Question

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 $x_{50} = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, find $m + n.$

Answer

Answer: 173

Concepts:system of equations summation

Difficulty rating: 2330

Solution:

Let $S = x_1 + x_2 + \cdots + x_{100}.$ The condition says $x_k = (S - x_k) - k,$ so $x_k = \frac{S - k}{2}$ for every $k.$ Summing over $k = 1, \ldots, 100,$ $\begin{aligned} S &= \frac{100S - (1 + 2 + \cdots + 100)}{2} \\ &= \frac{100S - 5050}{2}, \end{aligned}$ so $98S = 5050$ and $S = \frac{2525}{49}.$

Then $\begin{aligned} x_{50} &= \frac{S - 50}{2} \\ &= \frac{2525 - 2450}{98} = \frac{75}{98}, \end{aligned}$ which is in lowest terms, so $m + n = 75 + 98 = 173.$

2000 AIME I Problem 10

Answer: 173

Difficulty rating: 2330

Solution:

Problem 10 in Other Years