What is the remainder after all?

What is the remainder when 1 2015 + 2 2015 + 3 2015 + + 201 4 2015 1^{2015} + 2^{2015} + 3^{2015} + \cdots + 2014^{2015} is divided by 2016?


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2 solutions

Takahiro Waki
May 27, 2018

k 2015 + ( 2016 k ) 2015 0 , m o d 2016 k^{2015}+(2016-k)^{2015}≡0, \mod {2016} Then 1 2015 + 100 8 2015 1 1^{2015}+1008^{2015}≡1

Giorgos K.
May 25, 2018

M a t h e m a t i c a Mathematica

Mod[Sum[n^2015,{n,2014}],2016]

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