3. Find all prime numbers of the form 1010101...0101 (i.e., whose decimal representation consists of the digit 1 followed by a block "01" repeated an arbitrary number of times).
3. The numbers in question are of the form $\frac{10^{2 k}-1}{99}$ for some natural number $k$. For $k=1$, we get the number 1, which is not a prime number. For $k=2$, we get the number 101, which is a prime number. For $k \geqslant 3$, we have

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

One of the two factors in the numerator must be divisible by 11 (since the observed number is a natural number). If $11 \mid 10^{k}+1$, then $\frac{10^{k}-1}{9}$ is a factor of the observed number (greater than 1 and different from the number itself), so the observed number is composite; if $11 \left\lvert\, \frac{10^{k}-1}{9}\right.$, then $\frac{10^{k}-1}{99}$ is a factor of the observed number (greater than 1 and different from the number itself), so the observed number is again composite.

Thus, the only solution is the number 101.