# Difference between revisions of "2013 AMC 12B Problems/Problem 9"

## Problem

What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$ ?

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

## Solution

Looking at the prime numbers under 12, we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of 2, $\lfloor\frac{12}{3}\rfloor+\lfloor\frac{12}{3^2}\rfloor=4+1=5$ factors of 3, and $\lfloor\frac{12}{5}\rfloor=2$ factors of 5. All greater primes are represented once or not at all in $12!$, so they cannot be part of the square. Since we are looking for a perfect square, the exponents on its prime factors must be even, so we can only use $4$ of the $5$ factors of $3$. The prime factorization of the square is therefore $2^{10}*3^4*5^2$. To find the square root of this, we halve the exponents, leaving $2^5*3^2*5$. The sum of the exponents is $\boxed{\textbf{(C) }8}$