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Everything posted by plainglazed

  1. another seven and I think an example of six? still working on five...
  2. yeah, think I agree with that statement, dealer.
  3. I think I see the pattern here and am encouraged by the beauty of binary symmetry so here goes:
  4. @aiemdao - Yes, nice solve indeed Aiem. Well done. @bonanova - thanks.
  5. hey bn - Thanks for getting this one started. You are correct in all you say above including your initial summary of the problem. Have edited the OP to include that explanation. I assure you there is a scheme in which one can guarantee finding seven heavy coins. Perhaps start off by finding one guaranteed heavy coin in two uses of a balance scale. Or not. I have no doubt you or others here will discover the method for finding seven heavy dimes.
  6. You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes. EDIT: for clarification
  7. okay, not too proud to start this off with some probably way simplistic maths:
  8. had thought perhaps then perhaps rats, still thinking
  9. ooh, nice one Cyget. Spoiler for some additional thoughts:
  10. 1. EDIT: yeah, as I continue to wake up here, am less convinced of my logic above.
  11. Not a generalized formula but better than strategy for three?
  12. dang, like that answer MM - nice
  13. Silently be in doubt and subtle...
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