Clever me, I gave away the solution in the original puzzle, but with two sectors switched. It makes the image editing easier.

See the solution disks

There's a slight bit of trickiness in the solution: one of the disks has been flipped over.

If you're curious how to deduce the solution, there are any number of ways, all of which are difficult to explain briefly. But here are a few facts:

- The full sequence of numbers will repeat every twelve numbers.
- The sum of the twelve numbers must be the square of the number of transparent sectors.
- A zero in the sequence proves that no more than 6 sectors are transparent.
- If the disks are
*not*flipped relative to each other, then the sequence will look like A,B,C,D,E,F,G,F,E,D,C,B. 'A' must equal the number of transparent sectors. - Let's say that each disk has 6 transparent sectors, the disks are flipped relative to each other, and that part of the sequence looks like 0,A,B,C,D,... Then there are exactly A transparent sectors such that if we go 1 sector clockwise, the sector is opaque. There are exactly B transparent sectors such that if we go 2 sectors clockwise, the sector is opaque. And so forth.

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