If n is a positive integer and p is a prime number, is p a f

[hazelnut01]

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!

Source : https://gmatclub.com/forum/devil-s-doze ... l#p1063847

Answer : C

[Jay@ManhattanReview]

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!

Anyway, how to edit with the proper formula in this forum?

I have a hard time to solve this problem. Please help!

Hi ziyuenlau,

We have n: {1, 2, 3, ....}; p: {2, 3, 5, 7, ...}

We have to see if n! / p is integer or not.

Let's rephrase each statement.

S1: p is a factor of (n+2)! - n!

(n+2)! - n! = n!*(n+1)*(n+2) - n! = n!*[(n+1)*(n+2) - 1]

So we have p is a factor of n!*[(n+1)*(n+2) - 1]

Let's test this with a couple of test values.

Case 1: n = 1 and p = 5

n!*[(n+1)*(n+2) - 1] = 1!*[(1+1)*(1+2) - 1] = 5.

We see that 1! / 5 is not an integer. The answer is NO.

Case 2: n = 2 and p = 2

n!*[(n+1)*(n+2) - 1] = 2!*[(2+1)*(2+2) - 1] = 22.

We see that 2! / 2 is an integer. The answer is Yes. No unique answer.

S2: p is a factor of (n+2)! / n!

(n+2)! / n! = [n!*(n+1)*(n+2)] / n! = (n+1)*(n+2)

So we have p is a factor of (n+1)*(n+2).

Case 1: n = 1 and p = 2 or 3

(n+1)*(n+2) = (1+1)*(1+2) = 6.

We see that 1! / 2 is not an integer. The answer is NO.

Case 2: n = 2 and p = 2

(n+1)*(n+2) = (2+1)*(2+2) = 12.

We see that 2! / 2 is an integer. The answer is YES. No unique answer.

S1 and S2:

From S1, we know that n!*[(n+1)*(n+2) - 1] is a factor of p and from S2, we know that (n+1)*(n+2) is a factor of p.

Thus,

n!*[(n+1)*(n+2) - 1] / p = integer

=> n!*[(n+1)*(n+2)] / p - n! / p = integer; we know that (n+1)*(n+2) / p is an integer,

=> Integer - n! / p = Integer

=> n! / p = Integer - Integer = Integer

Thus, p is a factor of n!. Sufficient.

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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[Brent@GMATPrepNow]

Anyway, how to edit with the proper formula in this forum?

That's about as good as you can do with the posting functionality.
The important thing is that you have added the appropriate brackets to avoid ambiguity.

Aside: one my biggest pet peeves are ambiguous posts like x2+2x-15/x-3+4x2 = 4x. So, kudos for you ambiguity-free post

Cheers,
Brent

[ceilidh.erickson]

Please remember to cite your source and include the OA! It's a copyright violation to print a question without citing the original author.