Sum them all

To generalize the problem, consider a positive integer $n$

$n=p_1^{e_1}\cdot p_2^{e_2}\cdot\ldots\cdot p_n^{e_n}$

Where the $p_i\phantom{c} i\in\{1,2,\ldots,n\}$ are $n$ 's prime factors

Now consider the following product

$P=(1+p_1+p_1^2+\ldots+p_1^{e_1})(1+p_2+p_2^2+\ldots+p_2^{e_2})\dots(1+p_n+p_n^2+\ldots+p_n^{e_n})$

$\textbf{Claim:}$

When we expand $P$ we get the sum of all the divisors of $n$

$\textbf{Proof:}$

All the numbers we get when we expand $P$ are different and they are all of the form

$p_1^{a_1}\cdot p_2^{a_2}\cdot\ldots\cdot p_n^{a_n}$

Where $0\leq a_i \leq e_1 \phantom{i} \forall i\in\{1,2,\ldots ,n\}$

Since when we expand we get $(e_1+1)(e_2+1)\ldots(e_n+1)$ terms, they must indeed be all the divisors of $n$ $\square$

In a compact form

$P=\displaystyle\prod_{i=1}^n \displaystyle\sum_{j=0}^{e_i} p_i^j=\displaystyle\prod_{i=1}^n \dfrac{p_i^{e_i+1}-1}{p_i-1}$

In this case $n=12600=2^3\cdot 3^2\cdot 5^2 \cdot 7$ , so

$\begin{aligned} P&=\dfrac{2^4-1}{2-1}\cdot\dfrac{3^3-1}{3-1}\cdot\dfrac{5^3-1}{5-1}\cdot\dfrac{7^2-1}{7-1}\\ &=15\cdot 13 \cdot 31 \cdot 8=\boxed{48360} \end{aligned}$