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        riddles >> easy >> Sum of 99th powers
        
(Message started by: NickH on Apr 5^th, 2003, 10:07am)

Title: Sum of 99th powers
Post by NickH on Apr 5^th, 2003, 10:07am

Is 1⁹⁹ + 2⁹⁹ + ... + 99⁹⁹ divisible by 99?

Title: Re: Sum of 99th powers
Post by Icarus on Apr 5^th, 2003, 11:49am

My answer is...[hide]Yes - more generally 1^k + ... + n^k is divisable by n whenever n and k are odd. It is also divisable by n if n is divisible by 4 and k >= 3 is odd.[/hide]

Title: Re: Sum of 99th powers
Post by SWF on Apr 16^th, 2003, 5:24pm

For integer n, if (99-n)⁹⁹ is multiplied out, it should be obvious that every term is a multiple of 99 except the -n⁹⁹ term. Or you can write out the whole binomal expansion if this is not clear. Therefore, n⁹⁹+(99-n)⁹⁹ is a multiple of 99 because the n⁹⁹ terms cancel out leaving a multiple of 99.

http://www.ai.rug.nl/~towr/PHP/FORMULA/formula.php?md5=21b8920d96fc77ad426b829f52134aef

Each term on the right side is a multiple of 99, so the original sum must be too.

Title: Re: Sum of 99th powers
Post by NickH on Apr 18^th, 2003, 9:08am

Another solution is to note that 1⁹⁹ + 98⁹⁹, ... , 49⁹⁹ + 50⁹⁹ are divisible by 99, since (a+b) is a factor of (aⁿ+bⁿ), for odd n. As 99⁹⁹ is divisible by 99, this completes the proof.