When reading Mario Livio's beautiful book about the golden ratio (see below), I came across the following problem taken from Leonardo of Pisa's book Liber abaci, which was published in 1202 and introduced Arabic numerals in Europe (Leonardo of Pisa was posthumously known as Fibonacci ):
A man whose end was approaching summoned his sons and said: "Divide my money as I shall prescribe." To his eldest son, he said, "You are to have 1 gold coin and a seventh of what is left." To his second son he said, "Take 2 gold coins and a seventh of what remains." To the third son, "You are to take 3 gold coins and a seventh of what is left." Thus he gave each son 1 gold coin more than the previous son and a seventh of what remained, and to the last son all that was left. After following their father's instructions with care, the sons found that they had shared their inheritance equally. How many sons were there, and how large was the estate?
At a first glance, this sounds like a weird problem to solve, but with the help of some little algebra, its solution can be easily found. Let's note with a small x the amount that each son receives (which is the same amount for every son) and with a capital X the amount of the estate.
The eldest son receives 1 gold coin and a seventh of what is left, hence we can write the following equation :
where the amount that is left after the 1 coin has been taken is obviously
The second son receives 2 gold coins and a seventh of what is left:
where the amount that is left after the first son has taken his part x and the second son has taken his 2 coins is
Now let's look at the the third son and check if we can generalise the procedure. He receives 3 gold coins and a seventh of what is left:
where the amount that is left after the first and second son have taken their parts x and the third son has taken his 3 coins is
We start to see a pattern and it is not difficult to verify that this partitioning rule can be readily extended to the n-th son, who gets n gold coins and a seventh of what is left:
which gives
where the amount that is left generalises to
From the previous equation concerning the n-th son, we can derive the total amount of the estate X in terms of the amount that each son gets x :
Since the total amount X must be constant, i.e. it cannot depend on n, the solution is simply given by x = 6:
and hence X = 36, from which it follows that the total number of sons is 6.
This way of dividing the estate can be actually generalised. For example, if the estate amounts to 49 gold coins and there are 7 sons, then by giving each son 1 gold coin more than the previous one and an eighth of what remains, results in equally sharing the estate, i.e. each son gets 7 gold coins. If the estate amounts to 100 gold coins and there are 10 sons, then by giving each son 1 gold coin more than the previous one and an eleventh of what remains, results again in equally sharing the estate, i.e. each son gets 10 gold coins.
In general, if the estate amounts to N² gold coins and there are N sons, then by giving each son 1 gold coin more than the previous one and an (N+1)-th of what remains, results in equally sharing the estate, i.e. each son gets N gold coins.
This can be easily proved by noting that the amount that is left is given by
where X = N² and x = N.
Therefore by applying the rule, it follows that the n-th son receives:
by definition.
As we can see, the problem takes advantage of this mathematical property which guarantees that each son gets the same amount even though one might be deceived by the partitioning rule. Clever indeed!
Here is the link to Mario Livio's nice book about the Golden Ratio:
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