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

by : $12$

 * A Weird Game
 * Seven men engaged in play.
 * Whenever a player won a game he doubled the money of each of the other players.
 * That is, he gave each player just as much money as each had in his pocket.
 * They played $7$ games and, strange to say, each won a game in turn in the order of their names,
 * which began with the letters $\text A$, $\text B$, $\text C$, $\text D$, $\text E$, $\text F$, and $\text G$.
 * When they had finished it was found that each man had exactly $2$ shillings and $8$ pence in his pocket.


 * How much had each man had in his pocket before play?

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 $\dfrac {225} 4 = 112 \tfrac 1 4 \oldpence = 9 \shillings 4 \tfrac 1 4 \oldpence$