   
Puzzle #476  The Voyage of the Monkeys
Puzzle #476  The Voyage of the Monkeys
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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.


Puzzle #476  The Voyage of the Monkeys
Originally Posted by Bhawini 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.


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 B2 and C3.
Trip 4: A, D and 1 goes; Only D returns, leaving A1, B2 and C3.
Trip 5: D and 4 goes together which is the last ride.


Puzzle #476  The Voyage of the Monkeys
Originally Posted by Bhawini 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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