>>10018428Fine then
B is the answer.
Proof:
Let R1=(A,B,C) and R4=(A+B-C,2(B-A),2(C-(B-A)))=(A4,B4,C4) Moreover, fix A4=B4=C4.
Then the condition that B4=C4 implies that C=2(B-A). One obtains upon inputting this into A4 and equating with B4 the expression:
A4=3A-B
A=3B/5.
This further implies that R1=(3B/5,B,4B/5) and R4=(4B/5,4B/5,4B/5).
The constraint that each element A1,B1, and C1 of R1 are natural numbers implies that B mod 5=0 or B=5n where n is an element of the natural numbers. Substituting, we have:
R1: (3n,5n,4n)
R4: (4n,4n,4n).
Naturally, any n will produce an integer value for all associated elements of R1 an R4; this satisfies our first constraint. Our next constraint is that A, B, or C =50.
Looking at R1, we see that neither A nor C can satisfy this constraint because neither 3n=50 or 4n=50 produces a natural number.
Hence we conclude that n=10 and B=50 with the initial round:
R1: (30,50,40).
This completes the proof.
