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Daily Sudoku — 28 June 2026

The puzzle

  • 28 June 2026
  • Medium
  • 34 clues
Puzzle
7··69···1
5··3···4·
··34275·9
481···3··
··295·4··
3··1···27
·········
9··5···84
·3·8·297·
Solution
724695831
596318742
813427569
481276395
672953418
359184627
248769153
967531284
135842976

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 — 47 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. R1C6 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C5 and R2C6 cannot hold 5.So 5 must go in R1C6.
  2. R1C8 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R1C7 cannot hold 3.- Because row 2 already has a 3, R2C7 and R2C9 cannot hold 3.- Because row 3 already has a 3, R3C8 cannot hold 3.So 3 must go in R1C8.
  3. R2C7 = 7 · Hidden SingleIn the top-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 7, R1C7 cannot hold 7.- Because column 9 already has a 7, R2C9 cannot hold 7.- Because row 3 already has a 7, R3C8 cannot hold 7.So 7 must go in R2C7.
  4. R5C2 = 7 · Hidden SingleIn the middle-left 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 7, R5C1 cannot hold 7.- Because row 6 already has a 7, R6C2 and R6C3 cannot hold 7.So 7 must go in R5C2.
  5. R4C4 = 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, R4C5 cannot hold 2.- Because column 6 already has a 2, R4C6 cannot hold 2.- Because row 5 already has a 2, R5C6 cannot hold 2.- Because row 6 already has a 2, R6C5 and R6C6 cannot hold 2.So 2 must go in R4C4.
  6. R7C4 = 7 · Naked SingleLooking at R7C4:- Column 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  7. R5C6 = 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, R4C5 and R4C6 cannot hold 3.- Because row 6 already has a 3, R6C5 and R6C6 cannot hold 3.So 3 must go in R5C6.
  8. R4C5 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 7, R4C6 cannot hold 7.- Because row 6 already has a 7, R6C5 and R6C6 cannot hold 7.So 7 must go in R4C5.
  9. R5C8 = 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, R4C8 and R4C9 cannot hold 1.- Because column 9 already has a 1, R5C9 cannot hold 1.- Because row 6 already has a 1, R6C7 cannot hold 1.So 1 must go in R5C8.
  10. R4C8 = 9 · Hidden SingleIn the middle-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 9, R4C9 cannot hold 9.- Because row 5 already has a 9, R5C9 cannot hold 9.- Because column 7 already has a 9, R6C7 cannot hold 9.So 9 must go in R4C8.
  11. R4C9 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 5, R5C9 cannot hold 5.- Because column 7 already has a 5, R6C7 cannot hold 5.So 5 must go in R4C9.
  12. R4C6 = 6 · Naked SingleLooking at R4C6:- Row 4 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  13. R8C3 = 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, R7C2 and R7C3 cannot hold 7.- Because column 2 already has a 7, R8C2 cannot hold 7.- Because row 9 already has a 7, R9C1 and R9C3 cannot hold 7.So 7 must go in R8C3.
  14. R7C6 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 9, R7C5 cannot hold 9.- Because row 8 already has a 9, R8C5 and R8C6 cannot hold 9.- Because row 9 already has a 9, R9C5 cannot hold 9.So 9 must go in R7C6.
  15. R7C9 = 3 · Hidden SingleIn the bottom-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 3, R7C7 and R8C7 cannot hold 3.- Because column 8 already has a 3, R7C8 cannot hold 3.- Because row 9 already has a 3, R9C9 cannot hold 3.So 3 must go in R7C9.
  16. R8C5 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C5 cannot hold 3.- Because column 6 already has a 3, R8C6 cannot hold 3.- Because row 9 already has a 3, R9C5 cannot hold 3.So 3 must go in R8C5.
  17. R7C8 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 5, R7C7 cannot hold 5.- Because row 8 already has a 5, R8C7 cannot hold 5.- Because column 9 already has a 5, R9C9 cannot hold 5.So 5 must go in R7C8.
  18. R3C8 = 6 · Naked SingleLooking at R3C8:- Row 3 already has 2, 3, 4, 5, 7 and 9 → ruled out.- Column 8 already has 1 and 8 → ruled out.So this cell can only be 6.
  19. R9C3 = 5 · Hidden SingleIn the bottom-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C1, R7C2 and R7C3 cannot hold 5.- Because row 8 already has a 5, R8C2 cannot hold 5.- Because column 1 already has a 5, R9C1 cannot hold 5.So 5 must go in R9C3.
  20. R6C2 = 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 cannot hold 5.- Because column 3 already has a 5, R6C3 cannot hold 5.So 5 must go in R6C2.
  21. R6C3 = 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 cannot hold 9.So 9 must go in R6C3.
  22. R5C1 = 6 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 4, 5, 7 and 9 → ruled out.- The middle-left 3×3 box already has 8 → ruled out.So this cell can only be 6.
  23. R5C9 = 8 · Naked SingleLooking at R5C9:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  24. R6C7 = 6 · Naked SingleLooking at R6C7:- Row 6 already has 1, 2, 3, 5, 7 and 9 → ruled out.- Column 7 already has 4 → ruled out.- The middle-right 3×3 box already has 8 → ruled out.So this cell can only be 6.
  25. R2C2 = 9 · Hidden SingleIn the top-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C2 and R1C3 cannot hold 9.- Because column 3 already has a 9, R2C3 cannot hold 9.- Because row 3 already has a 9, R3C1 and R3C2 cannot hold 9.So 9 must go in R2C2.
  26. R1C2 = 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, R1C3 and R2C3 cannot hold 2.- Because row 3 already has a 2, R3C1 and R3C2 cannot hold 2.So 2 must go in R1C2.
  27. R1C3 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 4, R2C3 cannot hold 4.- Because row 3 already has a 4, R3C1 and R3C2 cannot hold 4.So 4 must go in R1C3.
  28. R1C7 = 8 · Naked SingleLooking at R1C7:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  29. R2C9 = 2 · Naked SingleLooking at R2C9:- Row 2 already has 3, 4, 5, 7 and 9 → ruled out.- Column 9 already has 1 and 8 → ruled out.- The top-right 3×3 box already has 6 → ruled out.So this cell can only be 2.
  30. R9C9 = 6 · Naked SingleLooking at R9C9:- Row 9 already has 2, 3, 5, 7, 8 and 9 → ruled out.- Column 9 already has 1 and 4 → ruled out.So this cell can only be 6.
  31. R2C3 = 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 and R3C2 cannot hold 6.So 6 must go in R2C3.
  32. R7C3 = 8 · Naked SingleLooking at R7C3:- Row 7 already has 3, 5, 7 and 9 → ruled out.- Column 3 already has 1, 2, 4 and 6 → ruled out.So this cell can only be 8.
  33. R3C1 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 8, R3C2 cannot hold 8.So 8 must go in R3C1.
  34. R3C2 = 1 · Naked SingleLooking at R3C2:- Row 3 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  35. R7C1 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 2, R7C2 and R8C2 cannot hold 2.- Because row 9 already has a 2, R9C1 cannot hold 2.So 2 must go in R7C1.
  36. R9C1 = 1 · Naked SingleLooking at R9C1:- Row 9 already has 2, 3, 5, 6, 7, 8 and 9 → ruled out.- Column 1 already has 4 → ruled out.So this cell can only be 1.
  37. R9C5 = 4 · Naked SingleLooking at R9C5:- Row 9 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  38. R6C6 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 4, R6C5 cannot hold 4.So 4 must go in R6C6.
  39. R6C5 = 8 · Naked SingleLooking at R6C5:- Row 6 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  40. R2C6 = 8 · Hidden SingleIn the top-centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 8, R2C5 cannot hold 8.So 8 must go in R2C6.
  41. R2C5 = 1 · Naked SingleLooking at R2C5:- Row 2 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  42. R7C5 = 6 · Naked SingleLooking at R7C5:- Row 7 already has 2, 3, 5, 7, 8 and 9 → ruled out.- Column 5 already has 1 and 4 → ruled out.So this cell can only be 6.
  43. R8C6 = 1 · Naked SingleLooking at R8C6:- Row 8 already has 3, 4, 5, 7, 8 and 9 → ruled out.- Column 6 already has 2 and 6 → ruled out.So this cell can only be 1.
  44. R7C2 = 4 · Hidden SingleIn the bottom-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 4, R8C2 cannot hold 4.So 4 must go in R7C2.
  45. R7C7 = 1 · Naked SingleLooking at R7C7:- Row 7 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  46. R8C2 = 6 · Naked SingleLooking at R8C2:- Row 8 already has 1, 3, 4, 5, 7, 8 and 9 → ruled out.- Column 2 already has 2 → ruled out.So this cell can only be 6.
  47. R8C7 = 2 · Naked SingleLooking at R8C7:- Row 8 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.

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