Difference between revisions of "1997 PMWC Problems/Problem T5"
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Thus it was either B or C. Since A is telling the truth, C is lying. If D is telling the truth, then B is lying, and E is lying, but that's impossible. Thus D is lying. Then E is telling the truth, and so is B. Thus two people are lying (C and D), and the other three are telling the truth. Since it was one of B or C, and it wasn't B, it must have been C. | Thus it was either B or C. Since A is telling the truth, C is lying. If D is telling the truth, then B is lying, and E is lying, but that's impossible. Thus D is lying. Then E is telling the truth, and so is B. Thus two people are lying (C and D), and the other three are telling the truth. Since it was one of B or C, and it wasn't B, it must have been C. | ||
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==See also== | ==See also== |
Revision as of 14:45, 15 May 2012
Problem
During recess, one of five pupils wrote something nasty on the chalkboard. When questioned by the class teacher, the following ensued:
'A': It was 'B' or 'C'
'B': Neither 'E' nor I did it.
'C': You are both lying.
'D': No, either A or B is telling the truth.
'E': No, 'D', that's not true.
The class teacher knows that three of them never lie while the other two cannot be trusted. Who was the culprit?
Solution
Case 1: A is telling the truth
Thus it was either B or C. Since A is telling the truth, C is lying. If D is telling the truth, then B is lying, and E is lying, but that's impossible. Thus D is lying. Then E is telling the truth, and so is B. Thus two people are lying (C and D), and the other three are telling the truth. Since it was one of B or C, and it wasn't B, it must have been C.
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See also
1997 PMWC (Problems) | ||
Preceded by Problem T4 |
Followed by Problem T6 | |
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 |