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Daily Sudoku — 10 July 2026

The puzzle

  • 10 July 2026
  • Easy
  • 40 clues
Puzzle
35481·926
9·2·4··1·
········5
···7·2·3·
······79·
729··458·
·3·4·8179
1·7··3··8
··51·9263
Solution
354817926
962345817
871926345
518792634
643581792
729634581
236458179
197263458
485179263

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. R1C6 = 7 · Naked SingleLooking at R1C6:- Row 1 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  2. R2C9 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 7, R2C7 and R3C7 cannot hold 7.- Because column 8 already has a 7, R3C8 cannot hold 7.So 7 must go in R2C9.
  3. R3C2 = 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.- Because column 1 already has a 7, R3C1 cannot hold 7.- Because column 3 already has a 7, R3C3 cannot hold 7.So 7 must go in R3C2.
  4. R3C3 = 1 · Hidden SingleIn the top-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C2 cannot hold 1.- Because column 1 already has a 1, R3C1 cannot hold 1.So 1 must go in R3C3.
  5. R5C3 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C1, R4C2 and R4C3 cannot hold 3.- Because column 1 already has a 3, R5C1 cannot hold 3.- Because column 2 already has a 3, R5C2 cannot hold 3.So 3 must go in R5C3.
  6. R5C6 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 1, R4C5, R5C5 and R6C5 cannot hold 1.- Because column 4 already has a 1, R5C4 and R6C4 cannot hold 1.So 1 must go in R5C6.
  7. R4C2 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 1, R4C1 cannot hold 1.- Because column 3 already has a 1, R4C3 cannot hold 1.- Because row 5 already has a 1, R5C1 and R5C2 cannot hold 1.So 1 must go in R4C2.
  8. R4C5 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C4 and R5C5 cannot hold 9.- Because row 6 already has a 9, R6C4 and R6C5 cannot hold 9.So 9 must go in R4C5.
  9. R3C4 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 9, R2C4 and R2C6 cannot hold 9.- Because column 5 already has a 9, R3C5 cannot hold 9.- Because column 6 already has a 9, R3C6 cannot hold 9.So 9 must go in R3C4.
  10. R3C5 = 2 · Hidden SingleIn the top-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 2, R2C4 and R2C6 cannot hold 2.- Because column 6 already has a 2, R3C6 cannot hold 2.So 2 must go in R3C5.
  11. R2C4 = 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, R2C6 and R3C6 cannot hold 3.So 3 must go in R2C4.
  12. 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, R3C6 cannot hold 5.So 5 must go in R2C6.
  13. R3C6 = 6 · Naked SingleLooking at R3C6:- Row 3 already has 1, 2, 5, 7 and 9 → ruled out.- Column 6 already has 3, 4 and 8 → ruled out.So this cell can only be 6.
  14. R2C2 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 6, R3C1 cannot hold 6.So 6 must go in R2C2.
  15. R2C7 = 8 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  16. R3C1 = 8 · Naked SingleLooking at R3C1:- Row 3 already has 1, 2, 5, 6, 7 and 9 → ruled out.- Column 1 already has 3 → ruled out.- The top-left 3×3 box already has 4 → ruled out.So this cell can only be 8.
  17. R3C7 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 3, R3C8 cannot hold 3.So 3 must go in R3C7.
  18. R3C8 = 4 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  19. R8C8 = 5 · Naked SingleLooking at R8C8:- Row 8 already has 1, 3, 7 and 8 → ruled out.- Column 8 already has 2, 4, 6 and 9 → ruled out.So this cell can only be 5.
  20. R8C7 = 4 · Naked SingleLooking at R8C7:- Row 8 already has 1, 3, 5, 7 and 8 → ruled out.- Column 7 already has 2 and 9 → ruled out.- The bottom-right 3×3 box already has 6 → ruled out.So this cell can only be 4.
  21. R4C7 = 6 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 7 and 9 → ruled out.- Column 7 already has 4, 5 and 8 → ruled out.So this cell can only be 6.
  22. R5C1 = 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, R4C1 and R4C3 cannot hold 6.- Because column 2 already has a 6, R5C2 cannot hold 6.So 6 must go in R5C1.
  23. R4C1 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 5, R4C3 cannot hold 5.- Because column 2 already has a 5, R5C2 cannot hold 5.So 5 must go in R4C1.
  24. R5C2 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R4C3 cannot hold 4.So 4 must go in R5C2.
  25. R4C3 = 8 · Naked SingleLooking at R4C3:- Row 4 already has 1, 2, 3, 5, 6, 7 and 9 → ruled out.- Column 3 already has 4 → ruled out.So this cell can only be 8.
  26. R4C9 = 4 · Naked SingleLooking at R4C9:- Row 4 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  27. R7C3 = 6 · Naked SingleLooking at R7C3:- Row 7 already has 1, 3, 4, 7, 8 and 9 → ruled out.- Column 3 already has 2 and 5 → ruled out.So this cell can only be 6.
  28. R6C5 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C4 and R5C5 cannot hold 3.- Because column 4 already has a 3, R6C4 cannot hold 3.So 3 must go in R6C5.
  29. R6C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 6, R5C4 and R5C5 cannot hold 6.So 6 must go in R6C4.
  30. 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.
  31. R5C9 = 2 · Naked SingleLooking at R5C9:- Row 5 already has 1, 3, 4, 6, 7 and 9 → ruled out.- Column 9 already has 5 and 8 → ruled out.So this cell can only be 2.
  32. R5C5 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 8, R5C4 cannot hold 8.So 8 must go in R5C5.
  33. R5C4 = 5 · Naked SingleLooking at R5C4:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  34. R8C4 = 2 · Naked SingleLooking at R8C4:- Row 8 already has 1, 3, 4, 5, 7 and 8 → ruled out.- Column 4 already has 6 and 9 → ruled out.So this cell can only be 2.
  35. R7C1 = 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 cannot hold 2.- Because row 9 already has a 2, R9C1 and R9C2 cannot hold 2.So 2 must go in R7C1.
  36. R7C5 = 5 · Naked SingleLooking at R7C5:- Row 7 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  37. R9C1 = 4 · Naked SingleLooking at R9C1:- Row 9 already has 1, 2, 3, 5, 6 and 9 → ruled out.- Column 1 already has 7 and 8 → ruled out.So this cell can only be 4.
  38. R9C2 = 8 · Hidden SingleIn the bottom-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 8, R8C2 cannot hold 8.So 8 must go in R9C2.
  39. R9C5 = 7 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R8C2 = 9 · Naked SingleLooking at R8C2:- Row 8 already has 1, 2, 3, 4, 5, 7 and 8 → ruled out.- Column 2 already has 6 → ruled out.So this cell can only be 9.
  41. R8C5 = 6 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.

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