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

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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).

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

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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.

If anything remains unclear please continue discussion there.

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