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Re: Number Properties and Inequalities Question [#permalink]
04 Oct 2009, 10:26

1

This post received 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?

Re: Number Properties and Inequalities Question [#permalink]
04 Oct 2009, 14:06

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?

Re: Number Properties and Inequalities Question [#permalink]
04 Oct 2009, 15:53

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.

Re: Number Properties and Inequalities Question [#permalink]
11 Oct 2009, 04:07

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?

So the answer is B. ___________________________- 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..

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 )