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