Difference between revisions of "2002 AIME II Problems/Problem 9"
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We want the remainder when <math>n</math> is divided by 1000, so we find the last three digits of each. | We want the remainder when <math>n</math> is divided by 1000, so we find the last three digits of each. | ||
− | {{ | + | ==Solution 2== |
+ | Let the two disjoint subsets be <math>A</math> and <math>B</math>, and let <math>C = S-(A+B)</math>. | ||
+ | For each <math>i \in S</math>, either <math>i \in A</math>, <math>i \in B</math>, or <math>i \in C</math>. So there are <math>3^{10}</math> ways to organize the elements of <math>S</math> into disjoint <math>A</math>, <math>B</math>, and <math>C</math>. | ||
+ | |||
+ | However, there are <math>2^{10}</math> ways to organize the elements of <math>S</math> such that <math>A = \emptyset</math> and <math>S = B+C</math>, and there are <math>2^{10}</math> ways to organize the elements of <math>S</math> such that <math>B = \emptyset</math> and <math>S = A+C</math>. | ||
+ | But, the combination such that <math>A = B = \emptyset</math> and <math>S = C</math> is counted twice. | ||
+ | |||
+ | Thus, there are <math>3^{10}-2\cdot2^{10}+1</math> ordered pairs of sets <math>(A,B)</math>. But since the question asks for the number of unordered sets <math>\{ A,B \}</math>, <math>n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=II|num-b=8|num-a=10}} | {{AIME box|year=2002|n=II|num-b=8|num-a=10}} |
Revision as of 13:24, 24 June 2008
Contents
[hide]Problem
Let be the set Let be the number of sets of two non-empty disjoint subsets of . (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when is divided by .
Solution
For simplicity, let's call the sets and . Now if we choose members from to be in , then we have elements to choose for . Thus
.
We want the remainder when is divided by 1000, so we find the last three digits of each.
Solution 2
Let the two disjoint subsets be and , and let . For each , either , , or . So there are ways to organize the elements of into disjoint , , and .
However, there are ways to organize the elements of such that and , and there are ways to organize the elements of such that and . But, the combination such that and is counted twice.
Thus, there are ordered pairs of sets . But since the question asks for the number of unordered sets , .
See also
2002 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |