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

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# 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 $b$.

## Mathematics Subject Classification

11A63*no label found*

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