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How many are the natural numbers (whole positive integers ≥ 1) that are equal to the product of their figures?

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closed as off-topic by Parcly Taxel, lulu, Learnmore, Lee David Chung Lin, John Omielan Mar 9 at 15:41

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  • $\begingroup$ What have you tried? And by "figures" you mean "digits"? $\endgroup$ – Parcly Taxel Mar 9 at 12:58
  • $\begingroup$ Have you found any (other than single digits)? $\endgroup$ – lulu Mar 9 at 13:07
  • $\begingroup$ I mean how many numbers? $\endgroup$ – Galy Mar 9 at 13:11
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    $\begingroup$ The question is clear, but the comments are asking you what you have tried. It's easy to search, say, two or three digit numbers. Are there any examples amongst those? Does your answer in those cases begin to suggest something? $\endgroup$ – lulu Mar 9 at 13:12
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    $\begingroup$ As you are new to the site: people here tend to not respond well (or at all) to questions, like this, that show no effort at all. What do you think the answer is? Surely, you've searched for examples...what have you found? What does that suggest to you? $\endgroup$ – lulu Mar 9 at 13:27
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Trivially, all single-digit numbers have the desired property.

Let $n$ is a $k$-digit number with the desired property and $d$ is its smallest digit and assume $k\ge2$. We have $n>d\cdot 10^{k-1}$ because the leading digit is at least $d$ and the contribution of the lower digits must not be $0$. We also have $n\le d\cdot 9^{k-1}$ from the product of all digits. Hence $d\cdot 9^{k-1}>d\cdot 10^{k-1}$, which is absurd.

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I don't know how to work it out on the general case, only considering two digit numbers in base 10 (this can easily be generalized to other bases):

A two digit number is in the form of $10a + b$, with $a$ and $b$ being the two digits. Since we want it to be equal to the product of its digits $ab$, we have: $$10a+b = a*b \Leftrightarrow 10a = (a-1)b .$$

Since $a$ a and $b$ are integers, we must have either: $$b=2 \ \Rightarrow \ 5a=a-1 \Leftrightarrow a=-1/4 , \textrm{or}$$ $$b=5 \ \Rightarrow \ 2a=a-1 \Leftrightarrow a=-1/2 . \ \ \ $$ Since both cases are impossible, we conclude that no two digit numbers are equal to the product of their digits.

Furthermore, if there is any number with more than two digits that is equal to the product of their digits, is must be smaller than $10^{21}$ because these numbers have at least $n \geq 22$ digits and since $10^\frac{n-1}{n} > 9$ for $n \geq 22$, we would need all digits to be strictly larger than $9$, which is impossible.

For numbers that have between 3 and 21 digits it might be possible to prove something similarly to the 2 digits case, or use a brute force algorithm.

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