Qn. “Suppose you had eight apparently identical billiard balls, but one of the eight is slightly heavier than the others. The only tool you have to measure is a balance scale. What’s the fewest number of times you’d have to use the scale to find the heavier ball?”
Ans. Read the puzzle carefully. The puzzle did not say the eight billiard balls are of different weight. It says only one ball is slightly heavier. Comparing one ball to other is not efficient. You will use the scale too many times in that case.
The solution to this problem is by using the “divine and conquer” logic. Divide the 8 balls into three groups. The first group A have 3 balls. The second group B has three balls. The third group C has two balls.
(1) Use the scale once to compare the group A and B.
(2) If the group A weighs heavier than group B, take the group A into consideration. Ignore group B. Take any two balls in the group A and compare each other on the scale. If they are the same, the third ball in the group A is heavier. You don’t have to weigh the third ball to figure this out. If they are different, one of them is heavier. So, you have used the scale only twice to find the heavier – scale is used once in the step 1 and the second time in the step 2.
(3) Now consider, the groups A and B weigh the same from the step 1. This brings the conclusion that they both don’t have the heavier ball. Take the group C, compare two balls to find the heavier. So, you have used the scale only twice to find the heavier – scale is used once in the step 1 and the second time in the step 3.
In just two steps and using the scale twice, you can find the heavier ball.