Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Weighing the balls

  1. Feb 16, 2006 #1
    Ýou have 12 balls.
    All of them have the same mass.
    Except one.
    You don't know if it weighs more or less than the other 11 balls.
    Your mission is to find the odd one out, and say if it weighs more or less than the 11 others.
    You have a balancescale.
    You may use it for three weighings.
    The technique must work no mmater where the odd ball is.

    EDIT: Added the balance part
    Last edited: Feb 16, 2006
  2. jcsd
  3. Feb 16, 2006 #2


    User Avatar
    Gold Member

    I think there's one item missing from the givens. Do you know the weight of the balls, or is that unknown?
  4. Feb 16, 2006 #3
    There is no need to know the masses of the balls in order to complete the brain teaser.

    The correct answer will fill many A4:s of possible weighings.


    If i weigh those balls against those, which one will weigh the most? How and in which weighingss do I get controlballs etc.
  5. Feb 16, 2006 #4
    I think the missing item could be that it's not a *scale* per se, so much as a balance. So, one weighing might be to have ball A vs. ball B, the result of which would tell you which was heavier, A or B. Doing that, then yes, it's possible within 3 weighs, and no other given information.

    If it's really a *scale*... then... hm. Not sure how many weighs you'd need, but I'd bet a bit more than 3. I get between 3 and 6 weighs using a scale, 3 if you make lucky picks, 6 if you're unlucky. Better than 6, anyone?

    Last edited: Feb 16, 2006
  6. Feb 16, 2006 #5
    Oh, you meant it that way! Yes it is a balance scale. You get the masses of the balls relative to each other and nothing else. Sorry for the confusion.
  7. Feb 16, 2006 #6
    Came back to this and noticed it was possible in 5 (assuming a scale, not a balance). Considering that by effectively doing a binary-search-type weighing, you could do weighs of 6 balls, then 3, then 2, then 1, so maybe it's possible in 4 scale-weighs? But 5 might be the minimum considering that you need at least 4 weighs to *find* what you're looking for, plus 1 weigh to find out whether the one you're looking for weighs more or less than the remaining 12.

  8. Feb 16, 2006 #7
    It is a balance.
  9. Feb 16, 2006 #8
    But the scale weighing is much more intersting to me... The balance ones have been way overdone.
  10. Feb 17, 2006 #9
    The question is with a balance. Solve it. If you want a scale post another brain teaser
  11. Feb 17, 2006 #10
    It can't be both?

    The "find-the-1-of-12-objects-that-weighs-a-different-amount" problem has been done a bunch of times on this forum, so it's entirely possible that you won't see much of a response. The last time someone posted it, you can see the response they got:


    So, don't feel too bad if there isn't a reply.

    I don't remember seeing a *scale* question before, though, so I know at least I was interested (and apparently Moo, too) in seeing a new problem. But, anyway, I'm done. Unless someone wants to discuss the scale problem further...

  12. Feb 17, 2006 #11


    User Avatar
    Science Advisor
    Homework Helper

    If all you care about is worst case, then the scale is a pretty straightforward binary search.
  13. Feb 17, 2006 #12
    Ok, i didn't realize that it had been posted to death.

    Ok it is a scale instead of a balance
  14. Feb 17, 2006 #13
    I have just worked it out in 3 steps using a plain balance so you don't really need a scale. (No, I didn't look up the solution.) It's a bit tricky, not just a binary search.
  15. Feb 17, 2006 #14
    ok, post the answer in the colour ;P
  16. Feb 17, 2006 #15
    Ok, I'll try COLOR="#e9e9e9" to match the "PF Prime" color scheme. I hope it works well enough with other schemes too...

    Eleven balls have mass M, the target ball has a greater or lesser mass.

    First, balance four balls against four balls.

    Case 1: the balance is level so the target is in the remaining goup of 4. Label these four balls 1, 2, 3 and 4. Set aside ball 4 and balance group {1,2,3} against known balls {M,M,M} from the remaining eight.

    Case 1.1: the balance is level so ball 4 is the target. A third balance determines if it is heavy or light.

    Case 1.2: the balance tips towards {1,2,3} so one of these balls is the target, which is now known to be heavy. Balance ball 1 against 2 to find out which of these three is the heavy one.

    Case 1.3: the balance tips towards {M,M,M} so one of {1,2,3} is the target, which is now known to be light. Balance ball 1 against 2 to find out which of these three is the light one.

    Case 2: the first balance was not level so the target is in this group of 8 balls. Label balls on the heavier side {1,2,3,4} and balls on the lighter side {5,6,7,8}.

    Now balance balls {1,2,M} against balls {3,4,5}.

    Case 2.1: the balance is level so the target is in remaining set {6,7,8} and is known to be light (from the first balance). Balance 6 against 7 to know which of these three is the light one.

    Case 2.2: the balance tips towards {1,2,M} so either a ball in {1,2} is heavy or ball 5 is light. Balance ball 1 againt ball 2 to know if one is heavy and which one, or to know that ball 5 is light if the balance is level.

    Case 2.3: the balance tips towards {3,4,5} so either ball 3 or 4 is heavy. Balance 3 against 4 to know which one.

    I hope I didn't screw up anywhere. :uhh: Someome will let me know for sure.
  17. Feb 19, 2006 #16
    i propose a solution shud be right ...enjoy

    i will use W for the word weigh..u remove 2 coins .so u W the 2 piles of 5 coins .if they r the same then the 2 coins contain 1 odd coin so u take them and wiegh them against each other now u no 1 is light or normal or the other is heavy or normal, so now u choose any of them and weigh them against a normal coin.

    if the piles of 5 r difirent u name 1 pile heavy pile and 1 pile light similar to wat we did with the 2 coins. we remove a light coin put it aside and we form groups HHL HHL HLL we weigh HHL vs HHL if they weigh diffirent then odd coin is either H ,H or L , ( call normal coin N) so u do HL vs NN leaving out L .. if HHL vs HHL shows they weigh same then odd coin in HLL with 1 wiehing left u do HL vs NN leaving out L ...QED

    tell me if u find me incorect.(btw i didnt write all details in last part u can easily fill in the gaps)
    Last edited: Feb 19, 2006
  18. Jul 1, 2006 #17

    gosh! :surprised
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook