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Question 442481: show that a number and it's cube leaves the same remainder when divided by 6
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! If x^3 and x leave the same remainder upon dividing by 6, then x^3 ≡ x (mod 6) --> x^3 - x ≡ 0 (mod 6).
The left expression factors to
x(x^2 - 1) = x(x+1)(x-1). At least one of these numbers must be even, and exactly one of them is divisible by 3, so x(x+1)(x-1) ≡ 0 (mod 6), so our original claim is true.
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