99 Set $A=\left\{z \mid z^{18}=1\right\}, B=\left\{w \mid w^{48}=1\right\}$, are both sets of complex unit roots of 1. $C=$ $\{z w \mid z \in A, w \in B\}$ is also a set of complex unit roots of 1, then the set $C$ contains $\qquad$ elements.
99144 . From the problem, we have
$$
(z w)^{144}=\left(z^{18}\right)^{8} \cdot\left(w^{48}\right)^{3}=1 .
$$
$=w \cdot z, w=\mathrm{e}^{\frac{2 m \cdot 3 k}{8 d}} \in B, z \in A$. Therefore, all 144th roots of unity are in $C$.