Man, I hate this problem! I saw it a few weeks ago and it took me over an hour of research (particularly on BTG) to finally understand it. I'm praying I get something like this on my GMAT, though, as I get it inside and out now!
Anyhow, the single most valuable piece of info you need to understand is the concept of co-primes (and a well deserved shout-out to a post by Mitch which broke this down and finally made it click for me).
Co-primes are any two integers that share no common positive factors other than 1. And that is the definition of any two consecutive numbers.
Once you get your head around that, a question like this can be broken down really easily.
We know that h(100) is the sum of the evens between 2 and 100, inclusive. And we know that once we know that number, we just add 1 to get the sum of h(100)+1.
Look at the pattern in the formula for h(100):
h(100) = 2 * 4 * 6 * 8 * 10.... etc. They're all increasing by an increment of 2, so we can extract a 2 from this and we get:
h(100) = 2^50(1*2*3*4*5...*48*49*50)
And that's it. The greatest prime number in h(100) is, therefore, 47. And that means the SMALLEST prime in h(100)+1 cannot be less than or equal to 47.
And a quick look at the answer choices reveals E.
But since we only need to know the smallest prime number of h(100)+1, all we have to do is find the LARGEST prime of h(100), because we'll know that it can't be equal t
