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Daily Sudoku — 27 September 2026

The puzzle

  • 27 September 2026
  • Medium
  • 34 clues
Puzzle
··3···1··
87··41···
9·1·2···8
3····2·81
65····4··
1····42··
·18·3·5··
7·5··89·2
·397·58··
Solution
523876149
876941325
941523678
394652781
652187493
187394256
418239567
765418932
239765814

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 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. R1C1 = 5 · Hidden SingleIn the top-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 5, R1C2 and R3C2 cannot hold 5.- Because column 3 already has a 5, R2C3 cannot hold 5.So 5 must go in R1C1.
  2. R5C3 = 2 · Hidden SingleIn the middle-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 2, R4C2 and R4C3 cannot hold 2.- Because row 6 already has a 2, R6C2 and R6C3 cannot hold 2.So 2 must go in R5C3.
  3. 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, R2C3 cannot hold 2.- Because row 3 already has a 2, R3C2 cannot hold 2.So 2 must go in R1C2.
  4. R3C2 = 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.So 4 must go in R3C2.
  5. R2C3 = 6 · Naked SingleLooking at R2C3:- Row 2 already has 1, 4, 7 and 8 → ruled out.- Column 3 already has 2, 3, 5 and 9 → ruled out.So this cell can only be 6.
  6. R2C8 = 2 · Hidden SingleIn the top-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C8 and R1C9 cannot hold 2.- Because column 7 already has a 2, R2C7 cannot hold 2.- Because column 9 already has a 2, R2C9 cannot hold 2.- Because row 3 already has a 2, R3C7 and R3C8 cannot hold 2.So 2 must go in R2C8.
  7. R4C3 = 4 · Hidden SingleIn the middle-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 4, R4C2 cannot hold 4.- Because row 6 already has a 4, R6C2 and R6C3 cannot hold 4.So 4 must go in R4C3.
  8. R6C3 = 7 · Naked SingleLooking at R6C3:- Row 6 already has 1, 2 and 4 → ruled out.- Column 3 already has 3, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  9. R6C2 = 8 · Hidden SingleIn the middle-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has an 8, R4C2 cannot hold 8.So 8 must go in R6C2.
  10. R4C2 = 9 · Naked SingleLooking at R4C2:- Row 4 already has 1, 2, 3, 4 and 8 → ruled out.- Column 2 already has 5 and 7 → ruled out.- The middle-left 3×3 box already has 6 → ruled out.So this cell can only be 9.
  11. R8C2 = 6 · Naked SingleLooking at R8C2:- Row 8 already has 2, 5, 7, 8 and 9 → ruled out.- Column 2 already has 1, 3 and 4 → ruled out.So this cell can only be 6.
  12. R7C4 = 2 · Hidden SingleIn the bottom-centre 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 2, R7C6 cannot hold 2.- Because row 8 already has a 2, R8C4 and R8C5 cannot hold 2.- Because column 5 already has a 2, R9C5 cannot hold 2.So 2 must go in R7C4.
  13. R9C1 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 2, R7C1 cannot hold 2.So 2 must go in R9C1.
  14. R7C1 = 4 · Naked SingleLooking at R7C1:- Row 7 already has 1, 2, 3, 5 and 8 → ruled out.- Column 1 already has 6, 7 and 9 → ruled out.So this cell can only be 4.
  15. R8C4 = 4 · Hidden SingleIn the bottom-centre 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 4, R7C6 cannot hold 4.- Because column 5 already has a 4, R8C5 and R9C5 cannot hold 4.So 4 must go in R8C4.
  16. R7C6 = 9 · Hidden SingleIn the bottom-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 9, R8C5 cannot hold 9.- Because row 9 already has a 9, R9C5 cannot hold 9.So 9 must go in R7C6.
  17. R9C5 = 6 · Hidden SingleIn the bottom-centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 8 already has a 6, R8C5 cannot hold 6.So 6 must go in R9C5.
  18. R8C5 = 1 · Naked SingleLooking at R8C5:- Row 8 already has 2, 4, 5, 6, 7, 8 and 9 → ruled out.- Column 5 already has 3 → ruled out.So this cell can only be 1.
  19. R8C8 = 3 · Naked SingleLooking at R8C8:- Row 8 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  20. R5C4 = 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 and R4C5 cannot hold 1.- Because column 5 already has a 1, R5C5 cannot hold 1.- Because column 6 already has a 1, R5C6 cannot hold 1.- Because row 6 already has a 1, R6C4 and R6C5 cannot hold 1.So 1 must go in R5C4.
  21. R5C5 = 8 · Hidden SingleIn the centre 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has an 8, R4C4 and R4C5 cannot hold 8.- Because column 6 already has an 8, R5C6 cannot hold 8.- Because row 6 already has an 8, R6C4 and R6C5 cannot hold 8.So 8 must go in R5C5.
  22. R1C4 = 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 an 8, R1C5 cannot hold 8.- Because column 6 already has an 8, R1C6 cannot hold 8.- Because row 2 already has an 8, R2C4 cannot hold 8.- Because row 3 already has an 8, R3C4 and R3C6 cannot hold 8.So 8 must go in R1C4.
  23. R9C8 = 1 · Hidden SingleIn the bottom-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 1, R7C8 and R7C9 cannot hold 1.- Because column 9 already has a 1, R9C9 cannot hold 1.So 1 must go in R9C8.
  24. R9C9 = 4 · Naked SingleLooking at R9C9:- Row 9 already has 1, 2, 3, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 4.
  25. R1C8 = 4 · Hidden SingleIn the top-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 4, R1C9 cannot hold 4.- Because row 2 already has a 4, R2C7 and R2C9 cannot hold 4.- Because row 3 already has a 4, R3C7 and R3C8 cannot hold 4.So 4 must go in R1C8.
  26. R2C7 = 3 · Naked SingleLooking at R2C7:- Row 2 already has 1, 2, 4, 6, 7 and 8 → ruled out.- Column 7 already has 5 and 9 → ruled out.So this cell can only be 3.
  27. R4C5 = 5 · Hidden SingleStep 1 · Pick a digit and a box, and find where it can goThe top-centre 3×3 box has no 5 yet; 5 can only go in R2C4 and R3C4.Step 2 · Places on one column: point along it (pointing)All of them lie on column 4, so the box's 5 is on column 4. The rest of column 4 loses 5: R4C4 and R6C4.Step 3 · Pick a digit and a box, and find where it can goThe middle-right 3×3 box has no 5 yet; 5 can only go in R6C8 and R6C9.Step 4 · 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: R6C5.Step 5 · Collect what the eliminations unlockIn the centre 3×3 box, 5 has one place left, R4C5 (lost 5: R4C4 in step 2, R6C4 in step 2 and R6C5 in step 4) → place 5 in R4C5.
  28. R5C6 = 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 6 already has a 7, R6C4 and R6C5 cannot hold 7.So 7 must go in R5C6.
  29. 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 and R3C6 cannot hold 7.- Because row 2 already has a 7, R2C4 cannot hold 7.- Because column 4 already has a 7, R3C4 cannot hold 7.So 7 must go in R1C5.
  30. R6C5 = 9 · Naked SingleLooking at R6C5:- Row 6 already has 1, 2, 4, 7 and 8 → ruled out.- Column 5 already has 3, 5 and 6 → ruled out.So this cell can only be 9.
  31. R2C4 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 9, R1C6 cannot hold 9.- Because row 3 already has a 9, R3C4 and R3C6 cannot hold 9.So 9 must go in R2C4.
  32. R2C9 = 5 · Naked SingleLooking at R2C9:- Row 2 already has 1, 2, 3, 4, 6, 7, 8 and 9 → ruled out.So this cell can only be 5.
  33. R3C4 = 5 · Hidden SingleIn the top-centre 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 5, R1C6 cannot hold 5.- Because column 6 already has a 5, R3C6 cannot hold 5.So 5 must go in R3C4.
  34. R3C6 = 3 · Hidden SingleIn the top-centre 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 3, R1C6 cannot hold 3.So 3 must go in R3C6.
  35. R1C6 = 6 · Naked SingleLooking at R1C6:- Row 1 already has 1, 2, 3, 4, 5, 7 and 8 → ruled out.- Column 6 already has 9 → ruled out.So this cell can only be 6.
  36. R1C9 = 9 · Naked SingleLooking at R1C9:- Row 1 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  37. R6C4 = 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 cannot hold 3.So 3 must go in R6C4.
  38. R4C4 = 6 · Naked SingleLooking at R4C4:- Row 4 already has 1, 2, 3, 4, 5, 8 and 9 → ruled out.- Column 4 already has 7 → ruled out.So this cell can only be 6.
  39. R4C7 = 7 · Naked SingleLooking at R4C7:- Row 4 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  40. R3C7 = 6 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 3, 4, 5, 8 and 9 → ruled out.- Column 7 already has 7 → ruled out.So this cell can only be 6.
  41. R3C8 = 7 · Naked SingleLooking at R3C8:- Row 3 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  42. R5C9 = 3 · Hidden SingleIn the middle-right 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 8 already has a 3, R5C8 cannot hold 3.- Because row 6 already has a 3, R6C8 and R6C9 cannot hold 3.So 3 must go in R5C9.
  43. R5C8 = 9 · Naked SingleLooking at R5C8:- Row 5 already has 1, 2, 3, 4, 5, 6, 7 and 8 → ruled out.So this cell can only be 9.
  44. R6C8 = 5 · Hidden SingleIn the middle-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because column 9 already has a 5, R6C9 cannot hold 5.So 5 must go in R6C8.
  45. R6C9 = 6 · Naked SingleLooking at R6C9:- Row 6 already has 1, 2, 3, 4, 5, 7, 8 and 9 → ruled out.So this cell can only be 6.
  46. R7C8 = 6 · Naked SingleLooking at R7C8:- Row 7 already has 1, 2, 3, 4, 5, 8 and 9 → ruled out.- Column 8 already has 7 → ruled out.So this cell can only be 6.
  47. R7C9 = 7 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.

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