2. Let $n(A)$ denote the number of subsets of $A$. Given $|A|=|B|=100, n(A)+$ $n(B)+n(C)=n(A \cup B \cup C)$. Find the minimum value of $|A \cap B \cap C|$.
2. Let $|C|=c,|A \cup B \cup C|=d$, then $2^{200}+2^{100}+2^{c}=2^{d}$, i.e., $2^{101}+2^{c}=2^{d}$. Clearly, $d>c$ and 101, so $2^{101}|2^{c}, 2^{c}|2^{101}$, thus $c=101$, $d=$ 102. $A \cap B$ has at least $100+100-102=98$ elements, of which at most $102-$ $101=1$ element does not belong to $C$. The minimum value sought is $98-1=97$.