Henry Ernest Dudeney/Puzzles and Curious Problems/11 - Distribution/Solution

by : $11$

 * Distribution

Solution

 * $10$, $19$, $37$, $73$, $145$, $289$, $577$, $1153$, $2305$.

$A$ is the largest holder, progressing to $K$ being the one whose holding is $10$.

At the end of the game, all hold $2^9 = 512$ pence.

Proof
The question does not state this, but it is implicit in the nature of the answer that everybody starts with a whole number of pence.

This is another instance of Henry Ernest Dudeney: Modern Puzzles 12: A Weird Game.

The smallest number originally held is $1$ more than the number of persons.

The others are obtained by doubling and subtracting $1$.