# One Hundred Fowls

## Classic Problem

$100$ fowls were sold for $100$ shillings.

- The cockerels were sold for $5$ shillings each.
- The hens were sold for $3$ shillings each.
- The chicks were sold for $\frac 1 3$ of a shilling each.

How many of each were sold?

## Solution

There were either:

- $0$ cockerels, $25$ hens and $75$ chicks;
- $4$ cockerels, $18$ hens and $78$ chicks;
- $8$ cockerels, $11$ hens and $81$ chicks;
- $12$ cockerels, $4$ hens and $84$ chicks.

## Proof

Let $x$, $y$ and $z$ denote the number of cockerels, hens and chicks respectively.

We are to solve for $x, y, z \in \N$:

\(\text {(1)}: \quad\) | \(\ds 5 x + 3 y + \dfrac z 3\) | \(=\) | \(\ds 100\) | |||||||||||

\(\text {(2)}: \quad\) | \(\ds x + y + z\) | \(=\) | \(\ds 100\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds 14 x + 8 y\) | \(=\) | \(\ds 200\) | $(1) \times 3 - (2)$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 7 x + 4 y\) | \(=\) | \(\ds 100\) | dividing both sides by $2$ |

$7 x$ must be divisible by $4$.

Hence so must $x$.

Write $x = 4 n$.

Then the problem is reduced to:

- $7 n + y = 25$

Hence $0 \le n \le 3$.

We list the solutions $\tuple {n, x, y, z}$ for each $n$:

- $\tuple {0, 0, 25, 75}, \tuple {1, 4, 18, 78}, \tuple {2, 8, 11, 81}, \tuple {3, 12, 4, 84}$

$\blacksquare$

## Version from Bakhshali Manuscript

$20$ men, women and childen earn $20$ coins between them as follows:

- Each man earns $3$ coins.
- Each woman earns $1 \frac 1 2$ coins.
- Each child earns $\frac 1 2$ a coin.

How many men, women and children are there?

## Historical Note

This is one of the earliest known example of this kind of problem, dating back as it does to the $5$th century.

However, it is predated by the example in the *Bakhshali Manuscript*, if that does indeed date from the $3$rd century, as is possible.

## Sources

- c. 466 -- c. 485: Zhang Qiujian:
*Zhang Qiujian Suanjing* - 1913: Yoshio Mikami:
*The Development of Mathematics in China and Japan* - 1992: David Wells:
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