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Difference between revisions of "2000 AIME I Problems/Problem 10"
(Added another way to solve this question)
 
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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>
+
<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>
  
−
<math>x_2=\mathbb{S}-x_2-2</math>
+
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>.
  
−
<math>\vdots</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>  
  
−
<math>x_{100}=\mathbb{S}-x_{100}-100</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>.
  
−
<math>2x_n=\mathbb{S} - n</math>
+
Solving, we get <math>x_{50}=\dfrac{75}{98}.</math> The answer is <math>75+98=\boxed{173}.</math>
  
−
<math>2\mathbb{S}=100\mathbb{S}-5050</math>
+
- JZ
  
−
<math>98\mathbb{S}=5050</math>
+
- edited by erinb28lms
  
−
<math>\mathbb{S}=\frac{2525}{49}</math>
+
==Solution 3 (Sum of equations)==
 +
Like Solution 1, let the sum of all of the terms in this sequence be <math>\mathbb{S}</math>. By definition:
  
−
<math>2x_{50}=\frac{2525}{49}-50=\frac{75}{49}</math>
+
<cmath>x_1 + 1 = x_2+x_3+x_4+...+x_{100}</cmath>
 +
<cmath>x_2 + 2 = x_1+x_3+x_4+...+x_{100}</cmath>
 +
<cmath>x_3 + 3 = x_1+x_2+x_4+...+x_{100}</cmath>
 +
<cmath>...</cmath>
 +
<cmath>x_{99} + 99 = x_1+x_2+x_3+...+x_{98}+x_{100}</cmath>
 +
<cmath>x_{100} + 100 = x_1+x_2+x_3+...+x_{98}+x_{99}</cmath>.
  
−
<math>x_{50}=\frac{75}{98}</math>
+
Adding up all of these equations yields:
  
−
<math>75+98=\boxed{173}</math>
+
<cmath>\mathbb{S} + TR(100) = 99\mathbb{S}</cmath>
 +
 
 +
Here <math>TR(100)</math> represents the <math>100</math>th triangular number, which is <math>5050</math>. Solving for <math>\mathbb{S}</math> yields:
 +
 
 +
<cmath>\mathbb{S} = \frac{2525}{49}</cmath>.
 +
 
 +
<math>\mathbb{S}</math> can also be written as <math>x_{50} + (x_{50} + 50)</math>. Solving for <math>x_{50}</math>,
 +
 
 +
<cmath>2x_{50} = \frac{2525-2450}{49}</cmath>
 +
<cmath>2x_{50} = \frac{75}{49}</cmath>
 +
<cmath>x_{50} = \frac{75}{98}</cmath>
 +
 
 +
The requested sum is therefore <math>75+98 = \boxed{173}</math>.
 +
 
 +
~mathwizard123123
 +
 
 +
==Video solution==
 +
 
 +
https://www.youtube.com/watch?v=TdvxgrSZTQw
  
 
== 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]]
 +
{{MAA Notice}}

Latest revision as of 14:13, 17 July 2025

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

Solution 3 (Sum of equations)

Like Solution 1, let the sum of all of the terms in this sequence be $\mathbb{S}$. By definition:

\[x_1 + 1 = x_2+x_3+x_4+...+x_{100}\] \[x_2 + 2 = x_1+x_3+x_4+...+x_{100}\] \[x_3 + 3 = x_1+x_2+x_4+...+x_{100}\] \[...\] \[x_{99} + 99 = x_1+x_2+x_3+...+x_{98}+x_{100}\] \[x_{100} + 100 = x_1+x_2+x_3+...+x_{98}+x_{99}\].

Adding up all of these equations yields:

\[\mathbb{S} + TR(100) = 99\mathbb{S}\]

Here $TR(100)$ represents the $100$th triangular number, which is $5050$. Solving for $\mathbb{S}$ yields:

\[\mathbb{S} = \frac{2525}{49}\].

$\mathbb{S}$ can also be written as $x_{50} + (x_{50} + 50)$. Solving for $x_{50}$,

\[2x_{50} = \frac{2525-2450}{49}\] \[2x_{50} = \frac{75}{49}\] \[x_{50} = \frac{75}{98}\]

The requested sum is therefore $75+98 = \boxed{173}$.

~mathwizard123123

Video solution

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

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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