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Daily Sudoku — 26 September 2026

The puzzle

  • 26 September 2026
  • Hard
  • 30 clues
Puzzle
···4·7··9
······5··
·18····74
·29·3·4·7
····92·8·
6··7··9··
···5·····
·3····698
·418·973·
Solution
365427819
472981563
918356274
129638457
754192386
683745921
896573142
537214698
241869735

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 — 51 steps using 3 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
  • Locked Candidates (pointing/claiming) — a digit inside a box is confined to one row or column, so it is eliminated from the rest of that row or column
  1. R1C7 = 8 · Hidden SingleIn the top-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 8 already has an 8, R1C8 and R2C8 cannot hold 8.- Because column 9 already has an 8, R2C9 cannot hold 8.- Because row 3 already has an 8, R3C7 cannot hold 8.So 8 must go in R1C7.
  2. R7C6 = 3 · Hidden SingleIn the bottom-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R7C5 cannot hold 3.- Because row 8 already has a 3, R8C4, R8C5 and R8C6 cannot hold 3.- Because row 9 already has a 3, R9C5 cannot hold 3.So 3 must go in R7C6.
  3. R7C8 = 4 · Hidden SingleIn the bottom-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 4, R7C7 cannot hold 4.- Because column 9 already has a 4, R7C9 cannot hold 4.- Because row 9 already has a 4, R9C9 cannot hold 4.So 4 must go in R7C8.
  4. R9C9 = 5 · Hidden SingleIn the bottom-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 5, R7C7 and R7C9 cannot hold 5.So 5 must go in R9C9.
  5. R9C5 = 6 · Hidden SingleIn row 9, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R9C1 cannot hold 6.So 6 must go in R9C5.
  6. R9C1 = 2 · Naked SingleLooking at R9C1:- Row 9 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  7. R6C2 = 8 · Naked SingleStep 1 · Places in one box: clear the rest of it (claiming)Row 5 has 5 only in R5C1, R5C2 and R5C3, all in the middle-left 3×3 box. The rest of that box loses 5: R4C1, R6C2 and R6C3.Step 2 · Collect what the eliminations unlockR6C2 has one candidate left, 8 (lost 5 in step 1) → place 8 in R6C2.
  8. R4C6 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 4 already has an 8, R4C4 cannot hold 8.- Because row 5 already has an 8, R5C4 cannot hold 8.- Because row 6 already has an 8, R6C5 and R6C6 cannot hold 8.So 8 must go in R4C6.
  9. R2C5 = 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, R1C5 cannot hold 8.- Because column 4 already has an 8, R2C4 cannot hold 8.- Because column 6 already has an 8, R2C6 cannot hold 8.- Because row 3 already has an 8, R3C4, R3C5 and R3C6 cannot hold 8.So 8 must go in R2C5.
  10. R7C1 = 8 · Hidden SingleIn the bottom-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because column 2 already has an 8, R7C2 cannot hold 8.- Because column 3 already has an 8, R7C3 cannot hold 8.- Because row 8 already has an 8, R8C1 and R8C3 cannot hold 8.So 8 must go in R7C1.
  11. R7C2 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 9, R7C3 cannot hold 9.- Because row 8 already has a 9, R8C1 and R8C3 cannot hold 9.So 9 must go in R7C2.
  12. R7C3 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 6, R8C1 and R8C3 cannot hold 6.So 6 must go in R7C3.
  13. R7C5 = 7 · Hidden SingleIn row 7, 7 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 7, R7C7 cannot hold 7.- Because column 9 already has a 7, R7C9 cannot hold 7.So 7 must go in R7C5.
  14. R4C8 = 5 · Hidden SingleStep 1 · Pick a digit and a box, and find where it can goThe centre 3×3 box has no 5 yet; 5 can only go in R6C5 and R6C6.Step 2 · Places on one row: point along it (pointing)All of them lie on row 6, so the box's 5 is on row 6. The rest of row 6 loses 5: R6C3 and R6C8.Step 3 · Collect what the eliminations unlockIn the middle-right 3×3 box, 5 has one place left, R4C8 (lost 5: R6C8 in step 2) → place 5 in R4C8.
  15. R5C9 = 6 · Hidden SingleIn the middle-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 6, R5C7 cannot hold 6.- Because row 6 already has a 6, R6C8 and R6C9 cannot hold 6.So 6 must go in R5C9.
  16. R4C4 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 6, R5C4 cannot hold 6.- Because row 6 already has a 6, R6C5 and R6C6 cannot hold 6.So 6 must go in R4C4.
  17. R4C1 = 1 · Naked SingleLooking at R4C1:- Row 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  18. R3C6 = 6 · Hidden SingleIn row 3, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R3C1 cannot hold 6.- Because column 4 already has a 6, R3C4 cannot hold 6.- Because column 5 already has a 6, R3C5 cannot hold 6.- Because column 7 already has a 6, R3C7 cannot hold 6.So 6 must go in R3C6.
  19. R6C6 = 5 · Hidden SingleIn column 6, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C6 cannot hold 5.- Because the bottom-centre 3×3 box already has a 5, R8C6 cannot hold 5.So 5 must go in R6C6.
  20. R6C5 = 4 · Hidden SingleIn the centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R5C4 cannot hold 4.So 4 must go in R6C5.
  21. R5C4 = 1 · Naked SingleLooking at R5C4:- Row 5 already has 2, 6, 8 and 9 → ruled out.- Column 4 already has 4, 5 and 7 → ruled out.- The centre 3×3 box already has 3 → ruled out.So this cell can only be 1.
  22. R8C6 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 4 already has a 4, R8C4 cannot hold 4.- Because column 5 already has a 4, R8C5 cannot hold 4.So 4 must go in R8C6.
  23. R2C6 = 1 · Naked SingleLooking at R2C6:- Row 2 already has 5 and 8 → ruled out.- Column 6 already has 2, 3, 4, 6, 7 and 9 → ruled out.So this cell can only be 1.
  24. R1C8 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 1, R2C8 and R2C9 cannot hold 1.- Because row 3 already has a 1, R3C7 cannot hold 1.So 1 must go in R1C8.
  25. R2C8 = 6 · Hidden SingleIn the top-right 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 6, R2C9 cannot hold 6.- Because row 3 already has a 6, R3C7 cannot hold 6.So 6 must go in R2C8.
  26. R6C8 = 2 · Naked SingleLooking at R6C8:- Row 6 already has 4, 5, 6, 7, 8 and 9 → ruled out.- Column 8 already has 1 and 3 → ruled out.So this cell can only be 2.
  27. R1C2 = 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 2 already has a 6, R2C1, R2C2 and R2C3 cannot hold 6.- Because row 3 already has a 6, R3C1 cannot hold 6.So 6 must go in R1C2.
  28. R6C9 = 1 · Hidden SingleIn the middle-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 1, R5C7 cannot hold 1.So 1 must go in R6C9.
  29. R6C3 = 3 · Naked SingleLooking at R6C3:- Row 6 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  30. R5C7 = 3 · Naked SingleLooking at R5C7:- Row 5 already has 1, 2, 6, 8 and 9 → ruled out.- Column 7 already has 4, 5 and 7 → ruled out.So this cell can only be 3.
  31. R2C9 = 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, R3C7 cannot hold 3.So 3 must go in R2C9.
  32. R7C9 = 2 · Naked SingleLooking at R7C9:- Row 7 already has 3, 4, 5, 6, 7, 8 and 9 → ruled out.- Column 9 already has 1 → ruled out.So this cell can only be 2.
  33. 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.
  34. R3C7 = 2 · Naked SingleLooking at R3C7:- Row 3 already has 1, 4, 6, 7 and 8 → ruled out.- Column 7 already has 3, 5 and 9 → ruled out.So this cell can only be 2.
  35. R3C4 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 3, R1C5 and R3C5 cannot hold 3.- Because row 2 already has a 3, R2C4 cannot hold 3.So 3 must go in R3C4.
  36. R1C1 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 3, R1C3 cannot hold 3.- Because row 2 already has a 3, R2C1, R2C2 and R2C3 cannot hold 3.- Because row 3 already has a 3, R3C1 cannot hold 3.So 3 must go in R1C1.
  37. R2C4 = 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, R1C5 cannot hold 9.- Because column 5 already has a 9, R3C5 cannot hold 9.So 9 must go in R2C4.
  38. R8C4 = 2 · Naked SingleLooking at R8C4:- Row 8 already has 3, 4, 6, 8 and 9 → ruled out.- Column 4 already has 1, 5 and 7 → ruled out.So this cell can only be 2.
  39. R8C5 = 1 · Naked SingleLooking at R8C5:- Row 8 already has 2, 3, 4, 6, 8 and 9 → ruled out.- Column 5 already has 7 → ruled out.- The bottom-centre 3×3 box already has 5 → ruled out.So this cell can only be 1.
  40. R3C1 = 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, R1C3 cannot hold 9.- Because row 2 already has a 9, R2C1, R2C2 and R2C3 cannot hold 9.So 9 must go in R3C1.
  41. R3C5 = 5 · Naked SingleLooking at R3C5:- Row 3 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  42. R1C5 = 2 · Naked SingleLooking at R1C5:- Row 1 already has 1, 3, 4, 6, 7, 8 and 9 → ruled out.- Column 5 already has 5 → ruled out.So this cell can only be 2.
  43. R1C3 = 5 · Naked SingleLooking at R1C3:- Row 1 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  44. R2C3 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 2, R2C1 cannot hold 2.- Because column 2 already has a 2, R2C2 cannot hold 2.So 2 must go in R2C3.
  45. R2C1 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 4, R2C2 cannot hold 4.So 4 must go in R2C1.
  46. R2C2 = 7 · Naked SingleLooking at R2C2:- Row 2 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  47. R5C2 = 5 · Naked SingleLooking at R5C2:- Row 5 already has 1, 2, 3, 6, 8 and 9 → ruled out.- Column 2 already has 4 and 7 → ruled out.So this cell can only be 5.
  48. 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.
  49. R5C1 = 7 · Naked SingleLooking at R5C1:- Row 5 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  50. R8C1 = 5 · Naked SingleLooking at R8C1:- Row 8 already has 1, 2, 3, 4, 6, 8 and 9 → ruled out.- Column 1 already has 7 → ruled out.So this cell can only be 5.
  51. R8C3 = 7 · Naked SingleLooking at R8C3:- Row 8 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.

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