8. For a 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 three sets $A, B, C$ satisfy $n(A)+n(B)+n(C)=n(A \cup B \cup C), |A|=|B|=100$, find the minimum possible value of $|A \cap B \cap C|$.
8. Since $n(x)=2^{|x|}$, combining with the given conditions we have: $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$, and $|A|=|B|=100$. Therefore,
(1) If $|A \cap B|=100$, and $|A|=|B|=100,|C|=101$, and $|A \cup B \cup C|=102$, then $C$ has exactly two elements not in $A \cap B$, thus $|A \cap B \cap C|=99$;
(2) If $|A \cap B|=99$, then $|A \cup B|=101$, and by $|C|=101,|A \cup B \cup C|=102$, then $C$ has exactly one element not in $A \cup B$. To minimize $|A \cap B \cap C|$, then $|C \cap[A \backslash(A \cap B)]| \geqslant 1$, and $|C \cap[B \backslash(A \cap B)]| \geqslant 1$, at this time $|A \cap B \cap C| \geqslant 98$;
(3) If $|A \cap B|=98$, to minimize $|A \cap B \cap C|$, it should be that $|C \cap(A \backslash(A \cap B))|=2, \mid C \cap (B \backslash(A \cap B)) \mid=2$, and since $|A \cup B \cup C|=102$, and $|C|=101$, thus $|A \cap B \cap C| \geqslant 97$;
(4) If $|A \cap B| \leqslant 97$, then $|A \backslash(A \cap B)| \geqslant 3,|B \backslash(A \cap B)| \geqslant 3$, thus $|A \cup B| \geqslant 100+3=103$, which is a contradiction.
Combining (1)(2)(3)(4), we know that the minimum possible value of $|A \cap B \cap C|$ is 97.