China Mathematical Competition (Extra Test)

Since $$\frac{3^l}{10^4}-\lfloor \frac{3^l}{10^4}\rfloor =\frac{3^m}{10^4}-\lfloor \frac{3^m}{10^4}\rfloor =\frac{3^n}{10^4}-\lfloor \frac{3^n}{10^4}\rfloor ,$$ we have $$\begin{array}{rl}{3}^l& \equiv 3^m\equiv 3^n(\mathrm{mod}10^4)\\ & ⟺\{\begin{array}{l}{3}^l\equiv 3^m\equiv 3^n(\mathrm{mod}{2}^4),\\ 3^l\equiv 3^m\equiv 3^n(\mathrm{mod}{5}^4).\end{array}\end{array}\text{\hspace{1em}(1)}$$ As $(3,2)=1$, we then have from $(1)$ $3^{l-n}\equiv 3^{m-n}\equiv 1(\mathrm{mod}{2}^4)$. Let $u$ be the minimum positive integer satisfying $3^u\equiv 1(\mathrm{mod}{2}^4)$. Then for every positive integer $v$ satisfying $3^v\equiv 1(\mathrm{mod}{2}^4)$, we must have $u\mid v$. Otherwise, if $u\nmid v$, then using division with a remainder we could get two non-negative integers $a$ and $b$ satisfying $v=au+b$ with $0<b<u$. Then $3^b\equiv 3^{au+b}\equiv 3^v\equiv 1(\mathrm{mod}{2}^4)$, contradicting the definition of $u$. Therefore $u\mid v$. Notice that $$3\equiv 3(\mathrm{mod}{2}^4),3^2\equiv 9(\mathrm{mod}{2}^4),$$ $$3^3\equiv 27\equiv 11(\mathrm{mod}{2}^4),3^4\equiv 1(\mathrm{mod}{2}^4),$$ then $u=4$. We may assume that $m-n=4k$, where $k$ is a positive integer. In the same way, we get from $(2)$ $3^{m-n}\equiv 1(\mathrm{mod}{5}^4)$, that is, $3^{4k}\equiv 1(\mathrm{mod}{5}^4)$. Now, we are going to find number $k$. As $3^{4k}-1=(1+5\times 2^4{)}^k-1=0(\mathrm{mod}{5}^4)$, i.e. $$\begin{array}{rl}& 5k\times 2^4+\frac{k(k-1)}{2}\times 5^2\times 2^8+\frac{k(k-1)(k-2)}{6}\times 5^3\times 2^{12}\\ & =5k+5^{2}k[3+(k-1)\times 2^7]+\frac{k(k-1)(k-2)}{3}\times 5^3\times 2^{11}\\ & =0(\mathrm{mod}{5}^4),\end{array}$$ so $k=5t$. Substituting in the above expression, we get $$t+5t[3+(5t-1)\times 2^7]\equiv 0(\mathrm{mod}{5}^2).$$ Then $k=5t=5^{3}s$, and $m-n=500s$, where $s$ is a positive integer. In the same way, we get $l-n=500r$, where $r$ is a positive integer and $r>s$ since $l>m>n$. So the three sides of the required triangle are $l=500r+n$, $m=500s+n$ and $n$, respectively, satisfying $n>l-n=500(r-s)$. When $s=1$, $r=2$ the perimeter reaches the minimum which equals $(1000+501)+(500+501)+501=3003$.