Dudeney's Modern Puzzles/Arithmetical and Algebraical Problems/Money Puzzles/2 - Pocket-money

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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:

$20$ shillings make one pound.

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


Sources