Find the number of ways to express 1050 as sum of consecutive integers

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 $(2)$.

The sum of the first $n$ natural numbers is $$\sum_{i=0}^ni=\frac12n(n+1).$$ So by subtracting the first $m-1$ terms we get the sum of all consecutive integers from $m$ to $n$; $$\sum_{i=m}^ni=\frac12n(n+1)-\frac12(m-1)m=\frac12(n+m)(n-m+1).$$ To count the number of ways to write a number $k$ as a sum of consecutive integers, we want to find natural numbers $m$ and $n$ with $m<n$ such that $$(n+m)(n-m+1)=2k.$$ In particular this gives a factorization of $2k$. Conversely, if $2k=a\times b$ is a factorization where $a\not\equiv b\pmod{2}$ then setting $$m:=\frac{a-b+1}{2}\qquad\text{ and }\qquad n:=\frac{a+b-1}{2},$$ gives $(n+m)(n-m+1)=2k$. This shows that if $k=2^lk'$ with $l\in\Bbb{N}$ and $k'$ odd, then the expressions of $k$ as a sum of consecutive integers correspond $2$-to-$1$ to the divisors of $k'$; for each divisor $d$ of $k'$ we have the two factorizations $$2k=d\times\left(2^l\frac{k'}{d}\right)=\left(2^ld\right)\times\frac{k'}{d},$$ of $2k$ into an even and an odd number. The corresponding sums include the trivial sum $k=\sum_{i=k}^ki$, as well as sums with negative integers. This shows that the total number of ways to represent a number $k$ as a sum of consecutive integers, is twice the number of divisors of $k'$.

The number of expressions of $k$ as a sum of positive integers is the number of factorizations for which $m\geq0$, or equivalently $a+1\geq b$. Of course for every factorization $2k=a\times b$ with $a\neq b$ we have either $a+1\geq b$ or $b+1\geq a$ exclusively, so if $2k$ is not a square then precisely half of all expressions involve only positive integers.

In this particular case $k=1050=2\cdot3\cdot5^2\cdot7$ and so $k'=525=3\cdot5^2\cdot7$, and the number of divisors of $k'$ equals $2\times3\times2=12$, so there are $12$ expressions of $k$ as a sum of positive integers. The factors of $k'$ and corresponding sums are \begin{eqnarray*} \text{factor}&&\qquad&&\text{sums}\\ \hline 1&&\qquad&&k=\sum_{i=1050}^{1050}i &&\qquad&&k=\sum_{i=261}^{264}i\\ 3&&\qquad&&k=\sum_{i=349}^{352}i &&\qquad&&k=\sum_{i=82}^{93}i\\ 5&&\qquad&&k=\sum_{i=208}^{212}i &&\qquad&&k=\sum_{i=43}^{62}i\\ 7&&\qquad&&k=\sum_{i=147}^{153}i &&\qquad&&k=\sum_{i=24}^{51}i\\ 15&&\qquad&&k=\sum_{i=63}^{77}i &&\qquad&&k=\sum_{i=-12}^{47}i\\ 21&&\qquad&&k=\sum_{i=40}^{60}i &&\qquad&&k=\sum_{i=-29}^{54}i\\ 25&&\qquad&&k=\sum_{i=30}^{54}i &&\qquad&&k=\sum_{i=-39}^{60}i\\ 35&&\qquad&&k=\sum_{i=13}^{47}i &&\qquad&&k=\sum_{i=-62}^{77}i\\ 75&&\qquad&&k=\sum_{i=-23}^{51}i &&\qquad&&k=\sum_{i=-146}^{153}i\\ 105&&\qquad&&k=\sum_{i=-42}^{62}i &&\qquad&&k=\sum_{i=-207}^{212}i\\ 175&&\qquad&&k=\sum_{i=-81}^{93}i &&\qquad&&k=\sum_{i=-348}^{351}i\\ 525&&\qquad&&k=\sum_{i=-260}^{264}i &&\qquad&&k=\sum_{i=-1049}^{1050}i\\ \end{eqnarray*} We see that indeed $12$ out of these $24$ expressions involve only positive integers.

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:

1. 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 $n$.

2. 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.

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)"];