Prove or disprove that the sum of any 5 consecutive integers is divisible by 5

please provide full example fully worked out.

n +(n+1) +(n+2)+(n+3) +(n+4) = 5n + 10
(5n + 10 )/5 = n + 2
for all n

Prove that the sum of any 5 consecutive integers is divisible by 5.
In mathematical terms, that means prove that:
n + (n + 1) + (n + 2) + (n + 3) + (n + 4) is divisible by 5, for any integer n.
Rearranging and combining like terms gives us:
5n + 10 is divisible by 5.
Factor this:
5(n + 2) is divisible by 5.
This is obviously divisible by 5, because dividing it by 5 yields:
n + 2
And since n is an integer, n + 2 is an integer.

Let's first make sure this is true.

0+1+2+3+4 = 10. Yup!

So now, let's think why.

Mod 5, any five consecutive integers must be (0, 1, 2, 3, 4) mod 5 or (1, 2, 3, 4, 0) mod 5 or (2, 3, 4, 0, 1) mod 5, etc. Essentially there will always be that mix of remainders.

So the sum of any five consecutive numbers, will be congruent mod 5 to the sum 0+1+2+3+4, which is indeed 0 mod 5. Yay!

Example: 4+5+6+7+8 = 30. 4=4 mod 5, 5=0 mod 5. 6=1 mod 5, 7=2 mod 5, 7 = 3 mod 5. So
4+5+6+7+8 = 4+0+1+2+3 = 0 mod 5, as desired.

Five Numbers : (x+1)+(x+2)+(x+3+(x+4)+(x)
Result = 5x + 10
No matter what x is, 5x+15 is always divisible by 5

False.

6+4+3+7+2 = 22
22 is not divisible by 5.