Integers a and n

[carro103]

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


A little confused on this one. Any help is appreciated. Thanks![/i]

[joymukhi]

IMO C . Please post OA.

[Robinmrtha]

IMO C
Statement 1...
There can be three values for both a and n
i.e. 8^2=64
or 2^6=64
or 4^3=64
Not sufficient
Statement 2..does not say anything about a
Both statement combined we get 2^6=64
therefore a=2
So the answer is C

[carro103]

C is correct. Thanks for the help!

[Ian Stewart]

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.

[Robinmrtha]

thank you...

[mehravikas]

Answer should be 'B'

[Domnu]

Er... actually, the answer should be B..

8! = 2^7 * 3^2 * 5 * 7

The only factor with n = 6 is 2, so we get a = 2.