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12 balls

  1. Jan 26, 2008 #1
    You have 12 balls, and a weighing machine with no standard weighs. One of those 12 balls is either lighter or heavier than the others. Spot the ball in 3 chances with the weighing machine.
    Last edited: Jan 26, 2008
  2. jcsd
  3. Jan 26, 2008 #2
    2x6, 2x3, 2x1+1
    or 2x4+4, 2x2, 2x1
    Last edited: Jan 26, 2008
  4. Jan 27, 2008 #3
    Sorry, I forgot to mention that you must also be able to say whether the ball is heavy or light.
  5. Jan 27, 2008 #4
    This "puzzle" has been posted a zillion times here...
  6. Jan 27, 2008 #5
  7. Jan 27, 2008 #6
    Maybe not a zillion times, but more than 5 times.
  8. Jan 28, 2008 #7
    I've had that dream before. Kind of a nightmare, actually. It was impossible to walk or sit down.

    Oh! Okay.

    6 vs. 5 + 1 lighter. Get the lighter set.

    3 vs 2 + 1 lighter. Get the lighter set.

    Weigh any 2 of them. If they are equal, the 3rd one is lighter.
  9. Jan 28, 2008 #8


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    Thank you. I actually have spittle on my screen.
  10. Jan 29, 2008 #9
    It is not ok: the defective ball could be in the heavier set .
  11. Jan 29, 2008 #10
    Oh craps. I didn't notice that it could be lighter or heavier.
  12. Jan 29, 2008 #11
    The puzzle is new to me, so here's my answer:

    Weigh 4 against 4.

    If they even out, select three and weigh against 3 from the remainder. If they cancel out still, and since there's 1 ball left, one more weighing will give away the answer. If not, then the foreign ball is within those 3 balls selected from the remainder. At this step, it should be clear whether the foreign ball is lighter or heavier than the others. Now say the said balls are A B C, and other random balls now known to be normal are X X X. Weigh A X against B X. If they even out, then obviously it's C, if not, then either A or B.

    Now if 4 against 4 doesn't cancel out, substitute one side with the remainder. Depending on what the scale reads, determine whether the foreign ball is lighter or heavier. In any case, you'll end with a 2 against 2 weighing, which will easily give away the foreign ball by swapping two balls from each side, and substituting for 1 ball from the remainder on one side.
    Last edited: Jan 29, 2008
  13. Jan 29, 2008 #12


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    I've told you a million times: don't hyperbolize!
  14. Jan 29, 2008 #13
    I am pretty positive the bolded part would not work.
  15. Jan 30, 2008 #14
    You're right, what I wrote is senseless. Impossible to determine a foreign ball out of 4 balls with only one weighing. I'll try rectifying the solution, tomorow
    Last edited: Jan 30, 2008
  16. Jan 30, 2008 #15
    You are right, Dave!
    Even my mother has already told me this a billion times...
  17. Jan 31, 2008 #16
    Ok here's the correction:

    4 against 4 doesn't even out

    Say we have on the heavier side A B C D and on the lighter side S T U V, and X represent a ball from the remainder. Now weigh S X X X against A T U V. If they even, out, the foreign ball is among B C D, and is heavier. If S X X X > A T U V, then the foreign ball is among T U V and is lighter. If S X X X < A T U V, weigh A against X. If A = X, then S is the foreign ball, and is lighter. If A > X, then A is the foreign ball and is heavier.

    Solved :biggrin:
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