# Difference between revisions of "2004 AIME II Problems/Problem 8"

## Problem

How many positive integer divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?

## Solution

The prime factorization of 2004 is $2^2\cdot 3\cdot 167$. Thus the prime factorization of $2004^{2004}$ is $2^{4008}\cdot 3^{2004}\cdot 167^{2004}$.

We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization. For example, the number of divisors of $2004=2^2\cdot 3^1\cdot 167^1$ is $(2+1)(1+1)(1+1)=12$.

A positive integer divisor of $2004^{2004}$ will be of the form $2^a\cdot 3^b\cdot 167^c$. Thus we need to find how many $(a,b,c)$ satisfy

$(a+1)(b+1)(c+1)=2^2\cdot 3\cdot 167.$

We can think of this as partitioning the exponents to $a+1,$ $b+1,$ and $c+1$. So let's partition the 2's first. There are two 2's so this is equivalent to partitioning two items in three containers. We can do this in ${4 \choose 2} = 6$ ways. We can partition the 3 in three ways and likewise we can partition the 167 in three ways. So we have $6\cdot 3\cdot 3 = 054$ as our answer.