Find the heavier ball in 2 weighs—8 balls weight puzzle
In this first 8 balls weight puzzle, find the single heavier ball in 2 weighs. All 8 balls look alike and you are given a pan balance with no weights.
In how many ways you can solve the puzzle is the second part of the puzzle.
Total time to solve: 12 minutes.
Do give it a try before going ahead for the solution.
Solution to the 8 balls weight puzzle: find the heavier ball in 2 weighs
With no weights, you have the only option of weighing a few balls in one pan and same number of balls on the second pan.
In simple words, you have to compare weights of equal number of balls.
Question is: How many should be the number balls in each pan for the first weighing?
Let's see the consequence of weighing all 8 split into two groups of 4 plus 4 in the first weighing.
Whichever pan goes down, that group of 4 balls must have the heavier ball.
You have only one weighing left. Would you again take the simple path of splitting the 4 balls into 2 plus 2 groups and weigh? What would be the result? The heavier pan with 2 balls would certainly have one of them the ball you are trying to find. But you have finished your quota of weighing chances. This simple way won't work.
You may try other ways to find the heavier ball among 4 balls in a single weighing, but, you may find it impossible.
First stage analysis: Strategic approach of working backwards
So taking a more promising path now, you would decide to check whether you would be able to find the heavier ball among 3 balls that have been identified as suspect balls in the first weighing.
You have now reduced the number in the suspect set of balls by 1. At this point don't think how you would isolate a set of 3 balls in the first weighing.
This is what we call—Working backwards technique.
With 3 balls containing the heavier ball, naturally you would weigh 1 ball against 1 other from the three keeping aside the 3rd ball. As a result.
Whichever pan goes down has the heavier ball, and if both sides are balanced equally, the 3rd ball kept aside must be the heavier ball.
So you would certainly be able to identify the heavier ball if after the first weighing you are able to identify 3 balls containing the heavier ball.
Now turn your attention to the first weighing. Your objective would be to isolate a set of at most 3 balls that contain the heavier ball.
Second stage: How to isolate 3 balls containing the heavier ball among 8 balls in the first weighing
Now you know what you have to do—you would just split 6 balls, not 8, into two groups of 3 each keeping 2 balls aside and weigh the two groups of 3 balls against each other.
This again is applying the technique of reducing number of balls in the action set.
The possibilities are,
- Whichever pan goes down contains the heavier ball. And you already know how to find the heavier ball among suspect 3 balls.
- If the pans are equally balanced, the 2 balls left out must have the heavier ball. You know by now, finding the heavier ball among 2 balls would be dead easy.
Okay, first part of the puzzle is solved. Now you'll have to take up the second part.
In how many ways can the puzzle be solved?
At first thought, there seems to be no other way to solve the puzzle—if you reduce the total number of balls at first weighing to 4, 2 plus 2, and then if the two pans are balanced, then you would still be left with the 4 balls kept aside as suspect set for the second weighing.
Remember when you first met with this situation? When you attempted to weigh all 8 balls split into two groups of 4 plus 4.
So it boils down to the challenge of identifying the heavier ball among 4 suspect balls in 1 weighing.
How to identify 1 heavier ball among 4 suspect set of balls in 1 weighing
First action is natural—you would keep 1 ball aside reducing the number of balls to be weighed to 3. This is same as the strategic approach of reducing number of variables we have often used in solving algebraic problems quickly.
You may think of this approach in the form of action in this case as,
Reduce the total number of balls in suspect set of balls to be weighed as much as possible.
And you may call this now simply as the strategy of reducing number of balls. In essence, it is same as the strategy in algebraic problem solving.
Okay, what more would you do with the three balls left?
Naturally, you would fill up the empty slot of 1 ball caused by keeping aside 1 ball by a good ball.
You know you have at least 4 good balls as resources that can be used. Now you weigh 2 suspect balls against 1 suspect ball and 1 good ball.
Possible results are,
- If the pan with the good ball goes down, its companion single ball must be the heavier one,
- If the two pans are equally balanced, the left out fourth ball must be the heavier one, but,
- If the pan with 2 suspect balls goes down you would still have 2 suspect balls left and your weighing limit exhausted.
So this won't work.
Thinking new—thinking out of the box
This is the time you have to innovate and think out of the box—imagine that you have exchanged 1 ball between the two pans and then introduced the good ball. This is a new technique of exchanging balls. But would it work?
Unfortunately no—if the pan with two suspect balls goes down, you are again left with 2 balls with weighing quota exhausted.
Now you know for sure that if you are left with 4 suspect balls after the first weighing, it would really be impossible to identify the heavier ball in 1 weighing.
So there can be only one way to solve the puzzle that we have already found.
You have exhausted all possibilities following the principle of exhaustivity.
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