factorization of is . This yields a total of divisors of In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that and can not be in the prime factorization of a perfect square because there is only one of each in Thus, there are perfect squares. (For , you can have , , , , , or s, etc.) The probability that the divisor chosen is a perfect square is
~mshell214, edited by Rzhpamath
Video Solution
Education, The Study of Everything
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.