How many factors of 3 can be in this product? The smallest possible is 1,2,3,4,6,9 (which is a solution) and has 4. There are only 6 possible numbers that can contribute factors of 3, so if all of these are selected, we’d have 3,6,9,12,18,24 (which is not a solution) and there’d be 8. Since there’s only one way to get all the 3’s and it’s not a solution, the maximum number of 3’s is 7. So there are between 4 and 7 factors of 3 in any solution, and of course the same number of factors of 2.

Suppose there are 7. Then exactly one of [3,6,12,24] is missing from the solution, and is replaced by a power of 2. The total number of powers of 2 present in each case where we omit one and BEFORE adding the final number is [7,6,5,4] so we’d need to add [0,1,2,3] to get to parity. Each of these is a solution:

6,12,24,9,18,1 => [1,6,9,12,18,24]

3,12,24,9,18,2 => [2,3,9,12,18,24]

3,6,24,9,18,4 => [3,4,6,9,18,24]

3,6,12,9,18,8 => [3,6,8,9,12,18]

Now suppose there are 6 factors of 3. Then either all of the 3-contributors except either 9 or 18 is present, or both 9 and 18 are present and TWO of the others are absent. 

The first case requires 3,6,12,24 (6 2’s) plus 9 or 18. But if we use 18 then there are more 2’s than 3’s so that’s no solution, and if we use 9 then we have parity and so use 1 for the final factor giving [1,3,6,9,12,24]

All of the rest have both 9 and 18, and two of [3,6,12,24]. As a result, they share 4 of 5 3-contributors with at least one of the 7 solutions and so can be constructed from one of these by dividing one number by 6, or by dividing one number by 3 and the power of 2 by two, subject to the requirement that all numbers be unique.

We can see that the prior case is also formed by dividing the 18 in the first 7-solution by 6, so this pattern holds. For the first solution, we can also divide the 12 or the 24, giving [1,2,6,9,18,24] and [1,4,6,9,12,18] respectively. The second solution gives only one more solutions from dividing the 24: [2,3,4,9,12,18]. The third gives [1,3,4,9,18,24] from dividing the 6 and the 4th gives [1,3,8,9,12,18] and [2,3,6,8,9,18] from dividing the 6 and the 12 respectively. The third also allows for dividing the 3 by 3 and the 4 by 2 to yield [1,2,6,9,18,24] (which we already had), and similarly the 4th gives [1,4,6,9,12,18] and [2,3,4,9,12,18] by dividing the 8 by 2 and the 3 or 6 by 3 respectively. Both of these are duplicates of ones we’ve already seen.

There is a final way to construct a 6-solution from a 7-solution: by dividing one number by 12 and multiplying another by 2 or by dividing one number by 24 and multiplying another by 4.  The 4th of these produces one new non-duplicate solution: [3,6,16,9,1,18] = [1,3,6,9,16,18]

We’ve already established a single solution with 4 factors of 3. Any solution with 5 must be derivable from the 4 solution by adding exactly one factor of 6 to [1,2,3,4,6,9]. If we add it all to one number, our only options are [1,3,4,6,9,12] and [1,2,4,6,9,18] and [1,2,3,6,9,24]. If we add a factor of 3 and of 4 to separate numbers, we also gain [1,2,3,8,9,18], [1,2,4,6,9,18] (dup) and [1,3,4,6,9,12] (dup).

Summarizing by product then, we have:

6^4:

[1,2,3,4,6,9]

6^5:

[1,2,3,6,9,24]

[1,2,3,8,9,18]

[1,2,4,6,9,18]

[1,3,4,6,9,12]

6^6:

[1,2,6,9,18,24]

[1,3,4,9,18,24]

[1,3,6,9,12,24]

[1,3,6,9,16,18]

[1,3,8,9,12,18]

[1,4,6,9,12,18]

[2,3,4,9,12,18]

[2,3,6,8,9,18]

6^7:

[1,6,9,12,18,24]

[2,3,9,12,18,24]

[3,4,6,9,18,24]

[3,6,8,9,12,18]

That makes a total of 17 solutions

NOTE - I saw after writing this that Charlie interpreted “power of 6” differently, as “some number to the 6th power” where I used “6 raised to some power”. I’m not sure which was the intent, but I figured I’d post anyway.