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?
The knights always tell the truth and liars always lie, but the knaves can either tell the truth and then lie or lie then tell the truth. So there are four possibilites as to what groups of individuals may do:
Possibility (type): A B C D
First question: T T L L
Second question: T L T L
Type A is the knight, D is the liar, and B and C are knaves that are out of phase with each other.
Using the above letters to represent the numbers of those types, we are given that:
A+B+C+D = 25
A+B = 17
A+C = 6
The last equation imposes the most severe restrictions. Any given value of A will determine a value for C, which can't be negative. The value of A will provide a value for B by the middle equation, and, knowing A, B and C, the first equation tells us D.
So the following table is based on the values of A:
A B C D
0 17 6 2
1 16 5 3
2 15 4 4
3 14 3 5
4 13 2 6
5 12 1 7
6 11 0 8
So there may be as many as 6 knights or as few as none, and there are two more liars than there are knights.
If it is taken that the group does indeed contain at least one of each type, then the minimum number of knights becomes 1 and the minimum number of liars becomes 3, which is the necessary 2 higher.
If the wording of "knights" in the plural is taken to mean there are at least 2 knights, then of course there are at least 4 liars.
It doesn't matter if type C has zero members, as there are plenty (11 in this case) of the other phase of knave.
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Posted by Charlie
on 2007-05-31 11:52:52 |
solution
