13 ball - 1 lighter OR heavier - in 3 moves with only a balance scales identify which one

13 balls in total similar in every way except:
12 are of equal weight, 1 is lighter OR heavier.

With only the use of a balance scale and 3 measurements how can I identify which one is different?
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lemmeCConnect With a Mentor Commented:

Divide them up into 2 groups of 4 and one group of 5 and weigh the two 4 ball groups against each other.
Case 1: The scales balance -- All balls on the scale are normal, and the group not on the scale contains the odd ball. Leave two of these balls aside and weigh the other three against three normal balls.

     Case 1.1: The scales balance. One of the 2 balls left aside is the odd ball. Take one of these and weigh it against one of the normal balls.

          Case 1.1.1: If the scales balance the one left is the odd ball, you do not know whether or not it is heavy or light.

          Case 1.1.2: If the scales don't balance you also know the odd ball, but also know if it's heave or light.

     Case 1.2: The scales don't balance. The odd ball is in the group of three being weighed against the normal balls, and its being light or heavy
     is indicated by whether the group is lighter or heavier than the normal group. Weigh any two of these three against each other to determine
     which ball it is.

Case 2: The scales don't balance -- one of the two groups of on the scale contains the odd ball, either a havy ball on the side that is down, or a light ball on the side that is up. Remove two balls from the heavy group and one ball from the light group, replacing them with normal balls. Also switch a ball from the heavy side with a ball from the light side.

     Case 2.1: The scales balance. The group of three balls removed from the scale contain the odd ball. Weigh the two from the heavy side against
     each other. If the scales balance, the third ball is the odd ball and it's light. If the scales don't balance, the odd ball is the heavier ball on the
     balance and it's heavy.

     Case 2.2: The scales don't balance and the heavy side stays the same. Either the single ball remaining on the heavy side from the previous
     weighing is heavy, or the two balls remaining in the right pan from the previous weighings are light. Determine which is which by weighing the
     two light balls against each other.

     Case 2.3: The scales don't balance and the heavy side changes. One of the two switched balls is the odd ball. Determine which it is and
     whether it is heavy or light by weighing the ball from the now heavy side against a normal ball.
Melt the balance scale.  Drop in all 13 balls.  Observe closely which ball floats higher/lower or sinks faster/slower than the others.  That is the odd ball.
QUIVLABAuthor Commented:
Thank you, but that was not the answer for which I was looking. Short of having the capability of melting the scale, and without any objects not listed, how can I identify the odd ball?
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Ah, well, if there's no other objects around, then you must be in deep space, with little net gravity.  So if you arrange the balls evenly around the balance equidistant from the center of mass, then they should all accelerate towards the center at the same rate, except for the odd ball which will accelerate slower or faster, as we see from F=(G*m1*m2)/(r^2).

Naturally, you're looking for the kind of answer in which you weigh some number of balls on each side, and rearrange if each set does or does not weigh the same, to which you can find several answers recently posted on this site, as well as a good explanation at http://www.mazes.com/scales/Scale301.html 
QUIVLABAuthor Commented:
Thanks again, but I can not find the answer to my question on this site or the other. The puzzle, as it is typically posed, involves 12 objects. I am looking to solve the same puzzle with the addition of 1 object for a total of THIRTEEN.
Measure 1:
Put 6 on one side and 6 on the other.  Leave one ball out.

If they both weight the same, heavy ball is the ball that is left out.  If not, go to Measure 2:

Measure 2:
Take balls from heavier side on measure one and split them into groups of 3 and measure.  Take heavier side and go to Measure 3

Measure 3:
Put one ball and each side and leave one out.

If they are even, the heavy one is the one you left out.
Measure 1:

Put 6 balls on each side

If both sides balance you have your lighter ball

else take the 6 balls from the lighter side

Measure 2:

Put 3 balls on each side

take the 3 balls from the lighter side

Measure 3:

Put one ball on each side

If both sides balance you have your lighter ball

else take the lighter ball
oldskool79's solution does not work because if you take the heavier side to the next measure, the heavier group does not contain the lightest ball
Label the coins with the letters A,C,D,E,F,I,K,L,M,N,O,T,Z and then weigh
them as follows:


You don't need to memorize the letter list, just the phrase "Ma do like
me to find fake coin."  Alternatively, you can use the phrase "MIND

The following table of outcomes tells you which coin is heavy or
light ("L" = left down, "R" = right down, "=" = balance):

LLL I light
LL= M heavy
LLR O heavy
L=L A heavy
L== L light
L=R K light
LR= D heavy
LRR E light
=LL N light
=L= T heavy
=LR F light
==L C light
=== Z heavy or light
==R C heavy
=RL F heavy
=R= T light
=RR N heavy
RLL E heavy
RL= D light
R=L K heavy
R== L heavy
R=R A light
RRL O light
RR= M light
RRR I heavy
This question appeared in the Puzzle Corner of Technolgy Review (the MIT Alumni magazine) within the past year or so.

There was an additional restriction (that is actually a hint):  The second and third weighings can not depend on the first.
                                                                                      The third weighing can not depend on the first or second.

                                                                                       You must determine what you are going to do on all three weighings before you start.

This is important because you must use the three weighings efficiently.

There are 26 possibilities:  (Ball 1 is H or L) or (Ball 2 is H or L) ... or (Ball 13 is H or L)

And there are 27 ways the weighings can turn out: (Tilt Left, Tilt Right, or Ballance) x 3 weighings.
Darn.  All these weighing problems sound alike.

The puzzle I was remembering in my last post was different.  But the solution method might still help.


ozo's solution (where the weighings are determined in advance) looks like the right approach, but he never weighs Z.

So if it's the odd ball, he can't tell if it's heavy or light.
>> weighings are determined in advance
There is no such restriction in the original question.
Ignore all my prior posts on this question.

              lemmeC has the solution.
> So if it's the odd ball, he can't tell if it's heavy or light.
nor can lemmeC in case 1.1.1
>can't tell if it's heavy or light.
Such determination is not a requirement stated, just to find which one is different.
ozo, how do you do it?
QUIVLABAuthor Commented:
Thank you. After working on the problem for far too many hours, I reached the same answer provided by lemmeC. Unfotunately, the answers provided by oldskool73 and dubbs2003 do not work, because they require you to know if the odd ball is heavy or light.  With only a cursory look, I believe that ozo's answer also works, however my conclusion followed the same thought process demonstrated by lemmeC's answer, so without having to spend another minute checking answers I accepted lemmeC's as the solution.
QUIVLABAuthor Commented:
PS. There was some discussion,  in reference to d-glitch's post, "So if it's the odd ball, he can't tell if it's heavy or light.".  Using lemmeC's answer allows one to determine if the odd ball is heavy or light with 12 of the 13, however if the first, second, and third weigh are equal, the remaining ball can be identified as odd without knowing whether it's lighter or heavier. Snoyes jw is correct, such a determination is not required as the question was posed.
Read somewhere that the maximum number of balls that can be used to find both
(1) which ball is odd, and  
(2) whether the odd ball is heavier or lighter
in all cases is (3^w - 3)/2 , where w is the number of weighings.
This works out to 12 balls for 3 weighings.

A general solution for n balls can be found at http://www.creativepuzzels.nl/spel/speel1/speel2/12bcoin.htm.
If you don't need to find whether the odd ball is heavier or lighter, and you have an extra ball know to be genuine, you can do (3^W+1)/2 = 14 unknown balls for 3 weighings.
Unless you hjave many balls, when the scales will break...
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