As the shaded squares sum to 12, then the possibilities are (3, 9), (4, 8) or (5, 7).

Of these, 3, 5 and 7 are prime, so they can’t be placed next to the 2 given. Equally, 8 + 2 or 9 + 2 breaks the second condition, so the number next to 2 must be 4, and the other shaded sector must be 8.

The sector on the other side of 4 cannot be 1 or 9 (as these are square), 3 or 5 (consecutive numbers) or 6, 7 (sum to 10, 11, 12). Therefore, it must be 0.

Between 2 and 8, we now have a choice of 1, 3, 5, 6, 7 or 9. 1, 3, 7 and 9 are adjacent, leaving a choice of 5 or 6. 5 is prime (as is 2), so the sector must equal 6.

On the other side of the circle, we need to fit in the odd numbers 1, 3, 5, 7, 9. Being prime 3, 5 and 7 need to occupy alternate sectors – one of which must be next to the number 8. This cannot be 3 (sum to 11) or 7 (adjacent), so it must be 5. 9 cannot border 3, so it must be next to 5. 7, 1, 3 fill in easily from there.