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We think of an island inhabited by rogues and knights.

Rogues are always lying, knights always speak the truth.

One day 12 islanders come together among which is at least one knight and at least one rogue.

2 of them say: "Exactly 2 of us 12 are rogues".

4 of the remaining ones say:" Exactly 4 among us 12 are rogues."

And the remaining 6 say:" Exactly 6 among us 12 are rogues."

How many of these 12 are actually rogues?

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If 6 knights say "Exactly 6 among us 12 are rogues", then the other 6 would be lying about the number of rogues, which Rogues must do. So 6 rogues.

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Here is a somewhat expanded explanation of the accepted answer.

It is given that there is at least one knight. He speaks one of the sentences, so he says "Exactly x among us 12 are rogues" for some x. All the other knights will agree with him, but the rogues will not. The sentence "Exactly x among us 12 are rogues" is uttered exactly x times, so there are x knights and they all claim the other x people are rogues. That means exactly half the people are knights and the other half rogues, i.e. 6 each.
Note that I needed to use the fact that there is at least one knight, because they could all be rogues and all statements would be false and that would be a second solution. The fact that there is at least one rogue follows from the fact that there are contradictory statements, so that does not need to be given up front.

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My way of solving this:

  1. Knights have to give same answers as they don't lie, so if there is more than one knight their answers coincide.

    1. if there is only one knight then hell say that there are 11 rogues. As nobody told that its not the case.

  2. If there is more than one knight it comes to check

    1. if at least two of them said that 2 are rouges (but then 10 knights of who only two said truth contradiction)

    2. if at least two of them said that there are 4 rogues(but then there are 8 knights and only 4 people said that there are 4 rogues)

    3. if at least two of them said that there are 6 rogues but then there are 6 knights, so all 6 knight told true and there are 6 rogues is the only solution in this puzzle

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Each separate group MUST have all Rogues or all Knights. Knights can't lie and Rogues must so they can't say the same thing within the group. So it's inconclusive. If the 1st group is telling the truth, then the remaining 2 groups would be groups of lying Rogues validating the 1st group making it 2 Rogues. If the 2nd group of 4 doesn't lie, then there are 4 Rogues which reconciles with what the 1st and 3rd group said when lying. If the 3rd group of 6 doesn't lie, then there are 6, which reconciles with the 1st and 3rd group that are lying.

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  • $\begingroup$ "If the 1st group is telling the truth, then the remaining 2 groups would be groups of lying Rogues validating the 1st group making it 2 Rogues." That doesn't make sense. The first group has two people, and if they're both knights, the other two groups (collectively 10 people) are all rogues. If there are 10 rogues, that invalidates the knights saying there are 2 rogues. $\endgroup$ – DarthFennec Jun 27 at 22:57

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