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

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

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New post 25 Jun 2011, 23:23
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Question Stats:

67% (00:31) correct 33% (01:11) wrong based on 12 sessions

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Does the integer k have a factor p such that 1 = 4!
2. 13! + 2 =< k =< 13! + 13

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

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New post 25 Jun 2011, 23:24
can somebody please explain me this?
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Re: Does the integer k have a factor p such that 1 < p < k [#permalink]

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New post 26 Jun 2011, 00:21
harshavmrg wrote:
can somebody please explain me this?

It's asked whether K is a prime number or no.
from 1) K might equal \(27\) (and it has such factors as 3 or 9) or it can be \(29\) (prime number without factors we need)
ins
(2)
write it in this way:
\(2*(1*3*4*...*13+1)<=K<=13*(1*2*3*...*12+1)\)
Now we see you can write any number fro this interval in such form.
Example: \(13!+7=7*(1*2*3*4*5*6*8*...*13+1)\)
we do have a factor that we ask to find.
So, it's (B).
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Re: Does the integer k have a factor p such that 1 < p < k [#permalink]

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New post 04 Dec 2017, 03:32
Does the integer k have a factor p such that 14![/m] --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient.

(2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient.

Answer: B.

Check similar question: http://gmatclub.com/forum/factor-factorials-100670.html

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Re: Does the integer k have a factor p such that 1 < p < k   [#permalink] 04 Dec 2017, 03:32
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