Billiard-balls - logic puzzle
|
Nine billiard-balls are lying in front of you. Except for one they all have the same weight. The exception is a little bit heavier than the others. What is the minimal number of times you have to use a balance in order to find the wrong billiard ball? |
|
Explanation
Only two times!
Divide all nine billiard-balls in groups of three balls. Put two of these groups on the balance. Now there are two possibilities. Either the balance indicates a difference or it is indicates that both groups have equal weight. In both cases you can find the heavy group, in the firstcase it is one the balance, in the second case it is not.
From the heavy group you select two balls, and you put one ball apart. Put these two selected balls on the balance. If the balance indicates an equality, the heavy ball lies apart, if it does not the balance points you to the heavy ball.
Divide all nine billiard-balls in groups of three balls. Put two of these groups on the balance. Now there are two possibilities. Either the balance indicates a difference or it is indicates that both groups have equal weight. In both cases you can find the heavy group, in the firstcase it is one the balance, in the second case it is not.
From the heavy group you select two balls, and you put one ball apart. Put these two selected balls on the balance. If the balance indicates an equality, the heavy ball lies apart, if it does not the balance points you to the heavy ball.
|
|