Which sets of base 10 digits have the property that, for every $n$, there is a $n$-digit number made up of these digits that is divisible by $5^n$? Which sets of non-zero base 10 digits have the property that, for every $n$, there is a $n$-digit number made up of these digits that is divisible by $5^n$?
This is an extension of
Prove that for any integer $n>0$, there exists a number consisting of 1's and 2's only, which is divisible by $2^n$.
Here is my answer:
Any set of digits 
which form
a complete residue set
modulo 5.
In particular,
any 5 consecutive digits.
A proof by induction
is not too difficult.
 A: Partial result . . .

Claim:

If $S \subseteq \{1,2,3,4,5,6,7,8,9\}$ contains a complete residue system, mod $5$, then for all positive integers $n$, there is an $n$-digit number $x$ such that 


*

*All digits of $x$ are elements of $S$.$\\[4pt]$

*$5^n{\mid}x$.


Proof:

Assume $S \subseteq \{1,2,3,4,5,6,7,8,9\}$ contains a complete residue system, mod $5$.

Necessarily $5 \in S$, hence the claim holds for $n=1$.

Proceed by induction on $n$.

Fix $n \ge 1$, and let $x$ be an $n$-digit number such that all digits of $x$ are elements of $S$, and $5^n{\mid}x$.

Let $y ={\large{\frac{x}{5^n}}}$.

Choose $d \in S$ such that $d(2^n) + y \equiv 0\;(\text{mod}\;5)$.

Let $x'=d(10^n)+x$. 

Then $x'$ is an $(n+1)$-digit number, all of whose digits are elements of $S$.
\begin{align*}
\text{Also,}\;\;x'&=d(10^n)+x\\[4pt]
&=(5^n)(d(2^n)+y)\\[4pt]
\end{align*}
which is a multiple of $5^{n+1}$.

Thus, the induction is complete.

Update:

For $S \subseteq \{1,2,3,4,5,6,7,8,9\}$, call $S$ qualifying if for every positive integer $n$, there is an $n$-digit number $x$ such that


*

*All digits of $x$ are in $S$.$\\[4pt]$

*$5^n{\mid}x$.


As was shown, if $S \subseteq \{1,2,3,4,5,6,7,8,9\}$, and $S$ contains a complete residue system, mod $5$, then $S$ is qualifying.

For $S \subseteq \{1,2,3,4,5,6,7,8,9\}$, call $S$ a minimal exception if


*

*$S$ is a qualifying set.$\\[4pt]$

*$S$ does not contain a complete residue system, mod $5$.$\\[4pt]$

*No proper subset of $S$ is qualifying.


Conjecture:

There are exactly $11$ minimal exceptions, as listed below:
\begin{align*}
&\{1, 2, 3, 5, 6, 7\}\\[4pt]
&\{1, 2, 3, 5, 6, 8\}\\[4pt]
&\{1, 2, 3, 5, 7, 8\}\\[4pt]
&\{1, 2, 5, 6, 7, 8\}\\[4pt]
&\{1, 2, 5, 6, 7, 9\}\\[4pt]
&\{1, 3, 5, 6, 7, 8\}\\[4pt]
&\{2, 3, 4, 5, 7, 9\}\\[4pt]
&\{2, 3, 5, 6, 7, 8\}\\[4pt]
&\{2, 3, 5, 7, 8, 9\}\\[4pt]
&\{2, 4, 5, 6, 7, 9\}\\[4pt]
&\{3, 4, 5, 7, 8, 9\}\\[4pt]
\end{align*}
Remarks: All of the above sets survived testing for $1 \le n \le 10000$.
