2.11 There are three piles of stones with the numbers being $19, 8, 9$. Now, the following operation is performed: each time, 1 stone is taken from any two of the three piles, and then these 2 stones are added to the other pile. The question is: can the following be achieved after several such operations?
(1) The number of stones in the three piles are $2, 12, 22$;
(2) The number of stones in each of the three piles is 12.
If the requirements can be met, complete it with the minimum number of operations. If not, explain the reason.
(China Guangzhou, Chongqing, Luoyang, Fuzhou, Wuhan Junior High School Mathematics Competition, 1989)
［Solution］(1) It can be achieved.
Since the original smallest pile has 8 stones, to make it 2, at least 6 operations are required. In fact, 6 times can succeed:
$$
(19,8,9) \rightarrow(21,7,8) \rightarrow(23,6,7,) \rightarrow(22,5,9) \rightarrow(24,4,8) \rightarrow
$$
$(23,3,10) \rightarrow(22,2,12)$.
(2) It cannot be achieved.
Since after each operation, the number of stones in each pile either decreases by 1 or increases by 2, we can write it as
$$
(a, b, c) \rightarrow(a-1, b-1, c+2) \text {. }
$$

If $a, b, c$ have different remainders when divided by 3, let's say 0, 1, 2, then after the operation, the remainders of the three numbers $a-1, b-1, c+2$ when divided by 3 are 2, 0, 1.
Thus, after each operation, the relationship that the remainders of the three numbers when divided by 3 are all different does not change.
Since the initial number of stones given is $19,8,9$, their remainders when divided by 3 are $1,2,0$, and if they become $12,12,12$, then their remainders when divided by 3 are all 0, therefore, no matter how the operations are performed, it is impossible to make the number of stones in the three piles all 12.