# Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins/Solution

*Modern Puzzles* by Henry Ernest Dudeney: $13$

- Find the Coins

*Three men, Abel, Best and Crewe, possessed money, all in silver coins.**Abel had one coin fewer than Best and one more than Crewe.**Abel gave Best and Crewe as much money as they already had,**then Best gave Abel and Crewe the same amount of money as they they held,**and finally Crewe gave Abel and Best as much money as they then had.*

*Each man then held exactly $10$ shillings.*

*To find what amount each man started with is not difficult.**But the sting of the puzzle is in the tail.**Each man held exactly the*same coins*(the fewest possible) amounting to $10$ shillings.*

*What were the coins and how were they originally distributed?*

## Solution

Abel:

- $2$ crowns
- $2$ half crowns
- $1$ shilling
- $1$ threepenny bit

Best:

- $1$ crown
- $3$ shillings
- $3$ threepenny bit

Crewe:

- $1$ half crown
- $2$ shillings
- $2$ threepenny bit.

## Proof

Recall that:

Hence $10$ shillings is $120$ pence.

Also recall that the "silver" coins available at the time of Dudeney were:

- The threepenny bit: $\tfrac 1 4 \shillings$
- The sixpence: $\tfrac 1 2 \shillings$
- The shilling
- The florin: $2 \shillings$
- The half-crown: $2 \tfrac 1 2 \shillings$
- The crown: $5 \shillings$
- The half sovereign: $10 \shillings$

Let $a$, $b$ and $c$ respectively be the amounts of money each of Abel, Best and Crewe started with.

Let $a_n$, $b_n$, $c_n$ be the amounts held after the $n$th transaction, where $n \in \set {1, 2, 3}$.

We have:

\(\ds a_1\) | \(=\) | \(\ds a - b - c\) | ||||||||||||

\(\ds b_1\) | \(=\) | \(\ds 2 b\) | ||||||||||||

\(\ds c_1\) | \(=\) | \(\ds 2 c\) |

\(\ds a_2\) | \(=\) | \(\ds 2 a_1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \paren {a - b - c}\) | ||||||||||||

\(\ds b_2\) | \(=\) | \(\ds b_1 - a_1 - c_1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 b - \paren {a - b - c} - 2 c\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 3 b - a - c\) | ||||||||||||

\(\ds c_2\) | \(=\) | \(\ds 2 c_1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 4 c\) |

\(\ds a_3\) | \(=\) | \(\ds 2 a_2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \paren {2 \paren {a - b - c} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 4 \paren {a - b - c}\) | ||||||||||||

\(\ds b_3\) | \(=\) | \(\ds 2 b_2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \paren {3 b - a - c}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 6 b - 2 a - 2 c\) | ||||||||||||

\(\ds c_3\) | \(=\) | \(\ds c_2 - a_2 - b_2\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 4 c - 2 \paren {a - b - c} - \paren {3 b - a - c}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 4 c - 2 a + 2 b + 2 c - 3 b + a + c\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 7 c - a - b\) |

But:

- $a_3 = b_3 = c_3 = 120$

Hence we have the three simultaneous equations, expressed in matrix form as:

- $\begin {bmatrix} 4 & -4 & -4 \\ -2 & 6 & -2 \\ -1 & -1 & 7 \end {bmatrix} \begin {bmatrix} a \\ b \\ c \end {bmatrix} = \begin {bmatrix} 120 \\ 120 \\ 120 \end {bmatrix}$

In reduced echelon form, this gives:

- $\begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix} \begin {bmatrix} a \\ b \\ c \end {bmatrix} = \begin {bmatrix} 195 \\ 105 \\ 60 \end {bmatrix}$

Hence converting to shillings and pence:

\(\ds a\) | \(=\) | \(\ds 16 \shillings \ 3 \oldpence\) | ||||||||||||

\(\ds b\) | \(=\) | \(\ds 8 \shillings \ 9 \oldpence\) | ||||||||||||

\(\ds c\) | \(=\) | \(\ds 5 \shillings\) |

Because all men hold the same coins at the end, the total number of each coin must be a multiple of $3$.

Both Abel and Best have an odd number of threepenny bits, that is, $3 \oldpence$ coins.

Crewe has an even number of threepenny bits.

Thus the number of threepenny bits in total must be a multiple of $6$.

Hence the fewest possible threepenny bits is $6$.

Let us recall who had what after each transaction.

After Abel paid Best and Crewe, they had:

- Abel: $30 \oldpence$
- Best: $210 \oldpence$
- Crewe: $120 \oldpence$

After Best paid Abel and Crewe, they had:

- Abel: $60 \oldpence$
- Best: $60 \oldpence$
- Crewe: $240 \oldpence$

and of course after Crewe paid Abel and Best, they had:

- Abel: $120 \oldpence$
- Best: $120 \oldpence$
- Crewe: $120 \oldpence$

The sum paid by Abel to Best of $105 \oldpence = 8 \shillings \ 9 \oldpence$ in the fewest possible coins is:

- $1$ crown: $5 \shillings$ or $60 \oldpence$
- $1$ half crown: $2 \tfrac 1 2 \shillings$ or $30 \oldpence$
- $1$ shilling or $12 \oldpence$
- $1$ threepenny bit.

The sum paid by Abel to Crewe of $60 \oldpence = 5 \shillings$ in the fewest possible coins is $1$ crown.

Abel is now left with $195 - 105 - 60 = 30 \oldpence$, which, in the fewest possible coins, is $1$ half crown.

So, the smallest number of coins Abel could have started with was:

- $2$ crowns, one of which went to Best and the other to Crewe
- $2$ half crowns, one of which went to Best and the other he kept
- $1$ shilling, which went to Best
- $1$ threepenny bit, which went to Best.

Thus Abel had $6$ coins.

Best had $1$ more coin than Abel, that is, $7$ coins.

Crewe had $1$ coin less than Abel, that is, $5$ coins.

The number of crowns must be a multiple of $3$.

Hence Best started with $1$ crown, and $6$ more coins adding to $45 \oldpence$

Suppose Best started with one half crown.

Then his remaining $5$ coins add up to $45 \oldpence$, and so were threepenny bit.

This tells us where all $6$ threepenny bits are.

So Best must have started with:

- $1$ crown
- $1$ half crown
- $5$ threepenny bits.

By the rule of $3$, no half crown then remains for Crewe, and Crewe must have started with $5$ shillings.

After receiving $105 \oldpence$ from Abel, Best now has:

- $2$ crowns
- $2$ half crowns
- $1$ shilling
- $6$ threepenny bits.

Best now has to give:

- at least $1$ of his crowns
- at least $1$ of his half crowns
- at least $4$ of his threepenny bits.

$10 \shillings = 120 \oldpence$ of this must go to Crewe.

Of this money, in order for Crewe to get his share of the coins:

- at least $1$ of those coins must be a half crown, in order for Crewe to have his share
- at least $2$ of those coins must be threepenny bits, in order for Crewe to have his share

The only way to make up $120 \oldpence$ with $1$ crown, $1$ half crown and threepenny bits, Best must give Crewe:

- $1$ crown
- $1$ half crown
- $1$ shilling
- $6$ threepenny bits.

This leaves $1$ half crown to give to Abel, leaving Best with just $1$ crown.

But now Abel has $2$ half crowns and Best has none, which cannot be redressed by Crewe.

Hence it cannot be the case that Best started with a half crown.

So, to make $45 \oldpence$ with $6$ coins, all shillings and threepenny bits, Best must have started with:

- $1$ crown
- $3$ shillings
- $3$ threepenny bits.

So to make up the coins so as for there to be a total of a multiple of $3$ each, Crewe must have started with:

- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

Thus at the end, each has:

- $1$ crown
- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

After Abel has given to Best and Crewe:

Abel has:

- $1$ half crown

Best has:

- $2$ crowns
- $1$ half crown
- $4$ shillings
- $4$ threepenny bits.

Crewe has:

- $1$ crown
- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

Best now gives $30 \oldpence$ to Abel.

He cannot give his half crown as Abel now has $2$.

So he has to give to Abel:

- $2$ shillings
- $2$ threepenny bits

This leaves Best with:

- $2$ crowns
- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

of which he has to give $120 \oldpence$ to Crewe.

He does this by giving him his $2$ crowns.

Hence after Best has given to Abel and Crewe:

Abel has:

- $1$ half crown
- $2$ shillings
- $2$ threepenny bits

Best has:

- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

Crewe has:

- $3$ crowns
- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

Crewe now gives $1$ crown each to Abel and Best.

Each now has:

- $1$ crown
- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

In summary, they all started with:

Abel:

- $2$ crowns
- $2$ half crowns
- $1$ shilling
- $1$ threepenny bit

Best:

- $1$ crown
- $3$ shillings
- $3$ threepenny bits

Crewe:

- $1$ half crown
- $2$ shillings
- $2$ threepenny bits.

$\blacksquare$

## Sources

- 1926: Henry Ernest Dudeney:
*Modern Puzzles*... (previous) ... (next): Solutions: $13$. -- Find the Coins