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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 (PIE)

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

Because we're given that $n(A)+n(B)+n(C)=n(A \cup B \cup C)$, we know that$2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$, so we can write

\[2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}\] \[\Rightarrow 2^{101}+2^{|C|}=2^{|A \cup B \cup C|}\]

Because $|C|$ and $|A \cup B \cup C|$ are integers, $|C|=101$ and $|A \cup B \cup C| = 102.$


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

Since $|A|=|B| \le |A \cup B| \le |A \cup B \cup C|$, we know $100 \le |A \cup B| \le 102$, so $98 \le |A \cap B| \le 100.$


By the Principle of Inclusion-Exclusion, $|A \cup C| = |A| + |C| - |A \cap C| = 201 - |A \cap C|.$

Since $|C| \le |A \cup C| \le |A \cup B \cup C|$, we know $101 \le |A \cup C| \le 102$, so $99 \le |A \cap C| \le 100.$


By the Principle of Inclusion-Exclusion, $|B \cup C| = |B| + |C| - |B \cap C| = 201 - |B \cap C|.$

Since $|C| \le |B \cup C| \le |A \cup B \cup C|$, we know $101 \le |B \cup C| \le 102$,so $99 \le |B \cap C| \le 100.$


By the Principle of Inclusion-Exclusion,

\begin{align} |A \cap B \cap C| &= |A \cup B \cup C| - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| \\ &= 102 - 100 - 100 - 101 + |A \cap B| + |A \cap C| + |B \cap C| \\ &= |A \cap B| + |A \cap C| + |B \cap C| - 199 \end{align}


Therefore, \[98 + 99 + 99 - 199 \le |A \cap B \cap C| \le 100+100+100-199\]

\[\Rightarrow \boxed{\textbf{97}} \le |A \cap B \cap C| \le 101.\]

~isabelchen

~formatting- growingdaisy

Solution 3

Represent the elements of $A, B, C$ as an ordered $102$-tuple of $0$'s and $1$'s. $A$ and $B$ contain exactly $100$ $1$'s, while $C$ contains $101$ $1$'s. We want to minimize the number of $3$'s after summing the numbers in the respective positions of these $102$-tuples. In the most optimal situation, all positions of the resultant tuple contains at least a $2$; this leaves $100+100+101-2\times102=97$ positions with a $3$. Thus, the minimum value is $\boxed{B: 97}$, with construction given in the above solutions.

-cretinouscretin

See also

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

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