Find the fake ball in 3 weighs—8 ball weight puzzle
In this second 8 balls weight puzzle, find the fake ball in 3 weighs. All 8 balls look alike and you are given a pan balance without any weights.
Recommended time to solve: 15 minutes.
Note: Fake means heavier or lighter.
Give it a try before going ahead.
Solution to the 8 balls weight puzzle: Find the fake ball in 3 weighs
Experienced now, you decide to weigh first, 3 balls against 3 others. 2 balls are kept aside. You have divided the suspect set of 8 balls into 3 suspect sets numbering 3, 3 and 2 balls. An application of divide and conquer rule.
Note: To gain experience in solving ball weighing puzzles, start with the puzzle of finding the heavier among 8 balls in 2 weighing.
Three possible results of first weighing are,
The first possible result: The right pan descends:
Conclusion 1: All six balls are suspect. Either the left side up-going three balls has the fake lighter ball, or the right side descending pan has the fake heavier ball. The figure shows the situation.
The second possible result: The left pan descends:
Conclusion 2: All six balls are suspect. Either the left side descending three balls has the fake heavier ball, or the right side up-going pan has the fake lighter ball.
Both results will need similar treatment.
The third possible result: The pans are equally balanced:
Conclusion 3: The two kept aside balls must have the lighter or the heavier fake ball. All six balls weighed are good balls.
Let us take up the third result first.
Third result in 1st weighing: To find the fake ball among 2 suspect balls in 2 weighing
Pick 1 suspect ball from the 2 and weigh it against 1 of 6 good balls. 1 of the two suspect balls is kept aside. This is the second weighing.
The ball in the pan will be the heavier or lighter fake ball if the pan with the suspect ball descends or ascends, respectively.
The kept aside ball in the second weighing must be the fake ball if the two pans are equally balanced, though you won’t know whether it is heavier or lighter.
You will know whether the defective ball is lighter or heavier in the third weighing against a good ball.
Question: Do you know how the fake ball can be identified out of 2 suspect balls in a different way way?
You could then identify the fake ball in the third result of first weighing when the two pans were balanced.
The job left is to analyze and resolve the first or the second result of first weighing when one of the pans with 3 balls moved up or down.
First and second case of 1st weighing: Analysis and decision for second weighing with two groups of 3 suspect balls
Analysis of first and second case of one pan descending and the other going up will be equivalent. Let us take up the case when the right pan descends and the left pan goes up.
The number of suspect balls is as large as 6, but we have two favorable points,
- Two balls kept aside in first weighing are good balls, and,
- Relative nature of the two sets of balls in the pans is known. Either the 3 up-going balls has the lighter fake ball or the descending set of 3 balls has the heavier fake ball.
To exploit this result, mark the three balls going up as UP balls and the other three as DOWN balls. Keep track of the nature of each ball in the two sets of balls. This is important, as you will exchange the balls between the two sets in second weighing now.
The suspect set of balls being as large as 6, a special technique is needed. Take an extreme measure.
Replace two of the 3 UP marked balls by two good balls and keep aside the replaced two UP marked balls. Further, exchange the remaining UP marked ball with 1 of the 3 DOWN marked balls.
What are the ball combinations of the two pans now?
The earlier UP marked pan has 2 good balls and 1 DOWN marked ball. 2 UP marked balls are kept aside. The earlier DOWN marked pan has 2 DOWN marked balls and 1 UP marked ball.
Now do the second weighing.
Analysis of results of second weighing and actions
Three outcomes are possible,
First outcome: The two pans are equally balanced.
Conclusion 4: Defective ball is one of the two UP marked balls kept aside in the second weighing and it is lighter. All 6 balls in 2nd weighing are good balls.
3rd weighing: Weigh 1 kept aside ball against a good ball leaving aside the second ball. If the pans are balanced, the ball left aside is the lighter fake ball. Otherwise, the ball in the up-going pan in 3rd weighing is the lighter ball.
Second outcome: The right pan that descended before, descends again.
Conclusion 5: The three balls on the left pan that goes up this time, cannot have the (lighter) culprit. Two of them are good balls and the third was in the DOWN marked group in first weighing.
The fake ball must be a heavier ball and is in the pan that descended this time.
Among the three balls in the descending pan, 1 is an UP marked ball. So the heavier fake ball must be one of the two DOWN marked balls.
3rd weighing: Weigh again one of the two suspected heavier balls against a good ball leaving aside the second suspect ball. The left aside ball must be the heavier fake ball if the pans are balanced. Otherwise, the ball in the descending pan must be the heavier fake ball.
Third outcome: The left pan descends. The reverse of earlier result.
Conclusion 6 is clear: Either the single DOWN marked ball in the left pan is the heavier fake ball, or the single UP marked ball in the right pan is the lighter fake ball.
You cannot weigh one against the other. Decide why.
Instead, you will weigh in the 3rd weighing, say the single UP marked ball with a good ball, keeping aside the other suspect heavier ball. In case of equal balancing, the left out ball is the heavier fake ball. In case of good ball descending, the suspect lighter UP marked ball must be the fake lighter ball.
The pan with UP marked ball cannot descend, isn’t it? There can only be two possible results in this weighing.
Parting Question: Can you solve the puzzle in any other way?
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