Let $a$ be a positive integer. The sum of $a$ consecutive integers is divisible by $a$ if and only if $a$ is odd.

Question

How would one prove this? Other than using cases to prove the if and only if part, how would I prove each case to complete the proof?

Answer 1

Hints:

$$1+2+...+n=n\cdot\frac{(n+1)}2$$

so the above is divisible by $\;n\;$ iff...

Answer 2

If the first term of the $a$ consecutive integers is $b$

the sum will be $$\frac a2\left[2\cdot b+(a-1)\cdot1\right]=ab+\frac{a(a-1)}2$$

which will be divisible by $a$ iff $2\mid(a-1)$

Answer 3

Given any set of $a$ numbers, the sum of them is divisible by $a$ iff the average of them is an integer. What's the average value of the $a$ consecutive numbers $n+1, n+2, \ldots, n+a$? (Hint: it's halfway between the two ends.)