proof that number of sum-product numbers in any base is finite

Let $b$ be the base of numeration.

Suppose that an integer $n$ has $m$ digits when expressed in base $b$(not counting leading zeros, of course). Then $n\ge b^{m-1}$.

Since each digit is at most $b-1$, we have that the sum of the digitsis at most $m(b-1)$ and the product is at most ${(b-1)}^m$, hence thesum of the digits of $n$ times the product of the digits of $n$ is atmost $m{(b-1)}^{m+1}$.

If $n$ is a sum-product number, then $n$ equals the sum of its digitstimes the product of its digits. In light of the inequalities of thelast two paragraphs, this implies that $m{(b-1)}^{m+1}\ge n\ge b^{m-1}$, so $m{(b-1)}^{m+1}\ge b^{m-1}$. Dividing both sides, weobtain ${(b-1)}^{2}m\ge {(b/(b-1))}^{m-1}$. By the growth of exponentialfunction, there can only be a finite number of values of $m$ for whichthis is true. Hence, there is a finite limit to the number of digitsof $n$, so there can only be a finite number of sum-product numbers toany given base $b$.