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