## Puzzle : 9 balls

Well,

Problem was given by manoj, which he in turn found from some website.

Problem : 9 balls, 8 identical and 1 is lighter. You need to find out in 4 weights, which one is lighter. Huh, sounds simpler?? well, hold on. You are given 3 balance weight machines, out of which one is defective. You do not know which one is defective. Also, defective machine can give any outcome ( >, =, < ) irrespective of what you put on the machine. Now, can you find out which ball is lighter weighing only 4 times in total???

Solution : None of us could figure out the solution. But, once we read the solution from the website, deepanjan gave a very nice explanation. I am lazy to draw the figure, so trying out a text based table.

Divide the balls in 3 sets of 3 balls, A, B & C.

On machine m1, measure A and B. Without loss of generality, say it tells that C is the lighter set.

Now, measure A1,B1,C1 vs A2, B2, C2 on machine m2. Again, without loss of generality say it tells us that one of the A3,B3,C3 is lighter.

So we have this

1   |    2   |  3

————————-

A |   A1 |  A2  |A3

————————-

B |   B1| B2 |B3

————————

C |  C1| C2 |C3

On machine m1 we measured columns and on m2 we measured row. Intersection in this case is C3 ( it could be any cell, but C3 is just a representative )

Now, we measure ( C1,C2) vs (A3,B3) on the third machine. Or in other words, measure two remaining balls of the columns vs 2 remaining  balls of the rows on machine 3.

Case 1 : (C1,C2) = (A3,B3)

In this case, C3, indeed is the lighter ball. Because, two out of three machines are correct. If m1, m2 are correct, they already told C3 is lighter. If, m1 and m3 are correct then m1 says C is a lighter set and m3 says that C1 and C2 are not lighter. so C3 must be the lighter ball. Similarly if m2 and m3 are correct, C3 will still come out as the lighter ball.

Case 2 : (C1,C2) < (A3,B3)

Machine 3, says one of the C1 or C2 is lighter, which is a contradiction to machine 2 which said that one of the A3,B3 or C3 is lighter. Hence, machine 1 must be correct which said C is a lighter set.

Now measure C1 and C2 on machine 1 and find out which one is the lighter ball.

Case 3 : (C1,C2) > ( A3,B3)

Here machine m3 contradictions with machine 1 which said that C is the lighter set. Hence, m2 must be the correct machine. Measure A3,B3 on m2 and find out which one of A3,B3 or C3 is lighter.

Phew!!!! No wonder, it is a math olympiad question.

— Saurabh Joshi

Related post : here

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### 3 Responses to “Puzzle : 9 balls”

1. Prudhvi Pavan Says:

Hi..nice puzzle. I got an alternate solution for this!

Divide the balls in 3 sets of 3 balls, say A, B & C. And select a weight machine randomly, say M having two plates P1 and P2.

Step 1: Weigh the sets A & B in M. (A in P1 and B in P2)
Step 2: Weigh the sets A & B in same M, but exchanging the plates.( A in P2 and B in P1).

case 1: If the M is not defective and A=B. Plates are balanced both the times. Hence C has the lighter ball and can be found in next 2 steps.

case 2: If the M is not defective and A has lighter ball. Balance favors A both the times and Lighter ball can be identified in next one step.

Case 3: If the M is defective and A = B. Plates tilt towards same side both the times. hence C has the lighter ball and can be found in next 2 steps.

Case 4: If M is defective and A has lighter ball. Balance tilts toward A( or equally balanced) and second time it doesn’t. Using the other balance lighter ball can be identified in next 2 steps.

Let me know what you think.

• Saurabh Joshi Says:

Your solution makes an assumption about defective machine which is wrong. You are assuming that defective machine, even though defective is consistent with its result. The defective machine defined in the problem behaves arbitrarily. That is if you put A,B on M (and M is defective) and weigh them twice it can give any of the (<<,<=,,=,>=,>>) for the two weighing irrespective of which plate you put it on.

So your solution does not work.

• Prudhvi Pavan Says:

yes! I thought so! 🙂 My solution works with real defective machine. Not the arbitrary and imaginary one. Thanks for your reply. 🙂