December 2004:

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December 2004:
Three men go fishing. After catching n fish, they sleep under a tree. One of the three wakes up after a while. He doesn't want to wake the others, so he divides the fish equally into three and throws the remaining one back in the sea. He takes his share and then leaves the other two. The second one wakes up after a while. Not knowing that his friend has left, he also divides the fish equally into three, gets rid of the remaing one fish, takes his share and leaves. This also happens for the last man after he wakes up, and he also gets rid of one remaining fish. At the end, there remains a number of fish divisible by three. What is the general solution for n?

There are n fish to begin. The first man divides the fish into three equal parts and gets rid of one. Therefore, there remains 2(n-1)/3 for the second man. The second man does the same with the fish. Therefore, there remains 2((2(n-1)/3)-1)/3 for the third man. This simplifies to (4n-10)/9 after some algebra. The third man does the similar procedure with this quantity of fish. Hence, the last remaining quantity is 2(((4n-10)/9)-1)/3 which simplifies to (8n-38)/27. We are told that this quantity is in fact divisible by three. Algebraically, then (8n-38)/27 = 3m where m is a positive integer. Solving for n, we obtain n = (38+81m)/8. Notice that this simplifies to an integer for n only when m = 2, 10, 18, ... or mathematically m = 2 + 8k where k = 0 , 1 , 2 ,... By substituting this expression for m in (38+81m)/8, we get n = 25 + 81k. This is the general solution for n.
The answers, therefore are {25, 106, 187, ...} The most practically applicable answer for this problem is probably 25.
Mr. Andy Young also showed by further analysis that the final remaining quantity of fish has the expression 6 + 24k. This can be achieved by substituting our final expression for n, that is 25+81k, in (8n-38)/27.
Topic: Algebra

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