5 Coins Puzzle
[ Issue 24 ]

# The Five Coins Puzzle

There are five coins lying in a straight row, all touching — three 20-cent coins and two 10-cent coins.  When you start, they are positioned so that the coin types alternate, with a 20-cent coin on the left.  Your goal is, in as few moves as possible, to arrange them (again in a straight line), so that the 20-cent coins end up together on the left, then the 10-cent coins, all touching, according to these rules:

coins must be slid, not picked up;
two adjacent coins (always of different types) must be moved each time;
at least one of the pair moved must touch a third coin after the move;
the coins moved must not change their left-to-right orientation
(e.g., 20 10 must not be twisted to 10 20).

Look here for the solution — but only after you've given up.

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 To Phrase a Coin or Two — Tony Rogers Copyright The “Five Coin Puzzle” states: There are five coins lying in a straight row, all touching — three 20-cent and two 10-cent pieces. When you start, they are positioned so that the coin types alternate, with a 20-cent coin on the left. Your goal is to arrange them in the fewest moves so that the 20-cent coins end up together on the left, then the 10-cent coins, again with no gaps between. Thus, starting with this: 20 10 20 10 20, you have to get to this: 20 20 20 10 10. Here are the rules for moving: coins must be slid, not picked up, two adjacent coins (always of different types) must be moved each time, at least one of the pair moved must touch a third coin after the move, the coins moved must not change their left-to-right orientation (e.g. 20 10 must not be twisted to 10 20). As you will see from the diagrammatic solution below, as few as five moves are required. The fourth and fifth moves are the crucial ones. Readers interested enough can extend the puzzle to any similar combination — say, four 20-cent coins alternating with three 20-cent coins, or five 20¢ and four 10¢, or even six 20¢ and five 10¢. And then, if you’re really enthused, you might like to predict how many moves it will take for any given case. Will the seven-coin case take seven moves? (Sorry, it will take 14 moves.) What about the nine-coin and eleven-coin cases? (Answer: 30 and 55 moves, respectively.) Is there a pattern here? Yes there is, and for those of you who remember any algebra, this is the elegant formula that embodies it: number of moves = n(n+1)(2n+1)/6 where n is the number of 10-cent coins (not the total number of coins). Contents  Read Next Item  Read Previous Item