Example 2 Let the positive integer $n>4$ satisfy that the decimal representation of $n!$ ends with exactly $k$ zeros, and the decimal representation of $(2n)!$ ends with exactly $3k$ zeros. Find such an $n$.
(Adapted from the 2015 National High School Mathematics Joint Competition Additional Problem)
【Analysis】This problem is a typical application of Lucas' Theorem, the essence of which is to consider the number of factor 5s in two factorials.

Solution: Since the product of one factor 5 and one factor 2 produces a 0, and the number of factor 2s is greater than the number of factor 5s, we only need to consider the multiple relationship of the number of factor 5s in the two factorials.
By Lucas' Theorem, we have
$$
\begin{array}{l}
v_{5}(n)=\frac{n-S_{5}(n)}{4}=k, \\
v_{5}(2 n)=\frac{2 n-S_{5}(2 n)}{4}=3 k,
\end{array}
$$

which means $3 \times \frac{n-S_{5}(n)}{4}=\frac{2 n-S_{5}(2 n)}{4}$.
Let $n$ be a $r$-digit number in base 5, and the number of carries when $n+n$ is $t(t \leqslant r)$. Then
$$
S_{5}(2 n)=2 S_{5}(n)-4 t \text {. }
$$

Substituting into equation (1) gives
$$
n=S_{5}(n)+4 t \Rightarrow 4^{r-1} \leqslant 8 r \Rightarrow r \leqslant 3 \text {. }
$$

Since $n>4$, we have $r>1$.
When $r=2$, let
$$
n=5 a+b(1 \leqslant a \leqslant 4,0 \leqslant b \leqslant 4) \text {. }
$$

Then $5 a+b=a+b+4 t$
$\Rightarrow a=t=1$ or 2
$$
\begin{array}{l}
\Rightarrow n=\overline{(13)_{5}}, \overline{(14)_{5}}, \overline{(23)_{5}}, \overline{(24)_{5}} \\
\Rightarrow n=8,9,13,14 .
\end{array}
$$

When $r=3$, let
$$
n=25 a+5 b+c(1 \leqslant a \leqslant 4,0 \leqslant b 、 c \leqslant 4) \text {. }
$$

Then $25 a+5 b+c=a+b+c+4 t$, which has no solution.