Difference between revisions of "2000 AIME II Problems/Problem 10"
m |
|||
Line 1: | Line 1: | ||
== 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}} | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2000|n=II|num-b=9|num-a=11}} |
Revision as of 19:14, 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 II (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 |