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 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 $m{(b-1)}^{m+1}\ge n\ge b^{m-1}$, so $m{(b-1)}^{m+1}\ge b^{m-1}$. Dividing both sides, we obtain ${(b-1)}^{2}m\ge {(b/(b-1))}^{m-1}$. 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
Canonical name	ProofThatNumberOfSumproductNumbersInAnyBaseIsFinite
Date of creation	2013-03-22 15:47:06
Last modified on	2013-03-22 15:47:06
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	7
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 11A63