Can someone explain the line of reasoning to arrive at answer (e) to the following question in the first diagnostic exam?

For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is:

a) between 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Thanks.

Hi Emily,

The function h(n) can be expressed like this h(n)=2*4*6*8...*n.

From that, we can write h(100) + 1 like this: h(100)+1=(2*4*6*...*98*100) + 1

The key here is knowing that all terms in h(n) are even. This means that we can factor out a 2, leaving us with h(100)=2(1*2*3*4...*49*50). If 1 through 50 are all factors of h(100), they can't be factors of h(100)+1.

Thus, we know that any prime factor of h(100)+1 must be greater than 50.

Let me know if you need any further explanation!

Bill