In the previous post, we described Fibonacci’s “problem of the birds” (“the problem of the man who buys thirty birds of three kinds for 30 *denari*”). In English:

“A man buys 30 birds of three kinds (partridges, doves, and sparrows) for 30

denari. He buys a partridge for 3denari, a dove for 2denari, and 2 sparrows for 1denaro, that is, 1 sparrow for ½denaro. How many birds of each kind does he buy?”

The man must buy at least one of each kind of bird, or he wouldn’t be buying “birds of three kinds.” Also, he must buy less than 10 partridges, because 10 partridges (at 3 denari each) would use up all his money. Similarly, he must buy less than 15 doves. We can thus make up a table of possible solutions:

Of those 126 possible solutions, only one works out correctly in terms of cost, and that’s the answer. But that’s an unbelievably tedious way of getting the answer, and you’d be rather foolish to try to solve the problem that way. The obvious approach is to use algebra. Write *p* for the number of partridges bought, *d* for the number of doves, and *s* for the number of sparrows. Because the man buys 30 birds, we have the equation:

*p*+

*d*+

*s*= 30

And because the costs add up to 30 *denari*, we have:

*p*+ 2

*d*+ ½

*s*= 30

Doubling that second equation gets rid of the annoying fraction:

*p*+ 4

*d*+

*s*= 60

If you’ve done any high school algebra, no doubt you want to subtract the first equation from this, which will eliminate the variable *s*:

*p*+ 3

*d*= 30

But now what? That gives a relationship between the variables *p* and *d*, but there doesn’t seem to be enough information to get specific values for those variables.

Fibonacci solves the problem a different way. His solution is based on a key insight – the man buys 30 birds for 30 *denari*, so that the birds cost, on average, 1 *denaro* each. Fibonacci then makes up “packages” of birds averaging 1 *denaro* each. There are only two ways of doing this. Package A has 1 partridge and 4 sparrows (5 birds for 5 *denari*), and package B has 1 dove and 2 sparrows (3 birds for 3 *denari*). The solution will be a combination of those two packages.

Now the man can take 1, 2, 3, 4, or 5 copies of package A, leaving 25, 20, 15, 10, or 5 birds to be made up of package B. But the birds making up package B must be multiple of 3, so that the only possible answer is 3 copies of package A and 5 copies of package B. This means that the man buys 3 partridges, 5 doves, and 3×4 + 5×2 = 22 sparrows. That’s 30 birds and 3×3 + 2×5 + ½×22 = 30 *denari*.

Now it turns out that, had we kept on going with the algebraic approach, we would have gotten the same answer. We had:

*p*+ 3

*d*= 30

Given that the numbers of partridges and doves (*p* and *d*) had to be positive whole numbers, that meant that *p* had to be a multiple of 3, and *d* a multiple of 5. That could only be achieved with *p* = 3 and *d* = 5.

We can also return to the diagrammatic approach. The equation:

*p*+ 3

*d*= 30

describes the diagonal red line in the diagram below. That line only crosses one of the possible solutions, namely the dot corresponding to 3 partridges and 5 doves.

In mathematics, there’s more than one way to skin a cat. Or, in this case, a bird.