1991 AHSME Problems/Problem 30

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 1

$n(A)=n(B)=2^{100}$ , 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 $\fbox{B=97}$

Solution 2

As $|A|=|B|=100$ , $n(A)=n(B)=2^{100}$

As $n(A)+n(B)+n(C)=n(A \cup B \cup C)$ , $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$ , $2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}$

$2^{101}+2^{|C|}=2^{|A \cup B \cup C|}$ as $|C|$ and $|A \cup B \cup C|$ are integers, $|C|$ =101 and $|A \cup B \cup C| = 102$

By Principle of Inclusion-Exclusion, $|A \cup B| = |A| + |B| - |A \cap B| = 200 - A \cap B|$

$|A| + |B| \le |A \cup B| \le |A \cup B \cup C|$ , $100 \le |A \cup B| \le 102$ , $98 \le |A \cap B| \le 100$

1991 AHSME (Problems • Answer Key • Resources)
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1991 AHSME Problems/Problem 30

Contents

Problem

Solution 1

Solution 2

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