“Five Coin Puzzle” states:
five coins lying in a straight row, all touching — three
20-cent and two 10-cent pieces.
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.
starting with this:
10 20 10 20,
to get to this:
20 20 10 10.
the rules for moving:
must be slid, not picked up,
adjacent coins (always of different types) must be moved each time,
least one of the pair moved must touch a third coin after the move,
coins moved must not change their left-to-right orientation
20 10 must not be twisted to 10 20).
will see from the diagrammatic solution below, as few as five moves are
required. The fourth and fifth moves are the crucial ones.
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¢.
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.)
a pattern here? Yes there is, and for those of you who remember any
algebra, this is the elegant formula that embodies it:
of moves = n(n+1)(2n+1)/6
where n is the number of 10-cent coins (not the total
number of coins).