Timeline for Convoluted Logic Puzzles 101 and 102
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 13, 2017 at 20:41 | comment | added | Gareth McCaughan♦ | Handwavily, if I ask p+q+r=10 questions then there are 2^p+2^q+2^r-2 possibilities for which questions if any are answered wrong, and that's at least 2^3+2^3+2^4-2=30, and 2^10/30 is much smaller than 64, which feels as if this shouldn't actually be possible. But that falls some way short of being an actual proof :-). | |
| Mar 13, 2017 at 20:11 | history | edited | Gareth McCaughan♦ | CC BY-SA 3.0 |
oops
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| Mar 13, 2017 at 20:10 | comment | added | Gareth McCaughan♦ | Oooops, of course it would. I don't know what I was thinking. | |
| Mar 13, 2017 at 19:53 | comment | added | Wesley Situ | For the first part of 102, wouldn't doing the same thing for bits 2,3,4, and 5 be a total of 9 questions? | |
| Mar 13, 2017 at 19:49 | history | edited | Gareth McCaughan♦ | CC BY-SA 3.0 |
102
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| Mar 13, 2017 at 19:16 | history | answered | Gareth McCaughan♦ | CC BY-SA 3.0 |