The integer 99 is unique in that it is equal to its digital sum plus its digital product. Are there infinitely many integers that are equal to their digital sum times their digital product?
2026-03-25 19:04:05.1774465445
Are there infinitely many elements $\in \mathbb Z$ equal to digital sum times digital product
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No, because for an $n$ digit number the digit sum is $\le 9n$ and the digit product is $\le 9^n$. The product of the two is therefore $\le 9^{n+1}n$.
An $n$ digit number is $\ge 10^{n-1}$
You would therefore need $81n\cdot9^{n-1}\gt 10^{n-1}$ or $81n\ge \left(\frac {10}{9}\right)^{n-1}$.
This is true for only finitely many values of $n$.