Because the function of h(100) is equal to the product of all evens between 2 and 100, inclusive, and because we need to know the smallest prime of h(100)+1, and because we already know that any two consecutive integers are co-primes and therefore don't share any factors except for 1, all we need to do is figure out which primes belong to h(100), and we'll know which ones DON'T belong to h(100)+1.
So, again, if you follow the function, h(100) = 2 * 4 * 6 * 8.... They're all multiples of two, though, so if we want to see which primes make up h(100), we need to break down every single even number into its primes... OR we can notice the pattern.
When you factor out a two from every single multiple in h(100), then you get 2^50(1 * 2* 3 * 4 * 5....). So you see that every single number between 2 and 50 is a factor of h(100). Therefore, NONE of those prime numbers can be factors of h(100)+1. And since the greatest prime between 2 and 50 is 47, this means the the smallest prime of H(100)+1 can be any prime greater than 47.
Does that help any?
