How many factors of the number 2520 are perfect squares?

We can prime factorize 2520:

2520 = 10*252 = (2)(5)(2)(126) = (2)(5)(2)(2)(9)(7) = 2^3 * 3^2 * 5 * 7

Every factor of 2520 thus looks like: 2^a * 3^b * 5^c * 7^d, where a can be 0, 1, 2, or 3, b can be 0, 1 or 2, c can be 0 or 1, and d can be 0 or 1. The only factors that are perfect squares are the ones that have exclusively even exponents. So to make a factor of 2520 that is a perfect square, a can be 0 or 2, for two choices, b can be 0 or 2, for two choices, and c and d must be 0, for one choice each. Multiplying our choices, we have 2*2*1*1 = 4 factors that are perfect squares. If we wanted to list them all, they are (I'll ignore the 5 and 7, since we'll never use them to make a perfect square divisor here) :

2^0 *3^0 = 1
2^2 * 3^0 = 4
2^0 * 3^2 = 9
2^2 * 3^2 = 36