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Daily Sudoku — 29 September 2026

The puzzle

  • 29 September 2026
  • Easy
  • 40 clues
Puzzle
1325····6
·······45
··5··71·2
3··1···6·
45·96·3··
8··7···51
2·8315···
69·8··513
51·4·627·
Solution
132549786
786231945
945687132
327154869
451968327
869723451
278315694
694872513
513496278

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. R3C2 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 4, R2C1, R2C2 and R2C3 cannot hold 4.- Because column 1 already has a 4, R3C1 cannot hold 4.So 4 must go in R3C2.
  2. R2C2 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 1 already has an 8, R2C1 and R3C1 cannot hold 8.- Because column 3 already has an 8, R2C3 cannot hold 8.So 8 must go in R2C2.
  3. R2C3 = 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, R2C1 and R3C1 cannot hold 6.So 6 must go in R2C3.
  4. R2C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 7, R3C1 cannot hold 7.So 7 must go in R2C1.
  5. R3C1 = 9 · Naked SingleLooking at R3C1:- Row 3 already has 1, 2, 4, 5 and 7 → ruled out.- Column 1 already has 3, 6 and 8 → ruled out.So this cell can only be 9.
  6. R2C6 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C5 and R1C6 cannot hold 1.- Because column 4 already has a 1, R2C4 cannot hold 1.- Because column 5 already has a 1, R2C5 cannot hold 1.- Because row 3 already has a 1, R3C4 and R3C5 cannot hold 1.So 1 must go in R2C6.
  7. R3C4 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C5 and R1C6 cannot hold 6.- Because row 2 already has a 6, R2C4 and R2C5 cannot hold 6.- Because column 5 already has a 6, R3C5 cannot hold 6.So 6 must go in R3C4.
  8. R2C4 = 2 · Naked SingleLooking at R2C4:- Row 2 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 4 already has 3 and 9 → ruled out.So this cell can only be 2.
  9. 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, R1C7 and R1C8 cannot hold 3.- Because column 7 already has a 3, R2C7 cannot hold 3.So 3 must go in R3C8.
  10. R3C5 = 8 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  11. 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, R1C5 and R1C6 cannot hold 3.So 3 must go in R2C5.
  12. R2C7 = 9 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  13. R1C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 7, R1C8 cannot hold 7.So 7 must go in R1C7.
  14. R1C8 = 8 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 3, 5, 6 and 7 → ruled out.- Column 8 already has 4 → ruled out.- The top-right 3×3 box already has 9 → ruled out.So this cell can only be 8.
  15. R5C3 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C2 and R4C3 cannot hold 1.- Because row 6 already has a 1, R6C2 and R6C3 cannot hold 1.So 1 must go in R5C3.
  16. R6C2 = 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, R4C2 and R4C3 cannot hold 6.- Because column 3 already has a 6, R6C3 cannot hold 6.So 6 must go in R6C2.
  17. R4C2 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 2, R4C3 and R6C3 cannot hold 2.So 2 must go in R4C2.
  18. R7C2 = 7 · Naked SingleLooking at R7C2:- Row 7 already has 1, 2, 3, 5 and 8 → ruled out.- Column 2 already has 4, 6 and 9 → ruled out.So this cell can only be 7.
  19. R4C3 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 7, R6C3 cannot hold 7.So 7 must go in R4C3.
  20. R6C3 = 9 · Naked SingleLooking at R6C3:- Row 6 already has 1, 5, 6, 7 and 8 → ruled out.- Column 3 already has 2 → ruled out.- The middle-left 3×3 box already has 3 and 4 → ruled out.So this cell can only be 9.
  21. R6C6 = 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, R4C5 and R4C6 cannot hold 3.- Because row 5 already has a 3, R5C6 cannot hold 3.- Because column 5 already has a 3, R6C5 cannot hold 3.So 3 must go in R6C6.
  22. R4C5 = 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 cannot hold 5.- Because row 5 already has a 5, R5C6 cannot hold 5.- Because row 6 already has a 5, R6C5 cannot hold 5.So 5 must go in R4C5.
  23. R5C8 = 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, R4C7 and R4C9 cannot hold 2.- Because column 9 already has a 2, R5C9 cannot hold 2.- Because column 7 already has a 2, R6C7 cannot hold 2.So 2 must go in R5C8.
  24. R7C8 = 9 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 5, 7 and 8 → ruled out.- Column 8 already has 4 and 6 → ruled out.So this cell can only be 9.
  25. 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, R5C6 cannot hold 2.So 2 must go in R6C5.
  26. 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.
  27. R4C6 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 4, R5C6 cannot hold 4.So 4 must go in R4C6.
  28. R5C6 = 8 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 3, 4, 5, 6 and 9 → ruled out.- Column 6 already has 7 → ruled out.So this cell can only be 8.
  29. R5C9 = 7 · Naked SingleLooking at R5C9:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  30. R1C5 = 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 R1C5.
  31. R1C6 = 9 · Naked SingleLooking at R1C6:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  32. R8C6 = 2 · Naked SingleLooking at R8C6:- Row 8 already has 1, 3, 5, 6, 8 and 9 → ruled out.- Column 6 already has 4 and 7 → ruled out.So this cell can only be 2.
  33. R4C9 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 9, R4C7 cannot hold 9.So 9 must go in R4C9.
  34. R4C7 = 8 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  35. R7C7 = 6 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 5, 7, 8 and 9 → ruled out.- Column 7 already has 4 → ruled out.So this cell can only be 6.
  36. R7C9 = 4 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  37. R9C9 = 8 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 4, 5, 6 and 7 → ruled out.- Column 9 already has 3 and 9 → ruled out.So this cell can only be 8.
  38. R9C3 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 3, R8C3 cannot hold 3.So 3 must go in R9C3.
  39. R9C5 = 9 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  40. R8C3 = 4 · Naked SingleLooking at R8C3:- Row 8 already has 1, 2, 3, 5, 6, 8 and 9 → ruled out.- Column 3 already has 7 → ruled out.So this cell can only be 4.
  41. R8C5 = 7 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.

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