The Number Mysteries: A Mathematical Odyssey through Everyday Life. Marcus Sautoy du

which of the twin primes 71 and 73 you place the counter on. One is a winning choice; the other will lose you the game because every move from that point on is forced on you. Whoever is on 67 can win the game, and it seems that 89 is not so important. So how can you make sure you get there?

      If you carry on tracing your way back through the game you’ll find that there’s a crucial decision to be made for anyone on the prime 37. From there, you can reach my daughters’ twin primes, 41 and 43. Move to 41, and you can guarantee winning the game. So now it looks as if the game is decided by whoever can force their opponent to move them to the prime 37. Continuing to wind the game back in this way reveals that there is indeed a winning opening move. Put the counter on 5, and from there you can guarantee that you get all the crucial decisions that ensure you get to move the stone to 89 and win the game because then your opponent can’t move.

      What if we continue to make the maximum permitted jump even bigger: can we be always be sure that the game will end? What if we allow each player to move a maximum of 99 steps—can we be sure that the game won’t just go on for ever because you can always jump to another prime within 99 of the last one? After all, we know that there are infinitely many primes, so perhaps at some point you can simply jump from one prime to the next.

      It is actually possible to prove that the game does always end. However far you set the maximum jump, there will always be a stretch of numbers greater than the maximum jump containing no primes, and there the game will end. Let’s look at how to find 99 consecutive numbers, none of which is prime. Take the number 100×99×98×97×…×3×2×1. This number is known as 100 factorial, and written as 100! We’re going to use an important fact about this number: if you take any number between 1 and 100, then 100! is divisible by this number.

      Look at this sequence of consecutive numbers:

      100!+2, 100!+3, 100!+4, …, 100!+98, 100!+99, 100!+100

      100!+2 is not prime because it is divisible by 2. Similarly, 100!+3 is not prime because it is divisible by 3. (100! is divisible by 3, so if we add 3 it’s still divisible by 3.) In fact, none of these numbers is prime. Take 100!+53, which is not prime because 100! is divisible by 53, and if we add 53 the result is still divisible by 53. Here are 99 consecutive numbers, none of which is prime. The reason we started at 100!+2 and not 100!+1 is that with this simple method we can deduce only that 100!+1 is divisible by 1, and that won’t help us to tell whether it’s prime. (In fact it isn’t.)

      This website has information about where the hopscotch game will end for larger and larger jumps: http://bit.ly/Primehopscotch You can use your smartphone to scan this code.

      So we know for certain that if we set the maximum jump to 99, our prime number hopscotch game will end at some point. But 100! is a ridiculously large number. The game actually finished way before this point: the first place where a prime is followed by 99 non-primes is 396,733.

      Playing this game certainly reveals the erratic way in which the primes seem to be scattered through the universe of numbers. At first sight there’s no way of knowing where to find the next prime. But if we can’t find a clever device for navigating from one prime to the next, can we at least come up with some clever formulas to produce primes?

      Could rabbits and sunflowers be used to find primes?

      Count the number of petals on a sunflower. Often there are 89, a prime number. The number of pairs of rabbits after 11 generations is also 89. Have rabbits and flowers discovered some secret formula for finding primes? Not exactly. They like 89 not because it is prime, but because it is one of nature’s other favourite numbers: the Fibonacci numbers. The Italian mathematician Fibonacci of Pisa discovered this important sequence of numbers in 1202 when he was trying to understand the way rabbits multiply (in the biological rather than the mathematical sense).

      Fibonacci started by imagining a pair of baby rabbits, one male, one female. Call this starting point month 1. By month 2, these rabbits have matured into an adult pair, which can breed and produce in month 3 a new pair of baby rabbits. (For the purposes of this thought experiment, all litters consist of one male and one female.) In month 4 the first adult pair produce another pair of baby rabbits. Their first pair of baby rabbits has now reached adulthood, so there are now two pairs of adult rabbits and a pair of baby rabbits. In month 5 the two pairs of adult rabbits each produce a pair of baby rabbits. The baby rabbits from month 4 become adults. So by month 5 there are three pairs of adult rabbits and two pairs of baby rabbits, making five pairs of rabbits in total. The number of pairs of rabbits in successive months is given by the following sequence:

      FIGURE 1.22 The Fibonacci numbers are the key to calculating the population growth of rabbits.

      1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

      Keeping track of all these multiplying rabbits was quite a headache until Fibonacci spotted an easy way to work out the numbers. To get the next number in the sequence, you just add the two previous numbers. The bigger of the two is of course the number of pairs of rabbits up to that point. They all survive to the next month, and the smaller of the two is the number of adult pairs. These adult pairs each produce an extra pair of baby rabbits, so the number of rabbits in the next month is the sum of the numbers in the two previous generations.

      Some readers might recognize this sequence from Dan Brown’s novel The Da Vinci Code. They are in fact the first code that the hero has to crack on his way to the Holy Grail.

      It isn’t only rabbits and Dan Brown who like these numbers. The number of petals on a flower is often a Fibonacci number. Trillium has three, a pansy has five, a delphinium has eight, marigolds have 13, chicory has 21, pyrethrum 34, and sunflowers often have 55 or even 89 petals. Some plants have flowers with twice a Fibonacci number of petals. These are plants, like some lilies, that are made up of two copies of a flower. And if your flower doesn’t have a Fibonacci number of petals, then that’s because a petal has fallen off … which is how mathematicians get round exceptions. (I don’t want to be inundated with letters from irate gardeners, so I’ll concede that there are a few exceptions which aren’t just examples of wilting flowers. For example, the starflower often has seven petals. Biology is never as perfect as mathematics.)

      As well as in flowers, you can find the Fibonacci numbers running up and down pine cones and pineapples. Slice across a banana and you’ll find that it’s divided into 3 segments. Cut open an apple with a slice halfway between the stalk and the base, and you’ll see a 5-pointed star. Try the same with a Sharon fruit, and you’ll get an 8-pointed star. Whether it’s populations of rabbits or the structures of sunflowers or fruit, the Fibonacci numbers seem to crop up whenever there is growth happening.

      The way shells evolve is also closely connected to these numbers. A baby snail starts off with a tiny shell, effectively a little one-by-one square house. As it outgrows its shell it adds another room to the house and repeats the process as it continues to grow. Since it doesn’t have much to go on, it simply adds a room whose dimensions are based on those of the two previous rooms, just as Fibonacci numbers are the sum of the previous two numbers. The result of this growth is a simple but beautiful spiral.

      Actually these numbers shouldn’t be named after Fibonacci at all, because he was not the first to stumble across them. In fact they weren’t discovered by mathematicians at all, but by poets and musicians in medieval India. Indian poets and musicians were keen to explore all the possible rhythmic structures you can generate by using combinations of short and long rhythmic units. If a long sound is twice the length of a short sound, then how many different patterns are there with a set number of beats? For example, with eight beats you could do four long sounds or eight short ones. But there are lots of combinations between these two extremes.

      FIGURE 1.23 How to build a shell using