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The Riddle Of The 9 Balls

  1. Mar 17, 2003 #1
    Ok, here is a good riddle (waiting for a Riddles thread to be made).
    You have 9 balls that look totally identical.
    They all weight the same, except for one of them (that weight either less or more, we don't know).
    You are supposed to get the odd one out by using a two-sided weight scale only for 3 times .
    Think about it, it is a bit tricky :wink:
  2. jcsd
  3. Mar 17, 2003 #2
    I know the first step is three on each side, logically...
  4. Mar 17, 2003 #3


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    The next step is either 4 - 4 or 1 - 1,
    not too hard to guess the last... :wink:

    Live long and prosper.
  5. Mar 17, 2003 #4
    how about if there were twelve billiard balls? that was the question asked to me when i was in 3rd year highschool? still can't think of the answer.
  6. Mar 17, 2003 #5


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    Here is my solution to the 12 ball problem

    Seperate the 12 balls into 3 sets, A, B, & C, of 4 balls each
    1st measurement: compare sets A and B.

    IF Sets A and B balance
    the oddball is in set C, we also know that all of the balls from sets A and B are the same and may use them for comparison, any of these balls may be considered standard.

    2nd measurement
    3 balls from C against 3 balls (standards) from A or B.
    The odd ball is the remaing 1 of C

    3rd measure
    compare 1 of the remaining balls from C to 1 of the balls from A or B to
    determine relative weight
    Not Balance
    The oddball is one of the 3 from set C on the scale. We know wethere it is light
    or heavy

    3rd measure
    compare 2 of the 3 balls in question
    The remaining ball is the oddball and we know its relative wight.

    Else IF Sets A and B do not Balance
    The oddball is on the scale so the balls of set C are standard
    2nd Measure
    1. Remove 3 balls from the low side (leave one behind).
    2. Move 3 balls from the high side to low side(leave one behind).
    3. Place 3 balls from set C (standard) on the high side.
    If Balance then
    Oddball is heavy and is one of the 3 removed from low side.
    3rd measure
    Compare 2 of the balls removed from the heavy side.
    Balance => the other ball is the oddball
    no Balance => the oddball is the one on the low side of the scale
    Else IF The Balance Shifts (previous high side is now low)
    The oddball was moved and is light.
    3rd measure
    Compare 2 of the moved balls.
    Balance => other is oddball
    No Balance => oddball is on the high side of scale
    Else if The Balance does not Shift (previous high side remains high)
    The oddball is one of the 2 balls which were not moved
    3rd measure
    compare one of these balls with a standard.
    Balance => other ball is oddball
    No Balance => oddball is on the scale.

  7. Mar 18, 2003 #6
    no offence...but we were solving problems like this in primary school...
    Integral -> wasn't it a real challenge for your intelligence ?
    Try this way :
    1st : 5-5;2 remain...
    2nd:2-2(from the odd 5-group)
    3rd:1-1(from the odd 2-group)
    Last edited: Mar 18, 2003
  8. Mar 18, 2003 #7
    Good logic, but try to continue :smile:.
    Well if not, try to do it :wink:, i don't think you will be able to continue from here (i was unable to continue when taking a group of 4).
    Can you addapt it for 9 balls ? :wink:
    Sorry, but i don't think this is quite right.
    It seems that you forgot that we do not know wether the odd ball is lighter or heavier than the other balls.
  9. Mar 18, 2003 #8
    I'm so sorry...I was wrong...can you forgive me ?
    As for the 9 balls...divide them into 3 groups of 3 and the rest is quite the same...
    sorry again...
  10. Mar 19, 2003 #9


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    Must've been awfully hard... (Sorry, just couldn't
    resist. )

    BTW, the problem can also be solved for 13
    balls. (But not 14+... :wink:)

    Peace and long life.
  11. Mar 19, 2003 #10
    that is a cool one STAii, i have never thought about it before. the best i have been able to manage is three steps but could take between two and four uses of the scale.

    9 balls, 3 groups of 3; a, b and c

    if a=b; c=x
    if a!=b; a~c
    if a=c; b=x
    else a=x
    repeate process for x, 3 balls; a, b and c
    x = the odd ball

    so how do you improve upon that[?]
    Last edited: Mar 19, 2003
  12. Mar 20, 2003 #11


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    Greetings !

    If you'd like, here's the solution for 13 balls :

    1. measure 4 and 4.

    2.1 If equal - measure 3 of the remaining with
    3 measured (normal).

    2.2 If not equal - measure 3 of one side + 1
    from the other side with the 3 normal balls +
    1 from the other side.

    3.1 Following from 2.1 if the balls are equal
    then measure 1 of two yet unknown balls with
    a normal ball.

    3.2 Following from 2.1 if the balls are not equal
    measure 1 - 1 (2 balls) from the 3 "unnormal".
    Since from 2.1 we know that the ball is either
    heavier or lighter this 1-1 can allow us to find
    it, or if they're equal it's the remaining 1.

    3.3 If 2.2 turns out eual then we have 3 balls
    left and we know we're looking for a certain -
    heavier or lighter ball. We now do it like 3.2.

    3.4 If 2.2 turns out to have changed previous
    heavier/lighter sides then it's either the one we
    transfered to the normal 3 or the one transfered
    sides to the other unknown 3. So, we measure
    one of them with a normal ball and find out.

    3.5 If the heavier/lighter after 2.2 stays the same
    then we know it's one of the 3 unknown on the
    left and we know it's either heavier or lighter.
    So, we do it like in 3.2.

    Live long and prosper.
  13. Mar 25, 2003 #12
    bump for the answer.

    and i mean the one for 9 balls in 3 trys, none of this off topic junk!
  14. Apr 11, 2003 #13
    Well kyleb, good answer, but not good enough.
    You see if when you repeat the proccess, you go to the point (2) again, then you will use the scale 4 times, and you are supposed to use it 3 times in the maximum.
    Here if the solution.
    Make the 9 balls into 3 sets, a, b and c.
    The balls are also numbered from 1 to 9, (iow set a containst balls 1,2 and 3 .... etc)
    ->a&b on scale
    -->if a=b, then c is odd group, put ball7 and ball8 on the scale.
    --->if ball7=ball8 then ball9 is the odd one (used the scale only twice)
    --->if ball7[x=]ball8, then leave the heavier on the scale, and put instead of the other one the last of the balls in set c, if the scale is still not in equilibrum, then the heavier ball is the odd one, otherwise the one you took out (between ball7 and ball8) is the heavier. (founded using the scale 3 times)
    -->if a[x=]b, either a or b is heavier, suppose a is heavier. Put a and c on the scale.
    --->if a=c then b is odd set, and the odd ball is lighter than all other balls (since a is heavier than b), if a[x=]c then a is the odd set, and the odd ball is heavier than all other balls.
    ---->Put any two balls of the odd set onto the scale, if the scale stays in equilibrum then the third ball is the odd one, if not then you will know which one is the odd one (from the last step you know if the odd ball is heavier or lighter, so if at this step the scale is not in equilibrum, then the heavier ball is the odd one if the odd ball is heavier than the other balls (From the last step), or the lighter ball is the odd one if the odd ball is lighter than the other balls (From the last step too)).

    Maybe it is not clear enough, so feel free to ask any questions on it.
  15. Apr 11, 2003 #14
    than you STAii, it makes sense now. i just failed to realize that pervious measurements could be used to determine if the odd ball was heavier or lighter.
  16. Apr 12, 2003 #15
    Well, this is the tricky part in it :smile:
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