Is there a divisibility rule of 5?

Suppose first digit is $x$ .

$= x + (x + 1) + (x + 2) + (x + 3) + (x + 4)$

$= 5x + 10$

$= 5(x + 2)$

So, we can conclude that the statement is true.

We can always assign the middle integer as $k$ ; then the sum of the 5 consecutive integers is $S = (k-2)+(k-1)+k+(k+1)+(k+2) = 5k$ , which is divisible by 5.

Bonus: Proof that if $n$ is a positive odd number, then the sum of any $n$ consecutive integers is always divisible by $n$ .

Proof: The sum of $n$ consecutive integers starting with $a$ is given by $S = \dfrac {n(2a+n-1}2 = n \left(a+\dfrac {n-1}2 \right)$ . If $n$ is odd, then $n-1$ is even and divisible by 2 then $S$ is divisible by $n$ . $\square$

Assume the five numbers are: a, a+1, a+2, a+3, a+4, a+5 The sum of these 5 numbers is : S = 5a + 1 + 2 +3 +4 S= 5a + 10 S = 5 (a + 2)

So S has 5 times something, it means it is divisible by 5

Yes, since the sum of those five numbers will always be the same as five times the third number in the sequence.

More rigorously, let $n$ be the third number and $S$ be the sum of the sequence.

Then $S = n - 2 + n - 1 + n + n + n + 1 + n + 2$ .

Adding/subtracting the constants gives $S = n + n + n + n + n = 5n$ .

Therefore the sum of any 5 consecutive numbers is divisible by 5.

Let 5 consecutives integers are: n-2, n-1, n, n+1, n+2. Their sum is: n-2+n-1+n+n+1+n+2 = 5n and this is divisible by 5.

Yes. 0+1+2+3+4 = 10. Every consecutive sum starting at n = 5n + 10 = 5(n+2).

The sum of any N consecutive digits, where N is odd, is the middle number times N.