Correct.
You have 9 eggs to present to the king. However one of the eggs is rotten, and presenting it to the king would result in your death. The only way to discern the rotten egg is by its weight, it will be heavier than the other eggs (the rest of which weigh the same).
You have one set of scales, but you may only place eggs on either side of the scales. You may place as many or few on either side as you wish.
What is the minimum number of weighings required to ALWAYS find the rotten egg?
Angelus said:
No it's not. If two people have the same number, then the third person immediately knows his number is the sum of those two, because the alternative, that he has the difference of those two, is impossible, since 0 is not positive. For any other combination of numbers, A cannot determine whether his number is the sum or difference of two numbers in the first round. So, the only ways for B to know what his number is when A did not are when the two numbers B sees are the same, and they are in the ratio 2:1 with the larger number on A. In the latter case, if B has the difference, or the same as C, then A would have known immediately when he saw two identical numbers, so he knows A:B:C are in ratio 2:3:1. The reasoning continues in this fashion, each turn the logician being able to know what his number is only when the other alternative would have given it away to another logician previously. So when C goes, he knows his hat is neither the same as A's, nor the same as B's, nor half A's. He determines his number if A and B are identical, or if the ratio A:B:C is 1:2:3, 2:1:3, or 2:3:5. So when A goes again, he knows his number is not equal to or twice either of the others, not half of B's, and not in a 2:3 ratio with B's. He determines his number when A:B:C are in any of these ratios: 3:2:1, 3:1:2, 4:3:1, 4:1:3, 5:2:3, 8:3:5. (In each case, the alternative is a combination somebody else would have already solved.) Out of these, only 5:2:3 allows A's number to be 50. |
Sorry about that. I should have redone the problem before posting.
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Two. First, take two groups of three, if one is different, it is in that one, if they are the same, it is in the third group.
Do this again with the three remaining eggs.
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This one is a bit harder, but quite fun:
110 pirates want to split a treasure of 50 gold coins (they can't be split any further.) This is done in the following manner: the highest ranked pirate (no.110) proposes a way to split the treasure. This is voted upon democratically by all pirates including himself, and is accepted if he has AT LEAST 50% (so 50 % included. I screwed up on the terminology last time, don't want that again.)
In order, the priorities of a pirate are:
1. Their own life
2. Gold
3. Space on the boat (They rather have their comrades dead than alive.)
Everyone has perfect logic.
Who is the highest ranked pirate to survive, and how does he do it?
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An easier one in the meantime: you have a dice, and three colours, how many ways can one colour it?
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Correct.
It is now time to prepare the king hIs breakfast. However the King is very picky, and failure to cook his egg as required will result in death.
His majesty's egg must be boiled for exactly 9 minutes. You have no other means of measuring time except for a 7-minute and 4-minute hourglass.
What is the minimum amount of time required to prepare a 9-minute-boiled egg using just these two hourglasses?
Start 4 minutes and 7 minutes at the same time, immediately turn over 4 when finished, when 7 is finished, put in egg, turn over 4 when finished, again, stake out egg when 4 is finished. 16 minutes.
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Try one of mine! (the two unsolved.)
Bet with PeH:
I win if Arms sells over 700 000 units worldwide by the end of 2017.
Bet with WagnerPaiva:
I win if Emmanuel Macron wins the french presidential election May 7th 2017.
16 minutes is incorrect :(
Are there constraints for the dice colouring problem? I'm assuming that one may colour each die face one solid colour only, and that the die pips aren't to be coloured.
I would say 4096 combinations (assuming that the die is a different colour to the paints)
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