15.16 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 $n(A)+n(B)+n(C)=n(A \cup B \cup C)$, and $|A|=|B|=100$, then the minimum possible value of $|A \cap B \cap C|$ is
(A) 96.
(B) 97.
(C) 98.
(D) 99.
(E) 100.
(42nd American High School Mathematics Examination, 1991)
［Solution］The number of subsets of set $X$ is $n(X)=2^{|X|}$. Given $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$. Also, $|A|=|B|=100$, thus $2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}$, which means $2^{101}+2^{|C|}=2^{|A \cup B \cup C|}$. Therefore, $|C|=101$, and thus $|A \cup B \cup C|=102$.
Now, consider the following cases:
(1) If $|A \cap B|=100$, and $|C|=101$, and $|A \cup B \cup C|=102$, then $C$ must have at least two elements different from $A \cap B$, so $|A \cap B \cap C|=99$.
(2) If $|A \cap B|=99$, then $|A \cup B|=101$.
Given $|C|=101$ and $|A \cup B \cup C|=102$, then $C$ can have at most one element different from $A \cup B$.
To minimize $|A \cap B \cap C|$, then
$$
C \cap(A \backslash A \cap B) \neq \varnothing \text {, and } \quad C \cap(B \backslash A \cap B) \neq \varnothing \text {. }
$$

In this case,
$$
|A \cap B \cap C| \geqslant 98 \text {. }
$$
(3) If $|A \cap B|=98$, to minimize $|A \cap B \cap C|$, then
$$
|C \cap(A \backslash A \cap B)|=2, \quad|C \cap(B \backslash A \cap B)|=2,
$$

Given $|A \cup B \cup C|=102$ and $|C|=101$,
then
$$
|A \cap B \cap C| \geqslant 97 \text {. }
$$
(4) If $|A \cap B| \leqslant 97$, to minimize $|A \cap B \cap C|$, then $|A \backslash A \cap B| \geqslant 3$, and $|B \backslash A \cap B| \geqslant 3$.

But this would mean $|A \cup B| \geqslant 97+3+3=103$, which contradicts the given conditions. Therefore, the minimum value of $|A \cap B \cap C|$ is 97.
Hence, the answer is $(B)$.