I need to begin with an apology – I have not been able to create a video solution to this puzzle as I seem to have misplaced the writing instrument for my iPad. Hence a written solution to this one, hopefully it is simple enough for everyone to follow.

Solution to Part (1) – All one needs to do is to take different number of balls from each of the machine and weigh them once. For example, if one takes 1, 2 and 3 balls respectively from the three machines, then the “fair” weight would be 60 grams. If the weight is 62 grams, it means that it is the second machine which is faulty, and is producing balls which are 1 grams heavier. Similarly, if the weight is 57 grams, then it is the 3rd machine which is producing balls which are 1 gram lighter.

Solution to part (2) is similar to (1) – in this case at a minimum one need 1,2,3,….10 balls from Â machine 1, machine 2, …., machine 10 respectively. Faulty machine can be identified by variation from the “fair weight” of 550 grams.

The basic principle for part (3) is similar. Only Alok Kuchlous sent me the correct answer for this. In this case, since the fault can be either 1 or 2 grams, one needs to ensure that if one has selected n(k) balls for machine(k), then one cannot chose either n(k) or 2n(k) balls from any other machine as there could be a possible overlap. The minimum such sequence that one can select is as follows:

1,3,4,5,7,9,11,12,13,15……which adds up to 80 balls.

Hope you all enjoyed the problem and the variation!

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