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

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Modern Puzzles by Henry Ernest Dudeney: $12$

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

$\blacksquare$


This theorem requires a proof.
In particular: Dudeney says: "The answer may be found by laboriously working backwards, but a simpler method is as follows: $7 + 1 = 8$; $2 \times 7 + 1 = 15$; $4 \times 7 + 1 = 29$; and so on, where the multiplied increases in powers of $2$, that is, $2$, $4$, $8$, $16$, $32$ and $64$." Hence a proof is needed in which a recurrence relation is constructed and solved.
You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof.
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Sources

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