Sum logic

Relevant wiki: Sum of Factors

We know that $\mathbb{N}=a^{p}\cdot b^{q}\cdot c^{r}\cdots$

Hence the sum of all factors of $\mathbb{N}=\dfrac{a^{p+1}-1}{a-1}\times\dfrac{b^{q+1}-1}{b-1}\times\dfrac{c^{r+1}-1}{c-1}\cdots$

Therefore $40000$ can be written as $5^{4}\times 2^{6}$ .

Now, sum of all factors of $40,000$ can be written as: $\begin{aligned} & =\dfrac{5^{4+1}-1}{5-1}\times\dfrac{2^{6+1}-1}{2-1}\\ & = \dfrac{5^{5}-1}{4}\times\dfrac{2^{7}-1}{1}\\ & =\dfrac{3125-1}{4}\times\frac{128-1}{1}\\ & =\dfrac{3124}{4}\times\frac{127}{1}\\ & =781\times 127\\ & =99,187 \end{aligned}$

$40000=2^6\times5^4,sum=(1+2+4+8+16+32+64)(1+5+25+125+625)=\boxed{99187}$