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.

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PhD in Applied Mathematics

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