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by neha.katyal22 » Sun Feb 06, 2011 8:27 pm
Fof every positive even integer n, the function H(n) is equal to the product of all even integers from 2 to n inclusive. if p is the smallest prime factor of H(100) + 1, the p is:
1)bw 2 and 10
2)bw 10 and 20
3)bw 20 and 30
4)bw 30 and 40
5)greater than 40
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Source: — Problem Solving |

by Anurag@Gurome » Sun Feb 06, 2011 9:16 pm
H(100) = (2*4*6*8*....*100) = ((2^50)*(1*2*......*50)) = (2^50)*(50!).
So all the prime number from 2 to 50, would divide (2^50 * 50!).
Thus none of the numbers between 2 and 50 would NOT divide [(2^50)*(50!) + 1] as they will always leave a reminder of 1.

Therefore, the required prime number would be greater than 50.

The correct answer is E.
Anurag Mairal, Ph.D., MBA
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