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Thalia

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Everything posted by Thalia

  1. Thanks for the clarification. After some googling, I see apartment house is another term for apartment building. . . So if you live in an apartment, you live in a "house". Haven't done the puzzle yet so not sure how relevant that is but it would explain the perceived discrepancy. It was noted that this was published in 1959. I'm sure reality hasn't changed since then in terms of children and bachelors but I'm wondering if there were some different assumptions back then about recognizing children outside of marriage. I get the feeling that unless you're a politician, it might not be quite as big of a deal for a bachelor (or bachelorette) to have a child now as it was in the 50s. I don't have any evidence of that though. How does recognizing divorce contradict engaged men being bachelors? Google shows a bachelor as being a man who isn't and has never been married. So an engaged man would be a bachelor because he's not actually married, a divorced man would not be a bachelor because he used to be married, but as written, it doesn't matter anyway because Green is still married until the suit is settled.
  2. "3. The third baseman lives across the corridor from Jones in the same apartment house." Normally, I would consider an apartment and a house to be different. But clue number 3 seems to indicate that they are interchangeable.
  3. flamebirde - last 2 digits
  4. Think this one has been posted before. EDIT:
  5. Lol. That's what lead to one of the other guesses. Plain to see--> crystal clear. Crystal would be fancier drinking cups though. And I couldn't quite call it white.
  6. Lol. I saw people making it on tv once. The process is pretty impressive. Think it was a specialized kind though. Not good for mass production. Nice riddle.
  7. Still struggling with the last line.
  8. Assuming the 99% refers to weight...
  9. Well, there goes your curve estimation. I don't remember the situation but there were certain combinations that I could calculate with some multiplication. I think one of them was a CE combination. But of course the other combinations didn't feel like cooperating. I'll try to work on this over the weekend. Agree on your pink combinations except for XMEE. Should be the same count as EEM, right?
  10. Sorry to say I didn't save my counts. Slightly burnt out from counting, recounting, and rerecounting the previous ones. lol. Maybe I'll try again when I've had more time to recover.
  11. I did some quick counting for the easier ones. I've noticed that at least up to this point, the number of combinations of C, E, M, and X is (# cubelets removed+1)^2. There are 25 combinations for 4 cubelets removed. The ones including X would be the same as the counts for 3 cubelets without X. I did not count the combinations that have 2 or more E's or CCEM, CCMM, AND CEMM. I got 191 for the rest. I'm guessing the ones I didn't count are going to be pretty high so maybe around 500?
  12. When you remove a cubelet, the kind you remove matters because it changes the shape. When you take away 26, it doesn't matter which one is left. There may be 4 different labels but in the end, when it comes to the remaining shape, all you see is a cube. Another thing. For T27, there is one possibility. But given the phrasing of the original question, does nothing count as a shape?
  13. A couple more revisions because I can't tell the difference between counting to 4 and counting to 6 at night. Thanks for checking all these. That's a lot of numbers!
  14. Full guess with numbered cubelets and revision to CEE.
  15. Not sure if these are supposed to go together or be 6 separate things.
  16. New number for EEE. But I'm still getting the same result for CEE and CEM.
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