Solution to the 12 coin problem

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The following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. Suppose after the first weighing that the set 1,2,3,4 balances with 5,6,7,8.

Now weigh 9,10,11 against 1,2,3. If they balance, then coin 12 is the unequal coin. Weigh coin 12 against coin 1 to determine whether coin 12 is heavier or lighter.

If instead the set 9,10,11 is *heavier* than 1,2,3, then any one of coins 9,10,11 could be heavier. Weigh coin 9 against coin 10; if they balance, then coin 11 is heavier. If they do not balance, then the coin that weighs more is the heavier coin. If the set 9,10,11 is *lighter* than 1,2,3, then any one of coins 9,10,11

could be lighter. Weigh coin 9 against coin 10; if they balance, then coin 11 is lighter. If they do not balance, then the coin that weighs less is the lighter coin.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.