Difference between revisions of "1991 AHSME Problems/Problem 30"
Isabelchen (talk | contribs) m (→Solution 2) |
Isabelchen (talk | contribs) m (→Solution 2 (PIE)) |
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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> | ||
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<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| = 201 - |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 | + | <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| = 201 - |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 |B \cup C| \le |A \cup B \cup C|</math>, <math>101 \le |B \cup C| \le 102</math>, <math>99 \le | | + | <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 11:29, 6 May 2023
Problem
For any set , let
denote the number of elements in
, and let
be the number of subsets of
, including the empty set and the set
itself. If
,
, and
are sets for which
and
, then what is the minimum possible value of
?
Solution 1
, so
and
are integral powers of
and
. Let
,
, and
where
Thus, the minimum value of
is
Solution 2 (PIE)
As ,
As ,
,
as
and
are integers,
and
By the Principle of Inclusion-Exclusion,
,
,
By the Principle of Inclusion-Exclusion,
,
,
By the Principle of Inclusion-Exclusion,
,
,
By the Principle of Inclusion-Exclusion,
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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