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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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Solution solving for the original numbers with MATLAB | Comment 4 of 7 |
Call the set A, B, C and D, from smallest to largest.

We can assume that x and y are B+D and C+D.

If that's the case, the smallest value we're given is A+B = 189 and the next smallest is A+C = 264. The question remains:

Do we have:

B+C = 320, A+D = 287

or

B+C = 287, A+D = 320

?

We try both:

syms a b c d
eqns=[a+b==189,a+c==264,b+c==320,a+d==287];
s=solve(eqns,[a b c d]);
[s.a s.b s.c s.d]
>> s.c+2*s.d+s.b
ans =
761

eqns=[a+b==189,a+c==264,b+c==287,a+d==320];
s=solve(eqns,[a b c d]);
[s.a s.b s.c s.d]
>> s.c+2*s.d+s.b
ans =
761

Each results in the same total for C + D: 761.

The respective values of A, B, C and D are 

133/2, 245/2, 395/2, 441/2

and

83, 106, 181, 237


As further verification, other set of assumptions about the relations of the given sums results in an empty solution set.

Edited on November 30, 2020, 10:40 am
  Posted by Charlie on 2020-11-30 10:28:10

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