3. Among the subsets of the set $\{1,2, \cdots, 100\}$, some subsets satisfy the condition: no number is twice another number. The maximum number of elements in such a subset is ( ).
(A) 67
(B) 68
(C) 69
(D) 70
3. A.

Let $A_{1}=\{51,52, \cdots, 100\}, A_{2}=\{26,27, \cdots, 50\}$, $A_{3}=\{13,14, \cdots, 25\}, A_{4}=\{7,8, \cdots, 12\}$,
$$
A_{5}=\{4,5,6\}, A_{6}=\{2,3\}, A_{7}=\{1\} \text {. }
$$

Then $A_{1} \cup A_{3} \cup A_{5} \cup A_{7}$ contains $50+13+3+1=67$ elements, and none of these elements is twice another.

If $B \subseteq\{1,2, \cdots, 100\}$, where each element is not twice another, when $a \in B \cap A_{2}$, $2 a \notin B$. Therefore,
$$
\left|B \cap A_{2}\right|+\left|B \cap A_{1}\right| \leqslant 50 .
$$

Similarly, $\left|B \cap A_{4}\right|+\left|B \cap A_{3}\right| \leqslant 13$,
$$
\left|B \cap A_{6}\right|+\left|B \cap A_{5}\right| \leqslant 3,\left|B \cap A_{7}\right| \leqslant 1 .
$$

Thus, $|B| \leqslant 67$.