I was recently working on finding $m$ that has $n$-digit polydivisible pandigital number $m$ in base $n$. I have found that $n$ can't be odd number and I have proved it. however, for even $n$, I have only checked $n=2,4,6,8,10$ and they are possible. $(m,n)=(10,2),(1230,4),(3210,4),(143250,6),(543210,6),(32541670,8),(52347610,8),(3816547290,10)$
For the bigger even numbers, as it contains a lot of possibilities, I can't do anymore. Is there anyone can find more possible $m$ or the first even $n$ such that there is no $m$? Thanks!
There’s no such number for $n=12$. For $n=14$ there is
9c3a5476b812d0(with letters standing for the digits above $9$ as usual). There are no further such numbers up to $n=38$. That’s interesting, as the Wikipedia article shows that the estimate that about every $n!$-th number is polydivisible works well for ordinary (not necessarily pandigital) polydivisible numbers, and this would suggest that there should on average be $1$ of the numbers you’re looking for for each $n$, independent of $n$, but apparently this is not the case and they are much rarer for larger $n$.I’m not sure whether you intended your list to be complete; if so, you missed the $8$-digit octal number $56743210$.
Here’s the code I used to find these numbers.