Steel balls

May 3, 2001

The problem

You have 8 steel balls. One is slightly heavier than the rest. You have a pair of scales to weigh the balls. How can you find the heavy ball, if you are only allowed to use the scales twice?

Solution

(There is an implied assumption: that 7 of the balls weigh the same.)

Set aside 2 balls.

Place the remaining 6 balls on the scales – three each side.

If the balls balance then the heavy ball is one of the 2 balls set aside. Place the 2 set-aside balls on the balance; the heavy ball is then the heavier of the two.

Otherwise the heavy ball is one of the heavier set of 3 balls. Of these 3, set one aside and weigh the remaining 2. Either these balls balance and the one set aside is the heavy ball, or the balls do not balance and the heavy ball is the heavier of the two weighed.

Extensions

1. Can the problem be solved if there are 9 balls (instead of 8)?

2. What is the least number of weighings if there are n balls, where n is 2? … 3?

3. What are the possible strategies if the number of balls n is 4? … 5? … 6? … 7?

4. Can you generalise? … Given n balls (instead of 8) how many weighings would you need to be allowed to be sure of isolating the heavy ball? What strategy do you need to follow to attain this number of weighings.

5. In reality no two balls are likely to have exactly the same mass. Comment on how practical the strategy of the previous exercise would be? … where the uncertainty of the mass of any ball is ±u g, the difference in mass between the heavy ball and one of the remaining balls is d g, the scales can be used to detect differences in mass that are larger than D g and there are n balls.