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Daily Sudoku — 7 October 2026

The puzzle

  • 7 October 2026
  • Hard
  • 30 clues
Puzzle
94······1
···1···4·
7·····836
·6·8···57
4·7······
·1···536·
···4··183
1··2··694
·9·····7·
Solution
945683721
836172945
721594836
369821457
457936218
218745369
572469183
183257694
694318572

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 — 51 steps using 3 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
  • Locked Candidates (pointing/claiming) — a digit inside a box is confined to one row or column, so it is eliminated from the rest of that row or column
  1. R3C3 = 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, R1C3 cannot hold 1.- Because row 2 already has a 1, R2C1, R2C2 and R2C3 cannot hold 1.- Because column 2 already has a 1, R3C2 cannot hold 1.So 1 must go in R3C3.
  2. R5C2 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C1 and R4C3 cannot hold 5.- Because row 6 already has a 5, R6C1 and R6C3 cannot hold 5.So 5 must go in R5C2.
  3. R5C8 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 1, R4C7 and R5C7 cannot hold 1.- Because column 9 already has a 1, R5C9 cannot hold 1.- Because row 6 already has a 1, R6C9 cannot hold 1.So 1 must go in R5C8.
  4. R1C8 = 2 · Naked SingleLooking at R1C8:- Row 1 already has 1, 4 and 9 → ruled out.- Column 8 already has 3, 5, 6, 7 and 8 → ruled out.So this cell can only be 2.
  5. R4C7 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 4, R5C7 and R5C9 cannot hold 4.- Because column 9 already has a 4, R6C9 cannot hold 4.So 4 must go in R4C7.
  6. R6C5 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C5 and R4C6 cannot hold 4.- Because row 5 already has a 4, R5C4, R5C5 and R5C6 cannot hold 4.- Because column 4 already has a 4, R6C4 cannot hold 4.So 4 must go in R6C5.
  7. R3C6 = 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, R1C4, R1C5 and R1C6 cannot hold 4.- Because row 2 already has a 4, R2C5 and R2C6 cannot hold 4.- Because column 4 already has a 4, R3C4 cannot hold 4.- Because column 5 already has a 4, R3C5 cannot hold 4.So 4 must go in R3C6.
  8. R6C4 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C5 and R4C6 cannot hold 7.- Because row 5 already has a 7, R5C4, R5C5 and R5C6 cannot hold 7.So 7 must go in R6C4.
  9. 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, R7C1, R7C2 and R7C3 cannot hold 4.- Because row 8 already has a 4, R8C2 and R8C3 cannot hold 4.- Because column 1 already has a 4, R9C1 cannot hold 4.So 4 must go in R9C3.
  10. R5C9 = 8 · Hidden SingleIn row 5, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has an 8, R5C4 cannot hold 8.- Because the centre 3×3 box already has an 8, R5C5 and R5C6 cannot hold 8.- Because column 7 already has an 8, R5C7 cannot hold 8.So 8 must go in R5C9.
  11. R3C2 = 2 · Naked SingleLooking at R3C2:- Row 3 already has 1, 3, 4, 6, 7 and 8 → ruled out.- Column 2 already has 5 and 9 → ruled out.So this cell can only be 2.
  12. R7C2 = 7 · Naked SingleLooking at R7C2:- Row 7 already has 1, 3, 4 and 8 → ruled out.- Column 2 already has 2, 5, 6 and 9 → ruled out.So this cell can only be 7.
  13. R3C4 = 5 · Hidden SingleStep 1 · Pick a digit and a box, and find where it can goThe bottom-right 3×3 box has no 5 yet; 5 can only go in R9C7 and R9C9.Step 2 · Places on one row: point along it (pointing)All of them lie on row 9, so the box's 5 is on row 9. The rest of row 9 loses 5: R9C1, R9C4 and R9C5.Step 3 · Places in one box: clear the rest of it (claiming)Row 3 has 5 only in R3C4 and R3C5, all in the top-centre 3×3 box. The rest of that box loses 5: R1C4, R1C5 and R2C5.Step 4 · Collect what the eliminations unlockIn column 4, 5 has one place left, R3C4 (lost 5: R1C4 in step 3 and R9C4 in step 2) → place 5 in R3C4.
  14. R3C5 = 9 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  15. R7C6 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 9, R7C5 cannot hold 9.- Because row 8 already has a 9, R8C5 and R8C6 cannot hold 9.- Because row 9 already has a 9, R9C4, R9C5 and R9C6 cannot hold 9.So 9 must go in R7C6.
  16. R5C4 = 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 and R5C5 cannot hold 9.- Because column 6 already has a 9, R4C6 and R5C6 cannot hold 9.So 9 must go in R5C4.
  17. R6C9 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C7 cannot hold 9.So 9 must go in R6C9.
  18. R5C7 = 2 · Naked SingleLooking at R5C7:- Row 5 already has 1, 4, 5, 7, 8 and 9 → ruled out.- Column 7 already has 3 and 6 → ruled out.So this cell can only be 2.
  19. R2C7 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C7 cannot hold 9.- Because column 9 already has a 9, R2C9 cannot hold 9.So 9 must go in R2C7.
  20. R1C7 = 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.So 7 must go in R1C7.
  21. R9C7 = 5 · Naked SingleLooking at R9C7:- Row 9 already has 4, 7 and 9 → ruled out.- Column 7 already has 1, 2, 3, 6 and 8 → ruled out.So this cell can only be 5.
  22. R2C9 = 5 · Naked SingleLooking at R2C9:- Row 2 already has 1, 4 and 9 → ruled out.- Column 9 already has 3, 6, 7 and 8 → ruled out.- The top-right 3×3 box already has 2 → ruled out.So this cell can only be 5.
  23. R9C9 = 2 · Naked SingleLooking at R9C9:- Row 9 already has 4, 5, 7 and 9 → ruled out.- Column 9 already has 1, 3, 6 and 8 → ruled out.So this cell can only be 2.
  24. R1C3 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C1, R2C2 and R2C3 cannot hold 5.So 5 must go in R1C3.
  25. R4C3 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 9, R4C1 cannot hold 9.- Because row 6 already has a 9, R6C1 and R6C3 cannot hold 9.So 9 must go in R4C3.
  26. R4C1 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 3, R6C1 and R6C3 cannot hold 3.So 3 must go in R4C1.
  27. R7C1 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 5, R7C3 and R8C3 cannot hold 5.- Because column 2 already has a 5, R8C2 cannot hold 5.- Because row 9 already has a 5, R9C1 cannot hold 5.So 5 must go in R7C1.
  28. R7C3 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 2, R8C2 and R8C3 cannot hold 2.- Because row 9 already has a 2, R9C1 cannot hold 2.So 2 must go in R7C3.
  29. R7C5 = 6 · Naked SingleLooking at R7C5:- Row 7 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  30. R6C1 = 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, R6C3 cannot hold 2.So 2 must go in R6C1.
  31. R6C3 = 8 · Naked SingleLooking at R6C3:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  32. R5C6 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 6, R4C5 and R4C6 cannot hold 6.- Because column 5 already has a 6, R5C5 cannot hold 6.So 6 must go in R5C6.
  33. R5C5 = 3 · Naked SingleLooking at R5C5:- Row 5 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  34. R1C4 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 6, R1C5 and R2C5 cannot hold 6.- Because column 6 already has a 6, R1C6 and R2C6 cannot hold 6.So 6 must go in R1C4.
  35. R9C4 = 3 · Naked SingleLooking at R9C4:- Row 9 already has 2, 4, 5, 7 and 9 → ruled out.- Column 4 already has 1, 6 and 8 → ruled out.So this cell can only be 3.
  36. R9C1 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 6, R8C2 and R8C3 cannot hold 6.So 6 must go in R9C1.
  37. R2C1 = 8 · Naked SingleLooking at R2C1:- Row 2 already has 1, 4, 5 and 9 → ruled out.- Column 1 already has 2, 3, 6 and 7 → ruled out.So this cell can only be 8.
  38. R2C3 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 6, R2C2 cannot hold 6.So 6 must go in R2C3.
  39. R8C3 = 3 · Naked SingleLooking at R8C3:- Row 8 already has 1, 2, 4, 6 and 9 → ruled out.- Column 3 already has 5, 7 and 8 → ruled out.So this cell can only be 3.
  40. R2C2 = 3 · Naked SingleLooking at R2C2:- Row 2 already has 1, 4, 5, 6, 8 and 9 → ruled out.- Column 2 already has 2 and 7 → ruled out.So this cell can only be 3.
  41. R8C2 = 8 · Naked SingleLooking at R8C2:- Row 8 already has 1, 2, 3, 4, 6 and 9 → ruled out.- Column 2 already has 5 and 7 → ruled out.So this cell can only be 8.
  42. R1C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R1C5 cannot hold 3.- Because row 2 already has a 3, R2C5 and R2C6 cannot hold 3.So 3 must go in R1C6.
  43. R1C5 = 8 · Naked SingleLooking at R1C5:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  44. R8C5 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 5, R8C6 cannot hold 5.- Because row 9 already has a 5, R9C5 and R9C6 cannot hold 5.So 5 must go in R8C5.
  45. R8C6 = 7 · Naked SingleLooking at R8C6:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  46. R2C5 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 7, R2C6 cannot hold 7.So 7 must go in R2C5.
  47. R2C6 = 2 · Naked SingleLooking at R2C6:- Row 2 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  48. 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.So 2 must go in R4C5.
  49. R4C6 = 1 · Naked SingleLooking at R4C6:- Row 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  50. R9C5 = 1 · Naked SingleLooking at R9C5:- Row 9 already has 2, 3, 4, 5, 6, 7 and 9 → ruled out.- Column 5 already has 8 → ruled out.So this cell can only be 1.
  51. R9C6 = 8 · Naked SingleLooking at R9C6:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.

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