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by dpatwa » Sun Oct 14, 2007 8:26 pm
For every positive even integer n, the function h(n) is defined to be the product of all the event integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is

A. between 2 and 10
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40

[spoiler]OA: E[/spoiler]
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by argus79 » Mon Oct 15, 2007 1:58 pm
According to Wilson theorm (P-1)!+1 is only divisible by P if P is prime.
Here H(100)+1 can be written as 100!+1 or (101-1)!+1. Since 101 is a prime number, (101-1)!+1 will be divisible by 101.
Hence lowest prime factor of H(100)+1 is 101.
Answer E
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by Badri » Tue Oct 16, 2007 8:16 am
Thanks for the theorem!
I think you made a small error ..

H(100) is not 100! but 2* 50!
I got p=51..can u please recheck
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by jangojess » Tue Oct 16, 2007 8:17 am
argus...the function h(n) is the prod of even numbers till 100..so
h(100) + 1 = 2^50 * 50! + 1 = 2^50 (51-1)! + 1. at this stage can we apply wilson's theorem...

argus, this new theorem seems handy..think theorem will find an application in some Q related to prime numbers...thnx dude for sharing with us...btw do u have any other handy tips...
Trying hard!!!
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by dpatwa » Tue Oct 16, 2007 10:26 pm
Argus says that:
According to Wilson theorm (P-1)!+1 is only divisible by P if P is prime.
which means that P must be prime, but according to "h(100) + 1 = 2^50 * 50! + 1 = 2^50 (51-1)! + 1" by jangojess, P=51, which is not prime.
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by wongee » Thu Nov 15, 2007 5:32 pm
hey,

Patwa and argus,

can you explain how h(100) + 1=2^50 X50! +1 (the bold portion especially).

Also, it seems like 101 is more likely to be the prime number vs 51 - correct?
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