There should be a problem with the answer on the first floor, right? This kind of question can be 1 or 2. If you want to take the last one, it must be a multiple of 3. Let the other party choose first. If the other party chooses 1, you choose 2 and the other party chooses 1. So let the opponent choose between 3, 6 and 9. If it is not a multiple of 3, for example, this question has 10 block, you should choose first, and the remainder of 10/3 is 1 block.

The general solution to this kind of problem, for example, there are m coins, and two people take any one between 1 and n in turn, and whoever gets the last one wins.

Then if m is a multiple of (1+N), the latter hand wins, and if it is not a multiple of (1+N), the former hand wins.