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

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Propositiones ad Acuendos Juvenes by Alcuin of York: 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 Alcuin 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:

\(\ds 3 m + 2 w + \dfrac c 2\) \(=\) \(\ds 100\) apportioning the measures of grain
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 6 m + 4 w + c\) \(=\) \(\ds 200\)
\(\text {(2)}: \quad\) \(\ds m + w + c\) \(=\) \(\ds 100\) the total number of people
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 5 m + 3 w\) \(=\) \(\ds 100\) $(1) - (2)$

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 $w$:

\(\ds w = 0\) \(,\) \(\ds m = 20\)
\(\ds w = 5\) \(,\) \(\ds m = 17\)
\(\ds w = 10\) \(,\) \(\ds m = 14\)
\(\ds w = 15\) \(,\) \(\ds m = 11\)
\(\ds w = 20\) \(,\) \(\ds m = 8\)
\(\ds w = 25\) \(,\) \(\ds m = 5\)
\(\ds w = 30\) \(,\) \(\ds m = 2\)


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:

\(\ds m\) \(=\) \(\ds 20 - 3 n\)
\(\ds w\) \(=\) \(\ds 5 n\)
\(\ds c\) \(=\) \(\ds 80 - 2 n\)

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

$\blacksquare$


Sources

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