# 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. Thus, there are 5 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, etc. 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!.) The probability that the divisor chosen is a perfect square is 1/22. m + n = 23.