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Daily Sudoku — 6 August 2026

The puzzle

  • 6 August 2026
  • Hard
  • 30 clues
Puzzle
····5·8·4
·2·1·····
··6··8··1
·7·5·4·8·
··1···4··
·6···1·73
81··9·2··
6··8··59·
··5·42·1·
Solution
137956824
428137659
956428731
379564182
281379465
564281973
813795246
642813597
795642318

Solve it in the app to get on the board

Step-by-step solution

Every cell below is filled by the same hint engine the app uses, in the order it can be deduced — 51 steps using 2 techniques.

Techniques used

  • Hidden Single — inside one row, column or box this digit fits in only one cell, so it must go there
  • Naked Single — every other digit is already used in this cell's row, column or box, so only one candidate is left
  1. R1C1 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 1, R1C2 cannot hold 1.- Because column 3 already has a 1, R1C3 cannot hold 1.- Because row 2 already has a 1, R2C1 and R2C3 cannot hold 1.- Because row 3 already has a 1, R3C1 and R3C2 cannot hold 1.So 1 must go in R1C1.
  2. R2C3 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 8, R1C2 and R1C3 cannot hold 8.- Because column 1 already has a 8, R2C1 cannot hold 8.- Because row 3 already has a 8, R3C1 and R3C2 cannot hold 8.So 8 must go in R2C3.
  3. R3C4 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C4 and R1C6 cannot hold 4.- Because column 5 already has a 4, R2C5 and R3C5 cannot hold 4.- Because column 6 already has a 4, R2C6 cannot hold 4.So 4 must go in R3C4.
  4. R2C1 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C2 and R1C3 cannot hold 4.- Because row 3 already has a 4, R3C1 and R3C2 cannot hold 4.So 4 must go in R2C1.
  5. R6C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C1 and R4C3 cannot hold 4.- Because row 5 already has a 4, R5C1 and R5C2 cannot hold 4.- Because column 1 already has a 4, R6C1 cannot hold 4.So 4 must go in R6C3.
  6. R5C2 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 8, R4C1 and R4C3 cannot hold 8.- Because column 1 already has a 8, R5C1 and R6C1 cannot hold 8.So 8 must go in R5C2.
  7. R6C5 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 8, R4C5 cannot hold 8.- Because row 5 already has a 8, R5C4, R5C5 and R5C6 cannot hold 8.- Because column 4 already has a 8, R6C4 cannot hold 8.So 8 must go in R6C5.
  8. R4C7 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 1, R4C9 cannot hold 1.- Because row 5 already has a 1, R5C8 and R5C9 cannot hold 1.- Because row 6 already has a 1, R6C7 cannot hold 1.So 1 must go in R4C7.
  9. R8C3 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C3 cannot hold 2.- Because column 2 already has a 2, R8C2 cannot hold 2.- Because row 9 already has a 2, R9C1 and R9C2 cannot hold 2.So 2 must go in R8C3.
  10. R8C2 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R7C3 cannot hold 4.- Because row 9 already has a 4, R9C1 and R9C2 cannot hold 4.So 4 must go in R8C2.
  11. R8C5 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C4 and R7C6 cannot hold 1.- Because column 6 already has a 1, R8C6 cannot hold 1.- Because row 9 already has a 1, R9C4 cannot hold 1.So 1 must go in R8C5.
  12. R7C6 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 5, R7C4 cannot hold 5.- Because row 8 already has a 5, R8C6 cannot hold 5.- Because row 9 already has a 5, R9C4 cannot hold 5.So 5 must go in R7C6.
  13. R7C8 = 4 · Hidden SingleIn the bottom-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 4, R7C9 cannot hold 4.- Because row 8 already has a 4, R8C9 cannot hold 4.- Because row 9 already has a 4, R9C7 and R9C9 cannot hold 4.So 4 must go in R7C8.
  14. R9C7 = 3 · Hidden SingleIn the bottom-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R7C9, R8C9 and R9C9 cannot hold 3.So 3 must go in R9C7.
  15. R7C3 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 3, R9C1 and R9C2 cannot hold 3.So 3 must go in R7C3.
  16. R9C1 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 7, R9C2 cannot hold 7.So 7 must go in R9C1.
  17. R9C2 = 9 · Naked SingleLooking at R9C2:- Row 9 already has 1, 2, 3, 4, 5 and 7 → ruled out.- Column 2 already has 6 and 8 → ruled out.So this cell can only be 9.
  18. R1C3 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 7, R1C2 and R3C2 cannot hold 7.- Because column 1 already has a 7, R3C1 cannot hold 7.So 7 must go in R1C3.
  19. R4C3 = 9 · Naked SingleLooking at R4C3:- Row 4 already has 1, 4, 5, 7 and 8 → ruled out.- Column 3 already has 2, 3 and 6 → ruled out.So this cell can only be 9.
  20. R3C1 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 9, R1C2 and R3C2 cannot hold 9.So 9 must go in R3C1.
  21. R3C2 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C2 cannot hold 5.So 5 must go in R3C2.
  22. R1C2 = 3 · Naked SingleLooking at R1C2:- Row 1 already has 1, 4, 5, 7 and 8 → ruled out.- Column 2 already has 2, 6 and 9 → ruled out.So this cell can only be 3.
  23. R8C6 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C4 cannot hold 3.- Because row 9 already has a 3, R9C4 cannot hold 3.So 3 must go in R8C6.
  24. R8C9 = 7 · Naked SingleLooking at R8C9:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  25. R7C4 = 7 · Hidden SingleIn the bottom-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 7, R9C4 cannot hold 7.So 7 must go in R7C4.
  26. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  27. R9C4 = 6 · Naked SingleLooking at R9C4:- Row 9 already has 1, 2, 3, 4, 5, 7 and 9 → ruled out.- Column 4 already has 8 → ruled out.So this cell can only be 6.
  28. R9C9 = 8 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  29. R5C8 = 6 · Hidden SingleIn the middle-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 6, R4C9 and R5C9 cannot hold 6.- Because row 6 already has a 6, R6C7 cannot hold 6.So 6 must go in R5C8.
  30. R2C7 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 6, R1C8 and R2C8 cannot hold 6.- Because column 9 already has a 6, R2C9 cannot hold 6.- Because row 3 already has a 6, R3C7 and R3C8 cannot hold 6.So 6 must go in R2C7.
  31. R1C6 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 6, R1C4 cannot hold 6.- Because row 2 already has a 6, R2C5 and R2C6 cannot hold 6.- Because row 3 already has a 6, R3C5 cannot hold 6.So 6 must go in R1C6.
  32. R3C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C8 cannot hold 7.- Because column 8 already has a 7, R2C8 and R3C8 cannot hold 7.- Because column 9 already has a 7, R2C9 cannot hold 7.So 7 must go in R3C7.
  33. R6C7 = 9 · Naked SingleLooking at R6C7:- Row 6 already has 1, 3, 4, 6, 7 and 8 → ruled out.- Column 7 already has 2 and 5 → ruled out.So this cell can only be 9.
  34. R2C9 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R1C8 and R2C8 cannot hold 9.- Because row 3 already has a 9, R3C8 cannot hold 9.So 9 must go in R2C9.
  35. R1C4 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 9, R2C5 and R2C6 cannot hold 9.- Because row 3 already has a 9, R3C5 cannot hold 9.So 9 must go in R1C4.
  36. R1C8 = 2 · Naked SingleLooking at R1C8:- Row 1 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  37. R3C5 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 2, R2C5 and R2C6 cannot hold 2.So 2 must go in R3C5.
  38. R3C8 = 3 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  39. R2C8 = 5 · Naked SingleLooking at R2C8:- Row 2 already has 1, 2, 4, 6, 8 and 9 → ruled out.- Column 8 already has 3 and 7 → ruled out.So this cell can only be 5.
  40. R2C5 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 3, R2C6 cannot hold 3.So 3 must go in R2C5.
  41. R2C6 = 7 · Naked SingleLooking at R2C6:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  42. R5C6 = 9 · Naked SingleLooking at R5C6:- Row 5 already has 1, 4, 6 and 8 → ruled out.- Column 6 already has 2, 3, 5 and 7 → ruled out.So this cell can only be 9.
  43. R5C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R4C5 and R5C5 cannot hold 3.- Because row 6 already has a 3, R6C4 cannot hold 3.So 3 must go in R5C4.
  44. R6C4 = 2 · Naked SingleLooking at R6C4:- Row 6 already has 1, 3, 4, 6, 7, 8 and 9 → ruled out.- Column 4 already has 5 → ruled out.So this cell can only be 2.
  45. R6C1 = 5 · Naked SingleLooking at R6C1:- Row 6 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  46. R4C1 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C1 cannot hold 3.So 3 must go in R4C1.
  47. R5C1 = 2 · Naked SingleLooking at R5C1:- Row 5 already has 1, 3, 4, 6, 8 and 9 → ruled out.- Column 1 already has 5 and 7 → ruled out.So this cell can only be 2.
  48. R4C5 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 6, R5C5 cannot hold 6.So 6 must go in R4C5.
  49. R4C9 = 2 · Naked SingleLooking at R4C9:- Row 4 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  50. R5C5 = 7 · Naked SingleLooking at R5C5:- Row 5 already has 1, 2, 3, 4, 6, 8 and 9 → ruled out.- Column 5 already has 5 → ruled out.So this cell can only be 7.
  51. R5C9 = 5 · Naked SingleLooking at R5C9:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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