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.

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- Sep 3rd 2007, 04:03 AMchibuike1divisor
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.

- Sep 3rd 2007, 04:23 AMtopsquark
- Sep 3rd 2007, 05:40 AMtopsquark
$\displaystyle H(100) = 2 \cdot 4 \cdot 6 \cdot ~ ... ~ \cdot 98 \cdot 100$

$\displaystyle = 2^{50} \cdot (1 \cdot 2 \cdot 3 \cdot ~ ... ~ \cdot 49 \cdot 50 )$

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 - Sep 3rd 2007, 07:19 AMchibuike1Math solution reqd
*Sorry the right question is:*

*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.* - Sep 3rd 2007, 09:16 PMtopsquark