Problem:
Let x1​≤x2​≤⋯≤x100​ be real numbers such that ∣x1​∣+∣x2​∣+⋯+∣x100​∣=1 and x1​+x2​+⋯+x100​=0. Among all such 100-tuples of numbers, the greatest value that x76​−x16​ can achieve is nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let s be the sum of all the positive numbers in the list. Then the sum of the negative numbers in the list is −s and the sum of all the absolute values is 2s. Hence s=21​. Because there cannot be more than 25 numbers greater than or equal to 501​, it follows that x76​≤501​. Similarly, because there cannot be more than 16 numbers less than or equal to −321​, it follows that x16​≥−321​. Thus x76​−x16​≤501​+321​=80041​.
To see that the bound 80041​ can be achieved, let xi​=−321​ for i≤16, let xi​=0 for 17≤i≤75, and let xi​=501​ for i≥76. Then all the conditions in the problem are satisfied and x76​−x16​=501​+321​=80041​. Hence the greatest value that x76​−x16​ can achieve is 80041​. The requested sum is 41+800=841​.
The problems on this page are the property of the MAA's American Mathematics Competitions