Difference between revisions of "2000 AIME I Problems/Problem 10"
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== Problem == | == Problem == | ||
| + | A sequence of numbers <math>x_{1},x_{2},x_{3},\ldots,x_{100}</math> has the property that, for every integer <math>k</math> between <math>1</math> and <math>100,</math> inclusive, the number <math>x_{k}</math> is <math>k</math> less than the sum of the other <math>99</math> numbers. Given that <math>x_{50} = m/n,</math> where <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>m + n</math>. | ||
== Solution == | == Solution == | ||
{{solution}} | {{solution}} | ||
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== See also == | == See also == | ||
| − | + | {{AIME box|year=2000|n=I|num-b=9|num-a=11}} | |
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Revision as of 18:32, 11 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
This problem needs a solution. If you have a solution for it, please help us out by adding it.
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 | ||
