The
“Five Coin Puzzle” states:
There are
five coins lying in a straight row, all touching — three
20cent and two 10cent pieces.
When you
start, they are positioned so that the coin types alternate, with a
20cent coin on the left. Your goal is to arrange them in the fewest moves
so that the 20cent coins end up together on the left, then the 10cent
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 lefttoright 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 20cent coins alternating with three 20cent 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 sevencoin case take seven moves?
(Sorry, it will take 14 moves.) What about the ninecoin and elevencoin
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 10cent coins (not the total
number of coins).
