Liber Abaci/Problems/Bezants and a Seventh

Classic Problem

 * A man left to his oldest son $1$ bezant and a seventh of what was left;
 * then, from the remainder, to his next son he left $2$ bezants and a seventh of what was left;
 * then, from the new remainder, to his third son he left $3$ bezants and a seventh of what was left.
 * He continued in this way, giving each son $1$ bezant more than the previous son and a seventh of what remained.
 * By this division it developed that the last son received all that was left, and all the sons shared equally.


 * How many sons were there and how large was the man's estate?

Solution
There were $6$ sons and $36$ bezants.

Proof
Let the last son receive $n$ bezants.

Then the last son but one received $n - 1$ bezants plus $\dfrac 1 7$ of what remained.

But that also equals $n$ bezants.

So the last son received $\paren {n - 1} + 1$ bezant.

Therefore what remained after his having taken $n - 1$ bezants must have been $7 \times 1 = 7$ bezants.

He took $1$ more bezant, leaving $6$.

Because the $n$th son took $n$ bezants, it must be the case that there were $6$ sons, all of whom took $6$ bezants.

To check this, we confirm what each son receives:


 * Son $1$ took $1$ bezant and $\frac 1 7 \times 35 = 5$ bezants, leaving $30$ bezants.


 * Son $2$ took $2$ bezants and $\frac 1 7 \times 28 = 4$ bezants, leaving $24$ bezants.


 * Son $3$ took $3$ bezants and $\frac 1 7 \times 21 = 3$ bezants, leaving $18$ bezants.


 * Son $4$ took $4$ bezants and $\frac 1 7 \times 14 = 2$ bezants, leaving $12$ bezants.


 * Son $5$ took $5$ bezants and $\frac 1 7 \times 7 = 1$ bezants, leaving $6$ bezants.


 * Son $6$ took $6$ bezants, and there was nothing left over.