<s>[INST] 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)
 [/INST] 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 $(2)$.
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