The exponents in the prime factorization of a perfect cube must all be divisible by 3, so a number like 2^15 * 3^6 is a perfect cube (it is the cube of 2^5 * 3^2), while 2^8 * 3^7 is not
If we prime factorize 72^72, we get (2^3 * 3^2 )^72 = 2^216 * 3^144
For a divisor of this to be a perfect cube, it needs to look like this:
2^a * 3^b
where a and b must both be divisible by 3 (and either exponent could be zero). So a must be in this list:
0, 3, 6, 9, ..., 213, 216
and b must be in this list:
0, 3, 6, ...., 141, 144
There are 73 numbers in the first list (if you just divide everything by 3, the list becomes 0, 1, 2, 3, .., 72, which has 73 numbers in it) and similarly there are 49 numbers in the second list, so we have 73*49 choices in total for a and b together, and since the units digit of 73*49 is 7, the only possible answer among the choices is 3577.
Overall it's too inelegant a question to be a realistic GMAT problem, but the ingredients in the solution can all be tested in simpler ways.
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