Judging from the title, the number of balls taken by three people in each round is A-A=0, B-A, C-A, that is, the total number of balls taken by three people in each round is B+C-2A, so let a * * * take N rounds, so finally, because A has 20 balls, B has 10 and C has 9, there is (B+.

First, suppose n= 13, then B+C-2A=3, because C >；; B& gt； A, that is, C-A is greater than or equal to 2 and B-A is greater than or equal to 1, so B+C-2A is greater than or equal to 3. If and only if C-A is 2 and B-A is 1, the number of balls taken in each round is 0, 1, 2. Next, you can analyze

However, when I saw that your question was filled three times, I directly analyzed the situation of n=3. The analysis is as follows: B+C-2A= 13, 1 3 = 12 = 2+16 = 3+10 = 4+9 = 5+8.

If it is 0, 1, 12, we can know that after three rounds, we can't get the number of A-take 20.

If it is 0, 2, 1 1, you can't get 20 of A either;

If it is 0,3, 10, we know that when A takes two rounds of C and one round of A, it is exactly 20; B takes a round of C, and two rounds of A are exactly 10, which meets the condition of taking C in the last round, so C takes three rounds of B, that is, 3*(B-A)=9, 3B= 18, so it is calculated that A = 3, B=6 and C = 18.

Similarly, we can analyze other situations and find that they have nothing to do with the problem.

Therefore, A = 3, B = 6 and C = 13.