When rotten PF members were rotten PF kids

[tribdog]

Many years ago I was a substitute teacher for the PF Academy for Incorrigible Youths. On my first day teaching I was pulled aside by the principal who told me a little bit about the kids in my class. “You have a rotten bunch of kids to look after, I feel sorry for you. These are the bottom of the barrel, three boys Artman, Gokul43201, and Greg B. and three girls Evo, Math Is Hard, and Moonbear. About the only thing these kids have in common, other than their bad attitudes, is the fact that only one third of the things they say are the truth, the other two-thirds are pure fabrications. They are quite adamant about it. Good Luck.”
I was full of nervous energy after hearing this, but my class didn’t start for another twenty minutes so I passed the time by rearranging the desks and placing a name tag on each one. I formed a circle out of the desks, in alphabetical order, faced them all towards a chair in the middle where I would sit. All of a sudden the bell rang and 6 rotten looking kids walked in and took a seat. I immediately knew I was in trouble because there was a girl sitting at Artman’s desk. Oh well, substitute teachers have had to endure this hazing ritual for centuries. I sat down in my chair and faced the girl sitting at Artman’s desk.
“Please raise your hand if you are in the correct desk according to the nametags I placed in front of you.
The girl “Artman” (call that desk 1) did not raise her hand. Going clockwise the kids sitting at desks 2, 3 and 6 did.
“How many of you are lying?” I joked. Surprisingly hands went up at desks 1, 3, 5 and 6.
Okay, I thought. I can figure this out. “Raise your hand if you are an odd number of chairs away from the desk with your name on it” the only hands that did not go up were the kids positioned at 4 and 6.
“How many of you are facing your desk?” The first four kids raised their hands.
“Gotcha!” I yelled and picked up the name tags and put them in front of the correct kid. Which kid sat at each desk?

 

[tribdog]

Last tribdog original for a while. I think this one is pretty hard, it took a long time for me to figure it out. Let me know if it is any good.

 

[arildno]

Yes, it's good. Haven't figured it out yet, though.

 

[tribdog]

legal disclaimer: I should say I don't really think these people are rotten, it is only a joke. I'm sure they are good people in real life.
Except for Gokul, I'm pretty sure he's rotten.

 

[Artman]

<While tribdog is explaining how we are not really rotten people, Artman, Gokul43201, Greg B., Evo, and Math Is Hard tape a kick me sign on his back, put a tack on his chair, and clap an eraser on his jacket while Moonbear distracts him by bending over in her short schoolgirl skirt. >

 

[Bartholomew]

You asked them 4 questions. How can 2 out of 3 of those be answered falsely? Or is it as a group instead of individually?

The premise is impossible anyway since assuming that some student or group of studenst have answered exactly 1/3 of the questions asked them truthfully, the answer any of them gives to the next question will wreck the balance so they either have answered more than 1/3 or less than 1/3 truthfully.

 

[Artman]

Sorry guys, I went back and put the aswers in white. Select to view.
I think these all work. I don't know if there is another solution or not.

1. MIH is in Artman's chair
2. Moonbear is in Evo's chair
3. Gokul is in his own chair
4. Artman is in Greg B's chair
5. Evo is in MIH's chair
6. Greg B is in Moonbear's chair

 

[tribdog]

Sorry Bartholomew, maybe I should reword it to say that 1 out of three are true instead of 1/3.
Artman, that isn't what I came up with. Here's here is a hint
the only people who can not raise their hands on the second question are lying AND lied on the first question. Therefore their next answer will be true. everyone else either lied then told truth or truth then lie so they will lie next time. seat 4 lied saying he wasn't in his correct seat, Greg B is the 4th name alphabetically so seat 4 is Greg B.

 

[tribdog]

come on, someone can figure this out can't you?

 

[Bartholomew]

Saying they answer 1 out of 3 truthfully says nothing about these particular 4 questions... what if one of them had just answered FFFFFFFF and is now ready to answer TTTT?

 

[Bartholomew]

Or did you mean one out of every _consecutive_ three so that each student must answer either TFFT, FTFF, or FFTF?

 

[Bartholomew]

Also, the problem only gives the truth-value pattern of things they _say_. Here they only raise hands.

 

[Bartholomew]

Huh, I run into a contradiction trying to solve it assuming you meant "one out of every consecutive three." The first student answered the first question truthfully so she must be TFFT, so from the last question she sits at desk 4. From the first two questions you know that the second student must be FFTF. The third and sixth students must be FTFF or TFFT from the first two questions. So the third answers of 3 and 6 must be F. So 6 is an even number of chairs away, 3 is an odd number of chairs away, and 2 is an even number of chairs away. So 2 must sit at 2 or 6 (can't sit at 4, it's occupied), as must 3, as must 6. Two chairs, three people.

So just what DO you mean?

 

[tribdog]

you are right, lol. I confused myself. sorry I do apologize. damn, I had it all worked out in my head. should have wrote it down.