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Daily Sudoku — 2 October 2026

The puzzle

  • 2 October 2026
  • Hard
  • 30 clues
Puzzle
·9···2·85
6····8··9
45·7·162·
··49··3··
·8·····4·
····2·97·
·7·2·5·9·
··5·4····
8··1···5·
Solution
791632485
632458719
458791623
524987361
987316542
316524978
173265894
265849137
849173256

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
  • Naked Pair — two cells hold exactly the same two digits, so those digits are eliminated from every other cell in the unit
  1. R3C3 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has an 8, R1C1 and R1C3 cannot hold 8.- Because row 2 already has an 8, R2C2 and R2C3 cannot hold 8.So 8 must go in R3C3.
  2. 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.- Because row 2 already has a 9, R2C4 and R2C5 cannot hold 9.So 9 must go in R3C5.
  3. R3C9 = 3 · Naked SingleLooking at R3C9:- Row 3 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  4. 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, R4C8 cannot hold 5.- Because column 9 already has a 5, R4C9, R5C9 and R6C9 cannot hold 5.So 5 must go in R5C7.
  5. R9C2 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R7C1 cannot hold 4.- Because column 3 already has a 4, R7C3 and R9C3 cannot hold 4.- Because row 8 already has a 4, R8C1 and R8C2 cannot hold 4.So 4 must go in R9C2.
  6. R8C8 = 3 · Hidden SingleIn the bottom-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R7C7, R8C7 and R9C7 cannot hold 3.- Because column 9 already has a 3, R7C9, R8C9 and R9C9 cannot hold 3.So 3 must go in R8C8.
  7. R6C6 = 4 · Hidden SingleIn column 6, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C6 cannot hold 4.- Because row 5 already has a 4, R5C6 cannot hold 4.- Because row 8 already has a 4, R8C6 cannot hold 4.- Because row 9 already has a 4, R9C6 cannot hold 4.So 4 must go in R6C6.
  8. R4C8 = 6 · Hidden SingleIn column 8, 6 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 6, R2C8 cannot hold 6.So 6 must go in R4C8.
  9. R2C8 = 1 · Naked SingleLooking at R2C8:- Row 2 already has 6, 8 and 9 → ruled out.- Column 8 already has 2, 3, 4, 5 and 7 → ruled out.So this cell can only be 1.
  10. R7C9 = 4 · Hidden SingleIn column 9, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C9 cannot hold 4.- Because row 5 already has a 4, R5C9 cannot hold 4.- Because row 6 already has a 4, R6C9 cannot hold 4.- Because row 8 already has a 4, R8C9 cannot hold 4.- Because row 9 already has a 4, R9C9 cannot hold 4.So 4 must go in R7C9.
  11. R4C6 = 7 · Naked SingleLooking at R4C6:- Row 4 already has 3, 4, 6 and 9 → ruled out.- Column 6 already has 1, 2, 5 and 8 → ruled out.So this cell can only be 7.
  12. R9C5 = 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, R7C5 cannot hold 7.- Because column 4 already has a 7, R8C4 cannot hold 7.- Because column 6 already has a 7, R8C6 and R9C6 cannot hold 7.So 7 must go in R9C5.
  13. R8C9 = 7 · Hidden SingleIn column 9, 7 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 7, R4C9 cannot hold 7.- Because the middle-right 3×3 box already has a 7, R5C9 cannot hold 7.- Because row 6 already has a 7, R6C9 cannot hold 7.- Because row 9 already has a 7, R9C9 cannot hold 7.So 7 must go in R8C9.
  14. R9C9 = 6 · Hidden SingleIn the bottom-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 6, R7C7, R8C7 and R9C7 cannot hold 6.So 6 must go in R9C9.
  15. R9C7 = 2 · Naked SingleLooking at R9C7:- Row 9 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 7 already has 3 and 9 → ruled out.So this cell can only be 2.
  16. R5C5 = 1 · Naked SingleStep 1 · Find two cells in one unit with the same two candidatesIn row 5, R5C4 and R5C6 both hold only {3, 6}.Step 2 · See why the two digits are used up, and whereThese two cells must take 3 and 6, one each, so row 5 has no other place for 3 and 6.Step 3 · Remove both digits from the rest of the unitIn row 5: R5C1 loses 3; R5C3 loses 3 and 6; R5C5 loses 3 and 6.Step 4 · When the pair also shares a box, clear the box tooThe two cells also share the centre 3×3 box. In that box: R5C5 loses 3 and 6; R6C4 loses 3 and 6.Step 5 · Collect what the clean-up unlocksR5C5 has one candidate left, 1 (lost 3 in step 3 and 6 in step 3) → place 1 in R5C5.
  17. R5C9 = 2 · Naked SingleLooking at R5C9:- Row 5 already has 1, 4, 5 and 8 → ruled out.- Column 9 already has 3, 6, 7 and 9 → ruled out.So this cell can only be 2.
  18. R2C3 = 2 · Hidden SingleIn column 3, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C3 cannot hold 2.- Because row 5 already has a 2, R5C3 cannot hold 2.- Because row 6 already has a 2, R6C3 cannot hold 2.- Because row 7 already has a 2, R7C3 cannot hold 2.- Because row 9 already has a 2, R9C3 cannot hold 2.So 2 must go in R2C3.
  19. R2C7 = 7 · Hidden SingleIn row 2, 7 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 7, R2C2 cannot hold 7.- Because column 4 already has a 7, R2C4 cannot hold 7.- Because column 5 already has a 7, R2C5 cannot hold 7.So 7 must go in R2C7.
  20. R1C7 = 4 · Naked SingleLooking at R1C7:- Row 1 already has 2, 5, 8 and 9 → ruled out.- Column 7 already has 3, 6 and 7 → ruled out.- The top-right 3×3 box already has 1 → ruled out.So this cell can only be 4.
  21. R2C4 = 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 and R1C5 cannot hold 4.- Because column 5 already has a 4, R2C5 cannot hold 4.So 4 must go in R2C4.
  22. R2C5 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C4 and R1C5 cannot hold 5.So 5 must go in R2C5.
  23. R2C2 = 3 · Naked SingleLooking at R2C2:- Row 2 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  24. R6C4 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 5, R4C5 cannot hold 5.- Because row 5 already has a 5, R5C4 and R5C6 cannot hold 5.So 5 must go in R6C4.
  25. R4C1 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 5, R4C2 cannot hold 5.- Because row 5 already has a 5, R5C1 and R5C3 cannot hold 5.- Because row 6 already has a 5, R6C1, R6C2 and R6C3 cannot hold 5.So 5 must go in R4C1.
  26. R4C2 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 2, R5C1 and R5C3 cannot hold 2.- Because row 6 already has a 2, R6C1, R6C2 and R6C3 cannot hold 2.So 2 must go in R4C2.
  27. R4C5 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 5 already has an 8, R5C4 and R5C6 cannot hold 8.So 8 must go in R4C5.
  28. R4C9 = 1 · Naked SingleLooking at R4C9:- Row 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  29. R6C9 = 8 · Naked SingleLooking at R6C9:- Row 6 already has 2, 4, 5, 7 and 9 → ruled out.- Column 9 already has 1, 3 and 6 → ruled out.So this cell can only be 8.
  30. R8C1 = 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 R7C3 cannot hold 2.- Because column 2 already has a 2, R8C2 cannot hold 2.- Because row 9 already has a 2, R9C3 cannot hold 2.So 2 must go in R8C1.
  31. R9C3 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 9, R7C1 and R7C3 cannot hold 9.- Because column 2 already has a 9, R8C2 cannot hold 9.So 9 must go in R9C3.
  32. R9C6 = 3 · Naked SingleLooking at R9C6:- Row 9 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  33. R5C1 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R5C3 cannot hold 9.- Because row 6 already has a 9, R6C1, R6C2 and R6C3 cannot hold 9.So 9 must go in R5C1.
  34. R5C3 = 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, R6C1, R6C2 and R6C3 cannot hold 7.So 7 must go in R5C3.
  35. R1C1 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 7, R1C3 cannot hold 7.So 7 must go in R1C1.
  36. R1C3 = 1 · Naked SingleLooking at R1C3:- Row 1 already has 2, 4, 5, 7, 8 and 9 → ruled out.- The top-left 3×3 box already has 3 and 6 → ruled out.So this cell can only be 1.
  37. R5C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 3, R5C6 cannot hold 3.So 3 must go in R5C4.
  38. R5C6 = 6 · Naked SingleLooking at R5C6:- Row 5 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  39. R8C6 = 9 · Naked SingleLooking at R8C6:- Row 8 already has 2, 3, 4, 5 and 7 → ruled out.- Column 6 already has 1, 6 and 8 → ruled out.So this cell can only be 9.
  40. R1C5 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 3, R1C4 cannot hold 3.So 3 must go in R1C5.
  41. R1C4 = 6 · Naked SingleLooking at R1C4:- Row 1 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  42. R8C4 = 8 · Naked SingleLooking at R8C4:- Row 8 already has 2, 3, 4, 5, 7 and 9 → ruled out.- Column 4 already has 1 and 6 → ruled out.So this cell can only be 8.
  43. R7C5 = 6 · Naked SingleLooking at R7C5:- Row 7 already has 2, 4, 5, 7 and 9 → ruled out.- Column 5 already has 1, 3 and 8 → ruled out.So this cell can only be 6.
  44. R8C2 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C1 and R7C3 cannot hold 6.So 6 must go in R8C2.
  45. R8C7 = 1 · Naked SingleLooking at R8C7:- Row 8 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  46. R6C2 = 1 · Naked SingleLooking at R6C2:- Row 6 already has 2, 4, 5, 7, 8 and 9 → ruled out.- Column 2 already has 3 and 6 → ruled out.So this cell can only be 1.
  47. R7C7 = 8 · Naked SingleLooking at R7C7:- Row 7 already has 2, 4, 5, 6, 7 and 9 → ruled out.- Column 7 already has 1 and 3 → ruled out.So this cell can only be 8.
  48. R6C3 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R6C1 cannot hold 6.So 6 must go in R6C3.
  49. R6C1 = 3 · Naked SingleLooking at R6C1:- Row 6 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  50. R7C1 = 1 · Naked SingleLooking at R7C1:- Row 7 already has 2, 4, 5, 6, 7, 8 and 9 → ruled out.- Column 1 already has 3 → ruled out.So this cell can only be 1.
  51. R7C3 = 3 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.

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