carro103 wrote:C is correct. Thanks for the help!
C is not correct. You should be quite suspicious of the form of this question; when you use both statements (n = 6 and a^n = 64) you can immediately solve for a without even using all of the information in the question stem ("the product of the first 8 positive integers is a multiple of a^n"). Why would they bother mentioning this fact if it weren't used in the solution? If you notice this, you can be quite certain that the answer will be A, B or D.
And indeed the answer is B here. From S1 alone, a can be 2, 4 or 8, so S1 is insufficient. From S2 alone, we know that 8! is a multiple a^6, and that a is an integer greater than 1. This is a question about multiples; we should prime factorize:
8! = 8*7*6*5*4*3*2 = (2^7)*(3^2)*(5)*(7)
Now, a^6 is a divisor of this number. Looking at the powers in the prime factorization, we can see that a can only be equal to 2 (we can see that 8! is certainly divisible by 2^6, but isn't divisible by, say, 3^6 or 4^6). So Statement 2 is sufficient.
