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

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For any set <math>S</math>, let <math>|S|</math> denote the number of elements in <math>S</math>, and let <math>n(S)</math> be the number of subsets of <math>S</math>, including the empty set and the set <math>S</math> itself. If <math>A</math>, <math>B</math>, and <math>C</math> are sets for which <math>n(A)+n(B)+n(C)=n(A\cup B\cup C)</math> and <math>|A|=|B|=100</math>, then what is the minimum possible value of <math>|A\cap B\cap C|</math>?
 
For any set <math>S</math>, let <math>|S|</math> denote the number of elements in <math>S</math>, and let <math>n(S)</math> be the number of subsets of <math>S</math>, including the empty set and the set <math>S</math> itself. If <math>A</math>, <math>B</math>, and <math>C</math> are sets for which <math>n(A)+n(B)+n(C)=n(A\cup B\cup C)</math> and <math>|A|=|B|=100</math>, then what is the minimum possible value of <math>|A\cap B\cap C|</math>?
  
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(A) 96 (B) 97 (C) 98 (D) 99 (E) 100
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<math>(A) 96 \ (B) 97 \ (C) 98 \ (D) 99 \ (E) 100</math>
  
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== Solution ==
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== Solution 1==
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<math>\fbox{}</math>
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<math>n(A)=n(B)=2^{100}</math>, so  <math>n(C)</math> and <math>n(A \cup B \cup C)</math> are integral powers of <math>2</math> <math>\Longrightarrow</math> <math>n(C)=2^{101}</math> and <math>n(A \cup B \cup C)=2^{102}</math>. Let <math>A=\{s_1,s_2,s_3,...,s_{100}\}</math>, <math>B=\{s_3,s_4,s_5,...,s_{102}\}</math>, and <math>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}\}</math> where <math>s_k \in A \cap B</math>
 +
Thus, the minimum value of <math>|A\cap B \cap C|</math> is <math>\fbox{B=97}</math>
 +
 
 +
==Solution 2 (PIE) ==
 +
 
 +
As <math>|A|=|B|=100</math>, <math>n(A)=n(B)=2^{100}.</math>
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 +
Because we're given that <math>n(A)+n(B)+n(C)=n(A \cup B \cup C)</math>, we know that<math>2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}</math>, so we can write
 +
 
 +
<cmath>2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}</cmath>
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<cmath>\Rightarrow 2^{101}+2^{|C|}=2^{|A \cup B \cup C|}</cmath>
 +
 
 +
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}
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|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| \\
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&= |A \cap B| + |A \cap C| + |B \cap C| - 199
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\end{align}
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 +
 
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Therefore,
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<cmath>98 + 99 + 99 - 199 \le |A \cap B \cap C| \le 100+100+100-199</cmath>
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 +
<cmath>\Rightarrow \boxed{\textbf{97}} \le |A \cap B \cap C| \le 101.</cmath>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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 +
~formatting- growingdaisy
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 +
==Solution 3==
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Represent the elements of <math>A, B, C</math> as an ordered <math>102</math>-tuple of <math>0</math>'s and <math>1</math>'s. <math>A</math> and <math>B</math> contain exactly <math>100</math> <math>1</math>'s, while <math>C</math> contains <math>101</math> <math>1</math>'s. We want to minimize the number of <math>3</math>'s after summing the numbers in the respective positions of these <math>102</math>-tuples. In the most optimal situation, all positions of the resultant tuple contains at least a <math>2</math>; this leaves <math>100+100+101-2\times102=97</math> positions with a <math>3</math>. Thus, the minimum value is <math>\boxed{B: 97}</math>, with construction given in the above solutions.
 +
 
 +
-cretinouscretin
  
 
== See also ==
 
== See also ==

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

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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