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Daily Sudoku — 28 July 2026

The puzzle

  • 28 July 2026
  • Hard
  • 30 clues
Puzzle
··86·2·4·
··354·7·1
·5·78····
··2··53··
·8·······
7·1·2··95
······273
··5·7····
32·9····8
Solution
178632549
263549781
954781632
642895317
589317426
731426895
496158273
815273964
327964158

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. R1C2 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 7, R1C1 cannot hold 7.- Because row 2 already has a 7, R2C1 and R2C2 cannot hold 7.- Because row 3 already has a 7, R3C1 and R3C3 cannot hold 7.So 7 must go in R1C2.
  2. R3C8 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R1C7 and R3C7 cannot hold 3.- Because column 9 already has a 3, R1C9 and R3C9 cannot hold 3.- Because row 2 already has a 3, R2C8 cannot hold 3.So 3 must go in R3C8.
  3. R1C5 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 3, R2C6 cannot hold 3.- Because row 3 already has a 3, R3C6 cannot hold 3.So 3 must go in R1C5.
  4. R3C6 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C6 cannot hold 1.So 1 must go in R3C6.
  5. R2C6 = 9 · Naked SingleLooking at R2C6:- Row 2 already has 1, 3, 4, 5 and 7 → ruled out.- Column 6 already has 2 → ruled out.- The top-centre 3×3 box already has 6 and 8 → ruled out.So this cell can only be 9.
  6. R1C1 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C1 and R2C2 cannot hold 1.- Because row 3 already has a 1, R3C1 and R3C3 cannot hold 1.So 1 must go in R1C1.
  7. R1C7 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 5, R1C9 cannot hold 5.- Because row 2 already has a 5, R2C8 cannot hold 5.- Because row 3 already has a 5, R3C7 and R3C9 cannot hold 5.So 5 must go in R1C7.
  8. R1C9 = 9 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  9. R2C8 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 8, R3C7 and R3C9 cannot hold 8.So 8 must go in R2C8.
  10. R3C9 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R3C7 cannot hold 2.So 2 must go in R3C9.
  11. R3C7 = 6 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 5, 7 and 8 → ruled out.- The top-right 3×3 box already has 4 and 9 → ruled out.So this cell can only be 6.
  12. R2C1 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R2C2 cannot hold 2.- Because row 3 already has a 2, R3C1 and R3C3 cannot hold 2.So 2 must go in R2C1.
  13. R2C2 = 6 · Naked SingleLooking at R2C2:- Row 2 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  14. R6C2 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C1 and R4C2 cannot hold 3.- Because column 1 already has a 3, R5C1 cannot hold 3.- Because column 3 already has a 3, R5C3 cannot hold 3.So 3 must go in R6C2.
  15. R5C1 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C1 and R4C2 cannot hold 5.- Because column 3 already has a 5, R5C3 cannot hold 5.So 5 must go in R5C1.
  16. R5C6 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 7, R4C4 and R5C4 cannot hold 7.- Because column 5 already has a 7, R4C5 and R5C5 cannot hold 7.- Because row 6 already has a 7, R6C4 and R6C6 cannot hold 7.So 7 must go in R5C6.
  17. R5C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C4 and R4C5 cannot hold 3.- Because column 5 already has a 3, R5C5 cannot hold 3.- Because row 6 already has a 3, R6C4 and R6C6 cannot hold 3.So 3 must go in R5C4.
  18. R5C8 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C8 and R4C9 cannot hold 2.- Because column 7 already has a 2, R5C7 cannot hold 2.- Because column 9 already has a 2, R5C9 cannot hold 2.- Because row 6 already has a 2, R6C7 cannot hold 2.So 2 must go in R5C8.
  19. R4C9 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 7, R4C8 cannot hold 7.- Because row 5 already has a 7, R5C7 and R5C9 cannot hold 7.- Because row 6 already has a 7, R6C7 cannot hold 7.So 7 must go in R4C9.
  20. R6C7 = 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 5 already has a 8, R5C7 and R5C9 cannot hold 8.So 8 must go in R6C7.
  21. R4C4 = 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, R4C5 cannot hold 8.- Because row 5 already has a 8, R5C5 cannot hold 8.- Because row 6 already has a 8, R6C4 and R6C6 cannot hold 8.So 8 must go in R4C4.
  22. R9C3 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 7, R7C1, R7C2 and R7C3 cannot hold 7.- Because row 8 already has a 7, R8C1 and R8C2 cannot hold 7.So 7 must go in R9C3.
  23. R8C4 = 2 · Hidden SingleIn the bottom-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C4, R7C5 and R7C6 cannot hold 2.- Because column 6 already has a 2, R8C6 cannot hold 2.- Because row 9 already has a 2, R9C5 and R9C6 cannot hold 2.So 2 must go in R8C4.
  24. 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, R7C5 and R7C6 cannot hold 3.- Because row 9 already has a 3, R9C5 and R9C6 cannot hold 3.So 3 must go in R8C6.
  25. R7C6 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 8, R7C4 cannot hold 8.- Because column 5 already has a 8, R7C5 cannot hold 8.- Because row 9 already has a 8, R9C5 and R9C6 cannot hold 8.So 8 must go in R7C6.
  26. R8C1 = 8 · Hidden SingleIn the bottom-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 8, R7C1, R7C2 and R7C3 cannot hold 8.- Because column 2 already has a 8, R8C2 cannot hold 8.So 8 must go in R8C1.
  27. R9C8 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 5, R8C7, R8C8 and R8C9 cannot hold 5.- Because column 7 already has a 5, R9C7 cannot hold 5.So 5 must go in R9C8.
  28. R7C5 = 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 9 already has a 5, R9C5 and R9C6 cannot hold 5.So 5 must go in R7C5.
  29. R8C7 = 9 · Hidden SingleIn the bottom-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R8C8 cannot hold 9.- Because column 9 already has a 9, R8C9 cannot hold 9.- Because row 9 already has a 9, R9C7 cannot hold 9.So 9 must go in R8C7.
  30. R6C6 = 6 · Hidden SingleIn row 6, 6 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 6, R6C4 cannot hold 6.So 6 must go in R6C6.
  31. R6C4 = 4 · Naked SingleLooking at R6C4:- Row 6 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  32. R7C4 = 1 · Naked SingleLooking at R7C4:- Row 7 already has 2, 3, 5, 7 and 8 → ruled out.- Column 4 already has 4, 6 and 9 → ruled out.So this cell can only be 1.
  33. R9C6 = 4 · Naked SingleLooking at R9C6:- Row 9 already has 2, 3, 5, 7, 8 and 9 → ruled out.- Column 6 already has 1 and 6 → ruled out.So this cell can only be 4.
  34. R9C5 = 6 · Naked SingleLooking at R9C5:- Row 9 already has 2, 3, 4, 5, 7, 8 and 9 → ruled out.- The bottom-centre 3×3 box already has 1 → ruled out.So this cell can only be 6.
  35. R9C7 = 1 · Naked SingleLooking at R9C7:- Row 9 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  36. R5C7 = 4 · Naked SingleLooking at R5C7:- Row 5 already has 2, 3, 5, 7 and 8 → ruled out.- Column 7 already has 1, 6 and 9 → ruled out.So this cell can only be 4.
  37. R4C8 = 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, R5C9 cannot hold 1.So 1 must go in R4C8.
  38. R8C8 = 6 · Naked SingleLooking at R8C8:- Row 8 already has 2, 3, 5, 7, 8 and 9 → ruled out.- Column 8 already has 1 and 4 → ruled out.So this cell can only be 6.
  39. R5C9 = 6 · Naked SingleLooking at R5C9:- Row 5 already has 2, 3, 4, 5, 7 and 8 → ruled out.- Column 9 already has 1 and 9 → ruled out.So this cell can only be 6.
  40. R8C9 = 4 · Naked SingleLooking at R8C9:- Row 8 already has 2, 3, 5, 6, 7, 8 and 9 → ruled out.- Column 9 already has 1 → ruled out.So this cell can only be 4.
  41. R8C2 = 1 · Naked SingleLooking at R8C2:- Row 8 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  42. R4C1 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 6, R4C2 cannot hold 6.- Because row 5 already has a 6, R5C3 cannot hold 6.So 6 must go in R4C1.
  43. 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, R5C3 cannot hold 4.So 4 must go in R4C2.
  44. R4C5 = 9 · Naked SingleLooking at R4C5:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  45. R7C2 = 9 · Naked SingleLooking at R7C2:- Row 7 already has 1, 2, 3, 5, 7 and 8 → ruled out.- Column 2 already has 4 and 6 → ruled out.So this cell can only be 9.
  46. R5C5 = 1 · Naked SingleLooking at R5C5:- Row 5 already has 2, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 5 already has 9 → ruled out.So this cell can only be 1.
  47. R5C3 = 9 · Naked SingleLooking at R5C3:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  48. R3C1 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R3C3 cannot hold 9.So 9 must go in R3C1.
  49. R3C3 = 4 · Naked SingleLooking at R3C3:- Row 3 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  50. R7C1 = 4 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 3, 5, 7, 8 and 9 → ruled out.- Column 1 already has 6 → ruled out.So this cell can only be 4.
  51. R7C3 = 6 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.

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