2. (18th Irish Mathematical Olympiad) Two people $A$ and $B$ form a 2005-digit number $N$ using only the digits $1,2,3,4,5$, each time choosing only one digit. $A$ goes first, then $A$ and $B$ take turns choosing digits. $A$ wins if and only if the number $N$ is divisible by 9. Who has a winning strategy?
2. $B$ has a winning strategy.
$B$ can adopt the following strategy:
If $A$ chose $k$ in the previous step, then $B$ chooses 6-k.
$N$ is divisible by 9 if and only if the sum of its digits is divisible by 9. When the 2004th digit is chosen, the sum of the first 2004 digits of $N$ is $6 \times \frac{2004}{2}=6012$, which is a multiple of 9. No matter how $A$ chooses the 2005th digit, it cannot make the sum of the digits of $N$ a multiple of 9. Therefore, $B$ wins.