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# If Z is an integer, is Z prime?

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Manager
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If Z is an integer, is Z prime?  [#permalink]

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07 Mar 2012, 11:51
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35% (medium)

Question Stats:

64% (00:57) correct 36% (01:14) wrong based on 454 sessions

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If Z is an integer, is Z prime?

(1) 15! < Z
(2) 17! + 2 < Z < 17! + 17
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Joined: 02 Sep 2009
Posts: 49251
If Z is an integer, is Z prime?  [#permalink]

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07 Mar 2012, 12:01
8
7
If Z is an integer, is Z prime?

(1) $$15!<z$$ --> $$z$$ is more than some number ($$15!$$). $$z$$ may or may not be a prime. Not sufficient.

(2) $$17!+2\leq{z}\leq{17!+17}$$ --> $$z$$ cannot be a prime. For instance if $$z=17!+13=13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17+1)$$, then $$z$$ is a multiple of 13, so not a prime. Same for all other numbers in this range. So, $$z=17!+x$$, where $$2\leq{x}\leq{17}$$ will definitely be a multiple of $$x$$ (as we would be able to factor out $$x$$ out of $$17!+x$$, the same way as we did for 13). Sufficient.

Similar questions to practice:
if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html
does-the-integer-k-have-a-factor-p-such-that-1-p-k-126735.html

Hope it's clear.
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Re: If Z is an integer, is Z prime?  [#permalink]

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12 Jun 2013, 04:27
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Number Properties: math-number-theory-88376.html

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Re: If Z is an integer, is Z prime?  [#permalink]

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08 Jan 2014, 03:25
1
If the statement B was rephrased to ===> 17! + 2 < Z < 17! + 19 then will the answer to this question change to E ?
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Re: If Z is an integer, is Z prime?  [#permalink]

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09 Nov 2014, 11:03
For the above question, this will help (its a matched question and all the posts there make things clearer)
if-x-is-an-integer-does-x-have-a-factor-n-such-that
regards
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Re: If Z is an integer, is Z prime?  [#permalink]

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15 Nov 2016, 09:02
I read the Kaplan explanation and read explanations above in the thread and I can't find a clue....
For instance, why 17!+13 is equal to 13∗(2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1)? Isn't (2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1) = 17!? So, why 17! multiplied by 13 would be equal to 17! + 13?
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Re: If Z is an integer, is Z prime?  [#permalink]

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15 Nov 2016, 09:08
1
Erjan_S wrote:
I read the Kaplan explanation and read explanations above in the thread and I can't find a clue....
For instance, why 17!+13 is equal to 13∗(2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1)? Isn't (2∗4∗5∗6∗7∗8∗9∗10∗11∗12∗14∗15∗16∗17+1) = 17!? So, why 17! multiplied by 13 would be equal to 17! + 13?

17! + 13 = 2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17 + 13

Factor out 13: 13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17 + 1)
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Re: If Z is an integer, is Z prime?  [#permalink]

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15 Nov 2016, 09:30
Ok, now I can see.. But honestly, it is kind of hard question for somebody who did not specialize in math...
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Re: If Z is an integer, is Z prime?  [#permalink]

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02 May 2017, 07:20
Amazing question & Amazing explanation by Bunuel!
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Re: If Z is an integer, is Z prime?  [#permalink]

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25 Aug 2018, 06:39
Bunuel wrote:
If Z is an integer, is Z prime?

(1) $$15!<z$$ --> $$z$$ is more than some number ($$15!$$). $$z$$ may or may not be a prime. Not sufficient.

(2) $$17!+2\leq{z}\leq{17!+17}$$ --> $$z$$ cannot be a prime. For instance if $$z=17!+13=13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17+1)$$, then $$z$$ is a multiple of 13, so not a prime. Same for all other numbers in this range. So, $$z=17!+x$$, where $$2\leq{x}\leq{17}$$ will definitely be a multiple of $$x$$ (as we would be able to factor out $$x$$ out of $$17!+x$$, the same way as we did for 13). Sufficient.

Similar questions to practice:
http://gmatclub.com/forum/if-x-is-an-in ... 00670.html
http://gmatclub.com/forum/does-the-inte ... 26735.html

Hope it's clear.

Bunuel can you please give similar example (with factoring out number) where Z is prime number ?
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Re: If Z is an integer, is Z prime?  [#permalink]

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26 Aug 2018, 04:01
1
dave13 wrote:
Bunuel wrote:
If Z is an integer, is Z prime?

(1) $$15!<z$$ --> $$z$$ is more than some number ($$15!$$). $$z$$ may or may not be a prime. Not sufficient.

(2) $$17!+2\leq{z}\leq{17!+17}$$ --> $$z$$ cannot be a prime. For instance if $$z=17!+13=13*(2*3*4*5*6*7*8*9*10*11*12*14*15*16*17+1)$$, then $$z$$ is a multiple of 13, so not a prime. Same for all other numbers in this range. So, $$z=17!+x$$, where $$2\leq{x}\leq{17}$$ will definitely be a multiple of $$x$$ (as we would be able to factor out $$x$$ out of $$17!+x$$, the same way as we did for 13). Sufficient.

Similar questions to practice:
http://gmatclub.com/forum/if-x-is-an-in ... 00670.html
http://gmatclub.com/forum/does-the-inte ... 26735.html

Hope it's clear.

Bunuel can you please give similar example (with factoring out number) where Z is prime number ?

Similar questions to practice:
http://gmatclub.com/forum/if-x-is-an-in ... 00670.html
http://gmatclub.com/forum/does-the-inte ... 26735.html
http://gmatclub.com/forum/for-any-integ ... 68575.html
http://gmatclub.com/forum/does-integer- ... 65983.html
http://gmatclub.com/forum/if-z-is-an-in ... 28732.html
https://gmatclub.com/forum/dose-positiv ... 90858.html
http://gmatclub.com/forum/does-p-have-a ... 88773.html

Hope this helps.
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Re: If Z is an integer, is Z prime? &nbs [#permalink] 26 Aug 2018, 04:01
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# If Z is an integer, is Z prime?

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