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OG11 Question 153

Expert replies
by cliou21 » Fri May 09, 2008 1:31 pm
Can someone please explain this better then the OG?

153. Does the integer k have a factor p such that 1>p>k?
(1) k>4!
(2) 13! + 2 <= k <= 13! + 13

I understand that (1) is not sufficient.
With (2), how can you be sure that k is not a prime number between 13!+2 and 13! + 13?

Thanks!
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Source: — Data Sufficiency |

by netigen » Fri May 09, 2008 2:08 pm
Ok this one is simple if you look at it this way

Lets see what 13! is: 13x12x11 ....x2x1

So it can be written as 13! = 2xQ where Q is an integer
Similarly, it can be written as 3xW where W is an integer

Above listed equations are true because numbers from 2,3,4 to 13 are all factors of 13! so we can write it in the form nR where n is the known factor and R is an integer

K = integer
13! + 2 <= k <= 13! + 13

so if K = 13!+2 = 2xQ+2 = 2(Q+1) not a prime
similarly, if K = 13!+3 = 3W+3 = 3(W+1) not a prime
you can do this upto 13 because you know for sure that all numbers from 2 to 13 are factors of 13!

hence (b) will not be a prime for all given possibilities of K
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