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)
Best Answer
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 $(2)$.