# Recreational Problems in Medieval Mathematics

Author(s):
Victor J. Katz (University of the District of Columbia)

### Introduction

Recreational problems have been a fixture in mathematics problem solving from antiquity. It has long been known that the same problems reappear in cultures all over the world – from ancient Egypt and Babylonia through Greece, medieval China and India, and on into medieval Europe and the Renaissance, as well as modern times. What is surprising, perhaps, is that often the exact same problems reappear, even with the same numerical values, in cultures separated by hundreds of years and thousands of miles. For many years, David Singmaster has been gathering a data base of common recreational problems, a database which is now huge [Singmaster, 1996]. Similarly, there was an earlier book by Johannes Tropfke [Tropfke, 1980] that mentioned many of these problems from different cultures.

In this article, we will consider some appearances of two of these classic recreational problem types in sources not mentioned either by Singmaster or Tropfke. And then we will speculate a bit about the paths these problems traveled.

# Recreational Problems in Medieval Mathematics - Men Buying a Horse

Author(s):
Victor J. Katz (University of the District of Columbia)

The first problem is that of men buying a horse (or perhaps some other animal). The basic problem outline is that three (or more) men want to buy a horse in common. The first says to the other two that if they give him some fixed fraction (say 1/3) of their money, then, with the money he already has, he can buy the horse. The second says to the first and third that if they give him some other fixed fraction (say 1/4) of their money, then, with the money he already has, he can buy the horse. And the third says to the first and second that, if they give him a still different fixed fraction (say 1/5) of their money, then, with the money he already has, he can buy the horse. The question then is how much each man has to begin with and how much the horse costs.

We can easily write down equations representing this problem: $x+r(y+z)=y+s(z+x)=z+t(x+y)=p,$ where $$x, y,$$ and $$z$$ represent the money held respectively by the three men, $$p$$ represents the price of the horse, and $$r, s,$$ and $$t$$ represent the given fractions. Of course, this is a modern representation; to solve this today, we would probably rewrite it as four equations in three unknowns and then use substitution to end up with the solution to each of the unknowns written in terms of a single parameter. But nowhere in the ancient or medieval civilizations involved does this kind of representation appear; rather, all discussions are carried out in words. However, there is a common thread to all of the solutions produced, namely, that this is an indeterminate problem with multiple solutions. Unlike today, however, these are not expressed in terms of a parameter. In fact, in most cases there is only one solution given. Sometimes, this results from choosing a value for one of the unknowns or for some combination of them, thus reducing this to a set of three equations in three unknowns, which possesses a single solution.

### Diophantus

The earliest appearance of this problem of which I am aware is in the work of Diophantus in the third century CE. Diophantus, however, does not discuss the problem in terms of men buying a horse, but simply as an abstract numerical problem, one of many such problems in his thirteen-book Arithmetica. And since this is a linear problem, it is one that he proposes in Book I. As he puts it in problem 24 of that book, we are:

To find three numbers such that, if each receives a given fraction of the sum of the other two, the results are all equal. [Heath, 1964, 139]

Here is his solution in the outline form given by Heath, where the given fractions he uses are 1/3, 1/4, and 1/5. Diophantus basically uses the relations to calculate the second and third amounts in terms of the first as well as determine the total amount in terms of the first.

Let the first receive 1/3 of second + third, the second 1/4 of third + first, and third 1/5 of first + second. Assume the first $$= x$$ and, for convenience sake, take for the sum of second and third a number of units divisible by 3, say 3. [Here is where Diophantus, as usual, converted his indeterminate problem to a determinate one.]

In modern notation, Diophantus noted that the sum of the three numbers is $$x + 3,$$ while all the results are equal to $$x + 1.$$ Since $y+{\tfrac{1}{4}}(z+x)=x+1,$ we have $$4y + z + x = 4x + 4.$$ So $$3y + x + 3 = 4x + 4$$ and $$y=x+{\tfrac{1}{3}}.$$ Similarly, $$z=x+{\tfrac{1}{2}}.$$ Therefore, $x+\left(x+{\tfrac{1}{3}}\right) +\left(x+{\tfrac{1}{2}}\right)=x+3$ and $$x = 13/12.$$ So, $$y = 17/12$$ and $$z = 19/12.$$ The single integral solution that Diophantus presents, then, is 13, 17, 19.

Our question here is why Diophantus included this problem. Of course, one could ask that question about many of Diophantus’s problems. We do not know what he was thinking here. Did he know the problem of men buying a horse? Or did the word problem come up later?

### Abu Bakr al-Karaji

The problem shows up again in the eleventh century in the work of Abū Bakr al-Karaji (c. 953-1029) in Baghdad. In problem 26 of section III of the al-Fakhri [Woepcke, 1853, 95-96], we see Diophantus’s problem repeated, except that the sum of all the three expressions is given as 20. By this time, Diophantus’s work had been translated into Arabic, but al-Karaji does not refer to him. In symbols, al-Karaji’s problem, now determinate, is:

$x+{\tfrac{1}{3}}(y+z)=20;\,\,\,\,y+{\tfrac{1}{4}}(x+z)=20;\,\,\,\,z+{\tfrac{1}{5}}(x+y)=20.$

Al-Karaji solves this via a relatively modern procedure, using the first and second equations to get $$z$$ in terms of $$x,$$ then the first and third equations to do this a second way, then solving for $$x.$$ He finds that $$x=10{\tfrac{2}{5}},$$ and then calculates that $$y=13{\tfrac{3}{5}}$$ and $$z=15{\tfrac{1}{5}}.$$

But he then repeats the problem by supposing that the sum of the three expressions is any number:

If the second member of the proposed equation is not given, one may assign to it any given number and proceed as above. One may also give a value to any unknown or put x + z = 4, because one needs to take a quarter of this sum, or equal to any other number.

In making the supposition that x + z = 4, a supposition similar to that of Diophantus, he solves it in a way similar to that of his Greek predecessor, in fact finding that x = 13/8, y = 17/8, and z = 19/8, or, in integers, x = 13, y = 17, z = 19.

Again, al-Karaji is presenting an abstract problem, one of numerous problems in his text, many of which are similar to those of Diophantus. We do not know whether he was aware of the word problem of the horse. The first time this word problem actually appears, as far as I know, is two centuries later in the Liber Abbaci of Leonardo of Pisa (c. 1170-1240), often called Fibonacci, with the exact same fractions. The solution, however, is much less “algebraic” than that of Diophantus or al-Karaji.

### Leonardo of Pisa

According to Leonardo:

There are three men having bezants who desire to buy a horse. And as none of them can buy it, the first proposes to take from the other two men 1/3 of their bezants. And the second proposes to take 1/4 of the bezants of the other two men. And similarly the third proposes to take 1/5 of the others, and thus each proposes to buy the horse. [Sigler, 2002, 357]

Leonardo presented a recipe for solving this, which could be used whatever fractions are given. But rather than give that, we will consider Fibonacci’s logic, since he explained the recipe in an earlier similar problem. If we let $$s$$ be the sum of $$x, y,$$ and $$z,$$ the problem results in the equations $s-p={\tfrac{2}{3}}(y+z) ={\tfrac{3}{4}}(x+z) ={\tfrac{4}{5}}(x+y).$

It follows that since the second and third keep 2/3 of their money, they give away 1/2 of what they keep; similarly, the first and third give up 1/3 of what they keep; and the first and second give 1/4 of what they keep. Since 12 is the least common multiple of the three denominators, Leonardo first assumes that what remains to any two men after they have given up part of their money is 12, i.e. s – p = 12. So, if the second and third keep 12, then they gave 6. Therefore, y + z = 18. Similarly, if the first and third keep 12, then they gave 4, so x + z = 16. And if the first and second keep 12, then they gave 3, so x + y = 15. Therefore, adding these three equations together, we get 2s = 49. But 2 does not divide 49 and we would like integral answers, so we double all the numbers and find that s = 49. Since now we are assuming that s – p = 24, we find that p = 25. Then, since the first takes from the second and third one half of what they keep, and they keep 24, it follows that the first takes 12 to reach the price of 25, so x = 13. By similar arguments, y = 17 and z = 19.

### Jordanus de Nemore

We see a similar problem in De Numeris Datis, the algebra work of Jordanus de Nemore, sometime later in the thirteenth century, this time with four numbers [Hughes, 1981, 154-155]. But again, since Jordanus was writing a theoretical book, he did not present the problem in terms of buying horses, but simply as an abstract numerical problem. He presented a numerical example and then a general abstract justification using many letters as he usually did. Unfortunately, his methodology is quite unclear, both in the abstract formulation and in the example, although it seems that he is using a form of false position.

Figure 1. An annotated image of the anatomy and pathology of a horse from Kitab al-baytara (Book on Veterinary Medicine), a 15th century work. The manuscript in which this image appears was copied in Egypt in 1766. For more information, see the article, "Anatomy of the Horse in the 15th Century," by Rania Elsayed, in MuslimHeritage.com, in which this image appears. (Source: The manuscript is held by the Bibliothèque nationale de France in Paris in its Manuscrits orientaux as Arabe 2817.)

Moving to North Africa, we find the horse problem in the algebraic work of Aḥmad ibn al-Bannā (1256-1321), around 1300 [Berggren, 2016, 419-420]. In his problem, instead of a horse, there was just an “animal”. And since the problem was presented in a book on algebra, ibn al-Bannā solved it using unknowns. In fact, he used two unknowns explicitly, calling the first a “thing” and the second a “dinar”, while he assumed that the third person has a particular number of dirhams. Diophantus and al-Karaji had, in effect, used only one unknown and expressed the other values in terms of that one. Ibn al-Bannā, however, found two separate expressions for the price in terms of the “thing”, then equated them to find the “thing”.

According to ibn al-Bannā:

Three men want to buy an animal together. If the first says to the second and third, “If one takes half of what you have and adds it to what I have, I will have the price of the animal,” and if the second says to the first and the third, “If one takes a third of what you have and adds it to what I have, I will have the price of the animal,” and if the third says to the first and the second, “If one takes a fourth of what you have and adds it to what I have, I will have the price of the animal,” how much does each have?

[In what follows, I will “translate” each of ibn al-Bannā’s statements into a modern algebraic expression. We start by assuming that the amount the first has is $$x,$$ the amount the second has is $$y,$$ and that the third has 3.]

You take, then, the half of what the second and third have, you add it to what the first has and you will have that the price of the animal is a thing plus half a dinar plus a dirham and a half.

$\left[x+{\tfrac{1}{2}}(y+3)= x+{\tfrac{y}{2}}+1{\tfrac{1}{2}}=p.\right]$

Then you take a third of what the first and third have and add it to the second. The price of the animal is then a dinar plus one-third of a thing plus a dirham.

$\left[p={\tfrac{x}{3}}+1+y.\right]$

So this is equal to the thing plus a half dinar plus a dirham and a half, the first [expression for the] price.

$\left[{\tfrac{x}{3}}+1+y= x+{\tfrac{y}{2}}+1{\tfrac{1}{2}}.\right]$

You simplify and you have: a half dinar is equal to two-thirds of a thing plus a half dirham.

$\left[{\tfrac{y}{2}}={\tfrac{2}{3}}x+{\tfrac{1}{2}}.\right]$

So the dinar is equal to a thing and a third of a thing plus one dirham.

$\left[y=1{\tfrac{1}{3}}x+1.\right]$

So the price of the animal is equal to a thing and two-thirds of the thing plus two dirhams.

$\left[p=1{\tfrac{2}{3}}x+2.\right]$

Then you take the fourth of what the first and the second have and you add it to what the third has. This will be three dirhams and a fourth of a dirham and three-sixths plus a half of a sixth of a thing, and this is the price of the animal.

$\left[{\tfrac{1}{4}}(x+y)+3={\tfrac{1}{4}}\left(x+1{\tfrac{1}{3}}x+1\right)+3={3\tfrac{1}{4}}+\left({\tfrac{1}{2}}+{\tfrac{1}{12}}\right)x=p.\right]$

It is equal to a thing and two-thirds of a thing plus two dirhams, and this is the first price.

$\left[1{\tfrac{2}{3}}x+2={3\tfrac{1}{4}}+\left({\tfrac{1}{2}}+{\tfrac{1}{12}}\right)x.\right]$

You simplify and you come to the third type [of equation]. The thing will be equal to a dirham and two-thirteenths of a dirham, and this is what the first has.

$\left[{\tfrac{5}{4}}={\tfrac{13}{12}}x\implies x={\tfrac{15}{13}}=1{\tfrac{2}{13}}.\right]$

What the second has is equal to two dirhams and seven-thirteenths of a dirham,

$\left[y={\tfrac{4}{3}}x+1={\tfrac{20}{13}}+1=2{\tfrac{7}{13}},\right]$

and that of the third is three dirhams. The price of the animal is three dirhams plus twelve-thirteenths of a dirham.

$\left[p=1{\tfrac{2}{3}}x+2=\left(1{\tfrac{2}{3}}\right)\left(1{\tfrac{2}{13}}\right)+2=3{\tfrac{12}{13}}.\right]$

If you wish to get rid of the fractions, multiply the price of the animal and what each one has by any number whatever having a thirteenth and you will have whole number [solutions]. This problem … is indeterminate. Think about it and take it as a model.

### Levi ben Gershon

Now we move across the Mediterranean to southern France a few years later. In 1321, Levi ben Gershon (1288-1344) produced the Ma‘ase Ḥoshev (The Art of the Calculator), probably the most substantial mathematics work written in Hebrew during the Middle Ages. The book had three parts, a first, theoretical part dealing with various arithmetic results, including the basic combinatorial theorems, all proved with Euclidean rigor. The second part had more practical results, and the third part was a selection of problems. We first look at proposition 53 from the theoretical part, a proposition that is the same as the one of Diophantus. Levi, however, did not derive a single solution like Diophantus but instead gave a general solution.

To find three numbers such that the first, increased by a given part of the sum of the other two, equals the second, increased by another, also given, smaller part of the sum of the other two, and equals the third, increased by a given third part of the sum of both others that is smaller than the other two parts. [Wagner, 2016, 259]

In symbols, the problem is to find three numbers $$h, t, k$$ such that

$h+{\tfrac{1}{a}}(t+k)=t+{\tfrac{1}{b}}(h+k)=k+{\tfrac{1}{c}}(h+t),$

where $$a<b<c.$$ Levi, using Hebrew letters for the unknowns and various combinations of them, states the solution in the form we would write as

$h=(a-2)bc+c+b-a,\,\,t=h+2(b-a)(c-1),\,\,k=t+2(c-b)(a-1).$

However, he did not give any derivation of this result. He just gave a detailed proof that this is correct. [One can derive this algebraically from the three equations with a bit of work. And what I mean by “using Hebrew letters for the unknowns and various combinations” is that, beginning with a letter representing a, he gives another letter to represent a – 2; similarly, with letters to represent b and c, he lets an additional letter represent bc, and so on.]

Levi certainly understood that this problem was indeterminate. For in the problem section of his book, he presented a problem that uses this result, but requires a definite answer.

Problem 21. The first number plus a given part of the sum of the second and third numbers equals the second number plus some other given part of the sum of the first and third numbers. It also equals the third number plus another given part of the sum of the first and the second numbers. One of the numbers is given. What are the rest of the numbers? [Simonson, 2000, 403]

Levi gave an example and expected the reader to be able to substitute the numbers into the formula and then apply proportions:

The first with a fourth of the rest equals the second with a sixth of the rest. This also equals the third with a ninth of the rest. …. The second number is 20. We want to know, what are the rest of the numbers?

Substituting into the formula with a = 4, b = 6, and c = 9 gives the first number as 119, the second as 151, and the third as 169. But since the second number should be 20 rather than 151, we use the ratio 20:151 to calculate the first and third numbers. Thus, the first number is $$15{\tfrac{115}{151}}$$ and the third is $$22{\tfrac{58}{151}}.$$ As Levi notes, “you can check this if you wish.”

### Elijah Mizrahi

Perhaps 200 years after Levi, Elijah Mizrahi (1455-1525) wrote an arithmetic book in Constantinople.  Among the problems he presented was the problem of three men buying a fish (I guess that Jews in Constantinople then did not buy horses). But unlike Fibonacci and others, Mizrahi named the men. And he also specifically asked for the ratio among their moneys, rather than a particular answer.

Reuven, Simon and Levi went to the fish market and found a fish.
Reuven said to his friends: If I gave all my money, and each of you gave half of yours, we could buy the fish.
Simon answered and said: If I gave all my money, and each of you gave a third of yours, we could buy the fish.
Levi answered and said: If I gave all my money, and each of you gave a quarter of yours, we could buy the fish.
What is the ratio between their money, that is, the ratio of each to each? [Wagner, 2016, 252-253]

Here, if $$a$$ is the money of Reuven, $$b$$ is Simon's money, $$c$$ is Levi's money, and $$d$$ is the price of the fish, we can write three equations for this problem:

$a+{\tfrac{1}{2}}b+{\tfrac{1}{2}}c=d,\,\,b+{\tfrac{1}{3}}a+{\tfrac{1}{3}}c=d,\,\,c+{\tfrac{1}{4}}a+{\tfrac{1}{4}}b=d.$

We say that the statements of Reuven and Simon determine that half of Simon's money equals two thirds of Reuven's and one sixth of Levi's. Therefore, Simon's entire money equals one and a third of Reuven's and a third of Levi's. That is because all of Reuven's money together with half of the others', as Reuven says, equals all of Simon's money together with a third of the others', as Simon says. Therefore, what Simon added to Reuven's statement – a half of his [Simon's] money – equals what he removed from Reuven's total – two thirds of Reuven's – and to what he removed from half of Levi's money – one sixth of Levi's.

[In symbolic notation, $a+{\tfrac{1}{2}}b+{\tfrac{1}{2}}c=b+{\tfrac{1}{3}}a+{\tfrac{1}{3}}c,$ so $${\tfrac{2}{3}}a+{\tfrac{1}{6}}c$$ equals $${\tfrac{1}{2}}b.$$ We rescale by two and obtain: $$b=1{\tfrac{1}{3}}a+{\tfrac{1}{3}}c.$$]

In the same way, according to Reuven's and Levi's statements, it is determined that half of Levi's money equals three quarters of Reuven's and one quarter of Simon's. Therefore, in the same ratio, a third of Levi's money equals half of Reuven's and a sixth of Simon's.

[By comparing the first and third identities, we get: $${\tfrac{1}{2}}c ={\tfrac{3}{4}}a+{\tfrac{1}{4}}b.$$ Rescaling the last identity by two thirds we get: $${\tfrac{1}{3}}c ={\tfrac{1}{2}}a+{\tfrac{1}{6}}b.$$]

We already know that Simon's money equals one and a third of Reuven's and a third of Levi's. Therefore, Simon's money equals one and five sixths of Reuven's and one sixth of Simon's own. We remove one sixth of Simon's which is common [to both sides], and remain with five sixths of Simon's money being equal to one and five sixths of Reuven's. According to the same ratio, all of Simon's money equals twice Reuven's and its fifth.

[By substitution, $b={1\tfrac{1}{3}}a+{\tfrac{1}{3}}c =1{\tfrac{1}{3}}a+\left({\tfrac{1}{2}}a+ {\tfrac{1}{6}}b\right).$ Rearranging we obtain: $${\tfrac{5}{6}}b =1{\tfrac{5}{6}}a,$$ or $$b=2{\tfrac{1}{5}}a$$.]

Therefore, it is determined that if Reuven had five, Simon would have eleven. We already saw that half of Levi's money equals three quarters of Reuven's and one quarter of Simon's. This ratio determines that all of Levi's money will be one and a half of Reuven's and half of Simon's. So it is determined that if Reuven had five and Simon had eleven, as we have already seen, Levi would have thirteen. This determines the price of the fish to be seventeen.

### Summary: Men Buying a Horse

In summary, the procedures for solving this problem vary quite a bit in detail, although they generally have the same goal of eliminating some unknowns so as to have a linear relationship between two of them. One can then figure out the ratios of the answers and, if one wishes, give integral answers to the problem. But since no two solution procedures are exactly alike, there is no evidence that the various authors simply copied from one another. But did they read each other’s work? That is unfortunately a very difficult question to answer.

# Recreational Problems in Medieval Mathematics - Men Finding a Purse

Author(s):
Victor J. Katz (University of the District of Columbia)

The second problem we want to discuss is that of men finding a purse. The basic problem outline is that three (or two or more) men find a purse. The first says that if he took the purse, then, with the money he already had, he would have a certain multiple of the sum of what the others had. The second says that if he took the purse, then, with the money he had, he would have another multiple of the sum of what the first and third had. Similarly, the third says that if he took the purse, then, with the money he had, he would have a third multiple of the sum of what the first and second had. The question then is to determine how much money each person had originally and to determine the amount of money in the purse.

In modern symbolism, the problem can be written in the form of three equations in four unknowns: $x+p=a(y+z),\,\,\,y+p=b(z+x),\,\,\,z+p=c(x+y).$

### Mahavira

The earliest appearance of this problem, as far as I know, is in the work of Mahāvīra (c. 800-870) in India in the ninth century. Mahāvīra gave a rule for finding the solution to such a problem in general, but it appears that he did not quite know what he was doing:

The rule for arriving at the value of the contents of a purse which, when added to what is on hand with each of certain persons, becomes a specified multiple of the sum of what is on hand with the others:

The quantities obtained by adding one to each of the specified multiple numbers in the problem and then multiplying these sums with each other, giving up in each case the sum relating to the particular specified multiple, are to be reduced to their lowest terms by the removal of common factors. These reduced quantities are then to be added. Thereafter the square root of this resulting sum is to be obtained from which one is to be subtracted. [This is a really strange statement, because this is in no way a quadratic problem. In fact, what is wanted here is that you take the number of men, instead of the square root. In the numerical example given, this produces the same result.] Then the reduced quantities referred to above are to be multiplied by this square root as diminished by one. Then these are to be separately subtracted from the sum of these same reduced quantities. Thus the moneys on hand with each of the several persons are arrived at. These quantities measuring the moneys on hand have to be added to one another, excluding from the addition in each case the value of the money on the hand of one of the persons; and the several sums so obtained are to be written down separately. These are then to be respectively multiplied by the specified multiple quantities mentioned above; from the several products so obtained the already found out values of the moneys on hand are to be separately subtracted. Then the same value of the money in the purse is obtained separately in relation to each of the several moneys on hand. [Katz, 2016, 563-564]

Thus, Mahāvīra presented a recipe, but he did not explain where the recipe came from. Of course, that is usual in Indian mathematics of the time. He then assumed the reader would use the recipe to solve the following problem:

Example: Three merchants saw dropped on the way a purse containing money. One said to the others, “If I secure this purse, I shall become twice as rich as both of you with your moneys on hand.” Then the second said, “I shall become three times as rich.” Then the other said, “I shall become five times as rich.” What is the value of the money in the purse, as also the money on hand with each of the three merchants? [Answer is 1, 3, 5, 15]

Here is the solution, according to the recipe:

2 + 1 = 3;  3 + 1 = 4;  5 + 1 = 6.

3 $$\times$$ 4 = 12 (c);  3 $$\times$$ 6 = 18 (b);  4 $$\times$$ 6 = 24 (a).

Reduce to 2, 3, 4;  2 + 3 + 4 = 9;  Square root is 3;  3 – 1 = 2.

2 $$\times$$ 2 = 4;  2 $$\times$$ 3 = 6;  2 $$\times$$ 4 = 8;

9 – 4 = 5 (c);  9 – 6 = 3 (b);  9 – 8 = 1 (a).

Then 5 + 3 = 8,  2 $$\times$$ 8 = 16,  and 16 – 1 = 15, the purse.

### Leonardo of Pisa

The purse problem next occurs in the work of Leonardo of Pisa (Fibonacci). As in the horse problem, Fibonacci presented several versions of the problem. Here is the simplest one:

Two men who had denari found a purse with denari in it; thus found, the first man said to the second: If I take these denari of the purse, then with the denari I have I shall have three times as many as you have. Alternately the other man responded: And if I shall have the denari of the purse with my denari, then I shall have four times as many as you have. It is sought how many denari each has, and how many denari they found in the purse. [Sigler, 2002, 317-318]

Leonardo’s first solution is a completely arithmetic one:

It is indeed noted that because the first, having the purse, has three times as many as the second, that if he has with the purse 3, then the second has 1; therefore among them both and the purse they have 4; therefore as the first with the purse has 3, he has 3/4 the entire sum of their denari and the purse.  And for the same reason, as the second with the purse has four times as many as the first, it is necessary for him to have 4/5 of the same sum. Therefore you find the least common denominator of 4/5 and 3/4, and it will be 20. Therefore you put the sum of the denari to be 20, of which the first with the purse has 3/4, namely 15. And the second with the purse has 4/5, namely 16; therefore among them both with the purse counted twice they have 31; the difference between 31 and 20, namely 11, is truly the denari of the purse. Because the purse is counted twice, and as one should only count it once, the purse is therefore counted once more than it should be. Whence the denari difference between the 20 and the 31, namely 11, is one times that which is found in the purse. Therefore you subtract the 11 from the 15; there remains 4, and this many the first man has; next you subtract the 11 from the 16; there remains 5, and this many the second has; therefore the first has 4, and the second 5, which added to the 11 of the purse makes 20 which we can put for the sum.

Leonardo solved this problem a second way, using algebra:

You put the first to have the thing; therefore with the purse he has the thing and the purse, which are triple the denari of the second; therefore the second has one third of the thing and one third of the purse. Therefore, if he has the purse, he will have the purse and a third of a purse, and a third of the thing, which equal IIII things, namely quadruple the denari of the first, as the second with the purse has four times as many as the first. You therefore subtract from both parts one third thing; there will remain the purse, and a third of the purse, that are equal to IIII things minus one third thing. Therefore, triple one and one third of a purse, namely 4 purses, are equal to triple IIII things minus triple one third of a thing, namely 11 things, and because four times 11 is equal to eleven times IIII, the proportion of denari of the purse to denari of the first man will be as 11 to 4. Whence if there are 11 denari in the purse, the first man has 4 denari, of which a number of thirds, namely 5, the second necessarily has, as the first with the purse has triple it.

### Levi ben Gershon

Levi ben Gershon also presented the purse problem, but, as before, abstractly. First, he stated a theorem equivalent to the purse problem in the abstract portion of his text:

58 Problem: To find three numbers such that the sum of the first and third contains the second as a factor as many times as a given number and such that the sum of the second and third contains the first as a factor as many times as a second given number. [Wagner, 2016, 259]

Levi represented the given numbers by a, b. Then he claimed that the first number turns out to be a + 1, the second number b + 1, and the third number ab – 1. He then demonstrated his claim by a detailed argument.

As before, he presented this theorem as a problem in the problem section of the book, asking to get a definite answer if one of the numbers is known:

Problem 18.  We add one number to a second number; and the ratio of the result to a third number is given. When we add the first number to the third number, the ratio of the result to the second number is a second given number. One of the three numbers is known. What are each of the remaining numbers? [Wagner, 2016, 267-268]

For example, when you add the first number to the second number, its ratio to the third equals 3 wholes and 2 fifths and a seventh $$\left[3{\tfrac{19}{35}}\right].$$ When the first is added to the third, its ratio to the second equals 7 wholes and 2 thirds and a fourth $$\left[7{\tfrac{11}{12}}\right].$$ The second number is 30. We want to know: what is the value of each remaining number?

Levi used his formula to get one set of three numbers and then used ratios to find the others. The first (i.e., the purse) turns out to be $$178{\tfrac{98}{159}},$$ while the third number is $$58{\tfrac{281}{318}}.$$ Note that Levi did not hesitate to give his readers practice in calculating with fractions.

Figure 2. These images of a valuable "mistake coin" appeared in London's Daily Mail (DailyMail.com) under the headline "800-year-old coin minted by Henry III but then scrapped when blundering officials realised it was worth more as gold than its face value is expected to make £500,000 at auction" (Dec. 26, 2017). Henry III (1207-1272) reigned in England from 1216 to his death in 1272. Henry's unfortunate goldsmith was William of Gloucester.

# Recreational Problems in Medieval Mathematics - Conclusions and Questions

Author(s):
Victor J. Katz (University of the District of Columbia)

Finally, we should note that the purse problem as well as the horse problem turn up in other places as well, from India to early modern Europe. The question that is difficult to answer, however, is how did these problems travel?

There is no direct evidence to answer this question. We see that frequently the problems are essentially identical in works separated by time and distance – albeit sometimes with different constants.  Yet the solution methods have considerable variation. Some authors attack these problems in what we would call an algebraic fashion, while others use more arithmetic methods. Sometimes the problems are abstract numerical problems, while at other times they are tied to a “story”. It is difficult to believe, however, that each of these people invented the problem from scratch, since whether one is in the abstract or story mode, the problem seems too specialized for independent invention.  Thus, somehow the idea of the problems, at least, must have traveled. It is certainly possible to imagine travelers between east and west carrying knowledge of these problems, perhaps even without the solution. We know, for example, that Diophantus’s Arithmetica was translated into Arabic by the tenth century, so would have been available for al-Karaji and later writers in Arabic. However, we know of no translation of Diophantus into Hebrew in the Middle Ages, so perhaps Levi ben Gershon learned of this through Arabic sources. On the other hand, we have no direct evidence of Indian problems being taken to western Europe, although, of course, the Indian decimal place-value system did make this journey. But given that Mahāvīra himself did not have an accurate idea of the solution procedure of the problem, it would seem that this problem must have come to him from elsewhere – whether from India or some other country. So there are lots of puzzles out there that these puzzle problems engender. It seems that we will have to wait for more research to get good answers.

Victor J. Katz is Professor Emeritus of Mathematics of the University of the District of Columbia; founding co-editor (with Frank J. Swetz) of MAA Convergence; author of History of Mathematics: An Introduction (3rd ed., 2009), widely recognized as the definitive general history of mathematics for professionals, instructors, and students; co-author with Karen Parshall of Taming the Unknown: A History of Algebra from Antiquity to the Early Twentieth Century (2014); and editor of two sourcebooks for history of mathematics, The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook (2007) and Sourcebook in the Mathematics of Medieval Europe and North Africa (2016). – Janet Beery, Editor, MAA Convergence

# Recreational Problems in Medieval Mathematics - References

Author(s):
Victor J. Katz (University of the District of Columbia)

Berggren, J. Lennart, 2016. Mathematics in the Islamic World in Medieval Spain and North Africa, in Victor J. Katz, et al, 2016.

Heath, Thomas L., 1964. Diophantus of Alexandria: A Study in the History of Greek Algebra. New York: Dover.

Hughes, Barnabas, 1981. Jordanus de Nemore: De numeris datis. Berkeley: University of California Press.

Katz, Victor J., Menso Folkerts, Barnabas Hughes, Roi Wagner, & J. Lennart Berggren, eds., 2016. Sourcebook in the Mathematics of Medieval Europe and North Africa. Princeton: Princeton University Press.

Sigler, L. E., 2002. Fibonacci’s Liber Abaci: A Translation into Modern English of Leonardo Pisano’s Book of Calculation. New York: Springer.

Simonson, Shai, 2000. “The Missing Problems of Gersonides: A Critical Edition,” Historia Mathematica 27, 243-302, 384-431.

Singmaster, David, 1996. Chronology of Recreational Mathematics. Accessed at http://anduin.eldar.org/~problemi/singmast/recchron.html

Tropfke, Johannes, 1980. Geschichte der Elementarmathematik. Bd. 1 Arithmetik und Algebra. Edited by Kurt Vogel, Karin Reich, Helmuth Gericke. Walter de Gruyter: Berlin.

Wagner, Roi, 2016. Mathematics in Hebrew, in Victor J. Katz, et al, 2016.

Woepcke, Franz, 1853. Extrait du Fakhrī, traité d’algèbre par AboùBkr Mohammed ben Alhaçan al-Karkhī. Paris: L’imprimerie Impériale.

Editor's note: For a discussion of Catholic, Jewish, and Islamic mathematical cultures in medieval Europe and the mathematics they produced, be sure to see Victor Katz's "The Mathematical Cultures of Medieval Europe," also published in Convergence in December of 2017.