Solution

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 arbitrary 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 \(\overline{a_{k-1} a_{k-2} \ldots a_{0}} \in\left{1, \ldots, 10^{k}-1\right}\) 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

\[10 \cdot \overline{a_{k-1} a_{k-2} \ldots a_{0}}-\overline{a_{k-2} \ldots a_{0} a_{k-1}}=\left(10^{k}-1\right) \cdot a_{k-1}\]

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 \(#\left{d_{i}\left(10^{k}-1\right)\right} \leqslant 2\). **Comment.** More careful analysis shows that \(#\left{d_{i}\left(10^{k}-1\right): 1 \leqslant i \leqslant 9\right}=1\) if and only if \(n \equiv 1\) \((\bmod 10)\), which is not the case for \(n=1829\).