3. Determine the smallest natural number $n$ for which each of the fractions

$$
\frac{7}{n+9}, \frac{8}{n+10}, \frac{9}{n+11}, \ldots, \frac{30}{n+32}, \frac{31}{n+33}
$$

is irreducible.
## Solution.

The fractions are irreducible exactly when their reciprocal fractions are irreducible

$$
\frac{n+9}{7}, \frac{n+10}{8}, \frac{n+11}{9}, \ldots, \frac{n+32}{30}, \frac{n+33}{31}
$$

or (reduced by 1)

$$
\frac{n+2}{7}, \frac{n+2}{8}, \frac{n+2}{9}, \ldots, \frac{n+2}{30}, \frac{n+2}{31}
$$

All these fractions are irreducible exactly when $n+2$ is relatively prime to all numbers $7,8,9, \ldots, 30,31$.

Obviously, $n+2$ is not a number less than 32, nor $32,33,34,35$ or 36, as these numbers have common divisors with some of the numbers $7,8,9, \ldots, 30,31$. Therefore, the smallest such $n+2$ is the number 37, i.e., $n=35$.