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

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== Problem ==
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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>?
  
(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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<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>
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Thus, the minimum value of <math>|A\cap B \cap C|</math> is <math>\fbox{B=97}</math>
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== See also ==
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{{AHSME box|year=1991|num-b=29|num-a=30}} 
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 15:18, 13 February 2018

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 $\fbox{B=97}$

See also

1991 AHSME (ProblemsAnswer KeyResources)
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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