Ian Stewart Read online

  Mathematics transfigures the fortuitous concourse of atoms into the tracery of the finger of God. Herbert Westren Turnbull

  In many cases, mathematics is an escape from reality. The mathematician finds his own monastic niche and happiness in pursuits that are disconnected from external affairs. Stanislaw Ulam

  God exists since mathematics is consistent, and the Devil exists since we cannot prove it. Andre Weil

  Mathematics as a science commenced when first someone, probably a Greek, proved propositions about ‘any’ things or about ‘some’ things, without specifications of definite particular things. Alfred North Whitehead

  Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives. Anonymous

  Wittgenstein’s Sheep

  This story is told by the Cambridge analyst John Edensor Littlewood in his lovely little book A Mathematician’s Miscellany:

  Schoolmaster: ‘Suppose x is the number of sheep in the problem.’

  Schoolboy: ‘But, Sir, suppose x is not the number of sheep.’

  Littlewood says that he asked the Cambridge philosopher Ludwig Wittgenstein whether this was a profound philosophical joke, and he said it was.

  Leaning Tower of Pizza

  It was early afternoon in Geronimo’s Pizzeria, and business was slow. Angelina, one of the servers, was amusing herself by piling pizza delivery boxes on top of each other on the edge of a table. It all looked rather precarious, and Luigi said as much.

  ‘I’m trying to see how far out I can make the pile go without the boxes actually falling off,’ Angelina explained. ‘I’ve discovered that with just three boxes, I can almost get the top one outside the line of the table.’

  If the boxes are 1 unit long, the top one pokes out 11/12 of a unit.

  ‘How did you figure that out?’ Luigi asked.

  ‘Well, I put the top one on the second one, so that its centre was poised exactly on the edge. So it poked out a unit. Then it was obvious that the centre of mass of the top two boxes was in the middle, so I placed them with the centre of mass exactly over the edge of the third box. If you do the sums, that makes it poke out another of a unit. Then I placed the three of them so that their combined centre of mass was right on the edge of the table, and that turned out to add a further of a unit to the overhang.’

  ‘And ,’ Luigi said. ‘You’re right, it does poke out almost 1 unit.’

  Alert readers will observe that Angelina and Luigi are assuming the boxes are identical and they are uniform, that is, the mass is evenly distributed. Real pizza boxes, full or empty, are not like that, but for this puzzle you should pretend that they are.

  ‘What happens if you add more boxes?’ Luigi asked.

  ‘I think the pattern continues. I could replace the table by a fourth box, and then slide the pile out until it is just about to topple, adding a further to the overhang. Then the top box does poke out over the edge of the table: the overhang is . And with more boxes still, I could do the same again, adding , and so on.’

  ‘So you’re saying,’ said Luigi, ‘that with n boxes you can get an overhang of

  units. Which I instantly recognise as Hn, where Hn is the nth harmonic number:

  Isn’t that right?’

  Angelina agreed that it was. As you do.

  This is a time-honoured puzzle, and the biggest overhang you can get with n boxes using this method is indeed Hn, so Angelina and Luigi are right. You can find the details nicely worked out in many sources, and I would have included them, but for one thing: this traditional answer is valid only with the extra assumption that exactly one box occurs at each level. And that raises a very interesting question: what happens without that assumption?

  In 1955, R. Sutton noticed that, even with just three boxes, you can do better than Angelina: an overhang of 1 instead of . With four boxes, the biggest possible overhang is

  Sutton discovered how to make the top one poke out 1 unit with three boxes

  With four boxes, the biggest overlap involves leaving a gap in the second layer.

  What happens for n boxes, with as many on each layer as you wish? (There is an even more general question, where the boxes can be tilted, but let’s restrict ourselves to layers, like the courses of a brick wall.)

  You might like to try your hand at this puzzle before reading any further. What is the biggest overhang you can get with 5 or 6 boxes?

  Answers on page 297

  To avoid misunderstandings, let me make the conditions clear. All boxes are identical and uniform, and everything is idealised to exact rectangles and all the usual stuff we assume in Euclidean geometry. The problem is posed in the plane, because in three-dimensional space you could also rotate boxes, without violating the ‘layers’ condition. The arrangement must be in equilibrium: that is, if you work out all the forces that act on any box, they all balance each other out. Boxes must be arranged in layers, but you can leave gaps. And one other important condition: you do not have to be able to build the arrangement by adding one box at a time. Intermediate stages might topple if left unsupported. Only the final arrangement must be in equilibrium. (This equilibrium condition turns out not to be terribly intuitive; it can be turned into equations and checked by computer. When there aren’t too many boxes, though, it should be intuitive enough for you to tackle this puzzle.)

  The answers for 4, 5 and 6 boxes were worked out by J. F. Hall in 2005. In fact, he proposed some general patterns, and suggested that they should always maximise the overhang. But, in 2009, Mike Paterson and Uri Zwick showed that Hall’s stacks maximise the overhang only for 19 boxes or fewer (see page 297 for the reference). Finding exact arrangements with a lot of boxes is extremely complicated, but they proposed some near-optimal arrangements for up to 100 boxes.

  One very interesting question is: how fast can the biggest overhang grow as the number n of boxes increases? For the classic ‘one box per layer’ solution, the answer is Hn. There doesn’t seem to be a simple formula for this number, but Hn is very closely approximated by the natural logarithm log n. So the ‘asymptotic’ size of the largest overhang is log n.

  Paterson and Zwick proved that, when layers can contain many boxes, the maximal overhang is approximately proportional to the cube root of n. More precisely, there are constants c and C for which the maximal overhang always lies between and . They exhibited explicit arrangements with an overhang of at least

  units, using what they call ‘parabolic stacks’. The picture shows such a stack with 111 boxes and an overhang of exactly 3 units. (The approximate formula gives only 2.50069 instead of 3 when n = 111, but it still gives the best-known overhang for very large n.)

  Early in 2009, Peter Winkler, Yuval Peres and Mikkel Thorup joined the team, and took the question further. They proved that C is at most 6: the overhang can never be greater than . Their proof uses the probability theory of ‘random walks’, in which a person takes a step forward or backward with specified probabilities. Each new brick spreads the forces that act in a similar manner to the way probabilities spread as a random walk proceeds.

  A parabolic stack with 111 boxes, and overhang 3.

  PieThagoras’s World-Famous Mince πs

  Alvin, Brenda and Casimir went to the pie-shop and bought three of PieThagoras’s world-famous perfectly circular mince pies. They bought one mini-pie with diameter 6 centimetres, one midi-pie with diameter 8 centimetres, and one maxi-pie with diameter 10 centimetres, because those were the only pies left.

  Three pies.

  They could have settled for one pie each, but they wanted to share the pies fairly. Now, as everyone knows, PieThagoras’s world-famous mince pies consist of two flat layers of pastry, of uniform thickness, with a uniform layer of mince sandwiched in between. The thicknesses of the pastry and the mince are the same for all sizes of pie. So ‘fair’ means ‘having equal area’ when viewed from above as in my picture.

  They decided that sharing the pies fairly would be quite complicated, and
had just settled on dividing each pie separately into thirds when Desdemona turned up and demanded her fair share too. Fortunately they had not started cutting the pies. After some thought, they discovered that now they could divide the pies more easily, by cutting two of them into two pieces each, and leaving the third pie uncut. How?

  Answer on page 297

  Diamond Frame

  Innumeratus’s attempt at a magic frame.

  Innumeratus had taken the ace to 10 of diamonds from a pack of cards, and was arranging them to make a rectangular frame.

  ‘Look!’ he shouted to Mathophila. ‘I’ve arranged them so that the total number of pips along each side of the frame is the same!’

  Mathophila had learned to take such statements with a pinch of salt, and she quickly pointed out that the sums concerned were 19 (top), 20 (left), 22 (right) and 16 (bottom).

  ‘Well, I’ve arranged them so that the total number of pips along each side of the frame is different, then.’

  Mathophila agreed with that, but felt it was a silly puzzle. She really liked the first version better.

  Can you solve the original version? You can turn cards through a right angle if you wish.

  Answer on page 298

  Pour Relations

  This is a traditional puzzle that goes back to the Renaissance Italian mathematician Tartaglia in the 1500s, but its solution has systematic features that were not noticed until 1939, discussed in the Answer. There are many similar puzzles.

  You have three jugs, which respectively hold 3 litres, 5 litres, and 8 litres of water. The 8-litre jug is full, the other two are empty. Your task is to divide the water into two parts, each of 4 litres, by pouring water from one jug into another. You are not allowed to estimate quantities by eye, so you can only stop pouring when one of the jugs involved becomes either full or empty.

  Divide the water into two equal parts.

  Answer on page 299

  Alexander’s Horned Sphere

  If you draw a closed curve in the plane which doesn’t cross itself, then it seems pretty obvious that it must divide the plane into two regions: one inside the curve, the other outside it. But mathematical curves can be very wiggly, and it turns out that this obvious statement is difficult to prove. Camille Jordan gave an attempted proof, more than 80 pages long, in a textbook published in several volumes between 1882 and 1887, but it turned out to be incomplete. Oswald Veblen found the first correct proof of this ‘Jordan curve theorem’ in 1905. In 2005, a team of mathematicians developed a proof suitable for computer verification - and verified it. The proof was 6,500 lines long.

  A closed curve, with the inside shaded.

  A subtler topological feature of such a closed curve is that the regions inside and outside the curve are topologically equivalent to the regions inside and outside an ordinary circle. This too may seem obvious, but, remarkably, the corresponding statement in three dimensions, which seems equally obvious, is actually false. That is: there is a surface in space, topologically equivalent to an ordinary sphere, whose inside is topologically equivalent to the inside of an ordinary sphere, but whose outside is not topologically equivalent to the outside of an ordinary sphere! Such a surface was discovered by James Waddell Alexander in 1924, and is called Alexander’s horned sphere. It is like a sphere that has sprouted a pair of horns, which divide repeatedly and intertwine.

  Alexander’s horned sphere.

  The Sacred Principle of Mat

  The daredevil adventurer and treasure-hunter Colorado Smith, who is not like a real archaeologist at all, ducked a passing shower of blazing war-arrows to check the crude sketch-map scribbled in his father’s battered notebook.

  ‘The holy sanctuary of Pheedme-Pheedme the goddess of eating and sleeping,’ he read, ‘is formed from 64 identical square cushions stuffed with ostrich-down, arranged in an 8×8 array. The five sacred avatars of Pheedme-Pheedme, represented in overstuffed fabric, are to be placed on the cushions so that they “watch over” every other cushion: that is, every other cushion must be in line with one occupied by an avatar. This line can be horizontal, vertical or diagonal, relative to the array, where “diagonal” means “sloping at 45° ”.’

  ‘Look out!’ shrieked his sidekick Brunnhilde, taking cover beneath a large stone altar.

  ‘I wouldn’t do that if I were you,’ said Smith, and hauled her out a split second before the supporting slabs exploded in puffs of dust and the 10-tonne altar stone crashed to the ground. ‘Now, Dad’s notebook says something about the principles of—uh - Mat?’

  ‘Ma’at was the Egyptian concept of justice and rightful place,’ Brunnhilde pointed out. ‘But this temple is Burmalayan.’

  ‘True. Can’t be ma’at . . . No, it’s definitely the Principle of Mat. Apparently the goddess reclines on a mat, surrounded by her sacred avatars. We have to leave a space for the holy reclining mat, which is square. Hmm . . . maybe this would do.’

  Is this how to lay out the five sacred avatars and Pheedme-Pheedme’s reclining mat?

  ‘That seems suspiciously easy,’ said Brunnhilde. ‘What else do we have to do?’

  Smith quietly removed a deadly kamikaze-scorpion from her hair, hoping she wouldn’t notice. ‘Uh, we have to arrange the avatars to leave the largest possible space for a square mat. Bearing in mind that they must watch over every cushion. I doubt we can do better than my picture.’

  ‘Those ancient priests were sneaky, though,’ said Brunnhilde. She tried not to listen to the approaching bloodcurdling cries, and racked her brains. If they could solve the riddle of the sacred mat, they could proceed to the enigma of the potted dormice, and then only 17 more puzzles would lie between them and the Hoard of Treasures. ‘Does the mat have to be arranged with its sides parallel to those of the cushions? Could it be tilted?’

  ‘I don’t see anything in the 999 pages of The Book of the Ninth Life to prohibit that,’ said Smith. ‘The only restriction is that the mat can’t overlap any cushion bearing one of the sacred avatars. The edges of the mat and the cushion can touch, but there mustn’t be a genuine overlap.’

  How can the biggest mat be fitted in without breaking the sacred rules?

  Answer on page 301

  Perfectly Abundantly Amicably Deficient

  If n is a whole number, then the sum of its divisors, including n itself, is the divisor-sum σ(n). So, for example,

  σ(24) = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60

  The divisor-sum is the key to a very ancient recreation, the search for perfect numbers. A number is abundant if it is smaller than the sum of its ‘proper’ divisors - those that exclude the number itself. It is deficient if it is greater than that sum, and perfect if it is equal to it. In terms of the divisor-sum, these conditions become

  σ(ν) > 2ν σ(ν) < 2ν σ(ν) = 2ν

  Here we see 2n, rather than n, because σ(n) includes the divisor n as well as all the others. This is done so that the nice formula σ(mn) = σ(m)σ(n) holds when m and n have no common factor greater than 1.

  Lots of numbers are deficient; for example 10 has proper divisors 1, 2, 5, which sum to 8. Abundant numbers are rarer: 12 has proper divisors 1, 2, 3, 4, 6, with sum 16. Perfect numbers are very rare; the first few are:

  6 = 1+2+3

  28 = 1 + 2 + 4 + 7 + 14

  followed by 496 and 8,128. Euclid discovered a pattern in these perfect numbers: he proved that whenever 2p - 1 is a prime, the number 2p-1(2p - 1) is perfect. Much later, Euler proved that every even perfect number must be of this form. Primes of the form 2p - 1 are called Mersenne primes (Cabinet, page 151).

  No one knows whether any odd perfect numbers exist; however, Carl Pomerance has given a non-rigorous but plausible argument that they don’t. There is a solid proof that if an odd perfect number does exist then it must be at least 10300, and have at least 75 prime factors. Its largest prime factor must be greater than 108.

  A related, equally ancient pastime is to find pairs of amicable numbers - each equal to the sum of the proper
divisors of the other. That is,

  m = σ(ν) - ν

  n = σ(μ) - μ

  so σ(n) = σ(m) = m + n. For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, adding to 284; the proper divisors of 284 are 1, 2, 4, 71, 142, adding to 220. The next few amicable pairs are (1184, 1210), (2620, 2924), (5020, 5564) and (6232, 6368).

  In all known examples, the numbers in an amicable pair are either both even or both odd. Every known pair shares at least one common factor; it is not known whether a pair of amicable numbers with no common factor can exist. If there is such pair, then their product is at least 1067.

  An integer is multiply perfect if it divides the sum of its divisors exactly; the multiplicity is the quotient. Here it makes no difference whether we include the number itself, or not, except that the multiplicity goes down by 1 if we don’t. But it’s normal to include it. If we do, then ordinary perfect numbers have multiplicity 2, triperfect numbers have multiplicity 3, and so on. The smallest triperfect number is 120, as Robert Recorde knew in 1557: the sum of its divisors is