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?

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

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.

$\blacksquare$