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

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

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...

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.

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?