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Topic: Sum of 99th powers (Read 804 times) |
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Icarus
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Re: Sum of 99th powers
« Reply #1 on: Apr 5th, 2003, 11:49am » |
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My answer is...Yes - more generally 1k + ... + nk 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.
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SWF
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Re: Sum of 99th powers
« Reply #2 on: Apr 16th, 2003, 5:24pm » |
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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.
Each term on the right side is a multiple of 99, so the original sum must be too.
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NickH
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Re: Sum of 99th powers
« Reply #3 on: Apr 18th, 2003, 9:08am » |
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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.
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| « Last Edit: Jan 31st, 2004, 6:04am by NickH » |
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