The second question: Let the three sides of a triangle be integers $l$, $m$, and $n$, and $l>m>n$. It is known that $\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=$ $\left\{\frac{3^{n}}{10^{4}}\right\}$, where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the minimum perimeter of such a triangle.
Proof: Since $l, m, n$ are the lengths of the sides of a triangle, hence $m+n>l$.
From the given information,
$$
3^{l-n} \equiv 3^{m-n} \equiv 3^{l-m} \equiv 1\left(\bmod 10^{4}\right).
$$

Without loss of generality, let $\left\{\begin{array}{l}l-m=s, \\ m-n=t\end{array}\left(s, t \in \mathbf{N}_{+}\right)\right.$, then $3^{s} \equiv 3^{t} \equiv 1\left(\bmod 10^{4}\right)$.
The perimeter of the triangle is
$$
c=n+n+t+n+s+t=3 n+2 t+s.
$$

Let the smallest positive integer $x$ that satisfies $3^{x} \equiv 1\left(\bmod 10^{4}\right)$ be $k$, so $3^{k} \equiv 1\left(\bmod 10^{4}\right)$. Therefore, $s \geqslant k, t \geqslant k$.
From (1) we get $n+n+t>n+s+t \Rightarrow n>s$.
Thus, $n \geqslant s+1 \geqslant k+1$,
$$
c \geqslant 3(k+1)+2 k+k=6 k+3.
$$

Now we analyze the minimum value of $k$.
According to Euler's theorem, since $\left(3,10^{4}\right)=1$, we have
$$
3^{\varphi\left(10^{4}\right)} \equiv 1\left(\bmod 10^{4}\right).
$$

Here, $\varphi\left(10^{4}\right)=10^{4}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=4000$.
Thus, $k \mid 4000$.
It is easy to see that the last digit of $3^{k}$ cycles through 3, 9, 7, 1. Since $3^{k} \equiv 1\left(\bmod 10^{4}\right)$, hence $4 \mid k$.
Let $k=4 h\left(h \in \mathbf{N}_{+}\right)$, then
$$
\begin{aligned}
3^{k} & =9^{2 h}=(10-1)^{2 h} \\
= & 10^{2 h}-C_{2 h}^{2 h-1} 10^{2 h-1}+\cdots+C_{2 h}^{4} 10^{4}- \\
& C_{2 h}^{3} 10^{3}+C_{2 h}^{2} 10^{2}-C_{2 h}^{1} 10+1.
\end{aligned}
$$

Therefore, $3^{k} \equiv 1\left(\bmod 10^{4}\right)$
$$
\begin{array}{l}
\Leftrightarrow 3^{k} \equiv-\mathrm{C}_{2 h}^{3} 10^{3}+\mathrm{C}_{2 h}^{2} 10^{2}-\mathrm{C}_{2 h}^{1} 10+1 \\
\equiv 1\left(\bmod 10^{4}\right) \\
\Leftrightarrow-\mathrm{C}_{2 h}^{3} 10^{3}+\mathrm{C}_{2 h}^{2} 10^{2}-\mathrm{C}_{2 h}^{1} 10 \equiv 0\left(\bmod 10^{4}\right) \\
\Leftrightarrow \frac{2 h(2 h-1)}{2} \times 100 \\
\equiv 2 h \cdot 10+\frac{2 h(2 h-1)(2 h-2)}{3 \times 2} \times \\
1000\left(\bmod 10^{4}\right) \text {. } \\
\Leftrightarrow 20 h[50(2 h-1)(2 h-2)+3-15 h(2 h-1)] \\
\equiv 0\left(\bmod 10^{4}\right) \\
\Leftrightarrow 20 h\left(170 h^{2}-285 h+103\right) \equiv 0\left(\bmod 10^{4}\right) \\
\Leftrightarrow h\left(170 h^{2}-285 h+103\right) \equiv 0(\bmod 500) \text {. } \\
\end{array}
$$

Since 5 does not divide $170 h^{2}-285 h+103$, we have $125 \mid h$, i.e., $h=125 p, p \in \mathbf{N}_{+}$, and $p \equiv 1(\bmod 4)$ or $p \equiv 0(\bmod 4)$.
When $p=1$, $h_{\min }=125$. Therefore,
$$
k_{\text {min }}=4 h_{\text {min }}=500.
$$

Thus, the minimum perimeter $c_{\min }=3003$. At this time,
$$
\begin{array}{l}
\left\{\begin{array}{l}
l-m=s=k=500, \\
m-n=t=k=500, \\
n=k+1=501
\end{array}\right. \\
\Rightarrow l=1501, m=1001, n=501.
\end{array}
$$
(Dang Zhongliang, Chongqing Yucai High School, 400050)