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 $\mathbb{S}$ . Then for each integer $k$ , $x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k$ . Summing this up for all $k$ from $1, 2, \ldots, 100$ ,

$\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\ 100\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\ \mathbb{S}&=\frac{2525}{49}\end{align*}$

Now, substituting for $x_{50}$ , we get $2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}$ , and the answer is $75+98=\boxed{173}$ .

Solution 2

Consider $x_k$ and $x_{k+1}$ . Let $S$ be the sum of the rest 98 terms. Then $x_k+k=S+x_{k+1}$ and $x_{k+1}+(k+1)=S+x_k.$ Eliminating $S$ we have $x_{k+1}-x_k=-\dfrac{1}{2}.$ So the sequence is arithmetic with common difference $-\dfrac{1}{2}.$

In terms of $x_{50},$ the sequence is $x_{50}+\dfrac{49}{2}, x_{50}+\dfrac{48}{2},\cdots,x_{50}+\dfrac{1}{2}, x_{50}, x_{50}-\dfrac{1}{2}, \cdots, x_{50}-\dfrac{49}{2}, x_{50}-\dfrac{50}{2}.$ Therefore $x_{50}+50=99x_{50}-\dfrac{50}{2}$ . We are done by solving for $x_{50}$ .

-JZ

Video solution

https://www.youtube.com/watch?v=TdvxgrSZTQw

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

Page

Toolbox

Search

2000 AIME I Problems/Problem 10

Contents

Problem

Solution

Solution 2

Video solution

See also