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Daily Sudoku — 5 August 2026

The puzzle

  • 5 August 2026
  • Medium
  • 34 clues
Puzzle
·5·4·3·9·
8·1·5··2·
··4·98···
··7·12···
·13···54·
··87···1·
·7···4·8·
·352816·9
···97··3·
Solution
756423891
891657324
324198756
547312968
213869547
968745213
679534182
435281679
182976435

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 — 47 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. R3C1 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C1 and R1C3 cannot hold 3.- Because column 2 already has a 3, R2C2 and R3C2 cannot hold 3.So 3 must go in R3C1.
  2. R1C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 7, R1C3 cannot hold 7.- Because column 2 already has a 7, R2C2 and R3C2 cannot hold 7.So 7 must go in R1C1.
  3. R2C2 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C3 cannot hold 9.- Because row 3 already has a 9, R3C2 cannot hold 9.So 9 must go in R2C2.
  4. R3C4 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 1, R1C5 cannot hold 1.- Because row 2 already has a 1, R2C4 and R2C6 cannot hold 1.So 1 must go in R3C4.
  5. R1C5 = 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, R2C4 and R2C6 cannot hold 2.So 2 must go in R1C5.
  6. R3C2 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C3 cannot hold 2.So 2 must go in R3C2.
  7. R1C3 = 6 · Naked SingleLooking at R1C3:- Row 1 already has 2, 3, 4, 5, 7 and 9 → ruled out.- Column 3 already has 1 and 8 → ruled out.So this cell can only be 6.
  8. R2C6 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 7, R2C4 cannot hold 7.So 7 must go in R2C6.
  9. R2C4 = 6 · Naked SingleLooking at R2C4:- Row 2 already has 1, 2, 5, 7, 8 and 9 → ruled out.- Column 4 already has 4 → ruled out.- The top-centre 3×3 box already has 3 → ruled out.So this cell can only be 6.
  10. R6C5 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R4C4 cannot hold 4.- Because row 5 already has a 4, R5C4, R5C5 and R5C6 cannot hold 4.- Because column 6 already has a 4, R6C6 cannot hold 4.So 4 must go in R6C5.
  11. R4C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C4, R5C5 and R5C6 cannot hold 3.- Because column 6 already has a 3, R6C6 cannot hold 3.So 3 must go in R4C4.
  12. R6C6 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 5, R5C4, R5C5 and R5C6 cannot hold 5.So 5 must go in R6C6.
  13. R4C1 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 5, R4C2 cannot hold 5.- Because row 5 already has a 5, R5C1 cannot hold 5.- Because row 6 already has a 5, R6C1 and R6C2 cannot hold 5.So 5 must go in R4C1.
  14. R4C2 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 4, R5C1 cannot hold 4.- Because row 6 already has a 4, R6C1 and R6C2 cannot hold 4.So 4 must go in R4C2.
  15. R5C4 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 8, R5C5 cannot hold 8.- Because column 6 already has a 8, R5C6 cannot hold 8.So 8 must go in R5C4.
  16. R7C4 = 5 · Naked SingleLooking at R7C4:- Row 7 already has 4, 7 and 8 → ruled out.- Column 4 already has 1, 2, 3, 6 and 9 → ruled out.So this cell can only be 5.
  17. R5C6 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 9, R5C5 cannot hold 9.So 9 must go in R5C6.
  18. R9C6 = 6 · Naked SingleLooking at R9C6:- Row 9 already has 3, 7 and 9 → ruled out.- Column 6 already has 1, 2, 4, 5 and 8 → ruled out.So this cell can only be 6.
  19. R5C5 = 6 · Naked SingleLooking at R5C5:- Row 5 already has 1, 3, 4, 5, 8 and 9 → ruled out.- Column 5 already has 2 and 7 → ruled out.So this cell can only be 6.
  20. R7C5 = 3 · Naked SingleLooking at R7C5:- Row 7 already has 4, 5, 7 and 8 → ruled out.- Column 5 already has 1, 2, 6 and 9 → ruled out.So this cell can only be 3.
  21. R6C1 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C1 cannot hold 9.- Because column 2 already has a 9, R6C2 cannot hold 9.So 9 must go in R6C1.
  22. R5C1 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R6C2 cannot hold 2.So 2 must go in R5C1.
  23. R5C9 = 7 · Naked SingleLooking at R5C9:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  24. R6C2 = 6 · Naked SingleLooking at R6C2:- Row 6 already has 1, 4, 5, 7, 8 and 9 → ruled out.- Column 2 already has 2 and 3 → ruled out.So this cell can only be 6.
  25. R9C2 = 8 · Naked SingleLooking at R9C2:- Row 9 already has 3, 6, 7 and 9 → ruled out.- Column 2 already has 1, 2, 4 and 5 → ruled out.So this cell can only be 8.
  26. R4C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R4C8 cannot hold 9.- Because column 9 already has a 9, R4C9 cannot hold 9.- Because row 6 already has a 9, R6C7 and R6C9 cannot hold 9.So 9 must go in R4C7.
  27. R4C9 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 8, R4C8 cannot hold 8.- Because row 6 already has a 8, R6C7 and R6C9 cannot hold 8.So 8 must go in R4C9.
  28. R4C8 = 6 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  29. R3C9 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C7 and R1C9 cannot hold 6.- Because row 2 already has a 6, R2C7 and R2C9 cannot hold 6.- Because column 7 already has a 6, R3C7 cannot hold 6.- Because column 8 already has a 6, R3C8 cannot hold 6.So 6 must go in R3C9.
  30. R3C8 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C7 and R1C9 cannot hold 5.- Because row 2 already has a 5, R2C7 and R2C9 cannot hold 5.- Because column 7 already has a 5, R3C7 cannot hold 5.So 5 must go in R3C8.
  31. R3C7 = 7 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  32. R8C8 = 7 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 5, 6, 8 and 9 → ruled out.- Column 8 already has 4 → ruled out.So this cell can only be 7.
  33. R8C1 = 4 · Naked SingleLooking at R8C1:- Row 8 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  34. R1C7 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 8, R1C9 cannot hold 8.- Because row 2 already has a 8, R2C7 and R2C9 cannot hold 8.So 8 must go in R1C7.
  35. R1C9 = 1 · Naked SingleLooking at R1C9:- Row 1 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  36. R7C1 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 6, R7C3 cannot hold 6.- Because row 9 already has a 6, R9C1 and R9C3 cannot hold 6.So 6 must go in R7C1.
  37. R9C1 = 1 · Naked SingleLooking at R9C1:- Row 9 already has 3, 6, 7, 8 and 9 → ruled out.- Column 1 already has 2, 4 and 5 → ruled out.So this cell can only be 1.
  38. R7C3 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 9, R9C3 cannot hold 9.So 9 must go in R7C3.
  39. R9C3 = 2 · Naked SingleLooking at R9C3:- Row 9 already has 1, 3, 6, 7, 8 and 9 → ruled out.- Column 3 already has 4 and 5 → ruled out.So this cell can only be 2.
  40. R7C7 = 1 · Hidden SingleIn the bottom-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 1, R7C9 cannot hold 1.- Because row 9 already has a 1, R9C7 and R9C9 cannot hold 1.So 1 must go in R7C7.
  41. R7C9 = 2 · Naked SingleLooking at R7C9:- Row 7 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  42. R6C7 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 2, R6C9 cannot hold 2.So 2 must go in R6C7.
  43. R6C9 = 3 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  44. R2C7 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R2C9 cannot hold 3.So 3 must go in R2C7.
  45. R2C9 = 4 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  46. R9C7 = 4 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 6, 7, 8 and 9 → ruled out.- Column 7 already has 5 → ruled out.So this cell can only be 4.
  47. R9C9 = 5 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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