From 2010 to 2026: the inheritance problem generalized

From the inheritance problem discussed in 2010 to its general form in 2026: finding the numbers is not enough; we must show that the rule works for every child. This article continues “The solution is not enough: Fibonacci’s inheritance”.

Introduction - the starting problem

A father leaves an inheritance to his children: the first receives 1 000 euros and then one tenth of what remains; the second receives 2 000 euros and then one tenth of the remainder; the third receives 3 000 euros and then one tenth of the remainder; and so on. In the end they all receive the same amount and the inheritance is completely distributed. How many children are there, how much does each receive, and what is the total inheritance?

Let n be the number of children and S their common share. The initial inheritance is nS. The conditions for the first two children are:

S = 1 000 + (nS − 1 000) / 10
S = 2 000 + [(n − 1)S − 2 000] / 10
(10 − n)S = 9 000
(11 − n)S = 18 000
S = 9 000
n = 9
E = nS = 9 · 9 000 = 81 000 euros

The system gives 9 children, 9 000 euros each, and 81 000 euros in total. But this solution uses only the first two children: it finds necessary values without yet explaining why the third, fourth, and all later children really receive the same share. We must examine the k-th child.

1. The starting point

The problem is interesting because of its hidden structure: at each turn a fixed amount increases, then the same fraction of what remains is added. In 2010 the idea was already to look beyond the numerical case for a general law. Fixed amounts 1, 2, 3, … and a fraction 1/d suggested n = d − 1 and a quadratic inheritance. In 2026 the argument is completed: we prove the relation for all children and replace 1, 2, 3, … with a, 2a, 3a, …, namely ka.

2. General statement

A father leaves an inheritance to n children. The first receives a and then 1/d of what remains; the second receives 2a and then 1/d of the remainder; the third receives 3a and then 1/d; in general, the k-th receives ka and then 1/d of what remains. All receive the same amount S and the inheritance is exhausted.

For the nontrivial case proved here, assume a > 0, d an integer greater than 1, and at least two children. With only one child, the sole share may equal the initial inheritance a for any d: one cannot deduce n = d − 1 by comparing turns that do not exist.

3. The k-th child: the decisive formula

Before the k-th child’s turn, the preceding k − 1 children have each received S. Of the inheritance nS, (n − k + 1)S remains. After removing the fixed amount ka, (n − k + 1)S − ka remains. The child also receives 1/d of this remainder. For the total to equal S:

S = ka + [(n − k + 1)S − ka] / d
dS = dka + (n − k + 1)S − ka
(d − n + k − 1)S = k(d − 1)a
(d − n − 1)S = k[(d − 1)a − S]

4. Why this identity proves everything

With at least two children, the final identity holds for both k = 1 and k = 2. Subtracting the two equations makes the left-hand side disappear and gives (d − 1)a − S = 0. Hence S = (d − 1)a. Substitution gives (d − n − 1)S = 0; because S > 0, necessarily n = d − 1.

(d − 1)a − S = 0
S = (d − 1)a
(d − n − 1)S = 0
n = d − 1

This is more than a necessary condition: substituting n = d − 1 and S = (d − 1)a into the k-th child’s identity makes both sides zero for every k from 1 to n. Thus every turn really yields the same share; at the final turn the remaining inheritance equals the fixed share na and is exhausted.

5. The total inheritance

The inheritance E is the number of children times their common share:

E = nS
E = (d − 1)²a

6. The complete generalization

• Fixed amounts: a, 2a, 3a, …, ka, …
• Fraction of the remainder: 1/d
• Number of children: n = d − 1
• Each child’s share: S = (d − 1)a
• Total inheritance: E = (d − 1)²a

The original case with 1, 2, 3, … simply has a = 1. The parameter d governs the structure; a only changes the scale of the amounts.

7. Examples: state the problem, then use the formula

In each example we apply the general formulas and then follow the distribution concretely, starting from the values found for n, S, and E.

Example 1 - a = 100, d = 10

Problem. The first child receives 100 euros and then one tenth of the remainder, the second 200 euros and then one tenth, the third 300 euros and then one tenth, and so on. All receive the same amount and the inheritance is exhausted. Find the number of children, each share, and the total inheritance.

n = 9, S = 900 euros, E = 8 100 euros.

Check. First child: 100 + (8 100 − 100) / 10 = 100 + 800 = 900 euros. Second: 7 200 euros remain; 200 + (7 200 − 200) / 10 = 200 + 700 = 900 euros. Fifth: 5 · 900 = 4 500 euros remain; 500 + (4 500 − 500) / 10 = 500 + 400 = 900 euros. Ninth: 900 euros remain; 900 + (900 − 900) / 10 = 900 euros. The inheritance is exhausted. The general law guarantees the same outcome at every other turn.

Example 2 - a = 11, d = 10

Problem. The first child receives 11 euros and then one tenth of the remainder, the second 22 euros and then one tenth, the third 33 euros and then one tenth; in general the k-th receives 11k euros and then one tenth of the remainder. All receive the same amount. Find n, S, and E.

n = 9, S = 99 euros, E = 891 euros.

Check. First: 11 + (891 − 11) / 10 = 11 + 88 = 99 euros. Second: 792 euros remain; 22 + (792 − 22) / 10 = 22 + 77 = 99 euros. Fifth: 5 · 99 = 495 euros remain; 55 + (495 − 55) / 10 = 55 + 44 = 99 euros. Ninth: 99 euros remain; 99 + (99 − 99) / 10 = 99 euros. The inheritance is again exhausted. a need not be 1, 10, 100, or 1000: it may be any positive value.

Example 3 - a = 13, d = 11

Problem. The first child receives 13 euros and then one eleventh of the remainder, the second 26 euros and then one eleventh, the third 39 euros and then one eleventh; the k-th receives 13k euros and then one eleventh. Find n, S, and E if all receive the same amount and the inheritance is exhausted.

n = 10, S = 130 euros, E = 1 300 euros.

Check. First: 13 + (1 300 − 13) / 11 = 13 + 117 = 130 euros. Second: 1 170 euros remain; 26 + (1 170 − 26) / 11 = 26 + 104 = 130 euros. Fifth: 6 · 130 = 780 euros remain; 65 + (780 − 65) / 11 = 65 + 65 = 130 euros. Tenth: 130 euros remain; 130 + (130 − 130) / 11 = 130 euros. The inheritance is exhausted.

Example 4 - a = 10, d = 11

Problem. The first child receives 10 euros and then one eleventh of the remainder, the second 20 euros and then one eleventh, the third 30 euros and then one eleventh, and so on. All receive the same amount. What is the inheritance?

n = 10, S = 100 euros, E = 1 000 euros.

Check. First: 10 + (1 000 − 10) / 11 = 10 + 90 = 100 euros. Second: 900 euros remain; 20 + (900 − 20) / 11 = 20 + 80 = 100 euros. Fifth: 6 · 100 = 600 euros remain; 50 + (600 − 50) / 11 = 50 + 50 = 100 euros. Tenth: 100 euros remain; 100 + (100 − 100) / 11 = 100 euros. The inheritance is exhausted.

8. From 2010 to 2026

In 2010 the numerical solution was not an endpoint: the task was to recognize the underlying law. In 2026 the argument is completed in two directions. The behavior of the k-th child directly proves n = d − 1 in the nontrivial case; and the sequence 1, 2, 3, … becomes a, 2a, 3a, …, namely ka. The denominator determines the number of children; a determines the scale of the amounts. The essential structure remains unchanged:

n = d − 1      S = (d − 1)a      E = (d − 1)²a

9. Final observation

A numerical solution answers a question; a generalization explains why the answer has that form. The formula for the k-th child turns an intuition into a proof, while a shows that the phenomenon does not depend on the particular choice 1, 2, 3, … . To revisit the problem that began this journey, read the 2010 article on Fibonacci’s inheritance.