2. Represent the fraction $\frac{93}{91}$ as a sum of two positive fractions whose denominators are 7 and 13.
Solution. The fraction $\frac{93}{91}$ needs to be represented as the sum of two fractions $\frac{a}{7}$ and $\frac{b}{13}$, that is, $\frac{93}{91}=\frac{a}{7}+\frac{b}{13}=\frac{13 a+7 b}{91}$. Therefore, $93=13 a+7 b$, or $13 a=93-7 b, a, b \in \mathbb{N}$. Clearly, $b \geq 1$, so $13 a \leq 93-7 \cdot 1=86$, from which we get $a \leq 6$. From $93=13 a+7 b$ it follows that $7 b=93-13 a$, and since 7 is a divisor of $7 b$, 7 must also be a divisor of 93-13a. By direct verification, for $a \leq 6$ only when $a=5$ the number $93-13 a=28$ is divisible by 7. Finally, for $a=5$ we have $b=4$, so the desired representation is $\frac{93}{91}=\frac{5}{7}+\frac{4}{13}$.