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

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

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1991 AHSME Problems/Problem 30

Problem

Solution

See also