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In the OG's explanation for Data Sufficiency $ 153 it says

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In the OG's explanation for Data Sufficiency $ 153 it says [#permalink]

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New post 15 Nov 2008, 03:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This is more of a math question than a reasoning one.

In the OG's explanation for Data Sufficiency $ 153 it says

"Note that 2 is a factor of 13! + 2, since it is a factor of both 13! and 2. Similarly, 3 is a factor of 13! + 3; 4 is a factor of 13! + 4..."

This is probably daft - but how do they know that 2 is a factor of 13!, or that 3 is a factor of 13! ?

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Re: Factorials Question -OG 11th Edition, Data Sufficiency # 153 [#permalink]

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New post 15 Nov 2008, 11:26
I got B as well for #153..

Question stem asks whether integer k has a factor p such that 1 < p < k

Statement # 1 says K > 4!...not sufficient coz if K is a prime number > 4!, then K can't have a factor p that satisfies above. But if K is non prime, then you can find a factor. Hence #1 not sufficient. So A and D are out.

#2 says 13! + 2 <= k <= 13! + 13. And since K is an integer...there are discrete values for K between the 2 ranges given by #2. Now for your question mmace1...

n! by definition is n x (n-1) x (n-2).....x 3 x 2 x 1

Whenever you see n!, you should automatically assume integers 1 thru n-1 are factors...as you can draw from the definiton of factorial stated above. So from #2 one can conclude that 13! + 2 definitely has 2 as a factor. 13! + 3 has 3 as its factor and so on. HTH..
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Re: Factorials Question -OG 11th Edition, Data Sufficiency # 153 [#permalink]

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New post 15 Nov 2008, 12:00
mmace1 wrote:
This is more of a math question than a reasoning one.

In the OG's explanation for Data Sufficiency $ 153 it says

"Note that 2 is a factor of 13! + 2, since it is a factor of both 13! and 2. Similarly, 3 is a factor of 13! + 3; 4 is a factor of 13! + 4..."

This is probably daft - but how do they know that 2 is a factor of 13!, or that 3 is a factor of 13! ?

13! = 13x12x11x10x9x8x7x6x5x4x3x2x1

so 2 and 3 both are in 13!. no only these 13 individual integers but also their different combinations, such as 6, 8, 20, 50, 100 etc., are each factors of 13!.
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Re: Factorials Question -OG 11th Edition, Data Sufficiency # 153 [#permalink]

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New post 16 Nov 2008, 16:15
Oh duh right...thank you.

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Re: Factorials Question -OG 11th Edition, Data Sufficiency # 153   [#permalink] 16 Nov 2008, 16:15
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In the OG's explanation for Data Sufficiency $ 153 it says

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