1. Let the set $A=\{1,2,3, \cdots, 99\}, B=\{2 x \mid x \in A\}, C=\{x \mid 2 x \in A\}$, then the number of elements in $B \cap C$ is $\qquad$ ـ.
Answer: 24.
Solution: From the given conditions, $B \cap C=\{2,4,6, \cdots, 198\} \cap\left\{\frac{1}{2}, 1, \frac{3}{2}, 2, \cdots, \frac{99}{2}\right\}=\{2,4,6, \cdots, 48\}$, thus the number of elements in $B \cap C$ is 24.