In a letter to Master Theodorus, Fibonacci (Leonardo of Pisa) posed and solved three variants of a problem on birds. These are particularly interesting since one problem has a unique solution, one had two solutions and the other has no solution. The letter is contained in B Boncompagni (ed.), Scritti di Leonardo Pisano Vol. 2 (Rome, 1862), 247-252. We gave below versions of the three problems, our own modern solution to each, and Fibonacci's solution to each problem. We note that we can use algebraic notation and manipulation while Fibonacci did not have access to these tools so had to use an arithmetical argument written entirely in words. However, as the reader will see, our modern solution basically comes down to the same as Fibonacci's solution. We chose to give our modern solution before Fibonacci's solution since this will, we hope, help the reader to understand what Fibonacci is doing. |
1. The thirty pound bird problem.
Here is the first problem.
A certain man buys 30 birds which are sparrows, turtle-doves, and doves, for 30 pounds. Three sparrow cost 1 pound, two turtle-doves cost 1 pound, and a dove costs 2 pounds. It is required to find how many birds he buys of each kind.
1.1. A modern solution.
Let x be the number of sparrows, y the number of turtle-doves, and z the number of doves. Using the information in the problem we get two equations:
x + y + z = 30 (1)
^{1}/_{3} x + ^{1}/_{2} y + 2 z = 30 (2)
Now, as is well known, two equations don't give enough information to determine the solution to a problem in three unknowns. But, never fear, we have more information. The unknowns x, y and z must be positive integers and, moreover, none can be zero since the man purchases all three types of bird.
Start by multiplying the equation (2) by 6 to get rid of the fractions:
x + y + z = 30
2 x + 3 y + 12 z = 180
Subtract 2 times equation (1) from equation (2) to eliminate x:
y + 10 z = 120
Since 10 z and 120 are both divisible by 10, so is y. Now y = 10 gives a solution
y = 10, z = 11, x = 9.
Also y = 20 gives z = 10 and x = 0 which is not a solution (we have to have a positive number of each kind of bird) and clearly no larger multiple of 10 will give a solution.
1.2. Fibonacci's solution.
I first suppose that the man buys 30 sparrows for 10 pounds, and keeps back 20 pounds, which is the difference between 30 pounds and 10 pounds. Now suppose I changed one of the sparrows into a turtle-dove which would increase the money spent by this change by ^{1}/_{6} of a pound since the sparrow was worth ^{1}/_{3} of a pound and the turtle-dove ^{1}/_{2} of a pound, that is ^{1}/_{6} more than the price of a sparrow. Suppose now that I changed one of the sparrows into a dove and by that change increased my spending by 1 ^{2}/_{3} of a pound, that is the difference between 2 pounds and ^{1}/_{3} of a pound. If I made six of these 1^{2}/_{3} changes then these six would make in total an increase of 10 pounds. Now I must change sparrows into turtle-doves and into doves until I have used up the 20 pounds that I kept back earlier. So I multiplied by six and so obtained 120 which I split into two parts, one of which could be divided exactly by 10 and the other by 1. The total of these two divisions was not to be as large as 30. So the first part is 110 and the other 10; and I divided the first part, that is 110, by 10, and the second by 1. This gives 11 doves and 10 turtle-doves; taking these from 30 there remain 9 for the number of sparrows. So there are 9 sparrows worth 3 pounds, and 10 turtle-doves worth 5 pounds, and 11 doves worth 22 pounds. So from these three kinds of birds we shall have 30 for 30 pounds as was required.
2. The twenty-nine pound bird problem
This problem is a slight variant of the first problem.
A certain man buys 29 birds which are sparrows, turtle-doves, and doves, for 29 pounds. Three sparrow cost 1 pound, two turtle-doves cost 1 pound, and a dove costs 2 pounds. It is required to find how many birds he buys of each kind.
2.1. A modern solution.
As before, let x be the number of sparrows, y the number of turtle-doves, and z the number of doves. Using the information in the problem we get two equations:
x + y + z = 29 (1)
^{1}/_{3} x + ^{1}/_{2} y + 2 z = 29 (2)
Again multiply (2) by 6 and subtract 2 times (1) to get
y + 10 z = 116
Now y must be congruent to 6 mod 10. We get two solutions
y = 6, z = 11, x = 12
and
y = 16, z = 10, x = 3.
2.2. Fibonacci's solution.
If we wish to have 29 birds for 29 pounds we may proceed in the same way as before, that is we take the price of 29 sparrows, the cheapest birds, from the 29 pounds, and the remaining money is taken 6 times and this gives 116. Again we split this into two parts, one of which is to be exactly divisible by 10 and the other by 1, and the sum of the two divisions is not as large as 29. These parts can be constructed in two ways; firstly so that the first part is 110, and the second is 6; and when 110 is divided by 10 we obtain 11 doves, and when 6 is divided by 1 we obtain 6 turtle-doves. Taking their sum from 29, there remains 12 for the number of sparrows. Secondly we shall split 116 into 100 and 16; and we shall divide 100 by 10, and 16 by 1, and we shall have 10 doves and 16 turtle-doves. The remainder, to make up the total number to 29, that is 3, will be sparrows. Thus we have solved this problem in two ways.
3. The fifteen pound bird problem
This problem is a slight variant of the two problems above.
A certain man buys 15 birds which are sparrows, turtle-doves, and doves, for 15 pounds. Three sparrow cost 1 pound, two turtle-doves cost 1 pound, and a dove costs 2 pounds. It is required to find how many birds he buys of each kind.
3.1. A modern solution.
As before, let x be the number of sparrows, y the number of turtle-doves, and z the number of doves. Using the information in the problem we get two equations:
x + y + z = 15 (1)
^{1}/_{3} x + ^{1}/_{2} y + 2 z = 15 (2)
Again multiply (2) by 6 and subtract 2 times (1) to get
y + 10 z = 60.
As in the first problem above, y must be divisible by 10. But y = 10 gives z = 5 and x = 0. This cannot be a solution since all of x, y, z must be greater than 0. All larger multiples of 10 are greater than 15 so we have no chance of finding positive values for y and z.
3.2. Fibonacci's solution.
If we wish to have 15 birds for 15 pounds I shall show this is not possible without a fractional number of birds. For example, if I were to subtract the price of 15 sparrows from 15 pounds and I take the remaining pounds and multiply by six, this gives 60 which cannot be divided into two parts, one of which is to be divisible by 10 and the other by 1 so that from these divisions we obtain a number less than 15. For instance, if I divide 60 into 50 and 10; and divide 50 by 10, and 10 by 1, the result of the two divisions are 5 and 10 which together add to 15. That is to be the sum of all the birds and so there will be no sparrows in this purchase because 5 doves are worth 10 pounds and 10 turtle-doves are worth 5 pounds and so from these two kinds of birds alone we have 15 birds for 15 pounds. Nor is there any number less that 60 and greater than 50 which can be divided by 10 and a smaller number has no place here for if we put 40 for one part there would remain 20 for the other part. From this we have 40 divided by 10 and 20 divided by 1 and the total is 24 birds which has no place here since the number must be 15 birds. However, if we wished to consider fractions of birds there is a solution by dividing 60 into 55 and 5. The 55 divided by 10 gives 5^{1}/_{2} doves and dividing 5 by 1 gives 5 turtle-doves. Subtracting 5^{1}/_{2} doves and 5 turtle-doves from 15 gives 4^{1}/_{2} sparrows. Now the sparrows cost 1^{1}/_{2} pounds, the 5 turtle-doves cost 2^{1}/_{2} pounds and 5^{1}/_{2} doves cost 11 pounds. Thus, with fractions of birds, we have 15 birds for 15 pounds.