Old avatar of poivre13.

Difficulty : ****

Tested, logical.
If you are stuck:
- When counting the cells which will be still to blacken above row R10, notice that exactly eight cells will be to blacken aside from C3, C5 and C7. Which means that, either there are the 6 and the 2, or there are the 5 and the 3 above this row. Because all of the others combination of 2, 3, 4, 5 and 6 can't give a sum equal to 8 (hurray for maths ^^).
- Thus, there is the 6 or the 5 in the upper part, and then the 1 from R5 is necessarily in C1 or in C19. So we can place crosses elsewhere in this row and place the 4 and the 2-1 from C3-7.
- We can now easily know if this is 6-2 or 5-3 which are to place in the upper part.
- Finally, there is three fewer cells to blacken in R12 than in R13. Wave-logic allows us to place the 3 and to raise the 6 from C3 and the 12 one cell upper.
Good blackenage.