This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with plain old algebra, which yields the shortest, simplest answers, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

A man dies and leave his estate to his sons.

The estate is divided as follows:

$1$st son gets 100 crowns $+$ $\frac{1}{10}$ of remainder of estate.

$2$nd son gets 200 crowns $+$ $\frac{1}{10}$ of remainder of estate ...

$(n)$$th$ son gets $100 × (n)$ crowns + $\frac{1}{10}$ of the remainder.

Each son receives same amount.

How many sons were there, what did each receive, and what was the estate?


Assuming all of his property was 'crowns', let the total amount he had be $\mathbf{T}$ crowns. Now the first son gets $\mathbf{100}$ crowns, so $\mathbf{T-100}$ crowns are left, of which he gets one-tenth, so his total inheritance is $\mathbf{\frac{T-100}{10} + 100}$ crowns. The second son first takes $\mathbf{200}$ crowns from the remaining estate, leaving us with $\mathbf{T-(200 + 100 + \frac{T-100}{10}) = \frac{9T - 2900}{10}}$ crowns, of which he takes one-tenth again. Gievn that each son receives an equal share, we equate these amounts with $\mathbf T$ being unknown $$\mathbf{\frac{T-100}{10} + 100 = \frac{9T - 2900}{100} + 200}$$ On solving, you are left with $\mathbf{T = 8100}$, so each son receives $\mathbf{900}$ crowns, i.e. there are $\mathbf9$ sons


Here is a slightly different approach than the one taken by @ClementGuerin.

Let $n$ be the total number of sons. Since we cannot have money left over after giving him $100n$ (as if there were money to give $1/10$ of what remained, then $9/10$ of what remained would still be there), we can compare the $n$th son to the $n-1$st.

$100n=100(n-1)+100$, and $100=1000/10$, and so before giving away the extra 100 to the $(n-1)$st son, there was $1000$ left. Therefore, the $n$th son got $900$ and so there were $9$ total sons. The only question is if the total $($9*900=8100)$ is actually split evenly when divided in the described manner. It can be verified that it is.

  • $\begingroup$ After I read your solution and Umashankar's one, I think you too are right about the interpretation. $\endgroup$ – Clément Guérin Jun 23 '15 at 10:45
  • $\begingroup$ Oh, I didn't think about your solution carefully enough to notice that we were interpreting the problem differently! But with your interpretation, the last son gets $x=100n+x/10$, making x=1000n/9$, so the total estate is $1000n^2/9$, and we can compare the first son to the last (or the second to last to the last). My intention was not to question your interpretation, only to show that one could reason some things more simply by looking at the last son. $\endgroup$ – Aaron Jun 23 '15 at 10:58

I get that the estate is $9000$, the amout received by each son is $1000$ and there are $9$ sons (however the solution is simple, I am not sure I have understood properly the problem).

Take $E$ to be the quantity of the estate. Set $x_i$ to be the amount received by the $i$-th son. You have :

$$x_1=100+\frac{E}{10}\text{ and } x_2=200+\frac{E-x_1}{10} $$

By assumption $x_1=x_2$ hence :


This gives an order $1$ equation in $E$, easily solvable, the rest of the solution follows easily.

  • 2
    $\begingroup$ The first son would get 100 crowns + 8900/10, as I understand it. Not 100 crowns + 9000/10. Perhaps @anonymous can clarify? $\endgroup$ – TonyK Jun 23 '15 at 10:38
  • $\begingroup$ @TonyK You are right, I will upvote Umashankar S'answer, it is your interpretation he is using, I think. $\endgroup$ – Clément Guérin Jun 23 '15 at 10:44

Each son get 1100, and the total estate is 10100. To find 10100 just resolve the equation $$100+0.1(x-100)=200+0.09(x-100)$$


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