So I've been thinking about the pirates for a minute or two, and although i haven't seen it before, the method of solution is the same as many similar problems. That is to say, the question is only hard because there are 5 of them. So you solve the puzzle by starting with 1 pirate and making it successively harder.

If there's 1 pirate he clearly get's the whole pile. If there are two pirates the second pirate can't give the 1st pirate more than 100 and so no matter what he says he dies. If there are three pirates he thus doesn't need anything to buy off pirate 2 (who just wants to live), and can keep the remaining 100. If there are four pirates, the fourth one can buy off the first pirate's vote for 1 coin and the second pirates vote for 1 coin and keep the remaining 98. Finally the littlest pirate can buy off the first pirate's vote for 2 coins and the third pirate's vote for 1 coin and keep the remaining 97.

However, pirates are clearly NOT rational in this manner. Take the 3 pirate case for example. If the third pirate offered to take all the money himself, I can assure you the second oldest pirate is going to kill the annoying little cheap bastard. The first pirate out of gratitude will thus be willing to take the vast majority of the money instead of all of it, since he'd have gotten nothing if the second pirate hadn't helped him out.

Furthermore, how do you think the oldest two pirates got to live that long? They've clearly been playing this little game with all their new pirate buddies and must have some trick up their sleeve. So the point is, unless they get a great deal from pirate 5, they'll kill him. Since pirate 4 thinks he's about to win he won't be bought off and that leaves 1,2,4, vs 3,5 and 5 dies. But then 1 and 2 have a majority and can kill 3 and 4 and then split the treasure 60-40.