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1

I have an algorithm to achieve the task with no more than There are some interesting solutions here, most of them do not appear to be complete, however, so I have decided to add mine as well. First, divide the 30 coins into 5 groups of 6 coins each. Label them A, B, C, D, and E. The gist of the algorithm is to find out which of these groups have a fake and ...


4

There are ${30 \choose 2} = 435$ ways of selecting 2 coins out of 30. Since the fakes can be heavy or light, that yields $435 \times 2 = 870$ possible combinations. Since each test yeilds one of 3 outcomes, and $3^6 = 729 < 870 < 2187 = 3^7$, at least 7 tests are needed. Below is proof that Update with more details


1

Here is my simple methodology; Let's call then Step 12v34 and the possible outcomes become; $$\begin{array}{c|c|c|c|} & \text{12} & \text{34}& \text{Result} \\ \hline \text{I} & GG & GF& NE\\ \hline \text{II} & GG & FF& NE\\ \hline \text{III} & GG & GG& E\\ \hline \text{IV} & GF & GF &E\\ \...


0

At first sight I would say it is: reasoning: EDIT Not because I think the above written is entirely bullshit, but I have an intuition, that it is very arbitrary and it is far from the optimality (both in theoretical and practical terms). I have a different idea. With this type of scale after k measurements we have 3^k outcomes. If there would be one ...


4

These steps detail a full algorithm to figuring out which two coins are fake. Worst case, this takes $10$ comparisons. Edit: The algorithm was fixed to account for the error noted in the comments. This didn't affect the worst case scenario. Step 1: Split the $30$ coins into $3$ piles of $10$, which we will call $A$, $B$, and $C$. Compare $A$ and $B$. If ...


0

30 weighings. Minimum (if you somehow luckily weigh all 30 fake coins first you can determine them to all to be fake) The minimum weight of a fake coin is always 1gram lighter, the maximum weight of a fake coin is always 1gram heavier. Hence a comparison of 2 fake coins will always be 2 grams. Once the fake minimum, and fake maximum are determined the ...


13

You only need Assume the genuine coins weigh $x$ grammes. Credit to Hexomino and Jaap for the corrections!


2

The number of weighings will be: What will you do is to: Mathematically speaking: So the next step is:


7

In 5 moves :


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