The sum of $n$ consecutive numbers is divisible by the greatest prime factor of $n$.

Question

I facilitated the following task with pre-service math teachers:

1. Take the sum of any three consecutive numbers. Do you notice anything special? Write a clear conjecture. Then write a clear proof for your conjecture.

2. Now, take the sum of any amount of consecutive numbers. Can you broaden your conjecture from problem 1? Prove your conjecture.

I left the task open because I wanted students to create a variety of conjectures and proofs for whole class discussion. For task 2, one student came up with the following conjecture: "The sum of $n$ consecutive integers is divisible by the greatest prime factor of $n$". I'm curious if anyone has a proof or counterexample for this claim as I do not.

Answer 1

This is an excellent conjecture. It is not quite true, as it fails for $n=2$. The sum of two consecutive numbers is odd. We can say more. The sum of $n$ consecutive numbers is divisible by $n$ if $n$ is odd and by $\frac n2$ if $n$ is even. This implies the student's conjecture for $n \gt 2$.

To see this, reduce all the numbers $\bmod n$. We will then have one each congruent to $0,1,2,\ldots n-1 \bmod n$. The sum of the numbers from $0$ to $n-1$ is $\frac 12(n-1)n$, which is divisible by $n$ or $\frac n2$ as required.

Answer 2

If $n=2$ the statement is false. Let's look at $n>2$.

The sum of $n$ consecutive numbers starting with $a$ is

$$ z=\frac{n}{2}(2a+n-1) $$ If $n$ is even, $n/2$ is an integer containing the largest prime factor of $n$, hence $z$ is divisible by that prime factor.

If $n$ is odd, $2a-1+n$ is even and $(2a+n-1)/2$ is an integer. Therefore $z$ is divisible by $n$ and by all of its prime factors.

Answer 3

If the first of the $n$ summands is $m+1$, then the sum is $$(m+1)+(m+2)+\ldots+(m+n)=nm+1+2+\ldots+n=nm+\frac{n(n+1)}{2}. $$

• If $n$ is odd, say $n=2k-1$, this is even a multiple of $n$, namely $n\cdot(m+k)$. Then even more so, it is a multiple of e.g. the largest prime divisor of $n$.
• If $n$ is even, say $n=2k$, then it is at least a multiple of $k$, namely $k\cdot(m+n+1)$. This is still a multiple of the largest prime divisor of $n$, unless $k=1$.

Hence the conjecture fails only for $n=2$ (and is meaningless for $n=1$).

Answer 4

One simple way of seeing it is the sum of $n $ consecutive integers is $n $ times the average of the consecutive integers.

If $n $ is odd then the average of the consecutive integers is the middle integer and ... is an integer... so the sum is an integer times $n $

If $n $ is even then the average of the consecutive integers is this mid point between two integers. This is an integer if multiply by $2$ so the sum is So the sum is an integer times $\frac n2$ which is an integer..

So long as $n>2$ ($n =2$ is a simple oversight; $1$ divides $2$ but.....) this stronger result necessitates the student's result as $n $ and $\frac n2$ are divisible by all the primes except maybe $2$ (and $2$ is only a factor if $n $ is even and then $2$ isn't the largest prime unless $n $ is a power of $2$ and then $2$ does divide unless $n =2^1$ and then... oops.).

Answer 5

This is an additive form of Wilson's theorem: a set of $\, n\,$ consectutive integers forms a complete system of residues $\bmod n\,$ so it is closed under negation. Its non-fixed points $ -k\not\equiv k\:$ pair up so contribute zero to the sum $\,S,\,$ leaving the sum of the fixed points. But $\, - k\equiv k \iff 2k\equiv 0,\, $ hence if $\, n\,$ is odd then $\, k\equiv 0\, $ so $\, S\equiv 0,\,$ and if $\: n\:$ is even then $\ k \equiv 0,\ n/2,\,$ so $\,S\equiv n/2.\,$ The same proof works for any subset of residues closed under negation, e.g. for reduced residues.

Remark $ $ This is a special case of Wilson's theorem for groups - see this answer - which highlights the key role played by symmetry (here a negation reflection / involution).

See also Gauss's grade-school trick for summing an arithmetic progression.