h(100) + 1
= 2 * 4 * ... * 100 + 1
= 2*50! + 1

2*50! can be expressed as a multiple of X (2 <= X <= 50) and so
(2*50! + 1) cannot be a multiple of of that same number X.

Since p is a factor of h(100) +1, p should be > X and so option
"E" looks like the correct answer.

Thoughts ?

this is an official gmat queionst and the answer is indeed E. i still haven't figured out why though.

We know that h(100) contains the product of even integers.

h(100) -> 100.98.96.....2.

We know that h(100) is divisible by 2.because it has a 2 in it.But if we add 1 to a 2 we will not get to the next multiple of 2.Eg lets say h(100)= 222.Then adding1 to it will give us 223,which is not a multiple of 2

Same case for 3.We know h(100) is divisible by 3 because it has 6.But if we add 1 to a multiple of 3 we will not get to the next multiple of 3.

Similarly h(100) is a multiple of 5 for it has 10. same for other prime number 7(14),11 (22),13(26),17(34),19(38),23(46),29(58),31(62),37(74),41(82).In this way I have eliminated all answer choices except E.

Therefore,the answer is E.

Basically the concept tested by this questions is - if you add a multiple of a number say X to another multiple of the same number X,you will get another multiple of X.

For eg : if you have a multiple of a number say (7) -> 84.If you add another multiple of this number say 70 to 84.The resulting number (154) too will be a multiple of 7.

Hope this helps.