Question

What is the sum of all natural numbers n such that the product of the digits of n(in base $10$) is equal to $n^2-10n-36$?

Answer 1

$\begin{array}{c}{n}^2-10n-36\\ {\sum }_{k=1}^n{n}^2-10{\sum }_{k=1}^{n}n-36{\sum }_{k=1}^{n}1\\ \frac{n\left(n+1\right)\left(2n+1\right)}{6}-10\left[\frac{n\left(n+1\right)}{2}\right]-36n\\ \left[\frac{n\left(n+1\right)\left(2n+1\right)}{6}-5\left(n+1\right)-36\right]\\ \frac{n}{6}\left[\left(n+1\right)\left(2n+1\right)-30n-216\right]=\frac{n}{6}\left[\left(n+1\right)\left(2n+1\right)-30n-246\right]Ans.\end{array}$