1. Determine all prime numbers of the form $\frac{11 \ldots 1}{11}$, where $n$ is a natural number.
Solution. We use the equality: $11 \ldots 1=\frac{k}{9}$. 10 . We get

$$
\frac{11 \ldots 1}{11}=\frac{10^{2 n}-1}{11 \cdot 9}=\frac{\left(10^{n}-1\right)\left(10^{n}+1\right)}{11 \cdot 9}
$$

$2 n$

For $n=1, \frac{11 \ldots 1}{11}=\frac{11}{11}=1$ which is not a prime number. For $n=2$,

$$
\frac{11 \ldots 1}{11}=\frac{\left(10^{2}-1\right)\left(10^{2}+1\right)}{11 \cdot 9}=\frac{101 \cdot 99}{99}=101
$$

We will show that for $n>2$, the number $\frac{11 \ldots 1}{11}$ is composite. We consider two cases: when $n$ is even and when $n$ is odd.

Case 1. Let $n$ be even. Then $10^{2 k}$ has a remainder of 1 when divided by 9 and 11. Since $\operatorname{GCD}(9,11)=1$, we get that $10^{n}-1$ is divisible by 99 and the number

$$
\frac{11 \ldots 1}{11}=\frac{\left(10^{n}-1\right)\left(10^{n}+1\right)}{11 \cdot 9}=\frac{10^{n}-1}{99}\left(10^{n}+1\right)
$$

is the product of two natural numbers greater than 1.

Case 2. Let $n$ be odd. Then $10^{2 k+1}$ has a remainder of -1 when divided by 11. The number $10^{2 k+1}+1$ is divisible by 11, and $10^{2 k+1}-1$ is divisible by 9. The number

$$
\frac{11 \ldots 1}{11}=\frac{\left(10^{n}-1\right)\left(10^{n}+1\right)}{11 \cdot 9}=\frac{10^{2 k+1}-1}{9} \cdot \frac{10^{2 k+1}+1}{11}
$$

is the product of two natural numbers greater than 1, so the number is composite.