# Difference between revisions of "2020 AMC 10A Problems/Problem 15"

## Problem

A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23$

## Solution

The prime factorization of $12!$ is $2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11$. This yields a total of $11 \cdot 6 \cdot 3 \cdot 2 \cdot 2$ divisors of $12!.$ In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that $7$ and $11$ can not be in the prime factorization of a perfect square because there is only one of each in $12!.$ Thus, there are $6 \cdot 3 \cdot 2$ perfect squares. (For $2$, you can have $0$, $2$, $4$, $6$, $8$, or $10$ $2$s, etc.) The probability that the divisor chosen is a perfect square is $$\frac{6\cdot 3\cdot 2}{11\cdot 6\cdot 3\cdot 2\cdot 2}=\frac{1}{22} \implies \frac{m}{n}=\frac{1}{22} \implies m\ +\ n = 1\ +\ 22 = \boxed{\textbf{(E) } 23 }$$

~mshell214, edited by Rzhpamath

~IceMatrix

## See Also

 2020 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 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

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