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# M27-01 - Anyone got a more expansive answer?

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M27-01 - Anyone got a more expansive answer? [#permalink]

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13 Aug 2012, 13:22
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
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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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13 Aug 2012, 15:01
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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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Last edited by EvaJager on 14 Aug 2012, 07:23, edited 1 time in total.

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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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14 Aug 2012, 01:11
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Expert's post
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

The second statement should read: p is a factor of (n+2)!/n!.

Below is OE for this question:

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! --> if $$n=2$$ then $$(n+2)!-n!=22$$ and for $$p=2$$ then answer will be YES but for $$p=11$$ the answer will be NO. Not sufficient.

(2) p is a factor of (n+2)!/n! --> $$\frac{(n+2)!}{n!}=(n+1)(n+2)$$ --> if $$n=2$$ then $$(n+1)(n+2)=12$$ and for $$p=2$$ the answer will be YES but for $$p=3$$ the answer will be NO. Not sufficient.

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$. Now, $$(n+1)(n+2)-1$$ and $$(n+1)(n+2)$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) $$p$$ is a factor of $$(n+1)(n+2)$$ then it cannot be a factor of $$(n+1)(n+2)-1$$, thus in order $$p$$ to be a factor of $$n!*((n+1)(n+2)-1)$$, from (1), then it should be a factor of the first multiple of this expression: $$n!$$. Sufficient.

Answer: C.

Discussed here: devil-s-dozen-129312.html

Hope it helps.
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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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15 Aug 2012, 11:02
Bunuel wrote:

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$.

didnt get it. how comes (n+1)(n+2)?
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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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15 Aug 2012, 11:23
LalaB wrote:
Bunuel wrote:

(1)+(2) $$(n+2)!-n!=n!((n+1)(n+2)-1)$$.

didnt get it. how comes (n+1)(n+2)?

$$(n+2)!=n!(n+1)(n+2)$$
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Kudos [?]: 1041 [0], given: 43

Senior Manager
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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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15 Aug 2012, 11:27
EvaJager I meant WHY (n+2)!=n!(n+1)(n+2)
perhaps I am tired and need some rest to get it
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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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15 Aug 2012, 11:38
LalaB wrote:
EvaJager I meant WHY (n+2)!=n!(n+1)(n+2)
perhaps I am tired and need some rest to get it

$$(n+2)!=1*2*...*n*(n+1)*(n+2)=n!*(n+1)*(n+2)$$

By definition, $$n!=1*2*...*n$$, for any positive integer $$n$$, and $$0!=1.$$
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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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15 Aug 2012, 11:39
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LalaB wrote:
EvaJager I meant WHY (n+2)!=n!(n+1)(n+2)
perhaps I am tired and need some rest to get it

$$(n+2)!=1*2*...*n*(n+1)*(n+2)=n!*(n+1)*(n+2)$$

By definition, $$n!=1*2*...*n$$, for any positive integer $$n$$, and $$0!=1.$$
_________________

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Love GMAT Quant questions and running.

Kudos [?]: 1041 [1], given: 43

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Re: M27-01 - Anyone got a more expansive answer? [#permalink]

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15 Aug 2012, 11:42
EvaJager ,
ah got it. thnx
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Re: M27-01 - Anyone got a more expansive answer?   [#permalink] 15 Aug 2012, 11:42
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# M27-01 - Anyone got a more expansive answer?

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