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

Revision as of 21:24, 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

Solution 3

Let $n$ be the value of $x_k$ , $n+k$ be the sum of the 99 other numbers, and $S$ be the sum of $\textbf{all}$ 100 numbers. Thus, $S = 2n + k$ , so $n = \frac{S-k}{2}$ . We then obtain the following sequence:

$\frac{S-1}{2} + \frac{S-2}{2} + ... + \frac{S-100}{2} = S$

Solving for the left hand side, we get $\frac{100S - 5050}{2} = 50S - 2525 = S$ . This means $S = \frac{2525}{49}$ . Since $x_{50}$ = $\frac{S-50}{2}$ , we get $\frac{\frac{2525-2450}{49}}{2} = \frac{\frac{75}{49}}{2} = \frac{75}{98}$ . Thus, the answer is $75 + 98 = \boxed{173}$ .

Video solution

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

@@ Line 25: / Line 25: @@
 == Solution 3 ==
-Let <math>n</math> be the value of <math>x_k</math>, <math>n+k</math> be the sum of the 99 other numbers, and <math>S</math> be the sum of <math>\textbf{all}</math> 100 numbers. Thus,
+Let <math>n</math> be the value of <math>x_k</math>, <math>n+k</math> be the sum of the 99 other numbers, and <math>S</math> be the sum of <math>\textbf{all}</math> 100 numbers. Thus, <math>S = 2n + k</math>, so <math>n = \frac{S-k}{2}</math>. We then obtain the following sequence:
-<center>
-<math>S = 2n + k \\
+<center>
-n = \frac{S-k}{2}
+<math>\frac{S-1}{2} + \frac{S-2}{2} + ... + \frac{S-100}{2} = S</math>
-</math>
+</center>
+Solving for the left hand side, we get <math>\frac{100S - 5050}{2} = 50S - 2525 = S</math>. This means <math>S = \frac{2525}{49}</math>. Since <math>x_{50}</math> = <math>\frac{S-50}{2}</math>, we get <math>\frac{\frac{2525-2450}{49}}{2} = \frac{\frac{75}{49}}{2} = \frac{75}{98}</math>. Thus, the answer is <math>75 + 98 = \boxed{173}</math>.
 ==Video solution==

2000 AIME I (Problems • Answer Key • Resources)
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