1A. The numbers $m$ and $n$ are coprime. The fraction $\frac{3 n-m}{5 n+2 m}$ can be reduced by some natural number.

Determine the number by which it can be reduced.
Solution. Let's assume that $k, k>1$ is the number by which the fraction can be reduced. According to this, there exist natural numbers $p$ and $s$, such that $(p, s)=1$ and $3 n-m=k p, 5 n+2 m=k s$. If we solve the system

$$
\left\{\begin{array}{l}
3 n-m=k p \\
5 n+2 m=k s
\end{array}\right.
$$

for $n$ and $m$, we will get $n=\frac{k(2 p+s)}{11}$ and $m=\frac{k(3 s-5 p)}{11}$. The numbers $m$ and $n$ are coprime, so according to this $k=11$. Indeed, if we assume that $k \neq 11$, then for any divisor $d$ of $k$ greater than 1 and different from 11 we have $(m, n) \geq d>1$. But this contradicts the assumption $(m, n)=1$. Therefore, $k=11$.