All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic > Liars and Knights
The Prime Club (Posted on 2004-02-03) Difficulty: 2 of 5
There is a nightclub in Truth town called the Truth Club which is made up entirely of knights and liars.

Sometimes they start singing a song. One person sings "At least one of us is a liar", the next person sings "At least two of us are liars", continuing on like this such that each person says one more person than the last person; each person singing exactly one line. (If there were 10 people in the club, the only person who hadn't sung a line would sing the last line, "at least 10 of us are liars" and then the song would be done.)

One day when you know there was a prime number of people in the club, you hear the start of the song "At least...", but don't hear the middle; all you know is that they sang the song through completely. Even though you only hear those two words at the start of the song, you can tell how many people are in the club. How many people were there?

  Submitted by RoyCook    
Rating: 3.7500 (4 votes)
Solution: (Hide)
Two (one knight and one liar).

The key is to realize that it has to be an even number of people, or else the song is paradoxical.

If an even number 2m are singing, then one can easily verify that the first m are Knights and the second m are liars.

If there is an odd number 2n+1, however, then consider the n+1th singer. He sings "At least n+1 of us are liars". If he is telling the truth, then so was everyone before him (since they all sang "At least x of us are liars" for some x less than n), so there are n+1 knights, which means there could be at most n liars. Thus, he lied, resulting in a contradiction. Similarly, if he were lying, then everyone after him would also be lying, but this would mean that there were n+1 liars, resulting in his having told the truth. Again, contradiction.

Once we know that it has to be an even number of singers, the behaviour of the bouncers guarantees that it must be two singers, the only even prime.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionAnswerMath Man2012-02-26 12:04:29
No SubjectJohn2004-02-05 17:07:56
re(4): SolutionPenny2004-02-03 17:55:01
re(3): SolutionSilverKnight2004-02-03 16:27:18
re(2): SolutionPenny2004-02-03 16:15:29
SolutionSolutione.g.2004-02-03 11:48:56
re: SolutionCharlie2004-02-03 10:27:09
SolutionPenny2004-02-03 09:34:24
Quiet night at the clubretiarius2004-02-03 09:23:58
SolutionSOLUTION??? MAY BEAdy TZIDON2004-02-03 09:09:43
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information