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Topic: Loopy Logic (Read 2126 times) |
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ThudnBlunder
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Loopy Logic
« on: May 29th, 2011, 6:40pm » |
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I found this on my hard drive but can't remember posting it.
Which of the folliwing statements are true and which are false?
1) The answers to 6 and 7 are the same.
2) 1 is false.
3) The answers to 4 and 20 are different.
4) The answers to 3 and 20 are different.
5) The answer to this statement is different from the answer to 19.
6) 2 is true.
7) 15 is true.
8) The answers to 11 and 19 are the same.
9) 10 is true.
10) 13 is false.
11 Ms Smith is allergic to penicillin.
12) 16 is true.
13) 12 is true.
14) The answer to 11 is the same as the answer to this statement.
15) At least half the statements in this puzzle are false.
16) At least half the statements in this puzzle are true.
17) The answers to 9 and 4 are the same.
18) 7 is true.
19) Ms Smith's first name is Jill.
20) The answers to 3 and 4 are different.
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| « Last Edit: May 30th, 2011, 8:11pm by ThudnBlunder » |
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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SMQ
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Re: Loopy Logic
« Reply #1 on: May 30th, 2011, 5:07pm » |
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1) and 2) are mutually exclusive, 6) implies 2), therefore 1) and 6) are mutually exclusive.
Thus if 1) is true then 6) is false and by 1), 7) must be false as well. Conversely, if 1) is false then 6) is true and by not 1), 7) must be false. So either way, 7) is false.
Therefore 18) is false, and by not 7), 15 is false.
But if 15 is false, then 16) must be true. Moreover, there must be no more then 9 false statements (or no less than 11 true statements).
16) implies 12) implies 13) implies not 10) implies not 9).
3), 4), and 20) admit four solutions among them, either they're all false, or exactly one of them is false.
Since 9) is false, 17) is true if 4) is false and false if 4) is true.
If 5) is true then 19) is false; if 5) if false then 19) is false; therefore 19) is false.
Similarly, if 14) is true then 11) is true; if 14) if false then 11) is true; therefore 11) is true.
Therefore 8) is false.
We now have:
7)F, 8)F, 9)F, 10)F, 11)T, 12)T, 13)T, 15)F, 16)T, 18)F, 19)F
either 1)T, 2)F, 6)F; or 1)F, 2)T, 6)T
either 3)F, 4)F, 17)T, 20)F; 3)F, 4)T, 17)F, 20)T; 3)T, 4)F, 17)T, 20)T; or 3)T, 4)T, 17)F, 20)F
5) and 14) free choices.
Since there are 7 known false answers and at least two more false answers among 1), 2), 3), 4), 6), 17), and 20), the only way 15) can be false is if the choices are made to minimize the number of false answers.
Therefore, the only consistent solution is:
1) F
2) T
3) T
4) F
5) T
6) T
7) F
8) F
9) F
10) F
11) T
12) T
13) T
14) T
15) F
16) T
17) T
18) F
19) F
20) T
--SMQ
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--SMQ
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ThudnBlunder
wu::riddles Moderator
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Re: Loopy Logic
« Reply #2 on: May 31st, 2011, 12:20pm » |
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Nice work, SMQ. Your solution indicates that the problem is more interesting than I thought.
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