B1. Find all pairs of natural numbers $k$ and $n$ for which the fraction $\frac{4^{k-1}}{n}$ has a decimal representation of the form $0, \bar{n}$, where $\bar{n}$ denotes the repeating part. For example, if $n=720$, then $0, \bar{n}=0,720720720 \ldots$
B1. From the given equation $\frac{4^{k-1}}{n}=0, \bar{n}$. Let $m$ be the number of digits in the number $n$. If we multiply the equation by $10^{m}$, we get $\frac{4^{n}-1 \cdot 10^{m}}{n}=n, \bar{n}$. Subtract the first equation from the second to get $\frac{4^{k-1} \cdot 10^{m}}{n}-\frac{4^{k-1}}{n}=n$, which simplifies to

$$
4^{k-1} \cdot\left(10^{m}-1\right)=n^{2}
$$

Since $10^{m}-1=\underbrace{99 \ldots 99}_{m}=9 \cdot \underbrace{11 \ldots 11}_{m}$, it follows that

$$
4^{k-1} \cdot 9 \cdot \underbrace{11 \ldots 11}_{m}=n^{2}
$$

The right side of the equation is a perfect square, and $4^{k-1} \cdot 9=\left(2^{k-1} \cdot 3\right)^{2}$ is also a perfect square. Therefore, the number $\underbrace{11 \ldots 11}_{m}$ must also be a perfect square. If $m \geq 2$, the remainder of the number $\underbrace{11 \ldots 11}_{m}$ when divided by 4 is the same as the remainder of the number 11 when divided by 4, which is 3. However, a perfect square can only have a remainder of 0 or 1 when divided by 4. Therefore, $m$ must be 1, meaning $n$ is a single digit. We get the equation $4^{k-1} \cdot 9=n^{2}$. Since $n$ is a single digit, $n^{2} \leq 81$, so $4^{k-1} \leq 9$ or $k \leq 2$. The solutions are $k=1$ and $n=3$, and $k=2$ and $n=6$. In the first case, we have $\frac{1}{3}=0, \overline{3}$, and in the second case, $\frac{4}{6}=0, \overline{6}$.

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Observation that the number $11 \ldots .11$ must be a perfect square .................................... 1 point.

Proof that the number $11 \ldots 11$ is not a perfect square if it has more than one digit . . . . . . . . . . . . . 3 points.

Final handling of the remaining cases ....................................................................