Sum of n consecutive numbers is divisible by n

[Matt]

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]

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.

 

[HallsofIvy]

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.

 

[Matt]

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.

 

[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]

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]

In your derivation you assumed n((2k- n- 1)/2) is an integer divisible by n. Therefore (2k-1-n)/2 must be an integer.

2k--n-1: 2k is even, if n is even, n+1 is odd. Even number minus odd number is odd.