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Permuted sums (Posted on 2020-11-30) Difficulty: 2 of 5
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.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

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Set up to solution; Part 1 | Comment 1 of 7
If the six pairwise sums are a,b,c,d,x,y in ascending order (not necessarily distinct)
then x = b+d and y = c+d.
The others are then a+b, a+c, a+d, b+c
But of a+d and b+c, we can't be sure which is larger or if they are equal.

189 a+b
264 a+c
287 b+c or a+d
320 a+d or b+c
x   b+d
y   c+d
     case1   case 2
a+b   189     189
a+c   264     264
a+d   320     287
b+c   287     320
b+d    x       x
c+d    y       y


Either:
a+d = 287 and b+c = 320    or
a+d = 320 and b+c = 287
So:
a+b = 189
a+c = 264
c = b + 75
y = x + 75

a = a
b = 189 - a
c = 264 - a
d = either 320 - a or 287 - a
x = b+d = either 189 - a + 320 - a   OR 189 - a + 287 - a
x = either 509 - 2a  OR 476 - 2a
y = x + 75

x+y = 2x + 75 = either 1093 - 4a   OR 1027 - 4a   

I'll suggest the larger 1093 - 4a
not sure about the other 2
So, either d-b = y - 320 and b-a = x - 287  OR
    either c-a = y - 320 and d-c = x - 287
and either b-a = x - 320 or

see part 2 for two programs, one for each possibility and the solution
  Posted by Larry on 2020-11-30 09:56:22
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