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Daily Sudoku — 14 October 2026

The puzzle

  • 14 October 2026
  • Easy
  • 40 clues
Puzzle
5·36··7··
8·6427···
7·4·1·869
3·21·6···
·89······
·51·843··
1···6·284
248····56
··58··1·3
Solution
513698742
896427531
724315869
372156498
489273615
651984327
137569284
248731956
965842173

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

  • Naked Single — every other digit is already used in this cell's row, column or box, so only one candidate is left
  • Hidden Single — inside one row, column or box this digit fits in only one cell, so it must go there
  1. R7C3 = 7 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 4, 6 and 8 → ruled out.- Column 3 already has 3, 5 and 9 → ruled out.So this cell can only be 7.
  2. R1C6 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has an 8, R1C5 cannot hold 8.- Because row 3 already has an 8, R3C4 and R3C6 cannot hold 8.So 8 must go in R1C6.
  3. R1C5 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 9, R3C4 and R3C6 cannot hold 9.So 9 must go in R1C5.
  4. R2C2 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C2 cannot hold 9.- Because row 3 already has a 9, R3C2 cannot hold 9.So 9 must go in R2C2.
  5. R1C2 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 1, R3C2 cannot hold 1.So 1 must go in R1C2.
  6. R3C2 = 2 · Naked SingleLooking at R3C2:- Row 3 already has 1, 4, 6, 7, 8 and 9 → ruled out.- Column 2 already has 5 → ruled out.- The top-left 3×3 box already has 3 → ruled out.So this cell can only be 2.
  7. R2C8 = 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, R1C8 and R1C9 cannot hold 3.- Because column 7 already has a 3, R2C7 cannot hold 3.- Because column 9 already has a 3, R2C9 cannot hold 3.So 3 must go in R2C8.
  8. R2C9 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 1, R1C8 and R1C9 cannot hold 1.- Because column 7 already has a 1, R2C7 cannot hold 1.So 1 must go in R2C9.
  9. R2C7 = 5 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  10. R1C8 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 4, R1C9 cannot hold 4.So 4 must go in R1C8.
  11. R1C9 = 2 · Naked SingleLooking at R1C9:- Row 1 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  12. R5C1 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 4, R4C2 cannot hold 4.- Because row 6 already has a 4, R6C1 cannot hold 4.So 4 must go in R5C1.
  13. R6C1 = 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 cannot hold 6.So 6 must go in R6C1.
  14. R9C1 = 9 · Naked SingleLooking at R9C1:- Row 9 already has 1, 3, 5 and 8 → ruled out.- Column 1 already has 2, 4, 6 and 7 → ruled out.So this cell can only be 9.
  15. R4C2 = 7 · Naked SingleLooking at R4C2:- Row 4 already has 1, 2, 3 and 6 → ruled out.- Column 2 already has 4, 5, 8 and 9 → ruled out.So this cell can only be 7.
  16. R6C4 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 9, R4C5 cannot hold 9.- Because row 5 already has a 9, R5C4, R5C5 and R5C6 cannot hold 9.So 9 must go in R6C4.
  17. R5C8 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C7, R4C8 and R4C9 cannot hold 1.- Because column 7 already has a 1, R5C7 cannot hold 1.- Because column 9 already has a 1, R5C9 cannot hold 1.- Because row 6 already has a 1, R6C8 and R6C9 cannot hold 1.So 1 must go in R5C8.
  18. 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, R4C7, R4C8 and R4C9 cannot hold 2.- Because column 7 already has a 2, R5C7 cannot hold 2.- Because column 9 already has a 2, R5C9 and R6C9 cannot hold 2.So 2 must go in R6C8.
  19. 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.
  20. R4C7 = 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 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 R4C7.
  21. R5C7 = 6 · Hidden SingleIn the middle-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C8 and R4C9 cannot hold 6.- Because column 9 already has a 6, R5C9 cannot hold 6.So 6 must go in R5C7.
  22. R8C7 = 9 · Naked SingleLooking at R8C7:- Row 8 already has 2, 4, 5, 6 and 8 → ruled out.- Column 7 already has 1, 3 and 7 → ruled out.So this cell can only be 9.
  23. R9C8 = 7 · Naked SingleLooking at R9C8:- Row 9 already has 1, 3, 5, 8 and 9 → ruled out.- Column 8 already has 2, 4 and 6 → ruled out.So this cell can only be 7.
  24. R4C8 = 9 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 4, 6 and 7 → ruled out.- Column 8 already has 5 and 8 → ruled out.So this cell can only be 9.
  25. R4C9 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 5 already has an 8, R5C9 cannot hold 8.So 8 must go in R4C9.
  26. R4C5 = 5 · Naked SingleLooking at R4C5:- Row 4 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  27. R5C9 = 5 · Naked SingleLooking at R5C9:- Row 5 already has 1, 4, 6, 8 and 9 → ruled out.- Column 9 already has 2, 3 and 7 → ruled out.So this cell can only be 5.
  28. R7C2 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 3, R9C2 cannot hold 3.So 3 must go in R7C2.
  29. R9C2 = 6 · Naked SingleLooking at R9C2:- Row 9 already has 1, 3, 5, 7, 8 and 9 → ruled out.- Column 2 already has 2 and 4 → ruled out.So this cell can only be 6.
  30. R8C6 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C4 and R7C6 cannot hold 1.- Because column 4 already has a 1, R8C4 cannot hold 1.- Because column 5 already has a 1, R8C5 cannot hold 1.- Because row 9 already has a 1, R9C5 and R9C6 cannot hold 1.So 1 must go in R8C6.
  31. R9C6 = 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 and R7C6 cannot hold 2.- Because row 8 already has a 2, R8C4 and R8C5 cannot hold 2.- Because column 5 already has a 2, R9C5 cannot hold 2.So 2 must go in R9C6.
  32. R9C5 = 4 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  33. R5C4 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 2, R5C5 cannot hold 2.- Because column 6 already has a 2, R5C6 cannot hold 2.So 2 must go in R5C4.
  34. R5C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 7, R5C6 cannot hold 7.So 7 must go in R5C5.
  35. 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.
  36. R8C5 = 3 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 4, 5, 6, 8 and 9 → ruled out.- Column 5 already has 7 → ruled out.So this cell can only be 3.
  37. R8C4 = 7 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  38. R3C4 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 3, R3C6 cannot hold 3.So 3 must go in R3C4.
  39. R3C6 = 5 · Naked SingleLooking at R3C6:- Row 3 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  40. R7C4 = 5 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 3, 4, 6, 7 and 8 → ruled out.- Column 4 already has 9 → ruled out.So this cell can only be 5.
  41. R7C6 = 9 · Naked SingleLooking at R7C6:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.

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