did the 10th century one have to be 3rd?
Game 4: Parking LotsThe possible combinations were XYYZZ and XYZZZ. I was not too sure about the question about Z being $12 and so which day would Y need to go. I think that the answer was either Monday or Tuesday.
XYZZZ can NOT be a combination for game four because Thursday was a 15 dollar lot and X must have a greater cost than Z, so Z can never be 15 dollars and therefore can never be Thursday.Not trying to be rude, just trying to make sure correct info is out there.
Quote from: cascagrossa on October 02, 2004, 03:23:08 PMdid the 10th century one have to be 3rd?There could be more than one 10th century one. There could only be one G, though, and G had to be 10th century. Also, the third had to be 10th century.
How many questions would someone who thought that it was BC as oppossed to AD automatically get wrong? I'm thinking 4/5, all except for the last one, the answer was definitely 4. I think I screwed up royally now that I realized misinterpreted this.
gooooo .... when you say the answer was definitely 4 r u referring to question 17 game three about max number of slots for f? i reread the instructions so many times I really think it didnt explicitly state each of the three (FGO) had to be represented. All it said was that that the sites were each one of the three and eachone of three centuries...it did not say (I dont think) that one of EACH OF FGO had to be in them or that one of 8 9 or 10 had ot be in them. This was the second twist of this game...because u nly know the distribution through your deductions -- only one G and you have to have at least 2 f's because you cant have o's on slot 4 and 5 bu becuase neither of slot 1 or 2 has to be 8, either or all could be f -- it could all 4 fs 1 g...does this sound right to you? Or did i re read it over and over and still screw it up?