From chunchen@usc.edu Wed Jun 2 10:01:18 2004 Subject: A clear answer to your puzzle To: propp@math.wisc.edu MIME-version: 1.0 Content-type: text/plain; charset=us-ascii Content-language: en Content-transfer-encoding: 7BIT Content-disposition: inline X-Accept-Language: en X-UWMath-MailScanner: Found to be clean X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on lcyoung.math.wisc.edu X-Spam-Level: X-Spam-Status: No, hits=0.0 required=1.4 tests=none autolearn=no version=2.63 Content-Length: 2193 Dear Professor Propp, I was directed to your puzzle at http://www.drunkmenworkhere.org/170 I spent couple hours figuring out the solution, and got it right. however i checked the one route for deriving the answers, and found that its actually much more work than the one i came up with. I solved it using pretty simple logic rules. Here is my solution : 1. Q6 = D, Q17 = B 2. Q10 = A, Q16 = D 3. Q20 = E 4. Q8 = E (from Q7) 5. Q12 = A (Originally I dont count Q2, Q7 as constant, but that leaves me with 10 which is answer A and E, so its wrong), If i include Q2, then I will get D, but that means Q7 = E. Looking at Q1, it states Q1 != A, B. so Q2 != B, therefore, Q7 cannot be E. Thus, there are 12 constants. 6. Q5 = E 7. Q3 or Q4 has to be B, but Q3 cannot be B (we have already 2 E), so Q4 has to be B. 8. Q1 = D (from 7) 9. Q3 = D (Because Q8 = E, so A + E = 8. We know A = 5, so E = 3) 10. Q15 = A (from 5) 11. Q13 = D (from 10) (this indicates that Q9 cannot be A, so Q2 has to be A. (B is wrong from Q1, C is wrong because Q8 is E, D is wrong because 10 is A, E is wrong because 11 cannot be A - there are already a B from Q4) 12. Q18 = A (Cannot be E , already have 3 Es, cannot be D, E = 3, A = 5, Cannot be C, because Q14 indicates that there are more than 5 Ds. B is not a choice because there are no possible 5 Cs in the remaining selections. So it has to be A) 13. Q9 = D (from 11, Q9 cannot be A. Q9 cannot be E-because we have 5 A, 5 B and 3 E, so D is is less than or equal to 7, Q9 cannot be B, because if so, then 11 has to be C, Q9 cannot be C ( C is A)) 14. Q14 = B (We only have Q7, Q11, Q14, Q19 left, but we have to have 3Bs in the 4 choices) Q14 can only be A or B, since all 5 As are taken, so it has to be B 15. Q7 = D, Q7 cannot be A (all 5 As are taken) Q7 cannot be B (if so, then 11 has to be C) Q7 cannot be C (0 Cs allowed from Q14) Q7 cannot be E (from Q2) 16. Q11 = B 17. Q19 = B (need 5 Bs) This is a very interesting logic game, I hope you can make more of these type of games :) Thanks for the challenging fun Best wishes Chun Yu James Chen Department of Engineering University of Southern California chunchen@usc.edu From krazybob42@gmail.com Mon May 23 13:03:41 2005 DomainKey-Signature: a=rsa-sha1; q=dns; c=nofws; s=beta; d=gmail.com; h=received:message-id:date:from:reply-to:to:subject:mime-version:content-type:content-transfer-encoding:content-disposition; b=G06DsP4OBUuvStw6Gxv3DCAcoC411hPAwbfsFjWiG/S60R+J5uj9qqw2bI9qGUc23hN02jtrLzCdXCqyXOrxqhwXkOtmWmaLxsb5kEIYuzO6il/BLb76Mo6WWVHFY83Sm04Tn+zvvjjE66s3uvCBkJuN3oR4ZZC6yPWNBor72ag= To: propp@math.wisc.edu Subject: self-referential aptitude test Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Disposition: inline X-UWMath-MailScanner: Found to be clean Content-Transfer-Encoding: 8bit X-MIME-Autoconverted: from quoted-printable to 8bit by lcyoung.math.wisc.edu id j4NI3Bw02126 X-Spam-Checker-Version: SpamAssassin 2.63 (2004-01-11) on lcyoung.math.wisc.edu X-Spam-Level: X-Spam-Status: No, hits=0.9 required=1.0 tests=FROM_ENDS_IN_NUMS autolearn=no version=2.63 Content-Length: 360 In Chun Yu James Chen's solution to the self-referential aptitude test ( at http://jamespropp.org/srat-C ), his second step is incorrect. He states that Q10 must = A and Q16 must = D, but at that stage, it could also be the case that Q10 = D and Q16 = 10. The solution's fine, but I just wanted to point out that the logic at that point is flawed.