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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Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $11$. -- Distribution
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $18$. Distribution