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

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Intern
Joined: 18 May 2017
Posts: 48
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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09 Jul 2017, 07:08
Bunuel - i have subtracted 13! from each side of the inequality and got that k is equal or between 0 to 11. From this inequality i can't got the answer. Where am i wrong? By the way - i looked at the math book for inequality theory but haven't find something relevant. Is there any post that covers this topic?

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Joined: 10 Jul 2017
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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12 Nov 2017, 09:22
Hi Bunuel,

I have the same question as oryahalom.

Cant we subtract 13! from both sides of the inequality?

Thanks,
RD
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Joined: 02 Sep 2009
Posts: 53698
Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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12 Nov 2017, 09:28
2
rudjlive wrote:
Hi Bunuel,

I have the same question as oryahalom.

Cant we subtract 13! from both sides of the inequality?

Thanks,
RD

Are there only two parts in $$13! + 2 \leq k \leq 13!+13$$? No, there are three. If you subtract 13! you should subtract from all three parts and you'll end up with $$2 \leq k -13! \leq 13$$, which gives you nothing.
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Re: Does the integer k have a factor p such that 1 < p < k ?  [#permalink]

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19 Dec 2017, 07:46
gmatcracker2010 wrote:
Does the integer k have a factor p such that 1<p<k?

(1) k > 4!

(2) $$13! + 2 \leq k \leq 13!+13$$

We need to determine whether k has a factor p such that 1<p<k, or in other words, whether k is a prime number. If it is, then it doesn’t have a factor between 1 and itself. If it isn’t, then it does.

Statement One Alone:
k > 4!

Since there are prime numbers greater than 4! and composite (non-prime) numbers greater than 4!, statement one alone is not sufficient to answer the question.

Statement Two Alone:
13! + 2 ≤ k ≤ 13! + 13

13! will have a factor of any integers from 2 to 13 inclusive. For any number k is, in the range of integers from 13! + 2 to 13! + 13 inclusive, k will have a factor from 2 to 13 inclusive. In particular, if k = 13! + n where (2 ≤ n ≤ 13), k will have n as a factor. For example, if k = 13! + 5, then k will have a factor of 5, since 5 divides into 13! and 5. If k = 13! + 8, then k will have a factor of 8 and hence a factor of 2.

Thus, statement two is sufficient to answer the question.

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

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16 Feb 2019, 14:26
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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 16 Feb 2019, 14:26

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