This story leads to the following general mathemati-
cal problem:
Suppose Nparticipants, numbered 1, 2, …, N,
sit in a circle and begin counting off every nth
person. Each person so selected is called
“out” and leaves the game. Counting contin-
ues until the last participant is declared the
winner of the game. Given the size of the
group, is it possible to predict beforehand
who the winner will be? Assume the count
begins with player 1.
The following table shows the position of the winner,
counting every third person, for games involving N= 1
through N = 15 people:
N1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
W(N) 1 2 2 1 4 1 4 7 1 4 7 10 13 2 5
The numbers arising can be analyzed as follows: A
game with Nplayers becomes a game with N–1 players
as soon as the first person is called out, except that the
start of the N–1 game is moved three places along.
Thus, if W(N) denotes the winner of an N-player game,
we have W(N) = W(N–1) + 3. One may need to adjust
this formula, however, to take into account possible
“counting around the full circle.” For example:
Clearly W(1) = 1.
Adding 3 gives W(2) = 4. But, counting around
the circle, we see that position four in a two-
player game is really player 2. Thus W(2) = 2.
Adding three gives W(3) = 5. But, counting
around the circle, we see that position five in a
three-player game is really player 2. Thus W(3)
= 2,
and so on.
In this way we can compute all the entries of the table
without having to perform the game. Other versions of
the game can be analyzed similarly.
Jourdain’s paradox In 1913 French mathematician
Philip Jourdain proposed a variation of the famous
LIAR
’
S PARADOX
, now sometimes called the card para-
dox. On one side of a card is printed: “The statement
on the other side of this card is true,” and on the
reverse side: “The statement on the other side of this
card is false.” Neither sentence can be true or false, for
the statement on the reverse side implies the opposite.
It is worth pointing out that if, instead, both sen-
tences read: “The statement on the other side of this
card is false,” then no paradox occurs; it is simply the
case that just one of the sentences is actually false, and
the other is true. If, on the other hand, both sentences
read: “The statement on the other side of this card is
true,” then both statements could be true, or both
could be false.
See also
SELF
-
REFERENCE
.
jug-filling problem A famous category of decanting
problems contains puzzlers of the following ilk:
Given a 3-gallon jug and a 5-gallon jug (with-
out any markings), is it possible to draw
exactly 1 gallon of water from a well?
As one is not given the means to measure the exact
contents of a partially filled jug, we are left with only
three allowable maneuvers:
1. Completely fill an empty jug from the well
2. Completely empty a full jug into the well
3. Pour water from one jug into another, completely
filling or emptying one jug in the process
This particular problem has an easy solution: com-
pletely fill the 3-gallon jug and pour its contents into
the 5-gallon jug. Refill the 3-gallon jug and pour part
of its contents to fill up the 5-gallon jug. This leaves
(and we are certain of this) precisely 1 gallon of water
in the 3-gallon jug and 5 gallons of water in the sec-
ond. Now empty the 5-gallon jug.
In this solution the 3-gallon jug was filled twice
and the 5-gallon jug emptied once. If we count +1 each
time a jug is filled and –1 for each time it is emptied,
the solution described above can thus be represented by
the equation:
2 × 3 + (–1) × 5 = 1
Surprisingly any solution to the D
IOPHANTINE EQUA
-
TION
x × 3 + y × 5 = 1 corresponds to a solution to
the jug-filling problem. (For example, (–3)×3 + 2×5 =
1 represents a solution in which the 5-gallon jug is
jug-filling problem 289