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12 red snooker balls

  1. Aug 24, 2004 #1
    We have 12 red snooker balls and a scale balance.
    One of the snooker balls has a different weight but we do not know whether it is heavier or lighter.
    The scale balance may be only used three times by putting an equal number of balls on each side and noting the balancing result.
    What 3 balancing arrangements would be required to ensure the identity of the odd weight ball and whether it was heavy or light.
     
  2. jcsd
  3. Aug 24, 2004 #2
    hmm. I am not trying to be rude here.
    But this q is asked before.
     
  4. Aug 24, 2004 #3
    i dunno if this is asked before but nice *twist* on the old *ionc* problem :wink:
     
  5. Aug 24, 2004 #4
    I don't think this is solveable with the alotted measurements and information.
     
  6. Aug 24, 2004 #5
    Surely it is!
    Consider 1234 x 5678 at first...
     
  7. Aug 24, 2004 #6
    youre sure you read right? that we dont know if the unique ball is heaver OR lighter? I think that creates a problem.
     
  8. Aug 24, 2004 #7
    Healey,
    It can be solved .....
    infact it can be shown that
    In n weighings, [(3^n-3)/2 + 1] snooker balls can be weighed and the odd ball can be found.

    Placing n = 3, we get 13 .. so ofcourse it is possible even for 12.

    I won't spoil this one for anyone .... i already received too many insinuating glares from Gokul already..... :D

    -- AI

    P.S -> So what does this post mean? It means that once the original problem is solved some followers will be,
    Find a solution when we have 13 snooker balls?
    Show that the general statement of n weighings given by me is true?
     
  9. Aug 24, 2004 #8

    NateTG

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    Odd -- according to your formula, I can't find it in 1 ball with 0 weighings, when the number for n=0 should be 1.
     
  10. Aug 24, 2004 #9
    Unfortunately, in just 3 weighings it's not possible for 13 balls...
    (12 is the max)
     
  11. Aug 24, 2004 #10

    NateTG

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    Really, I'm pretty sure it's possible to find the odd ball out with 13 balls - you may not be able to tell whether it's light or heavy though.
     
  12. Aug 24, 2004 #11

    Hurkyl

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    Then you're not looking hard enough. :biggrin:

    If you have one ball, and you know one ball has the incorrect weight, then you don't need to do any weighings to find it!

    The impossible case is where you have two balls.

    (Oh, and incidentally, the formula says you can find 0 balls in 1 weighing)
     
  13. Aug 24, 2004 #12
    Someone post the answer in white. I've tried it numerous ways for a few minutes, and i can get pretty close, but for every possibility theres one where I cant determine if the odd ball was heavier or lighter than the others, but just different.


    Hmm by that equation you should be able to find the odd ball out of 4 in 2 tries, and tell me if its heavier or lighter?? Please show me (its a lot simpler than the stated riddle).

    EDIT : nevermind, i got it. But my simple question still stands.
     
    Last edited: Aug 24, 2004
  14. Aug 24, 2004 #13

    NateTG

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    You're usually so on the ball (no pun intended) that I'm guessing you misread, or I wasn't clear enough, since we appear to (in principle) agree that the formula is incomplete.

    However, the formula indicates that one ball can be identified in one weighing:

    In the formula [tex]n[/tex] is the number of weighings, and it gives the number of balls.
    My post indicated that the formula was too low in the case of 0 weigings.

    [tex]\frac{3^n-3}{2}+1 = \frac{3^1-3}{2}+1 = \frac{0}{2}+1 = 1[/tex]

    But 0 balls in 0 weighings.
     
  15. Aug 24, 2004 #14

    Hurkyl

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    Yes, I meant 0 in 0.
     
  16. Aug 24, 2004 #15

    NateTG

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    My post in the thread
    https://www.physicsforums.com/showthread.php?t=22391&highlight=Brain+Teaser+#90

    Should give you a good idea on how to find a solution to this problem. Notably, using that method, the 10 coin case might be more difficult that the 12 coin one.
     
  17. Aug 24, 2004 #16

    Gokul43201

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    How about , <select to see>



    first : L = 5, 6, 8, 10 R = 7, 9, 11, 12

    second : L = 2, 3, 4, 7 R = 5, 6, 11, 12

    third : L = 1, 4, 10, 11 R = 2, 5, 7, 8 ?


    I think this gives unique outcomes for each number.

    Defining outcomes as <, > or = on the basis of the L pan being heavier, R pan being heavier, or both equal.


    1 : ==< or ==>
    2 : =<> or =><
    3 : =<= or =>=
    4 : =<< or =>>
    5 : <>> or ><<
    6 : <>= or ><=
    7 : ><> or <><
    8 : <=> or >=<
    9 : >== or <==
    10: <=< or >=>
    11: >>< or <<>
    12: >>= or <<=

    All look different to me...
     
  18. Aug 24, 2004 #17
    Both are related observations and well observed indeed ....

    Ofcourse it was my mistake that i forgot to mention one tiny 'precursor' reqd to relax the problem.....

    Hurkyl,
    Now with 1 ball u know which is the odd one but u don't know whether it is heavier or lighter so u need one weighing (but weighed against what ??? that will be ur question ... wait i will address it shortly)

    NateTG,
    You are pretty correct indeed and we will be able to find the lighter or heavier condition too .....

    The relaxation given to the problem,
    "You can add one true weight ball to the existing set of balls"
    (Hurkyl and NateTG would have got the entire general solution by now :wink: )

    -- AI
     
  19. Aug 25, 2004 #18
    Now it's easy (and possible): compare 12345 x 6789T at first...:-)
     
  20. Aug 25, 2004 #19

    BobG

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    1. Left scale= 1, 2, 3, 4
    Right scale = 5, 6, 7, 8

    2. Left scale = 1, 4, 7, 10
    Right scale = 2, 5, 8, 11

    3. Left scale = 3, 6, 12
    Right scale = 1, 8, 9
     
  21. Aug 26, 2004 #20

    BobG

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    Revised solution (forgot to check opposites)

    1. Left scale= 1, 2, 3, 4
    Right scale = 9, 10, 11, 12

    2. Left scale = 1, 4, 7, 10
    Right scale = 2, 5, 8, 11

    3. Left scale = 1, 3, 9, 11, 12
    Right scale = 2, 5, 6, 7, 10
     
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