A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are 189, 320, 287, 264, x, and y. Find the greatest possible value of: x + y.
Let the numbers be a,b,c,d in ascending order.
The smallest sum is a+b, the second smallest sum is a+c.
The third and fourth sums are a+d and b+c in some order.
The second largest sum is b+d, and the largest sum is c+d.
The sum of the smallest and largest sums is the sum of all four numbers a,b,c,d.
The sum of the second smallest and second largest sums is the sum of all four numbers a,b,c,d.
The sum of the third and fourth sums is the sum of all four numbers a,b,c,d.
Then sum of the smallest, second smallest, second largest, and largest sums is twice the sum of all four numbers.
The third and fourth sums are 287 and 320, which makes the sum a+b+c+d equal to 607.
The smallest, second smallest, second largest and largest sums are 189, 264, x, y; which makes 453+x+y = 2*(a+b+c+d).
Then a simple substitution yields the unique answer x+y=761.
Solution
