Henry Ernest Dudeney/Modern Puzzles/12 - A Weird Game/Solution

== by : $12$ ==
;A Weird Game
== Solution ==
:$A$ had $9 \shillings 4 \tfrac 1 4 \oldpence$
:$B$ had $4 \shillings 8 \tfrac 1 4 \oldpence$
:$C$ had $2 \shillings 4 \tfrac 1 4 \oldpence$
:$D$ had $1 \shillings 2 \tfrac 1 4 \oldpence$
:$E$ had $7 \tfrac 1 4 \oldpence$
:$F$ had $3 \tfrac 3 4 \oldpence$
:$G$ had $2 \oldpence$
== Proof ==
Recall that one shilling is $12$ pence.
We convert to pence.
Thus at the end, each person had exactly $32 \oldpence$
Because we know what direction this is going in, we decide to work entirely in farthings.
Recall that there are $4$ farthings to the old penny.
So, each starts with $4 \times 32 = 128$ farthings.
From here on in, all numbers are quantities of farthings. 
Before the last game, which $\text G$, won:
:Each of $\text A$ to $\text F$ had $\dfrac {128} 2 = 64$
:$\text G$ had $128 + 6 \times 64 = 512$
Before the game which $\text F$ won:
:$\text G$ had $\dfrac {512} 2 = 256$ 
:Each of $\text A$ to $\text E$ had $\dfrac {64} 2 = 32$
:$\text F$ had $64 + 5 \times 32 + 256 = 480$
Before the game which $\text E$ won:
:$\text G$ had $\dfrac {256} 2 = 128$ 
:$\text F$ had $\dfrac {480} 2 = 240$
:Each of $\text A$ to $\text D$ had $\dfrac 32 2 = 16$
:$\text E$ had $32 + 4 \times 16 + 240 + 128 = 464$
Before the game which $\text D$ won:
:$\text G$ had $\dfrac {128} 2 = 64$ 
:$\text F$ had $\dfrac {240} 2 = 120$
:$\text E$ had $\dfrac {464} 2 = 232$
:Each of $\text A$ to $\text C$ had $\dfrac 16 2 = 8$
:$\text D$ had $16 + 3 \times 8 + 232 + 120 + 64 = 456$
Before the game which $\text C$ won:
:$\text G$ had $\dfrac {64} 2 = 32$ 
:$\text F$ had $\dfrac {120} 2 = 60$
:$\text E$ had $\dfrac {232} 2 = 116$
:$\text D$ had $\dfrac {456} 2 = 228$
:$\text A$ and $\text B$ had $\dfrac 8 2 = 4$
:$\text C$ had $8 + 2 \times 4 + 228 + 116 + 60 + 32 = 452$
Before the game which $\text B$ won:
:$\text G$ had $\dfrac 32 2 = 16$ 
:$\text F$ had $\dfrac {60} 2 = 30$
:$\text E$ had $\dfrac {116} 2 = 58$
:$\text D$ had $\dfrac {228} 2 = 114$
:$\text C$ had $\dfrac {452} 2 = 226$
:$\text A$ had $\dfrac 4 2 = 2$
:$\text B$ had $4 + 2 + 226 + 114 + 58 + 30 + 16 = 450$
Before the game which $\text A$ won:
:$\text G$ had $\dfrac {16} 2 = 8$ 
:$\text F$ had $\dfrac {30} 2 = 15$
:$\text E$ had $\dfrac {58} 2 = 29$
:$\text D$ had $\dfrac {114} 2 = 57$
:$\text C$ had $\dfrac {226} 2 = 113$
:$\text B$ had $\dfrac {450} 2 = 225$
:$\text A$ had $2 + 225 + 113 + 57 + 29 + 15 + 8 = 449$
It remains to convert back to shillings and pence.
We have that:
:$8$ farthings is $\dfrac 8 4 = 2 \oldpence$
:$15$ farthings is $\dfrac {15} 4 = 3 \tfrac 3 4 \oldpence$
:$29$ farthings is $\dfrac {29} 4 = 7 \tfrac 1 4 \oldpence$
:$57$ farthings is $\dfrac {57} 4 = 14 \tfrac 1 4 \oldpence = 1 \shillings 2 \tfrac 1 4 \oldpence$
:$113$ farthings is $\dfrac {113} 4 = 28 \tfrac 1 4 \oldpence = 2 \shillings 4 \tfrac 1 4 \oldpence$
:$225$ farthings is $\dfrac {225} 4 = 56 \tfrac 1 4 \oldpence = 4 \shillings 8 \tfrac 1 4 \oldpence$
:$449$ farthings is $\dfrac {225} 4 = 112 \tfrac 1 4 \oldpence = 9 \shillings 4 \tfrac 1 4 \oldpence$
== Sources ==
* : Solutions: $12$. -- A Weird Game
* : Answers: $8$. A Weird Game
Category:Henry Ernest Dudeney/Modern Puzzles/Solutions