Question : One of twelve pool balls is a bit lighter or heavier (you do not know) than the others. At least how many times do you have to use an old pair of scales to identify this ball?
(a pair of scales = a scale consisting of a lever resting on a fulcrum with weighing pans at each end of the lever equidistant from the fulcrum)
Answer :
It is enough to use the pair of scales just 3 times. Let’s mark the balls using numbers from 1 to 12 and these special symbols:
x? means I know nothing about ball number x;
x< means that this ball is maybe lighter then the others;
x> means that this ball is maybe heavier then the others;
x. means this ball is “normal”.
At first, I lay on the left pan balls 1? 2? 3? 4? and on the right pan balls 5? 6? 7? 8?.
If there is equilibrium, then the wrong ball is among balls 9-12. I put 1. 2. 3. on the left and 9? 10?11? on the right pan.
If there is equilibrium, then the wrong ball is number 12 and comparing it with another ball I find out if it is heavier or lighter.
If the left pan is heavier, I know that 12 is normal and 9< 10< 11<. I weigh 9< and 10<.
If they are the same weight, then ball 11 is lighter then all other balls.
If they are not the same weight, then the lighter ball is the one up.
If the right pan is heavier, then 9> 10> and 11> and the procedure is similar to the former text.
If the left pan is heavier, then 1> 2> 3> 4>, 5< 6< 7< 8< and 9. 10. 11. 12. Now I lay on the left pan 1> 2> 3> 5< and on the right pan 4> 9. 10. 11.
If there is equilibrium, then the suspicious balls are 6< 7< and 8<. Identifying the wrong one is similar to the former case of 9< 10< 11<
If the left pan is lighter, then the wrong ball can be 5< or 4>. I compare for instance 1. and 4>. If they weigh the same, then ball 5 is lighter the all the others. Otherwise ball 4 is heavier (is down).
If the left pan is heavier, then all balls are normal except for 1> 2> and 3>. Identifying the wrong ball among 3 balls was described earlier.