Difference between revisions of "1991 AHSME Problems/Problem 30"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | <math>n(A)=n(B)=2^{100}</math>, so as <math>n(C)</math> and <math>n(A \cup B \cup C)</math> are integral powers of <math>2</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> | ||
== See also == | == See also == | ||
Revision as of 14:15, 13 February 2018
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 
?
(A) 96 (B) 97 (C) 98 (D) 99 (E) 100
Solution
, so as 
and 
are integral powers of 
, 
and 
. Let 
, 
, and 
where 
Thus, the minimum value of 
is 
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. 
