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

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Puzzles and Curious Problems by Henry Ernest Dudeney: $11$

Distribution
Nine persons in a party, $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $K$, did as follows:
First $A$ gave each of the others as much money as he (the receiver) already held;
then $B$ did the same; then $C$; and so on to the last,
$K$ giving to each of the other eight persons the amount the receiver then held.
Then it was found that each of the nine persons held the same amount.
Can you find the smallest amount in pence that each person could have originally held?


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 $1$.


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