Example 7. There are three piles of stones with the numbers being 19, 8, and 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. Can the following be achieved after several such operations?
(1) The number of stones in the three piles are 2, 12, and 22,
(2) The number of stones in the three piles are all 12.

If the requirements can be met, complete it with the minimum number of operations. If not, explain the reason. (1989-1990 Guangzhou, Wuhan, Chongqing, Luoyang, Fuzhou Junior High School Mathematics Competition)
(1) It can be achieved. Since the original minimum pile has 8 stones, to make it 2, it requires at least six operations. In fact, it can succeed in six operations:
\[
\begin{aligned}
& (19,8,9) \rightarrow(21,7,8) \\
\rightarrow & (23,6,7) \rightarrow(22,5,9) \rightarrow(24,4,8) \\
\rightarrow & (23,3,10) \rightarrow(22,2,12) .
\end{aligned}
\]
(2) It cannot be achieved. Each operation changes the number of stones in each pile by either subtracting 1 or adding 2. Let's denote this 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 assume the remainders are \(0, 1, 2\), respectively. After the operation, the remainders of the three numbers \(a-1, b-1, c+2\) when divided by 3 will be \(2, 0, 1\). This shows that each operation does not change the property that the remainders of the three numbers when divided by 3 are all different.
\[
\begin{array}{l}
19=3 \times 6+1,8=3 \times 2+2, \\
9=3 \times 3+0,12=3 \times 4+0,
\end{array}
\]

Therefore, no matter how many operations are performed, it is impossible to make all three piles of stones equal to 12.