Propositiones ad Acuendos Juvenes/Problems/38 - De Quodam Emptore in Animalibus Centum

From ProofWiki
Jump to navigation Jump to search

Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $38$

De Quodam Emptore in Animalibus Centum
A Man Buying a Hundred Animals
A man wanted to buy $100$ assorted animals for $100$ shillings.
He was willing to pay $3$ shillings for a horse,
$1$ shilling for an ox,
and $1$ shilling for $24$ sheep.
How many horses, oxen and sheep did he buy?


Solution

$23$ horses, $29$ oxen and $48$ sheep.


Proof

Let $h$, $o$ and $s$ denote the number of horses, oxen and sheep respectively.

We have:

\(\ds 3 h + o + \dfrac s {24}\) \(=\) \(\ds 100\) the amount spent
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 72 h + 24 o + s\) \(=\) \(\ds 2400\)
\(\text {(2)}: \quad\) \(\ds h + o + s\) \(=\) \(\ds 100\) the total number of animals
\(\ds \leadsto \ \ \) \(\ds 71 h + 23 o\) \(=\) \(\ds 2300\) $(1) - (2)$
\(\ds \leadsto \ \ \) \(\ds 2300 - 71 h\) \(=\) \(\ds 23 o\)

Note that both $h$ and $o$ need to be (strictly) positive.

We need to find possible values of $h$ such that $2300 - 71 h$ is divisible by $23$.

This can happen only when $h$ itself is divisible by $23$.

\(\ds h = 0: \, \) \(\ds 2300 - 71 \times 0\) \(=\) \(\ds 2300\) \(\ds = 23 \times 100\)
\(\ds h = 23: \, \) \(\ds 2300 - 71 \times 23\) \(=\) \(\ds 667\) \(\ds = 23 \times 29\)


It is implicit that there are at least some horses are bought, so the solution:

$h = 0, o = 100, s = 0$

is usually ruled out.

Hence we have:

$h = 23, o = 29, s = 48$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Propositiones_ad_Acuendos_Juvenes/Problems/38_-_De_Quodam_Emptore_in_Animalibus_Centum&oldid=546938"