12. Use the properties of the greatest common divisor from the previous question to find the greatest common divisor of the following:
(i) $216,64,1000$.
(ii) $24000,36000,144000$.
12.
(i) Solution: The property of the greatest common divisor (GCD) from the previous problem can obviously be extended to more than two numbers, so
$$\begin{array}{c}
(216,64,1000)=\left(6^{3}, 4^{3}, 10^{3}\right) \\
=(6,4,10)^{3}=2^{3}=8
\end{array}$$
(ii) Solution:
$$\begin{array}{l}
(24000,36000,144000) \\
=1000 \times(24,36,144) \\
=1000 \times 12 \times(2,3,12) \\
=1000 \times 12=12000
\end{array}$$