Solution:

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 the coins 100 : 0 (giving all to the other pirate). Otherwise, and perhaps even then, Pirate One (the youngest) would vote against him and over he goes!

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

4 Pirates: Pirate Four splits the coins 1 : 2 : 0 : 97. This way, Pirate One will vote for him, and so will Pirate Two - they're getting more than they would under 3 pirates.

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