Fishing for Thirds
Puzzler time.
Here it is.
Three guys go out fishing. They decide in advance that whatever they catch, they're going to divvy up three ways evenly.
So they finish fishing for the day, they pull back into port and they're going to sleep on the boat overnight, get up in the morning, divvy up the fish to go home.
In the middle of the night, however, one of the guys has a severe hemorrhoidal flare up, and he's got to get to the drugstore right away to buy some stuff.
He goes to take his third of the fish, and he notices that the number that they caught is not divisible by three, unless he throws one of the fish overboard.
So he does. He throws one of the fish overboard. So he takes his third of the remaining fish and he leaves.
Then, in the middle of the night, another guy wakes up with horrible stomach pains and he has to go home right away.
He says, "I'll take my third of the fish and then I'll go home, because I can't stay here like this."
So he goes to take his third, and he notices, interestingly, the same thing, that he can't take a third unless he throws one fish overboard. So he throws one fish overboard, takes his third and goes home.
The third guy who isn't sick, he gets up in the morning and he realizes that he hasn't taken his third of fish, but he's got to go home anyway. He figures the other guys are still sleeping.
So he says, "I'll just take my third and I'll go and when they wake up, they can take their third." But he realizes he can't take a third without throwing one fish overboard.
He throws one fish overboard, takes his third and leaves.
And here is the puzzler question.
What is the smallest number of fish by which this little scenario could have taken place? The smallest number that they could have possibly had when they pulled into port that evening, in order for this whole thing to work?
Good luck.
