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Daily Sudoku — 19 September 2026

The puzzle

  • 19 September 2026
  • Easy
  • 40 clues
Puzzle
·198···2·
··7··24··
·2··91··7
2·3·4·658
·7·6·5·93
······17·
·427·8···
7··5···42
53692478·
Solution
419876325
867352419
325491867
293147658
174685293
658239174
942718536
781563942
536924781

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. R9C9 = 1 · Naked SingleLooking at R9C9:- Row 9 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  2. R3C4 = 4 · Hidden SingleIn the top-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 4, R1C5 cannot hold 4.- Because column 6 already has a 4, R1C6 cannot hold 4.- Because row 2 already has a 4, R2C4 and R2C5 cannot hold 4.So 4 must go in R3C4.
  3. R1C1 = 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, R2C1 and R2C2 cannot hold 4.- Because row 3 already has a 4, R3C1 and R3C3 cannot hold 4.So 4 must go in R1C1.
  4. R2C8 = 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 and R1C9 cannot hold 1.- Because column 9 already has a 1, R2C9 cannot hold 1.- Because row 3 already has a 1, R3C7 and R3C8 cannot hold 1.So 1 must go in R2C8.
  5. R3C7 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has an 8, R1C7 and R1C9 cannot hold 8.- Because column 9 already has an 8, R2C9 cannot hold 8.- Because column 8 already has an 8, R3C8 cannot hold 8.So 8 must go in R3C7.
  6. R2C9 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 9, R1C7 and R1C9 cannot hold 9.- Because row 3 already has a 9, R3C8 cannot hold 9.So 9 must go in R2C9.
  7. R6C4 = 2 · Hidden SingleIn the centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C4 and R4C6 cannot hold 2.- Because column 5 already has a 2, R5C5 and R6C5 cannot hold 2.- Because column 6 already has a 2, R6C6 cannot hold 2.So 2 must go in R6C4.
  8. R4C6 = 7 · Hidden SingleIn the centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 7, R4C4 cannot hold 7.- Because row 5 already has a 7, R5C5 cannot hold 7.- Because row 6 already has a 7, R6C5 and R6C6 cannot hold 7.So 7 must go in R4C6.
  9. R1C5 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 7, R1C6 cannot hold 7.- Because row 2 already has a 7, R2C4 and R2C5 cannot hold 7.So 7 must go in R1C5.
  10. R2C5 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 5, R1C6 cannot hold 5.- Because column 4 already has a 5, R2C4 cannot hold 5.So 5 must go in R2C5.
  11. R3C3 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C1 and R2C2 cannot hold 5.- Because column 1 already has a 5, R3C1 cannot hold 5.So 5 must go in R3C3.
  12. R1C6 = 6 · Hidden SingleIn the top-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 6, R2C4 cannot hold 6.So 6 must go in R1C6.
  13. R2C4 = 3 · Naked SingleLooking at R2C4:- Row 2 already has 1, 2, 4, 5, 7 and 9 → ruled out.- Column 4 already has 6 and 8 → ruled out.So this cell can only be 3.
  14. R4C4 = 1 · Naked SingleLooking at R4C4:- Row 4 already has 2, 3, 4, 5, 6, 7 and 8 → ruled out.- Column 4 already has 9 → ruled out.So this cell can only be 1.
  15. R4C2 = 9 · Naked SingleLooking at R4C2:- Row 4 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  16. R3C1 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 3, R2C1 and R2C2 cannot hold 3.So 3 must go in R3C1.
  17. R3C8 = 6 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  18. R7C8 = 3 · Naked SingleLooking at R7C8:- Row 7 already has 2, 4, 7 and 8 → ruled out.- Column 8 already has 1, 5, 6 and 9 → ruled out.So this cell can only be 3.
  19. R1C7 = 3 · Hidden SingleIn the top-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 3, R1C9 cannot hold 3.So 3 must go in R1C7.
  20. R1C9 = 5 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  21. 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 and R5C3 cannot hold 5.- Because column 1 already has a 5, R6C1 cannot hold 5.- Because column 3 already has a 5, R6C3 cannot hold 5.So 5 must go in R6C2.
  22. R6C1 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 6, R5C1 and R5C3 cannot hold 6.- Because column 3 already has a 6, R6C3 cannot hold 6.So 6 must go in R6C1.
  23. 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, R2C1 cannot hold 6.So 6 must go in R2C2.
  24. R2C1 = 8 · Naked SingleLooking at R2C1:- Row 2 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  25. R8C2 = 8 · Naked SingleLooking at R8C2:- Row 8 already has 2, 4, 5 and 7 → ruled out.- Column 2 already has 1, 3, 6 and 9 → ruled out.So this cell can only be 8.
  26. R6C6 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 9, R5C5 cannot hold 9.- Because column 5 already has a 9, R6C5 cannot hold 9.So 9 must go in R6C6.
  27. R8C6 = 3 · Naked SingleLooking at R8C6:- Row 8 already has 2, 4, 5, 7 and 8 → ruled out.- Column 6 already has 1, 6 and 9 → ruled out.So this cell can only be 3.
  28. R6C5 = 3 · Hidden SingleIn the centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C5 cannot hold 3.So 3 must go in R6C5.
  29. R5C5 = 8 · Naked SingleLooking at R5C5:- Row 5 already has 3, 5, 6, 7 and 9 → ruled out.- Column 5 already has 2 and 4 → ruled out.- The centre 3×3 box already has 1 → ruled out.So this cell can only be 8.
  30. R6C3 = 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 an 8, R5C1 and R5C3 cannot hold 8.So 8 must go in R6C3.
  31. R6C9 = 4 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  32. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 2, 3, 4, 7 and 8 → ruled out.- Column 9 already has 1, 5 and 9 → ruled out.So this cell can only be 6.
  33. R5C7 = 2 · Naked SingleLooking at R5C7:- Row 5 already has 3, 5, 6, 7, 8 and 9 → ruled out.- Column 7 already has 1 and 4 → ruled out.So this cell can only be 2.
  34. R5C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 4, R5C1 cannot hold 4.So 4 must go in R5C3.
  35. R5C1 = 1 · Naked SingleLooking at R5C1:- Row 5 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  36. R7C1 = 9 · Naked SingleLooking at R7C1:- Row 7 already has 2, 3, 4, 6, 7 and 8 → ruled out.- Column 1 already has 1 and 5 → ruled out.So this cell can only be 9.
  37. R8C3 = 1 · Naked SingleLooking at R8C3:- Row 8 already has 2, 3, 4, 5, 7 and 8 → ruled out.- Column 3 already has 6 and 9 → ruled out.So this cell can only be 1.
  38. R7C5 = 1 · Hidden SingleIn the bottom-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 1, R8C5 cannot hold 1.So 1 must go in R7C5.
  39. 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.
  40. R8C5 = 6 · Naked SingleLooking at R8C5:- Row 8 already has 1, 2, 3, 4, 5, 7 and 8 → ruled out.- Column 5 already has 9 → ruled out.So this cell can only be 6.
  41. R8C7 = 9 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.

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