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Recreational Problems in Medieval Mathematics - Men Finding a Purse

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: \[x+p=a(y+z),\,\,\,y+p=b(z+x),\,\,\,z+p=c(x+y).\]


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 ( 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. 

Victor J. Katz (University of the District of Columbia), "Recreational Problems in Medieval Mathematics - Men Finding a Purse," Convergence (December 2017)