Example 5: A and B take turns picking distinct numbers from $0,1, \cdots, 81$, with A starting first, and each person picks one number from the remaining numbers each time. After all 82 numbers are picked, let $A$ and $B$ be the sums of all numbers chosen by A and B, respectively. During the process of picking numbers, A hopes that the greatest common divisor (GCD) of $A$ and $B$ is as large as possible, while B hopes that the GCD of $A$ and $B$ is as small as possible. Under the best strategies for both A and B, find the GCD of $A$ and $B$ after all numbers are picked. ${ }^{[4]}$
【Analysis】Answer: The greatest common divisor of $A$ and $B$ is
41.
Let $d=(A, B)$ denote the greatest common divisor of $A$ and $B$.
Notice that, $d \perp (A+B)$.
And $A+B=0+1+\cdots+81=41 \times 81$, then
$d \mid 81 \times 41$.
First, prove: Player A can ensure $41 \mid d$.
Pair the numbers $0,1, \cdots, 81$ into 41 pairs:
$\{x, x+41\}(x=0,1, \cdots, 40)$.
Player A can ensure that exactly one number is chosen from each pair.
In fact, in the first round, Player A can choose any number $a$. If Player B chooses the paired number $a'$, then in the second round, Player A can choose any unselected number $b$; if Player B chooses a number $c \neq a'$ in the first round, then in the second round, Player A can choose the paired number $c'$ of $c$. This continues until the last round.
At this point, Player A has chosen exactly one number from each pair, so
$A \equiv 0+1+\cdots+40 \equiv 0(\bmod 41)$.
Thus, $41 \mid B \Rightarrow 41 \mid d$.
Next, prove: Player B can ensure $d$ is not a multiple of 3.
Pair the numbers $0,1, \cdots, 81$ into 41 pairs:
$$
\begin{array}{l}
\{x, x+39\}(x=0,1, \cdots, 38), \\
\{78,81\},\{79,80\} .
\end{array}
$$

Player B can ensure that exactly one number is chosen from each pair.
In fact, Player B only needs to choose the paired number of the number chosen by Player A in each round. Except for the last pair $\{79,80\}$, the numbers chosen by Player A and Player B are congruent modulo 3, while $79$ and $80$ are not congruent modulo 3, so $A-B$ is not a multiple of 3. Therefore, $d$ is not a multiple of 3.
Combining $d \mid 81 \times 41$, we know $d \mid 41$.
In summary, under the best strategies of both players, $d=41$.
【Summary】Notice that, $0+1+\cdots+81=3^{4} \times 41$, Player A can adopt an appropriate grouping to make the greatest common divisor $d$ a multiple of 41, while Player B can ensure $d$ is not a multiple of 3 through another grouping method, thus the balanced state of the game is $d=41$.