Let the lengths of the three sides of a triangle be integers $l$, $m$, $n$, and $l > m > n$. It is known that
$$
\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\},
$$

where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the minimum perimeter of such a triangle.
1 Another Solution to the Test Question
Solution: From the given, we have
$$
\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 10^{4}\right) . \\
\text { Equation (1) } \Leftrightarrow\left\{\begin{array}{l}
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 2^{4}\right), \\
3^{l} \equiv 3^{m} \equiv 3^{n}\left(\bmod 5^{4}\right)
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
3^{l-n} \equiv 3^{m-n} \equiv 1\left(\bmod 2^{4}\right), \\
3^{l-n} \equiv 3^{m-n} \equiv 1\left(\bmod 5^{4}\right) .
\end{array}\right.
\end{array}
$$

Since $\left(3,2^{4}\right)=\left(3,5^{4}\right)=1$, according to Euler's theorem, we have $3^{\varphi\left(2^{4}\right)} \equiv 1\left(\bmod 2^{4}\right), 3^{\varphi\left(5^{4}\right)} \equiv 1\left(\bmod 5^{4}\right)$, where
$$
\varphi\left(2^{4}\right)=2^{4}\left(1-\frac{1}{2}\right)=8, \varphi\left(5^{4}\right)=5^{4}\left(1-\frac{1}{5}\right)=500 \text {. }
$$

Let $k_{1}$ and $k_{2}$ be the smallest positive integers that make $3^{k} \equiv 1\left(\bmod 2^{4}\right)$ and $3^{k} \equiv 1\left(\bmod 5^{4}\right)$, respectively. Then $k_{1}\left|8, k_{2}\right| 500$. It is easy to verify that $k_{1}=4$. Now we verify that $k_{2}=500$.
The possible values of $k_{2}$ are
$500,250,125,100,50,25,20,10,5,4,2,1$.
Since $3^{k_{2}} \equiv 1\left(\bmod 5^{4}\right)$, then $3^{k_{2}} \equiv 1\left(\bmod 5^{2}\right)$.
Because 20 is the smallest positive integer that makes $3^{k} \equiv 1\left(\bmod 5^{2}\right)$, we have $20 \mid k_{2}$. Therefore, $k_{2}$ can only take the values 500, 100, 20. Since $3^{100} \equiv 126\left(\bmod 5^{4}\right)$, $k_{2}$ cannot be 100 or 20. Thus, $\left[k_{1}, k_{2}\right]=500$.
From the above discussion, we know that $500 \mid (l-n)$ and $500 \mid (m-n)$.
Let $l-n=500 u, m-n=500 v$, and $u>v$. Then the lengths of the three sides of the triangle are $500 u+n, 500 v+n$, and
$$
500 u+n<(500 v+n)+n .
$$

Therefore, $500(u-v)<n$.
To minimize the perimeter of the triangle, take $u=2, v=1$, and $n=501$. At this time, the perimeter of the triangle is 3003.