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I've received recently a problem from my friend (and I really find it a hard one), it's about numbers that equal the product of their digits with a constant.

Well, to make it clear: Let $m \in\mathbb N$ and let $(a_n)_{n \in \mathbb N^{\ast}}$ where $n$ is the number of digits and $a_0,a_1,a_2\ldots ,a_{n-1}$ are the digits, we have: $$m=\sum_{i=0}^{n-1} {10^ia_i}\qquad\text{and}\qquad m=k\prod_{i=0}^{n-1}a_i\quad\text{for some }k\in \mathbb N^{\ast}$$

  1. Find such $m$ in cases where the constant $k$ is known (he gave me $k\in{\{5,7,13\}}$ as examples).
  2. Is there infinitely many $m$s satisfying the second equation for every $k\in \mathbb N^\ast$.
  3. Does $m$ exist for every $k \in \mathbb P$ where $\mathbb P$ is the set of Prime numbers.

$m=0$ is clearly a trivial solution, and for cases where the constant equals $5$, $7$ and $13$ I found that the number can be $175$, $735$ and $\{624,3276,1886976\}$ respectively using an Algorithm in my PC that test the exitence of such numbers up to $10^{10}.$

And assuming $a_i \neq0\quad\forall i\in[0;n-1]\cap\mathbb N$ ($\exists a_i=0\Rightarrow\prod_{i=0}^{n-1}a_i=0$), I was able to prove that: $$\lfloor{\log {\left(k+1\right)}}\rfloor\le n \lt {{\ln k+\ln 9}\over{\ln 10-\ln 9}}$$ but it's too much bigger than the tested numbers with no more than $10$ digits. So it may exist some numbers with more digits... And it doesn't answer the second question too (so far)...

For the last question, the biggest prime that I have found was $8191$, if $k=8191\quad m =4831248384$. And other primes such as $43,47$ seem to disprove it for $m \lt 5\times10^{11}$. it's the same thing for other primes such as $71,73,127,131,137,139,179,181,193,197$ for $m \lt 10^{10}$

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