Ian Stewart
Page 26
• 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
The stated conditions do not require you to use the integers 1-9, and in fact no solution exists if you do, because then the even numbers must go in the corners. But by using fractions, you can solve the puzzle. The figure shows the traditional solution, perhaps the simplest, but there are infinitely many others, even if you restrict the entries to positive numbers.
An unorthodox magic square.
The Look and Say Sequence
The rule of formation is best stated in words. The first term is ‘1’, which can be read as ‘one one’, so the next term is 11. This reads ‘two ones’, leading to 21. Read this as ‘one two, one one’ and you see where 1211 comes from, and so on.
Conway proved that if L(n) is the length of the nth term in this sequence, then
L(n)≈(1.30357726903...)n
where 1.30357726903 . . . is the smallest real solution of the 71stdegree polynomial equation
x71 — x69 —2x68 —x67 + 2x66 + 2x65 —x63 —x62 —x61 —x 60
+ 2x58 + 5x57 + 3x56 —2x55 —10x54 —3x53 —2x52 + 6x51
+ 6x50 + x49 + 9x48 —3x47 —7x46 —8x45 —8x44 + 10x43
+ 6x42 + 8x41 —5x40 —12x39 + 7x38 —7x37 + 7x36 + x35
—3x34 + 10x33 + x32 —6x31 —2x30 —10x29 —3x28 + 2x27
+ 9x26 —3x25 + 14x24 —8x23 —7x21 + 9x20 + 3x19 —4x18
—10x17 — 7x16 + 12x15 + 7x14 + 2x13 — 12x12 — 4x11 — 2x10
+ 5x9 + x7 — 7x6 + 7x5 — 4x4 + 12x3 — 6x2 + 3x — 6
= 0
I don’t claim this is obvious.
The Millionth Digit
The millionth digit is 1.
The numbers 1-9 occupy the first 9 positions.
The numbers 10-99 occupy the next 2 × 90 = 180 positions.
The numbers 100-999 occupy the next 3 × 900 = 2,700 positions.
The numbers 1,000-9,999 occupy the next 4 × 9,000 = 36,000
positions.
The numbers 10,000-99,999 occupy the next
5 × 90,000 = 450,000 positions.
At this stage, we have reached the 488,889th digit altogether. Since 1,000,000 — 488,889 = 511,111, we want the digit in the 511,111th place, in the block that starts 100,000-100,001-100,002- and so on. Since these are grouped in sixes, we work out 511,111/6 = 85,185 . Therefore we are seeking the first digit of the 85,186th 6-digit block. That block must be 185,185, and its first digit is 1.
Piratical Pathways
Redbeard’s bank is at 19 Taxhaven Street.
Calculating the number of paths.
The number is small enough that you can just list the possible paths, but there’s a systematic method for this sort of question. The diagram shows the same map; I’ve removed superfluous connections that can never be used to keep thing simple, but it makes no difference to the method or the result if you leave them in.
I’ve written numbers beside the letters. These numbers tell us how many ways there are to reach that particular letter, and we calculate them in turn: P, the three I’s, the four R’s, the three A’s, the three T’s, and the final E.
• Start by writing 1 next to the P.
• There is exactly one path from the P to each of the I’s, so we write 1 next to each I.
• Look at each R in turn, see which I’s connect to it, and add up the numbers beside those letters. Here one of the R’s is connected to only one I, numbered 1, so it also gets the number 1. The other three are connected to two I’s, numbered 1, so they are given the number 1 + 1 = 2.
• Next, move on to the A’s. The leftmost A is connected to three R’s: one numbered 1 and two numbered 2, so we give that A the number 1 + 2 + 2 = 5. And so on.
• Continuing in this way, we eventually reach the final E. The T’s that connect to it are numbered 7, 7 and 5. So we number E with 7 + 7 + 5 = 19. And that’s the number of ways to get to E.
Trains That Pass in the Siding
Yes, they can pass each other - however long the trains might be.
Here’s how.
1. Initially, each train is on its side of the siding.
2. Train B backs away to the right. Train A runs right, past the siding, backs on to it, drops off four coaches, returns to the main line going right, and backs off to the far left.
3. Train A moves past the siding, going left, and joins the main part of train B.
4. Trains A+B move right, going along the siding, pick up the four coaches, and return to the main line on the right of the siding.
5. Then they back up on to the siding, drop off four more coaches, and return to the main line on the right of the siding.
6. The main part of the combined trains A+B moves left along the main line until it clears the siding.
7. Again trains A+B move right, going along the siding, pick up the four coaches, and return to the main line on the right of the siding.
8. Then they back up on to the siding, drop off one coach and locomotive A, and return to the main line on the right of the siding.
9. The main part of the combined trains A+B moves left along the main line until it clears the siding.
10. Finally, A+B goes right along the siding, rejoins with locomotive A and its coach. Then the trains split apart and each continues on its way.
The same method works no matter how long the trains are, provided the siding can contain at least one coach or locomotive.
Squares, Lists and Digital Sums
The next such sequence is
99,980,001, 100,000,000, 100,020,001, 100,040,004 100,060,009, 100,080,016, 100,100,025
which are the squares of the numbers 9,999-10,005.
A good place to look is the squares of numbers 100 ... 00, 100 . . . 01, 100 . . . 02, 100 . . . 03, 100 . . . 04, 100 . . . 05, which have lots of zeros, while the few remaining digits give digital sums that are the squares 1, 4, 9, 16, 16, 9. To extend this list of six consecutive squares to seven, we have to look at 999 . . . 9 and 100 . . . 06. The digits of 992 sum to 18, not a square; those of 9992 sum to 27, also not a square. But the digits of 99992 sum to 36, a square. Looking at the other end, the digits of 1062, 1,0062, and 10,0062 sum to 13, not a square.
To rule out anything in between 152 and 99992, we just have to find a sequence of squares of numbers that differ by at most 6, whose digits sum to non-squares. For example,
162 = 256 with digit-sum 13
192 = 361 with digit-sum 10
(202, 212, and 222 have square digit-sums so I can’t use those)
252 = 625 with digit-sum 13
292 = 841 with digit-sum 13
and so on. I’m sure there must be short cuts, and a computer can quickly check all possibilities in that range.
No one seems to know whether eight consecutive squares can all have square digit-sums.
Match Trick
It’s easy if you let the edges of the triangles overlap.
Slicing the Cake
You can get at most 16 pieces. Here’s one way to do it:
How to get 16 pieces with five cuts.
In general, using n cuts, the maximum number of pieces is + 1, the nth triangular number plus 1.
Sliding Coins
Like this.
Note that on the third move, coin 5 just slides out from between coins 2 and 4. The arrow doesn’t show the direction of the move, just which coin goes where.
Beat That!
One of the dice stopped rolling with a 6 uppermost. The other hit a rock, split in two, and the two pieces showed a 6 and a 1. So Olaf scored 13, beating the King of Sweden’s feeble 12.
In mathematical circles this kind of thing is called ‘expanding the state space’. That is, extending the range of possible outcomes. It is one of the reasons why mathematical models never match reality perfectly.
In gambling circles it’s called ‘rigging the dice’.
In political circles it’s called ‘politics’.
I learned this story from Ivar Ekeland’s The Broken Dice.
Euclid’s Puzzle
The donkey was carrying five sacks and the mule was carrying seven.
Suppose the donkey carries x sacks and the mule y. Then the mule tells us two things:
y + 1 = 2(x - 1)
x + 1 = y - 1
The second equation tells us that y = x + 2. Now the first equation tells us that x + 3 = 2x - 2, which implies that x = 5. So y = 7.
The Infinite Monkey Theorem
Each character has a probability of 1/36 of being selected on any given throw, so on average it takes 36 throws to get any given character. To get DEAR SIR, with 8 characters including the space, you need
36 × 36 × 36 × 36 × 36 × 36 × 36 × 36 = 368 = 2,821,109,907,456
throws. The complete works of Shakespeare would need 365000000 throws, which is roughly 102385606. If the monkey typed 10 characters per second, which is faster than a really good typist, it would take roughly 3 × 102385597 years to complete the task.
Dan Oliver ran a computer program in 2004, and after a simulated time of 42 octillion years, the digital monkey typed
VALENTINE. Cease toIdor:eFLP0FRjWK78aXzVOwm)-’;8.t
The first 19 letters appear in The Two Gentlemen of Verona. Similar results are reported at: en.wikipedia.org/wiki/Infinite_monkey_theorem
Snakes and Adders
If a square board with no missing corner has at least one even side, which is the case here, then the first player can always win. Imagine the board tiled with dominoes: 2×1 rectangles. Any tiling will do - for example, this one on the complete 8×8 board.
Domino grid for winning strategy on a complete board.
Whatever the second player does, other than losing by hitting the edge, the first player can always find a move for which the snake ends in the middle of a domino. That can’t be a losing move, since it hasn’t hit an edge, and eventually the second player runs out of options.
If both sides of the board are odd, the second player wins with a similar strategy, using a tiling that omits the initial square with + on it.
Cutting out the bottom right corner square interferes with these domino strategies. The first player can’t cover the modified board with dominos, since it has an odd number of squares. The second player can’t cover all but the starting square with dominos, either, but that’s less obvious since those squares are even in number. But if you imagine the usual chessboard pattern of alternating black and white squares, there are 30 of one colour and 32 of the other, again omitting the starting square marked with a +. However, any domino covers one square of each colour, so a tiling would have to cover 31 of each.
There must still be a winning strategy for one or other player, because this is a finite game and can’t be drawn. But it’s no longer clear what that strategy might be, or who should win.
Powerful Crossnumber
Across Down
2 7776 = 65 1 512 = 29
5 128 = 27 2 784 = 282
6 27 = 33 3 729 = 36
7 4096 = 212 4 676 = 262
Magic Handkerchiefs
If you’ve followed the instructions correctly, the two handkerchiefs will miraculously separate.
If not, try again and be more careful.
The mathematical aspect is topological: when you convert the handkerchiefs to closed loops by clasping their ends, the loops are not linked. They just look as though they are.
Digital Century Revisited
She should write down:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 × 9
to avoid losing her money.
A Century in Fractions
The solution Dudeney asked for is 3 .
The others, including the example I gave when posing the puzzle, are:
Proof That 2 + 2 = 4
This proof isn’t a joke - it’s how you do it in courses on the foundations of mathematics. It looks like the hard part is to prove the associative law, which of course I assumed. Actually the hard part is defining numbers and addition. That’s why Russell and Whitehead needed 379 pages to prove the simpler theorem 1 + 1 = 2 in Principia Mathematica. After that, 2 + 2 = 4 is a doddle.
Slicing the Doughnut
You can get nine pieces. Here are two possible ways.
Two ways to get nine pieces with three cuts.
Tippe Top Twister
When the tippe top turns over, it still spins clockwise, looking down from above.
If you imagine the tippe top spinning in space, unsupported, and then turn it upside down, it would be spinning anticlockwise. But that’s not what the top does. As it starts to tip over, the end of the stalk hits the ground, and itself starts spinning. This changes the behaviour when the top finally ends up standing on its stalk.
The physical point here is angular momentum (see page 30), a quantity associated with a rotating body, roughly equal to its mass times its rate of spin round an appropriate axis. The angular momentum of a moving body is conserved - does not change - unless some force such as friction is acting.
Most of the angular momentum of the tippe top arises from the spherical part, not the stalk. Since angular momentum must be conserved - subject to a small loss through friction - the final direction of spin has to be the same as the initial direction. Friction just slows the spin down a little.
Juniper Green
No other opening moves in JG-40 can force a win. There is a similar strategy for JG-100, which I’ll explain in a moment, and Mathophila wins. As for JG-n, let’s save that for a bit.
This game seems first to have appeared in a course on number theory given by the great mathematical physicist Eugene Wigner at Princeton in the late 1930s. More recently it was reinvented independently by Rob Porteous, a schoolteacher, to teach young children about multiplication and division. Porteous’s pupils discovered that Mathophila can always win JG-100 if (and only if) she starts with 58 or 62.
The analysis depends on prime numbers, which divide into several kinds: big primes greater than 100/2 (53, 59, 61, 67, 71, 73, 79, 83, 89, 97), medium large primes between 100/2 and 100/3 (37, 41, 43, 47), medium primes between 100/3 and 100/4 (29, 31), small primes less than 100/4 but not too small (17, 19), and very small primes (2, 3, 5, 7, 11). The winning openings are twice the medium primes. For example, here is the analysis if Mathophila opens with 58.
In his number theory course, Wigner solved the whole thing by providing a criterion for winning play in all cases. The answer for JG-n depends on whether the various powers of primes that occur in the factorisation of n! are odd or even.
Slade’s Braid
Slade’s braid trick relies on a topological curiosity: his leather strip can be deformed, in perfectly ordinary 3D space, to end up braided. So all he had to do was fiddle with it under the table until it reached the braided state. I’ve separated the three strips in my picture to make the method clearer.
This s
equence of moves introduces six extra crossings of the three strips, so by repeating it you can make really long braids.
Slade had a colourful career, and was exposed as a fraud by the Seybert Commission in 1885. See: www.answers.com/topic/henry-slade
Avoiding the Neighbours
How to keep the neighbours apart.
This is the only solution, aside from rotations and reflections.
A Rolling Wheel Gathers No Speed
The point on the rim of the wheel, where it touches the road, has instantaneous velocity zero. The ‘no slip’ condition means that the horizontal component of velocity at this point is 0; the ‘no bounce’ condition means that the vertical component here is also 0.
This is interesting, because the point concerned moves along the road at 10 metres per second. But as it moves, the point on the road corresponds to different points on the wheel. And the question was about points on the wheel, not points on the road.
A more detailed analysis using calculus shows that this is the only such point. Assume the wheel starts with its centre at (0, 1) and rolls along the x-axis to the right. Place a black dot on the rim, starting at the origin (0, 0) at time 0.
After time t the circle has rolled 10t to the right, and has also turned clockwise through an angle 10t. So the black dot will now be at the point
(10t - sin 10t, 1 - cos 10t)