Propositiones ad Acuendos Juvenes/Problems/5 - De Emptore in C Denariis

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Propositiones ad Acuendos Juvenes by Alcuin of York: Problem $5$

De Emptore in $\text C$ Denariis
A Merchant and $100$ Pence
A merchant wanted to buy $100$ pigs at $100$ pence.
For a boar he would pay $10$ pence;
and for a sow $5$ pence;
while he would pay $1$ penny for a couple of piglets.
How many boars, sows and piglets must there have been for him to have paid exactly $100$ pence for the $100$ animals?


Solution

$1$ boar
$9$ sows
$90$ piglets.


Proof

Let $x$, $y$ and $z$ denote the number of boars, sows and piglets respectively.

We are to solve for $x, y, z \in \N$:

\(\text {(1)}: \quad\) \(\ds 10 x + 5 y + \dfrac z 2\) \(=\) \(\ds 100\)
\(\text {(2)}: \quad\) \(\ds x + y + z\) \(=\) \(\ds 100\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 19 x + 9 y\) \(=\) \(\ds 100\) $(1) \times 2 - (2)$

Note that both $x$ and $y$ need to be (strictly) positive.

We need to find possible values of $x$ such that $100 - 19 x$ is divisible by $9$.

Inspecting possible contenders for such an $x$ individually:

\(\, \ds x = 1: \, \) \(\ds 100 - 19 x\) \(=\) \(\ds 81\)
\(\, \ds x = 2: \, \) \(\ds 100 - 19 x\) \(=\) \(\ds 62\)
\(\, \ds x = 3: \, \) \(\ds 100 - 19 x\) \(=\) \(\ds 43\)
\(\, \ds x = 4: \, \) \(\ds 100 - 19 x\) \(=\) \(\ds 24\)
\(\, \ds x = 5: \, \) \(\ds 100 - 19 x\) \(=\) \(\ds 5\)

Only one of these satisfies the condition, that is:

$x = 1$
$y = 9$

which leaves:

$z = 100 - 1 - 9 = 90$

Hence the result.

$\blacksquare$


Historical Note

David Singmaster remarks that this is the first European appearance of a one hundred fowls problem.


Sources

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