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?
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1You could start out by writing an equation in two variables to describe what you're looking for. – Daron Feb 25 '18 at 13:00
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1There are four of them: https://oeis.org/A038369 – aschepler Feb 25 '18 at 15:12
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Do "digital sum" and "digital product" mean the sum and product of the number's digits? Then 99 is hardly unique, isn't it? – ilkkachu Feb 25 '18 at 17:57
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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 $n$.
Mark Bennet
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