479*. For what least natural $n$ are each of the fractions

irreducible

$$
\frac{5}{n+6}, \quad \frac{6}{n+7}, \frac{7}{n+8}, \ldots, \quad \frac{24}{n+25}
$$
$\triangle$ Transform the given fractions as follows:

$$
\frac{5}{n+6}=\frac{5}{(n+1)+5}, \quad \frac{6}{n+7}=\frac{6}{(n+1)+6}, \ldots, \frac{24}{n+25}=\frac{24}{(n+1)+24}
$$

From this, it is clear that each of these fractions is irreducible if and only if the number $n+1$ is coprime with each of the numbers $5,6,7, \ldots, 24$. Then $n+1$ is also coprime with the numbers 25 (otherwise, the fractions with numerators $5,10,15,20$ are reducible by 5), 26, 27, and 28. Thus, none of the values of $n+1$ from 2 to 28 are suitable. However, the value $n+1=29$ is suitable, as the number 29 is coprime with each of the natural numbers from 5 to 24. Therefore, the smallest value of $n$ is 28.

Answer: 28.