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This question already has an answer here:

You have 20 bottles of pills. 19 bottles have 1.0 gram pills, but one bottle has 1.1 gram pills. There is a weighing machine which calculates exact weight, and you have to find the heavy bottle by using that machine only once?

Note:- You can take out the pills from bottles for measurement.

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marked as duplicate by boboquack, ffao, Rubio Sep 8 '17 at 5:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ puzzling.stackexchange.com/questions/1965/coin-weighing-problem duplicate $\endgroup$ – Oray Sep 8 '17 at 5:06
  • $\begingroup$ I don't think so, the example you are mentioning is very different and has 3 chances to find the solution but there is only chance to find out the exact solution. $\endgroup$ – Mohit Tyagi Sep 8 '17 at 5:06
  • $\begingroup$ i put the wrong link :) $\endgroup$ – Oray Sep 8 '17 at 5:07
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I believe the answer to this one is:

Take one pill out out of the first bottle, two pills out of the second bottle... etc. and weigh them all together. Then find out which bottle has the heavy pills by getting the total and finding the amount of the anomaly.
So if every bottle had 1.0g pills in it and the pills were added to the scale as described above, the total would be 210g. If the first bottle (the one we only used one pill for) had 1.1g pills, the total would be 210.1g. If the second bottle (two pills) had 1.1g pills, the total would be 210.2g... etc. all the way up to 212g.

To put it in (sort of) mathematical terms:

Bottle no. with heavy pills = (Weight - 210g) * 10
(Where bottle no. = amount of pills placed on scales)

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  • $\begingroup$ The second sentence doesn't make much sense even after reading elsewhere what the answer is. $\endgroup$ – Ouroborus Sep 8 '17 at 15:29
  • $\begingroup$ Was it edited? Makes sense to me $\endgroup$ – greenturtle3141 Sep 8 '17 at 18:54
  • $\begingroup$ @Ouroborus updated with some examples to help explain. Let me know if it's still too hard to understand. $\endgroup$ – Alpha Sep 10 '17 at 22:36

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