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

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Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $33$

De Alio Patrefamilias Erogante Suae Familiae Annonam
Another Landlord Apportioning Grain
A gentleman has a household of $30$ persons and directs that they be given $30$ 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?


$\text {33 a}$: Variant

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

$3$ men, $5$ women and $22$ children.


Proof

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

We have:

\(\ds 3 m + 2 w + \dfrac c 2\) \(=\) \(\ds 30\) apportioning the measures of grain
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 6 m + 4 w + c\) \(=\) \(\ds 60\)
\(\text {(2)}: \quad\) \(\ds m + w + c\) \(=\) \(\ds 30\) the total number of people
\(\ds \leadsto \ \ \) \(\ds 5 m + 3 w\) \(=\) \(\ds 30\) $(1) - (2)$

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 $w$:

\(\ds w = 0\) \(,\) \(\ds m = 6\)
\(\ds w = 5\) \(,\) \(\ds m = 3\)
\(\ds w = 10\) \(,\) \(\ds m = 0\)

It is implicit that there are at least some men and some women in the household, so the solutions:

$m = 6, w = 0, c = 24$
$m = 0, w = 10, c = 20$

are usually ruled out.

Hence we have:

$m = 3, w = 5, c = 22$

$\blacksquare$


Sources

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