Example 4 (Problem from the 5th American Invitational Mathematics Examination) Let $[r, s]$ denote the least common multiple of the positive integers $r, s$. Find the number of ordered triples $(a, b, c)$ such that $[a, b]=1000,[b, c]=2000,[c, a]=2000$.


Solution: Clearly, $a, b, c$ are all numbers of the form $2^{m} \cdot 5^{n}$. Let $a=2^{m_{1}} \cdot 5^{n_{1}}, b=2^{m_{2}} \cdot 5^{n_{2}}, c=2^{m_{3}} \cdot 5^{n_{3}}$. From $[a, b]=1000=2^{3} \cdot 5^{3}$, we know $\max \left\{m_{1}, m_{2}\right\}=3, \max \left\{n_{1}, n_{2}\right\}=3$. Similarly, $\max \left\{m_{2}, m_{3}\right\}=$ $4, \max \left\{n_{2}, n_{3}\right\}=3, \max \left\{m_{1}, m_{3}\right\}=4, \max \left\{n_{1}, n_{3}\right\}=3$.

Therefore, $m_{3}$ should be 4, and $m_{1}, m_{2}$ must include one 3, and the other can be any of $0,1,2$ or 3, so there are $C_{4}^{1}+C_{4}^{1}-1=7$ ways to choose $m_{1}, m_{2}$. Also, $n_{1}, n_{2}, n_{3}$ must include two 3s, and the other can be $0,1,2$ or 3, so there are $C_{3}^{2}+4=10$ ways to choose $n_{1}, n_{2}, n_{3}$. Therefore, the different ways to choose $m_{i}, n_{i}(i=1,2,3)$ are $7 \times 10=70$, meaning there are 70 triples.