Month Rabbits # Pairs

1. Fibonacci s Rabbit Problem. Fibonacci rabbits come in pairs. Once a pair is two months old, it bears another pair and from then on bears one pair every month. Starting with a newborn pair at the beginning of a year, how many pairs of rabbits will there be at the end of the year? I start the problem in the pictures below. A white pair if rabbits is one that is not yet mature enough to produce another pair. Pink pairs reproduce. It may be hard to know who the parents are after a while, but for every pink pair in a given row, there will be an additional white pair in the row below. Page from the Liber Abaci where the rabbit problem appears Month Rabbits # Pairs 1 1 2 1 3 2 4 3 5 5 After a year there will be rabbits.

2 Solution. One way of doing it is to see the pattern, or to just go on with the rabbit pictures until month 12 is reached. We then get the following pattern: Month J F M A M J J A S O N D # rabbit pairs 1 1 2 3 5 8 13 21 34 55 89 144 But suppose we want to go on for longer than a year, can we be sure the pattern stays the same? Here is a more precise justification. It may be a bit harder to follow, but I hope not incredibly hard. Let us look at some consecutive months and following the mathematical custom of calling things by letters, suppose that one month is called n, the next one n + 1. Suppose that in month n we have x pink rabbit pairs and y white ones. We write this out as p = x, w = y. The total number of rabbit pairs is then fn = x + y.(fn for Fibonacci at time n). Now every pink rabbit pair will produce a new pair that is white. The white pairs become pink. This means that in month n + 1 our equations are p = x + y (all pink stay pink, whites become pink) and w = x (all pink produce one white). So now we have for the total number of rabbit pairs f(n + 1) = x + y + x = 2x + y. This still doesn t tell us that much, so let s go to the next month, month n + 2. All pink stay pink, all white become pink, our equation is now p = x + y + x = 2x + y; all pink produce one white, so w = x + y, and f(n + 2) = 2x + y + x + y = 3x + 2y. Well, tis is still a bit obscure, but we might notice that f(n + 2) = 3x + 2y = (2x + y) + (x + y) = f(n + 1) + fn. The pattern is, in fact and forever: The number of rabbit pairs in any given month past the second one equals the sum of the number of pairs in the the two preceding months. That is, the pattern is 1, 1, 1 + 1 = 2, 1 + 2 = 3,, 2 + 3 = 5,, 3 + 5 = 8, 5 + 8 = 13, and so forth The sequence of numbers so created is known as the Fibonacci sequence. 2. A tiling problem. Suppose we have rectangular tiles; they are twice as long as they are wide.to be precise, each tile is 10 inches long and 5 inches wide. Mr. Snuckelbakker wants to use these tiles to tile a 10 inch wide rectangular area in a room of his house and he wonders in how many different ways he can lay the tiles, depending on how much floor he wants to cover. If all he wants to do is to cover 5 inches of floor, there is only one way of doing it, he uses a single tile: If he wants to cover 10 inches of floor, he has two choices:

3 If he wants to cover 15 inches of floor, he has three choices: However, he wants to cover a strip that is 10 feet = 24 5 long. In how many ways can he do it? Solution. The Fibonacci sequence appears again, except that the first term is missing. The sequence is 1, 2, 3, 5,.... So the answer is the 25th number in the series, namely f 25 = 75, 025. Here is a justification: Say Mr. S has laid out tiles covering a strip at least 15 long. The tiling either ended with a single vertical tile (as in the first and third picture of how to tile a 15 inch strip) or with two horizontal tiles (as in the middle picture of how to tile a 15 inch strip). In the first case, removing the last tile, tiles 5 inches less, in the second case, removing the horizontal tiles, tiles 10 inches less. From this one can see that the number of different ways to tile a strip of n times 5 inches is the sum of ways one can tile a strip of n 1 times 5 inches, plus the number of ways one can tile a strip of n 2 inches. Fibonacci again! 3. A problem posed to Fibonacci by a magister from near Constantinople. This is one of a whole series of problems involving two or more men, giving and taking money. You can replace the word denari by dollars. Also it is proposed that one man takes 7 denari from the other, and he will have five times the second man. And the second man takes 5 denari from the first, and he will have seven times the denari of the first. How much money does each have? Modern version: If A were to take $7 from B, then A would have five times what B has. If B were to take $5 from A, then B would have seven times what A has. How much money do A, B have? The correct answer involves fractions! Solution. This problem involves equations in addition to fractions. Let us call a the amount of money A has, b the amount B has. The problem states: b + 5 = 7(a 5). a + 7 = 5(b 7) The first equation can be worked on a bit to get first a + 7 = 5b 35, we can then subtract 7 from both sides to get a = 5b 42. We go with this to the second equation: b + 5 = 7a 35 = 7(5b 42) 35 = 35b 294 35 = 35b 329.

4 We can turn this equation around to get 35b 329 = b + 5; subtracting b from both sides gives 34b 329 = 5, adding 329 to both sides gives 34b = 334. From this: b = 334 34 = 167 Returning with this value of b to the equation a = 5b 42 we get a = 5 167 42 = 835 42 = 835 The answer is that A has 121/, B has 167/ denarii. 42 = 835 714 = 121 4. Fibonacci s Birds from two towers Problem On a certain ground there are two towers, one of which is 30 feet high, the other 40, and they are only 50 feet apart; two birds descending together from the heights of the two towers fly to the center of a fountain between the towers; the distances from the center to both towers are sought. In the picture below, the taller tower is represented by the line segment BA, the shorter by DG. The center of the fountain is at Z. Since both birds start at the same time, and fly at the same speed, the segments GZ and AZ have the same length. One has to find the lengths of ZD and ZB. Solution. Let us write x = DZ (the length of DZ) and y = ZB we know that x + y = 50, so y = 50 x. Let h be the length of GZ, the hypotenuse of the triangle DGZ. This is also the length of AZ, the hypotenuse of the triangle ABZ. By Pythagoras: Equating the two expressions for h 2 : h 2 = 30 2 + x 2 = 900 + x 2, h 2 = 40 2 + y 2 = 1600 + (50 x) 2 900 + x 2 = 1600 + (50 x) 2 We now use a very famous formula giving the square of a sum or difference of two terms (a sum or difference of two terms is known as a binomial), namely (a ± b) 2 = a 2 ± 2ab + b 2.

5 So (50 x) 2 = 2500 2(50)x + x 2 = 2500 100x + x 2. Returning to the last equation before this digression, 900 + x 2 = 1600 + 2500 100x + x 2 = 4100 10x + x 2. We can now: Subtract x 2 from both sides, add 100x to both sides, subtract 900 from both sides; we are left with 100x = 3200. Dividing both sides by 100 we get x = 32, the length of ZD. The length of ZB is then y = 50 x = 18. 5. A Tournament problem Fibonacci took part in a mathematical tournament where he solved three problems. Here is the third one. Three men possess a pile of money. One man owns 1/2 of the pile, another one owns 1/3 and the third man owns 1/6 of the pile. Each man takes some money from the pile, until nothing is left. But then the first man returns 1/2 of what he took, the second man returns 1/3 of what he took, and the third man returns 1/6 of what he took. If the returned total is divided evenly among the men, it is found that each then has what he is entitled to. How much money was in the original pile, and how much money did each man take? I am leaving this as a problem still to be solved.