Skudo
ENVI
Play on web

Daily Sudoku — 10 October 2026

The puzzle

  • 10 October 2026
  • Medium
  • 34 clues
Puzzle
·····42·8
·5·····6·
96··52··7
·9·4·5···
2·····5··
48··71693
5······4·
·48·27··1
1··648·2·
Solution
713964258
852713469
964852317
396485172
271396584
485271693
527139846
648527931
139648725

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. R2C3 = 2 · Hidden SingleIn the top-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 2, R1C1, R1C2 and R1C3 cannot hold 2.- Because column 1 already has a 2, R2C1 cannot hold 2.- Because row 3 already has a 2, R3C3 cannot hold 2.So 2 must go in R2C3.
  2. R3C3 = 4 · Hidden SingleIn the top-left 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 1 already has a 4, R1C1, R1C2 and R1C3 cannot hold 4.- Because column 1 already has a 4, R2C1 cannot hold 4.So 4 must go in R3C3.
  3. R2C1 = 8 · Hidden SingleIn the top-left 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 1 already has an 8, R1C1, R1C2 and R1C3 cannot hold 8.So 8 must go in R2C1.
  4. R1C5 = 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, R1C4 cannot hold 6.- Because row 2 already has a 6, R2C4, R2C5 and R2C6 cannot hold 6.- Because row 3 already has a 6, R3C4 cannot hold 6.So 6 must go in R1C5.
  5. R3C4 = 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, R1C4 cannot hold 8.- Because row 2 already has an 8, R2C4, R2C5 and R2C6 cannot hold 8.So 8 must go in R3C4.
  6. R1C8 = 5 · Hidden SingleIn the top-right 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 5, R2C7 and R2C9 cannot hold 5.- Because row 3 already has a 5, R3C7 and R3C8 cannot hold 5.So 5 must go in R1C8.
  7. R6C3 = 5 · Hidden SingleIn the middle-left 3×3 box, 5 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 5, R4C1 and R4C3 cannot hold 5.- Because row 5 already has a 5, R5C2 and R5C3 cannot hold 5.So 5 must go in R6C3.
  8. R6C4 = 2 · Naked SingleLooking at R6C4:- Row 6 already has 1, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 2.
  9. R5C6 = 6 · Hidden SingleIn the centre 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 6, R4C5 and R5C5 cannot hold 6.- Because column 4 already has a 6, R5C4 cannot hold 6.So 6 must go in R5C6.
  10. R4C9 = 2 · Hidden SingleIn the middle-right 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 2, R4C7 cannot hold 2.- Because column 8 already has a 2, R4C8 cannot hold 2.- Because row 5 already has a 2, R5C8 and R5C9 cannot hold 2.So 2 must go in R4C9.
  11. R5C9 = 4 · Hidden SingleIn the middle-right 3×3 box, 4 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 4, R4C7 and R4C8 cannot hold 4.- Because column 8 already has a 4, R5C8 cannot hold 4.So 4 must go in R5C9.
  12. R2C7 = 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, R2C9 cannot hold 4.- Because row 3 already has a 4, R3C7 and R3C8 cannot hold 4.So 4 must go in R2C7.
  13. R2C9 = 9 · Hidden SingleIn the top-right 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 3 already has a 9, R3C7 and R3C8 cannot hold 9.So 9 must go in R2C9.
  14. R1C4 = 9 · Hidden SingleIn the top-centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 2 already has a 9, R2C4, R2C5 and R2C6 cannot hold 9.So 9 must go in R1C4.
  15. R2C4 = 7 · Hidden SingleIn the top-centre 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 5 already has a 7, R2C5 cannot hold 7.- Because column 6 already has a 7, R2C6 cannot hold 7.So 7 must go in R2C4.
  16. R2C5 = 1 · Hidden SingleIn the top-centre 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 6 already has a 1, R2C6 cannot hold 1.So 1 must go in R2C5.
  17. R2C6 = 3 · Naked SingleLooking at R2C6:- Row 2 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  18. R7C6 = 9 · Naked SingleLooking at R7C6:- Row 7 already has 4 and 5 → ruled out.- Column 6 already has 1, 2, 3, 6, 7 and 8 → ruled out.So this cell can only be 9.
  19. R5C5 = 9 · Hidden SingleIn the centre 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 4 already has a 9, R4C5 cannot hold 9.- Because column 4 already has a 9, R5C4 cannot hold 9.So 9 must go in R5C5.
  20. R4C5 = 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, R5C4 cannot hold 8.So 8 must go in R4C5.
  21. R7C5 = 3 · Naked SingleLooking at R7C5:- Row 7 already has 4, 5 and 9 → ruled out.- Column 5 already has 1, 2, 6, 7 and 8 → ruled out.So this cell can only be 3.
  22. R5C4 = 3 · Naked SingleLooking at R5C4:- Row 5 already has 2, 4, 5, 6 and 9 → ruled out.- Column 4 already has 7 and 8 → ruled out.- The centre 3×3 box already has 1 → ruled out.So this cell can only be 3.
  23. R5C8 = 8 · Hidden SingleIn the middle-right 3×3 box, 8 fits in only one cell. Looking at the other empty cells:- Because row 4 already has an 8, R4C7 and R4C8 cannot hold 8.So 8 must go in R5C8.
  24. R7C2 = 2 · Hidden SingleIn the bottom-left 3×3 box, 2 fits in only one cell. Looking at the other empty cells:- Because column 3 already has a 2, R7C3 cannot hold 2.- Because row 8 already has a 2, R8C1 cannot hold 2.- Because row 9 already has a 2, R9C2 and R9C3 cannot hold 2.So 2 must go in R7C2.
  25. R9C3 = 9 · Hidden SingleIn the bottom-left 3×3 box, 9 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 9, R7C3 cannot hold 9.- Because column 1 already has a 9, R8C1 cannot hold 9.- Because column 2 already has a 9, R9C2 cannot hold 9.So 9 must go in R9C3.
  26. R7C4 = 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, R8C4 cannot hold 1.So 1 must go in R7C4.
  27. R8C4 = 5 · Naked SingleLooking at R8C4:- Row 8 already has 1, 2, 4, 7 and 8 → ruled out.- Column 4 already has 3, 6 and 9 → ruled out.So this cell can only be 5.
  28. 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.- Because row 8 already has a 5, R8C7 and R8C8 cannot hold 5.- Because column 7 already has a 5, R9C7 cannot hold 5.So 5 must go in R9C9.
  29. R7C9 = 6 · Naked SingleLooking at R7C9:- Row 7 already has 1, 2, 3, 4, 5 and 9 → ruled out.- Column 9 already has 7 and 8 → ruled out.So this cell can only be 6.
  30. R8C1 = 6 · Hidden SingleIn the bottom-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 6, R7C3 cannot hold 6.- Because row 9 already has a 6, R9C2 cannot hold 6.So 6 must go in R8C1.
  31. R4C3 = 6 · Hidden SingleIn the middle-left 3×3 box, 6 fits in only one cell. Looking at the other empty cells:- Because column 1 already has a 6, R4C1 cannot hold 6.- Because row 5 already has a 6, R5C2 and R5C3 cannot hold 6.So 6 must go in R4C3.
  32. R4C1 = 3 · Hidden SingleIn the middle-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 5 already has a 3, R5C2 and R5C3 cannot hold 3.So 3 must go in R4C1.
  33. R1C1 = 7 · Naked SingleLooking at R1C1:- Row 1 already has 2, 4, 5, 6, 8 and 9 → ruled out.- Column 1 already has 1 and 3 → ruled out.So this cell can only be 7.
  34. R9C2 = 3 · Hidden SingleIn the bottom-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because row 7 already has a 3, R7C3 cannot hold 3.So 3 must go in R9C2.
  35. R9C7 = 7 · Naked SingleLooking at R9C7:- Row 9 already has 1, 2, 3, 4, 5, 6, 8 and 9 → ruled out.So this cell can only be 7.
  36. R7C3 = 7 · Naked SingleLooking at R7C3:- Row 7 already has 1, 2, 3, 4, 5, 6 and 9 → ruled out.- Column 3 already has 8 → ruled out.So this cell can only be 7.
  37. R7C7 = 8 · Naked SingleLooking at R7C7:- Row 7 already has 1, 2, 3, 4, 5, 6, 7 and 9 → ruled out.So this cell can only be 8.
  38. R1C3 = 3 · Hidden SingleIn the top-left 3×3 box, 3 fits in only one cell. Looking at the other empty cells:- Because column 2 already has a 3, R1C2 cannot hold 3.So 3 must go in R1C3.
  39. R1C2 = 1 · Naked SingleLooking at R1C2:- Row 1 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  40. R5C2 = 7 · Naked SingleLooking at R5C2:- Row 5 already has 2, 3, 4, 5, 6, 8 and 9 → ruled out.- Column 2 already has 1 → ruled out.So this cell can only be 7.
  41. R5C3 = 1 · Naked SingleLooking at R5C3:- Row 5 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  42. R4C8 = 7 · Hidden SingleIn the middle-right 3×3 box, 7 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 7, R4C7 cannot hold 7.So 7 must go in R4C8.
  43. R4C7 = 1 · Naked SingleLooking at R4C7:- Row 4 already has 2, 3, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 1.
  44. R3C8 = 1 · Hidden SingleIn the top-right 3×3 box, 1 fits in only one cell. Looking at the other empty cells:- Because column 7 already has a 1, R3C7 cannot hold 1.So 1 must go in R3C8.
  45. R3C7 = 3 · Naked SingleLooking at R3C7:- Row 3 already has 1, 2, 4, 5, 6, 7, 8 and 9 → ruled out.So this cell can only be 3.
  46. R8C7 = 9 · Naked SingleLooking at R8C7:- Row 8 already has 1, 2, 4, 5, 6, 7 and 8 → ruled out.- Column 7 already has 3 → ruled out.So this cell can only be 9.
  47. 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.

First to finish

Ranked by when each player finished, not by how long they took — so looking up a solution never moves anyone up this board.

Open the app to see who finished first.

Recent days

Full archive →