(In reply to
Please!!!! by randy)
There are only 32 possible combinations of Trues (Ts) and Falses (Fs). Assume the following format for the 32 potential answers, ordered A,B,C,D,E. It follows a pattern.
1=TTTTT, 2=TTTTF, 3=TTTFT, 4=TTTFF, 5=TTFTT, 6=TTFTF, 7=TTFFT, 8=TTFFF, 9=TFTTT, 10=TFTTF, 11=TFTFT, 12=TFTFF, 13=TFFTT, 14=TFFTF, 15=TFFFT, 16=TFFFF.
17=FTTTT, 18=FTTTF, 19=FTTFT, 20=FTTFF, 21=FTFTT, 22=FTFTF, 23=FTFFT, 24=FTFFF, 25=FFTTT, 26=FFTTF, 27=FFTFT, 28=FFTFF, 29=FFFTT, 30=FFFTF, 31=FFFFT, 32=FFFFF.
Now to eliminate answers, one by one.
1) If A is T, then C is T (there can't be 3 consecutive Fs unless there are also 2 of them). Eliminate: 5,6,7,8,13,14,15,16. Remaining: 1-4,9-12,17-32.
2) If A is T, then there cannot be 2 consecutive Fs (given). Eliminate: 4,12. Remaining: 1-3,9-11,17-32.
3) If B is T, then E is F (can't be fewer Fs than Ts if there are 3 Fs, and therefore 2 Ts). Eliminate: 1,3,17,19,21,23. Remaining: 2,9-11,18,20,22,24-32.
4) If E is T, then there must be 3 Fs (given). Eliminate: 9,11,25,31. Remaining: 2,10,18,20,22,24,26-30,32.
5) If D is T, then C and E must be F (if either were true, then they would make consecutive Ts with D, therefore D couldn't be true). Eliminate: 2,10,18,26,29. Remaining: 20,22,24,27,28,30,32.
6) We have eliminated all cases of A being T. Therefore, A must be F. Therefore, there must be two consecutive Fs (given). Eliminate: 22. Remaining: 20,24,27,28,30,32.
7) If B is T, there must be at least 3 Ts (so there are more Ts than Fs). Eliminate: 24. Remaining: 20,27,28,30,32.
8) If D is F, there must be 2 consecutive Ts (given). Eliminate: 27,28,32. Remaining: 20,30.
9) If D is T, there cannot be 2 consecutive Ts (given). Eliminate: 20. Remaining: 30.
Answer 30 is FFFTF; in other words, only D is true. This is the answer randy posted.
I am sure this is not the only way to prove it.
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Posted by Kearns
on 2005-04-14 16:03:33 |
re: Please!!!! (long proof of randy's answer)
