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> | ||
| + | |||
| + | <math>x_2=\mathbb{S}-x_2-2</math> | ||
| + | |||
| + | <math>\vdots</math> | ||
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| + | <math>x_{100}=\mathbb{S}-x_{100}-100</math> | ||
| + | |||
| + | <math>2x_n=\mathbb{S} - n</math> | ||
| + | |||
| + | <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> | ||
| + | |||
| + | <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49}</math> | ||
| + | |||
| + | <math>x_{50}=\frac{75}{98}</math> | ||
| + | |||
| + | <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 | ||
