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What is the remainder when 1^2017+2^2017+3^2017+4^2017 is divided by 2017?

+5 votes
asked Aug 18, 2016 in Mathematics by Rahul Roy (7,895 points) 31 96 261

1 Answer

+3 votes
answered Aug 18, 2016 by vikash (21,277 points) 4 18 65
selected Sep 22, 2016 by sarthaks
Best answer

The remainder will be zero.

a^n + b^n is divisible by a + b (provided n is odd).

So, 1^2017+2^2017+3^2017+4^2017 will be divisible by (1+2+3+4)

So, it is divisible by 2017 as well.

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