8.6 Two people, A and B, are playing a math game. They start by writing down the numbers $0,1,2, \cdots, 1024$. A then crosses out 512 numbers, followed by B crossing out 256 numbers, then A crosses out 128 numbers, and so on. After the 10th step, when B crosses out 1 number, only two numbers remain. At this point, B must pay A a score, the value of which is the difference between the two remaining numbers. What strategy should A use to maximize the score, and what strategy should B use to minimize the score? If both play optimally, how many points will B pay to A?
[Solution] Since the larger the difference between the last two numbers is, the more advantageous it is for Player A, therefore, every time Player A crosses out numbers, the minimum difference between any two remaining numbers should be as large as possible. Every time Player B crosses out numbers, the maximum difference between any two remaining numbers should be as small as possible.

Player A first crosses out all 512 odd numbers. Then, regardless of which numbers Player B crosses out, Player A, in the second round, first crosses out all even numbers that are not multiples of 4. If not enough, Player A can randomly cross out some more numbers until the required number is reached. This continues until the fifth round, when Player A crosses out all numbers that are not multiples of 32. Thus, the difference between any two remaining numbers is no less than 32, so Player A will score at least 32 points.

When it is Player B's turn for the first time, Player A has already crossed out 512 numbers. Player B only needs to recognize which of the first 512 numbers and the last 512 numbers has more remaining numbers. Player B can then cross out all the numbers from the part with fewer remaining numbers, and if not enough, continue crossing out until the required number is reached. Thus, the maximum difference between any two remaining numbers does not exceed 512. Each subsequent time it is Player B's turn, the same strategy is followed. Each time, the maximum difference between any two remaining numbers can be reduced by at least half, until the 10th step when 1 number is crossed out, leaving the difference between any two remaining numbers no more than 32.
Thus, when both players play correctly, the score that Player B has to pay to Player A is 32.