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The puzzle is as follows:

Detective Mike has four bags with 100 coins of one cent each. All of these coins have the weight with the exception of one coin which weighs a little less than the rest because it is counterfeited. Assuming that he has a two pan scale. How many weigh trials at minimum has had to make to find with certainty the false coin?

The choices given are:

  1. 6 trials
  2. 5 trials
  3. 4 trials
  4. 7 trials

The source of this puzzle is an adaptation from a reprinted copy of an APA IQ test from early 1990. It appears to be based on the Weschler/Thurstone tests of psychometric exams from such time period.

Regarding the way on how to solve this it is some cryptic to me. Is it possible to use a two pan scale for this purpose?.

I guess one way to do it is by splitting in two halves. The side which is greater in weight will contain all real coins, whereas the lighter side will have the counterfeited one.

This accounts for the first trial.

Now that we have 25 coins where there is the false one we could split it in another two groups, but because this number is not even one of these coins must be subtracted from the rest, which leaves us with two scenarios.

If both sides balance out, then the coin that we removed is the false one. But this means the optimal and best chance. So let's assume that our mysterious false coin is among the group. Therefore as we did with the previous group, the side with the greatest weight will be real while in the other side we have the false one.

We return our subtracted real coin from this stack of 25 coins to the group of the 12 ones from which we're sure are real.

This accounts for the second trial.

Now we have 12 coins where there our mysterious coin remains.

We can repeat this cycle again and we end up with 6 real and a group of 6 where is our mysterious coin. Then we return the group of the 6 coins from which we're sure they are authentic and what we are left with is with 6 from which we don't know which is the false one.

This accounts for the third trial.

Then we do the same again, and we end up with 3 from which we know they're authentic and 3 from which we don't know. So we return the group from which we know they are authentic.

This accounts for a fourth trial.

Finally we end up with 3 coins. Because this is not even we subtract one. Now, regardless of the result we can be sure where is the false one. Assuming that we subtracted the false one, the two pan scale will balance even. Thus what we have is the false one. But if our false coin is in any of the two pans, it will be exposed by its weight as the side which is lighter will raise over the authentic one which is greater in mass.

Therefore this accounts for our final fifth trial.

This would work just okay for only one bag and assuming the false coin is in that specific bag. But in this case there are four bags. Thus there are 400 coins.

Thus I assume that a previous steps are needed and that is compare the weight of those bags first. Since there are four, we put these in the two pan scale.

The side which is lighter will have the bag with the false coin, thus this accounts for a weighing trial in other words 5+1=6 trials.

Then we do the same for the two remaining bags, which accounts for 6+1=7 trials.

This is my reasoning and I conclude it might be 7 trials or choice 4. But I don't know if this logic is the correct one. Could it be that it exist another way to solve this?. I'd appreciate someone could help me here, or perhaps to reassure that my strategy was the most appropiate.

I'd also like to notice that there's a similar question here but it doesn't specifically state to use a two pan scale such I made my question to reflect that requirement.

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  • $\begingroup$ Duplicate. Answer: puzzling.stackexchange.com/a/61638/74080 $\endgroup$ – infinigon Mar 11 at 11:10
  • $\begingroup$ @infinygon I disagree, this problem is different in nature from the way how it is described there. And there it means, how do we know this is a ternary search?. This part is not clear to me hence an explanation would be kindly appreciated. $\endgroup$ – Chris Steinbeck Bell Mar 11 at 11:33
  • $\begingroup$ The question I proposed as the duplicate is essentially asking the same question from the "other side". The ternary search is the procedure to answer this question, making this question "uninteresting". $\endgroup$ – Jeff Zeitlin Mar 11 at 11:57
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Against my better judgement, I am providing the procedure that will find the coin in six weighings:

Weighing 1: Divide the coins into groups of A=133, B=133, and C=134. Weigh A against B. If A and B are equal, take C and add one coin from A. Otherwise, take the lighter of the two and add two coins from C. You now have 135 coins, one of which is fake.

Weighing 2: Divide the coins into three groups of 45 each (D, E, F). Weigh D against E. If they are equal, take F, otherwise take the lighter of the two.

Weighing 3: Divide the coins into three groups of 15 each (G, H, I). Weigh G against H. If they are equal, take I, otherwise take the lighter of the two.

Weighing 4: Divide the coins into three groups of 5 each (J, K, L). Weigh J against K. If they are equal, take L and add one coin from J, otherwise take the lighter of the two and add one coin from L. You now have six coins, one of which is fake.

Weighing 5: Divide the coins into three groups of two each (M, N, O). Weigh M against N. If they are equal, take O, otherwise take the lighter of the two.

Weighing 6: You have two coins. Weigh them against each other. The lighter one is the fake.

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