Difference between revisions of "2000 AIME I Problems/Problem 10"

Latest revision as of 21:26, 29 December 2022

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, $x_{50}+50=99x_{50}-\dfrac{50}{2}$ .

Solving, we get $x_{50}=\dfrac{75}{98}.$ The answer is $75+98=\boxed{173}.$

- JZ

- edited by erinb28lms

Video solution

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

@@ Line 1: / Line 1: @@
 == 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 ==
+Let the sum of all of the terms in the sequence be <math>\mathbb{S}</math>. Then for each integer <math>k</math>, <math>x_k = \mathbb{S}-x_k-k \Longrightarrow \mathbb{S} - 2x_k = k</math>. Summing this up for all <math>k</math> from <math>1, 2, \ldots, 100</math>,
+<cmath>\begin{align*}100\mathbb{S}-2(x_1 + x_2 + \cdots + x_{100}) &= 1 + 2 + \cdots + 100\\
+\mathbb{S} - 2\mathbb{S} &= \frac{100 \cdot 101}{2} = 5050\\
+\mathbb{S}&=\frac{2525}{49}\end{align*}</cmath>
+Now, substituting for <math>x_{50}</math>, we get <math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49} \Longrightarrow x_{50}=\frac{75}{98}</math>, and the answer is <math>75+98=\boxed{173}</math>.
+== Solution 2 ==
+Consider <math>x_k</math> and <math>x_{k+1}</math>. Let <math>S</math> be the sum of the rest 98 terms. Then <math>x_k+k=S+x_{k+1}</math> and <math>x_{k+1}+(k+1)=S+x_k.</math> Eliminating <math>S</math> we have <math>x_{k+1}-x_k=-\dfrac{1}{2}.</math> So the sequence is arithmetic with common difference <math>-\dfrac{1}{2}.</math>
+In terms of <math>x_{50},</math> the sequence is <math>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}.</math> Therefore, <math>x_{50}+50=99x_{50}-\dfrac{50}{2}</math>.
+Solving, we get <math>x_{50}=\dfrac{75}{98}.</math> The answer is <math>75+98=\boxed{173}.</math>
+- JZ
+- edited by erinb28lms
+==Video solution==
+https://www.youtube.com/watch?v=TdvxgrSZTQw
 == See also ==
-* [[2000 AIME I Problems]]
+{{AIME box|year=2000|n=I|num-b=9|num-a=11}}
+[[Category:Intermediate Algebra Problems]]
+{{MAA Notice}}

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