# divisor

• Sep 3rd 2007, 04:03 AM
chibuike1
divisor
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 AM
topsquark
Quote:

Originally Posted by chibuike1
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.

Do you mean what is the lowest possible prime factor of H(100) - 1?

-Dan
• Sep 3rd 2007, 05:40 AM
topsquark
Quote:

Originally Posted by chibuike1
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
• Sep 3rd 2007, 07:19 AM
chibuike1
Math 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 PM
topsquark
Quote:

Originally Posted by chibuike1
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.

From my knowledge base, the answer would be the same as the one I gave for H(100) - 1.

-Dan