$5 \cdot 3$ If $\frac{3 n+1}{n(2 n-1)}$ can be written as a finite decimal. Find the natural number $n$.
(Kiev Mathematical Olympiad, 1983)
[Solution] Since $(n, 3 n+1)=1$.
If $3 n+1$ and $2 n-1$ are coprime, then the fraction $p=\frac{3 n+1}{n(2 n-1)}$ is in its simplest form.
If $3 n+1$ and $2 n-1$ are not coprime, let them have a common divisor $d$, and $d>1$. Set

then
$$
\begin{array}{c}
3 n+1=d a, 2 n-1=d b, \\
d(2 a-3 b)=2(3 n+1)-3(2 n-1)=5 .
\end{array}
$$

Hence, the common divisor of $3 n+1$ and $2 n-1$ is 5, and at this time, the numerator and denominator of the fraction $p$ have only the common divisor 5.

Since $p$ can be written as a finite decimal, after simplification, the denominator of $p$ has no other divisors except 2 and 5.
Therefore, $n(2 n-1)$ can only be divisible by 2 or 5, i.e.,
$$
n(2 n-1)=2^{k} \cdot 5^{m} \text {. }
$$

Since $2 n-1$ is odd, and $n$ and $2 n-1$ are coprime. Thus we have

i.e.,
$$
\left\{\begin{array}{l}
n=2^{k} \\
2 n-1=5^{m}
\end{array}\right.
$$

Since
$$
2^{k+1}=5^{m}+1 .
$$

Then
Thus, we can only have
$$
\begin{array}{c}
5^{m}+1 \equiv 2 \quad(\bmod 4) \\
2^{k+1} \equiv 2 \quad(\bmod 4) \\
k+1=1, \quad k=0
\end{array}
$$

Thus, $m=0$.
Therefore, only when $n=1$, $p$ satisfies the condition.