Let's set a = b+ca > b>；; c
There are six situations, which we can discuss in categories:
That is, the principals of Party A, Party B and Party C are: ≦.
1 abc 2 acb 3 bac 4 bca 5 cab 6 cba
Because we know the other two numbers, but we don't know their operation rules, the number on our head may be their sum or their difference (if we take the absolute value, it will be greatly discounted). If someone says he doesn't know for a while, at least prove that the difference can't be equal to the minimum of the other two numbers he sees. That is, some people can't see a large number twice as large as a decimal.
Discussion 1: In the first round, everyone said they didn't know.
Proof: b≠2c a≠2c a≠2b (all six situations are like this, which was revealed in the first round).
Let's sort it out: the results after the first round:
Communication ≠2c Communication ≠3c
b≠2a-2b a≠3b/2
The second round: A says he doesn't know, which proves that the two numbers A has still can't rule out the possibility that he is poor! ~ so it is proved that the two numbers seen by a satisfy that the big one is not equal to 2 times the small one, the big one is not equal to 3 times the small one, not 3/2 times the small one. B the same is true.
But C found it in the second round, which proves that he found that the two numbers he mastered were three times smaller, or three-thirds smaller.
Suppose what you see is three times as big as a small one. Obviously, if you are twice as small, you can't satisfy the meaning of the problem, so you can only sum it up.
In this case, a = 144b and c = 108 and 36.
Suppose the big one is 3/2 times smaller, obviously you can't be small 1/2, so you can only be the sum of them. You are a = 144b, and c is 216/5,288/5.
So these three numbers are 36, 108, 144.
We can't tell who is 36 and who is 108.