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From the the question we know that A and N>1 and 8! is a multiple of a^N
8!=40,320.

Clue one tells us that a^N is 64-
40,320 divided by 64 leaves 630 which is which is equal to 2^1*5^1*7^1*3^2 so the prime factors of 40,320 are 2^7,3^2,5^1,3^2 so the only integer greater than raised to the power of another integer greater than 1 that can equal 64 and be a factor of 40,320 is 2 so the data is sufficient. Discard BCE and go with AD

Clue 2 tells us that n=6, and from the work above the only prime factor raised to a power 6 or greater is 2. so statement 2 is sufficient as well.
Answer D

Please correct me if I am wrong.

8! (40,320) is a multiple of a^n

(1) 64 = 2^6 = 4^3 = 8^2 = a^n. Not sufficient.
(2) n = 6. Factorizing, 40320 = 2^6 * 2 * 3^2 * 5^1 * 7^1. So, a=2, n=6 is the only possibility.

Answer is (B)

If this were not a circular arrangement, then the number of different possibilities would simply be 5! = 120.

With a circular arrangement for 5, it is equivalent of choosing from 4 people without a circular arrangement which is 4! = 24.

(To imagine this, in a circular arrangement, if the first person is seated at table 1, then we have 4! = 24 different combos that other people can be seated. But when you make person 2 sit at table 1, all those arrangements are still possible with person 1 sitting at table 1 anyway).

Thx I didnt see that. Behold the dangers of reading too much into the 2nd clue and using that information in the first. I hope I dont do that tomorrow.

Good luck with your test tomorrow drhomler. Go kick some GMAT butt!