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# Does the integer k have a factor p such that 1 < p <

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Intern
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Does the integer k have a factor p such that 1 < p < [#permalink]

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04 Oct 2009, 11:13
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Does the integer k have a factor p such that 1 < p < k?

(1) k > 4!

(2) 13! +2 $$<=$$ k $$<=$$ 13! + 13

Can someone help me understand how to solve this problem?
[Reveal] Spoiler:
Senior Manager
Joined: 18 Aug 2009
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Re: Number Properties and Inequalities Question [#permalink]

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04 Oct 2009, 11:26
1
KUDOS
Navigator wrote:
Does the integer k have a factor p such that 1 < p < k? - Question is simply asking if K is a PRIME number or not.

(1) k > 4! - It is not possible to find if K is a prime number. Insufficient.

(2) 13! +2 $$<=$$ k $$<=$$ 13! + 13 -
13! has factors {2, 3, 5, 7, 11, 13}.
So (13!+2) is divisible by 2 and (13!+13) is divisible by 13. Also, any number between (13!+2) and (13!+13) must be divisible by one of the factors of 13!. So this is sufficient.

Can someone help me understand how to solve this problem?
[Reveal] Spoiler:

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Intern
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Re: Number Properties and Inequalities Question [#permalink]

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04 Oct 2009, 15:06
Thank you! I understand the logic you applied to the two statements, but what was it about the question stem that lets you know that the question was ultimately asking if K is prime?
Senior Manager
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Re: Number Properties and Inequalities Question [#permalink]

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04 Oct 2009, 16:53
Well, the question is: Does the integer k have a factor p such that 1 < p < k. That means is there a factor of k other than 1 and k itself? It indicates that k is a prime factor if there is no other factor than 1 and k itself.

Hope it is clear now.
Intern
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Re: Number Properties and Inequalities Question [#permalink]

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05 Oct 2009, 05:53
Much clearer now, thanks again.
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Re: Number Properties and Inequalities Question [#permalink]

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11 Oct 2009, 05:07
hgp2k wrote:
Navigator wrote:
Does the integer k have a factor p such that 1 < p < k? - Question is simply asking if K is a PRIME number or not.

(1) k > 4! - It is not possible to find if K is a prime number. Insufficient.

(2) 13! +2 $$<=$$ k $$<=$$ 13! + 13 -
13! has factors {2, 3, 5, 7, 11, 13}.
So (13!+2) is divisible by 2 and (13!+13) is divisible by 13. Also, any number between (13!+2) and (13!+13) must be divisible by one of the factors of 13!. So this is sufficient.

Can someone help me understand how to solve this problem?
[Reveal] Spoiler:

___________________________-
Consider KUDOS for good posts

Can you tell me please how did we know that any number between 13!+2 and 13!+13 must be divisble by one of the factors of 13! and is that applicable on all numbers with !
I mean i tried it with 3! and found that 3!+5 is for example prime number..

Director
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Name: Ronak Amin
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Re: Number Properties and Inequalities Question [#permalink]

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11 Oct 2009, 07:36
rshawabka,
hgp2k has given a good explanation, for more details you can refer
http://www.beatthegmat.com/ds-question-t1514.html

The rule is : In a sum of two numbers, if those two numbers have a common factor then the sum will also have that factor ( i.e. the sum will be divisible by the common factor )
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Re: Number Properties and Inequalities Question [#permalink]

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23 Oct 2009, 09:57
Thanks economist.....
Re: Number Properties and Inequalities Question   [#permalink] 23 Oct 2009, 09:57
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