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Daily Sudoku — 20 September 2026

The puzzle

  • 20 September 2026
  • Hard
  • 30 clues
Puzzle
·7·8·····
9·126··45
··4··3··7
625·3··7·
···1···52
·9···2···
·1·5·····
3·9·2····
·68····21
Solution
276845913
931267845
854913267
625438179
483179652
197652438
712586394
349721586
568394721

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. R1C3 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R1C1 and R3C1 cannot hold 6.- Because row 2 already has a 6, R2C2 cannot hold 6.- Because column 2 already has a 6, R3C2 cannot hold 6.So 6 must go in R1C3.
  2. R2C2 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 3, R1C1 cannot hold 3.- Because row 3 already has a 3, R3C1 and R3C2 cannot hold 3.So 3 must go in R2C2.
  3. R2C6 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C5 and R1C6 cannot hold 7.- Because row 3 already has a 7, R3C4 and R3C5 cannot hold 7.So 7 must go in R2C6.
  4. R2C7 = 8 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  5. R6C1 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 1, R5C1, R5C2 and R5C3 cannot hold 1.- Because column 3 already has a 1, R6C3 cannot hold 1.So 1 must go in R6C1.
  6. R6C5 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C4 and R4C6 cannot hold 5.- Because row 5 already has a 5, R5C5 and R5C6 cannot hold 5.- Because column 4 already has a 5, R6C4 cannot hold 5.So 5 must go in R6C5.
  7. R1C6 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 5, R1C5 and R3C5 cannot hold 5.- Because column 4 already has a 5, R3C4 cannot hold 5.So 5 must go in R1C6.
  8. R1C5 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 4, R3C4 and R3C5 cannot hold 4.So 4 must go in R1C5.
  9. R3C5 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 1, R3C4 cannot hold 1.So 1 must go in R3C5.
  10. R3C4 = 9 · Naked SingleLooking at R3C4:- Row 3 already has 1, 3, 4 and 7 → ruled out.- Column 4 already has 2, 5 and 8 → ruled out.- The top-centre 3×3 box already has 6 → ruled out.So this cell can only be 9.
  11. 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, R5C7 cannot hold 1.- Because row 6 already has a 1, R6C7, R6C8 and R6C9 cannot hold 1.So 1 must go in R4C7.
  12. R1C8 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 1, R1C7 cannot hold 1.- Because column 9 already has a 1, R1C9 cannot hold 1.- Because row 3 already has a 1, R3C7 and R3C8 cannot hold 1.So 1 must go in R1C8.
  13. R8C6 = 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, R7C5 and R7C6 cannot hold 1.- Because column 4 already has a 1, R8C4 cannot hold 1.- Because row 9 already has a 1, R9C4, R9C5 and R9C6 cannot hold 1.So 1 must go in R8C6.
  14. R9C4 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R7C5 and R9C5 cannot hold 3.- Because column 6 already has a 3, R7C6 and R9C6 cannot hold 3.- Because row 8 already has a 3, R8C4 cannot hold 3.So 3 must go in R9C4.
  15. R7C3 = 2 · Hidden SingleIn column 3, 2 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 2, R5C3 cannot hold 2.- Because row 6 already has a 2, R6C3 cannot hold 2.So 2 must go in R7C3.
  16. R7C8 = 9 · Hidden SingleIn column 8, 9 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 9, R3C8 cannot hold 9.- Because row 6 already has a 9, R6C8 cannot hold 9.- Because row 8 already has a 9, R8C8 cannot hold 9.So 9 must go in R7C8.
  17. R6C8 = 3 · Hidden SingleIn column 8, 3 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 3, R3C8 cannot hold 3.- Because row 8 already has a 3, R8C8 cannot hold 3.So 3 must go in R6C8.
  18. R5C3 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 3, R5C1 cannot hold 3.- Because column 2 already has a 3, R5C2 cannot hold 3.- Because row 6 already has a 3, R6C3 cannot hold 3.So 3 must go in R5C3.
  19. R6C3 = 7 · Naked SingleLooking at R6C3:- Row 6 already has 1, 2, 3, 5 and 9 → ruled out.- Column 3 already has 4, 6 and 8 → ruled out.So this cell can only be 7.
  20. R5C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C4 and R4C6 cannot hold 7.- Because column 6 already has a 7, R5C6 cannot hold 7.- Because row 6 already has a 7, R6C4 cannot hold 7.So 7 must go in R5C5.
  21. R8C4 = 7 · Hidden SingleIn the bottom-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R7C5 and R9C5 cannot hold 7.- Because column 6 already has a 7, R7C6 and R9C6 cannot hold 7.So 7 must go in R8C4.
  22. R7C6 = 6 · Hidden SingleIn the bottom-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 6, R7C5 cannot hold 6.- Because row 9 already has a 6, R9C5 and R9C6 cannot hold 6.So 6 must go in R7C6.
  23. R6C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C4 and R4C6 cannot hold 6.- Because column 6 already has a 6, R5C6 cannot hold 6.So 6 must go in R6C4.
  24. R4C4 = 4 · Naked SingleLooking at R4C4:- Row 4 already has 1, 2, 3, 5, 6 and 7 → ruled out.- Column 4 already has 8 and 9 → ruled out.So this cell can only be 4.
  25. R5C7 = 6 · Hidden SingleIn the middle-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C9 cannot hold 6.- Because row 6 already has a 6, R6C7 and R6C9 cannot hold 6.So 6 must go in R5C7.
  26. R3C8 = 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 column 7 already has a 6, R3C7 cannot hold 6.So 6 must go in R3C8.
  27. R8C8 = 8 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 7 and 9 → ruled out.- Column 8 already has 4, 5 and 6 → ruled out.So this cell can only be 8.
  28. R4C9 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 9, R6C7 and R6C9 cannot hold 9.So 9 must go in R4C9.
  29. R4C6 = 8 · Naked SingleLooking at R4C6:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  30. R5C6 = 9 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 3, 5, 6 and 7 → ruled out.- Column 6 already has 8 → ruled out.- The centre 3×3 box already has 4 → ruled out.So this cell can only be 9.
  31. R9C6 = 4 · Naked SingleLooking at R9C6:- Row 9 already has 1, 2, 3, 6 and 8 → ruled out.- Column 6 already has 5, 7 and 9 → ruled out.So this cell can only be 4.
  32. R1C7 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 9, R1C9 cannot hold 9.- Because row 3 already has a 9, R3C7 cannot hold 9.So 9 must go in R1C7.
  33. R3C7 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 2, R1C9 cannot hold 2.So 2 must go in R3C7.
  34. R1C9 = 3 · Naked SingleLooking at R1C9:- Row 1 already has 1, 4, 5, 6, 7, 8 and 9 → ruled out.- Column 9 already has 2 → ruled out.So this cell can only be 3.
  35. R1C1 = 2 · Naked SingleLooking at R1C1:- Row 1 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  36. R6C9 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 7 already has an 8, R6C7 cannot hold 8.So 8 must go in R6C9.
  37. R6C7 = 4 · Naked SingleLooking at R6C7:- Row 6 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  38. R7C5 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 9 already has an 8, R9C5 cannot hold 8.So 8 must go in R7C5.
  39. R9C5 = 9 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 4, 6 and 8 → ruled out.- Column 5 already has 5 and 7 → ruled out.So this cell can only be 9.
  40. R7C7 = 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 cannot hold 3.- Because row 8 already has a 3, R8C7 and R8C9 cannot hold 3.- Because row 9 already has a 3, R9C7 cannot hold 3.So 3 must go in R7C7.
  41. R8C9 = 6 · Hidden SingleIn the bottom-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C9 cannot hold 6.- Because column 7 already has a 6, R8C7 cannot hold 6.- Because row 9 already has a 6, R9C7 cannot hold 6.So 6 must go in R8C9.
  42. R7C9 = 4 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 5, 6, 8 and 9 → ruled out.- Column 9 already has 7 → ruled out.So this cell can only be 4.
  43. R7C1 = 7 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  44. R8C2 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 4, R9C1 cannot hold 4.So 4 must go in R8C2.
  45. R8C7 = 5 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  46. R9C7 = 7 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 6, 8 and 9 → ruled out.- Column 7 already has 5 → ruled out.So this cell can only be 7.
  47. R9C1 = 5 · Naked SingleLooking at R9C1:- Row 9 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  48. R3C2 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 5, R3C1 cannot hold 5.So 5 must go in R3C2.
  49. R3C1 = 8 · Naked SingleLooking at R3C1:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  50. R5C1 = 4 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 5, 6, 7 and 9 → ruled out.- Column 1 already has 8 → ruled out.So this cell can only be 4.
  51. R5C2 = 8 · Naked SingleLooking at R5C2:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.

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