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Daily Sudoku — 25 September 2026

The puzzle

  • 25 September 2026
  • Hard
  • 30 clues
Puzzle
·····56·4
2··6·4·3·
··9···2··
···8·····
9···46··1
·179·····
7···1····
5··4·9··3
·92763·18
Solution
173285694
258694137
649137285
364871952
925346871
817952346
736518429
581429763
492763518

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. R2C7 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 1, R1C8 and R3C8 cannot hold 1.- Because column 9 already has a 1, R2C9 and R3C9 cannot hold 1.So 1 must go in R2C7.
  2. R4C6 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 1, R4C5 cannot hold 1.- Because row 5 already has a 1, R5C4 cannot hold 1.- Because row 6 already has a 1, R6C5 and R6C6 cannot hold 1.So 1 must go in R4C6.
  3. R4C5 = 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, R5C4 cannot hold 7.- Because row 6 already has a 7, R6C5 and R6C6 cannot hold 7.So 7 must go in R4C5.
  4. R3C6 = 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, R1C4 and R3C4 cannot hold 7.- Because column 5 already has a 7, R1C5, R2C5 and R3C5 cannot hold 7.So 7 must go in R3C6.
  5. R8C3 = 1 · Hidden SingleIn the bottom-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C2 and R7C3 cannot hold 1.- Because column 2 already has a 1, R8C2 cannot hold 1.- Because row 9 already has a 1, R9C1 cannot hold 1.So 1 must go in R8C3.
  6. R7C4 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 5, R7C6 cannot hold 5.- Because row 8 already has a 5, R8C5 cannot hold 5.So 5 must go in R7C4.
  7. R6C5 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 5, R5C4 cannot hold 5.- Because column 6 already has a 5, R6C6 cannot hold 5.So 5 must go in R6C5.
  8. R5C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 3, R6C6 cannot hold 3.So 3 must go in R5C4.
  9. R6C6 = 2 · Naked SingleLooking at R6C6:- Row 6 already has 1, 5, 7 and 9 → ruled out.- Column 6 already has 3, 4 and 6 → ruled out.- The centre 3×3 box already has 8 → ruled out.So this cell can only be 2.
  10. R7C6 = 8 · Naked SingleLooking at R7C6:- Row 7 already has 1, 5 and 7 → ruled out.- Column 6 already has 2, 3, 4, 6 and 9 → ruled out.So this cell can only be 8.
  11. R8C5 = 2 · Naked SingleLooking at R8C5:- Row 8 already has 1, 3, 4, 5 and 9 → ruled out.- Column 5 already has 6 and 7 → ruled out.- The bottom-centre 3×3 box already has 8 → ruled out.So this cell can only be 2.
  12. R1C4 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 2, R1C5 cannot hold 2.- Because row 2 already has a 2, R2C5 cannot hold 2.- Because row 3 already has a 2, R3C4 and R3C5 cannot hold 2.So 2 must go in R1C4.
  13. R3C4 = 1 · Naked SingleLooking at R3C4:- Row 3 already has 2, 7 and 9 → ruled out.- Column 4 already has 3, 4, 5, 6 and 8 → ruled out.So this cell can only be 1.
  14. 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, R2C2 and R2C3 cannot hold 1.- Because row 3 already has a 1, R3C1 and R3C2 cannot hold 1.So 1 must go in R1C1.
  15. R8C2 = 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 an 8, R7C2 and R7C3 cannot hold 8.- Because row 9 already has an 8, R9C1 cannot hold 8.So 8 must go in R8C2.
  16. R9C7 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C7, R7C8 and R7C9 cannot hold 5.- Because row 8 already has a 5, R8C7 and R8C8 cannot hold 5.So 5 must go in R9C7.
  17. R9C1 = 4 · Naked SingleLooking at R9C1:- Row 9 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  18. R3C2 = 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 2 already has a 4, R2C2 and R2C3 cannot hold 4.- Because column 1 already has a 4, R3C1 cannot hold 4.So 4 must go in R3C2.
  19. R3C1 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C2 and R1C3 cannot hold 6.- Because row 2 already has a 6, R2C2 and R2C3 cannot hold 6.So 6 must go in R3C1.
  20. R4C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R4C1 and R6C1 cannot hold 4.- Because column 2 already has a 4, R4C2 cannot hold 4.- Because row 5 already has a 4, R5C2 and R5C3 cannot hold 4.So 4 must go in R4C3.
  21. R4C2 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R4C1 and R6C1 cannot hold 6.- Because row 5 already has a 6, R5C2 and R5C3 cannot hold 6.So 6 must go in R4C2.
  22. R5C2 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 2, R4C1 cannot hold 2.- Because column 3 already has a 2, R5C3 cannot hold 2.- Because row 6 already has a 2, R6C1 cannot hold 2.So 2 must go in R5C2.
  23. R5C3 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 5, R4C1 cannot hold 5.- Because row 6 already has a 5, R6C1 cannot hold 5.So 5 must go in R5C3.
  24. R2C2 = 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 and R1C3 cannot hold 5.- Because column 3 already has a 5, R2C3 cannot hold 5.So 5 must go in R2C2.
  25. R1C2 = 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 and R2C3 cannot hold 7.So 7 must go in R1C2.
  26. R7C2 = 3 · Naked SingleLooking at R7C2:- Row 7 already has 1, 5, 7 and 8 → ruled out.- Column 2 already has 2, 4, 6 and 9 → ruled out.So this cell can only be 3.
  27. R7C3 = 6 · Naked SingleLooking at R7C3:- Row 7 already has 1, 3, 5, 7 and 8 → ruled out.- Column 3 already has 2, 4 and 9 → ruled out.So this cell can only be 6.
  28. R1C3 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 3, R2C3 cannot hold 3.So 3 must go in R1C3.
  29. R2C3 = 8 · Naked SingleLooking at R2C3:- Row 2 already has 1, 2, 3, 4, 5 and 6 → ruled out.- Column 3 already has 7 and 9 → ruled out.So this cell can only be 8.
  30. R3C5 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C5 cannot hold 3.- Because row 2 already has a 3, R2C5 cannot hold 3.So 3 must go in R3C5.
  31. R1C5 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 2 already has an 8, R2C5 cannot hold 8.So 8 must go in R1C5.
  32. R1C8 = 9 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  33. R2C5 = 9 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 3, 4, 5, 6 and 8 → ruled out.- Column 5 already has 7 → ruled out.So this cell can only be 9.
  34. R2C9 = 7 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  35. R3C8 = 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 an 8, R3C9 cannot hold 8.So 8 must go in R3C8.
  36. R3C9 = 5 · Naked SingleLooking at R3C9:- Row 3 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  37. R6C1 = 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 an 8, R4C1 cannot hold 8.So 8 must go in R6C1.
  38. R4C1 = 3 · Naked SingleLooking at R4C1:- Row 4 already has 1, 4, 6, 7 and 8 → ruled out.- Column 1 already has 2, 5 and 9 → ruled out.So this cell can only be 3.
  39. R6C7 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C7, R4C8 and R4C9 cannot hold 3.- Because row 5 already has a 3, R5C7 and R5C8 cannot hold 3.- Because column 8 already has a 3, R6C8 cannot hold 3.- Because column 9 already has a 3, R6C9 cannot hold 3.So 3 must go in R6C7.
  40. R6C8 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C7, R4C8 and R4C9 cannot hold 4.- Because row 5 already has a 4, R5C7 and R5C8 cannot hold 4.- Because column 9 already has a 4, R6C9 cannot hold 4.So 4 must go in R6C8.
  41. R6C9 = 6 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  42. R4C8 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 5, R4C7 cannot hold 5.- Because column 9 already has a 5, R4C9 cannot hold 5.- Because row 5 already has a 5, R5C7 and R5C8 cannot hold 5.So 5 must go in R4C8.
  43. R4C9 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R4C7 cannot hold 2.- Because row 5 already has a 2, R5C7 and R5C8 cannot hold 2.So 2 must go in R4C9.
  44. R4C7 = 9 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  45. R7C9 = 9 · Naked SingleLooking at R7C9:- Row 7 already has 1, 3, 5, 6, 7 and 8 → ruled out.- Column 9 already has 2 and 4 → ruled out.So this cell can only be 9.
  46. R5C7 = 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 an 8, R5C8 cannot hold 8.So 8 must go in R5C7.
  47. R5C8 = 7 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  48. R7C8 = 2 · Hidden SingleIn the bottom-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R7C7 cannot hold 2.- Because row 8 already has a 2, R8C7 and R8C8 cannot hold 2.So 2 must go in R7C8.
  49. R7C7 = 4 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  50. R8C7 = 7 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 5, 8 and 9 → ruled out.- Column 7 already has 6 → ruled out.So this cell can only be 7.
  51. R8C8 = 6 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.

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