One Hundred Fowls/Bakhshali Version

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 man 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:

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:

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.