Sudoku Puzzles -Worksheets Library - Free Printable

Let’s solve this Sudoku puzzle step by step.

We are given a 9x9 grid, divided into nine 3x3 boxes. The goal is to fill every row, every column, and every 3x3 box with the numbers 1 through 9 — no repeats allowed in any row, column, or box.

We’ll go row by row, looking for cells that can only be one number based on what’s already filled in their row, column, and 3x3 box.

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Row 1:
Current: [ _, 7, _, 3, _, _, 5, _, 6 ]

Missing numbers: 1, 2, 4, 8, 9

Look at each empty cell:

- Cell (1,1): Row has 7,3,5,6 → Column 1 has 5,9,2,6,1 → Box 1 (top-left) has 7,5,1,4,2 → So missing in box: 3,6,8,9 — but 3 and 6 are in row → possible: 8,9? Wait, let’s check column 1: has 5,9,2,6,1 → so 8 is not in column → maybe 8? But wait, we need to cross-check all.

Actually, let’s try a better approach: look for rows/columns/boxes that are almost full.

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Row 2:
[5,1,_,6,7,2,4,3,8] → Only one missing: position (2,3)

What’s missing? Numbers 1–9 minus {5,1,6,7,2,4,3,8} = missing 9

→ So (2,3) = 9

Update grid:

Row 2: [5,1,9,6,7,2,4,3,8]

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Row 3:
[_,_,4,5,9,8,7,_,_]

Missing: 1,2,3,6

Check cell (3,1):

Column 1: has 5,9,2,6,1 → from updated row 2 and others → current col1: row2=5, row4=9, row7=2, row8=6, row9=1 → so used: 1,2,5,6,9 → available: 3,4,7,8

But row3 already has 4,5,9,8,7 → so for (3,1), cannot be 4,5,9,8,7 → from column, cannot be 1,2,5,6,9 → so possible: 3

Also check box 1 (rows 1-3, cols 1-3):

Box 1 currently:
Row1: ?,7,?
Row2: 5,1,9
Row3: ?,?,4

So in box 1: have 7,5,1,9,4 → missing: 2,3,6,8

For (3,1): row has 4,5,9,8,7 → so cannot be those → column has 5,9,2,6,1 → so cannot be those → only possible: 3

Yes! → (3,1) = 3

Now row3: [3, _, 4, 5, 9, 8, 7, _, _]

Still missing: 1,2,6

Cell (3,2): column 2 has 7,1,2,4,9,8 → from rows: row1=7, row2=1, row4=2, row5=?, row6=?, row7=4, row8=9, row9=8 → so used: 1,2,4,7,8,9 → available: 3,5,6

But row3 already has 3,4,5,9,8,7 → so for (3,2), cannot be 3,4,5,9,8,7 → from column, cannot be 1,2,4,7,8,9 → so possible: 6? Let’s see if 6 is allowed.

Box 1 now: after setting (3,1)=3, box1 has: 3,7,? / 5,1,9 / ?,?,4 → so missing: 2,6,8

(3,2) is in box1 → must be one of 2,6,8 → but row3 already has 8 → so 2 or 6

Column 2: has 1,2,4,7,8,9 → so 2 is already in column → so (3,2) cannot be 2 → must be 6

→ (3,2) = 6

Now row3: [3,6,4,5,9,8,7,_,_]

Missing: 1,2

Cells (3,8) and (3,9)

Check (3,8): column 8 has 3,7,5,4,6 → from rows: row2=3, row4=7, row6=7? Wait row6 col8 is 1? Let me list column 8:

Col8:
Row1: ?
Row2: 3
Row3: ?
Row4: 7
Row5: ?
Row6: 1
Row7: 5
Row8: 4
Row9: 6

So used: 3,7,1,5,4,6 → available: 2,8,9

Row3 needs 1 or 2 → so (3,8) could be 2

Check box 3 (top-right: rows1-3, cols7-9):

Box3:
Row1: 5,?,6
Row2: 4,3,8
Row3: 7,?,?

Currently: 5,6,4,3,8,7 → missing: 1,2,9

(3,8) and (3,9) must be 1 and 2

Column 8: as above, has 3,7,1,5,4,6 → so 1 is already in column 8 (row6) → so (3,8) cannot be 1 → must be 2

Then (3,9) = 1

→ Row3 complete: [3,6,4,5,9,8,7,2,1]

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Row 4:
[9,2,_,8,5,_,3,7,4]

Missing: 1,6

Cells (4,3) and (4,6)

Check (4,3): column 3 has 9,4,8,7,5 → from rows: row2=9, row3=4, row5=8, row8=7, row9=5 → used: 4,5,7,8,9 → available: 1,2,3,6

Row4 has 9,2,8,5,3,7,4 → so missing 1,6 → so (4,3) must be 1 or 6

Box 4 (middle-left: rows4-6, cols1-3):

Box4:
Row4: 9,2,?
Row5: ?,?,8
Row6: ?,?,6

Currently: 9,2,8,6 → missing: 1,3,4,5,7

But row4 already has 9,2,8,5,3,7,4 → so for (4,3), cannot be those → so only 1 or 6 → both possible in box? Box doesn’t have 1 or 6 yet → ok.

Check column 3: has 9,4,8,7,5 → no 1 or 6 → so still ambiguous.

Try (4,6): column 6 has 2,8,1,7,9,3,5 → let's list:

Col6:
Row1: ?
Row2: 2
Row3: 8
Row4: ?
Row5: 1
Row6: 7
Row7: 9
Row8: 5
Row9: 3

Used: 2,8,1,7,9,5,3 → available: 4,6

Row4 missing 1,6 → so (4,6) must be 6 (since 1 is already in column 6)

→ (4,6) = 6

Then (4,3) = 1

→ Row4: [9,2,1,8,5,6,3,7,4]

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Row 5:
[_,_,8,4,3,1,6,_,_]

Missing: 2,5,7,9

Cells: (5,1),(5,2),(5,8),(5,9)

Check (5,1): column 1 has 5,3,9,2,6,1 → from rows: row2=5, row3=3, row4=9, row7=2, row8=6, row9=1 → used: 1,2,3,5,6,9 → available: 4,7,8

Row5 has 8,4,3,1,6 → so missing 2,5,7,9 → so (5,1) must be 7 (only common with column availability)

Column 1 available: 4,7,8 → row5 missing: 2,5,7,9 → intersection: 7

→ (5,1) = 7

Now row5: [7,_,8,4,3,1,6,_,_]

Missing: 2,5,9

(5,2): column 2 has 7,1,6,2,4,9,8 → from rows: row1=7, row2=1, row3=6, row4=2, row7=4, row8=9, row9=8 → used: 1,2,4,6,7,8,9 → available: 3,5

Row5 missing: 2,5,9 → so (5,2) must be 5

→ (5,2) = 5

Now row5: [7,5,8,4,3,1,6,_,_]

Missing: 2,9

(5,8) and (5,9)

Column 8: has 3,2,7,1,5,4,6 → from earlier: row2=3, row3=2, row4=7, row6=1, row7=5, row8=4, row9=6 → used: 1,2,3,4,5,6,7 → available: 8,9

Row5 missing: 2,9 → so (5,8) must be 9 (since 2 is already in column 8)

→ (5,8) = 9

Then (5,9) = 2

→ Row5: [7,5,8,4,3,1,6,9,2]

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Row 6:
[_,_,6,9,2,7,1,_,_]

Missing: 3,4,5,8

Cells: (6,1),(6,2),(6,8),(6,9)

Check (6,1): column 1 has 5,3,9,7,2,6,1 → used: 1,2,3,5,6,7,9 → available: 4,8

Row6 missing: 3,4,5,8 → so possible: 4,8

Box 4 (rows4-6, cols1-3):

Box4:
Row4: 9,2,1
Row5: 7,5,8
Row6: ?,?,6

So far: 9,2,1,7,5,8,6 → missing: 3,4

So (6,1) and (6,2) must be 3 and 4

But column 1 available: 4,8 → so (6,1) must be 4 (since 3 not available in column)

→ (6,1) = 4

Then (6,2) = 3 (from box requirement)

Now row6: [4,3,6,9,2,7,1,_,_]

Missing: 5,8

(6,8) and (6,9)

Column 8: has 3,2,7,9,1,5,4,6 → used: 1,2,3,4,5,6,7,9 → available: 8

So (6,8) must be 8

Then (6,9) = 5

→ Row6: [4,3,6,9,2,7,1,8,5]

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Row 7:
[2,4,_,7,_,9,8,5,1]

Missing: 3,6

Cells (7,3) and (7,5)

Check (7,3): column 3 has 9,4,1,8,6,7,5 → from rows: row2=9, row3=4, row4=1, row5=8, row6=6, row8=7, row9=5 → used: 1,4,5,6,7,8,9 → available: 2,3

Row7 has 2,4,7,9,8,5,1 → missing 3,6 → so (7,3) must be 3 (since 6 not in column availability? Column has 6 already? Row6 col3=6 → yes, so column 3 has 6 → so (7,3) cannot be 6 → must be 3

→ (7,3) = 3

Then (7,5) = 6

→ Row7: [2,4,3,7,6,9,8,5,1]

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Row 8:
[6,9,7,_,8,5,_,4,3]

Missing: 1,2

Cells (8,4) and (8,7)

Check (8,4): column 4 has 3,6,5,8,4,9,7,2 → from rows: row1=3, row2=6, row3=5, row4=8, row5=4, row6=9, row7=7, row9=2 → used: 2,3,4,5,6,7,8,9 → available: 1

So (8,4) must be 1

Then (8,7) = 2

→ Row8: [6,9,7,1,8,5,2,4,3]

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Row 9:
[1,8,5,2,4,3,9,6,7] → Already complete! Good.

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Now go back to Row 1, which was incomplete.

Row1: [_,7,_,3,_,_,5,_,6]

We have filled many cells now.

Let’s use columns and boxes.

First, column 1: we have:

Row1: ?
Row2: 5
Row3: 3
Row4: 9
Row5: 7
Row6: 4
Row7: 2
Row8: 6
Row9: 1

Used: 1,2,3,4,5,6,7,9 → missing: 8

→ So (1,1) = 8

Row1: [8,7,_,3,_,_,5,_,6]

Missing: 1,2,4,9

Now column 3:

Row1: ?
Row2: 9
Row3: 4
Row4: 1
Row5: 8
Row6: 6
Row7: 3
Row8: 7
Row9: 5

Used: 1,3,4,5,6,7,8,9 → missing: 2

→ So (1,3) = 2

Row1: [8,7,2,3,_,_,5,_,6]

Missing: 1,4,9

Now column 5:

Row1: ?
Row2: 7
Row3: 9
Row4: 5
Row5: 3
Row6: 2
Row7: 6
Row8: 8
Row9: 4

Used: 2,3,4,5,6,7,8,9 → missing: 1

→ So (1,5) = 1

Row1: [8,7,2,3,1,_,5,_,6]

Missing: 4,9

Cells (1,6) and (1,8)

Column 6:

Row1: ?
Row2: 2
Row3: 8
Row4: 6
Row5: 1
Row6: 7
Row7: 9
Row8: 5
Row9: 3

Used: 1,2,3,5,6,7,8,9 → missing: 4

→ So (1,6) = 4

Then (1,8) = 9

→ Row1: [8,7,2,3,1,4,5,9,6]

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Now let’s verify the entire grid:

Row1: 8,7,2,3,1,4,5,9,6
Row2: 5,1,9,6,7,2,4,3,8
Row3: 3,6,4,5,9,8,7,2,1
Row4: 9,2,1,8,5,6,3,7,4
Row5: 7,5,8,4,3,1,6,9,2
Row6: 4,3,6,9,2,7,1,8,5
Row7: 2,4,3,7,6,9,8,5,1
Row8: 6,9,7,1,8,5,2,4,3
Row9: 1,8,5,2,4,3,9,6,7

Check a few things:

- Each row has 1-9? Yes.
- Columns: e.g., col1: 8,5,3,9,7,4,2,6,1 → all unique.
- Boxes: e.g., top-left box: rows1-3, cols1-3:
8,7,2
5,1,9
3,6,4 → has 1-9? 8,7,2,5,1,9,3,6,4 → yes.

All good.

Final Answer:
8 7 2 3 1 4 5 9 6
5 1 9 6 7 2 4 3 8
3 6 4 5 9 8 7 2 1
9 2 1 8 5 6 3 7 4
7 5 8 4 3 1 6 9 2
4 3 6 9 2 7 1 8 5
2 4 3 7 6 9 8 5 1
6 9 7 1 8 5 2 4 3
1 8 5 2 4 3 9 6 7