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Prove that the sum of five consecutive even numbers is divisible by five.

I know this can be proved easily using proof by examples but, the instructions forbid us from doing so. I was thinking of applying direct proofs but have no idea as to how to go about it.

Edit: Is there any way to prove the statement?

E Bas
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    Say the first even number is $n$. What is the rest of the numbers, and what is their sum? – player3236 Oct 16 '20 at 04:08
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    No, it cannot be proved by examples. You can check that it’s true for some finite collection of examples, but that provides no guarantee that it is true for *all examples. – Brian M. Scott Oct 16 '20 at 04:10
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    If you do the calculation player3236 suggests it's actually a bit nicer to label $n$ the number in the middle, so the five consecutive numbers are $n-4, n-2, n, n+2, n+4$. – Qiaochu Yuan Oct 16 '20 at 04:21
  • Re: prior comment, there are nice symmetry-inspired ways to compute such sums, going back to Gauss's grade school summation trick and continuing with proofs of Wilson's theorem, etc. Note: with $k = n/2,$ the sum is twice the sum of $, k-2,, k-1,, k,, k+1,, k+2,,$ i.e. $5$ consecutive integers. – Bill Dubuque Oct 16 '20 at 09:06

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You're asked to prove for even numbers, so start with an even number $n$.

The next four consecutive even numbers are $n+2$, $n+4$, $n+6$, and $n+8$.

The sum of these numbers is $5n+20 = 5(n+4)$.

Since the sum is an integer times $5$, the sum is divisible by $5$, and you're done.

As a bonus, $n$ can be odd also!

John
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