 Leonardo of Pisa is best known for discovering a remarkable recursive sequence and inventing discrete time modeling. Well, actually, he posed one problem about rabbits in a book whose modern printing is over 600 pages long. So if you don't read Latin, the translation is something like the following how many rabbits can be produced in a year from a pair of rabbits when rabbits mature after one month and begin producing a pair of rabbits each month thereafter. Leonardo's solution begins. In the first month, the original pair produces a pair of immature rabbits, so now there are two pairs. In the second month, the original pair bears another pair, so there are now three pairs. In the third month, two pairs give birth, so there are now five pairs. In the fourth month, three pairs give birth, so there are eight powers and so on. Ending with 377 pairs at the end of the year. Leonardo tabulates the results in a marginal note and in a last paragraph he observes that the terms can be formed by adding two consecutive terms. A much more interesting problem from Leonardo is the following. A man buys 30 birds for 30 dinari, a dinari being a unit of currency. A part which costs three dinari, a pigeon costs two dinari and sparrows are two for one dinaro. Similar problems appeared in Islamic, Indian and Chinese sources and Alcon included some in his problems to sharpen the minds of the young, but the sources only gave solutions without explaining how they found them. Leonardo not only solved this problem but gave a clear explanation that could be used on all problems of this type. Leonardo's solution is based on the alloy problems from book 11 of Libera Bacchai. In general, these problems require you to produce an amount of an alloy with a specific concentration from two or more alloys of given concentrations. For example, given an alloy that contains 3 ounces of silver per pound and another that contains 6 ounces of silver per pound, produce 10 pounds of an alloy containing 5 ounces of silver per pound. Or, given 3 alloys with 3, 4 and 6 ounces of silver per pound respectively, produce 20 pounds of an alloy containing 5 ounces of silver per pound. If there are only two different concentrations, there is a unique solution, but with 3 or more concentrations there are an infinite number of solutions. Leonardo recognized the bird problem is an alloy problem where the cost per bird is the concentration. Leonardo's solution went through the following steps. First, Leonardo combined the cheapest and most expensive birds to form an alloy that cost one dinaro per bird. Trial and error finds a solution. One partridge and four sparrows cost 3 plus 2, 5 dinaro. And so that's five birds for five dinaro. Next, Leonardo combined the second most expensive and cheapest birds, the sparrows and pigeons. And again, by trial and error, we find that one pigeon and two sparrows cost 2 plus 1, 3 dinaro. So that's again three birds for three dinaro. Since both alloys, both sets, give one bird for each dinaro, then any combination of the sets will also give one bird for each dinaro. So we look for a combination of these sets that will give us 30 birds. And so Leonardo used three of the first alloy and five of the second alloy. Leonardo went further and noted the following. One could solve the problem for any number more than 15 birds, say 17 birds for 17 dinaro. But for fewer birds, the only solvable situations were 3, 5, 8, 11, or 13 birds for the same number of dinaro.