Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2)
m (→‎Solution 2 (PIE))
 
(3 intermediate revisions by the same user not shown)
Line 9: Line 9:
 
Thus, the minimum value of <math>|A\cap B \cap C|</math> is <math>\fbox{B=97}</math>
 
Thus, the minimum value of <math>|A\cap B \cap C|</math> is <math>\fbox{B=97}</math>
  
−
==Solution 2==
+
==Solution 2 (PIE) ==
  
 
As <math>|A|=|B|=100</math>, <math>n(A)=n(B)=2^{100}</math>
 
As <math>|A|=|B|=100</math>, <math>n(A)=n(B)=2^{100}</math>
Line 17: Line 17:
 
<math>2^{101}+2^{|C|}=2^{|A \cup B \cup C|}</math> as <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>
 
<math>2^{101}+2^{|C|}=2^{|A \cup B \cup C|}</math> as <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 [[Principle of Inclusion-Exclusion]], <math>|A \cup B| = |A| + |B| - |A \cap B| = 200 - |A \cap B|</math>
+
By the [[Principle of Inclusion-Exclusion]], <math>|A \cup B| = |A| + |B| - |A \cap B| = 200 - |A \cap B|</math>
  
 
<math>|A|=|B| \le |A \cup B| \le |A \cup B \cup C|</math>, <math>100 \le |A \cup B| \le 102</math>, <math>98 \le |A \cap B| \le 100</math>
 
<math>|A|=|B| \le |A \cup B| \le |A \cup B \cup C|</math>, <math>100 \le |A \cup B| \le 102</math>, <math>98 \le |A \cap B| \le 100</math>
  
−
By [[Principle of Inclusion-Exclusion]], <math>|A \cup C| = |A| + |C| - |A \cap C| = 200 - |A \cap C|</math>
+
By the [[Principle of Inclusion-Exclusion]], <math>|A \cup C| = |A| + |C| - |A \cap C| = 201 - |A \cap C|</math>
  
−
<math>|C| \le |A \cup C| \le |A \cup B \cup C|</math>, <math>101 \le |A \cup C| \le 102</math>, <math>99 \le |A \cap B| \le 100</math>
+
<math>|C| \le |A \cup C| \le |A \cup B \cup C|</math>, <math>101 \le |A \cup C| \le 102</math>, <math>99 \le |A \cap C| \le 100</math>
  
−
By [[Principle of Inclusion-Exclusion]], <math>|B \cup C| = |B| + |C| - |B \cap C| = 200 - |B \cap C|</math>
+
By the [[Principle of Inclusion-Exclusion]], <math>|B \cup C| = |B| + |C| - |B \cap C| = 201 - |B \cap C|</math>
  
−
<math>|C| \le |A \cup C| \le |A \cup B \cup C|</math>, <math>101 \le |B \cup C| \le 102</math>, <math>99 \le |A \cap B| \le 100</math>
+
<math>|C| \le |B \cup C| \le |A \cup B \cup C|</math>, <math>101 \le |B \cup C| \le 102</math>, <math>99 \le |B \cap C| \le 100</math>
  
−
By [[Principle of Inclusion-Exclusion]], <math>|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|</math>
+
By the [[Principle of Inclusion-Exclusion]], <math>|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|</math>
 
<math>+|B \cap C|=|A \cap B| + |A \cap C|+|B \cap C| -199</math>
 
<math>+|B \cap C|=|A \cap B| + |A \cap C|+|B \cap C| -199</math>
  

Latest revision as of 10:29, 6 May 2023

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

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

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

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

$|C| \le |A \cup C| \le |A \cup B \cup C|$, $101 \le |A \cup C| \le 102$, $99 \le |A \cap C| \le 100$

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

$|C| \le |B \cup C| \le |A \cup B \cup C|$, $101 \le |B \cup C| \le 102$, $99 \le |B \cap C| \le 100$

By the Principle of Inclusion-Exclusion, $|A \cap B \cap C|=|A \cup B \cup C|- |A| - |B| - |C| + |A \cap B| + |A \cap C|+|B \cap C| = 102-100-100-101+ |A \cap B| + |A \cap C|$ $+|B \cap C|=|A \cap B| + |A \cap C|+|B \cap C| -199$

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

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

~isabelchen

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_30&oldid=192834"