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The function should result in 100! + 1, which is odd. But I can't move further.

You made a small mistake: h(n) is DIFFERENT than n!

Please note that h(n) is defined as the multiplication of the EVEN integers for 2 to n

so h(6)= 2*4*6 = 2^4 * 3

and h(100)=2*4*6*...*100

obviously h(100)+1 can not be divided for any non trivial factor of h(100) and this includes of course any prime number P from 1 to 50 which appears in h(100) at least as 2*P, for example 17 appears as factor in 34 and 68 in h(n). From that follows the right answer you showed.

Just for fun it would have been the same answer if h(n)=n! .

Thanx Jorge, excellent explanation. I already spot my error after posting, however it didn't make much difference because I was lacking the concept to solve this problem.

Let me just fix up a small error on your side:

so h(6)= 2*4*6 = 2^3 * 3! because 2*4*6 = 2*2(2)*2(3) = 2^3*3!

Stare at it really hard... it's in the quotient remainder form. Anytime a number can be written as \(Q + R\) where \(0<=R<=Q\) and \(R & Q are coprime\) (ie they have no common factors in common) then that number is NOT divisible by Q or any factor of Q.

Hence the largest prime factor has to be bigger than 50, and conversely, 40.