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There were three men in a boat. They received an amount of coins in a chest from 200 - 300 as a reward for surviving the storm without damaging anything. One night, the first man thought that he would his share first so he wouldn't have to bring up an argument. He split the money into thirds and took a piece. There was an extra so he threw it out into the sea. Later, the second man did the same exact thing as the first man, taking his share of one-third of the remaining money, not knowing the first man already got his part. There was one more left over, so he threw it into the sea. Finally the third man did the same exact same thing as the other two, taking his share of one-third of that remaining money. The tax collector found some coins left in the chest and divided in three parts giving one part to each sailor, all who are angry with the decision. There was one coin left again so the tax collector took it as payment for his services.

SO how many coins were there and how many coins did each sailor receive?

Hint: The number of coins a sailor got times three will always be one less than the number of coins there in the chest at that time. For example, Sailor Three got ten coins when he got his "share". So there were 31 coins before he divided the money. This problem comes from one of my favorite books made by Malba.

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  • $\begingroup$ This is very similar to puzzling.stackexchange.com/questions/376/… Aside from the requirement to start with $200-300$ coins, you can start with $-2$ coins. $\endgroup$ – Ross Millikan Feb 22 '15 at 23:29
  • $\begingroup$ I find the part about throwing money overboard to be completely unbelievable. ;-p $\endgroup$ – Hellion Feb 23 '15 at 21:35
  • $\begingroup$ Well they did that $\endgroup$ – Anthony Pham Feb 23 '15 at 22:37
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Each sailor received the following:

The first received 103 coins.
The second received 76 coins.
The third received 58 coins.
There were a grand total of 241 coins in the chest.

It can be solved using algebra, which is the formal method, but also by trial and error, which is how I ended up doing it.

If the chest has 241 coins, then the first sailor would take 1/3 and throw the one leftover coin overboard. One-third of 241 is 80r1, so the first sailor gets 80 coins and tosses the one left over. This leaves 160 coins in the chest.

The second sailor sees 160 coins and takes 1/3. One-third of 160 is 53r1, so the second sailor gets 53 coins and tosses the one left over. This leaves 106 coins in the chest.

The third sailor sees 106 coins and takes 1/3. One-third of 106 is 35r1, so the third sailor gets 35 coins and tosses the one left over. This leaves 70 coins in the chest.

Finally, the tax collector sees 70 coins and divides it into three parts to give to each sailor. One-third of 70 is 23r1, so each sailor gets an additional 23 coins added to their collection and the one left over goes to the tax collector.

Each sailor, therefore, has the following:
The first has 80 + 23 = 103 coins.
The second has 53 + 23 76 coins.
The third has 35 + 23 58 coins.
And the tax collector has 1 coin.

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