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

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Propositiones ad Acuendos Juvenes by Alcuin of York: 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 Alcuin 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:

\(\ds 3 m + 2 w + \dfrac c 2\) \(=\) \(\ds 90\) apportioning the measures of grain
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 6 m + 4 w + c\) \(=\) \(\ds 180\)
\(\text {(2)}: \quad\) \(\ds m + w + c\) \(=\) \(\ds 90\) the total number of people
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 5 m + 3 w\) \(=\) \(\ds 90\) $(1) - (2)$

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 $w$:

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

\(\ds w = 0\) \(,\) \(\ds m = 18\)
\(\ds w = 5\) \(,\) \(\ds m = 15\)
\(\ds w = 10\) \(,\) \(\ds m = 12\)
\(\ds w = 15\) \(,\) \(\ds m = 9\)
\(\ds w = 20\) \(,\) \(\ds m = 6\)
\(\ds w = 25\) \(,\) \(\ds m = 3\)
\(\ds w = 30\) \(,\) \(\ds m = 0\)


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:

\(\ds m\) \(=\) \(\ds 3 n\)
\(\ds w\) \(=\) \(\ds 30 - 5 n\)
\(\ds c\) \(=\) \(\ds 60 + 2 n\)

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

$\blacksquare$


Sources

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