Henry Ernest Dudeney/Modern Puzzles/19 - Market Transactions/Solution

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Modern Puzzles by Henry Ernest Dudeney: $19$

Market Transactions
A farmer goes to market and buys $100$ animals at a total cost of $\pounds 100$.
The price of cows being $\pounds 5$ each,
sheep $\pounds 1$ each,
and rabbits $1 \shillings$ each,
how many of each kind does he buy?


Solution

$19$ cows, $1$ sheep and $80$ rabbits.


Proof

Recall:

$1$ pound sterling ($\pounds 1$) is $20$ shillings ($20 \shillings$)

Let all prices be expressed, therefore, in shillings.

Thus:

the price of cows is $5 \times 20 = 100 \shillings$ each
the price of sheep is $20 \shillings$ each
the price of rabbits is $1 \shillings$ each

and:

the total amount of money to spend is $100 \times 20 = 2000 \shillings$.

Let $c$, $s$ and $r$ denote the number of cows, sheep and rabbits respectively.

We have:

\(\text {(1)}: \quad\) \(\ds 100 c + 20 s + r\) \(=\) \(\ds 2000\) the amount spent
\(\text {(2)}: \quad\) \(\ds c + s + r\) \(=\) \(\ds 100\) the total number of animals
\(\ds \leadsto \ \ \) \(\ds 99 c + 19 s\) \(=\) \(\ds 1900\) $(1) - (2)$
\(\ds \leadsto \ \ \) \(\ds 1900 - 99 c\) \(=\) \(\ds 19 s\)

Note that both $c$ and $s$ need to be positive.

We need to find possible values of $c$ such that $1900 - 99 c$ is divisible by $19$.

This can happen only when $c$ itself is divisible by $19$.

\(\ds c = 0: \, \) \(\ds 1900 - 99 \times 0\) \(=\) \(\ds 1900\) \(\ds = 19 \times 100\)
\(\ds c = 19: \, \) \(\ds 1900 - 99 \times 19\) \(=\) \(\ds 19\) \(\ds = 19 \times 1\)


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

$c = 0, s = 100, r = 0$

is usually ruled out.

Hence we have:

$c = 19, s = 1, r = 80$

$\blacksquare$


Sources

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