A naive upper bound is N = 243 = 3⁵, since you must distinguish among N possibilities using 5 ternary tests. However, you can tdo better.

Here is a proof that the maximum possible value of $N$ is

121

There are 3⁵ = 243 possible outcomes of the five weighings, which can be represented as a string of 5 letters, each of which is L, R or B, for left heavy, right heavy, or balanced. The outcome you see is determined by which coin is fake, its weight, and which scales it was put on. For example, if coin 7 was heavy, and it was placed on the right for weighings 1 and 2, on the left for 3 and 4, and not weighed during the 5^th, you would observe the outcome LLRRB. On the other hand, if coin 7 was light, you would observe the opposite outcome where L and R are interchanged, RRLLB.

This illustrates the fact that if an outcome causes you to conclude a coin is fake, its opposite must as well. Therefore, for each coin there must be two outcomes which cause you to conclude that coin is fake. The only exception is a coin which is never weighed, since this corresponds to the outcome BBBBB, which is its own opposite. This means there can be at most (243-1)/2+1 = 122 coins if there is a successful strategy.

To get the promised upper bound of 121, we have to be more careful. Suppose that the first weighing places K coins on each side of the scale, leaving L of them aside.

• If the scale doesn't balance, there are now 2K possibilities for the fake coin, to be determined in four tests which have three possibilities each. Therefore, it must be true 2K ≤ 3⁴ = 81. Since 2K is even and 81 is odd, this further implies 2K ≤ 80.

• If the scale does balance, the fake is one of the L remaining ones. By the same logic as before, you can only find the fake now if L ≤ (81-1)/2+1 = 41.

Combining the last two bullets, the total number of coins, 2K + L, is at most 80 + 41 = 121, as advertised.

A naive upper bound is N = 243 = 3⁵, since you must distinguish among N possibilities using 5 ternary tests. However, you can prove a better bound.

Here is a proof that the maximum possible value of $N$ is

121

There are 3⁵ = 243 possible outcomes of the five weighings, which can be represented as a string of 5 letters, each of which is L, R or B, for left heavy, right heavy, or balanced. The outcome you see is determined by which coin is fake, its weight, and which scales it was put on. For example, if coin 7 was heavy, and it was placed on the right for weighings 1 and 2, on the left for 3 and 4, and not weighed during the 5^th, you would observe the outcome LLRRB. On the other hand, if coin 7 was light, you would observe the opposite outcome where L and R are interchanged, RRLLB.

This illustrates the fact that if an outcome causes you to conclude a coin is fake, its opposite must as well. Therefore, for each coin there must be two outcomes which cause you to conclude that coin is fake. The only exception is a coin which is never weighed, since this corresponds to the outcome BBBBB, which is its own opposite. This means there can be at most (243-1)/2+1 = 122 coins if there is a successful strategy.

To get the promised upper bound of 121, we have to be more careful. Suppose that the first weighing places K coins on each side of the scale, leaving L of them aside.

• If the scale doesn't balance, there are now 2K possibilities for the fake coin, to be determined in four tests which have three possibilities each. Therefore, it must be true 2K ≤ 3⁴ = 81. Since 2K is even and 81 is odd, this further implies 2K ≤ 80.

• If the scale does balance, the fake is one of the L remaining ones. By the same logic as before, you can only find the fake now if L ≤ (81-1)/2+1 = 41.

Combining the last two bullets, the total number of coins, 2K + L, is at most 80 + 41 = 121, as advertised.