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Daily Sudoku — 8 August 2026

The puzzle

  • 8 August 2026
  • Easy
  • 40 clues
Puzzle
··8·····5
·175··48·
5·96··1··
49·1·6·2·
·······54
·7·49586·
76··23·48
9238··61·
··4761···
Solution
648217395
217539486
539648172
495186723
186372954
372495861
761923548
923854617
854761239

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. R1C5 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 1, R1C4 cannot hold 1.- Because column 6 already has a 1, R1C6 cannot hold 1.- Because row 2 already has a 1, R2C5 and R2C6 cannot hold 1.- Because row 3 already has a 1, R3C5 and R3C6 cannot hold 1.So 1 must go in R1C5.
  2. R2C9 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 6, R1C7 cannot hold 6.- Because column 8 already has a 6, R1C8 cannot hold 6.- Because row 3 already has a 6, R3C8 and R3C9 cannot hold 6.So 6 must go in R2C9.
  3. R1C1 = 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, R1C2 cannot hold 6.- Because row 2 already has a 6, R2C1 cannot hold 6.- Because row 3 already has a 6, R3C2 cannot hold 6.So 6 must go in R1C1.
  4. R2C1 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R1C2 and R3C2 cannot hold 2.So 2 must go in R2C1.
  5. R4C3 = 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 row 6 already has a 5, R6C1 and R6C3 cannot hold 5.So 5 must go in R4C3.
  6. R5C3 = 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, R5C1 cannot hold 6.- Because column 2 already has a 6, R5C2 cannot hold 6.- Because row 6 already has a 6, R6C1 and R6C3 cannot hold 6.So 6 must go in R5C3.
  7. R6C3 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 2, R5C1 and R6C1 cannot hold 2.- Because column 2 already has a 2, R5C2 cannot hold 2.So 2 must go in R6C3.
  8. R7C3 = 1 · Naked SingleLooking at R7C3:- Row 7 already has 2, 3, 4, 6, 7 and 8 → ruled out.- Column 3 already has 5 and 9 → ruled out.So this cell can only be 1.
  9. R6C9 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C7 and R4C9 cannot hold 1.- Because column 7 already has a 1, R5C7 cannot hold 1.So 1 must go in R6C9.
  10. 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.
  11. R5C1 = 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, R5C2 cannot hold 1.So 1 must go in R5C1.
  12. R9C1 = 8 · Naked SingleLooking at R9C1:- Row 9 already has 1, 4, 6 and 7 → ruled out.- Column 1 already has 2, 3, 5 and 9 → ruled out.So this cell can only be 8.
  13. R5C2 = 8 · Naked SingleLooking at R5C2:- Row 5 already has 1, 4, 5 and 6 → ruled out.- Column 2 already has 2, 7 and 9 → ruled out.- The middle-left 3×3 box already has 3 → ruled out.So this cell can only be 8.
  14. R9C2 = 5 · Naked SingleLooking at R9C2:- Row 9 already has 1, 4, 6, 7 and 8 → ruled out.- Column 2 already has 2 and 9 → ruled out.- The bottom-left 3×3 box already has 3 → ruled out.So this cell can only be 5.
  15. 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, R5C5 and R5C6 cannot hold 8.So 8 must go in R4C5.
  16. R3C6 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has an 8, R1C4 and R1C6 cannot hold 8.- Because row 2 already has an 8, R2C5 and R2C6 cannot hold 8.- Because column 5 already has an 8, R3C5 cannot hold 8.So 8 must go in R3C6.
  17. R5C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C7 and R4C9 cannot hold 9.So 9 must go in R5C7.
  18. R1C8 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 9, R1C7 cannot hold 9.- Because row 3 already has a 9, R3C8 and R3C9 cannot hold 9.So 9 must go in R1C8.
  19. R2C6 = 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 R1C6 cannot hold 9.- Because column 5 already has a 9, R2C5 cannot hold 9.- Because row 3 already has a 9, R3C5 cannot hold 9.So 9 must go in R2C6.
  20. R2C5 = 3 · Naked SingleLooking at R2C5:- Row 2 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  21. R5C4 = 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 column 6 already has a 3, R5C6 cannot hold 3.So 3 must go in R5C4.
  22. R5C6 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 2, R5C5 cannot hold 2.So 2 must go in R5C6.
  23. R5C5 = 7 · Naked SingleLooking at R5C5:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  24. R1C4 = 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 cannot hold 2.- Because column 5 already has a 2, R3C5 cannot hold 2.So 2 must go in R1C4.
  25. R7C4 = 9 · Naked SingleLooking at R7C4:- Row 7 already has 1, 2, 3, 4, 6, 7 and 8 → ruled out.- Column 4 already has 5 → ruled out.So this cell can only be 9.
  26. R7C7 = 5 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  27. R1C6 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R3C5 cannot hold 7.So 7 must go in R1C6.
  28. R8C6 = 4 · Naked SingleLooking at R8C6:- Row 8 already has 1, 2, 3, 6, 8 and 9 → ruled out.- Column 6 already has 5 and 7 → ruled out.So this cell can only be 4.
  29. R3C5 = 4 · Naked SingleLooking at R3C5:- Row 3 already has 1, 5, 6, 8 and 9 → ruled out.- Column 5 already has 2, 3 and 7 → ruled out.So this cell can only be 4.
  30. R8C5 = 5 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 3, 4, 6, 8 and 9 → ruled out.- Column 5 already has 7 → ruled out.So this cell can only be 5.
  31. R8C9 = 7 · Naked SingleLooking at R8C9:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  32. R1C2 = 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, R3C2 cannot hold 4.So 4 must go in R1C2.
  33. R1C7 = 3 · Naked SingleLooking at R1C7:- Row 1 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  34. R3C2 = 3 · Naked SingleLooking at R3C2:- Row 3 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.
  35. R3C9 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 2, R3C8 cannot hold 2.So 2 must go in R3C9.
  36. R3C8 = 7 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  37. R9C8 = 3 · Naked SingleLooking at R9C8:- Row 9 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.
  38. R4C9 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R4C7 cannot hold 3.So 3 must go in R4C9.
  39. R4C7 = 7 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R9C7 = 2 · Naked SingleLooking at R9C7:- Row 9 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.
  41. R9C9 = 9 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.

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