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Difference between revisions of "2000 AIME I Problems/Problem 10"
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== 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>.
+
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 ==
−
Let the sum of all of the terms in the sequence be <math>\mathbb{S}</math>.
+
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>,
  
−
<math>x_1=\mathbb{S}-x_1-1</math>
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<cmath>\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*}</cmath>
  
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<math>x_2=\mathbb{S}-x_2-2</math>
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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>.
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<math>\vdots</math>
 
−
 
 
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<math>x_{100}=\mathbb{S}-x_{100}-100</math>
 
−
 
 
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<math>2x_n=\mathbb{S} - n</math>
 
−
 
 
−
<math>2\mathbb{S}=100\mathbb{S}-5050</math>
 
−
 
 
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<math>98\mathbb{S}=5050</math>
 
−
 
 
−
<math>\mathbb{S}=\frac{2525}{49}</math>
 
−
 
 
−
<math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49}</math>
 
−
 
 
−
<math>x_{50}=\frac{75}{98}</math>
 
−
 
 
−
<math>75+98=\boxed{173}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2000|n=I|num-b=9|num-a=11}}
 +
 +
[[Category:Intermediate Algebra Problems]]

Revision as of 10:46, 1 January 2008

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 $75+98=\boxed{173}$.

See also

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
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