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Daily Sudoku — 23 September 2026

The puzzle

  • 23 September 2026
  • Easy
  • 40 clues
Puzzle
·68·2···3
52···1·6·
193·67···
68·74·93·
2·····6··
9·5·8··4·
8····4··2
··6·538·9
·5291847·
Solution
768425193
524391768
193867524
681742935
247539681
935186247
819674352
476253819
352918476

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 — 41 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. R1C8 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 9, R1C7 and R2C7 cannot hold 9.- Because column 9 already has a 9, R2C9 cannot hold 9.- Because row 3 already has a 9, R3C7, R3C8 and R3C9 cannot hold 9.So 9 must go in R1C8.
  2. R2C5 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C4 and R1C6 cannot hold 9.- Because column 4 already has a 9, R2C4 cannot hold 9.- Because row 3 already has a 9, R3C4 cannot hold 9.So 9 must go in R2C5.
  3. R2C4 = 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, R1C4 and R1C6 cannot hold 3.- Because row 3 already has a 3, R3C4 cannot hold 3.So 3 must go in R2C4.
  4. R3C4 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has an 8, R1C4 and R1C6 cannot hold 8.So 8 must go in R3C4.
  5. R1C4 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 4, R1C6 cannot hold 4.So 4 must go in R1C4.
  6. R1C6 = 5 · Naked SingleLooking at R1C6:- Row 1 already has 2, 3, 4, 6, 8 and 9 → ruled out.- Column 6 already has 1 and 7 → ruled out.So this cell can only be 5.
  7. R2C3 = 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, R1C1 cannot hold 4.So 4 must go in R2C3.
  8. R1C1 = 7 · Naked SingleLooking at R1C1:- Row 1 already has 2, 3, 4, 5, 6, 8 and 9 → ruled out.- Column 1 already has 1 → ruled out.So this cell can only be 7.
  9. R1C7 = 1 · Naked SingleLooking at R1C7:- Row 1 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  10. R3C9 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 4, R2C7 and R2C9 cannot hold 4.- Because column 7 already has a 4, R3C7 cannot hold 4.- Because column 8 already has a 4, R3C8 cannot hold 4.So 4 must go in R3C9.
  11. R2C9 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 7 already has an 8, R2C7 cannot hold 8.- Because row 3 already has an 8, R3C7 and R3C8 cannot hold 8.So 8 must go in R2C9.
  12. R2C7 = 7 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  13. R5C2 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C3 cannot hold 4.- Because column 3 already has a 4, R5C3 cannot hold 4.- Because row 6 already has a 4, R6C2 cannot hold 4.So 4 must go in R5C2.
  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, R4C3 cannot hold 3.- Because column 3 already has a 3, R5C3 cannot hold 3.So 3 must go in R6C2.
  15. R5C3 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C3 cannot hold 7.So 7 must go in R5C3.
  16. R4C3 = 1 · Naked SingleLooking at R4C3:- Row 4 already has 3, 4, 6, 7, 8 and 9 → ruled out.- Column 3 already has 2 and 5 → ruled out.So this cell can only be 1.
  17. R7C3 = 9 · Naked SingleLooking at R7C3:- Row 7 already has 2, 4 and 8 → ruled out.- Column 3 already has 1, 3, 5, 6 and 7 → ruled out.So this cell can only be 9.
  18. R5C5 = 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, R4C6 cannot hold 3.- Because column 4 already has a 3, R5C4 cannot hold 3.- Because column 6 already has a 3, R5C6 cannot hold 3.- Because row 6 already has a 3, R6C4 and R6C6 cannot hold 3.So 3 must go in R5C5.
  19. R7C5 = 7 · Naked SingleLooking at R7C5:- Row 7 already has 2, 4, 8 and 9 → ruled out.- Column 5 already has 1, 3, 5 and 6 → ruled out.So this cell can only be 7.
  20. R5C4 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 5, R4C6 and R5C6 cannot hold 5.- Because row 6 already has a 5, R6C4 and R6C6 cannot hold 5.So 5 must go in R5C4.
  21. R6C4 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C6 cannot hold 1.- Because column 6 already has a 1, R5C6 and R6C6 cannot hold 1.So 1 must go in R6C4.
  22. R6C6 = 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, R4C6 cannot hold 6.- Because row 5 already has a 6, R5C6 cannot hold 6.So 6 must go in R6C6.
  23. R4C6 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 2, R5C6 cannot hold 2.So 2 must go in R4C6.
  24. R4C9 = 5 · Naked SingleLooking at R4C9:- Row 4 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  25. R5C6 = 9 · Naked SingleLooking at R5C6:- Row 5 already has 2, 3, 4, 5, 6 and 7 → ruled out.- Column 6 already has 1 and 8 → ruled out.So this cell can only be 9.
  26. R6C7 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 2, R5C8 and R5C9 cannot hold 2.- Because column 9 already has a 2, R6C9 cannot hold 2.So 2 must go in R6C7.
  27. R6C9 = 7 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  28. R3C8 = 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 R3C8.
  29. R3C7 = 5 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  30. R7C7 = 3 · Naked SingleLooking at R7C7:- Row 7 already has 2, 4, 7, 8 and 9 → ruled out.- Column 7 already has 1, 5 and 6 → ruled out.So this cell can only be 3.
  31. R5C8 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 9 already has an 8, R5C9 cannot hold 8.So 8 must go in R5C8.
  32. R5C9 = 1 · Naked SingleLooking at R5C9:- Row 5 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  33. R9C9 = 6 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 4, 5, 7, 8 and 9 → ruled out.- Column 9 already has 3 → ruled out.So this cell can only be 6.
  34. 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.
  35. R8C1 = 4 · Naked SingleLooking at R8C1:- Row 8 already has 3, 5, 6, 8 and 9 → ruled out.- Column 1 already has 1, 2 and 7 → ruled out.So this cell can only be 4.
  36. R8C2 = 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, R7C2 cannot hold 7.So 7 must go in R8C2.
  37. R7C2 = 1 · Naked SingleLooking at R7C2:- Row 7 already has 2, 3, 4, 7, 8 and 9 → ruled out.- Column 2 already has 5 and 6 → ruled out.So this cell can only be 1.
  38. 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 cannot hold 2.So 2 must go in R8C4.
  39. R8C8 = 1 · Naked SingleLooking at R8C8:- Row 8 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  40. R7C4 = 6 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 3, 4, 7, 8 and 9 → ruled out.- Column 4 already has 5 → ruled out.So this cell can only be 6.
  41. R7C8 = 5 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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