Propositiones ad Acuendos Juvenes/Problems/34 - Altera de Patrefamilias Partiente Familiae Suae Annonam

by : Problem $34$

 * Altera de Patrefamilias Partiente Familiae Suae Annonam: Another Landowner Dividing Grain among his Household
 * A gentleman has a household of $200$ persons and orders that they be given $100$ 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 $6$ solutions:
 * $17$ men, $5$ women and $78$ children
 * $14$ men, $10$ women and $76$ children
 * $11$ men, $15$ women and $74$ children
 * $8$ men, $20$ women and $72$ children
 * $5$ men, $25$ women and $70$ children
 * $2$ men, $30$ women and $68$ children.

The solution given by is:
 * $11$ men, $15$ women and $74$ 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 $w$:

It is implicit that there are at least some women in the household, so the solution:
 * $m = 20, w = 0, c = 80$

is usually ruled out.

Hence we have the following solutions:
 * $m = 17, w = 5, c = 78$
 * $m = 14, w = 10, c = 76$
 * $m = 11, w = 15, c = 74$
 * $m = 8, w = 20, c = 72$
 * $m = 5, w = 25, c = 70$
 * $m = 2, w = 30, c = 68$

This can be expressed as:

where $n = 1, 2, \ldots, 6$.