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m_blooms
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 Posted: Wed Apr 16, 2008 1:57 pm Post subject: LAymans terms please---GMAT Prep Data Sufficiency |
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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"
I know its asking something about prime numbers...HELP PLEASE? |
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Neo2000
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| Posted: Wed Apr 16, 2008 10:54 pm Post subject: |
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From statement A you cannot say anything. K could be 25 and still satisfy the condition 1<5<25
From the 2nd statement: You know that 13! ends in a 0
If you add 2 to this, the number is divisible by 2 thus 1<2<13!+2
Similarly for 3,4,5,6,7,8,9,10,11,12 and 13
So 2 is sufficient.
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zacharyz
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| Posted: Thu Apr 17, 2008 5:26 am Post subject: |
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For any of the even numbers, you know that there exists a number
1<2<k for all even k between 13!+2 and 13!+13.
However, how do you know that none of the odd numbers between 13!2 and 13!+13 are not prime? |
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Stuart Kovinsky
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| Posted: Thu Apr 17, 2008 12:24 pm Post subject: Re: LAymans terms please---GMAT Prep Data Sufficiency |
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| m_blooms wrote: | Does the integer k have a factor p such that 1<p<k
(1) "k>4!"
(2) "13!+2 <= k <= 13!+13"
I know its asking something about prime numbers...HELP PLEASE? |
The question is really asking "is k prime?"
If k is prime, then there will be no factor of k between 1 and k.
If k is not prime, then there will be a factor of k between 1 and k.
The only exception is that if k is less than or equal to 1, of course there will be no factors between 1 and k.
(1) k > 4
Well, there are lots of primes and non-primes greater than 4: insufficient.
(2) 13! + 2 <= k <= 13! + 13
There will be no primes in this range, so this statement is sufficient.
Here's how we know:
Every number from 1 to 13 is a factor of 13!.
Since 2 is a factor of 13!, 2 will also be a factor of 13! + 2
Since 3 is a factor of 13!, 3 will also be a factor of 13! + 3
Since 4 is a factor of 13!, 4 will also be a factor of 13! + 4
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Since 13 is a factor of 13!, 13 will also be a factor of 13! + 13
(2) is sufficient and (1) isn't: choose (b).
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