Dudeney's Modern Puzzles/Arithmetical and Algebraical Problems/Money Puzzles/2 - Pocket-money
Modern Puzzles by Henry Ernest Dudeney: $2$
- Pocket-money
- I went down the street with a certain amount of money in my pocket,
- and when I returned home I discovered that I had spent just half of it,
- and that I now had just as many shillings as I previously had pounds,
- and half as many pounds as I then had shillings.
- How much money had I spent?
Solution
The amount spent was $\pounds 9 \ 19 \, \mathrm s.$
Proof
Let $S$ be the amount I started with: $S_L$ pounds and $S_s$ shillings.
Let $F$ be the amount I finished with: $F_L$ pounds and $F_s$ shillings.
We have that:
- $2 F = S$
and so the amount spent is equal to $2 F - F = F$, that is, the amount I finished with.
We recall the conversion factors:
Hence any shilling quantities in either $S$ or $F$ cannot be greater than $19$.
That is:
- $S_s < 20$
- $F_s < 20$
It is assumed that $S_s$ and $F_s$ are both integers, that is: both $S$ and $F$ are a whole number of shillings.
We are given that:
\(\ds S\) | \(=\) | \(\ds 2 F\) | ... and when I returned home I discovered that I had spent just half of it, | |||||||||||
\(\ds F_s\) | \(=\) | \(\ds S_L\) | and that I now had just as many shillings as I previously had pounds, | |||||||||||
\(\ds 2 F_L\) | \(=\) | \(\ds S_s\) | and half as many pounds as I then had shillings. |
We have that:
\(\ds S\) | \(=\) | \(\ds 20 S_L + S_s\) | where $S$ shillings is the money I started out with | |||||||||||
\(\ds F\) | \(=\) | \(\ds 20 F_L + F_s\) | where $F$ shillings is the money I came home with | |||||||||||
\(\ds \) | \(=\) | \(\ds 20 \dfrac {S_s} 2 + S_L\) | as $2 F_L = S_s$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 10 S_s + S_L\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 20 S_L + S_s\) | \(=\) | \(\ds 2 \paren {10 S_s + S_L}\) | as $2 F = S$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 18 S_L\) | \(=\) | \(\ds 19 S_s\) | simplifying |
The smallest values of $S_L$ and $S_s$ that satisfy the above equation are:
\(\ds S_L\) | \(=\) | \(\ds 19\) | ||||||||||||
\(\ds S_s\) | \(=\) | \(\ds 18\) |
As $S_s \le 19$ it follows that there can be no other solution.
Hence:
\(\ds F\) | \(=\) | \(\ds \dfrac {\pounds 19 \ 18 \, \mathrm s.} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \pounds 9 \ 19 \, \mathrm s.\) |
which is equal to the amount spent.
$\blacksquare$
Also see
- Dollars and Cents, which follows exactly the same pattern.
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Arithmetical and Algebraical Problems: Money Puzzles: $2$. -- Pocket-money