Puzzles/Logic puzzles/Gold Coins/Solution

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The eldest pirate will propose a 97, 0, 1, 0, 2 split.

Working backwards, splits in terms of younger to older:

2 Pirates: Pirate two splits at 100, 0. Otherwise, and perhaps even then, pirate 1 (the youngest) votes against him and over he goes!

3 Pirates: Pirate three splits at 0, 1, 99. Pirate 1 (the youngest) is going to vote against him no matter what (see above), but this way, pirate 2 will vote for him, to get at least one gold out of it.

4 Pirates: Pirate four splits of 1, 2, 0, 97. This way, pirate 1 will vote for him, and so will pirate 2 - they're getting more than they would under 3 pirates.

5 Pirates: Pirate five splits of 2, 0, 1, 0, 97. This way, pirate 1 will vote for him, and so will pirate 3 - they're both getting better than they would under 4.