. These problems just blow me away. I do need to get the hang of them one day I hope. Consider the following two-player game. Player One places between 50 and 60 toothpicks on the table. Player Two picks up at least one and no more than nine of the toothpicks. Player One then picks up at least one and no more than nine of the remaining toothpicks. Play alternates in this way until there are no more toothpicks remaining on the table. The person who picks up the last toothpick from the table loses. If you are Player One, how many toothpicks should you place on the table to guarantee a win for yourself?
In these types of puzzles, it is easier to determine the strategy if you work from the end to the beginning. If there are 2 toothpicks left, you take 1 forcing the other player to take the last one. How do you plan it so that you can ensure that there are 2 left when it is your turn?
Actually when it is your last turn - you should have 10 toothpicks - now you can pick 9 tooth-picks and force the last guy to pick the last one. So your opponent must face (before you) > 10 tooth picks (that way s/he cannot leave 1 on the table by picking 9 tooth-picks)
I guess I botched the orignal response. Let's try again. If you want to force the other player to take the last toothpick, you obviously need them to have only one to choose at the end. To get to that position, if there were 11 left and the opponent were picking, their pick plus yours would total 10. Continuing upward you would obviously want to arrive at a total of 51 for the first pick of the opponent. So any number above 51 up to 59 would do. Your first pick would bring the total to 51 and then you would just pick sufficient toothpicks so that each round (yours and theirs) would total 10 eventually winning the game.