bradfris wrote:

Hi GMATers

This question has stumped me. Has anyone got a more detailed explanation for this? I can't follow the OG to save myself!

If n is a positive integer and p is a prime number, is p a factor of n!?

(1) p is a factor of (n+2)!−n!

(2) p is a factor of (n+2)!n!

Thanks

B

(1) p is a factor of

(n+2)!-n!=n![(n+1)(n+2)-1]=n!(n^2+3n+1).p not necessarily a factor of

n!, it can be a factor of

n^2+3n+1, which is greater than 1.

Not sufficient.

(2) p is a factor of

(n+2)!n!=(n!)^2(n+1)(n+2)=(n!)^2(n^2+3n+2)Again, p not necessarily a factor of

n!,it can be a factor of

n^2+3n+2, which is greater than 1.

Not sufficient.

(1) and (2) If

p is not a factor of n! then necessarily p has to be a factor of both

n^2+3n+1 and

n^2+3n+2. But these are two consecutive positive integers, so only 1 can be a factor of both. p being a prime, is greater than 1. It follows that necessarily p must be a factor of

n!.Sufficient.

Answer C

If (2) should be

\frac{(n+2)!}{n!}, then the above should be replaced by:

(2)

\frac{(n+2)!}{n!}=(n+1)(n+2). For example

n+1 can be a prime, and if

p=n+1, certainly p is not a factor of

n!.Not sufficient.

(1) and (2) together - stays the same as above:

If

p is not a factor of n! then necessarily p has to be a factor of both

n^2+3n+1 and

n^2+3n+2. But these are two consecutive positive integers, so only 1 can be a factor of both. p being a prime, is greater than 1. It follows that necessarily p must be a factor of

n!.Sufficient.

_________________

PhD in Applied Mathematics

Love GMAT Quant questions and running.