15 Positive integers $a, b, c$ satisfy: $[a, b]=1000,[b, c]=2000,[c, a]=2000$. Find the number of such ordered positive integer triples $(a, b, c)$.
15. From the conditions, we can set
$$a=2^{\alpha_{1}} \cdot 5^{\beta_{1}}, b=2^{\alpha_{2}} \cdot 5^{\beta_{2}}, c=2^{\alpha_{3}} \cdot 5^{\beta_{3}}$$

Then
$$\begin{array}{l}
\max \left\{\alpha_{1}, \alpha_{2}\right\}=3, \max \left\{\alpha_{2}, \alpha_{3}\right\}=4, \max \left\{\alpha_{3}, \alpha_{1}\right\}=4 \\
\max \left\{\beta_{1}, \beta_{2}\right\}=\max \left\{\beta_{2}, \beta_{3}\right\}=\max \left\{\beta_{3}, \beta_{1}\right\}=3
\end{array}$$

Note that, the array $(a, b, c)$ is determined only when the non-negative integer arrays $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ and $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ are both determined. From the conditions above, we know that $\alpha_{3}=4$, and at least one of $\alpha_{1}$ or $\alpha_{2}$ is 3, indicating that $\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ has 7 possible combinations; at least two of $\beta_{1}$, $\beta_{2}$, and $\beta_{3}$ are equal to 3, so $\left(\beta_{1}, \beta_{2}, \beta_{3}\right)$ has 10 possible combinations.

In summary, the number of ordered arrays $(a, b, c)$ that satisfy the conditions is $7 \times 10 = 70$ (groups).