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Daily Sudoku — 24 July 2026

The puzzle

  • 24 July 2026
  • Easy
  • 40 clues
Puzzle
·9···85·4
5·429··38
1·856·29·
·····5·42
····7·6·5
685···3·1
·5·7·381·
8·6·5···3
·2··16··9
Solution
297138564
564297138
138564297
371685942
942371685
685942371
459723816
816459723
723816459

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. R2C2 = 6 · Hidden SingleIn the top-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R1C1 cannot hold 6.- Because column 3 already has a 6, R1C3 cannot hold 6.- Because row 3 already has a 6, R3C2 cannot hold 6.So 6 must go in R2C2.
  2. R3C6 = 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 row 2 already has a 4, R2C6 cannot hold 4.So 4 must go in R3C6.
  3. R2C6 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 7, R1C4 cannot hold 7.- Because column 5 already has a 7, R1C5 cannot hold 7.So 7 must go in R2C6.
  4. R2C7 = 1 · Naked SingleLooking at R2C7:- Row 2 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  5. R1C4 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 1, R1C5 cannot hold 1.So 1 must go in R1C4.
  6. R1C5 = 3 · Naked SingleLooking at R1C5:- Row 1 already has 1, 4, 5, 8 and 9 → ruled out.- Column 5 already has 6 and 7 → ruled out.- The top-centre 3×3 box already has 2 → ruled out.So this cell can only be 3.
  7. R3C2 = 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 and R1C3 cannot hold 3.So 3 must go in R3C2.
  8. R3C9 = 7 · Naked SingleLooking at R3C9:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  9. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 1, 3, 5, 7 and 8 → ruled out.- Column 9 already has 2, 4 and 9 → ruled out.So this cell can only be 6.
  10. R1C8 = 6 · Naked SingleLooking at R1C8:- Row 1 already has 1, 3, 4, 5, 8 and 9 → ruled out.- The top-right 3×3 box already has 2 and 7 → ruled out.So this cell can only be 6.
  11. R5C6 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 1, R4C4 and R5C4 cannot hold 1.- Because column 5 already has a 1, R4C5 cannot hold 1.- Because row 6 already has a 1, R6C4, R6C5 and R6C6 cannot hold 1.So 1 must go in R5C6.
  12. R4C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 6, R4C5 cannot hold 6.- Because row 5 already has a 6, R5C4 cannot hold 6.- Because row 6 already has a 6, R6C4, R6C5 and R6C6 cannot hold 6.So 6 must go in R4C4.
  13. 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, R4C5 cannot hold 3.- Because row 6 already has a 3, R6C4, R6C5 and R6C6 cannot hold 3.So 3 must go in R5C4.
  14. R4C5 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 6 already has a 8, R6C4, R6C5 and R6C6 cannot hold 8.So 8 must go in R4C5.
  15. R5C8 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 8, R4C7 cannot hold 8.- Because row 6 already has a 8, R6C8 cannot hold 8.So 8 must go in R5C8.
  16. R4C7 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 9, R6C8 cannot hold 9.So 9 must go in R4C7.
  17. R6C8 = 7 · Naked SingleLooking at R6C8:- Row 6 already has 1, 3, 5, 6 and 8 → ruled out.- Column 8 already has 4 and 9 → ruled out.- The middle-right 3×3 box already has 2 → ruled out.So this cell can only be 7.
  18. R8C2 = 1 · Hidden SingleIn the bottom-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C1 and R7C3 cannot hold 1.- Because row 9 already has a 1, R9C1 and R9C3 cannot hold 1.So 1 must go in R8C2.
  19. R4C3 = 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 2 already has a 1, R4C2 cannot hold 1.- Because row 5 already has a 1, R5C1, R5C2 and R5C3 cannot hold 1.So 1 must go in R4C3.
  20. R4C1 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 3, R4C2 cannot hold 3.- Because row 5 already has a 3, R5C1, R5C2 and R5C3 cannot hold 3.So 3 must go in R4C1.
  21. R4C2 = 7 · Naked SingleLooking at R4C2:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  22. R5C2 = 4 · Naked SingleLooking at R5C2:- Row 5 already has 1, 3, 5, 6, 7 and 8 → ruled out.- Column 2 already has 2 and 9 → ruled out.So this cell can only be 4.
  23. R9C3 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C1 and R7C3 cannot hold 3.- Because column 1 already has a 3, R9C1 cannot hold 3.So 3 must go in R9C3.
  24. R9C1 = 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, R7C1 and R7C3 cannot hold 7.So 7 must go in R9C1.
  25. R1C3 = 7 · Hidden SingleIn the top-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 7, R1C1 cannot hold 7.So 7 must go in R1C3.
  26. R1C1 = 2 · Naked SingleLooking at R1C1:- Row 1 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  27. R5C3 = 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 cannot hold 2.So 2 must go in R5C3.
  28. R5C1 = 9 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  29. R7C1 = 4 · Naked SingleLooking at R7C1:- Row 7 already has 1, 3, 5, 6, 7 and 8 → ruled out.- Column 1 already has 2 and 9 → ruled out.So this cell can only be 4.
  30. R7C3 = 9 · Naked SingleLooking at R7C3:- Row 7 already has 1, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 3 already has 2 → ruled out.So this cell can only be 9.
  31. R7C5 = 2 · Naked SingleLooking at R7C5:- Row 7 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  32. R6C5 = 4 · Naked SingleLooking at R6C5:- Row 6 already has 1, 3, 5, 6, 7 and 8 → ruled out.- Column 5 already has 2 and 9 → ruled out.So this cell can only be 4.
  33. R6C6 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 2, R6C4 cannot hold 2.So 2 must go in R6C6.
  34. R6C4 = 9 · Naked SingleLooking at R6C4:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  35. R8C6 = 9 · Naked SingleLooking at R8C6:- Row 8 already has 1, 3, 5, 6 and 8 → ruled out.- Column 6 already has 2, 4 and 7 → ruled out.So this cell can only be 9.
  36. R9C4 = 8 · Hidden SingleIn the bottom-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 8, R8C4 cannot hold 8.So 8 must go in R9C4.
  37. R8C4 = 4 · Naked SingleLooking at R8C4:- Row 8 already has 1, 3, 5, 6, 8 and 9 → ruled out.- Column 4 already has 2 and 7 → ruled out.So this cell can only be 4.
  38. R8C8 = 2 · Hidden SingleIn the bottom-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R8C7 cannot hold 2.- Because row 9 already has a 2, R9C7 and R9C8 cannot hold 2.So 2 must go in R8C8.
  39. R8C7 = 7 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R9C7 = 4 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 6, 7, 8 and 9 → ruled out.- Column 7 already has 5 → ruled out.So this cell can only be 4.
  41. R9C8 = 5 · Naked SingleLooking at R9C8:- Row 9 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.

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