Henry Ernest Dudeney/Modern Puzzles/5 - Doubling the Value

by : $5$

 * Doubling the Value


 * It is a curious fact that if you double $\pounds 6 \ 13 \, \mathrm s.$, you get $\pounds 13 \ 6 \, \mathrm s.$, which is merely changing the shillings and the pounds.


 * Can you find another sum of money that has the same peculiarity that, when multiplied by any number you may choose to select, will merely exchange the shillings and the pounds?


 * There is only one other multiplier and sum of money, besides the case shown, that will work.


 * What is it?

Solution

 * $\pounds 2 \ 17 \, \mathrm s.$

which, when multiplied by $6$, becomes:


 * $\pounds 17 \ 2 \, \mathrm s.$

Proof
It is assumed that when says number he means natural number

Let $k$ denote the shilling value of the original sum of money.

Recall there are $20$ shillings to the pound.

Let $s$ and pounds $p$ denote the number of shillings and pounds that $k$ consists of such that $s < 20$.

That is:
 * $k = \pounds p \ s \, \mathrm s.$

where:
 * $0 \le s < 20$

and of course trivially $s \ne 0$.

Let $n$ be a number which switches the shillings and pounds when you multiply $k$ by it.

Hence we want to find values for $n$ that will result in:
 * $k = 20 p + s$

and:
 * $n k = 20 s + p$

where both $p < 20$ and $s < 20$.

We have:

It remains to substitute values for $n$ and see where that gets us.

Clearly $n < 20$ in order for $\paren {20 - n} s > 0$.

$n = 2$:

which is the answer we have been provided.

$n = 3$:

$n = 4$:

$n = 5$:

$n = 6$:

Hence we have that:
 * $k = \pounds 2 \ 17 \, \mathrm s.$

and we check that:

It remains to be shown that this is the only solution.

$n = 7$:

$n = 8$:

$n = 9$:

$n = 10$:

$n = 10$:

and at this stage $\dfrac p s > 20$ and so there can be no solutions such that $p < 20$.

That exhausts all possibilities for $n$.

It is interesting to see what happens when $n = 1$ is entered:

which is of course obvious.