For every positive even integer n, the function H(n) is defined to be the product of all even integers from 2 to n. What is the lowest possible prime factor of H(100) =1.
$\displaystyle H(100) = 2 \cdot 4 \cdot 6 \cdot ~ ... ~ \cdot 98 \cdot 100$
$\displaystyle = 2^{50} \cdot (1 \cdot 2 \cdot 3 \cdot ~ ... ~ \cdot 49 \cdot 50 )$
So H(100) - 1 cannot be divisible by any prime less than 53, the first prime which does not appear on this list. I'm not a number theorist so there may be a way to prove that this is not divisible by a prime higher than this.
-Dan