Difference between revisions of "2000 AIME I Problems/Problem 10"
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== Solution == | == Solution == | ||
− | {{ | + | Let the sum of all of the terms in the sequence be <math>\mathbb{S}</math>. |
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+ | <math>x_1=\mathbb{S}-x_1-1</math> | ||
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+ | <math>x_2=\mathbb{S}-x_2-2</math> | ||
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+ | <math>\vdots</math> | ||
+ | |||
+ | <math>x_{100}=\mathbb{S}-x_{100}-100</math> | ||
+ | |||
+ | <math>2x_n=\mathbb{S} - n</math> | ||
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+ | <math>2\mathbb{S}=100\mathbb{S}-5050</math> | ||
+ | |||
+ | <math>98\mathbb{S}=5050</math> | ||
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+ | <math>\mathbb{S}=\frac{2525}{49}</math> | ||
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+ | <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49}</math> | ||
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+ | <math>x_{50}=\frac{75}{98}</math> | ||
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+ | <math>75+98=\boxed{173}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=9|num-a=11}} | {{AIME box|year=2000|n=I|num-b=9|num-a=11}} |
Revision as of 12:32, 13 November 2007
Problem
A sequence of numbers has the property that, for every integer
between
and
inclusive, the number
is
less than the sum of the other
numbers. Given that
where
and
are relatively prime positive integers, find
.
Solution
Let the sum of all of the terms in the sequence be .
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |