# 1991 AHSME Problems/Problem 30

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## Problem

For any set $S$, let $|S|$ denote the number of elements in $S$, and let $n(S)$ be the number of subsets of $S$, including the empty set and the set $S$ itself. If $A$, $B$, and $C$ are sets for which $n(A)+n(B)+n(C)=n(A\cup B\cup C)$ and $|A|=|B|=100$, then what is the minimum possible value of $|A\cap B\cap C|$?

$(A) 96 \ (B) 97 \ (C) 98 \ (D) 99 \ (E) 100$

## Solution

$n(A)=n(B)=2^{100}$, so $n(C)$ and $n(A \cup B \cup C)$ are integral powers of $2$ $\Longrightarrow$ $n(C)=2^{101}$ and $n(A \cup B \cup C)=2^{102}$. Let $A=\{s_1,s_2,s_3,...,s_{100}\}$, $B=\{s_3,s_4,s_5,...,s_{102}\}$, and $C=\{s_1,s_2,s_3,...,s_{k-2},s_{k-1},s_{k+1},s_{k+2},...,s_{100},s_{101},s_{102}\}$ where $s_k \in A \cap B$ Thus, the minimum value of $|A\cap B \cap C|$ is $\fbox{B=97}$