2000 AIME I Problems/Problem 10

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Problem

A sequence of numbers $x_{1},x_{2},x_{3},\ldots,x_{100}$ has the property that, for every integer $k$ between $1$ and $100,$ inclusive, the number $x_{k}$ is $k$ less than the sum of the other $99$ numbers. Given that $x_{50} = m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m + n$.

Solution

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See also

2000 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AIME Problems and Solutions