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Daily Sudoku — 24 September 2026

The puzzle

  • 24 September 2026
  • Easy
  • 40 clues
Puzzle
···79·2··
2564·1···
79·2·6·3·
··56·····
·23149·7·
149·75··3
36······2
9····2·64
58·364·1·
Solution
431798256
256431798
798256431
875623149
623149875
149875623
364917582
917582364
582364917

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. R1C2 = 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 column 3 already has a 3, R1C3 cannot hold 3.- Because row 3 already has a 3, R3C3 cannot hold 3.So 3 must go in R1C2.
  2. R2C5 = 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, R1C6 cannot hold 3.- Because row 3 already has a 3, R3C5 cannot hold 3.So 3 must go in R2C5.
  3. R3C5 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 5, R1C6 cannot hold 5.So 5 must go in R3C5.
  4. R1C6 = 8 · Naked SingleLooking at R1C6:- Row 1 already has 2, 3, 7 and 9 → ruled out.- Column 6 already has 1, 4, 5 and 6 → ruled out.So this cell can only be 8.
  5. R3C3 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has an 8, R1C1 and R1C3 cannot hold 8.So 8 must go in R3C3.
  6. R1C3 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 1, R1C1 cannot hold 1.So 1 must go in R1C3.
  7. R1C1 = 4 · Naked SingleLooking at R1C1:- Row 1 already has 1, 2, 3, 7, 8 and 9 → ruled out.- Column 1 already has 5 → ruled out.- The top-left 3×3 box already has 6 → ruled out.So this cell can only be 4.
  8. R3C7 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C8 and R1C9 cannot hold 4.- Because row 2 already has a 4, R2C7, R2C8 and R2C9 cannot hold 4.- Because column 9 already has a 4, R3C9 cannot hold 4.So 4 must go in R3C7.
  9. R3C9 = 1 · Naked SingleLooking at R3C9:- Row 3 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  10. R1C9 = 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, R2C8 and R2C9 cannot hold 6.So 6 must go in R1C9.
  11. R1C8 = 5 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  12. R5C1 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C1 and R4C2 cannot hold 6.So 6 must go in R5C1.
  13. R4C1 = 8 · Naked SingleLooking at R4C1:- Row 4 already has 5 and 6 → ruled out.- Column 1 already has 1, 2, 3, 4, 7 and 9 → ruled out.So this cell can only be 8.
  14. R4C2 = 7 · Naked SingleLooking at R4C2:- Row 4 already has 5, 6 and 8 → ruled out.- Column 2 already has 2, 3, 4 and 9 → ruled out.- The middle-left 3×3 box already has 1 → ruled out.So this cell can only be 7.
  15. R8C2 = 1 · Naked SingleLooking at R8C2:- Row 8 already has 2, 4, 6 and 9 → ruled out.- Column 2 already has 3, 5, 7 and 8 → ruled out.So this cell can only be 1.
  16. R4C5 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 2, R4C6 cannot hold 2.- Because column 4 already has a 2, R6C4 cannot hold 2.So 2 must go in R4C5.
  17. R4C6 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 3, R6C4 cannot hold 3.So 3 must go in R4C6.
  18. R7C6 = 7 · Naked SingleLooking at R7C6:- Row 7 already has 2, 3 and 6 → ruled out.- Column 6 already has 1, 4, 5, 8 and 9 → ruled out.So this cell can only be 7.
  19. R6C4 = 8 · Naked SingleLooking at R6C4:- Row 6 already has 1, 3, 4, 5, 7 and 9 → ruled out.- Column 4 already has 2 and 6 → ruled out.So this cell can only be 8.
  20. R4C7 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 1, R4C8 cannot hold 1.- Because column 9 already has a 1, R4C9 cannot hold 1.- Because row 5 already has a 1, R5C7 and R5C9 cannot hold 1.- Because row 6 already has a 1, R6C7 and R6C8 cannot hold 1.So 1 must go in R4C7.
  21. R6C8 = 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 row 5 already has a 2, R5C7 and R5C9 cannot hold 2.- Because column 7 already has a 2, R6C7 cannot hold 2.So 2 must go in R6C8.
  22. R6C7 = 6 · Naked SingleLooking at R6C7:- Row 6 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  23. R4C8 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 4, R4C9 cannot hold 4.- Because row 5 already has a 4, R5C7 and R5C9 cannot hold 4.So 4 must go in R4C8.
  24. R4C9 = 9 · Naked SingleLooking at R4C9:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  25. 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, R7C3 cannot hold 2.- Because row 8 already has a 2, R8C3 cannot hold 2.So 2 must go in R9C3.
  26. R7C3 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 4, R8C3 cannot hold 4.So 4 must go in R7C3.
  27. R8C3 = 7 · Naked SingleLooking at R8C3:- Row 8 already has 1, 2, 4, 6 and 9 → ruled out.- Column 3 already has 3, 5 and 8 → ruled out.So this cell can only be 7.
  28. R7C5 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 1, R7C4 cannot hold 1.- Because row 8 already has a 1, R8C4 and R8C5 cannot hold 1.So 1 must go in R7C5.
  29. R8C5 = 8 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 4, 6, 7 and 9 → ruled out.- Column 5 already has 3 and 5 → ruled out.So this cell can only be 8.
  30. R7C4 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 9, R8C4 cannot hold 9.So 9 must go in R7C4.
  31. R8C4 = 5 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 4, 6, 7, 8 and 9 → ruled out.- Column 4 already has 3 → ruled out.So this cell can only be 5.
  32. R8C7 = 3 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  33. R7C7 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 5, R7C8 cannot hold 5.- Because row 9 already has a 5, R9C7 and R9C9 cannot hold 5.So 5 must go in R7C7.
  34. R7C8 = 8 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  35. R2C8 = 9 · Naked SingleLooking at R2C8:- Row 2 already has 1, 2, 3, 4, 5 and 6 → ruled out.- Column 8 already has 7 and 8 → ruled out.So this cell can only be 9.
  36. R5C9 = 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, R5C7 cannot hold 5.So 5 must go in R5C9.
  37. R5C7 = 8 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  38. 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.So 8 must go in R2C9.
  39. 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.
  40. R9C7 = 9 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 5, 6 and 8 → ruled out.- Column 7 already has 7 → ruled out.So this cell can only be 9.
  41. R9C9 = 7 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.

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