$8 \cdot 19$ The blackboard has the integer 1000 written on it, and there are 1000 matches on the table. Two people, A and B, play a game: A starts and they take turns. The person whose turn it is can take no more than 5 matches from the pile, or they can put no more than 5 matches back into the pile (but at the start, neither has any matches, and when putting matches back, they can only use matches they have previously taken), and they write down the current number of matches in the pile on the blackboard. If a person writes down a number that is already on the blackboard, they lose. Assuming both players use the correct strategy, who will win?
[Solution] B will win. B can divide the 1000 integers from 0 to 999 into 125 groups, each containing 8 consecutive integers, and then proceed as follows: When A writes a number from a certain group for the first time, this number is certainly not the smallest in that group, so B writes the smallest possible number in that group (at this point, B may take out 3, 4, or 5 matches), and mentally divides the remaining 6 numbers in the group into 3 pairs in ascending order. This way, when it's A's turn, he either keeps the number within this group or changes it to the previous group (i.e., the smaller group), but cannot revert to the next group. If it's the former, B can write the other number in the same pair; if it's the latter, B handles it as described above. Since B takes out at least 3 matches when dealing with the first number, even if B has to return 1 match each time a number is written in this group, it is still sufficient. It is clear that B can handle every situation with ease.