Consecutive Conundrum

Clearly, $n$ consecutive numbers give the residues $0\to n-1\pmod{n}$ . Thus, their sum is simply $S\equiv \dfrac{n(n-1)}{2} \pmod{n}$ Clearly if $n$ is odd then $\dfrac{n(n-1)}{2}\equiv 0\pmod{n}$ . It remains to prove this is not true for even $n=2k$ . But $\dfrac{n(n-1)}{2} = k(2k-1)\not\equiv 0\pmod{2k}$ so this property is true for odd $n$ only.

Two solutions in one. Lucky you!

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Argument using congruencies

For any integer $n$ , the set of congruencies modulo $n$ of any $n$ consecutive integers is exactly the set $\{0,\,1,\,2,\,\dots,\,n-1\}$ . In other words:

Let $A=\{m,\,m+1,\,\dots,\,m+n-1\}$ , for some positive integer $m$ (so $A$ is a set of $n$ consecutive integers). Then:

$R=\{r\in\mathbb{Z}\,|\,0\le r\le n-1;\,\exists a\in A:\,a\equiv r\,\mathrm{mod}\,n\}=\{0,\,1,\,2,\,\dots,\,n-1\}.$

We can now partition $R$ , starting with subsets that sum to $n$ . Specifically, we make the following pairs:

$\{1,\,n-1\}$

$\{2,\,n-2\}$

$\{3,\,n-3\}$

$\dots$

Now:

1. If $n$ is odd, we are left only with $\{0\}$ . So all of the remainders $r$ sum to a multiple of $n$ , and the original sum must therefore be a multiple of $n$ .

2. If $n$ is even, we are left with $\{0\}$ and $\{\frac12n\}$ . The sum of the remainders $r$ is, therefore, congruent to $\frac12n\,\mathrm{mod}\,n$ . Therefore the original sum cannot be a multiple of $n$ .

So, we conclude:

The sum of $n$ consecutive integers is a multiple of $n$ exactly when $\boxed{n\textrm{ is odd}}$ .

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Argument using arithmetic series

A sequence $\{u_r\}_{r=0\rightarrow n-1}$ of $n$ consecutive integers may be written as follows:

$\left\{m,\,m+1,\,m+2,\,\dots,\,m+n-1\right\},$

where $m$ is the first integer, and $u_r=m+r,\quad r=0\rightarrow n-1$ .

The sum of these integers is as follows:

$\begin{array}{c}[rcl] .\sum_{r=0}^{n-1}u_r & = & \sum_{r=0}^{n-1}(m+r)\\ & = & nm+\sum_{r=0}^{n-1}r\\ & = & nm+\frac12n(n-1). \end{array}$

Clearly, $nm$ is a multiple of $n$ . So:

$\sum_{r=0}^{n-1}u_r\textrm{ is a multiple of }n\Leftrightarrow\frac12n(n-1)\textrm{ is a multiple of }n.$

Now:

1. If $n$ is odd, $\frac12(n-1)$ is an integer, so $\frac12n(n-1)=n\left(\frac12(n-1)\right)$ is a multiple of $n$ . Therefore, the original sum is a multiple of $n$ .

2. If $n$ is even, $n-1$ is odd. Therefore, $\frac12n(n-1)$ is an odd multiple of $\frac12n$ , which cannot be a multiple of $n$ . Therefore, the original sum is not a multiple of $n$ .

So, we conclude:

The sum of $n$ consecutive integers is a multiple of $n$ exactly when $\boxed{n\textrm{ is odd}}$ .