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

by : $13$

 * 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 piece

Best:
 * $1$ crown
 * $3$ shillings
 * $3$ threepenny pieces

Crewe:
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

Proof
Recall that:
 * $1$ shilling is $12$ pence.

Hence $10$ shillings is $120$ pence.

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


 * The threepenny piece: $\tfrac 1 4 \shillings$
 * 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:

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:

Hence converting to shillings and pence:

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 pieces, that is, $3 \oldpence$ coins.

Crewe has an even number of threepenny pieces.

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

Hence the fewest possible threepenny pieces 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 piece.

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 piece, 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 pieces.

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

So Best must have started with:
 * $1$ crown
 * $1$ half crown
 * $5$ threepenny pieces.

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 pieces.

Best now has to give:
 * at least $1$ of his crowns
 * at least $1$ of his half crowns
 * at least $4$ of his threepenny pieces.

$10 \shillings = 120 \oldpence$ 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 pieces, in order for Crewe to have his share

The only way to make up $120 \oldpence$ with $1$ crown, $1$ half crown and threepenny pieces, Best must give Crewe:
 * $1$ crown
 * $1$ half crown
 * $1$ shilling
 * $6$ threepenny pieces.

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 pieces, Best must have started with:
 * $1$ crown
 * $3$ shillings
 * $3$ threepenny pieces.

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 pieces.

Thus at the end, each has:
 * $1$ crown
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

After Abel has given to Best and Crewe:

Abel has:
 * $1$ half crown

Best has:
 * $2$ crowns
 * $1$ half crown
 * $4$ shillings
 * $4$ threepenny pieces.

Crewe has:
 * $1$ crown
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

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 pieces

This leaves Best with:
 * $2$ crowns
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

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 pieces

Best has:
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

Crewe has:
 * $3$ crowns
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

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

Each now has:
 * $1$ crown
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.

In summary, they all started with:

Abel:
 * $2$ crowns
 * $2$ half crowns
 * $1$ shilling
 * $1$ threepenny piece

Best:
 * $1$ crown
 * $3$ shillings
 * $3$ threepenny pieces

Crewe:
 * $1$ half crown
 * $2$ shillings
 * $2$ threepenny pieces.