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  1. #1
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    Post Puzzle #476 - The Voyage of the Monkeys

    Puzzle #476 - The Voyage of the Monkeys



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  2. #2
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    Post Puzzle #476 - The Voyage of the Monkeys

    A B C D can be assumed as the 4 humans and a b c d can be the monkeys...so to help them fleas einstenia in the least number of trips,the trips should be like...A B D would go in the first trip together than a b d monkey will go in 2nd trip together and the last bugler C along with his monkey c will go at last.In this we assume that ABCD all knows how to raft...and only 'd' knows how to raft..and since no different combination of juggler and monkey can go together so in first trip any of the human can raft..in 2nd trip d will take the other two monkeys and in trip 3 at last juggler C along with his monkey will raft and flee to Newtonian.

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    Post Puzzle #476 - The Voyage of the Monkeys

    Quote Originally Posted by Bhawini View Post
    A B C D can be assumed as the 4 humans and a b c d can be the monkeys...so to help them fleas einstenia in the least number of trips,the trips should be like...A B D would go in the first trip together than a b d monkey will go in 2nd trip together and the last bugler C along with his monkey c will go at last.In this we assume that ABCD all knows how to raft...and only 'd' knows how to raft..and since no different combination of juggler and monkey can go together so in first trip any of the human can raft..in 2nd trip d will take the other two monkeys and in trip 3 at last juggler C along with his monkey will raft and flee to Newtonian.
    You're assuming that the raft is coming automatically after the first trip and second trip.

  4. #4
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    Post Puzzle #476 - The Voyage of the Monkeys

    Answer is Minimum 5 trips

    Let A, B, C and D be the jugglers and 1, 2, 3 and 4 be their respective monkeys. (1 is the monkey who knows how to raft)

    Trip 1: A and 1 goes; A returns with boat leaving 1.
    Trip 2: A,B and 2 goes; A and B returns, leaving 1 and 2.
    Trip 3: B, C and 3 goes; 1 returns, leaving B-2 and C-3.
    Trip 4: A, D and 1 goes; Only D returns, leaving A-1, B-2 and C-3.
    Trip 5: D and 4 goes together which is the last ride.

  5. #5
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    Post Puzzle #476 - The Voyage of the Monkeys

    Quote Originally Posted by Bhawini View Post
    A B C D can be assumed as the 4 humans and a b c d can be the monkeys...so to help them fleas einstenia in the least number of trips,the trips should be like...A B D would go in the first trip together than a b d monkey will go in 2nd trip together and the last bugler C along with his monkey c will go at last.In this we assume that ABCD all knows how to raft...and only 'd' knows how to raft..and since no different combination of juggler and monkey can go together so in first trip any of the human can raft..in 2nd trip d will take the other two monkeys and in trip 3 at last juggler C along with his monkey will raft and flee to Newtonian.
    In first trip as jugglers A, B and D are going. In this case their monkeys will stay with jugller C and hence they will start creating noise and this violates the condition in the question.

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