H(100) + 1

mdoganay wrote:Every positive even integer n, the function h(n) is defined to be product of all 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

Can somebody explain this question, I can not even figure out how to attack this question.

Me neither, but I never let that stop me from finding more interesting ways of organizing the givens.

product(evens from 2 through n) = 2^(n/2) * (n/2)!

for n = 100, product = 2^50 * 50!

So, we need to find the smallest prime factor of 2^50 * 50! + 1

Now the key to this question starts to present itself: 50! is a multiple of all prime numbers less than 50, since 50! is the product of ALL integers from 1 through 50. Therefore, since we're adding 1 to a multiple of 50!, we know that none of the factors of 50! can be a factor of h(100)+1. This, in turn, means that none of the prime numbers less than 50 can be a factor of h(100)+1.

The smallest prime of h(100)+1 is certainly greater than 40.

E

E +1