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

Home > Logic > Liars and Knights
Counting the Types (Posted on 2007-05-31) Difficulty: 2 of 5
A group of 25 consists of knights, knaves and liars. Each was asked two questions. 17 answered truthfully to the first question and 6 answered truthfully to the second.

What can be determined about the number of knights and liars in the group?

See The Solution Submitted by Brian Smith    
Rating: 2.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 3 of 4 |

Let the group consist of n knights, v knaves and r liars.
Let v_1 of the knaves responded truthfully to the first question
and v_2 of the knaves responded truthfully to the second question.

Since a knight will always respond truthfully to both the questions, we must have:
(a) v= v_1 + v_2
(b) 2n + v_1 + v_2 = 17 +6 = 23
(c) n + v + r = 25
(d) n < = 6

From (a) and (b), we have 2n + v = 23, or v = 23 - 2n
So, 23 - 2n + n + r = 25
Or, r = n + 2

v = 23 - 2n implies that the number of knaves will always be an odd number.

Now the group must contain at least one knight, otherwise the group would consist entirely of Knaves and Liars. This is a contradiction.

Thus, 1<=n<=6, giving: 11< = v< = 21


(i) The total number of liars will be precisely 2 more than the
number of knights.

(ii) The total number of knights is at least 1 but at most 6

(iii) The total number of knaves is odd

(iv) The total number of knaves is at least 11 but at most 21.

Edited on May 31, 2007, 12:24 pm
  Posted by K Sengupta on 2007-05-31 12:17:25

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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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