Players 1, 2, 3, ..., n are seated around a table and each has a single penny. Player 1 passes a penny to Player 2, who then passes two pennies to Player 3. Player 3 then passes one penny to Player 4, who passes two pennies to Player 5, and so on, players alternately passing one penny or two to the next player who still has some pennies. A player who runs out of pennies drops out of the game and leaves the table. Find an infinite set of numbers n for which some player ends up with all n pennies.
Let's deal separately with odd and even numbers of players. Suppose first that the number is even, so we may write it as 2m(2r + 1) + 2, where m >= 1 and r >= 0.
Call a round a set of moves in which each remaining player plays once, starting with the lowest numbered remaining player (the low player). Thus the first round always leaves the (new) low player with 3 or 4 coins and the other remaining players with 2 each. Call an eliminating round a round in which one or more players drop out.
The k-th eliminating round, for k <= m, leaves 2m-k(2r + 1) players, the low player with 2k + 2 coins, and the others with 2k each, and the low player about to pass 1 coin. This is an easy induction. It is easily ...
This solution is comprised of a detailed explanation to find an infinite set of numbers n for which some player ends up with all n pennies.