2000 AIME I Problems/Problem 10

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

Let the sum of all of the terms in the sequence be $\mathbb{S}$ .

$x_1=\mathbb{S}-x_1-1$

$x_2=\mathbb{S}-x_2-2$

$\vdots$

$x_{100}=\mathbb{S}-x_{100}-100$

$2x_n=\mathbb{S} - n$

$2\mathbb{S}=100\mathbb{S}-5050$

$98\mathbb{S}=5050$

$\mathbb{S}=\frac{2525}{49}$

$2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$

$x_{50}=\frac{75}{98}$

$75+98=\boxed{173}$

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
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2000 AIME I Problems/Problem 10

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