Find the sum of the positive square free divisors of 5400.
Note: A natural number is square free if it is not the multiple of any perfect square greater than 1.
5400 is 2^3 * 3^3 * 5^2 .
The square free divisors should be in the form 2^a * 3^b * 5^c... right?
Anyways so a, b, c, must be 0 or 1, since if it's 2 they would be a perfect square, 3, a perfect square times something.
So the sum should be (1+2)(1+3)(1+5).
Which is 72.
Tell me if AoPS says this is correct, please, because I think it is but I might be wrong.