*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 buys30birds which are sparrows, turtle-doves, and doves, for30pounds. Three sparrow cost1pound, two turtle-doves cost1pound, and a dove costs2pounds. 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)

*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

*x*:

*y*+ 10

*z*= 120

*z*and 120 are both divisible by 10, so is

*y*. Now

*y*= 10 gives a solution

*y*= 10,

*z*= 11,

*x*= 9.

*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.**

^{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 buys29birds which are sparrows, turtle-doves, and doves, for29pounds. Three sparrow cost1pound, two turtle-doves cost1pound, and a dove costs2pounds. 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)

*y*+ 10

*z*= 116

*y*= 6,

*z*= 11,

*x*= 12

*y*= 16,

*z*= 10,

*x*= 3.

**2.2. Fibonacci's solution.**

**3. The fifteen pound bird problem**

This problem is a slight variant of the two problems above.

A certain man buys15birds which are sparrows, turtle-doves, and doves, for15pounds. Three sparrow cost1pound, two turtle-doves cost1pound, and a dove costs2pounds. 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)

*y*+ 10

*z*= 60.

*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.**

^{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.