Professor Stewart's Hoard of Mathematical Treasures (54 page)

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Authors: Ian Stewart

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BOOK: Professor Stewart's Hoard of Mathematical Treasures
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Again, a complete proof can be given by induction, thinking about the general case of n monks, m of whom have blobs. I’ll spare you the details.
Pickled Onion Puzzle
There were 31 pickled onions.
Suppose there are a pickled onions to start with, b after the first traveller has eaten, c after the second traveller has eaten, and d after the third traveller has eaten. Then
b
= 2(
a
- 1)/3,
c
= 2(
b
- 2)/3,
d
= 2(
c
- 3)/3
which we rewrite as
a
= 3
b
/2 + 1,
b
= 3
c/
2 + 2,
c
= 3
d
/2 + 3
We are told that d = 6. Working backwards,
c
= 12, b = 20, a = 31.
Guess the Card
At each stage, when Whodunni picks up the cards he makes sure that the selected pile is sandwiched between the other two. As a result, the chosen card works its way to the middle of the stack of cards. So on the final deal it is the middle card in the chosen pile.
And Now with a Complete Pack
The first part of the trick is equivalent to asking which column the chosen card is in on the second deal. Knowing the column, and now being told the row, Whodunni can easily spot the card.
This trick is a bit transparent, but it can be dressed up to make it less obvious. It may be better performed with 30 cards, first dealing 6 rows of 5, then 5 rows of 6. It works with ab cards dealt in a rows of b, then b rows of a, for any whole numbers a and b.
Halloween = Christmas
Because 31 Oct = 25 Dec. That is, 31 in base-8 numerals (octal) = 25 in base-10 numerals (decimal). In base 8 notation, 31 means 3 × 8 + 1, and this is 25.
Rectangling the Square
I know at least two distinct solution, not counting rotations or reflections as distinct. The first was found by M. den Hertog, the second by Bertle Smith. In the first rectangles have sides 1 × 6, 2 × 10, 3 × 9, 4 × 7 and 5 × 8. In the second, the sides are 1 × 9, 2 × 8,3 × 6, 4 × 7 and 5 ×10.
Solutions found by den Hertog and Smith.
X Marks the Spot
From Abandon Hope Point: 113. From Buccaneer Bay: 99. From Cutlass Hill: 85.
Distances for Redbeard’s map.
The figure shows the three distances required: a, b, c. Of these, we know b = 99. Let s = 140 be the side of the square. Consider two further lengths x and y as shown. Then Pythagoras’s theorem tells us that
a
2
= x
2
+ (s - y)
2
= x
2
+ s
2
- 2sy + y
2
c
2
= y
2
+ (s - x)
2
= y
2
+ s
2
- 2sx + x
2
b
2
= x
2
+ y
2
The first step is to get rid of x and y. Subtract the third equation from the first and second, to obtain
2sy = s
2
+ b
2
- a
2
2sx = s
2
+ b
2
- c
2
Therefore
(s
2
+ b
2
- a
2
)
2
+ (s
2
+ b
2
- c
2
)
2
= 4s
2
(x
2
+ y
2
) = 4s
2
b
2
This is the fundamental relation between a, b, c and s.
Substitute known values
s
= 140 and
b
= 99, to get
(29,401 -
a
2
)
2
+ (29,401 -
c
2
)
2
= 27,720
2
= (2
3
× 3
2
× 5 × 7 × 11)
2
The hint now tells us that 29,401 -
a
2
and 29,401 -
c
2
are both multiples of 7. (The corresponding statement is false for the factors 2 and 5, but true for 3 and 11.) Considering the factor 7 (a similar trick works for 3 and 11) we observe that
29,401 = 4,200×7 + 1
so 1 -
a
2
and 1 -
c
2
are both multiples of 7. That is,
a
2
and
c
2
are of the form 7
k
+ 1 or 7
k
- 1 for suitable integers k.
Now it is a matter of trying the possible values for a, and seeing first whether
23,800
2
- (29,401 -
a
2
)
2
is a perfect square, and, if so, whether the corresponding c is a whole number. The business about multiples of 7 shortens the work because the only values for a that we need check are
1, 6, 8, 13, 15, 20, 22
and so on. We can stop when c becomes less than a, because then we’ll be trying the same calculations but with a and c interchanged.
For the 7
k
+ 1 case, we find
a
= 85,
c
= 113 when
k
= 12; there is also the solution
a
= 113,
c
= 85, interchanging a and c, when
k
= 16. There is no solution for the 7
k
- 1 case.
Since the instructions on the back of the map say that the nearest marker is C, we want c <
a
, so
a
= 113 and
c
= 85.
That’s one way to get the answer, but the mathematical story goes further.
This puzzle is a special case of the four distance problem: does there exist a square whose side is a whole number, and a point whose distances from the four corners of the square are all whole numbers? No one knows the answer. For a long time no one even knew whether three of those distances could be whole numbers.
We’ve already derived a relation between s, a, b and c:
(
s
2
+
b
2
-
a
2
)
2
+ (
s
2
+
b
2
-
c
2
)
2
= (2
bs
)
2
The fourth distance d (shown dotted on my diagram) must satisfy
a
2
+
c
2
=
b
2
+
d
2
J. A. H. Hunter discovered a formula giving some (but not all) solutions of the first equation:
a
=
m
2
- 2
mn
+ 2
n
2
b
=
m
2
+ 2
n
2
c
=
m
2
+ 2
mn
+ 2
n
2
s
2
= 2
m
2
(
m
2
+ 4
n
2
)
and observed that s is a whole number provided that we take
m
= 2(
u
2
+ 2
uv
-
v
2
)
n
=
u
2
- 2
uv
-
v
2
for whole numbers u and v.
The choice u = 2,
v
= 1 leads to s = 280,
a
= 170,
b
= 198,
c
= 226, and we can remove the factor 2 to get
s
= 140,
a
= 85,
b
= 99,
c
= 113. The fourth side here is
d
=
which is not a whole number, indeed not rational. In fact, it is known that in Hunter’s formula the fourth distance d can never be rational, so this formula alone won’t solve the four distance problem. However, there are solutions of the three distance problem that don’t arise from Hunter’s formula.
This tantalising problem has deep connections with Kummer surfaces in algebraic geometry: see Richard K. Guy,
Unsolved Problems in Number Theory.
Whatever’s the Antimatter?
Dirac’s electron equation looks like this:
where
ψ is the quantum wave-function of the electron
A
0
is the scalar potential of the electromagnetic field
A
is the vector potential of the electromagnetic field
ρ
= (
ρ
0
,
ρ
1
,
ρ
2
,
ρ
3
) is the momentum vector of the electron
σ
= (
σ
1
,
σ
2
,
σ
3
) is a list of three 4×4 spin matrices
r
= (
r
1
,
r
2
,
r
3
) is a list of three 4×4 matrices which anticommute
with σ
1
, σ
2
, σ
3
e is the charge on the electron
m is the mass of the electron
c is the speed of light
Got that? I include it only to make the point that the equation is far from obvious, and because it seemed a cheat to leave it out. Everyone writes out
E
= mc
2
, even Stephen Hawking,
55
and it’s wrong to discriminate against formulas on grounds of complexity. Dirac spends nearly four pages of his book The Principles of Quantum Mechanics explaining how he derived it, and most of the previous 250 pages setting up the ideas that go into it.

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