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

== by : $13$ ==
;Find the Coins
== 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:
:$1$ shilling is $12$ pence.
Hence $10$ shillings is $120$ pence.
Also recall that the "silver" coins available at the time of were:
:The threepenny 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:
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 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.
== Sources ==
* : Solutions: $13$. -- Find the Coins
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions
Category:Henry Ernest Dudeney/Modern Puzzles/13 - Find the Coins