【Question 12】A and B take turns marking numbers from 1 to 17, with the rules: you cannot mark a number that has already been marked; you cannot mark a number that is twice a marked number; you cannot mark a number that is half of a marked number; the player who cannot mark a number loses. Now A marks 8 first, to ensure a certain win, B should mark _ next. $\qquad$
Analysis:
According to the condition “cannot mark the double of a marked number; cannot mark the half of a marked number,” we can categorize the numbers from 1 to 17 into the following four categories:
\begin{tabular}{|l|l|}
\hline Five elements & $(16,8,4,2,1)$ \\
\hline Three elements & $(12,6,3)$ \\
\hline Two elements & $(14,7) ;(10,5)$ \\
\hline One element & $(17) ;(15) ;(13) ;(11) ;(9)$ \\
\hline
\end{tabular}

From the table above, we can see that after Player A marks 8, 16 and 4 cannot be marked anymore. At this point, Player B's available choices become:
\begin{tabular}{|l|l|}
\hline Three elements & $(12,6,3)$ \\
\hline Two elements & $(14,7) ;(10,5) ;(2,1)$ \\
\hline One element & $(17) ;(15) ;(13) ;(11) ;(9)$ \\
\hline
\end{tabular}

Next, for Player B to win, they should choose a number that imposes the most restrictions on other numbers. Therefore, if Player B chooses 6 from the three elements (12, 6, 3), Player A's available choices become:
\begin{tabular}{|l|l|}
\hline Two elements & $(14,7) ;(10,5) ;(2,1)$ \\
\hline One element & $(17) ;(15) ;(13) ;(11) ;(9)$ \\
\hline
\end{tabular}

At this point, among the remaining 8 groups of numbers, no matter which one Player A chooses, the one who will choose the last group is Player B.
Therefore, Player B should mark 6.