Professor Stewart's Hoard of Mathematical Treasures (57 page)
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Authors: Ian Stewart
Tags: #Mathematics, #General
BOOK: Professor Stewart's Hoard of Mathematical Treasures
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There is a corresponding position at 5/6 of a cycle, obtained by reflecting the diagram in the line ES.
How Dudeney Cooked Loyd• The children’s puzzle seems impossible on grounds of parity (odd/ even). All the numbers are odd, and six odd numbers must add to an even number, not 21. Gardner’s answer was to cook his own puzzle, turn the page upside down, and circle the three 6’s and the three 1’s that then appear. But a reader, Howard Wilkerson, circled each of the 3’s, one of the 1’s, and then drew a big circle round the other two 1’s (giving 11). I think this is a more elegant quibble-cook.
• Loyd’s construction leads to a rectangle which, while almost square, has sides of differing lengths. If the mitre is made from a square of side 1, giving it an area of
, then Loyd’s ‘square’ measures 6/7 horizontally by 7/8 vertically.
, then Loyd’s ‘square’ measures 6/7 horizontally by 7/8 vertically.• Dudeney’s five-piece solution, which is exact if the lengths are chosen correctly, looks like this:
Dudeney’s solution.
No one knows a four-piece dissection from the mitre to the square, and there probably isn’t one, but its existence has not been ruled out.
Cooking with WaterWhat I really should have said, to be super-cautious, was ‘you are not allowed to pass the connections through a house or a utility company building’. As David Uphill pointed out, there is a way to solve the problem if the words are interpreted literally, even with pipes forbidden to pass through houses. I’ve modified his suggestion slightly to fit my question more closely. It uses two big water tanks to pass the water connections through two of the houses. No connecting pipe even enters a house, let alone passes through it.
The utilities puzzle quibble-cooked.
Hmm . . . If you feel that a tank used like this is just a large pipe by another name, which was my first reaction, then this layout doesn’t fit the conditions. That’s why I consider it a quibble-cook. But it’s an ingenious one and deserves to be better known.
Calculator Curiosity 2When you multiply 0588235294117647 by 2, 3, 4, 5, ... , 16, the same sequence of digits arises, in the same cyclic order. That is, you have to start at a different place, and when you get to the end, you continue from the beginning. Specifically,
0588235294117647 × 2 = 1176470588235294
0588235294117647 × 3 = 1764705882352941
0588235294117647 × 4 = 2352941176470588
0588235294117647 × 5 = 2941176470588235
0588235294117647 × 6 = 3529411764705882
0588235294117647 × 7 = 4117647058823529
0588235294117647 × 8 = 4705882352941176
0588235294117647 × 9 = 5294117647058823
0588235294117647 × 10 = 5882352941176470
0588235294117647 × 11 = 6470588235294117
0588235294117647 × 12 = 7058823529411764
0588235294117647 × 13 = 7647058823529411
0588235294117647 × 14 = 8235294117647058
0588235294117647 × 15 = 8823529411764705
0588235294117647 × 16 = 9411764705882352
0588235294117647 × 3 = 1764705882352941
0588235294117647 × 4 = 2352941176470588
0588235294117647 × 5 = 2941176470588235
0588235294117647 × 6 = 3529411764705882
0588235294117647 × 7 = 4117647058823529
0588235294117647 × 8 = 4705882352941176
0588235294117647 × 9 = 5294117647058823
0588235294117647 × 10 = 5882352941176470
0588235294117647 × 11 = 6470588235294117
0588235294117647 × 12 = 7058823529411764
0588235294117647 × 13 = 7647058823529411
0588235294117647 × 14 = 8235294117647058
0588235294117647 × 15 = 8823529411764705
0588235294117647 × 16 = 9411764705882352
For my second question:
0588235294117647 × 17 = 9999999999999999
The source of this remarkable number is the decimal expansion of the fraction 1/17, which is
repeating indefinitely.
Which is Bigger?0.0588235294117647 0588235294117647
0588235294117647 . . .
0588235294117647 . . .
By direct calculation, e
π
= 23.1407 whereas π
e
= 22.4592. So e
π
> π
e
.
π
= 23.1407 whereas π
e
= 22.4592. So e
π
> π
e
.
Actually, there’s a more general result: e
x
≥ x
e
for any number x ≥ 0, and equality holds if and only if
x
= e. So, not only is e
π
> π
e
, but e
2
> 2
e
, e
3
> 3
e
, e
4
> 4
e
, e
√2
> (√2)
e
, and indeed e
999
> 999
e
. The simplest proof uses calculus, and here it is for anyone who wants to see the details. It also helps to explain why e
π
and π
e
are so close together.
x
≥ x
e
for any number x ≥ 0, and equality holds if and only if
x
= e. So, not only is e
π
> π
e
, but e
2
> 2
e
, e
3
> 3
e
, e
4
> 4
e
, e
√2
> (√2)
e
, and indeed e
999
> 999
e
. The simplest proof uses calculus, and here it is for anyone who wants to see the details. It also helps to explain why e
π
and π
e
are so close together.
Let
y
=
x
e
e
-x
, where x ≥ 0. We find the stationary points (maxima, minima, and the like) by setting d
y
/d
x
= 0. Now
which is zero at
x
= 0 and
x
= e, and nowhere else. The value of y at
x
= 0 is
y
= 0, which is clearly a minimum; the value at
x
= e is
y
= 1. This is in fact a maximum. To see why, calculate the second derivative
at
x
= e. This works out as -1, which is negative, so
x
= e is a maximum, and the maximum value of y is 1.
y
=
x
e
e
-x
, where x ≥ 0. We find the stationary points (maxima, minima, and the like) by setting d
y
/d
x
= 0. Now
x
= 0 and
x
= e, and nowhere else. The value of y at
x
= 0 is
y
= 0, which is clearly a minimum; the value at
x
= e is
y
= 1. This is in fact a maximum. To see why, calculate the second derivative
x
= e. This works out as -1, which is negative, so
x
= e is a maximum, and the maximum value of y is 1.
Therefore
x
e
e
-
x
≤ 1 for all x ≥ 0, with equality only at the unique maximum
x
= e. Multiply both sides by e
x
, to get
for all x ≥ 0, with equality only when
x
= e. Done!
x
e
e
-
x
≤ 1 for all x ≥ 0, with equality only at the unique maximum
x
= e. Multiply both sides by e
x
, to get
x
e
≤ e
x
e
≤ e
x
x
= e. Done!
The graph of the function
y
=
x
e
e
-x
has a single peak at
x
= e, and tails off towards zero as x increases to infinity.
y
=
x
e
e
-x
has a single peak at
x
= e, and tails off towards zero as x increases to infinity.
Graph of
y
=
x
e
e
-x
.
y
=
x
e
e
-x
.
This helps to explain why e
π
and π
e
are so close together for it not to be immediately obvious which is bigger. The graph also shows that
if a number x is reasonably close to e, then
x
e
e
-x
is close to 1, so
x
e
is close to e
x
. For example, if x lies between 1.8 and 3.9, then
x
e
is at least 0.8e
x
. In particular, this holds for
x
= π.
Colorado Smith and the Solar Templeπ
and π
e
are so close together for it not to be immediately obvious which is bigger. The graph also shows that
if a number x is reasonably close to e, then
x
e
e
-x
is close to 1, so
x
e
is close to e
x
. For example, if x lies between 1.8 and 3.9, then
x
e
is at least 0.8e
x
. In particular, this holds for
x
= π.
The division shown solves the puzzle. So does its reflection in the diagonal.
Four regions of the same shape, each containing a sun-disc.
Why Can’t I Add Fractions Like I Multiply Them?The short response is that we can’t add fractions that way because we don’t get the right answer! Since
is nearly
, and so is
, then when we add them the result must be at least
. But
is less than
because half of 12 is 6. The error is even more glaring when we try it on
+
, because
makes no sense: since
it tells us that
.
is nearly, and so is, then when we add them the result must be at least. Butis less thanbecause half of 12 is 6. The error is even more glaring when we try it on+, becauseit tells us that.Other books
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