How many perfect squares greater than 1 are divisors of $60^{5}$ ?
The prime factorization of 60 is $2^{2} \times 3 \times 5$, so the prime factorization of $60^{5}$ is

$$
\left(2^{2} \times 3 \times 5\right)^{5}=2^{10} 3^{5} 5^{5} .
$$

Therefore, the divisors of $60^{5}$ are numbers of the form $2^{a} 3^{b} 5^{c}$ where $a, b$, and $c$ are integers with $0 \leq a \leq 10,0 \leq b \leq 5$, and $0 \leq c \leq 5$.

A divisor is a perfect square exactly when each of $a, b$, and $c$ is even.

Therefore, the divisors of $60^{5}$ that are perfect squares are the numbers of the form $2^{a} 3^{b} 5^{c}$ where
$a$ is equal to $0,2,4,6,8$, or $10, b$ is equal to 0,2 , or 4 , and $c$ is equal to 0,2 , or 4 .

This gives a total of $6 \times 3 \times 3=54$ ways of choosing $a, b$, and $c$ so that $2^{a} 3^{b} 5^{c}$ is a perfect square.

This total includes the divisor obtained by setting $a=b=c=0$, which is $2^{0} 3^{0} 5^{0}=1$.

We want the number of perfect-square divisors greater than 1 , so the answer is $54-1=53$.

ANSWER: 53