International Mathematical Olympiad

Let $n:=1829$. First, we choose some $k$ such that $n\mid 10^k-1$. For instance, any multiple of $\phi (n)$ 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 $\#\left\{d_i\left(10^k-1\right):1⩽i⩽9\right\}⩽2$. Indeed, if $$\#\left\{d_i\left(10^k-1\right):1⩽i⩽9\right\}=1$$ then, since $10^k-1$ which consists of all nines is a multiple of $n$, we have $$d_i\left(10^k-2\right)=d_i\left(10^k-1\right)\text{ for }i\in \{1,\dots ,8\},\text{ and }{d}_9\left(10^k-2\right)<d_9\left(10^k-1\right)$$ This means that $\#\left\{d_i\left(10^k-2\right):1⩽i⩽9\right\}=2$.

To prove that $\#\left\{d_i\left(10^k-1\right)\right\}⩽2$ we need an observation. Let $\overline{a_{k-1}{a}_{k-2}\dots a_0}\in \{1,\dots ,10^k-1\}$ be the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\overline{a_{k-1}{a}_{k-2}\dots a_0}$ is divisible by $n$ if and only if $\overline{a_{k-2}\dots a_0{a}_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that $$10\cdot \overline{a_{k-1}{a}_{k-2}\dots a_0}-\overline{a_{k-2}\dots a_0{a}_{k-1}}=(10^k-1)\cdot a_{k-1}$$ 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 $i\in \{1,\dots ,9\}$ 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 $\lfloor 10^{k-1}/n\rfloor$ or $1+\lfloor 10^{k-1}/n\rfloor$, we conclude that $\#\left\{d_i\left(10^k-1\right)\right\}⩽2$.