Problem Solving

h(n) is product of all even integers 2 to n, inclusive. if P is smallest prime factor of h(100) + 1 , the n P is":

between 2 to 10
between 10 to 20
between 20 to 30
between 30 to 40
greater than 40

What is h(100)?

h(100) = 2*4*6*...*96*98*100 = (2*1)*(2*2)*(2*3)*...*(2*48)*(2*49)*(2*50) = (2^50)*50!

So we need to know about the smallest prime factor of (2^50)*50! + 1.

Notice that 50! is divisible by every prime less than 50. That ensures that (2^50)*50! is divisible by every prime less than 50, which ensures that (2^50)*50! + 1 will be divisible by no prime less than 50: the remainder will be 1 each time. The smallest prime factor of h(100) + 1 must therefore be larger than 50 (and therefore certainly larger than 40). E.

I am completely agree with Anurag@Gurome the correct answer is E I have just solved it and found the same answer given by him,I always watch your explanation and its very good.
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