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Daily Sudoku — 13 October 2026

The puzzle

  • 13 October 2026
  • Easy
  • 40 clues
Puzzle
·4·6·219·
··8··76·2
·5···8·4·
5·6·71423
··1423···
4·3·8·9··
73······5
81··39··4
·6·815··9
Solution
347652198
198347652
652198347
586971423
971423586
423586971
739264815
815739264
264815739

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. R3C1 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C1 and R1C3 cannot hold 6.- Because row 2 already has a 6, R2C1 and R2C2 cannot hold 6.- Because column 3 already has a 6, R3C3 cannot hold 6.So 6 must go in R3C1.
  2. R2C1 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C1 and R1C3 cannot hold 1.- Because column 2 already has a 1, R2C2 cannot hold 1.- Because column 3 already has a 1, R3C3 cannot hold 1.So 1 must go in R2C1.
  3. R3C3 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C1 and R1C3 cannot hold 2.- Because row 2 already has a 2, R2C2 cannot hold 2.So 2 must go in R3C3.
  4. R1C1 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 3, R1C3 cannot hold 3.- Because column 2 already has a 3, R2C2 cannot hold 3.So 3 must go in R1C1.
  5. R1C3 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 7, R2C2 cannot hold 7.So 7 must go in R1C3.
  6. R2C2 = 9 · Naked SingleLooking at R2C2:- Row 2 already has 1, 2, 6, 7 and 8 → ruled out.- Column 2 already has 3, 4 and 5 → ruled out.So this cell can only be 9.
  7. R3C4 = 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 cannot hold 1.- Because row 2 already has a 1, R2C4 and R2C5 cannot hold 1.- Because column 5 already has a 1, R3C5 cannot hold 1.So 1 must go in R3C4.
  8. 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, R1C5 cannot hold 3.- Because column 5 already has a 3, R2C5 and R3C5 cannot hold 3.So 3 must go in R2C4.
  9. R2C5 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C5 cannot hold 4.- Because row 3 already has a 4, R3C5 cannot hold 4.So 4 must go in R2C5.
  10. R2C8 = 5 · Naked SingleLooking at R2C8:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  11. R1C5 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 5, R3C5 cannot hold 5.So 5 must go in R1C5.
  12. R1C9 = 8 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  13. R3C5 = 9 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 4, 5, 6 and 8 → ruled out.- Column 5 already has 3 and 7 → ruled out.So this cell can only be 9.
  14. R7C5 = 6 · Naked SingleLooking at R7C5:- Row 7 already has 3, 5 and 7 → ruled out.- Column 5 already has 1, 2, 4, 8 and 9 → ruled out.So this cell can only be 6.
  15. R3C7 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R3C9 cannot hold 3.So 3 must go in R3C7.
  16. R3C9 = 7 · Naked SingleLooking at R3C9:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  17. R6C2 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C2 cannot hold 2.- Because row 5 already has a 2, R5C1 and R5C2 cannot hold 2.So 2 must go in R6C2.
  18. R5C2 = 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.- Because column 1 already has a 7, R5C1 cannot hold 7.So 7 must go in R5C2.
  19. R4C2 = 8 · Naked SingleLooking at R4C2:- Row 4 already has 1, 2, 3, 4, 5, 6 and 7 → ruled out.- Column 2 already has 9 → ruled out.So this cell can only be 8.
  20. R4C4 = 9 · Naked SingleLooking at R4C4:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  21. R5C1 = 9 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 4 and 7 → ruled out.- Column 1 already has 5, 6 and 8 → ruled out.So this cell can only be 9.
  22. R9C1 = 2 · Naked SingleLooking at R9C1:- Row 9 already has 1, 5, 6, 8 and 9 → ruled out.- Column 1 already has 3, 4 and 7 → ruled out.So this cell can only be 2.
  23. R6C4 = 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, R6C6 cannot hold 5.So 5 must go in R6C4.
  24. R6C6 = 6 · Naked SingleLooking at R6C6:- Row 6 already has 2, 3, 4, 5, 8 and 9 → ruled out.- Column 6 already has 1 and 7 → ruled out.So this cell can only be 6.
  25. R7C6 = 4 · Naked SingleLooking at R7C6:- Row 7 already has 3, 5, 6 and 7 → ruled out.- Column 6 already has 1, 2, 8 and 9 → ruled out.So this cell can only be 4.
  26. R5C7 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 5, R5C8 cannot hold 5.- Because column 9 already has a 5, R5C9 cannot hold 5.- Because row 6 already has a 5, R6C8 and R6C9 cannot hold 5.So 5 must go in R5C7.
  27. R6C8 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 7, R5C8 and R5C9 cannot hold 7.- Because column 9 already has a 7, R6C9 cannot hold 7.So 7 must go in R6C8.
  28. R6C9 = 1 · Naked SingleLooking at R6C9:- Row 6 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  29. R5C9 = 6 · Naked SingleLooking at R5C9:- Row 5 already has 1, 2, 3, 4, 5, 7 and 9 → ruled out.- Column 9 already has 8 → ruled out.So this cell can only be 6.
  30. R5C8 = 8 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  31. R9C3 = 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, R7C3 cannot hold 4.- Because row 8 already has a 4, R8C3 cannot hold 4.So 4 must go in R9C3.
  32. R8C3 = 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, R7C3 cannot hold 5.So 5 must go in R8C3.
  33. R7C3 = 9 · Naked SingleLooking at R7C3:- Row 7 already has 3, 4, 5, 6 and 7 → ruled out.- Column 3 already has 1, 2 and 8 → ruled out.So this cell can only be 9.
  34. R8C4 = 7 · Hidden SingleIn the bottom-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 7, R7C4 cannot hold 7.So 7 must go in R8C4.
  35. R7C4 = 2 · Naked SingleLooking at R7C4:- Row 7 already has 3, 4, 5, 6, 7 and 9 → ruled out.- Column 4 already has 1 and 8 → ruled out.So this cell can only be 2.
  36. R7C8 = 1 · Hidden SingleIn the bottom-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 1, R7C7 cannot hold 1.- Because row 8 already has a 1, R8C7 and R8C8 cannot hold 1.- Because row 9 already has a 1, R9C7 and R9C8 cannot hold 1.So 1 must go in R7C8.
  37. R7C7 = 8 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  38. R8C7 = 2 · Hidden SingleIn the bottom-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 2, R8C8 cannot hold 2.- Because row 9 already has a 2, R9C7 and R9C8 cannot hold 2.So 2 must go in R8C7.
  39. R8C8 = 6 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  40. R9C7 = 7 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 4, 5, 6, 8 and 9 → ruled out.- Column 7 already has 3 → ruled out.So this cell can only be 7.
  41. R9C8 = 3 · Naked SingleLooking at R9C8:- Row 9 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.

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