One Hundred Fowls

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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

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