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

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

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New post 24 Jul 2009, 13:29
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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


Thanks a lot in advance!

Last edited by uzonwagba on 24 Jul 2009, 18:44, edited 1 time in total.

Kudos [?]: 28 [0], given: 6

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Kudos [?]: 78 [0], given: 15

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Re: (Please Help) DS Problem #6: Please take a stab, anyone! Thx [#permalink]

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New post 24 Jul 2009, 14:20
I guess the question is not complete assuming it says k have a factor p such that 1<p<k?

Its basically asking us to tell if k is a prime or not

Stmt1 ) k > 4! so not sufficient

stmts2) 13!+2 <= k <=13!+13
13!+2 factors are all the factors of 13!( which is 2,3,4,...13) and 2
13!+13 factors are all the factors of 13!( which is 2,3,4,...13) and 3

so any number between this is not a prime as it would go like 13!+2 + 13!+3 + 13!+4 .... till 13!+13

So B is sufficient

Kudos [?]: 78 [0], given: 15

Intern
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Joined: 23 Jun 2009
Posts: 47

Kudos [?]: 28 [0], given: 6

Re: (Please Help) DS Problem #6: Please take a stab, anyone! Thx [#permalink]

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New post 24 Jul 2009, 18:49
Hello! Could you please shed a little more light on the following statement you made:
"13!+2 factors are all the factors of 13!( which is 2,3,4,...13) and 2
13!+13 factors are all the factors of 13!( which is 2,3,4,...13) and 3"
?

I don't quite understand. I believe you. Your answer is right; but I don't understand.


Thanks,


Uzonwagba

Posted from my mobile device

Kudos [?]: 28 [0], given: 6

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Posts: 356

Kudos [?]: 78 [0], given: 15

Location: San Francisco
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Re: (Please Help) DS Problem #6: Please take a stab, anyone! Thx [#permalink]

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New post 24 Jul 2009, 19:08
2, 3, 4, 5, ..., 12, 13 is a factor of 13! as 13! = 1.2.3.4.5.6.7.8.9.10.11.12.13

The definition of a factorial: it's just the product of all the integers from 1 through 13. because all of those numbers are in the product, they're all factors.

consider the lowest number allowed by statement 2: 13! + 2.
note that 2 goes into 13! (as shown above), and 2 also goes into 2. therefore, 2 is a factor of this sum

consider the next number allowed by statement 2: 13! + 3.
note that 3 goes into 13! (as shown above), and 3 also goes into 3. therefore, 3 is a factor of this sum

etc.
all the way to 13! + 13.
works the same way each time.

Kudos [?]: 78 [0], given: 15

Re: (Please Help) DS Problem #6: Please take a stab, anyone! Thx   [#permalink] 24 Jul 2009, 19:08
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