Question

Let $x_1 \le x_2 \le \cdots \le x_{100}$ be real numbers such that $|x_1| + |x_2| + \cdots + |x_{100}| = 1$ and $x_1 + x_2 + \cdots + x_{100} = 0.$ Among all such $100$ -tuples of numbers, the greatest value that $x_{76} - x_{16}$ can achieve is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

Since the terms sum to $0$ while their absolute values sum to $1,$ the positive terms sum to $\frac{1}{2}$ and the negative terms sum to $-\frac{1}{2}.$ If $x_{16} \lt -\frac{1}{32},$ then $x_1, \ldots, x_{16}$ are all less than $-\frac{1}{32}$ and would sum below $-\frac{1}{2},$ a contradiction; hence $x_{16} \ge -\frac{1}{32}.$ Similarly, if $x_{76} \gt \frac{1}{50}$ then $x_{76}, \ldots, x_{100}$ are $25$ terms each exceeding $\frac{1}{50},$ summing above $\frac{1}{2};$ hence $x_{76} \le \frac{1}{50}.$

Therefore $x_{76} - x_{16} \le \frac{1}{50} + \frac{1}{32} = \frac{16 + 25}{800} = \frac{41}{800},$ and this is achieved by taking $x_1 = \cdots = x_{16} = -\frac{1}{32},$ $x_{17} = \cdots = x_{75} = 0,$ and $x_{76} = \cdots = x_{100} = \frac{1}{50}.$

Since $\gcd(41, 800) = 1,$ the answer is $41 + 800 = 841.$

2022 AIME II Problem 6

Solution:

Problem 6 in Other Years