Find the number of ways to express 1050 as sum of consecutive integers I have to solve this task: 

Find the number of ways to present $1050$ as sum of consecutive
positive integers.

I was thinking if factorization can help there:
$$1050 = 2 \cdot  3 \cdot  5^2 \cdot  7 $$
but I am not sure how to use that information (if there is a sense)
example
I can solve something similar but on smaller scale:
\begin{align} 15 &= 15  \\ &= 7+8 \\ &=4+5+6 \\ &= 1+2+3+4+5 \end{align}
($4$ ways)

For the sum of next integer we may use formula for the sum of arithmetic sequence
$$\sum_{k=1}^na_k=\frac n2(a_1+a_n)$$
So
\begin{aligned}
1050 &= \frac n2(a_1+a_n) \\
2100 &= n(a_1+a_n) \\
2100&= n(a_1+a_n) \\
2100&= n(a_1+a1+(n-1)) \\
2^2\cdot3\cdot5^2\cdot7&= n(2a_1+n-1)
\end{aligned}
Now:


*

*If $n$ is even, then $(2a_1+n-1)$ is odd, so
   $$n=2^2\cdot3^x\cdot5^y\cdot7^z$$
where $x \in \{0,1\},\; y \in \{0,1,2\}, \;z \in \{0,1\}$,
so there are $2 \times 3 \times 2 = 12$ possibilities for $n$.

*If $n$ is odd, then similarly
$$n=3^x\cdot5^y\cdot7^z$$
and we obtain other $12$ possibilities for $n$.
So there are $24$ solutions altogether, a half o them, i. e. $\color{red}{12}$, for only positive integers, because for positive integers must be $a_1 \ge 1$, and consequently $(2a_1+n-1) >n$, so in the product $n(2a_1+n-1)$ the first multiplier have be smaller than the second one.
