11. (10 points) Let $n$ be a natural number less than 50. Find all $n$ such that $3n+5$ and $5n+4$ have a common divisor greater than 1.
【Analysis】For natural numbers less than 50, let $d$ be a common divisor greater than 1 of $3 n+5$ and $5 n+4$, then $d$ divides $5(3 n+5)$ $-3(5 n+4)=13$, so $d=13$, and thus $\left\{\begin{array}{l}3 \mathrm{n}+5=13 \mathrm{k} \\ 5 n+4=131\end{array}\right.$, from $3 n+5=13 k$, we get $k=3(n-4 k)$ $+5=3 s+2$, where $s=n-4 k+1$, so $n=4 k+s-1=4(3 s+2)+s-1=13 s+7$, then find all $n$.

【Solution】For natural numbers less than 50,
let $d$ be a common divisor greater than 1 of $3 n+5$ and $5 n+4$,
then $d$ divides $5(3 n+5)-3(5 n+4)=13$,
so $d=13$, and thus
$$
\left\{\begin{array}{l}
3 n+5=13 k \\
5 n+4=13 m
\end{array}\right.
$$

From $3 n+5=13 k$,
we get $k=3(n-4 k)+5=3 s+2$, where $s=n-4 k+1$,
so $n=4 k+s-1=4(3 s+2)+s-1$,
$$
n=13 s+7 \text {. }
$$

Since $0 \leqslant n \leqslant 50$,
we have $0 \leqslant s \leqslant 3$, the corresponding $n$ are $7,20,33,46$.
Answer: The $n$ for which $3 n+5$ and $5 n+4$ have a common divisor greater than 1 are $7,20,33,46$.