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# Factor p

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08 Feb 2004, 08:44
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

257.

This is DS problem. Could anyone try it and show the workings?

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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08 Feb 2004, 09:23
A) k > 4! means k > 24 let K be 25 which has factors 1, 5 and 25
1 < 5 < 25
But 29 has factors 1 and 29 and does not satisfy the given condition
So A is insufficient.
B) 13! has all the integer mutiples from 1 and 13
13!+2 is divisible by 2
13!+3 is divisible by 3 and so on
13!+12 is divisible by 12
For all of these 1 < n(from 2 to 12) < 13!+n

B is suffiient.
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08 Feb 2004, 17:10
anand.. could u pls.. tell me how to crack the sencond sentence.. is any easy or diff way.. what u did is good but a bit confusing..
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shubhangi

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08 Feb 2004, 20:03
I wont be surprised if the answer is D. The stem of the problem is confusing. But I am pretty sure that B is the answer.

1) k>4! involves infinite numbers. This set holds primes and non primes.
Only primes do not satisfy the condition 1<p<k
2) Now 13! = 13*12*11*10*9*8*7*6*5*4*3*2*1 = X
This huge number is divisible by any integer between 1 and 13.
The second condition says that the number K is in between
13!+2 and 13!+13. So what are the possible numbers

13*12*11*10*9*8*7*6*5*4*3*2*1 + 3 divisible by p=3 1<3<(X+3)
13*12*11*10*9*8*7*6*5*4*3*2*1 + 4 divisible by p=4 1<4<(X+4)
----
----
13*12*11*10*9*8*7*6*5*4*3*2*1 + 12 divisible by p=3 1<12<(X+12)

So any number you take it has a factor > 1 and < that number.
Hence B is sufficient

I hope I made it clear.
08 Feb 2004, 20:03
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