AMC 12 Mock Exam 1 Problem 8

A divisor has the form $2^a3^b5^c$ with $0\le a\le4$, $0\le b\le3$, $0\le c\le2$.

Perfect squares: $a,b,c$ all even: $3\cdot2\cdot2=12$ choices ($a\in\{0,2,4\}$, $b\in\{0,2\}$, $c\in\{0,2\}$).

Perfect cubes: $a,b,c$ all multiples of $3$: $2\cdot2\cdot1=4$ choices ($a\in\{0,3\}$, $b\in\{0,3\}$, $c=0$).

Both (perfect sixth powers): $a,b,c$ multiples of $6$, so $a=b=c=0$: only the divisor $1$.

By inclusion–exclusion the count is $12+4-1=\boxed{15}$.