1. Find all prime numbers of the form $101010 \ldots 101$.
IV/1. We also observe that a number is divisible by 101 if the digit 1 appears an even number of times. $1010 \ldots 101=\sum_{k=0}^{l} 10^{2 k}$, where $l$ is equal to the number of ones minus 1. If $l$ is odd, we have the above case. Therefore, let's take $l$ to be even. Let's calculate

$$
\begin{aligned}
\sum_{k=0}^{l} 10^{2 k} & =\frac{100^{l+1}-1}{100-1}=\frac{\left(10^{l+1}-1\right)\left(10^{l+1}+1\right)}{99}= \\
& =\frac{(10-1)\left(10^{l}+\ldots 10+1\right)(10+1)\left(10^{l}-\cdots+1\right)}{99}= \\
& =\left(10^{l}+\ldots 10+1\right)\left(10^{l}-\cdots+1\right)
\end{aligned}
$$

where each factor is greater than 1, and therefore the number is not a prime. The only prime number of this form is thus 101.

Conclusion "If $l$ is odd, the sum is divisible by 101": 2 points. The only prime number is 101: 1 point. Factorization $\left(10^{l}+\ldots 10+1\right)\left(10^{l}-\cdots+1\right)$ if $l$ is even: 3 points. Conclusion "both factors are greater than 1, so we do not get a prime number": 1 point.