Background: There exists a rather simple proof on the fact that for any natural number $n$, there exists a multiple of $n$ such that the multiple only consists of the digits $1$ and $0$, found here. Trivially, this proof can be extended to prove that any series of digits can be found in a multiple of n, give or take some zeroes at the end. E.g., given the factor $12$ and the series of digits $34$, on finds that $12 * 286195 = 3434340.$
Question: Given a number $n$ and digits {d1, d2 ...} is there a method for finding the smallest multiple of $n$ such that the multiple only contains every single one of the specified digits (and any $0$s at the end)? With $n=12$, $D= $ {3,4} one may find that $12 * 287 = 3444$, which would be a valid solution, given you can prove it is the smallest such multiple. (Note that $12*25=300$ is not a solution, as $4$ is not in the multiple).
What I've tried: I haven't progressed far on this problem using the aforementioned proof, except that the above produces an upper bound for search. Of course, if you could omit digits in the final multiple, then one could simply find the smallest number that is produced when using the earlier proof on the 'digits' $3, 4, 34$ and $43$, (power set of $D$) which results in $300$ as the answer. Unfortunately, none of these results would fulfill the requirement of including all digits. Using brute force may be fine for small inputs, but for larger ones I am hoping there is a trick.
