Because 2017 is odd

Forgive me for repeatedly deleting this problem as I am currently having some problems with our internet connection.

Consider a number $p$ which you would like to split into $k$ consecutive elements. If $k$ is odd, these elements will have a form of $a$ , $a\pm1$ , $a\pm2$ ... $a\pm \frac{k-1}{2}$ . Adding them all leads to $p = ka$ for some integer $a$ , which means that to divide $p$ into $k$ consecutive integers, $k | p$ . So, any number that is divisible by a prime number $k>2$ can be split into $k$ consecutive integers.

If $k$ is even on the other hand, then we will have $p$ divided as $a$ , $a\pm1$ , $a\pm2$ ... $a+ \frac{k}{2}$ for some arbitrary integer $k$ . Now, let us let $j= \frac{k}{2}$ . Adding them will lead to the statement $p = ka + j$ . and since we know that $j|k$ , this becomes $p = j(2a+1)$ .

We now have the condition such that $j|p$ for $p$ to be divided into $k$ consecutive integers. Following that, we see that $62$ can be divided into $4$ consecutive integers, but not $6$ . Furthermore, it can be seen that $j$ must contain all the even factors of $p$ , as $2a + 1$ must definitely be odd for the sequence to consist of integers.

Now, note that $2016 = 2^5 \cdot 3^2 \cdot 7$ .

For it to be divided into an odd number of consecutive integers, it can be seen that $k = 3, 7, 9, 21, 63$ . That means $2016$ can be expressed into a sum of an odd number of consecutive integers for $5$ ways.

Now, for it to be expressed as a sum of an even number of consecutive integers, it must be that $k$ is a factor of $64$ , since $32| 2016$ and it is the highest power of $2$ that divides 2016. Now, we can see the following: $k = 64, 192, 448, 576, 1344, 4032$ .

And thus there are $11$ ways to express 2016 as a sum of consecutive integers!

Let $2016$ be the sum of $n$ consecutive integer, we can expressed it as

$2016=a+(a+1)+(a+2)+...+(a+(n-2))+(a+(n-1))$

By arithmetic sum, we have

$\begin{aligned} 2016&=\dfrac{n(2a+(n-1).1)}{2} \\ 4032&=n(2a+n-1) \end{aligned}$

Since $n+(2a+n-1)=2(a+n)-1$ , we know that $n$ and $2a+n-1$ must be (even and odd) or (odd and even).

We list all even and odd multiplication of $4032$

$(n,2a+n-1)=(1,4032),(3,1344),(7,576),(9,448),(21,192),(63,64),(64,63),(192,21),(448,9),(576,7),(1344,3),(4032,1)$

For $n > 1$ , we have $\boxed{11}$ solutions.

If $k$ numbers sum to $2016$ , then obviously, the average of these $k$ numbers is $\frac{2016}{k}$ . Also, notice that the smallest of the $k$ numbers is $\frac{2016}{k} - \frac{k-1}{2}$ . (This is because the difference of the largest and smallest numbers must be $\frac{k-1}{2}$ , and the sum of the largest and smallest numbers must be equal to their average. A little bit of algebraic manipulation does the trick.)

So, if the smallest of the numbers, i.e. $\frac{2016}{k} - \frac{k-1}{2}$ is an integer, then the rest of the numbers will be integers (since they are obtained by adding a certain integer value to the smallest number). So the problem is basically asking for the number of integers $k$ for which $\frac{2016}{k} - \frac{k-1}{2}$ is integral.

Now, if $k$ is odd, then $\frac{k-1}{2}$ will be integral, so we just need $\frac{2016}{k}$ to be integral. This is just asking for the number of odd divisors of 2016, or the odd number of divisors of $\frac{2016}{32} = 63$ , which turns out to be 6.

Now, if $k$ is even, then $\frac{k-1}{2}$ will have a fractiona part of $\frac{1}{2}$ , so we need $\frac{2016}{k}$ to have a fractional part of $\frac{1}{2}$ for $\frac{2016}{k} - \frac{k-1}{2}$ to be integral. We can see that this can happen if $k$ is divisible by 64, but not by 128. Also, we must have $\frac{k}{64}$ be an odd divisor of $2016$ . We can easily see that there are $6$ values of $\frac{k}{64}$ which are odd and divide $2016$ , so we can see that there are $6$ ways to do the case where $k$ is even.

We can see that the total way to express $2016$ as the sum of consecutive integers is $6+6 = 12$ , but one of these representations include $2016 = 2016$ , Removing this representation, we can see that the answer is $\boxed{11}$ .