33 For any set $S$, let $|S|$ denote the number of elements in $S$, and let $n(S)$ denote the number of subsets of $S$ (including the empty set and $S$ itself). If sets $A, B, C$ satisfy: $n(A)+n(B)+n(C)=n(A \cup$ $B \cup C)$, and $|A|=|B|=100$, then the minimum possible value of $|A \cap B \cap C|$ is （ ）
A. 97
B. 96
C. 98
D. 100
33 A. Let $|C|=c,|A \cup B \cup C|=d$, then $2^{d}=2^{100}+2^{100}+2^{c}=2^{101}+2^{c}$. By the uniqueness of binary representation or by directly considering that both sides can be divided by $2^{d}(d>c, d>101)$, we easily get $c=101$, $d=102$. Therefore, $A \cup B$ must have exactly one element not in $C$. Let $x \in A, x \notin C$, then $|A \cap C|=$ $99, C$ contains exactly two elements $y, z \notin A. B$ contains at most three elements not in $A \cap C$, namely $x, y, z$. Thus, the intersection of $B$ and $A \cup C$ contains at least 97 elements. Let $C=\{1,2,3, \cdots, 101\}, A=\{3$, $4, \cdots, 102\}, B=\{1,2, \cdots, 99,102\}$, then $|A \cap B \cap C|=97$.