Ms. Cooper was teaching a logic class which is very always popular with students and this semester was no exception. There was only one space left in the class and three students had applied. Each of the applicants was interviewed.
The last space must be taken by a student who makes either three true or three false statements. One of the other two applicants makes two true and one false statement. The other applicant makes one true and two false statements.
Here's what the three applicants said during their interview:
1. Cade is the oldest.
2. Kenny will not bring the teacher an apple.
3. I am the one with three true statements.
1. I would bring the teacher an apple.
2. Cade will be chosen.
3. Keith's first statement is false.
1. I am the oldest, so I should be selected.
2. Kenny will be selected.
3. Kenny's third statement is true.
Which student was selected as the last person chosen for the logic class?
Cade's third statement indirectly contridicts his first; therefore he mixes true statement(s) with lie(s).
That makes Kenny's second statement false.
In order for Kenny to be consistent and therefore the chosen one, his other statements must also be false: he would not bring the teacher an apple, and Keith's first stetement would be true and therefore Cade would be the oldest. This makes at least Keith's first two statements true and his third not immediately falsifyable. This scenario also would make Cade's statements, in order, True, True, False. If Keith's were also True, True, False, that would violate the premises of the puzzle.
It looks as if Keith is the winner of the seat: either all his statements are true or all false:
Cada's statements are True, False, False.
Kenny's would be False, False, False, violating the conditions.
Cada's: False, False, True.
Kenny's: True, False, True.
So this last scenario is the case: Keith was selected as all his statements were false. Cada had one true statement and two false, while Kenny had two true statements and one false.
Posted by Charlie
on 2013-12-05 15:34:39