Reply to Re: Two interesting puzzles
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That's not the solution. I made a mistake in my first post. I meant that he could call the prisoner more than 3 times but in this way:
1,1,1,2,2,3,4,7,8,1
He cannot call the same person more than three days consecutively but can call him again after 6 to 7 days. Now you have to tell when exactly all the prisoners have been to the room. Suggest a plan in which you can be 100% sure that all the prisoners have been to the room and yes this puzzle can't be solved in 5 minutes.
All these puzzles my brother had told me, so I don't know if they had come from a website. Try these two:
1. You have 12 coins. 11 of them are of the same weight whereas one of them is heavier than the others. You have a Balance . Find out the heavier coin by not taking more than three measurements.
The difficult version of this puzzle is to find out the heavier coin in not more than two measurements.
1,1,1,2,2,3,4,7,8,1
He cannot call the same person more than three days consecutively but can call him again after 6 to 7 days. Now you have to tell when exactly all the prisoners have been to the room. Suggest a plan in which you can be 100% sure that all the prisoners have been to the room and yes this puzzle can't be solved in 5 minutes.
All these puzzles my brother had told me, so I don't know if they had come from a website. Try these two:
1. You have 12 coins. 11 of them are of the same weight whereas one of them is heavier than the others. You have a Balance . Find out the heavier coin by not taking more than three measurements.
The difficult version of this puzzle is to find out the heavier coin in not more than two measurements.
