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Daily Sudoku — 30 July 2026

The puzzle

  • 30 July 2026
  • Easy
  • 40 clues
Puzzle
78·91···3
6··3···84
·23···6·9
3··254···
9·81··4·2
···8··361
84·7·1···
19··328·5
2···4··97
Solution
784916523
619325784
523487619
361254978
978163452
452879361
845791236
197632845
236548197

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 = 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, R1C3 cannot hold 9.- Because column 2 already has a 9, R2C2 cannot hold 9.- Because row 3 already has a 9, R3C1 cannot hold 9.So 9 must go in R2C3.
  2. R2C2 = 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 column 1 already has a 1, R3C1 cannot hold 1.So 1 must go in R2C2.
  3. R2C5 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 2, R1C6 and R2C6 cannot hold 2.- Because row 3 already has a 2, R3C4, R3C5 and R3C6 cannot hold 2.So 2 must go in R2C5.
  4. R3C4 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 4, R1C6 and R3C6 cannot hold 4.- Because row 2 already has a 4, R2C6 cannot hold 4.- Because column 5 already has a 4, R3C5 cannot hold 4.So 4 must go in R3C4.
  5. R1C3 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 4, R3C1 cannot hold 4.So 4 must go in R1C3.
  6. R3C1 = 5 · Naked SingleLooking at R3C1:- Row 3 already has 2, 3, 4, 6 and 9 → ruled out.- Column 1 already has 1, 7 and 8 → ruled out.So this cell can only be 5.
  7. R6C1 = 4 · Naked SingleLooking at R6C1:- Row 6 already has 1, 3, 6 and 8 → ruled out.- Column 1 already has 2, 5, 7 and 9 → ruled out.So this cell can only be 4.
  8. R1C6 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 6, R2C6 cannot hold 6.- Because row 3 already has a 6, R3C5 and R3C6 cannot hold 6.So 6 must go in R1C6.
  9. R2C6 = 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 and R3C6 cannot hold 5.So 5 must go in R2C6.
  10. R2C7 = 7 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  11. R3C8 = 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, R1C7 and R1C8 cannot hold 1.So 1 must go in R3C8.
  12. R4C3 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 1, R4C2 cannot hold 1.- Because row 5 already has a 1, R5C2 cannot hold 1.- Because row 6 already has a 1, R6C2 and R6C3 cannot hold 1.So 1 must go in R4C3.
  13. R6C3 = 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, R5C2 cannot hold 2.- Because column 2 already has a 2, R6C2 cannot hold 2.So 2 must go in R6C3.
  14. R5C6 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R5C5 cannot hold 3.- Because row 6 already has a 3, R6C5 and R6C6 cannot hold 3.So 3 must go in R5C6.
  15. R5C5 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 6, R6C5 and R6C6 cannot hold 6.So 6 must go in R5C5.
  16. R4C2 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 6, R5C2 cannot hold 6.- Because row 6 already has a 6, R6C2 cannot hold 6.So 6 must go in R4C2.
  17. R5C8 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C7, R4C8 and R4C9 cannot hold 5.So 5 must go in R5C8.
  18. R5C2 = 7 · Naked SingleLooking at R5C2:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  19. R6C2 = 5 · Naked SingleLooking at R6C2:- Row 6 already has 1, 2, 3, 4, 6 and 8 → ruled out.- Column 2 already has 7 and 9 → ruled out.So this cell can only be 5.
  20. R9C2 = 3 · Naked SingleLooking at R9C2:- Row 9 already has 2, 4, 7 and 9 → ruled out.- Column 2 already has 1, 5, 6 and 8 → ruled out.So this cell can only be 3.
  21. R1C7 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 5, R1C8 cannot hold 5.So 5 must go in R1C7.
  22. R1C8 = 2 · Naked SingleLooking at R1C8:- Row 1 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  23. R4C8 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 7, R4C7 cannot hold 7.- Because column 9 already has a 7, R4C9 cannot hold 7.So 7 must go in R4C8.
  24. R4C9 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 8, R4C7 cannot hold 8.So 8 must go in R4C9.
  25. R4C7 = 9 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  26. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 1, 4, 7 and 8 → ruled out.- Column 9 already has 2, 3, 5 and 9 → ruled out.So this cell can only be 6.
  27. R8C3 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 7, R7C3 cannot hold 7.- Because row 9 already has a 7, R9C3 cannot hold 7.So 7 must go in R8C3.
  28. R9C3 = 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, R7C3 cannot hold 6.So 6 must go in R9C3.
  29. R7C3 = 5 · Naked SingleLooking at R7C3:- Row 7 already has 1, 4, 6, 7 and 8 → ruled out.- Column 3 already has 2, 3 and 9 → ruled out.So this cell can only be 5.
  30. R9C4 = 5 · Hidden SingleIn the bottom-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C5 cannot hold 5.- Because row 8 already has a 5, R8C4 cannot hold 5.- Because column 6 already has a 5, R9C6 cannot hold 5.So 5 must go in R9C4.
  31. R8C4 = 6 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 3, 5, 7, 8 and 9 → ruled out.- Column 4 already has 4 → ruled out.So this cell can only be 6.
  32. R8C8 = 4 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  33. R7C8 = 3 · Naked SingleLooking at R7C8:- Row 7 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 8 already has 2 and 9 → ruled out.So this cell can only be 3.
  34. R9C6 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 8, R7C5 cannot hold 8.So 8 must go in R9C6.
  35. R9C7 = 1 · Naked SingleLooking at R9C7:- Row 9 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  36. R7C7 = 2 · Naked SingleLooking at R7C7:- Row 7 already has 1, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 7 already has 9 → ruled out.So this cell can only be 2.
  37. R7C5 = 9 · Naked SingleLooking at R7C5:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  38. R3C5 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 8, R3C6 cannot hold 8.So 8 must go in R3C5.
  39. R3C6 = 7 · Naked SingleLooking at R3C6:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R6C5 = 7 · Naked SingleLooking at R6C5:- Row 6 already has 1, 2, 3, 4, 5, 6 and 8 → ruled out.- Column 5 already has 9 → ruled out.So this cell can only be 7.
  41. R6C6 = 9 · Naked SingleLooking at R6C6:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.

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