Sum of n consecutive numbers is divisible by n

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Discussion Overview

The discussion revolves around the statement that the sum of n consecutive numbers is divisible by n. Participants explore various mathematical approaches and reasoning related to this statement, examining conditions under which it holds true, particularly focusing on whether n must be odd or even. The scope includes mathematical reasoning and exploration of formulas.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the sum of n consecutive numbers is only divisible by n when n is odd, citing that the median and mean are the same in this case.
  • One participant presents a formula for the sum of n consecutive numbers and argues that the remainders when dividing by n lead to divisibility when n is odd.
  • Another participant provides a derivation for the sum from k to k+n-1, suggesting that it is divisible by n, but notes that for n even, there is a remainder of n/2.
  • A later reply challenges the derivation by stating that the expression requires 2k-n-1 to be even, implying n must be odd for divisibility.
  • Participants express appreciation for contributions and offer to discuss further problems, indicating a collaborative atmosphere.

Areas of Agreement / Disagreement

There is no consensus on whether the sum of n consecutive numbers is always divisible by n, as participants present competing views regarding the conditions under which this holds true, particularly distinguishing between odd and even values of n.

Contextual Notes

Some assumptions regarding the parity of n and the properties of the derived expressions remain unresolved, particularly concerning the conditions under which divisibility is achieved.

Matt
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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.
 
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Matt said:
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
 
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:
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.
 
HallsofIvy said:
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.
 
mathman said:
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.
 
Matt said:
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.
 
mathman said:
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.
 
  • #10
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.
 

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