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Master | Next Rank: 500 Posts
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by KICKGMATASS123 » Fri May 29, 2009 1:22 pm
I know this question has been posted before but I'm trying out this approach that I think might lead me somewhere but I'm stuck..

Can ppl make an attempt at solving this??

h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?



Incomplete solution:

h(n)= 2*4*6*8....*94*96*98*100
= 2(1*2*3*4....*50)
= 2(50!)

h(n)+1 = 2(50!) +1

I dont know how to solve this question from this point onwards..

Can someone help!

OA is p is greater than 47
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by Domnu » Fri May 29, 2009 1:41 pm
Well, we know that 47 goes into 2*50!. If we add 1 to this, we can be absolutely sure that 47 doesn't go into 50!. This is also true for all primes below 47. So, we have to have p > 47.
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