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)
 A: 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.
A: Using the MiniZinc solver with Gecode, I got the following $12$ solutions:
13 .. 47
24 .. 51
30 .. 54
40 .. 60
43 .. 62
63 .. 77
82 .. 93
147 .. 153
208 .. 212
261 .. 264
349 .. 351
1050 .. 1050

The model:
var 1..1050: k0;
var 0..1050: k1;

constraint
  (1050 == sum([k0 + k | k in 0..k1]));

solve satisfy;

output ["\n\(k0) .. \(k0+k1)"];

A: We want to find the number of solutions of
$$n+(n+1)+\ldots + (n+k) = 1050,\ n\in\mathbb Z_{>0},\ k\in\mathbb Z_{\geq 0}.\tag{1}$$
Rewrite the sum as $$n(k+1) + 0 + 1 +\ldots + k = n(k+1) + \frac{k(k+1)}{2}= \frac 12(2n+k)(k+1).$$
Thus, the number of solutions to $(1)$ is the same as the number of solutions of
$$(2n+k)(k+1) = 2100,\ n\in\mathbb Z_{>0},\ k\in\mathbb Z_{\geq 0}.\tag{2}$$
Let $a$ and $b$ be divisors of $2100$ such that
\begin{align}
2n+k &= a,\\
k+1 &= b.\tag{3}
\end{align}
Solving it we get
\begin{align}
n &= \frac{a-b+1}2,\\
k &= b -1.\tag{4}
\end{align}
From here we see that not every choice of integers $a$ and $b$ such that $ab = 2100$ will give us a solution to $(2)$. Since $a-b+1$ must be even, $a$ and $b$ are of opposite parities. Also, $a\geq b > 0$ since $n> 0$ and $k \geq 0$.
First determine the number of ways to factor $2100 = 2^2\cdot 3\cdot 5^2 \cdot 7$ such that one of the factors is odd. For this to be fulfilled, we shouldn't allow $4 = 2^2$ to be factored, so consider $2100 = 4\cdot 3 \cdot 5^2 \cdot 7$ instead. Thus, there are $2\cdot 2\cdot 3\cdot 2 = 24$ positive integral solutions to $2100 = a'b'$ such that one factor is odd. Because of commutativity, it means there are $12$ distinct ways to factor $2100$ into product of two factors, one of which is odd, and for every such factorization there is a unique choice for $a$ and $b$ such that $a\geq b$.
Thus, there are $12$ positive integral solutions to $(2)$.
