Propositiones ad Acuendos Juvenes/Problems/33a - De Alio Patrefamilias Erogante Suae Familiae Annonam: Variant

by : Problem $\text {33 a}$

 * De Alio Patrefamilias Erogante Suae Familiae Annonam: Another Landlord Apportioning Grain
 * A gentleman has a household of $90$ persons and ordered that they be given $90$ measures of grain.
 * He directs that:
 * each man should receive $3$ measures,
 * each woman $2$ measures,
 * and each child $\frac 1 2$ a measure.


 * How many men, women and children must there be?

Solution
There are $5$ solutions:
 * $15$ men, $5$ women and $70$ children
 * $12$ men, $10$ women and $68$ children
 * $9$ men, $15$ women and $66$ children
 * $6$ men, $20$ women and $64$ children
 * $3$ men, $25$ women and $62$ children.

The solution given by is:
 * $6$ men, $20$ women and $64$ children.

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

We have:

We note that $5 m$ is a multiple of $5$.

Hence $3 w$ also has to be a multiple of $5$.

Thus $w$ has to be a multiple of $5$.

Hence the following possible solutions for $m$ and $w$:

Hence the following possible solutions for $m$ and $w$:

It is implicit that there are at least some men and some women in the household, so the solutions:
 * $m = 18, w = 0, c = 72$
 * $m = 0, w = 30, c = 60$

are usually ruled out.

Hence we have the following solutions:
 * $m = 15, w = 5, c = 70$
 * $m = 12, w = 10, c = 68$
 * $m = 10, w = 15, c = 66$
 * $m = 6, w = 20, c = 64$
 * $m = 3, w = 25, c = 62$

This can be expressed as:

where $n = 1, 2, \ldots, 5$.