Solution:

The first (oldest) person will get 98 coins, the second will get no coins, and one coin apiece will go to two of the last three people. If we write the distribution as an ordered 5-tuple, with leftmost being oldest, etc., then the three possible solutions are $(98, 0, 1, 0, 1)$, $(90, 0, 0, 1, 1)$, and $(98, 0, 1, 1, 0)$.

To see why, we first mention a few principles. Since the mathematicians are all intelligent, and equally intelligent, but only out for themselves and unable to cooperate or negotiate, each person must assume the worst. No one will approve a plan that gives them zero coins. But furthermore, no one will vote against a plan that gives them a positive number of coins, if it may be possible that they could end up with fewer coins. This is not a game of chance; a person will settle, for example, for just one coin, if by rejecting this offer he or she may end up with no coins.

This explains why the common wrong answer "everyone votes against the oldest in order to decrease the number of players" will not happen with rational players. If the number of players dropped to 4 people, then the first person would have the advantage, and need only propose something that one other person would agree to. For example, the proposal may be $(50,0,0,50)$. This would give players #2, #3 (originally #3, #4) no coins at all, but only 2 votes are needed for the proposal to pass. On the other hand, the proposal could be $(50,0,50,0)$. In other words, when there are 5 people present, it is not in the best interest of #3, #4, or #5 to reject the proposal, if they are offered non-zero amounts of coins, since they may end up with zero later, and have absolutely no control over what happens!

Indeed, by analyzing smaller cases, it becomes clear how much advantage the first person has, and how the second person is in the worst situation. With just two people, $(100,0)$ wins all for the first person. With three people, $(99,0,1)$ will be accepted, since #3 dare not reject it, for then #3 would be the loser in the 2-person situation. Consequently, with 4 people, the winning proposals will be $(99,0,1,0)$ or $(99,0,0,1)$. In both cases, the people with zero will vote against the proposal, and the one person who is offered a single coin will accept, since in no circumstances will they do better than this. (Conceivably, the youngest person can vote against $(99,0,0,1)$, reasoning that the next proposal will be $(99,0,1)$ which also offers him or her one coin; knowing this, the most likely 4-person proposal will be $(99,0,1,0)$ which will definitely not be rejected by person #3.)

Finally, looking at 5 people, it is now clear that the first person need only "pay off" two others, with the least amount. Person #2 cannot be paid off, since he or she will always be better off being the proposer. Knowing this, person #1 always offers nothing to person #2. Persons #3, #4, #5 all realize that they may end up with no coins at all, by the analysis above. Consequently, if any two of them are offered just a single coin, they will accept, realizing that failing to accept will result in just 4 players, with absolutely no guarantee of any coins at all. In particular, player #3 (in the 5-person situation) realizes that with 4 players, he will surely be offered (and will receive) no coins, so she will gladly accept one coin in the 5-person situation. Persons #4 and #5 (in the 5-person case) are in the situation of not knowing for sure whether they will be offered 1 or 0 coins in the 4-person case. Consequently, if either of them is offered just one coin, the offer is accepted, since a certainty of getting one coin is superior to a possibility of getting no coins.