SOLUTION: What is the remainder when the sum 1^99 + 2^99 + 3^99........+ 2014^99 is divided by 2015?
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Question 923017: What is the remainder when the sum 1^99 + 2^99 + 3^99........+ 2014^99 is divided by 2015?
Thank you for your answer :)
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
In general, ^{99} \equiv k^{99} + (-k)^{99} \equiv 0)
(mod 2015)
Therefore 1^99 + 2014^99 is divisible by 2015, 2^99 + 2013^99 is divisible by 2015, ..., 1007^99 + 1008^99 is divisible by 2015, so the total remainder is 0.
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