. Find the smallest value of k such that 3240k is a perfect cube.

The product of multiple perfect cubes is also a perfect cube.
Proof for two:
n^3 * p^3 = nnnppp = npnpnp = (np)^3
And any integer whose exponent is a multiple of 3 is a perfect cube.
We will use this here:

Prime factorize 3240:
3240 = 405 * 2^3 = 2^3 * 3^4 * 5^1
We need to multiply this by k, to make all the exponents divisible by 3
The exponents not divisible by 3 are 4 and 1.
So let's fix that:
2^3 * 3^4 * 3*2 * 5^1 * 5^2
So, k is 3^2 * 5^2 = 225
(3240*225)^(1/3) = 90

What are the answer choices?