Question 923017
In general, *[tex \large k^{99} + (2015 - k)^{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.