I can do the problem using the Chinese Remainder Theorem but I do not know if you heard of it?
The Ultimate Puzzle Site - Brain-Teasers
Same question as there, but this problem only deals with 3 sailors and instead of splitting 1/5, they split 1/3 each time. Does this method work? If so, can someone help me complete it; I don't like their way of solving it.
Let n = total # of coconuts. Then,
(n - 1) - 1/3(n - 1) is after the 1st sailor;
Then, let k = (n - 1) - 1/3(n - 1), and after the 2nd sailor will be:
(k - 1) - 1/3(k - 1)
Then, let m = (k - 1) - 1/3(k - 1), and after the 3rd sailor will be:
(m - 1) - 1/3(m - 1);
Finally, when they all wake up and give 1 away and then split the remaining, we have to do it one more time.
Let n = (m - 1) - 1/3(m - 1), and thus the final answer is:
(n - 1) - 1/3(n - 1)...
Ok, now the tricky part is substituting that in. I have a whole white board filled up with substitutions, and it gets extremely messy with parentheses etc. Can anyone help?
Parentheses never give no trouble even when there are many of them, because numbers are my friends. But there is one thing in group theory, if you ever take it, that I would imagine greatly confuses students even more than many paranthesis. It is called the factor group of a factor group. To explain basically you can think of a group as a set. A factor group is a set of sets. For example {{1,2,3},{4,2,1}}
Now the factor group of that is the set of sets of sets.
Thus,
{{{1,4},{1,2,3}},{{3,5},{4,5,6}}}
And you need to find how they add together.
I would imagine that gives students great difficutly.
Hello, Ideasman!
Ah, the "Monkey and the Coconuts" problem with three sailors.
I have an algebraic approach to this problem . . . lengthy but simple.
Let = number of coconuts.
Sailor #1 divides coconuts by 3 and there is a remainder of .
. . [1]
He hides his coconuts and puts coconuts back in the pile.
Sailor #2 divides coconuts by 3 and there is a remainder of .
. . [2]
He hides his coconuts and puts coconuts back in the pile.
Sailor #3 divides coconuts by 3 and there is a remainder of .
. . [3]
He hides his coconuts and puts coconuts back in the pile.
In the morning, they divide coconuts by 3 and there is a remainder of .
. . [4]
In [4]: since is an integer, must be odd: .
Substitute: .
Substitute into [3]: .
. . Since is an integer, must be even: .
Substitute: .
Substitute into [2]: .
. . Since is an integer, must be even: .
Substitute: .
Substitute into [1]: .
For the least positive value of , let .
Therefore: .
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