khan.academy wrote: Problem Section #1 a) 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 $a,b,c,d$ be the four numbers. We just have two cases : 1) $x=a+b$ and $y=c+d$ (no common number) Then $a+c,a+d,b+c,b+d$ are in some order $189,320,287,264$ Adding these four sums, we get $a+b+c+d=530$ And so $x+y=530$ 2) $x=a+b$ and $y=a+c$ (one common number) Then $a+d,b+c,b+d,c+d$ are in some order $189,320,287,264$ Note then that $x+y=2(a+d)+2(b+c)-(b+d)-(c+d)\le 2(320+287)-(264+189)=761$ Which indeed can be reached with for example $(a,b,c,d=237,181,106,83)$ Hence the answer $\boxed{761}$