perplexus.info :: Just Math : unequal booty

unequal booty (Posted on 2025-01-02)

Three pirates an their monkey split a chest of gold coins.

First, the captain makes three equal piles, giving a single leftover coin to the monkey. He takes one of the piles for himself and pours the other two piles back in the treasure chest. The second-in-command does the exact same, followed by the swabbie also doing the exact same. Finally, the last of the gold in the chest is split equally between the three pirates.

How few coins could have been in the chest?

No Solution Yet Submitted by Danish Ahmed Khan

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Working backwards (spoiler)

| Comment 1 of 2

Assume that the last of the gold is just 6 coins, since it must be a multiple of both 2 and 3.

Then, before the Swabbie does the split there were 10 coins

(3/2 * 6 +1).

Then before the 2nd-in command does the split there were 16 coins (3/2*10 + 1).

Then before the captain does the split there were 25 coins (3/2*16 +1).

This works. Final answer = 25 coins.

Posted by Steve Herman on 2025-01-02 15:17:50

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