Propositiones ad Acuendos Juvenes/Problems/33a - De Alio Patrefamilias Erogante Suae Familiae Annonam: Variant
Jump to navigation
Jump to search
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
- c. 800: Alcuin of York: Propositiones ad Acuendos Juvenes
- 1992: John Hadley/2 and David Singmaster: Problems to Sharpen the Young (Math. Gazette Vol. 76, no. 475: pp. 102 – 126) www.jstor.org/stable/3620384