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AIME 2022b Problem 6

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Let x1≤x2≤⋯≤x100x_1\leq x_2\leq \cdots\leq x_{100} be real numbers such that ∣x1∣+∣x2∣+⋯+∣x100∣=1|x_1| + |x_2| + \cdots + |x_{100}| = 1 and x1+x2+⋯+x100=0x_1 + x_2 + \cdots + x_{100} = 0. Among all such 100100-tuples of numbers, the greatest value that x76−x16x_{76} - x_{16} can achieve is mn\tfrac mn, where mm and nn are relatively prime positive integers. Find m+nm+n.

Dataset text AIME 1983 to 2026. Quoted with source.

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