Stage Four
A Question of
Balance
In a box there are 27 new red snooker balls all looking exactly alike. However, it is known that one of them is faulty and weighs more than the others.
Given that you have a balance but no weights, how would you find the faulty ball?
Can you find a more efficient way?
What is the most efficient way? In other words, what is the smallest number of balances you would need to compare to find the faulty ball?
Extension: A very clever person had 8 snooker balls which included a light ball. They found the light ball in only 2 weighings. Can you work out how they achieved this?
Reference: The Amazing Mathematical Amusement Arcade
Brian Bolt, Cambridge University Press, p66
This answer from William at Finley explains how to do this in the most efficient way, needing only 3 comparisons!
Place the snooker balls in groups of 9 and weigh one group against another, leaving one aside. If they weigh the same the faulty ball is in the group left aside. If one weighs more that is the group with the faulty ball.
Now divide the remaining nine balls into groups of three. Weigh one group against another, leaving one group aside. Now it justs repeats. The group that is heavier holds the faulty ball. If they weigh the same the faulty ball is in the group left aside.
Now take the three balls and weight one against another. If one weighs more it is the faulty ball. If they weigh the same the ball left aside is the faulty ball. Therefore the least weighs needed is three.
Extension:
Make up two groups of three and one of two out of the eight balls. Weigh the two groups of three against each other. If one is lighter it holds the lighter ball.
Take these three balls and weigh one against an other. If one is lighter it is the faulty ball. If they weigh the same the ball left aside is the lighter ball.
If when you weigh the two groups of three they weigh the same then the faulty ball is in the group of two. Weigh these two balls against each other and the lighter one is faulty.
An excellent answer from William!