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Solution to a 12-Coin Game

The trick is that you need to achieve a situation in which "either the fake coin is lighter and is one of these coins; or the fake coin is heavier and is one of those coins".

The first weigh is: 4 vs 4

There are two cases:

A) the left side is heavier than the right side
B) the two sides are of the same weight

For A), the fake coin is on the scale, which means the remaining coins are all real. For ease of explanation, the coins are named as follows:

H1, H2, H3, H4, L1, L2, L3, L4, R1, R2, R3, R4

where H1, H2, H3 and H4 are the coins on the heavier side, L1, L2, L3 and L4 are the coins on the lighter side, and R1, R2, R3 and R4 are the remaining real coins. In this situation, the fake coin is either heavier and among H1 to H4, or is lighter and among L1 to L4.

The second weigh should be as follows:

H1,H2,L1,L2 vs H3,L3,R1,R2

Now there are three cases:

C) the left side is heavier
D) the right side is heavier
E) the two sides are of the same weight

For C), the fake coin is H1, H2 or L3 (can you figure out why?). The final weigh will be H1 vs H2 and you can get the answer.

For D), the fake coin is L1, L2 or H3. The final weigh will be L1 vs L2 and you can get the answer.

For E), the fake coin is either H4 or L4. The final weigh will be H4 vs R1 and you can get the answer.

Since case A) is complete, case B) will now be dealt with. In this case, the coins on the scale are real, while one of the remaining coins is fake. The coins are named as follows:

R1, R2, R3, R4, R5, R6, R7, R8, U1, U2, U3, U4

where the R coins are real and the U coins are the remaining coins.

The second weigh should be as follows:

R1, R2, R3 vs U1, U2, U3

Now there are two cases:

F) the left side is heavier (or the right side, which gives the same case)
G) the two sides are of the same weight

For F), the fake coin is lighter and is U1, U2 or U3. The final weigh will be U1 vs U2 and you can get the answer.

For G), the fake coin is U4, but you don't know whether it is lighter or heavier than the real ones. You can find this out by weighing it against one of the real coins.

This is the solution for a 12-coin problem, in which the second weigh of case A) is the tricky weigh.

In general, if you are allowed to weigh n times, then the maximum number of coins that you can play with is ((3 ^ (n - 1)) - 1) * 3 / 2

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