From chunchen@usc.edu Wed Jun  2 10:01:18 2004
Subject: A clear answer to your puzzle
To: propp@math.wisc.edu
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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
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To: propp@math.wisc.edu
Subject: self-referential aptitude test
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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.