# Dudeney's Modern Puzzles/Arithmetical and Algebraical Problems/Money Puzzles/1 - Concerning a Cheque

*Modern Puzzles* by Henry Ernest Dudeney: $1$

- Concerning a Cheque

*A man went into a bank to cash a cheque.**In handing over the money the cashier, by mistake, gave him pounds for shillings and shillings for pounds.**He pocketed the money without examining it, and spent half a crown on his way home, when he found that he possessed exactly twice the amount of the cheque.**He had no money in his pocket before going to the bank, and it is an interesting puzzle to find out what was the exact amount of that cheque.*

## Solution

As Dudeney puts it:

*If you set to work under the notion that there were only pounds and shillings -- no pence --- in the amount, a solution is impossible.*

The amount on the cheque was $\pounds 5 \ 11 \, \mathrm s. \ 6 \, \mathrm d.$

He received $\pounds 11 \ 5 \, \mathrm s. \ 6 \, \mathrm d.$

After spending half a crown, which is $2 \, \mathrm s. \ 6 \, \mathrm d.$, he had $\pounds 11 \ 3 \, \mathrm s.$

This is twice as much as $\pounds 5 \ 11 \, \mathrm s. \ 6 \, \mathrm d.$.

## Proof

We recall the conversion factors:

- $1$ half crown consists of $2$ shillings and $6$ pence.

Let $C$ be the amount written on the cheque.

Let $C'$ be the amount paid out to the man by the bank.

Let $P$ be the final amount in the man's pocket.

Suppose $C$ were an integral number of shillings.

Then $C'$ would also be an integral number of shillings.

But $P$, which is $C'$ less $2 \tfrac 1 2$ shillings is then *not* an integral number of shillings.

That means $P$ could not be twice $C$.

Hence Dudeney's warning.

Let $C$ consist of $C_l$ pounds, $C_s$ shillings and $C_d$ pennies.

Let $P$ consist of $P_l$ pounds, $P_s$ shillings and $P_d$ pennies.

But what we *can* say about the pennies is that:

- $(1): \quad 2 \times \paren {C_d - 6} = P_d \pmod {12}$

because $2 \paren {C - 2 \, \mathrm s. \ 6 \, \mathrm d.} = P$

and also that:

- $(2): \quad C_d - 6 = P_d \pmod {12}$

because $P_d$ is the actual number of pennies.

from which it follows that $C_d = 6$.

Let $D_1$ be the value of the cheque in pennies.

Let $D_2$ be the money the man left the bank with in pennies.

Let $D_3$ be the money the man arrived home with in pennies.

We have:

- $D_1 = 240 C_l + 12 C_s + 6$

After coming out of the bank, the man has:

- $D_2 = 240 C_s + 12 C_l + 6$

After arriving home, the man has:

- $D_3 = 240 C_s + 12 \paren {C_l - 2}$

But we have:

- $D_1 \times 2 = D_3$

which leads us to:

\(\ds 480 C_l + 24 C_s + 12\) | \(=\) | \(\ds 240 C_s + 12 C_l - 24\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds 468 C_l + 36\) | \(=\) | \(\ds 216 C_s\) | rearranging | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 6 C_s - 13 C_l\) | \(=\) | \(\ds 1\) | dividing by $36$ |

By inspection, we arrive at:

- $13 \times 5 = 65 = 6 \times 11 - 1$

and the answer follows.

$\blacksquare$

## Sources

- 1926: Henry Ernest Dudeney:
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