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If Z is an integer, is Z prime?

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

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New post 12 May 2011, 04:49
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If Z is an integer, is Z prime?


(1) \(15! < Z\)

(2) \(17! + 2 \leq Z \leq 17! + 17\)

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

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New post 07 Mar 2012, 12:01
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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.

Answer: B.

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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New post 15 May 2011, 20:54
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@ akhileshgupta05:
Satment 1 is not sufficient

For statement 2 possible values of z are

17! + 2 is Divisible by 2 . Hence not prime
17! + 3 is Divisible by 3 . Hence not prime
17! + 4 is Divisible by 4 . Hence not prime
17! + 5 is Divisible by 5 . Hence not prime
17! + 6 is Divisible by 6 . Hence not prime
17! + 7 is Divisible by 7 . Hence not prime
17! + 8 is Divisible by 8 . Hence not prime
17! + 9 is Divisible by 9 . Hence not prime
17! + 10 is Divisible by 10 . Hence not prime
17! + 11 is Divisible by 11 . Hence not prime
17! + 12 is Divisible by 12 . Hence not prime
17! + 13 is Divisible by 13 . Hence not prime
17! + 14 is Divisible by 14 . Hence not prime
17! + 15 is Divisible by 15 . Hence not prime
17! + 16 is Divisible by 16 . Hence not prime
17! + 17 is Divisible by 17 . Hence not prime


In any case z cannot not be prime hence answer is B
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Re: If Z is an integer, is Z prime?  [#permalink]

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New post 12 May 2011, 05:14
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The answer is B:

17! + 2 <= Z <= 17! + 17 means that z is like 17! + 2, 17! + 3 etc. So you can factor out a common integer from the expressions that will divide 17! + n (2 <= n <=17), e.g. 2 divides 17! + 2 because 17! also includes 2.

So z is not a prime.

(1) is not sufficient because there are primes and non-primes above 15!

Please ask if you have any more queries.
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Re: If Z is an integer, is Z prime?  [#permalink]

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New post 12 Jun 2013, 04:27
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Re: If Z is an integer, is Z prime?  [#permalink]

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New post 08 Jan 2014, 03:25
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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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New post 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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New post 15 Nov 2016, 09:08
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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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New post 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.

Answer: B.

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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New post 26 Aug 2018, 04:01
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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.

Answer: B.

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?  [#permalink]

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Re: If Z is an integer, is Z prime?   [#permalink] 05 Sep 2019, 23:15
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