Problem 1821

Let $n := 1829$. First, we choose some $k$ such that $n \mid 10^{k} - 1$. For instance, any multiple of $\varphi(n)$ would work since $n$ is coprime to $10$. We will show that either $T = 10^{k} - 1$ or $T = 10^{k} - 2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrarily large.

For this it suffices to show that $\#\left\{ d_{i}\left(10^{k} - 1\right) : 1 \leqslant i \leqslant 9 \right\} \leqslant 2$. Indeed, if
$$\#\left\{ d_{i}\left(10^{k} - 1\right) : 1 \leqslant i \leqslant 9 \right\} = 1$$
then, since $10^{k} - 1$ which consists of all nines is a multiple of $n$, we have
$$d_{i}\left(10^{k} - 2\right) = d_{i}\left(10^{k} - 1\right) \text{ for } i \in \{1, \ldots, 8\}, \text{ and } d_{9}\left(10^{k} - 2\right) < d_{9}\left(10^{k} - 1\right)$$
This means that $\#\left\{ d_{i}\left(10^{k} - 2\right) : 1 \leqslant i \leqslant 9 \right\} = 2$.

To prove that $\#\left\{ d_{i}\left(10^{k} - 1\right) \right\} \leqslant 2$ we need an observation. Let $\overline{a_{k-1} a_{k-2} \ldots a_{0}} \in \{1, \ldots, 10^{k} - 1\}$ be the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\overline{a_{k-1} a_{k-2} \ldots a_{0}}$ is divisible by $n$ if and only if $\overline{a_{k-2} \ldots a_{0} a_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that
$$10 \cdot \overline{a_{k-1} a_{k-2} \ldots a_{0}} - \overline{a_{k-2} \ldots a_{0} a_{k-1}} = (10^{k} - 1) \cdot a_{k-1}$$
is divisible by $n$.

This observation shows that the set of multiples of $n$ between $1$ and $10^{k} - 1$ is invariant under simultaneous cyclic permutation of digits when numbers are written with leading zeroes. Hence, for each $i \in \{1, \ldots, 9\}$ the number $d_{i}\left(10^{k} - 1\right)$ is $k$ times larger than the number of $k$ digit numbers which start from the digit $i$ and are divisible by $n$. Since the latter number is either $\left\lfloor 10^{k-1} / n \right\rfloor$ or $1 + \left\lfloor 10^{k-1} / n \right\rfloor$, we conclude that $\#\left\{ d_{i}\left(10^{k} - 1\right) \right\} \leqslant 2$.