Professor Stewart's Hoard of Mathematical Treasures (61 page)

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Authors: Ian Stewart

Tags: #Mathematics, #General

BOOK: Professor Stewart's Hoard of Mathematical Treasures
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It’s About Time
Crossnumber solution.
Do I Avoid Kangaroos?
I avoid kangaroos.
Write the conditions symbolically, as on page 275. Let
A = avoided by me
C = cat
D = detested by me
E = eats meat
H = in this house
K = kangaroos
L = loves to gaze at the moon
M = kills mice
P = prowls by night
S = suitable for pets
T = takes to me
Then with ⇒ meaning ‘implies’ and ¬ meaning ‘not’, the statements (in order) become
H ⇒ C, L ⇒ S, D ⇒ A, E ⇒ P, C ⇒ M
T ⇒ H, K ⇒ ¬ S, M ⇒ E, ¬ T ⇒ D, P ⇒ L
Now we appeal to the laws of logic that I mentioned on page XXX:
X ⇒ Y is the same as ¬ Y ⇒ ¬ X
If X ⇒ Y ⇒ Z, then X ⇒ Z
Using these laws, we can rewrite these conditions as
¬ A ⇒ ¬ D ⇒ T ⇒ H⇒ C ⇒ M ⇒ E ⇒ P ⇒ L ⇒ S ⇒ ¬ K
so that ¬ A ⇒ ¬ K, or equivalently, K ⇒ A.
Therefore I avoid kangaroos.
The Klein Bottle
To cut a Klein bottle into two Möbius bands, slice it lengthwise, cutting through the ‘handle’ of the bottle and the body along the plane of mirror symmetry. A little thought shows that the two pieces are Möbius bands.
Cutting a Klein bottle into two Möbius bands.
Accounting the Digits
This is the only number that works.
As Long as I Gaze on Laplacian Sunrise
If Laplace’s figures are correct - which is highly debatable - the probability that the Sun will always rise is zero.
The probability of the Sun rising on day n is (n - 1)/n. So:
• The probability of the Sun rising on day 2 is
• The probability of the Sun rising on day 3 is
• The probability of the Sun rising on day 4 is
and so on. Therefore
• The probability of the Sun rising on days 2 and 3 is
• The probability of the Sun rising on days 2, 3 and 4 is
• The probability of the Sun rising on days 2, 3, 4 and 5 is
and so on. The pattern is clear (and easy to prove): the probability of the Sun rising on all of the days 2, 3, . . . , n is 1/n. As n becomes arbitrarily large, this tends to 0.
Strictly for Calculus Buffs
For the details, see:
D. P.Dalzell, ‘On 22/7’, Journal of the London Mathematical Society, vol. 19 (1944), pp. 133-34.
Stephen K. Lucas, ‘Approximations to π derived from integrals with nonnegative integrands’, American Mathematical Monthly, vol. 116 (2009), pp. 166-72.
The Statue of Pallas Athene
The statue contained 40 talents of gold.
The four fractions add up to give
so what’s left is
. Since this requires 9 talents, the total must have been 40 talents.
Calculator Curiosity 3
6 × 6 = 36
66 × 66 = 4356
666 × 666 = = 443556
6666 × 6666 = 44435556
66666 × 66666 = 4444355556
666666 × 666666 = 444443555556
6666666 × 6666666 = 44444435555556
66666666 × 66666666 = 4444444355555556
Completing the Square

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