# Sum of n consecutive numbers is divisible by n

## Main Question or Discussion Point

I'm trying to investigate this statement: The sum of n consecutive numbers is always divisible by n.

I've found already that it's only true when the total amount of numbers is an odd number. I've also found that the median and mean are the same with consecutive numbers. I can not prove that the it's only true with odd numbers.

lavinia
Gold Member
I'm trying to investigate this statement: The sum of n consecutive numbers is always divisible by n.

I've found already that it's only true when the total amount of numbers is an odd number. I've also found that the median and mean are the same with consecutive numbers. I can not prove that the it's only true with odd numbers.
try writing down a general formula for the sum

mathman
The remainders, after dividing by n, for n consecutive numbers are 0,1,...,n-1. Sum the remainders to get n(n-1)/2.

If n is odd, n-1 is even and (n-1)/2 is an integer, so this sum is divisible by n.

If n is even, n-1 is odd. n(n-1)/2 = n(n-2)/2 + n/2. (n-2)/2 is an integer, so n/2 is the remainder after dividing the sum by n.

Last edited:
HallsofIvy
Homework Helper
Another way to look at it: to find the sum of numbers from k to k+ n- 1, find the sum from 1 to k+ n- 1, (k+ n- 1)(k+ n)/2, then subtract the sum from 1 to k, k(k- 1)/2. That is ((k^2+ 2kn+ n^2- k- n)- (k^2- k))/2= (2kn+ n^2- n)/2= n((2k- n- 1)/2) so is divisible by n.

mathman
Another way to look at it: to find the sum of numbers from k to k+ n- 1, find the sum from 1 to k+ n- 1, (k+ n- 1)(k+ n)/2, then subtract the sum from 1 to k, k(k- 1)/2. That is ((k^2+ 2kn+ n^2- k- n)- (k^2- k))/2= (2kn+ n^2- n)/2= n((2k- n- 1)/2) so is divisible by n.
You missed the point. In your derivation 2k-n-1 must be even, so n must be odd. When n is even the expression has a remainder of n/2.

Thank you very much for your everyone. I really appreciate it.

Mathman, could I run a couple of problems past you if you have a moment?

Regards,
Matt.

You missed the point. In your derivation 2k-n-1 must be even, so n must be odd. When n is even the expression has a remainder of n/2.
Thank you very much for your everyone. I really appreciate it.
Mathman, could I run a couple of problems past you if you have a moment?

Regards,
Matt.

mathman
Thank you very much for your everyone. I really appreciate it.
Mathman, could I run a couple of problems past you if you have a moment?

Regards,
Matt.
If you submit them via the forum I will look at them. I prefer not to answer privately.

HallsofIvy
Homework Helper
You missed the point. In your derivation 2k-n-1 must be even,
How do you get that?

so n must be odd. When n is even the expression has a remainder of n/2.

mathman