2005 Alabama ARML TST Problems/Problem 6

Problem

How many of the positive divisors of 3,240,000 are perfect cubes?

Solution

$3240000=2^63^45^4$. We want to know how many numbers are in the form $2^{3a}3^{3b}5^{3c}$, $a\leq 2$,$b\leq 1$, $c\leq 1$; there are 12 such numbers, if we count 1 as a perfect cube.