5
$\begingroup$

The 16 inhabitants of Knightsland are either knights or knaves (always tell the truth or always lie). Some, or all of them, meet every Sunday at the local restaurant and sit around a circular table, in equal distances from each other, in order to discuss their pending matters and then have lunch. Not all of them are always present. Also in the current meeting, not everyone was present. Jason, who is impatient and always in a hurry to finish things off and start eating, says: “We are already 11, so let’s start”. Harry, who was sitting right opposite to him, says: “No, we are not 11, we are 12 – apparently you don’t know how to count!!” The two of them were just about to start fighting but were calmed down by their left and right neighbors. Brandon, who was sitting calmly, says: “Calm down gentlemen; there were times in the past when we were even fewer! In fact, our today’s number is a product of two numbers, of which the smaller is the number we were in our smallest meeting!” Finally their gathering went out of control and any of the present members was blaming his adjacent neighbors for being liars.
How many Knightslanders were gathered this time and what is the smallest number that they ever in the past?

I understand we have 3 statements that can be either true or false, so we have to make a truth table.

The only thing I managed to deduce is that since Harry is sitting opposite to Jason, their current number is even. Furthermore, since Harry and Jason were calmed down by their right and left neighbors, assuming these are different for each, their current number is minimum 6. If Brandon was sitting calmly, can we also assume he was not one of the neighbors (who were trying to calm down the others)?

Anyway, I don't know how to continue... logic is not my favorite topic!

Any help is much appreciated!

$\endgroup$
5
  • 1
    $\begingroup$ I think you are missing an important cue: "Finally their gathering went out of control and any of the present members was blaming his adjacent neighbors for being liars." No knight would blame a neighboring knight to be a liar and no knave would blame another knave. So there are sitting alternatingly knight and knave. $\endgroup$
    – quarague
    Commented Jul 2, 2019 at 9:07
  • $\begingroup$ Also, Jason and Harry must be of opposite kinds (if one is a knight, the other must be Knave). But I am not sure... They may both be knaves. $\endgroup$ Commented Jul 2, 2019 at 9:19
  • $\begingroup$ guarague: I am not sure about this - why can't a knave blame another knave? $\endgroup$ Commented Jul 2, 2019 at 12:08
  • $\begingroup$ I just made an edit to make it more clear that not everyone was present (so 16 is excluded). Sorry if it was not clear from the beginning. $\endgroup$ Commented Jul 2, 2019 at 12:13
  • 1
    $\begingroup$ @WangXiuYingZhang: If a knave says to another knave, "you're a knave", then that accusation would be true -- and knaves cannot speak truth! $\endgroup$ Commented Jul 2, 2019 at 12:33

1 Answer 1

1
$\begingroup$

An important observations from @quarague in comments:

I think you are missing an important cue: "Finally their gathering went out of control and any of the present members was blaming his adjacent neighbors for being liars." No knight would blame a neighboring knight to be a liar and no knave would blame another knave. So there are sitting alternatingly knight and knave.

This implies the number of people must be even.

If the number is 2 modulo 4, then one of Harry and Jason is a knight and the other is a knave. But that is impossible since each of them claims a number that is not 2 modulo 4.

So the number is a multiple of 4, which means that Harry and Jason are of the same kind. But as they disagree, both of them are knaves. This means that there are not 12.

4 itself is impossible, as the OP notes, because then there would be no room for Brandon to be calm.

This leaves 8 and 16. But both of those appear to be possible:

  • There might be 8, if Brandon is a knave (the 4 knights are busy calming down Jason and Harry) which means that "we have been even fewer" is a lie, so 8 is the fewest they have ever been. (This also makes Brandon's second statement a lie because 8 is indeed not a product of two numbers the smaller of which is 8).

  • If there are 16, then Brandon must be a knight because he says "we have been even fewer" which we know is true because the narration said "not all of them are always present". In that case the smallest gathering they've ever had must be 1 or 2 (because there is no "the smaller" of 4 and 4). But would a knight say "even fewer" if this is the largest possible gathering? He would need to be a somewhat mischievous sort of knight ...


On the other hand this analysis means that Harry must be right when he says "we are not 11", which makes everything inconsistent. Can a knave utter a comma splice between two statements of which one is true and the other is a lie?

$\endgroup$
14
  • $\begingroup$ Henning Makholm can you please explain the below "If the number is 2 modulo 4, then one of Harry and Jason is a knight and the other is a knave. But that is impossible since each of them claims a number that is not 2 modulo 4"? $\endgroup$ Commented Jul 2, 2019 at 12:05
  • $\begingroup$ @Sal.Cognato: Harry and Jason are directly opposite each other, so if there are $n$ people in total, then there are $n/2$ spaces between Harry and Jason. Now if $n\equiv 2\pmod 4$ then $n/2$ is an odd number, and therefore Harry and Jason are different kinds. And therefore either 11 or 12 must be right. But neither 11 nor 12 is 2 modulo 4. $\endgroup$ Commented Jul 2, 2019 at 12:18
  • $\begingroup$ There are only $16$ inhabitants of the island and not everyone is present at this meeting, so $16$ is ruled out. $\endgroup$
    – David K
    Commented Jul 2, 2019 at 12:24
  • $\begingroup$ @DavidK: That information was added to the story after I answered. $\endgroup$ Commented Jul 2, 2019 at 12:28
  • $\begingroup$ @HenningMakholm you missed it because I just added it!! :) Sorry, my initial wording didn't make this clear enough, so I had to add this clarification. $\endgroup$ Commented Jul 2, 2019 at 12:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .