Sum of Different Sets of Consecutive Integers

I'll give a general solution (with any positive integer $N$ generalizing $21600$ ).

Let $N$ be the sum of $k\geq1$ positive consecutive integers $(x+1),(x+2), \cdots{ }\cdots{ } \cdots{ }, (x+k)$ ; where $x$ is a non-negative integer.

Now, number of ways to express $N$ as a sum of $k\geq1$ consecutive positive integers is equal to number of ordered pairs $(x,k)$ with $x\geq0$ , $k\geq1$ such that

$N= \dfrac{k}{2}\{(x+1)+(x+k)\}$

$\iff 2N=k(2x+1+k)$

We'll exploit the property that: $k$ and $(2x+1+k)$ are of opposite parity. That means, one of them is odd and another is even; not both odd, not both even. And Both of $k$ and $(2x+1+k)$ are positive integers.

So, we have to find number of unordered ways to express $2N$ as a product of two positive integers $p$ and $q$ such that $p$ has the opposite parity of $q$ .

For convenience, assume $p<q$ .

We have already seen, for each $(x,k)$ with $x\geq0, k\geq1$ , there is exactly one such $(p,q)$ .

And you can easily verify that, for each such $(p,q)$ , letting $k=p$ and $(2x+1+k)=q$ , there is exactly one such $(x,k)$ with $x\geq0, k\geq1$ .

So, number of such $(x,k)$ is equal to number such $(p,q)$ .

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Consider the prime factorization of $2N$ : $2N=2^a\times p_1^{a_1}\times p_2^{a_2}\times\cdots{ }\cdots{ }\cdots{ }\times p_r^{a_r}$ .

So, number of such $(p,q)$ for $2N$ will be $\boxed{(a_1+1)(a_2+1)\cdots{ }\cdots{ }\cdots{ }(a_r+1)}$ .

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For $N=21600=2^5\times3^3\times5^2$ ; number of such $(p,q)$ for $2N=2^6\times3^3\times5^2$ is $(3+1)(2+1)=12$ .

Hence, $\boxed{12}$ is the answer.