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Daily Sudoku — 18 September 2026

The puzzle

  • 18 September 2026
  • Hard
  • 30 clues
Puzzle
··5·3····
···6···49
2····9··7
·2·49····
6···7·82·
·7···149·
·6···42··
··72·3·6·
····6·754
Solution
945137682
731682549
286549137
128496375
694375821
573821496
869754213
457213968
312968754

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. R3C3 = 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 cannot hold 6.- Because column 2 already has a 6, R1C2 and R3C2 cannot hold 6.- Because row 2 already has a 6, R2C1, R2C2 and R2C3 cannot hold 6.So 6 must go in R3C3.
  2. R3C5 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R1C4 and R3C4 cannot hold 4.- Because column 6 already has a 4, R1C6 cannot hold 4.- Because row 2 already has a 4, R2C5 and R2C6 cannot hold 4.So 4 must go in R3C5.
  3. R1C9 = 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, R1C7 and R2C7 cannot hold 2.- Because column 8 already has a 2, R1C8 cannot hold 2.- Because row 3 already has a 2, R3C7 and R3C8 cannot hold 2.So 2 must go in R1C9.
  4. R1C7 = 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 cannot hold 6.- Because row 2 already has a 6, R2C7 cannot hold 6.- Because row 3 already has a 6, R3C7 and R3C8 cannot hold 6.So 6 must go in R1C7.
  5. R6C5 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C6 cannot hold 2.- Because row 5 already has a 2, R5C4 and R5C6 cannot hold 2.- Because column 4 already has a 2, R6C4 cannot hold 2.So 2 must go in R6C5.
  6. R2C6 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C4 and R1C6 cannot hold 2.- Because column 5 already has a 2, R2C5 cannot hold 2.- Because row 3 already has a 2, R3C4 cannot hold 2.So 2 must go in R2C6.
  7. R4C6 = 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, R5C4 and R5C6 cannot hold 6.- Because column 4 already has a 6, R6C4 cannot hold 6.So 6 must go in R4C6.
  8. R6C4 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 5 already has an 8, R5C4 and R5C6 cannot hold 8.So 8 must go in R6C4.
  9. 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, R5C6 cannot hold 3.So 3 must go in R5C4.
  10. R5C6 = 5 · Naked SingleLooking at R5C6:- Row 5 already has 2, 3, 6, 7 and 8 → ruled out.- Column 6 already has 1, 4 and 9 → ruled out.So this cell can only be 5.
  11. R6C9 = 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, R4C7, R4C8 and R4C9 cannot hold 6.- Because row 5 already has a 6, R5C9 cannot hold 6.So 6 must go in R6C9.
  12. R4C8 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 7, R4C7 cannot hold 7.- Because column 9 already has a 7, R4C9 cannot hold 7.- Because row 5 already has a 7, R5C9 cannot hold 7.So 7 must go in R4C8.
  13. R9C3 = 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, R7C1 and R7C3 cannot hold 2.- Because row 8 already has a 2, R8C1 and R8C2 cannot hold 2.- Because column 1 already has a 2, R9C1 cannot hold 2.- Because column 2 already has a 2, R9C2 cannot hold 2.So 2 must go in R9C3.
  14. R7C4 = 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 cannot hold 7.- Because row 8 already has a 7, R8C5 cannot hold 7.- Because row 9 already has a 7, R9C4 and R9C6 cannot hold 7.So 7 must go in R7C4.
  15. R1C6 = 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 cannot hold 7.- Because column 5 already has a 7, R2C5 cannot hold 7.- Because row 3 already has a 7, R3C4 cannot hold 7.So 7 must go in R1C6.
  16. R9C6 = 8 · Naked SingleLooking at R9C6:- Row 9 already has 2, 4, 5, 6 and 7 → ruled out.- Column 6 already has 1, 3 and 9 → ruled out.So this cell can only be 8.
  17. R2C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C1 and R1C2 cannot hold 7.- Because column 2 already has a 7, R2C2 cannot hold 7.- Because column 3 already has a 7, R2C3 cannot hold 7.- Because row 3 already has a 7, R3C2 cannot hold 7.So 7 must go in R2C1.
  18. R2C5 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has an 8, R1C4 and R3C4 cannot hold 8.So 8 must go in R2C5.
  19. R3C4 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C4 cannot hold 5.So 5 must go in R3C4.
  20. R1C4 = 1 · Naked SingleLooking at R1C4:- Row 1 already has 2, 3, 5, 6 and 7 → ruled out.- Column 4 already has 4 and 8 → ruled out.- The top-centre 3×3 box already has 9 → ruled out.So this cell can only be 1.
  21. R9C4 = 9 · Naked SingleLooking at R9C4:- Row 9 already has 2, 4, 5, 6, 7 and 8 → ruled out.- Column 4 already has 1 and 3 → ruled out.So this cell can only be 9.
  22. R2C7 = 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, R1C8 cannot hold 5.- Because row 3 already has a 5, R3C7 and R3C8 cannot hold 5.So 5 must go in R2C7.
  23. R4C9 = 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 row 5 already has a 5, R5C9 cannot hold 5.So 5 must go in R4C9.
  24. R6C1 = 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 R4C3 cannot hold 5.- Because row 5 already has a 5, R5C2 and R5C3 cannot hold 5.- Because column 3 already has a 5, R6C3 cannot hold 5.So 5 must go in R6C1.
  25. R6C3 = 3 · Naked SingleLooking at R6C3:- Row 6 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  26. R4C7 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C9 cannot hold 3.So 3 must go in R4C7.
  27. R5C9 = 1 · Naked SingleLooking at R5C9:- Row 5 already has 2, 3, 5, 6, 7 and 8 → ruled out.- Column 9 already has 4 and 9 → ruled out.So this cell can only be 1.
  28. R3C8 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C8 cannot hold 3.- Because column 7 already has a 3, R3C7 cannot hold 3.So 3 must go in R3C8.
  29. R2C2 = 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 R1C2 cannot hold 3.- Because column 3 already has a 3, R2C3 cannot hold 3.- Because row 3 already has a 3, R3C2 cannot hold 3.So 3 must go in R2C2.
  30. R2C3 = 1 · Naked SingleLooking at R2C3:- Row 2 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  31. R3C7 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C8 cannot hold 1.So 1 must go in R3C7.
  32. R3C2 = 8 · Naked SingleLooking at R3C2:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  33. R8C7 = 9 · Naked SingleLooking at R8C7:- Row 8 already has 2, 3, 6 and 7 → ruled out.- Column 7 already has 1, 4, 5 and 8 → ruled out.So this cell can only be 9.
  34. R1C8 = 8 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 3, 5, 6 and 7 → ruled out.- Column 8 already has 4 and 9 → ruled out.So this cell can only be 8.
  35. R7C8 = 1 · Naked SingleLooking at R7C8:- Row 7 already has 2, 4, 6 and 7 → ruled out.- Column 8 already has 3, 5, 8 and 9 → ruled out.So this cell can only be 1.
  36. R4C1 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 1, R4C3 cannot hold 1.- Because row 5 already has a 1, R5C2 and R5C3 cannot hold 1.So 1 must go in R4C1.
  37. R4C3 = 8 · Naked SingleLooking at R4C3:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  38. R8C2 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 5, R7C1 and R8C1 cannot hold 5.- Because column 3 already has a 5, R7C3 cannot hold 5.- Because row 9 already has a 5, R9C1 and R9C2 cannot hold 5.So 5 must go in R8C2.
  39. R9C2 = 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, R7C1 and R7C3 cannot hold 1.- Because column 1 already has a 1, R8C1 and R9C1 cannot hold 1.So 1 must go in R9C2.
  40. R9C1 = 3 · Naked SingleLooking at R9C1:- Row 9 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  41. R8C1 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 4, R7C1 and R7C3 cannot hold 4.So 4 must go in R8C1.
  42. R1C2 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R1C1 cannot hold 4.So 4 must go in R1C2.
  43. R1C1 = 9 · Naked SingleLooking at R1C1:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  44. R7C1 = 8 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 4, 6 and 7 → ruled out.- Column 1 already has 3, 5 and 9 → ruled out.So this cell can only be 8.
  45. R5C2 = 9 · Naked SingleLooking at R5C2:- Row 5 already has 1, 2, 3, 5, 6, 7 and 8 → ruled out.- Column 2 already has 4 → ruled out.So this cell can only be 9.
  46. R5C3 = 4 · Naked SingleLooking at R5C3:- Row 5 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  47. R7C3 = 9 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 4, 6, 7 and 8 → ruled out.- Column 3 already has 3 and 5 → ruled out.So this cell can only be 9.
  48. 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, R7C5 cannot hold 1.So 1 must go in R8C5.
  49. R8C9 = 8 · Naked SingleLooking at R8C9:- Row 8 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  50. R7C5 = 5 · Naked SingleLooking at R7C5:- Row 7 already has 1, 2, 4, 6, 7, 8 and 9 → ruled out.- Column 5 already has 3 → ruled out.So this cell can only be 5.
  51. R7C9 = 3 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.

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