For every positive even 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

Can someone walk me through this one?

Guest, this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function,
h(100) = 2*4*6*8*...*100

By factoring a 2 from each term of our function, h(100) can be rewritten as
2^50*(1*2*3*...*50).

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.

Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.

Hope that helps
-Dan

I got stuck on this question as well. can you please tell me how 2^50 can be factored? I'm not sure I'm following the math.

Thank you!

the number h(100) is 2 x 4 x 6 x 8 x ... x 100
which is (2 x 1) x (2 x 2) x (2 x 3) x ... x (2 x 50)

if you pull out all those 2's and send them all to the front of the expression - something you're allowed to do, because you can multiply a group of numbers in any order you please - you get 2^50 in front, because there are fifty 2's.

thanks, ron!

Nice job.