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6 or less liars (Posted on 2015-11-13) Difficulty: 1 of 5
In a small village there are two kinds of people: liars and truthtellers.
Everybody knows everybody and everybody knows as well who is a liar and who’s a truth-teller.
I approach six villagers and pose the same question to each of them:
"How many liars are among you?"

I get six distinct answers (integers, of course) and deduce the true one.

How many liars are in that group?

Liars always lie and truthtellers never do.

See The Solution Submitted by Ady TZIDON    
Rating: 2.0000 (1 votes)

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Solution Summing up - spoiler | Comment 11 of 16 |

Short answer:  
The 6 distinct answers imply that there were
  either 5 or 6 liars in the group.
Although the asker could deduce the correct partition (T, L) the puzzle solvers can't determine the number of liars in the group - either 5 or 6.

Explanation (revised  following Dej Mar's comment):

The asker based his decision only upon presence of numbers 5 and/or 6 in the 6-tuple of answers:

  1. If there was neither   5 nor 6  - then  there were  6 liars. 

  2. Both   5 and 6  - 5 liars.

  3. 5 and no 6 -  no possibility to deduce.

  4. obviously 6 without 5 is not possible, forcing a liar to tell the truth - ...


  1. (0,1,2,3,4,11) or (3,4,7,8,55,99) …..6 liars. 

  2. (1,2,3,4,5,6) or  (3,4,5,6,7,8,) …..5 liars.

  3. (0,1,2,3,4,5) or (5,7,8,11,23,25) …5 or 6 liars.

  4.  (0,1,2,3,4,6)  or (1,6,7,9,22,77)

If one considers “logically consistent” liars who use neither negative numbers nor integers over 6 (the 1st sets in the above examples) as their answer then  in 7 possible cases of a “logically consistent” 6-tuple (one of the 0,1,2,3,4,5,6 absent):

a.(0,1,2,3,4,6) ….…..not an option.

b.(1,2,3,4,5,6); (0,2,3,4,5,6); (0,1,3,4,5,6); (0,1,2,4,5,6) ;(0,1,2,3,5,6) ........…5 liars.

(0,1,2,3,5,6)   no way to tell -   …5 or 6 liars.

So if the author deduced correctly and the consistent  6-tuple was a randomly chosen set then the odds are 5.5 to .5 in favor of 5 liars, (5 implied absolutely by logic and .5 by luck) i.e. 11 to one.
- If there is no consistency and any set of 6 distinct numbers is allowed  the odds move heavily in favor of 6 liars and no truth tellers.

Minor remark: it is beyond my comprehension why somebody (Steve ?) rated this puzzle as level 2, - IMHO it is certainly easy, interesting, thought provoking and should not be downgraded by someone who even did not solve it correctly.
No hard feelings, bro. Just wondering...



Edited on November 16, 2015, 1:52 pm

Edited on November 17, 2015, 10:53 am
  Posted by Ady TZIDON on 2015-11-16 13:42:45

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