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We have 12 coins, and only 11 of them have the same mass. It can be assumed that all coins have identical appearances.

We have a lever.

How can we identify the special coin, by taking measurements from the lever for only three times?


Example:

If we have 6 identical coins and one of them is special because it is lighter, we can identify it by taking measurements from the lever for only two times.

  1. Put three on one side on the lever and the remaining on the other side. The special coin is on the lighter side.

  2. Take arbitrarily two coins from the lighter side and place them respectively on two sides of the lever.

Case I: The lever is not balanced

Then, the special coin is on the lighter side.

Case II: The lever is balanced

Then, the special coin is the one on the lighter side (in the first measurement) that was not taken for the second measurement.


However, in the 12-coin case, we do not know if the special coin is heavier or lighter than the others.

No one in my class was able to solve this puzzle, which was given by our maths teacher as a small brain challenge.

Any idea?

Thanks in advance.

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marked as duplicate by Jamal Senjaya, Jaap Scherphuis, Keelhaul, JonMark Perry, El-Guest Oct 23 '18 at 10:33

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