What is the positive integer n ? (1) For every positive

The question should surely read "For every positive integer m..."; you aren't expected to know what it means to take the factorial of a negative number on the GMAT. With that adjustment the answer is C. The key is in analyzing the first statement:

(m+n)!/(m-1)! is just the product of the integers between m and m+n (that is, (m+n)!/(m-1)! = m*(m+1)*...*(m+n)). So it's just the product of n+1 consecutive integers. If you ever take the product of six or more consecutive integers, the product will be divisible by 16 because at least three of the numbers will be even (divisible by 2), and at least one will be divisible by 4. In fact, the product of six consecutive integers will always be divisible by 6!, but we don't need that here. However, the product of five consecutive integers will not always be divisible by 16- take 1,2,3,4,5 for example. So Statement 1 guarantees that n is at least 5. From Statement 2, we have that n can only be 4 or 5, so together the statements are sufficient.

nicely explained

+1

u deserve the highest kudos: posts ratio!

i believe factorials only exist for positive numbers..

having said that worst case scenario is that m=1

in that case..n has to be 5..or greater ..cause we dont have enough even numbers..

therefore i will go with C..

Agree, Ian. That'll teach me to pay attention!

Well, the OA is C. Also, the question doesn't mention whether m is positive. I just double checked with the question and it's correct the way it is. I got this from a pdf file, so i'm not sure how completely correct is the question, but this is what it says at least.

that's a good explanation! +1

but what about (m-1)!? you avoided it in your explanation. Would you please explain how (m-1)! should be be handled?
thanks

Good explanation and good question!!!

Lets say m=5 n=4

9!/4! = 9 X 8 X 7 X 6 X 5. this is divisible by 16 correct How can n not be 4?

Statement 1 says that (m+n)!/(m-1)! is divisible by 16 for *every* positive integer m, not just for some particular value of m like m=5. Try m = 3, and you'll see why n cannot be 4.

Tarek, I'm not sure I understand your question about (m-1)!. In the fraction (m+n)!/(m-1)!, the (m-1)! cancels with much of (m+n)! to leave us with (m+n)*(m+n-1)*...*(m+1)*m. Hope that answers your question.