May 2016
Problem of the Month

Inheritance Problem

Fibonacci posed the following problem:

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 1/7 of what is left."
To his second son he said, "Take 2 gold coins and 1/7 of what remains."
To the third son, "You are to take 3 gold coins and 1/7 of what is left."
Thus he gave each son 1 gold coin more than the previous son and 1/7 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?

Solution to the Problem:

There were six sons and the estate consisted of 36 coins.

Let "G" = total number of gold coins and "N" = total number of sons.

We know the first son gets 1 gold coin and 1/7 of what is left (which is now [G - 1] because he took one).
Therefore, the first son gets
{ 1 + (1/7)*(G - 1) } gold coins.

Simplify the above expression to
{ 6/7 + G/7 }

We also know the second son gets 2 gold coins and 1/7 of what is left, or
{ 2 + (1/7)*(what is NOW left) }

The key is realizing that what is NOW left for the second son is { G - [6/7 + G/7] - 2 } because it is the total amount minus what the 1st son took minus the 2 gold coins that the second son took.

Therefore the second son receives
{ 2 + (1/7)*(G - 6/7 - G/7 - 2) }, which simplifies to:
6*G/49 + 78/49

Set these two expressions equal to one another because we know that each son receives the same amount.
6*G/49 + 78/49 = 6/7 + G/7 ----> G = 36 gold coins.

Now we solve for the number of sons "N".
We calculate that the first son got 6 gold coins by plugging in G=36 into the first expression 6/7 + G/7.
Indeed, this means that every son received 6 gold coins.
36 total gold coins * 1 son / 6 gold coins = 6 sons.

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Danilo Calcinaro sent in the following solution:

The 1rst son gets 1 coin + 1/7 of what is left,
So he gets [1+1/7(X-1)]

The 2nd son gets 2 coins + 1/7 of what is left,
So he gets {2+1/7[X-1-1/7(X-1)-2]}

They had shared their inheritance equally so:
[1+1/7(X-1)]= {2+1/7[X-1-1/7(X-1)-2]}
X=36

So the 1rst son gets 1+5;
the 2nd son gets 2+4;
the 3rd son gets 3+3;
the 4th son gets 4+2;
the 5th son gets 5+1;
the 6th son gets 6;
This man had six sons and the estate was large 36 coins.

Correctly solved by:

1. Tom Laidlaw	Vancouver, Washington
2. James Alarie	Flint, Michigan
3. Chase L. Smith	Mountain View High School, Mountain View, Wyoming
4. Marivi Bungay-Domingo	Philippines
5. Britan Woody	Mountain View High School, Mountain View, Wyoming
6. Mashayla Hurdsman	Mountain View High School, Mountain View, Wyoming
7. Charity Harmon	Mountain View High School, Mountain View, Wyoming
8. Ailee Bugas	Mountain View High School, Mountain View, Wyoming
9. Keaton Hurdsman	Mountain View High School, Mountain View, Wyoming
10. Gabe Gardiner	Mountain View High School, Mountain View, Wyoming
11. Hannah Behunin	Mountain View High School, Mountain View, Wyoming
12. Danilo Calcinaro	Istituto Tecnico Tecnologico (ITT) "Montani", Fermo, Italy

Send any comments or questions to: David Pleacher