The sum of $n$ consecutive numbers is divisible by the greatest prime factor of $n$. I facilitated the following task with pre-service math teachers: 


*

*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.

*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.
 A: 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.
A: 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.
A: 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 $n=1$).
A: 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.).
A: This is an additive form of Wilson's theorem: a set of $\,\rm n\,$ consectutive integers forms a complete system of residues $\bmod n\,$ so it is closed under negation. Its non-fixed points $\rm -k\not\equiv k\:$ pair up so contribute zero to the sum $\rm\,S,\,$ leaving the sum of the fixed points. But  $\rm\, - k\equiv k \iff 2k\equiv 0,\, $ hence if $\,\rm n\,$ is odd then $\rm\, k\equiv 0\, $ so $\,\rm S\equiv 0,\,$ and if $\rm\: n\:$ is even then  $\rm\ k \equiv 0,\ n/2,\,$ so $\rm\,S\equiv n/2.$ 
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.
