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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]  19 Jun 2010, 02:49
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i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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

(1) k > 4!
(2) 13! + 2<= k <= 13!+13
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Re: gmat prep ds [#permalink]  19 Jun 2010, 03:51
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Does the integer k have a factor p such that 1<p<k?

Question basically asks whether $$k$$ is a prime number. If it is, then it won't have a factor $$p$$ such that $$1<p<k$$ (definition of a prime number).

(1) $$k>4!$$ --> $$k$$ 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.

Check similar question: factor-factorials-100670.html

Hope it's clear.
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Re: gmat prep ds [#permalink]  19 Jun 2010, 04:46
Thanks..can not be clearer than this.
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Re: gmat prep ds [#permalink]  03 Nov 2010, 14:20
Bunuel wrote:
Does the integer k have a factor p such that 1<p<k?

Question basically asks is $$k$$ a prime number. If it is, then it won't have a factor $$p$$ such that $$1<p<k$$ (definition of a prime number).

(1) $$k>4!$$ --> $$k$$ 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 $$13!+x$$). Sufficient.

Check similar question: factor-factorials-100670.html

Hope it's clear.

Thanks Bunuel. I have a question, what would happen if the second statement said this?

(2) $$13!+1\leq{k}\leq{13!+13}$$

I have seen some similar problems, in which they add 1. Could you post some links about it?

Thanks!
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Re: gmat prep ds [#permalink]  03 Nov 2010, 14:51
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metallicafan wrote:
Bunuel wrote:
Does the integer k have a factor p such that 1<p<k?

Question basically asks is $$k$$ a prime number. If it is, then it won't have a factor $$p$$ such that $$1<p<k$$ (definition of a prime number).

(1) $$k>4!$$ --> $$k$$ 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 $$13!+x$$). Sufficient.

Check similar question: factor-factorials-100670.html

Hope it's clear.

Thanks Bunuel. I have a question, what would happen if the second statement said this?

(2) $$13!+1\leq{k}\leq{13!+13}$$

I have seen some similar problems, in which they add 1. Could you post some links about it?

Thanks!

So basically we should determine whether $$13!+1$$ is a prime number (as shown above all other possible values of k are not are not prime), which can not be done without a computer. There are some particular values of $$k=n!+1$$ for which we can say whether it's a prime or not with help of Wilson's theorem, but again it's out of the scope of GMAT.

By the way: $$13!+1$$ is not a prime number, it has two distinct prime factors: $$13!+1=83*75,024,347=6,227,020,801$$, so the answer still will be B.
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Re: gmat prep ds [#permalink]  10 Feb 2011, 17:04
Bunuel wrote:
Does the integer k have a factor p such that 1<p<k?

Question basically asks whether $$k$$ is a prime number. If it is, then it won't have a factor $$p$$ such that $$1<p<k$$ (definition of a prime number).

(1) $$k>4!$$ --> $$k$$ 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 $$13!+x$$). Sufficient.

Check similar question: factor-factorials-100670.html

Hope it's clear.

Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters?
I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...
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Re: gmat prep ds [#permalink]  10 Feb 2011, 17:19
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ChenggongMAS wrote:
Bunuel wrote:
Does the integer k have a factor p such that 1<p<k?

Question basically asks whether $$k$$ is a prime number. If it is, then it won't have a factor $$p$$ such that $$1<p<k$$ (definition of a prime number).

(1) $$k>4!$$ --> $$k$$ 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 $$13!+x$$). Sufficient.

Check similar question: factor-factorials-100670.html

Hope it's clear.

Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters?
I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...

k=13!+8 means k=2*3*4*5*6*7*8*9*10*11*12*13+8, now factor out 8: k=8*(2*4*5*6*7*9*10*11*12*13+1) --> k is a multiple of 8 as k=8*something --> as k is a multiple of 8 it can not be a prime number. You can have any number from 2 to 13 inclusive instead of 8, and you'll be able to factor out this number the same way as you did with 8, so any number of a type 13!+x, wher x is from 2 to 13 inclusive will be a multiple of x, thus not a prime number.

Check the link in my first post for similar problem. Also check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it's clear.
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OG 11 [#permalink]  10 Dec 2011, 16:57
153. Does the integer k have a factor p such that 1>p>k?
(1) k>4!
(2) 13! + 2 <= k <= 13! + 13
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Re: OG 11 [#permalink]  13 Dec 2011, 14:37
Answer can't be E, it has to be B. I don't have OG with me, but I need to check.

1. By itself not sufficient. K could be prime.
2. 13! + 2 <= K <= 13! + 13 ensures that you have ever number from 2 - 13 dividing k.
That is:
13! + 2 -> div by 2
13! + 3 -> div by 3
... so on till 13.

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Re: OG 11 [#permalink]  17 Jan 2012, 07:30
Expert's post
ashiima wrote:
153. Does the integer k have a factor p such that 1>p>k?
(1) k>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]  22 Jan 2013, 02:40
gmatcracker2010 wrote:
i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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

(1) k > 4!
(2) 13! + 2<= k <= 13!+13

1.
if k=29 p=29= k NO!
if k =35 p could be 7 or 5 YES!
INSUFFICIENT!

2.
13! + 2 = 2(13!/2 + 1) = 2*odd
13! + 3 = 3(13!/3 + 1) = 3*odd
...
13! + 13 = 13 (12! +1) = 13*odd
Always there is a 1< p < 13! + n

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Re: Does the integer k have a factor p such that 1<p<k?   [#permalink] 22 Jan 2013, 02:40
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