$5 \cdot 2$ If $n$ is a natural number less than 50, find all values of $n$ such that the values of the algebraic expressions $4n+5$ and $7n+6$ have a common divisor greater than 1.
(Tianjin, China Junior High School Math Competition, 1991)
[Solution] Let $(4 n+5,7 n+6)=d>1$, then
$$
d|4 n+5, \quad d| 7 n+6,
$$

Thus,
$$
\begin{array}{c}
d \mid(7 n+6)-(4 n+5)=3 n+1, \\
d \mid(4 n+5)-(3 n+1)=n+4, \\
d \mid(3 n+1)-2(n+4)=n-7 \\
d \mid(n+4)-(n-7)=11 .
\end{array}
$$

Since 11 is a prime number, then $d=11$.
Let $n-7=11 k$, then

Solving for
$$
\begin{aligned}
0<n & =11 k+7<50 \\
k & =0,1,2,3 .
\end{aligned}
$$

Thus,
$$
n=7,18,29,40
$$