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Daily Sudoku — 5 October 2026

The puzzle

  • 5 October 2026
  • Easy
  • 40 clues
Puzzle
397··4···
·1·265·3·
2653·7···
···9·82··
142·····7
9·8142·6·
··1·2·953
4······7·
·397·164·
Solution
397814526
814265739
265397481
653978214
142653897
978142365
781426953
426539178
539781642

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. R2C3 = 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, R2C1 cannot hold 4.So 4 must go in R2C3.
  2. R2C1 = 8 · Naked SingleLooking at R2C1:- Row 2 already has 1, 2, 3, 4, 5 and 6 → ruled out.- Column 1 already has 9 → ruled out.- The top-left 3×3 box already has 7 → ruled out.So this cell can only be 8.
  3. R3C5 = 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 R1C5 cannot hold 9.So 9 must go in R3C5.
  4. R1C5 = 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, R1C4 cannot hold 1.So 1 must go in R1C5.
  5. R1C4 = 8 · Naked SingleLooking at R1C4:- Row 1 already has 1, 3, 4, 7 and 9 → ruled out.- Column 4 already has 2 → ruled out.- The top-centre 3×3 box already has 5 and 6 → ruled out.So this cell can only be 8.
  6. R1C9 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 6, R1C7 cannot hold 6.- Because column 8 already has a 6, R1C8 cannot hold 6.- Because row 2 already has a 6, R2C7 and R2C9 cannot hold 6.- Because row 3 already has a 6, R3C7, R3C8 and R3C9 cannot hold 6.So 6 must go in R1C9.
  7. R1C8 = 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 cannot hold 2.- Because row 2 already has a 2, R2C7 and R2C9 cannot hold 2.- Because row 3 already has a 2, R3C7, R3C8 and R3C9 cannot hold 2.So 2 must go in R1C8.
  8. R1C7 = 5 · Naked SingleLooking at R1C7:- Row 1 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  9. R2C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 7, R2C9 cannot hold 7.- Because row 3 already has a 7, R3C7, R3C8 and R3C9 cannot hold 7.So 7 must go in R2C7.
  10. R2C9 = 9 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  11. R4C3 = 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, R4C1 cannot hold 3.- Because column 2 already has a 3, R4C2 and R6C2 cannot hold 3.So 3 must go in R4C3.
  12. R8C3 = 6 · Naked SingleLooking at R8C3:- Row 8 already has 4 and 7 → ruled out.- Column 3 already has 1, 2, 3, 5, 8 and 9 → ruled out.So this cell can only be 6.
  13. R4C1 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 6, R4C2 cannot hold 6.- Because row 6 already has a 6, R6C2 cannot hold 6.So 6 must go in R4C1.
  14. R4C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 7, R5C4, R5C5 and R5C6 cannot hold 7.So 7 must go in R4C5.
  15. R6C2 = 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, R4C2 cannot hold 7.So 7 must go in R6C2.
  16. R4C2 = 5 · Naked SingleLooking at R4C2:- Row 4 already has 2, 3, 6, 7, 8 and 9 → ruled out.- Column 2 already has 1 and 4 → ruled out.So this cell can only be 5.
  17. R4C9 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 4, R4C8 cannot hold 4.- Because row 5 already has a 4, R5C7 and R5C8 cannot hold 4.- Because row 6 already has a 4, R6C7 and R6C9 cannot hold 4.So 4 must go in R4C9.
  18. R4C8 = 1 · Naked SingleLooking at R4C8:- Row 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  19. R3C7 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 4, R3C8 cannot hold 4.- Because column 9 already has a 4, R3C9 cannot hold 4.So 4 must go in R3C7.
  20. R3C9 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 1, R3C8 cannot hold 1.So 1 must go in R3C9.
  21. R3C8 = 8 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  22. R5C8 = 9 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 4 and 7 → ruled out.- Column 8 already has 3, 5, 6 and 8 → ruled out.So this cell can only be 9.
  23. R6C9 = 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 and R6C7 cannot hold 5.So 5 must go in R6C9.
  24. R6C7 = 3 · Naked SingleLooking at R6C7:- Row 6 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  25. R5C7 = 8 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 4, 7 and 9 → ruled out.- Column 7 already has 3, 5 and 6 → ruled out.So this cell can only be 8.
  26. R8C7 = 1 · Naked SingleLooking at R8C7:- Row 8 already has 4, 6 and 7 → ruled out.- Column 7 already has 2, 3, 5, 8 and 9 → ruled out.So this cell can only be 1.
  27. R8C2 = 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 R7C2 cannot hold 2.- Because column 1 already has a 2, R9C1 cannot hold 2.So 2 must go in R8C2.
  28. R7C2 = 8 · Naked SingleLooking at R7C2:- Row 7 already has 1, 2, 3, 5 and 9 → ruled out.- Column 2 already has 4, 6 and 7 → ruled out.So this cell can only be 8.
  29. R9C1 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C1 cannot hold 5.So 5 must go in R9C1.
  30. R7C1 = 7 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 3, 5, 8 and 9 → ruled out.- Column 1 already has 4 and 6 → ruled out.So this cell can only be 7.
  31. R7C4 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 4, R7C6 cannot hold 4.- Because row 8 already has a 4, R8C4, R8C5 and R8C6 cannot hold 4.- Because row 9 already has a 4, R9C5 cannot hold 4.So 4 must go in R7C4.
  32. R7C6 = 6 · Naked SingleLooking at R7C6:- Row 7 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  33. R5C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 6, R5C5 cannot hold 6.- Because column 6 already has a 6, R5C6 cannot hold 6.So 6 must go in R5C4.
  34. R8C4 = 5 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 4, 6 and 7 → ruled out.- Column 4 already has 3, 8 and 9 → ruled out.So this cell can only be 5.
  35. R5C5 = 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, R5C6 cannot hold 5.So 5 must go in R5C5.
  36. R5C6 = 3 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  37. R8C6 = 9 · Naked SingleLooking at R8C6:- Row 8 already has 1, 2, 4, 5, 6 and 7 → ruled out.- Column 6 already has 3 and 8 → ruled out.So this cell can only be 9.
  38. R8C5 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 3, R9C5 cannot hold 3.So 3 must go in R8C5.
  39. 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.
  40. R9C5 = 8 · Naked SingleLooking at R9C5:- Row 9 already has 1, 3, 4, 5, 6, 7 and 9 → ruled out.- Column 5 already has 2 → ruled out.So this cell can only be 8.
  41. R9C9 = 2 · Naked SingleLooking at R9C9:- Row 9 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.

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