Ian Stewart wrote: You are not free to pick a value of n here; the question asks specifically about h(100) + 1. That is, n = 100.

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.