One Hundred Fowls/Bakhshali Version

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Problem

$20$ 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?


Solution

$2$ men
$5$ women
$13$ children.


Proof

Let $m$, $w$ and $c$ denote the number of men, women and children respectively.

Then we have:

\(\text {(1)}: \quad\) \(\ds m + w + c\) \(=\) \(\ds 20\)
\(\ds 3 m + 1 \frac 1 2 w + \frac 1 2 c\) \(=\) \(\ds 20\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 6 m + 3 w + c\) \(=\) \(\ds 40\) clearing the fractions
\(\ds \leadsto \ \ \) \(\ds 5 m + 2 w\) \(=\) \(\ds 20\) $(2) - (1)$

We are of course subject to the condition that each of $m$, $w$ and $c$ is not fewer than $0$.

It is seen that $2 w$ is even.

Hence, in order for $5 m + 2 w$ also to be even, it is necessary for $m$ to be even.

If $m > 4$, then $5 m > 20$.

In that case $w < 0$ and so it must be that $m \le 4$.

This allows for $m$ to be equal to $0$, $2$ or $4$.

This gives the solutions:

\(\text {(1)}: \quad\) \(\ds m\) \(=\) \(\ds 0\)
\(\ds w\) \(=\) \(\ds 10\)
\(\ds c\) \(=\) \(\ds 10\)


\(\text {(2)}: \quad\) \(\ds m\) \(=\) \(\ds 2\)
\(\ds w\) \(=\) \(\ds 5\)
\(\ds c\) \(=\) \(\ds 13\)


\(\text {(3)}: \quad\) \(\ds m\) \(=\) \(\ds 4\)
\(\ds w\) \(=\) \(\ds 0\)
\(\ds c\) \(=\) \(\ds 16\)


However, let it be understood that there was at least one man, at least one woman and at least one child.

Then there is one solution remaining:

$2$ men
$5$ women
$13$ children.

$\blacksquare$


Historical Note

If the Bakhshali Manuscript actually dates from as long ago as the $3$rd century C.E., this is probably the oldest instance of a One Hundred Fowls problem in the world.


Sources

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