Question

数列 $x_1, x_2, x_3, \ldots, x_{100}$ 满足：对每个从 $1$ 到 $100$ （含端点）的整数 $k$ ， $x_k$ 比其余 $99$ 个数之和小 $k$ 。已知 $x_{50} = \frac{m}{n}$ ，其中 $m$ 和 $n$ 是互质的正整数。求 $m + n$ 。

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 $m + n.$

答案

答案：173

知识点：方程组求和

难度评级：2330

解答：

设 $S = x_1 + x_2 + \cdots + x_{100}$ 。条件给出 $x_k = (S - x_k) - k$ ，所以对每个 $k$ 都有 $x_k = \frac{S - k}{2}$ 。对 $k = 1, \ldots, 100$ 求和： $\begin{aligned} S &= \frac{100S - (1 + 2 + \cdots + 100)}{2} \\ &= \frac{100S - 5050}{2}, \end{aligned}$ 因此 $98S = 5050$ ，即 $S = \frac{2525}{49}$ 。

所以 $\begin{aligned} x_{50} &= \frac{S - 50}{2} \\ &= \frac{2525 - 2450}{98} = \frac{75}{98}, \end{aligned}$ 该分数已最简，故 $m + n = 75 + 98 = 173$ 。

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 第 10 题

答案：173

难度评级：2330

解答：

其他年份的第 10 题