Ian Stewart wrote:

VP_Tatiana wrote:For this type of problem, it is easiest to solve for the most simple case and then generalize. If they are asking us for a general answer, and there is not an answer choice like "E: None of the above", then we know that no matter what n we pick, P must be in the same range.

So, I pick n = 2 so that h(n) = 2.

h(100) + 1 = 201 = 3*67 (I quickly tell that 201 is divisible by 3 because 2+0+1 = 3)

So, the smallest prime factor of the equation is 3, and the answer is "between 2 and 10."

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.