2. For natural numbers $k, m$ and $n$ that satisfy the inequalities $k \geqslant 2$ and $m<n$, it holds that

$$
\left(\frac{m}{n}\right)^{k}=\overline{0, x_{1} x_{2} \ldots x_{9} x_{1} x_{2} \ldots x_{9} x_{1} x_{2} \ldots x_{9} \ldots}
$$

where $x_{1}, x_{2}, \ldots, x_{9}$ are some digits (not necessarily distinct), and the block $\overline{x_{1} x_{2} \ldots x_{9}}$ repeats periodically infinitely many times. Find all possible values for $\left(\frac{m}{n}\right)^{k}$.
2. Let us note:

$$
\begin{aligned}
10^{9}\left(\frac{m}{n}\right)^{k} & =\overline{x_{1} x_{2} \ldots x_{9}, x_{1} x_{2} \ldots x_{9} x_{1} x_{2} \ldots x_{9} \ldots} \\
& =\overline{x_{1} x_{2} \ldots x_{9}}+\overline{0, x_{1} x_{2} \ldots x_{9} x_{1} x_{2} \ldots x_{9} \ldots}=\overline{x_{1} x_{2} \ldots x_{9}}+\left(\frac{m}{n}\right)^{k}
\end{aligned}
$$

then we get $\left(10^{9}-1\right)\left(\frac{m}{n}\right)^{k}=\overline{x_{1} x_{2} \ldots x_{9}}$, i.e., $\left(10^{9}-1\right) m^{k}=\overline{x_{1} x_{2} \ldots x_{9}} \cdot n^{k}$. Clearly, we can assume that $m$ and $n$ are coprime. From this, it follows that

$$
n^{k} \mid 10^{9}-1=\underbrace{999 \ldots 999}_{9 \text { times }}
$$

We will try to factorize the number on the right side. Clearly, $\underbrace{999 \ldots 999}_{9 \text { times }}=9 \cdot \underbrace{111 \ldots 111}_{9 \text { times }}$, and the remaining number is also divisible by 9, and it is clearly divisible by 111, so the prime factor 37 appears. Thus, by direct division, we get

$$
\underbrace{999 \ldots 999}_{9 \text { times }}=3^{4} \cdot 37 \cdot 333667
$$

where we have not determined whether the number 333667 is prime. Now, since $n^{k}$ must divide the above number for $k \geqslant 2$ and $n>1$ (which follows from $n>m \geqslant 1$), it is possible that $n=3$, $n=9$, or that $n$ is some perfect power that divides the number 333667. We will prove that the last case is not possible.

First, we will determine that the number 333667 is not divisible by any prime number up to 70. This is possible by direct testing (with enough time), but we will show a faster method. Suppose $p$ is a prime number that divides 333667. Then $p$ divides $10^{9}-1$, so we have $10^{9} \equiv 1(\bmod p)$, and thus $o_{p}(10) \mid 9$ (where $o_{m}(a)$ denotes the order of the number $a$ modulo $m$, i.e., the smallest natural number $t$ such that $\left.a^{t} \equiv 1(\bmod m)\right)$, i.e., $o_{p}(10) \in\{1,3,9\}$. In the first two cases, we get $p \mid 10^{1}-1=9$ or $p \mid 10^{3}-1=999$, i.e., $p \in\{3,37\}$, but 333667 is not divisible by any of these numbers. The only remaining case is $o_{p}(10)=9$. Now, since $o_{p}(10) \mid p-1$, it follows that $p$ gives a remainder of 1 when divided by 9. The only prime numbers up to 70 that satisfy this are 19 and 37, and since 333667 is not divisible by 19 (and we have already seen that it is not divisible by 37), it follows that it is not divisible by any prime number up to 70.

Furthermore, the number 333667 is not a perfect square, as it ends in the digit 7. Therefore, if it were divisible by some perfect power, we could write $333667=p^{2} t$, for some prime number $p$ and some natural number $t, t>1$. However, then we would have $p, t \geqslant 70$ and thus $p^{2} t \geqslant 70^{3}=343000>333667$, a contradiction.

Given all of the above, the only remaining cases are $n=9$ (and then $k=2$), or $n=3$ (and then $k \in\{2,3,4\}$). Substituting all possible values of $m$ less than $n$ and coprime with $n$, we get:

$$
\left(\frac{m}{n}\right)^{k} \in\left\{\frac{1}{81}, \frac{4}{81}, \frac{16}{81}, \frac{25}{81}, \frac{49}{81}, \frac{64}{81}, \frac{1}{9}, \frac{4}{9}, \frac{1}{27}, \frac{8}{27}\right\}
$$