if the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the value of a?

(1) a^n = 64

(2) n = 6


In analyzing the question stem, you can see that if 8! is a multiple of a^n, the prime factorization will, at the very least, have to contain a^n. If you break down the prime factorization of 8!, you get 2^3 x 7 x 3 x 2 x 5 x 2^2 x 3 x 2 (8 x 7 x 6, etc....), or simplified 2^7 x 3^2 x 5 x 7. We can now look at the statements to see which one might work....

Starting with statement 2, we see the exponent of a^n equals 6. Looking back at our simplified prime factorization, we can see that only 1 term is raised to the power of 6 or greater (2^7). Therefore, we can definitively conclude that a MUST equal 2, since we are told a^n is a factor of 8! (note that A could also have been equal to 1, had it not been for the constraint in the question stem). So we have it down to B and D.

I found statement 1 to be bit trickier. We are told a^n equals 64. Since we've been dealing with 2's already, it may be easier to see a^n as 2^6. However, 64 can also be expressed as 4^3. We have two possible values for a, thus yielding it insufficient.

Ans : B