Solution
First, suppose a person Z sees numbers X and Y where Y > X
Then Z has 2 options X+Y and Y-X
The only way to Z to do not know is if X,Y,Y-X and X+Y are different.
X,Y,X+Y are clearly different and Y>Y-X, so the only possibility to be equal is Y-X = X this is Y = 2X
So, if Y=2X Z knows and if Y is different from 2X
Z does not know.
Since A and B do not know then C knows that his number is different from 2A,2B,(1/2)A,(1/2)B
The 2 options for C are A+B and B-A (wlog B>A)
Since C can find his number, then 1 of these options should be equal to 2A,2B,(1/2)A,(1/2)B.
It's easy to check that A+B cannot be equal to any of these options. So, B-A should be equal to 1 of these.
2B=B-A # since 2B>B-A
(1/2)B=B-A implies A = (3/2)B # since B>A
2A=B-A implies B=3A
but since C=A+B=50 then 4A=50 # since A is an integer
so the only option left is (1/2)A=B-A
then B = (3/2)A
Now C=A+B=50 therefore (5/2)A=50 and A=20
then B = (3/2)20 = 30
and the solution is unique.
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