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Daily Sudoku — 15 July 2026

The puzzle

  • 15 July 2026
  • Easy
  • 40 clues
Puzzle
·6··75···
175·93··8
···8···7·
426···913
··7··9·2·
···42678·
···6···4·
63··4·1·9
748951·36
Solution
864175392
175293468
293864571
426587913
587319624
319426785
951632847
632748159
748951236

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. R9C7 = 2 · Naked SingleLooking at R9C7:- Row 9 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  2. R1C1 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 8, R1C3 cannot hold 8.- Because row 3 already has a 8, R3C1, R3C2 and R3C3 cannot hold 8.So 8 must go in R1C1.
  3. R3C6 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R1C4 and R2C4 cannot hold 4.- Because column 5 already has a 4, R3C5 cannot hold 4.So 4 must go in R3C6.
  4. 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, R3C2 and R3C3 cannot hold 4.So 4 must go in R1C3.
  5. R3C5 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 6, R1C4 cannot hold 6.- Because column 4 already has a 6, R2C4 cannot hold 6.So 6 must go in R3C5.
  6. R1C4 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C4 cannot hold 1.So 1 must go in R1C4.
  7. R2C4 = 2 · Naked SingleLooking at R2C4:- Row 2 already has 1, 3, 5, 7, 8 and 9 → ruled out.- Column 4 already has 4 and 6 → ruled out.So this cell can only be 2.
  8. R3C9 = 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, R1C8 and R1C9 cannot hold 1.- Because row 2 already has a 1, R2C7 and R2C8 cannot hold 1.- Because column 7 already has a 1, R3C7 cannot hold 1.So 1 must go in R3C9.
  9. R1C9 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R1C7 and R3C7 cannot hold 2.- Because column 8 already has a 2, R1C8 cannot hold 2.- Because row 2 already has a 2, R2C7 and R2C8 cannot hold 2.So 2 must go in R1C9.
  10. R2C7 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C7 and R1C8 cannot hold 4.- Because column 8 already has a 4, R2C8 cannot hold 4.- Because row 3 already has a 4, R3C7 cannot hold 4.So 4 must go in R2C7.
  11. R2C8 = 6 · Naked SingleLooking at R2C8:- Row 2 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  12. R3C7 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C7 and R1C8 cannot hold 5.So 5 must go in R3C7.
  13. R1C7 = 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, R1C8 cannot hold 3.So 3 must go in R1C7.
  14. R1C8 = 9 · Naked SingleLooking at R1C8:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  15. R8C8 = 5 · Naked SingleLooking at R8C8:- Row 8 already has 1, 3, 4, 6 and 9 → ruled out.- Column 8 already has 2, 7 and 8 → ruled out.So this cell can only be 5.
  16. R5C2 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 8, R5C1 cannot hold 8.- Because row 6 already has a 8, R6C1, R6C2 and R6C3 cannot hold 8.So 8 must go in R5C2.
  17. R5C5 = 1 · Hidden SingleIn the centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 1, R4C4, R4C5 and R4C6 cannot hold 1.- Because column 4 already has a 1, R5C4 cannot hold 1.So 1 must go in R5C5.
  18. R5C4 = 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, R4C5 and R4C6 cannot hold 3.So 3 must go in R5C4.
  19. R4C4 = 5 · Hidden SingleIn the centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 5, R4C5 cannot hold 5.- Because column 6 already has a 5, R4C6 cannot hold 5.So 5 must go in R4C4.
  20. R8C4 = 7 · Naked SingleLooking at R8C4:- Row 8 already has 1, 3, 4, 5, 6 and 9 → ruled out.- Column 4 already has 2 and 8 → ruled out.So this cell can only be 7.
  21. R4C6 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R4C5 cannot hold 7.So 7 must go in R4C6.
  22. R4C5 = 8 · Naked SingleLooking at R4C5:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  23. R7C5 = 3 · Naked SingleLooking at R7C5:- Row 7 already has 4 and 6 → ruled out.- Column 5 already has 1, 2, 5, 7, 8 and 9 → ruled out.So this cell can only be 3.
  24. R5C9 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 4, R5C7 cannot hold 4.- Because row 6 already has a 4, R6C9 cannot hold 4.So 4 must go in R5C9.
  25. R6C9 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 5, R5C7 cannot hold 5.So 5 must go in R6C9.
  26. R7C9 = 7 · Naked SingleLooking at R7C9:- Row 7 already has 3, 4 and 6 → ruled out.- Column 9 already has 1, 2, 5, 8 and 9 → ruled out.So this cell can only be 7.
  27. R5C7 = 6 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 3, 4, 7, 8 and 9 → ruled out.- Column 7 already has 5 → ruled out.So this cell can only be 6.
  28. R5C1 = 5 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  29. R7C7 = 8 · Naked SingleLooking at R7C7:- Row 7 already has 3, 4, 6 and 7 → ruled out.- Column 7 already has 1, 2, 5 and 9 → ruled out.So this cell can only be 8.
  30. R7C2 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 5, R7C1 cannot hold 5.- Because column 3 already has a 5, R7C3 cannot hold 5.- Because row 8 already has a 5, R8C3 cannot hold 5.So 5 must go in R7C2.
  31. R7C3 = 1 · Hidden SingleIn the bottom-left 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 1, R7C1 cannot hold 1.- Because row 8 already has a 1, R8C3 cannot hold 1.So 1 must go in R7C3.
  32. R6C2 = 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, R6C1 cannot hold 1.- Because column 3 already has a 1, R6C3 cannot hold 1.So 1 must go in R6C2.
  33. R3C2 = 9 · Naked SingleLooking at R3C2:- Row 3 already has 1, 4, 5, 6, 7 and 8 → ruled out.- Column 2 already has 2 and 3 → ruled out.So this cell can only be 9.
  34. R7C1 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 9, R8C3 cannot hold 9.So 9 must go in R7C1.
  35. R7C6 = 2 · Naked SingleLooking at R7C6:- Row 7 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  36. R8C6 = 8 · Naked SingleLooking at R8C6:- Row 8 already has 1, 3, 4, 5, 6, 7 and 9 → ruled out.- Column 6 already has 2 → ruled out.So this cell can only be 8.
  37. R8C3 = 2 · Naked SingleLooking at R8C3:- Row 8 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  38. R3C1 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 2, R3C3 cannot hold 2.So 2 must go in R3C1.
  39. R3C3 = 3 · Naked SingleLooking at R3C3:- Row 3 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  40. R6C1 = 3 · Naked SingleLooking at R6C1:- Row 6 already has 1, 2, 4, 5, 6, 7 and 8 → ruled out.- Column 1 already has 9 → ruled out.So this cell can only be 3.
  41. R6C3 = 9 · Naked SingleLooking at R6C3:- 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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