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

Find the minimum perimeter of such a triangle.
Parse $\frac{3^{m}}{10^{4}}-\left[\frac{3^{m}}{10^{4}}\right]=\frac{3^{n}}{10^{4}}-\left[\frac{3^{n}}{10^{4}}\right]$, so $\frac{3^{m}-3^{n}}{10^{4}}=\left[\frac{3^{m}}{10^{4}}\right]-\left[\frac{3^{n}}{10^{4}}\right]=$ integer, i.e., $3^{m}-3^{n} \equiv 0\left(\bmod 10^{4}\right)$, i.e.,
$$
3^{m-n} \equiv 1\left(\bmod 10^{4}\right) .
$$
(1) is equivalent to $\left\{\begin{array}{l}3^{m-n} \equiv 1\left(\bmod 2^{4}\right), \\ 3^{m-n} \equiv 1\left(\bmod 5^{4}\right) .\end{array}\right.$

It is easy to see that $3 \equiv 3,3^{2} \equiv 9,3^{3} \equiv 11,3^{4} \equiv 1\left(\bmod 2^{4}\right)$. Let $m-n=4 q+r, 0 \leqslant r < 4$.
Therefore, the three sides of the triangle are $n, n+500 s, n+500 r$.
$$
n>(n+500 r)-(n+500 s)=500(r-s) \geqslant 500 .
$$

Thus, the minimum sides are $501,501+500,501+500 \times 2$. The minimum perimeter is
$$
501+(501+500)+(501+500 \times 2)=3 \times(500+501)=3003 .
$$

Comment: When $a$ and $m$ are coprime, there is always
$$
a^{\varphi(m)} \equiv 1(\bmod m),
$$

where $\varphi(m)$ is the Euler's totient function. For $m=5^{4}$, $\varphi(m)=5^{3} \times 4=500$.
If the smallest positive integer $k$ such that $a^{k} \equiv 1(\bmod m)$ is $\varphi(m)$, then $a$ is called a primitive root of $m$. The above derivation proves that $5^{1}$ has a primitive root, and 3 is a primitive root of $5^{4}$.