This problem is solved in a similar way. Only now the coins need to be divided into three parts so that their number is approximately the same in each of the piles, in two of them the coins should be equal. It turns out 67, 67 and 66 coins.
If you put 67 coins on the scales and one of the bowls outweighs, then the heavier pile must be divided into 22, 22 and 23 coins. Then weigh the pieces of 22 coins.
If the scales are not balanced again, then the group must be divided into piles of 7, 7 and 8 coins. Now you have to weigh the piles of 7 coins again; if one outweighs, then it must be divided into piles of 3, 3 and 1 coins.
Conditions are obtained, as in the first problem, and then it remains to do two more weighings to find a fake. As a result, you will need at least five weighings.
The same number of weighings will be required to check if the scales at the first step are in equilibrium. Then the fake will be in a pile of 66 coins, it must be divided into three equal parts of 22 coins. Then do another weighing and find a pile with a fake. It remains to be divided into piles of 7, 7 and 8 coins and then proceed in the same way as in the first case.
Answer: A counterfeit coin can be detected in five weighings.