The solution is not enough: Fibonacci’s inheritance

Abstract

I have always been impressed by the imagination of ancient mathematicians. They devised beautiful problems, yet their solutions are often not proved: they simply state the result with which the conditions of the problem can be checked. Using modern techniques, we can instead seek a solution with equations, systems, or other tools.

I happened, however, to read modern solutions to two Fibonacci problems from the Liber Abaci. Although correct, they left me unconvinced: they find the numbers but do not solve the problem in its essence, because they do not explain why the mechanism works. In this article I offer an explanation and a generalisation.

Introduction

While reading Mathematical Games of the Middle Ages, edited by Nando Geronimi, I was puzzled by the proposed solution to the following problem.

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/7 of what remains.”

To the second son he said: “You shall take 2 bezants and 1/7 of what remains.”

To the third son he said: “You shall take 3 bezants and 1/7 of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/7 of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Note. The problem appears in part seven of chapter XII of the Liber Abaci, under the title “The inheritance of a man’s property”, on folio 70 recto of the manuscript.

The proposed solution

The proposed solution calculates what the first son receives and what the second son receives, then requires those two amounts to be equal. Let N be the number of sons and S the amount received by each; N · S is the entire inheritance.

First son:  1 + (N · S − 1) / 7
Second son: 2 + ((N − 1) · S − 2) / 7

Because both receive S, the following system results:

1 + (N · S − 1) / 7 = S
2 + ((N − 1) · S − 2) / 7 = S

Solving it gives N = 6 and S = 6.

Checking an inheritance of 36 bezants

My concern is that, although the reasoning is correct, it does not begin with the complete hypothesis: there are not merely two sons. Moreover, why do the third, fourth, fifth, and sixth sons also receive the same amount?

A solution that uses all the data

Let us begin with the correct hypothesis. Let T be the entire inheritance and k the amount received by each son. Before son j takes his turn, (j − 1)k coins have already been distributed. Son j therefore receives:

j + [T − (j − 1)k − j] / 7

Because every son receives the same amount, for j = 1, 2, 3, … we can write:

j + [T − (j − 1)k − j] / 7 = k

Multiplying both sides by 7 and expanding:

7j + T − (j − 1)k − j = 7k
7j + T − jk + k − j = 7k
T + k − 7k = jk − 7j + j
T − 6k = jk − 6j
T − 6k = j(k − 6)

This equality must hold for every j. That happens only if T − 6k = 0 and k − 6 = 0. Thus k = 6 and T = 36.

The first generalisation

Notice that in the problem 7 = k + 1. To see whether the structure can be generalised, replace the denominator 7 by k + 1:

j + [T − (j − 1)k − j] / (k + 1) = k

Multiplying both sides by k + 1 gives:

(k + 1)j + T − (j − 1)k − j = (k + 1)k
(k + 1)j + T − jk + k − j = (k + 1)k
kj + j + T − jk + k − j = (k + 1)k
T + k = k² + k
T = k²

Because the total inheritance is also T = N · k, the equality T = k² implies N = k (for k > 0). The situation therefore occurs when the inheritance is the square of the number of sons.

Example with 8 sons and 64 bezants

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/9 of what remains.”

To the second son he said: “You shall take 2 bezants and 1/9 of what remains.”

To the third son he said: “You shall take 3 bezants and 1/9 of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/9 of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Example with 9 sons and 81 bezants

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/10 of what remains.”

To the second son he said: “You shall take 2 bezants and 1/10 of what remains.”

To the third son he said: “You shall take 3 bezants and 1/10 of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/10 of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Concise check. Before son j takes his turn, the remaining inheritance is 81 − (j − 1) · 9 = 9(10 − j). His share is therefore:

j + [9(10 − j) − j] / 10 = 9

This identity holds for every j = 1, …, 9; consequently, all nine sons receive 9 bezants.

Conclusions for the first form

In general, if there are n sons and an inheritance of n² bezants, the problem becomes:

Shortly before his death, a man called his eldest son and said: “Divide my modest possessions among yourselves. You shall keep 1 bezant and 1/(n + 1) of what remains.”

To the second son he said: “You shall take 2 bezants and 1/(n + 1) of what remains.”

To the third son he said: “You shall take 3 bezants and 1/(n + 1) of what remains.”

He called all his sons in decreasing order of age in this way, giving each one 1 bezant more than the preceding son and always 1/(n + 1) of what remained. Finally, the last son took everything left.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Answer: there were n sons, the inheritance was n² bezants, and every son received n bezants.

A second form of the problem

The same method also generalises the following problem, in which the fraction of the available money is calculated first and the fixed number of bezants is then added.

Shortly before his death, a man called his eldest son and said: “My sons, divide my modest possessions among yourselves. You shall keep 1/7 of all the bezants, plus 1 bezant.”

To the second son he said: “You shall take 1/7 of the remaining bezants, plus another 2 bezants.”

He called all his sons in decreasing order of age in this way, giving each one 1/7 of the remaining bezants and one bezant more than the preceding son.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Note. This problem also appears in part seven of chapter XII of the Liber Abaci, under the title “The inheritance of a man’s property”, on folio 70 recto of the manuscript.

The proposed solution again lets N be the number of sons and S the share of each:

First son:  (N · S) / 7 + 1 = S
Second son: ((N − 1) · S) / 7 + 2 = S

The solutions are N = 6 and S = 7.

Checking an inheritance of 42 bezants

By the same reasoning used above, it is easy to prove that, in general, if n is the number of sons, the inheritance is n(n + 1) and each son receives n + 1 bezants. The statement becomes:

Shortly before his death, a man called his eldest son and said: “My sons, divide my modest possessions among yourselves. You shall keep 1/(n + 1) of all the bezants, plus 1 bezant.”

To the second son he said: “You shall take 1/(n + 1) of the remaining bezants, plus another 2 bezants.”

He called all his sons in decreasing order of age in this way, giving each one 1/(n + 1) of the remaining bezants and one bezant more than the preceding son.

After the whole inheritance had been divided, the sons saw that each had received the same number of bezants.

How many sons were there, and how many bezants did each receive?

Proof of the second generalisation

Let T = n(n + 1) be the inheritance and k = n + 1 the common share. Before son j takes his turn, the amount left is:

T − (j − 1)k = (n − j + 1)(n + 1)

The amount received by son j is therefore:

[T − (j − 1)k] / (n + 1) + j
= n − j + 1 + j
= n + 1

The calculation is valid for every j = 1, …, n, so all n sons receive the same share.

Answer: there are n sons, each receives n + 1 bezants, and the inheritance is n(n + 1) bezants.

Prepared by Salvatore Mosaico.