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Daily Sudoku — 2 July 2026

The puzzle

  • 2 July 2026
  • Easy
  • 40 clues
Puzzle
···8·93·2
86925·1·7
·5·741·6·
324·7·5··
···9·5·23
····2···1
···6··295
5······16
·1·5·2734
Solution
741869352
869253147
253741968
324176589
176985423
985324671
437618295
592437816
618592734

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. R8C7 = 8 · Naked SingleLooking at R8C7:- Row 8 already has 1, 5 and 6 → ruled out.- Column 7 already has 2, 3 and 7 → ruled out.- The bottom-right 3×3 box already has 4 and 9 → ruled out.So this cell can only be 8.
  2. R3C3 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C1, R1C2 and R1C3 cannot hold 3.- Because column 1 already has a 3, R3C1 cannot hold 3.So 3 must go in R3C3.
  3. R3C1 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C1, R1C2 and R1C3 cannot hold 2.So 2 must go in R3C1.
  4. R2C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C5 cannot hold 3.So 3 must go in R2C6.
  5. R2C8 = 4 · Naked SingleLooking at R2C8:- Row 2 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  6. R1C5 = 6 · Naked SingleLooking at R1C5:- Row 1 already has 2, 3, 8 and 9 → ruled out.- Column 5 already has 4, 5 and 7 → ruled out.- The top-centre 3×3 box already has 1 → ruled out.So this cell can only be 6.
  7. R1C8 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 5, R3C7 and R3C9 cannot hold 5.So 5 must go in R1C8.
  8. R3C9 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 8, R3C7 cannot hold 8.So 8 must go in R3C9.
  9. R3C7 = 9 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  10. R4C9 = 9 · Naked SingleLooking at R4C9:- Row 4 already has 2, 3, 4, 5 and 7 → ruled out.- Column 9 already has 1, 6 and 8 → ruled out.So this cell can only be 9.
  11. R6C3 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 5, R5C1, R5C2 and R5C3 cannot hold 5.- Because column 1 already has a 5, R6C1 cannot hold 5.- Because column 2 already has a 5, R6C2 cannot hold 5.So 5 must go in R6C3.
  12. R6C4 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 3, R4C4 and R4C6 cannot hold 3.- Because row 5 already has a 3, R5C5 cannot hold 3.- Because column 6 already has a 3, R6C6 cannot hold 3.So 3 must go in R6C4.
  13. R6C6 = 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, R4C4 and R4C6 cannot hold 4.- Because column 5 already has a 4, R5C5 cannot hold 4.So 4 must go in R6C6.
  14. R4C6 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 6, R4C4 cannot hold 6.- Because column 5 already has a 6, R5C5 cannot hold 6.So 6 must go in R4C6.
  15. 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, R4C4 cannot hold 8.So 8 must go in R5C5.
  16. R4C4 = 1 · Naked SingleLooking at R4C4:- Row 4 already has 2, 3, 4, 5, 6, 7 and 9 → ruled out.- Column 4 already has 8 → ruled out.So this cell can only be 1.
  17. R4C8 = 8 · Naked SingleLooking at R4C8:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  18. R8C4 = 4 · Naked SingleLooking at R8C4:- Row 8 already has 1, 5, 6 and 8 → ruled out.- Column 4 already has 2, 3, 7 and 9 → ruled out.So this cell can only be 4.
  19. R6C8 = 7 · Naked SingleLooking at R6C8:- Row 6 already has 1, 2, 3, 4 and 5 → ruled out.- Column 8 already has 6, 8 and 9 → ruled out.So this cell can only be 7.
  20. R6C2 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 8, R5C1, R5C2 and R5C3 cannot hold 8.- Because column 1 already has a 8, R6C1 cannot hold 8.So 8 must go in R6C2.
  21. R6C1 = 9 · Hidden SingleIn the middle-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C1, R5C2 and R5C3 cannot hold 9.So 9 must go in R6C1.
  22. R6C7 = 6 · Naked SingleLooking at R6C7:- Row 6 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  23. R5C7 = 4 · Naked SingleLooking at R5C7:- Row 5 already has 2, 3, 5, 8 and 9 → ruled out.- Column 7 already has 1, 6 and 7 → ruled out.So this cell can only be 4.
  24. R8C3 = 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, R7C2 and R7C3 cannot hold 2.- Because column 2 already has a 2, R8C2 cannot hold 2.- Because row 9 already has a 2, R9C1 and R9C3 cannot hold 2.So 2 must go in R8C3.
  25. R8C2 = 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, R7C2 and R7C3 cannot hold 9.- Because column 1 already has a 9, R9C1 cannot hold 9.- Because column 3 already has a 9, R9C3 cannot hold 9.So 9 must go in R8C2.
  26. R7C2 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 3, R7C1 cannot hold 3.- Because column 3 already has a 3, R7C3 cannot hold 3.- Because row 9 already has a 3, R9C1 and R9C3 cannot hold 3.So 3 must go in R7C2.
  27. R7C1 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 4, R7C3 cannot hold 4.- Because row 9 already has a 4, R9C1 and R9C3 cannot hold 4.So 4 must go in R7C1.
  28. R1C2 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R1C1 cannot hold 4.- Because column 3 already has a 4, R1C3 cannot hold 4.So 4 must go in R1C2.
  29. R5C2 = 7 · Naked SingleLooking at R5C2:- Row 5 already has 2, 3, 4, 5, 8 and 9 → ruled out.- Column 2 already has 1 and 6 → ruled out.So this cell can only be 7.
  30. R7C3 = 7 · Hidden SingleIn the bottom-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 9 already has a 7, R9C1 and R9C3 cannot hold 7.So 7 must go in R7C3.
  31. 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.
  32. R1C3 = 1 · Naked SingleLooking at R1C3:- Row 1 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  33. R5C1 = 1 · Hidden SingleIn the middle-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 1, R5C3 cannot hold 1.So 1 must go in R5C1.
  34. R5C3 = 6 · Naked SingleLooking at R5C3:- Row 5 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  35. R9C1 = 6 · Naked SingleLooking at R9C1:- Row 9 already has 1, 2, 3, 4, 5 and 7 → ruled out.- Column 1 already has 8 and 9 → ruled out.So this cell can only be 6.
  36. R9C3 = 8 · Naked SingleLooking at R9C3:- Row 9 already has 1, 2, 3, 4, 5, 6 and 7 → ruled out.- Column 3 already has 9 → ruled out.So this cell can only be 8.
  37. R9C5 = 9 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  38. R7C5 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 1, R7C6 cannot hold 1.- Because row 8 already has a 1, R8C5 and R8C6 cannot hold 1.So 1 must go in R7C5.
  39. R7C6 = 8 · Naked SingleLooking at R7C6:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  40. 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.
  41. 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.

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