3. Find the number of ordered triples of natural numbers $(a, b, c)$ such that $\mathrm{LCM}(a, b)=1000, \mathrm{LCM}(b, c)=2000$ and $\mathrm{LCM}(c, a)=2000$.
Solution. Since the numbers 1000 and 2000 are of the form $2^{m}5^{n}$ with $m, n$ being natural numbers, we get that the numbers $a, b, c$ are also of the form $2^{m}5^{n}$. Let $a=2^{m_{1}}5^{n_{1}}, b=2^{m_{2}}5^{n_{2}}$, and $c=2^{m_{3}}5^{n_{3}}$ for $m_{i}, n_{i}, i=1,2,3$ non-negative integers. From the condition of the problem, we have:

$$
\begin{aligned}
& \max \left\{m_{1}, m_{2}\right\}=3, \max \left\{m_{2}, m_{3}\right\}=4, \max \left\{m_{3}, m_{1}\right\}=4 \\
& \max \left\{n_{1}, n_{2}\right\}=3, \max \left\{n_{2}, n_{3}\right\}=3, \max \left\{n_{3}, n_{1}\right\}=3
\end{aligned}
$$

From (1) we get that $m_{3}=4, m_{1}=3$ or $m_{2}=3$, if one is 3 while the other can be $0,1,2,3$. There are 7 such ordered triples

$$
(0,3,4),(1,3,4),(2,3,4),(3,0,4),(3,1,4),(3,2,4),(3,3,4)
$$

From (2) we get that two of $n_{1}, n_{2}, n_{3}$ are 3 while the other can be $0,1,2,3$. The number of such ordered triples is 10. These are

$$
(3,3,0),(3,3,1),(3,3,2),(3,0,3),(3,1,3),(3,2,3),(0,3,3),(1,3,3),(2,3,3),(3,3,3)
$$

Since the choice of triples $\left(m_{1}, m_{2}, m_{3}\right),\left(n_{1}, n_{2}, n_{3}\right)$ is independent, we get that the number of triples $(a, b, c)$ is $7 \cdot 10=70$.