15. Let 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.
15. Given that $3^{l}=3^{m}=3^{n}\left(\bmod 10^{4}\right)$, first propose a lemma;

Let $a \geqslant 2, (a, b)=1$, then (1) there exists a positive integer $d \leqslant a-1$ such that $a \mid b^{d}-1$; (2) Let $d_{0}$ be the smallest positive integer $d$ satisfying (1), then $a\left|b^{h}-1(h \geqslant 1) \Leftrightarrow d_{0}\right| h$.
Its proof can be found in "A Concise Introduction to Number Theory" on page 34, Example 11. (P. C. Pan and C. B. Pan, Peking University Press)
Let $3^{k} \equiv 1(\bmod 1000)$. Since $3^{4} \equiv 1(\bmod 10)$. By the lemma, $4 \mid k$, and $3^{4} \equiv 81(\bmod 100), 3^{8} \equiv 61(\bmod 100), 3^{12} \equiv 41(\bmod 100), 3^{16} \equiv 21(\bmod 100), 3^{20} \equiv 1(\bmod 100)$.

Thus, $k=20n, n \in \mathbf{N}^{*}$. Since $3^{20} \equiv 401(\bmod 1000), 3^{40} \equiv 801(\bmod 1000), 3^{60} \equiv 201(\bmod 1000), 3^{80} \equiv 601(\bmod 1000), 3^{100} \equiv 1(\bmod 1000), 3^{20} \equiv 4401\left(\bmod 10^{4}\right), 3^{40} \equiv 8801\left(\bmod 10^{4}\right), 3^{60} \equiv 3201\left(\bmod 10^{4}\right), 3^{80} \equiv 7601\left(\bmod 10^{4}\right), 3^{100} \equiv 2001\left(\bmod 10^{4}\right), 3^{200} \equiv 4001\left(\bmod 10^{4}\right), 3^{300} \equiv 6001\left(\bmod 10^{4}\right), 3^{400} \equiv 8001\left(\bmod 10^{4}\right), 3^{500} \equiv 1\left(\bmod 10^{4}\right)$, so $m \equiv n \equiv 1(\bmod 500)$. Therefore, $n>l-m \geqslant 500, n \geqslant 501$, $m \geqslant n+500 \geqslant 1001, l \geqslant m+500 \geqslant 1501$, so $l+m+n \geqslant 501+1001+1501=3003$. When $l=1501$, $m=1001$, $n=501$, a triangle can be formed, i.e., $C_{\text {min }}=3003$.