Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Spring 2004 MATH025.1001
Nine Stones and a Balance
From the book, we discuss the first two story problems, on pages 5 and 6.
Although these `fun and games' problems appear frivolous, in the hands of a
mathematician they may expand into more interesting mathematics.
We illustrate this with story problem one, by squeezing more interesting
mathematics out of it.
We are given nine stones and a balance.
One of the stones is slightly heavier; the others are of identical weight.
Find a way to use a balance twice to determine which is the heavier stone.
Solution: Number the stones 1 through 9.
At the first weighing, put stones 1, 2, and 3 on the left scale, and put stones
4, 5, and 6 on the right scale.
Raise the balance.
There are three possible outcomes.
Left hand scale down (L).
Then the heavier stone is among 1, 2, and 3.
So for the second weighing, put stone 1 on the left scale, and stone 2 on the
right scale.
There are three possible outcomes.
Left hand scale down (LL).
Then 1 is the heavier stone.
Done.
The balance remains level (LM).
Then 3 is the heavier stone.
Done.
Right hand scale down (LR).
Then 2 is the heavier stone.
Done.
The balance remains level (M).
Then the heavier stone is among 7, 8, and 9.
So for the second weighing, put stone 7 on the left scale, and stone 8 on the
right scale.
There are three possible outcomes.
Left hand scale down (ML).
Then 7 is the heavier stone.
Done.
The balance remains level (MM).
Then 9 is the heavier stone.
Done.
Right hand scale down (MR).
Then 8 is the heavier stone.
Done.
Right hand scale down (R).
Then the heavier stone is among 4, 5, and 6.
So for the second weighing, put stone 4 on the left scale, and stone 5 on the
right scale.
There are three possible outcomes.
Left hand scale down (RL).
Then 4 is the heavier stone.
Done.
The balance remains level (RM).
Then 6 is the heavier stone.
Done.
Right hand scale down (RR).
Then 5 is the heavier stone.
Done.
As promised, let us make the problem slightly harder:
Suppose again that we have a balance, and nine stones one of them heavier.
Harder problem: Specify two weighings in advance, such that if we execute
them in a row, then the outcome surely will let us determine which stone is the
heavier one.
Solution: We adapt the solution to the original problem a bit.
First weighing (W1): Put 1, 2, and 3 on the left scale, and 4, 5,
and 6 on the right scale.
Second weighing (W2): Combine the three possible weighings of the
solution to the earlier problem into one: Put 1, 4, and 7 on the left scale,
and 2, 5, and 8 on the right scale.
Does this work, and how?
We sketch the argument.
In the table below we write down where we place each of the nine stones for
weighing W1 or weighing W2, where L means the stone is on the left scale, R
means the stone is on the right scale, and M means the stone is not on either
scale.
1
2
3
4
5
6
7
8
9
W1
W2
L
L
L
R
R
R
M
M
M
L
R
M
L
R
M
L
R
M
Observe that all nine columns are different.
For example, the column for stone 4 shows RL, while the column for stone 7
shows ML.
Assume for a moment that stone 7 is the heavier one.
We perform weighing W1 and weighing W2.
The first time the balance remains level because stone 7 is not used.
The second time the left scale moves down because stone 7 lies on the left
scale.
The two outcomes `say' that the stone must have an M as first letter of its
column, and the stone must have L as second letter of its column.
There is exactly one column that reads ML, and that is the column for stone 7.
Assume that we perform weighing W1 and weighing W2, and the first
time the left scale goes down, and the second time the right scale goes down.
Then the heavier stone must be one which has LR as it column.
There is exactly one such column, the one of stone 2.
So stone 2 is the heavier one.
And so on.
At this moment a true mathematician will see interesting patterns.
Let us do something seemingly silly: In the table above we write 0 for L, we
write 1 for R, and we write 2 for M.
So the table now reads:
1
2
3
4
5
6
7
8
9
W1
W2
0
0
0
1
1
1
2
2
2
0
1
2
0
1
2
0
1
2
The columns are now like numbers.
For example, the column for stone 4 reads 10, while the column for stone 7
reads 20.
Let us rewrite the table again, by now listing the two weighings combined as
two-digit numbers:
1
2
3
4
5
6
7
8
9
W1-2
00
01
02
10
11
12
20
21
22
These are like all the two-digit numbers from a planet of people with three
fingers.
On their planet there are exactly 3 times 3 equals 9 two-digit numbers.
In the table, each two-digit number corresponds with a weighing pattern.
For example, below stone 8 we see the number 21.
This means that for the first weighing stone 8 is not used (M), while for the
second weighing stone 8 is on the right scale (R).
With our new mathematical insights, we can now invent a new puzzle:
Suppose that we have a balance, and 27 stones one of them heavier.
Specify three weighings in advance, such that if we execute them in a row, then
the outcome surely will let us determine which stone is the heavier one.
Solution: We apply the insights that we gained with the previous
problem.
In the table below we encode where we place each of the 27 stones during each
of the three weighings, where the three-digit number below a stone number
corresponds with the three weighings as follows: A 0 means putting the stone on
the left scale; a 1 means putting the stone on the right scale; and a 2 means
not using the stone.
For example, the code 012 below stone 6 means that for weighing W1 we put stone
6 on the left scale (L); for weighing W2 we put stone 6 on the right scale (R);
and for weighing W3 we don't use stone 6 (M).
1
2
3
4
5
6
7
...
...
24
25
26
27
W1-3
000
001
002
010
011
012
020
...
...
212
220
221
222
Observe that we did not display all 27 columns.
This is only for lack of convenient space.
You are expected to be able to fill in the pattern where we now see the dots.
Obviously all stones have a different weighing pattern: On the
planet of three-fingered people there are exactly 3 times 3 times 3 equals 27
three-digit numbers.
At each of the three weighings, there will be 9 stones on the left
scale, and 9 stones on the right scale.
Assume, for example, that at weighing W1 the balance remains level,
at weighing W2 the balance remains level, and and weighing W3 the left scale
goes down.
Then the column code of the heavier stone must be MML using the older notation,
or 220 in the current notation.
So stone 25 is the heavier one.
Mathematicians also look for appropriate elegance.
In the proof above, they would be tempted to number the stones from 0 to 26,
rather than from 1 to 27.
The table would look slightly different:
0
1
2
3
4
5
6
...
...
23
24
25
26
W1-3
000
001
002
010
011
012
020
...
...
212
220
221
222
What earlier would have been referred to as stone 6, would now be called stone
5.
Naturally this renumbering does not change the solution in any essential way.
However, there is a notational advantage.
The codes in the second row now are the numbers that the three-fingered people
would write down when they refer to the numbers that we ten-fingered people
would use, right above it.
We would write 5 or 05 for the quantity five, while the three-fingered people
would write 12 or 012 for the same quantity five.
We would write 23 for the quantity twenty three, while the three-fingered
people would write 212 for the same quantity twenty three.
Our notation is called the decimal notation; theirs is called the
ternary notation.
A true mathematician still continues to generalize, using the insights
obtained.
Here is the final version, for now.
First some notation.
We write 3^n as short for 3 times 3 times 3 times ... times 3 a total of n
times.
So 3^1 equals 3, and 3^2 equals 9, and 3^3 equals 27, and 3^4 equals 81, and so
on.
It is considered appropriate elegance to let 3^0 equal 1 (why?).
Puzzle: Let n be a positive integer.
Suppose we have a balance, and 3^n stones one of them heavier.
Specify n weighings in advance, such that if we execute them in a row, then
the outcome surely will let us determine which stone is the heavier one.
Solution: A reasonably complete solution will look too abstract.
So we only say a few words about it.
First number the stones from 0 through (3^n)-1, and write their numbers in
ternary notation using n symbols.
Add paddings of zeros if necessary, to make it of length n.
Perform the n weighings as follows.
Let k be a number in the range from 1 to n.
At weighing k, put all the stones with kth symbol equal to 0, on
the left scale, and put all the stones with kth symbol equal to 1, on
the right scale.
If you understand the sentence above, then you will understand how to write an
acceptable argument to complete the solution.
Last updated: January 2004
Comments & suggestions:
wimr@mscs.mu.edu