# 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 $n\geq b^{m-1}$.

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

If $n$ is a sum-product number, then $n$ equals the sum of its digits times the product of its digits. In light of the inequalities of the last two paragraphs, this implies that $m(b-1)^{m+1}\geq n\geq b^{m-1}$, so $m(b-1)^{m+1}\geq b^{m-1}$. Dividing both sides, we obtain $(b-1)^{2}m\geq(b/(b-1))^{m-1}$. By the growth of exponential function, there can only be a finite number of values of $m$ for which this is true. Hence, there is a finite limit to the number of digits of $n$, so there can only be a finite number of sum-product numbers to any given base $b$.

Title proof that number of sum-product numbers in any base is finite ProofThatNumberOfSumproductNumbersInAnyBaseIsFinite 2013-03-22 15:47:06 2013-03-22 15:47:06 rspuzio (6075) rspuzio (6075) 7 rspuzio (6075) Proof msc 11A63