Problem 2. Determine all pairs $(a, b)$ of natural numbers for which the fractions

$$
\frac{3 a+8 b+2}{10 a+2 b+1} \quad \text { and } \quad \frac{8 a+b+3}{2 a+7 b+3}
$$

represent, simultaneously, natural numbers.
Solution. For both fractions to be natural numbers simultaneously, it is necessary that $3a + 8b + 2 \geq 10a + 2b + 1$ and $8a + b + 3 \geq 2a + 7b + 3$, from which, by adding, $11a + 9b + 5 \geq 12a + 9b + 4$, which implies $a \in \{0, 1\}$.

For $a=0$ the second fraction is $\frac{b+3}{7b+3}$; for it to be a natural number, it is necessary that $b \geq 7b$, so $b=0$, in which case the first fraction is also a natural number................... $\mathbf{2 p}$

For $a=1$ the second fraction is $\frac{b+11}{7b+5}$; for it to be a natural number, it is necessary that $b+11 \geq 7b+5$, so $b \in \{0, 1\}$ and only $b=1$ works, in which case the first fraction is also a natural number.

Therefore, the pairs sought are $a=0, b=0$ and $a=1, b=1$