Difference between revisions of "1991 AHSME Problems/Problem 30"

Latest revision as of 13:11, 20 June 2025

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, $|A \cup C| = |A| + |C| - |A \cap C| = 201 - |A \cap C|.$

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

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

@@ Line 19: / Line 19: @@
 Because <math>|C|</math> and <math>|A \cup B \cup C|</math> are integers, <math>|C|=101</math> and <math>|A \cup B \cup C| = 102.</math>
 By the [[Principle of Inclusion-Exclusion]], <math>|A \cup B| = |A| + |B| - |A \cap B| = 200 - |A \cap B|.</math>
 Since <math>|A|=|B| \le |A \cup B| \le |A \cup B \cup C|</math>, we know <math>100 \le |A \cup B| \le 102</math>, so <math>98 \le |A \cap B| \le 100.</math>
 By the [[Principle of Inclusion-Exclusion]], <math>|A \cup C| = |A| + |C| - |A \cap C| = 201 - |A \cap C|.</math>
 Since <math>|C| \le |A \cup C| \le |A \cup B \cup C|</math>, we know <math>101 \le |A \cup C| \le 102</math>, so <math>99 \le |A \cap C| \le 100.</math>
 By the [[Principle of Inclusion-Exclusion]], <math>|B \cup C| = |B| + |C| - |B \cap C| = 201 - |B \cap C|.</math>
 Since <math>|C| \le |B \cup C| \le |A \cup B \cup C|</math>, we know <math>101 \le |B \cup C| \le 102</math>,so <math>99 \le |B \cap C| \le 100.</math>
 By the [[Principle of Inclusion-Exclusion]],
 \begin{align}
-<cmath>|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.</cmath>
+|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,
 <cmath>98 + 99 + 99 - 199 \le |A \cap B \cap C| \le 100+100+100-199</cmath>
-<cmath>\boxed{\textbf{97}} \le |A \cap B \cap C| \le 101</cmath>
+<cmath>\Rightarrow \boxed{\textbf{97}} \le |A \cap B \cap C| \le 101.</cmath>
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]

1991 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Problem 30
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