100 Free Printable Sudoku Puzzles - Free Printable
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Show Answer Key & Explanations
Step-by-step solution for: 100 Free Printable Sudoku Puzzles
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Show Answer Key & Explanations
Step-by-step solution for: 100 Free Printable Sudoku Puzzles
To solve this Sudoku puzzle, we need to fill in the empty cells so that every row, every column, and every 3x3 box contains the numbers 1 through 9 exactly once.
Here is the step-by-step logical deduction to find the solution:
Step 1: Fill in obvious single candidates
Let's look for rows, columns, or boxes that are almost full.
* Row 5 (the middle row): The numbers present are `8, 5, 2`. The missing numbers are `1, 3, 4, 6, 7, 9`. This is too open. Let's look at Box 5 (center box). It has `8, 5, 2`.
* Column 5: Contains `4, 8, 6, 5, 9, 3, 2`. Missing are `1, 7`.
* Row 4 already has a `1` (in cell R4C1). So R4C5 cannot be 1. Thus, R4C5 = 7 and R6C5 = 1.
* Column 4: Contains `4, 3, 6, 8, 9, 2`. Missing are `1, 5, 7`.
* Row 4 needs `1, 3, 4, 6, 9` (since we just placed 7). Wait, let's re-evaluate Row 4.
* Row 4 currently: `1, 2, [ ], 8, [7], [ ], 9, [ ], 3`. Missing: `4, 5, 6`.
* Cell R4C3: Column 3 has `5, 8, 6...` let's check Box 4.
* Let's look at Box 4 (middle left). Cells: `1, 2, ?`, `?, ?, 5`, `?, ?, ?`. Numbers present: `1, 2, 5, 8`. Missing: `3, 4, 6, 7, 9`.
* We determined R4C5=7. So Row 4 is `1, 2, ?, 8, 7, ?, 9, ?, 3`. Missing in Row 4: `4, 5, 6`.
* Cell R4C3: Column 3 has `5` (R6C3) and `8` (R3C3? No, R3C3 is empty). Let's look at Column 3.
* Col 3 values: `?, ?, ?, ?, ?, 5, ?, ?, ?`.
* Let's look at Row 6: `3, ?, 5, ?, 1, 2, ?, 6, 4`. Missing: `7, 8, 9`.
* R6C2: Col 2 has `6, ?, ?, 2, ?, ?, ?, ?, ?`. Box 4 needs `3, 4, 6, 7, 9`.
* R6C2 is in Box 4. Box 4 missing `3, 4, 6, 7, 9`. Row 6 missing `7, 8, 9`. Intersection for R6C2 is `7, 9`.
* R6C4: Col 4 missing `1, 5, 7`. Row 6 missing `7, 8, 9`. Intersection is `7`. So R6C4 = 7.
* Now Row 6 missing `8, 9`. Cells R6C2 and R6C7.
* Col 7 has `1, 9, 6, 4`. So R6C7 cannot be 9? No, 9 is in R3C9. Let's check Col 7.
* Col 7: `1, ?, ?, 9, ?, ?, ?, ?, ?`.
* Let's look at Box 6 (middle right). Cells: `9, ?, 3`, `?, 1, 2`, `?, 6, 4`. Present: `1, 2, 3, 4, 6, 9`. Missing: `5, 7, 8`.
* Empty cells in Box 6: R4C8, R5C7, R5C8.
* Row 4 missing `4, 5, 6`. But Box 6 needs `5, 7, 8`. The only common number is `5`. So R4C8 = 5.
* Now Row 4 missing `4, 6`. Cells R4C3 and R4C6.
* Box 6 remaining missing `7, 8`. Cells R5C7 and R5C8.
* Row 5 is `?, ?, ?, ?, 5, ?, ?, ?, ?`. We know R5C5=5.
* Let's update Row 4: `1, 2, [4/6], 8, 7, [6/4], 9, 5, 3`.
* Check Col 3 for R4C3. Col 3 has `5` in R6.
* Check Col 6 for R4C6. Col 6 has `2` in R6.
Let's restart with a more systematic approach on the easiest areas.
1. Analyze Column 5
Values: `4, 8, 6, ?, 5, 1, ?, 3, 2` (Wait, I deduced R6C5=1 earlier, let's verify).
Original Col 5: `4, 8, 6, blank, 5, blank, blank, 3, 2`.
Missing: `1, 7, 9`.
Cells: R4C5, R6C5, R7C5.
Row 4 has `1, 2, 8, 9, 3`. So R4C5 cannot be 1, 2, 3, 8, 9.
Row 6 has `3, 5, 2, 6, 4`. So R6C5 cannot be 2, 3, 4, 5, 6.
Row 7 has `8, 9, 4, 7`. So R7C5 cannot be 4, 7, 8, 9.
Since R7C5 cannot be 7 or 9, R7C5 must be 1.
Now Col 5 missing `7, 9`. Cells R4C5, R6C5.
Row 4 has a `9` (R4C7). So R4C5 cannot be 9. Thus R4C5 = 7 and R6C5 = 9.
*Correction from previous thought*: I previously said R6C5=1, but R7C5=1 is stronger because R7 already has 7 and 9? Let's check Row 7: `8, blank, blank, blank, blank, 4, blank, 7, blank`. It has 4, 7, 8. It does NOT have 9.
Let's re-read Row 7 from image: `8, _, _, _, _, 4, _, 7, _`.
Col 5 original: `4, 8, 6, _, 5, _, _, 3, 2`.
Missing in Col 5: `1, 7, 9`.
R4C5: Row 4 is `1, 2, _, 8, _, _, 9, _, 3`. Has 1, 2, 3, 8, 9. Cannot be 1, 9. Must be 7? Or could it be something else? If R4C5 is 7, then remaining are 1, 9 for R6C5, R7C5.
R6C5: Row 6 is `3, _, 5, _, _, 2, _, 6, 4`. Has 2, 3, 4, 5, 6. Cannot be 1? No, 1 is missing. Cannot be 9? 9 is missing.
R7C5: Row 7 is `8, _, _, _, _, 4, _, 7, _`. Has 4, 7, 8. Cannot be 7. Can be 1 or 9.
Let's look at Box 8 (bottom middle).
Cells: `_, _, 9`, `_, 3, 8`, `_, 2, _`.
Present: `2, 3, 8, 9`. Missing: `1, 4, 5, 6, 7`.
Col 5 cells in Box 8: R7C5, R8C5, R9C5.
We know Col 5 missing `1, 7, 9`.
R8C5 is `3` (from image). Wait, image says R8C5 is `3`.
Let's re-read Col 5 from image carefully:
R1C5: 4
R2C5: 8
R3C5: 6
R4C5: ?
R5C5: 5
R6C5: ?
R7C5: ?
R8C5: 3
R9C5: 2
So Col 5 contains `2, 3, 4, 5, 6, 8`. Missing `1, 7, 9`.
Cells to fill: R4C5, R6C5, R7C5.
Check constraints:
* R4C5: Row 4 is `1, 2, ?, 8, ?, ?, 9, ?, 3`. Contains `1, 2, 3, 8, 9`. So R4C5 cannot be 1 or 9. Therefore, R4C5 = 7.
* Remaining for Col 5: `1, 9`. Cells R6C5, R7C5.
* R7C5: Row 7 is `8, ?, ?, ?, ?, 4, ?, 7, ?`. Contains `4, 7, 8`. Does not contain 1 or 9.
* R6C5: Row 6 is `3, ?, 5, ?, ?, 2, ?, 6, 4`. Contains `2, 3, 4, 5, 6`. Does not contain 1 or 9.
Let's look at Row 6.
Missing numbers in Row 6: `1, 7, 8, 9`. (Present: 2,3,4,5,6).
Cells: R6C2, R6C4, R6C5, R6C7.
We know R6C5 is either 1 or 9.
Let's look at Col 4.
Col 4 values: `?, 3, ?, 8, ?, ?, ?, ?, 2`.
Image:
R1C4: ?
R2C4: 3
R3C4: ?
R4C4: 8
R5C4: ?
R6C4: ?
R7C4: ?
R8C4: ?
R9C4: 2
Wait, looking at Image again.
Row 1: `9 6 . | . 4 . | 1 . .` -> C4 is empty.
Row 2: `. . . | 3 8 . | . . .` -> C4 is 3.
Row 3: `7 . 8 | . 6 . | . . 9` -> C4 is empty.
Row 4: `1 2 . | 8 . . | 9 . 3` -> C4 is 8.
Row 5: `. . . | . 5 . | . . .` -> C4 is empty.
Row 6: `3 . 5 | . . 2 | . 6 4` -> C4 is empty.
Row 7: `8 . . | . . 4 | . 7 .` -> C4 is empty.
Row 8: `. . . | . 3 8 | . . .` -> C4 is empty.
Row 9: `. . . | 9 2 . | . 8 5` -> C4 is 9? No, R9C4 is 9? Image: ` . . . | 9 2 . | . 8 5 `. Yes, R9C4=9? No, R9C4 is the first cell of the middle block in row 9.
Let's map Row 9: `Col1, Col2, Col3 | Col4, Col5, Col6 | Col7, Col8, Col9`.
Image Row 9: `_ _ _ | 9 2 _ | _ 8 5`. So R9C4=9, R9C5=2, R9C8=8, R9C9=5.
Okay, back to Col 4.
Values present: `3` (R2), `8` (R4), `9` (R9), `2` (R9? No R9C4=9).
Let's list Col 4 knowns:
R2C4 = 3
R4C4 = 8
R9C4 = 9 (from `9 2 _` block? No, `9` is in R9C4 position? Let's check alignment.
Row 9: `_ _ _` (Box 7) `9 2 _` (Box 8) `_ 8 5` (Box 9).
Yes, R9C4 = 9.
So Col 4 has `3, 8, 9`.
Also R1C5=4, R2C5=8, R3C5=6, R5C5=5, R8C5=3, R9C5=2.
Col 5 missing `1, 7, 9` for R4, R6, R7.
We established R4C5=7.
So R6C5, R7C5 are `1, 9`.
Let's look at Box 5 (Center).
Cells:
R4C4=8, R4C5=7, R4C6=?
R5C4=?, R5C5=5, R5C6=?
R6C4=?, R6C5=?, R6C6=2
Present in Box 5: `2, 5, 7, 8`.
Missing: `1, 3, 4, 6, 9`.
Cells: R4C6, R5C4, R5C6, R6C4, R6C5.
We know R6C5 is 1 or 9.
We know Col 4 has `3, 8, 9`. So R5C4, R6C4 cannot be 9? R9C4=9, so Col 4 has 9.
Thus R6C4 cannot be 9.
In Box 5, where can 9 go?
R4C6? Row 4 has 9 (R4C7). So no.
R5C4? Col 4 has 9 (R9C4). So no.
R5C6?
R6C4? Col 4 has 9. So no.
R6C5? This is a candidate.
So R6C5 must be 9.
Consequently, R7C5 = 1.
Now Box 5 missing `1, 3, 4, 6`. (9 is placed at R6C5).
Cells: R4C6, R5C4, R5C6, R6C4.
R6C5=9.
Row 6 now: `3, ?, 5, ?, 9, 2, ?, 6, 4`. Missing `1, 7, 8`.
Cells: R6C2, R6C4, R6C7.
Col 4 has `3, 8, 9`. And R6C4 is in Col 4.
Row 6 missing `1, 7, 8`.
R6C4 cannot be 8 (Col 4 has 8 at R4C4? No, R4C4=8. Yes.). So R6C4 != 8.
R6C4 cannot be 1? Col 4 doesn't have 1 yet.
R6C4 cannot be 7? Col 4 doesn't have 7 yet.
Let's look at Col 6.
Values: `1` (R1), `?` (R2), `?` (R3), `?` (R4), `?` (R5), `2` (R6), `4` (R7), `8` (R8), `?` (R9).
Knowns: `1, 2, 4, 8`.
Missing: `3, 5, 6, 7, 9`.
R4C6 is in Box 5. Box 5 missing `1, 3, 4, 6`.
R4C6 cannot be 1 (Row 4 has 1).
R4C6 cannot be 4 (Col 6 has 4? No, R7C6=4. So yes, Col 6 has 4).
So R4C6 can be `3, 6`.
Row 4 missing `4, 6` (since R4C3, R4C6 were the last two spots for 4,6? Row 4: `1,2,?,8,7,?,9,5,3`. Missing `4,6`).
So R4C6 is 4 or 6.
But Col 6 has 4 (R7C6). So R4C6 = 6.
Then R4C3 = 4.
Now Row 4 is complete: `1, 2, 4, 8, 7, 6, 9, 5, 3`.
Back to Box 5.
Missing `1, 3, 4`. (We placed 6 at R4C6, 9 at R6C5, 7 at R4C5, 8 at R4C4, 5 at R5C5, 2 at R6C6).
Cells remaining in Box 5: R5C4, R5C6, R6C4.
Numbers missing in Box 5: `1, 3, 4`.
R6C4 is in Row 6. Row 6 missing `1, 7, 8`.
Intersection of Box 5 missing (`1,3,4`) and Row 6 missing (`1,7,8`) is `1`.
So R6C4 = 1.
Now Box 5 missing `3, 4`. Cells R5C4, R5C6.
Col 4 has `3` (R2C4). So R5C4 cannot be 3.
Thus R5C4 = 4 and R5C6 = 3.
Now Row 6 missing `7, 8`. Cells R6C2, R6C7.
Col 2 has `6` (R1), `2` (R4).
Col 7 has `1` (R1), `9` (R4), `6` (R6? No), `4` (R9? No).
Let's check Col 7.
R1C7=1, R4C7=9.
Row 6: `3, ?, 5, 1, 9, 2, ?, 6, 4`.
R6C2 and R6C7 are `7, 8`.
Check Col 2: `6, ?, ?, 2, ?, ?, ?, ?, ?`.
Check Col 7: `1, ?, ?, 9, ?, ?, ?, ?, ?`.
Look at Box 4 (Middle Left).
Cells:
R4: `1, 2, 4`
R5: `?, ?, ?` (R5C1, R5C2, R5C3)
R6: `3, ?, 5` (R6C1=3, R6C2=?, R6C3=5)
Present in Box 4: `1, 2, 3, 4, 5`. Missing `6, 7, 8, 9`.
Cells: R5C1, R5C2, R5C3, R6C2.
We know R6C2 is `7` or `8`.
Row 5 so far: `?, ?, ?, 4, 5, 3, ?, ?, ?`.
Missing in Row 5: `1, 2, 6, 7, 8, 9`.
Box 4 missing `6, 7, 8, 9`.
R5C1, R5C2, R5C3 must be from `6, 7, 8, 9` (since R6C2 takes one).
Actually, Box 4 cells are R4C1-3, R5C1-3, R6C1-3.
R4C1-3: `1, 2, 4`.
R6C1-3: `3, [7/8], 5`.
R5C1-3: `?, ?, ?`.
Missing in Box 4: `6, 7, 8, 9`.
R6C2 is `7` or `8`.
So R5C1, R5C2, R5C3 contain the remaining 3 numbers from `6,7,8,9` plus the one not used by R6C2? No.
The set `{R5C1, R5C2, R5C3, R6C2}` is `{6, 7, 8, 9}`.
Let's look at Col 2.
Values: `6` (R1), `?` (R2), `?` (R3), `2` (R4), `?` (R5), `?` (R6), `?` (R7), `?` (R8), `?` (R9).
Missing: `1, 3, 4, 5, 7, 8, 9`.
This is complex. Let's look at Row 5 again.
`R5C1, R5C2, R5C3, 4, 5, 3, R5C7, R5C8, R5C9`.
We found R5C4=4, R5C5=5, R5C6=3.
Box 6 (Middle Right) missing `7, 8` (from earlier: Box 6 had `9,1,2,3,4,6` present? Let's re-verify Box 6).
Box 6 cells:
R4C7=9, R4C8=5, R4C9=3.
R5C7=?, R5C8=?, R5C9=?.
R6C7=?, R6C8=6, R6C9=4.
Present: `3, 4, 5, 6, 9`.
Missing: `1, 2, 7, 8`.
Cells: R5C7, R5C8, R5C9, R6C7.
Row 6 missing `7, 8` for R6C2, R6C7.
So R6C7 is `7` or `8`.
Therefore, in Box 6, R6C7 is `7` or `8`.
Remaining cells R5C7, R5C8, R5C9 must contain `1, 2` and the other of `7,8`.
Row 5 missing `1, 2, 6, 7, 8, 9`.
R5C7, R5C8, R5C9 are in Row 5.
So R5C7, R5C8, R5C9 contain `1, 2, 7/8`.
This implies R5C1, R5C2, R5C3 contain `6, 9` and the other of `7,8`?
Row 5 missing 6 numbers.
Box 4 contributes 3 cells to Row 5.
Box 6 contributes 3 cells to Row 5.
Box 4 missing `6, 7, 8, 9`. R6C2 takes one. So R5C1-3 take three of `6,7,8,9`.
Box 6 missing `1, 2, 7, 8`. R6C7 takes one (`7` or `8`). So R5C7-9 take `1, 2` and the other of `7,8`.
Let's determine R6C2 and R6C7.
Check Col 7.
Col 7: `1` (R1), `?` (R2), `?` (R3), `9` (R4), `?` (R5), `?` (R6), `?` (R7), `?` (R8), `?` (R9).
R9C7 is empty. R9 is `_ _ _ | 9 2 _ | _ 8 5`.
R9C6 is empty. R9C7 is empty.
Row 9 missing `1, 3, 4, 6, 7`. (Present: 2,5,8,9).
Col 7 has `1, 9`.
Let's look at Box 9 (Bottom Right).
Cells:
R7C7, R7C8=7, R7C9
R8C7, R8C8, R8C9
R9C7, R9C8=8, R9C9=5
Present: `5, 7, 8`.
Missing: `1, 2, 3, 4, 6, 9`.
Let's solve Row 9.
`R9C1, R9C2, R9C3, 9, 2, R9C6, R9C7, 8, 5`.
Missing: `1, 3, 4, 6, 7`.
Col 6 has `1, 2, 3, 4, 6, 8`?
Col 6 values: `1` (R1), `?` (R2), `?` (R3), `6` (R4), `3` (R5), `2` (R6), `4` (R7), `8` (R8), `?` (R9).
Present: `1, 2, 3, 4, 6, 8`.
Missing: `5, 7, 9`.
Cells: R2C6, R3C6, R9C6.
R9C6 must be `5, 7, 9`.
But Row 9 missing `1, 3, 4, 6, 7`.
Intersection: `7`.
So R9C6 = 7.
Then Col 6 missing `5, 9`. Cells R2C6, R3C6.
Row 2: `. . . | 3 8 . | . . .`.
Row 3: `7 . 8 | . 6 . | . . 9`.
Row 3 has `6, 7, 8, 9`.
R3C6 cannot be 9? Row 3 has 9 at R3C9. So R3C6 = 5.
Then R2C6 = 9.
Now Row 9 missing `1, 3, 4, 6`.
Cells: R9C1, R9C2, R9C3, R9C7.
Col 7 missing `2, 3, 4, 5, 6, 7, 8`?
Col 7 knowns: `1` (R1), `9` (R4).
R9C7 is in Box 9.
Box 9 missing `1, 2, 3, 4, 6, 9`.
R9C7 cannot be 9 (Row 9 has 9).
R9C7 cannot be 2 (Row 9 has 2).
So R9C7 is `1, 3, 4, 6`.
Let's look at Col 1.
Values: `9` (R1), `?` (R2), `7` (R3), `1` (R4), `?` (R5), `3` (R6), `8` (R7), `?` (R8), `?` (R9).
Present: `1, 3, 7, 8, 9`.
Missing: `2, 4, 5, 6`.
Cells: R2C1, R5C1, R8C1, R9C1.
R9C1 is in Row 9. Row 9 missing `1, 3, 4, 6`.
Intersection for R9C1: `4, 6`.
Let's look at Box 7 (Bottom Left).
Cells:
R7C1=8, R7C2=?, R7C3=?
R8C1=?, R8C2=?, R8C3=?
R9C1=?, R9C2=?, R9C3=?
Present: `8`.
Missing: `1, 2, 3, 4, 5, 6, 7, 9`.
Row 7: `8, ?, ?, ?, 1, 4, ?, 7, ?`.
Row 7 missing `2, 3, 5, 6, 9`.
R7C2, R7C3 are in Box 7.
This is getting long. Let's fill the grid based on the strong deductions so far.
Current Grid State:
R1: `9 6 . | . 4 . | 1 . .`
R2: `. . . | 3 8 9 | . . .`
R3: `7 . 8 | . 6 5 | . . 9`
R4: `1 2 4 | 8 7 6 | 9 5 3`
R5: `. . . | 4 5 3 | . . .`
R6: `3 . 5 | 1 9 2 | . 6 4`
R7: `8 . . | . 1 4 | . 7 .`
R8: `. . . | . 3 8 | . . .`
R9: `. . . | 9 2 7 | . 8 5`
Deductions:
1. Row 6: Missing `7, 8`. R6C2, R6C7.
Col 2 has `6, 2`. Col 7 has `1, 9`.
Look at Box 4. Missing `6, 7, 8, 9`.
R6C2 is `7` or `8`.
R5C1, R5C2, R5C3 are `6, 9` and the other of `7,8`.
Look at Col 2.
R1C2=6, R4C2=2.
R6C2 is `7` or `8`.
R2C2, R3C2, R5C2, R7C2, R8C2, R9C2 remain.
2. Row 5: `R5C1, R5C2, R5C3, 4, 5, 3, R5C7, R5C8, R5C9`.
Missing `1, 2, 6, 7, 8, 9`.
Box 6 missing `1, 2, 7, 8`.
R6C7 is `7` or `8`.
So R5C7, R5C8, R5C9 are `1, 2` and `7/8`.
This means R5C1, R5C2, R5C3 are `6, 9` and `8/7`.
3. Col 6 is done: `., 9, 5, 6, 3, 2, 4, 8, 7`.
R1C6 is missing?
Col 6: R1=?, R2=9, R3=5, R4=6, R5=3, R6=2, R7=4, R8=8, R9=7.
Missing in Col 6: `1`.
So R1C6 = 1.
4. Row 1: `9 6 . | . 4 1 | 1 . .` -> Wait, R1C7=1. Duplicate 1 in Row 1?
Image R1: `9 6 . | . 4 . | 1 . .`
My deduction R1C6=1 conflicts with R1C7=1.
Let's re-check Col 6.
R1C6, R2C6, R3C6, R4C6, R5C6, R6C6, R7C6, R8C6, R9C6.
Values:
R4C6=6.
R5C6=3.
R6C6=2.
R7C6=4.
R8C6=8.
R9C6=7.
R2C6=9.
R3C6=5.
So Col 6 has `2,3,4,5,6,7,8,9`.
Missing `1`.
So R1C6 = 1.
But R1C7 is `1` in the image.
Contradiction!
*Re-evaluate R3C6 and R2C6.*
Col 6 missing `5, 7, 9` before placing R9C6.
R9C6=7.
Remaining `5, 9` for R2C6, R3C6.
R3 has `9` at R3C9. So R3C6 cannot be 9.
So R3C6=5, R2C6=9. This logic holds.
So R1C6 MUST be 1.
But R1C7 IS 1.
Did I misread the image?
Row 1: `9 6 [ ] | [ ] 4 [ ] | 1 [ ] [ ]`
Col 7 is `1`.
Col 6 is `[ ]`.
If R1C6=1, then Row 1 has two 1s.
Where is the error?
Check Box 2 (Top Middle).
Cells:
R1C4, R1C5=4, R1C6
R2C4=3, R2C5=8, R2C6
R3C4, R3C5=6, R3C6
Present: `3, 4, 6, 8`.
Missing: `1, 2, 5, 7, 9`.
Check Col 6 again.
R1C6, R2C6, R3C6, R4C6, R5C6, R6C6, R7C6, R8C6, R9C6.
R4C6=6.
R5C6=3.
R6C6=2.
R7C6=4.
R8C6=8.
R9C6=7.
These are fixed by previous steps.
R2C6 and R3C6 are `5, 9`.
R1C6 is `1`.
Is R1C7 really 1?
Image: `9 6 . | . 4 . | 1 . .`
Yes, R1C7 is 1.
So R1C6 cannot be 1.
This implies one of the values in Col 6 (R4-R9) is wrong.
Let's re-verify R9C6=7.
Row 9: `_ _ _ | 9 2 _ | _ 8 5`.
Col 6 missing `5, 7, 9` (assuming R1-R3 unknown).
Row 9 missing `1, 3, 4, 6, 7`.
Intersection `7`. Correct.
Re-verify R3C6=5.
Row 3: `7 . 8 | . 6 . | . . 9`.
Col 6 missing `5, 9`.
Row 3 has 9. So R3C6=5. Correct.
Re-verify R2C6=9.
Col 6 missing `9`. Correct.
So Col 6 has `5, 9, 6, 3, 2, 4, 8, 7`.
Missing `1`.
So R1C6=1.
Conflict with R1C7=1.
Is it possible R4C6 is not 6?
Row 4: `1 2 4 | 8 7 6 | 9 5 3`.
R4C6=6 derived from:
Box 5 missing `1,3,4,6,9`.
R4C6 cannot be 1 (Row 4).
R4C6 cannot be 4 (Col 6 has 4? R7C6=4. Yes).
R4C6 cannot be 9 (Row 4 has 9).
So R4C6 is `3` or `6`.
Row 4 missing `4, 6`?
Row 4: `1, 2, ?, 8, 7, ?, 9, 5, 3`.
Missing `4, 6`.
R4C3 and R4C6.
R4C3 is in Col 3.
R4C6 is in Col 6.
If R4C6=3, then Row 4 has two 3s (R4C9=3). Impossible.
So R4C6 MUST be 6 or 4.
But Row 4 missing `4, 6`.
If R4C6=4, then R4C3=6.
If R4C6=6, then R4C3=4.
Check Col 6 for 4.
R7C6=4.
So R4C6 cannot be 4.
So R4C6 MUST BE 6.
So the chain holds. R1C6=1.
Is R1C7 actually 1?
Looking at the image crop 1: `9 6 . | . 4 . | 1 . .`
Yes.
Is there an error in my Col 6 filling?
R7C6=4.
Image Row 7: `8 . . | . . 4 | . 7 .`
Yes, R7C6=4.
R8C6=8.
Image Row 8: `. . . | . 3 8 | . . .`
Yes, R8C6=8.
R9C6=7.
Derived.
R6C6=2.
Image Row 6: `3 . 5 | . . 2 | . 6 4`
Yes.
R5C6=3.
Derived from Box 5.
R4C6=6.
Derived.
R2C6=9, R3C6=5.
Derived.
So R1C6=1.
Conclusion: The puzzle as interpreted has a contradiction in Row 1 (1 at C6 and C7).
However, often in these puzzles, I might have misidentified a number.
Let's check R1C5. Image: `4`.
Let's check R1C7. Image: `1`.
Wait, look at Box 3 (Top Right).
Cells:
R1C7=1, R1C8, R1C9
R2C7, R2C8, R2C9
R3C7, R3C8, R3C9=9
If R1C6=1, then R1C7 cannot be 1.
Is it possible R1C7 is NOT 1?
Image: `1` is clearly in R1C7.
Is it possible R1C6 is NOT 1?
This requires Col 6 to have a 1 elsewhere.
R2-R9 Col 6: `9, 5, 6, 3, 2, 4, 8, 7`.
None of these are 1.
Did I place a number wrong in Col 6?
R9C6=7.
If R9C6 is not 7, what can it be?
Row 9: `_ _ _ 9 2 _ _ 8 5`.
Missing `1, 3, 4, 6, 7`.
Col 6 missing `1, 5, 7, 9` (initially).
If R9C6=1?
Then Col 6 has 1. R1C6 is not 1.
If R9C6=1, then Row 9 has 1.
Then Col 6 missing `5, 7, 9`.
R2C6, R3C6, R1C6.
R3C6 cannot be 9.
Why did I say R9C6=7?
Intersection of Row 9 missing `1,3,4,6,7` and Col 6 missing `1,5,7,9`.
Common: `1, 7`.
I eliminated 1 because... I didn't.
I said "Intersection: 7". That was a mistake. Intersection is `1, 7`.
If R9C6=1:
Then R1C6 is `5, 7, 9`.
Col 6 missing `5, 7, 9`.
R1C6, R2C6, R3C6.
Here is the step-by-step logical deduction to find the solution:
Step 1: Fill in obvious single candidates
Let's look for rows, columns, or boxes that are almost full.
* Row 5 (the middle row): The numbers present are `8, 5, 2`. The missing numbers are `1, 3, 4, 6, 7, 9`. This is too open. Let's look at Box 5 (center box). It has `8, 5, 2`.
* Column 5: Contains `4, 8, 6, 5, 9, 3, 2`. Missing are `1, 7`.
* Row 4 already has a `1` (in cell R4C1). So R4C5 cannot be 1. Thus, R4C5 = 7 and R6C5 = 1.
* Column 4: Contains `4, 3, 6, 8, 9, 2`. Missing are `1, 5, 7`.
* Row 4 needs `1, 3, 4, 6, 9` (since we just placed 7). Wait, let's re-evaluate Row 4.
* Row 4 currently: `1, 2, [ ], 8, [7], [ ], 9, [ ], 3`. Missing: `4, 5, 6`.
* Cell R4C3: Column 3 has `5, 8, 6...` let's check Box 4.
* Let's look at Box 4 (middle left). Cells: `1, 2, ?`, `?, ?, 5`, `?, ?, ?`. Numbers present: `1, 2, 5, 8`. Missing: `3, 4, 6, 7, 9`.
* We determined R4C5=7. So Row 4 is `1, 2, ?, 8, 7, ?, 9, ?, 3`. Missing in Row 4: `4, 5, 6`.
* Cell R4C3: Column 3 has `5` (R6C3) and `8` (R3C3? No, R3C3 is empty). Let's look at Column 3.
* Col 3 values: `?, ?, ?, ?, ?, 5, ?, ?, ?`.
* Let's look at Row 6: `3, ?, 5, ?, 1, 2, ?, 6, 4`. Missing: `7, 8, 9`.
* R6C2: Col 2 has `6, ?, ?, 2, ?, ?, ?, ?, ?`. Box 4 needs `3, 4, 6, 7, 9`.
* R6C2 is in Box 4. Box 4 missing `3, 4, 6, 7, 9`. Row 6 missing `7, 8, 9`. Intersection for R6C2 is `7, 9`.
* R6C4: Col 4 missing `1, 5, 7`. Row 6 missing `7, 8, 9`. Intersection is `7`. So R6C4 = 7.
* Now Row 6 missing `8, 9`. Cells R6C2 and R6C7.
* Col 7 has `1, 9, 6, 4`. So R6C7 cannot be 9? No, 9 is in R3C9. Let's check Col 7.
* Col 7: `1, ?, ?, 9, ?, ?, ?, ?, ?`.
* Let's look at Box 6 (middle right). Cells: `9, ?, 3`, `?, 1, 2`, `?, 6, 4`. Present: `1, 2, 3, 4, 6, 9`. Missing: `5, 7, 8`.
* Empty cells in Box 6: R4C8, R5C7, R5C8.
* Row 4 missing `4, 5, 6`. But Box 6 needs `5, 7, 8`. The only common number is `5`. So R4C8 = 5.
* Now Row 4 missing `4, 6`. Cells R4C3 and R4C6.
* Box 6 remaining missing `7, 8`. Cells R5C7 and R5C8.
* Row 5 is `?, ?, ?, ?, 5, ?, ?, ?, ?`. We know R5C5=5.
* Let's update Row 4: `1, 2, [4/6], 8, 7, [6/4], 9, 5, 3`.
* Check Col 3 for R4C3. Col 3 has `5` in R6.
* Check Col 6 for R4C6. Col 6 has `2` in R6.
Let's restart with a more systematic approach on the easiest areas.
1. Analyze Column 5
Values: `4, 8, 6, ?, 5, 1, ?, 3, 2` (Wait, I deduced R6C5=1 earlier, let's verify).
Original Col 5: `4, 8, 6, blank, 5, blank, blank, 3, 2`.
Missing: `1, 7, 9`.
Cells: R4C5, R6C5, R7C5.
Row 4 has `1, 2, 8, 9, 3`. So R4C5 cannot be 1, 2, 3, 8, 9.
Row 6 has `3, 5, 2, 6, 4`. So R6C5 cannot be 2, 3, 4, 5, 6.
Row 7 has `8, 9, 4, 7`. So R7C5 cannot be 4, 7, 8, 9.
Since R7C5 cannot be 7 or 9, R7C5 must be 1.
Now Col 5 missing `7, 9`. Cells R4C5, R6C5.
Row 4 has a `9` (R4C7). So R4C5 cannot be 9. Thus R4C5 = 7 and R6C5 = 9.
*Correction from previous thought*: I previously said R6C5=1, but R7C5=1 is stronger because R7 already has 7 and 9? Let's check Row 7: `8, blank, blank, blank, blank, 4, blank, 7, blank`. It has 4, 7, 8. It does NOT have 9.
Let's re-read Row 7 from image: `8, _, _, _, _, 4, _, 7, _`.
Col 5 original: `4, 8, 6, _, 5, _, _, 3, 2`.
Missing in Col 5: `1, 7, 9`.
R4C5: Row 4 is `1, 2, _, 8, _, _, 9, _, 3`. Has 1, 2, 3, 8, 9. Cannot be 1, 9. Must be 7? Or could it be something else? If R4C5 is 7, then remaining are 1, 9 for R6C5, R7C5.
R6C5: Row 6 is `3, _, 5, _, _, 2, _, 6, 4`. Has 2, 3, 4, 5, 6. Cannot be 1? No, 1 is missing. Cannot be 9? 9 is missing.
R7C5: Row 7 is `8, _, _, _, _, 4, _, 7, _`. Has 4, 7, 8. Cannot be 7. Can be 1 or 9.
Let's look at Box 8 (bottom middle).
Cells: `_, _, 9`, `_, 3, 8`, `_, 2, _`.
Present: `2, 3, 8, 9`. Missing: `1, 4, 5, 6, 7`.
Col 5 cells in Box 8: R7C5, R8C5, R9C5.
We know Col 5 missing `1, 7, 9`.
R8C5 is `3` (from image). Wait, image says R8C5 is `3`.
Let's re-read Col 5 from image carefully:
R1C5: 4
R2C5: 8
R3C5: 6
R4C5: ?
R5C5: 5
R6C5: ?
R7C5: ?
R8C5: 3
R9C5: 2
So Col 5 contains `2, 3, 4, 5, 6, 8`. Missing `1, 7, 9`.
Cells to fill: R4C5, R6C5, R7C5.
Check constraints:
* R4C5: Row 4 is `1, 2, ?, 8, ?, ?, 9, ?, 3`. Contains `1, 2, 3, 8, 9`. So R4C5 cannot be 1 or 9. Therefore, R4C5 = 7.
* Remaining for Col 5: `1, 9`. Cells R6C5, R7C5.
* R7C5: Row 7 is `8, ?, ?, ?, ?, 4, ?, 7, ?`. Contains `4, 7, 8`. Does not contain 1 or 9.
* R6C5: Row 6 is `3, ?, 5, ?, ?, 2, ?, 6, 4`. Contains `2, 3, 4, 5, 6`. Does not contain 1 or 9.
Let's look at Row 6.
Missing numbers in Row 6: `1, 7, 8, 9`. (Present: 2,3,4,5,6).
Cells: R6C2, R6C4, R6C5, R6C7.
We know R6C5 is either 1 or 9.
Let's look at Col 4.
Col 4 values: `?, 3, ?, 8, ?, ?, ?, ?, 2`.
Image:
R1C4: ?
R2C4: 3
R3C4: ?
R4C4: 8
R5C4: ?
R6C4: ?
R7C4: ?
R8C4: ?
R9C4: 2
Wait, looking at Image again.
Row 1: `9 6 . | . 4 . | 1 . .` -> C4 is empty.
Row 2: `. . . | 3 8 . | . . .` -> C4 is 3.
Row 3: `7 . 8 | . 6 . | . . 9` -> C4 is empty.
Row 4: `1 2 . | 8 . . | 9 . 3` -> C4 is 8.
Row 5: `. . . | . 5 . | . . .` -> C4 is empty.
Row 6: `3 . 5 | . . 2 | . 6 4` -> C4 is empty.
Row 7: `8 . . | . . 4 | . 7 .` -> C4 is empty.
Row 8: `. . . | . 3 8 | . . .` -> C4 is empty.
Row 9: `. . . | 9 2 . | . 8 5` -> C4 is 9? No, R9C4 is 9? Image: ` . . . | 9 2 . | . 8 5 `. Yes, R9C4=9? No, R9C4 is the first cell of the middle block in row 9.
Let's map Row 9: `Col1, Col2, Col3 | Col4, Col5, Col6 | Col7, Col8, Col9`.
Image Row 9: `_ _ _ | 9 2 _ | _ 8 5`. So R9C4=9, R9C5=2, R9C8=8, R9C9=5.
Okay, back to Col 4.
Values present: `3` (R2), `8` (R4), `9` (R9), `2` (R9? No R9C4=9).
Let's list Col 4 knowns:
R2C4 = 3
R4C4 = 8
R9C4 = 9 (from `9 2 _` block? No, `9` is in R9C4 position? Let's check alignment.
Row 9: `_ _ _` (Box 7) `9 2 _` (Box 8) `_ 8 5` (Box 9).
Yes, R9C4 = 9.
So Col 4 has `3, 8, 9`.
Also R1C5=4, R2C5=8, R3C5=6, R5C5=5, R8C5=3, R9C5=2.
Col 5 missing `1, 7, 9` for R4, R6, R7.
We established R4C5=7.
So R6C5, R7C5 are `1, 9`.
Let's look at Box 5 (Center).
Cells:
R4C4=8, R4C5=7, R4C6=?
R5C4=?, R5C5=5, R5C6=?
R6C4=?, R6C5=?, R6C6=2
Present in Box 5: `2, 5, 7, 8`.
Missing: `1, 3, 4, 6, 9`.
Cells: R4C6, R5C4, R5C6, R6C4, R6C5.
We know R6C5 is 1 or 9.
We know Col 4 has `3, 8, 9`. So R5C4, R6C4 cannot be 9? R9C4=9, so Col 4 has 9.
Thus R6C4 cannot be 9.
In Box 5, where can 9 go?
R4C6? Row 4 has 9 (R4C7). So no.
R5C4? Col 4 has 9 (R9C4). So no.
R5C6?
R6C4? Col 4 has 9. So no.
R6C5? This is a candidate.
So R6C5 must be 9.
Consequently, R7C5 = 1.
Now Box 5 missing `1, 3, 4, 6`. (9 is placed at R6C5).
Cells: R4C6, R5C4, R5C6, R6C4.
R6C5=9.
Row 6 now: `3, ?, 5, ?, 9, 2, ?, 6, 4`. Missing `1, 7, 8`.
Cells: R6C2, R6C4, R6C7.
Col 4 has `3, 8, 9`. And R6C4 is in Col 4.
Row 6 missing `1, 7, 8`.
R6C4 cannot be 8 (Col 4 has 8 at R4C4? No, R4C4=8. Yes.). So R6C4 != 8.
R6C4 cannot be 1? Col 4 doesn't have 1 yet.
R6C4 cannot be 7? Col 4 doesn't have 7 yet.
Let's look at Col 6.
Values: `1` (R1), `?` (R2), `?` (R3), `?` (R4), `?` (R5), `2` (R6), `4` (R7), `8` (R8), `?` (R9).
Knowns: `1, 2, 4, 8`.
Missing: `3, 5, 6, 7, 9`.
R4C6 is in Box 5. Box 5 missing `1, 3, 4, 6`.
R4C6 cannot be 1 (Row 4 has 1).
R4C6 cannot be 4 (Col 6 has 4? No, R7C6=4. So yes, Col 6 has 4).
So R4C6 can be `3, 6`.
Row 4 missing `4, 6` (since R4C3, R4C6 were the last two spots for 4,6? Row 4: `1,2,?,8,7,?,9,5,3`. Missing `4,6`).
So R4C6 is 4 or 6.
But Col 6 has 4 (R7C6). So R4C6 = 6.
Then R4C3 = 4.
Now Row 4 is complete: `1, 2, 4, 8, 7, 6, 9, 5, 3`.
Back to Box 5.
Missing `1, 3, 4`. (We placed 6 at R4C6, 9 at R6C5, 7 at R4C5, 8 at R4C4, 5 at R5C5, 2 at R6C6).
Cells remaining in Box 5: R5C4, R5C6, R6C4.
Numbers missing in Box 5: `1, 3, 4`.
R6C4 is in Row 6. Row 6 missing `1, 7, 8`.
Intersection of Box 5 missing (`1,3,4`) and Row 6 missing (`1,7,8`) is `1`.
So R6C4 = 1.
Now Box 5 missing `3, 4`. Cells R5C4, R5C6.
Col 4 has `3` (R2C4). So R5C4 cannot be 3.
Thus R5C4 = 4 and R5C6 = 3.
Now Row 6 missing `7, 8`. Cells R6C2, R6C7.
Col 2 has `6` (R1), `2` (R4).
Col 7 has `1` (R1), `9` (R4), `6` (R6? No), `4` (R9? No).
Let's check Col 7.
R1C7=1, R4C7=9.
Row 6: `3, ?, 5, 1, 9, 2, ?, 6, 4`.
R6C2 and R6C7 are `7, 8`.
Check Col 2: `6, ?, ?, 2, ?, ?, ?, ?, ?`.
Check Col 7: `1, ?, ?, 9, ?, ?, ?, ?, ?`.
Look at Box 4 (Middle Left).
Cells:
R4: `1, 2, 4`
R5: `?, ?, ?` (R5C1, R5C2, R5C3)
R6: `3, ?, 5` (R6C1=3, R6C2=?, R6C3=5)
Present in Box 4: `1, 2, 3, 4, 5`. Missing `6, 7, 8, 9`.
Cells: R5C1, R5C2, R5C3, R6C2.
We know R6C2 is `7` or `8`.
Row 5 so far: `?, ?, ?, 4, 5, 3, ?, ?, ?`.
Missing in Row 5: `1, 2, 6, 7, 8, 9`.
Box 4 missing `6, 7, 8, 9`.
R5C1, R5C2, R5C3 must be from `6, 7, 8, 9` (since R6C2 takes one).
Actually, Box 4 cells are R4C1-3, R5C1-3, R6C1-3.
R4C1-3: `1, 2, 4`.
R6C1-3: `3, [7/8], 5`.
R5C1-3: `?, ?, ?`.
Missing in Box 4: `6, 7, 8, 9`.
R6C2 is `7` or `8`.
So R5C1, R5C2, R5C3 contain the remaining 3 numbers from `6,7,8,9` plus the one not used by R6C2? No.
The set `{R5C1, R5C2, R5C3, R6C2}` is `{6, 7, 8, 9}`.
Let's look at Col 2.
Values: `6` (R1), `?` (R2), `?` (R3), `2` (R4), `?` (R5), `?` (R6), `?` (R7), `?` (R8), `?` (R9).
Missing: `1, 3, 4, 5, 7, 8, 9`.
This is complex. Let's look at Row 5 again.
`R5C1, R5C2, R5C3, 4, 5, 3, R5C7, R5C8, R5C9`.
We found R5C4=4, R5C5=5, R5C6=3.
Box 6 (Middle Right) missing `7, 8` (from earlier: Box 6 had `9,1,2,3,4,6` present? Let's re-verify Box 6).
Box 6 cells:
R4C7=9, R4C8=5, R4C9=3.
R5C7=?, R5C8=?, R5C9=?.
R6C7=?, R6C8=6, R6C9=4.
Present: `3, 4, 5, 6, 9`.
Missing: `1, 2, 7, 8`.
Cells: R5C7, R5C8, R5C9, R6C7.
Row 6 missing `7, 8` for R6C2, R6C7.
So R6C7 is `7` or `8`.
Therefore, in Box 6, R6C7 is `7` or `8`.
Remaining cells R5C7, R5C8, R5C9 must contain `1, 2` and the other of `7,8`.
Row 5 missing `1, 2, 6, 7, 8, 9`.
R5C7, R5C8, R5C9 are in Row 5.
So R5C7, R5C8, R5C9 contain `1, 2, 7/8`.
This implies R5C1, R5C2, R5C3 contain `6, 9` and the other of `7,8`?
Row 5 missing 6 numbers.
Box 4 contributes 3 cells to Row 5.
Box 6 contributes 3 cells to Row 5.
Box 4 missing `6, 7, 8, 9`. R6C2 takes one. So R5C1-3 take three of `6,7,8,9`.
Box 6 missing `1, 2, 7, 8`. R6C7 takes one (`7` or `8`). So R5C7-9 take `1, 2` and the other of `7,8`.
Let's determine R6C2 and R6C7.
Check Col 7.
Col 7: `1` (R1), `?` (R2), `?` (R3), `9` (R4), `?` (R5), `?` (R6), `?` (R7), `?` (R8), `?` (R9).
R9C7 is empty. R9 is `_ _ _ | 9 2 _ | _ 8 5`.
R9C6 is empty. R9C7 is empty.
Row 9 missing `1, 3, 4, 6, 7`. (Present: 2,5,8,9).
Col 7 has `1, 9`.
Let's look at Box 9 (Bottom Right).
Cells:
R7C7, R7C8=7, R7C9
R8C7, R8C8, R8C9
R9C7, R9C8=8, R9C9=5
Present: `5, 7, 8`.
Missing: `1, 2, 3, 4, 6, 9`.
Let's solve Row 9.
`R9C1, R9C2, R9C3, 9, 2, R9C6, R9C7, 8, 5`.
Missing: `1, 3, 4, 6, 7`.
Col 6 has `1, 2, 3, 4, 6, 8`?
Col 6 values: `1` (R1), `?` (R2), `?` (R3), `6` (R4), `3` (R5), `2` (R6), `4` (R7), `8` (R8), `?` (R9).
Present: `1, 2, 3, 4, 6, 8`.
Missing: `5, 7, 9`.
Cells: R2C6, R3C6, R9C6.
R9C6 must be `5, 7, 9`.
But Row 9 missing `1, 3, 4, 6, 7`.
Intersection: `7`.
So R9C6 = 7.
Then Col 6 missing `5, 9`. Cells R2C6, R3C6.
Row 2: `. . . | 3 8 . | . . .`.
Row 3: `7 . 8 | . 6 . | . . 9`.
Row 3 has `6, 7, 8, 9`.
R3C6 cannot be 9? Row 3 has 9 at R3C9. So R3C6 = 5.
Then R2C6 = 9.
Now Row 9 missing `1, 3, 4, 6`.
Cells: R9C1, R9C2, R9C3, R9C7.
Col 7 missing `2, 3, 4, 5, 6, 7, 8`?
Col 7 knowns: `1` (R1), `9` (R4).
R9C7 is in Box 9.
Box 9 missing `1, 2, 3, 4, 6, 9`.
R9C7 cannot be 9 (Row 9 has 9).
R9C7 cannot be 2 (Row 9 has 2).
So R9C7 is `1, 3, 4, 6`.
Let's look at Col 1.
Values: `9` (R1), `?` (R2), `7` (R3), `1` (R4), `?` (R5), `3` (R6), `8` (R7), `?` (R8), `?` (R9).
Present: `1, 3, 7, 8, 9`.
Missing: `2, 4, 5, 6`.
Cells: R2C1, R5C1, R8C1, R9C1.
R9C1 is in Row 9. Row 9 missing `1, 3, 4, 6`.
Intersection for R9C1: `4, 6`.
Let's look at Box 7 (Bottom Left).
Cells:
R7C1=8, R7C2=?, R7C3=?
R8C1=?, R8C2=?, R8C3=?
R9C1=?, R9C2=?, R9C3=?
Present: `8`.
Missing: `1, 2, 3, 4, 5, 6, 7, 9`.
Row 7: `8, ?, ?, ?, 1, 4, ?, 7, ?`.
Row 7 missing `2, 3, 5, 6, 9`.
R7C2, R7C3 are in Box 7.
This is getting long. Let's fill the grid based on the strong deductions so far.
Current Grid State:
R1: `9 6 . | . 4 . | 1 . .`
R2: `. . . | 3 8 9 | . . .`
R3: `7 . 8 | . 6 5 | . . 9`
R4: `1 2 4 | 8 7 6 | 9 5 3`
R5: `. . . | 4 5 3 | . . .`
R6: `3 . 5 | 1 9 2 | . 6 4`
R7: `8 . . | . 1 4 | . 7 .`
R8: `. . . | . 3 8 | . . .`
R9: `. . . | 9 2 7 | . 8 5`
Deductions:
1. Row 6: Missing `7, 8`. R6C2, R6C7.
Col 2 has `6, 2`. Col 7 has `1, 9`.
Look at Box 4. Missing `6, 7, 8, 9`.
R6C2 is `7` or `8`.
R5C1, R5C2, R5C3 are `6, 9` and the other of `7,8`.
Look at Col 2.
R1C2=6, R4C2=2.
R6C2 is `7` or `8`.
R2C2, R3C2, R5C2, R7C2, R8C2, R9C2 remain.
2. Row 5: `R5C1, R5C2, R5C3, 4, 5, 3, R5C7, R5C8, R5C9`.
Missing `1, 2, 6, 7, 8, 9`.
Box 6 missing `1, 2, 7, 8`.
R6C7 is `7` or `8`.
So R5C7, R5C8, R5C9 are `1, 2` and `7/8`.
This means R5C1, R5C2, R5C3 are `6, 9` and `8/7`.
3. Col 6 is done: `., 9, 5, 6, 3, 2, 4, 8, 7`.
R1C6 is missing?
Col 6: R1=?, R2=9, R3=5, R4=6, R5=3, R6=2, R7=4, R8=8, R9=7.
Missing in Col 6: `1`.
So R1C6 = 1.
4. Row 1: `9 6 . | . 4 1 | 1 . .` -> Wait, R1C7=1. Duplicate 1 in Row 1?
Image R1: `9 6 . | . 4 . | 1 . .`
My deduction R1C6=1 conflicts with R1C7=1.
Let's re-check Col 6.
R1C6, R2C6, R3C6, R4C6, R5C6, R6C6, R7C6, R8C6, R9C6.
Values:
R4C6=6.
R5C6=3.
R6C6=2.
R7C6=4.
R8C6=8.
R9C6=7.
R2C6=9.
R3C6=5.
So Col 6 has `2,3,4,5,6,7,8,9`.
Missing `1`.
So R1C6 = 1.
But R1C7 is `1` in the image.
Contradiction!
*Re-evaluate R3C6 and R2C6.*
Col 6 missing `5, 7, 9` before placing R9C6.
R9C6=7.
Remaining `5, 9` for R2C6, R3C6.
R3 has `9` at R3C9. So R3C6 cannot be 9.
So R3C6=5, R2C6=9. This logic holds.
So R1C6 MUST be 1.
But R1C7 IS 1.
Did I misread the image?
Row 1: `9 6 [ ] | [ ] 4 [ ] | 1 [ ] [ ]`
Col 7 is `1`.
Col 6 is `[ ]`.
If R1C6=1, then Row 1 has two 1s.
Where is the error?
Check Box 2 (Top Middle).
Cells:
R1C4, R1C5=4, R1C6
R2C4=3, R2C5=8, R2C6
R3C4, R3C5=6, R3C6
Present: `3, 4, 6, 8`.
Missing: `1, 2, 5, 7, 9`.
Check Col 6 again.
R1C6, R2C6, R3C6, R4C6, R5C6, R6C6, R7C6, R8C6, R9C6.
R4C6=6.
R5C6=3.
R6C6=2.
R7C6=4.
R8C6=8.
R9C6=7.
These are fixed by previous steps.
R2C6 and R3C6 are `5, 9`.
R1C6 is `1`.
Is R1C7 really 1?
Image: `9 6 . | . 4 . | 1 . .`
Yes, R1C7 is 1.
So R1C6 cannot be 1.
This implies one of the values in Col 6 (R4-R9) is wrong.
Let's re-verify R9C6=7.
Row 9: `_ _ _ | 9 2 _ | _ 8 5`.
Col 6 missing `5, 7, 9` (assuming R1-R3 unknown).
Row 9 missing `1, 3, 4, 6, 7`.
Intersection `7`. Correct.
Re-verify R3C6=5.
Row 3: `7 . 8 | . 6 . | . . 9`.
Col 6 missing `5, 9`.
Row 3 has 9. So R3C6=5. Correct.
Re-verify R2C6=9.
Col 6 missing `9`. Correct.
So Col 6 has `5, 9, 6, 3, 2, 4, 8, 7`.
Missing `1`.
So R1C6=1.
Conflict with R1C7=1.
Is it possible R4C6 is not 6?
Row 4: `1 2 4 | 8 7 6 | 9 5 3`.
R4C6=6 derived from:
Box 5 missing `1,3,4,6,9`.
R4C6 cannot be 1 (Row 4).
R4C6 cannot be 4 (Col 6 has 4? R7C6=4. Yes).
R4C6 cannot be 9 (Row 4 has 9).
So R4C6 is `3` or `6`.
Row 4 missing `4, 6`?
Row 4: `1, 2, ?, 8, 7, ?, 9, 5, 3`.
Missing `4, 6`.
R4C3 and R4C6.
R4C3 is in Col 3.
R4C6 is in Col 6.
If R4C6=3, then Row 4 has two 3s (R4C9=3). Impossible.
So R4C6 MUST be 6 or 4.
But Row 4 missing `4, 6`.
If R4C6=4, then R4C3=6.
If R4C6=6, then R4C3=4.
Check Col 6 for 4.
R7C6=4.
So R4C6 cannot be 4.
So R4C6 MUST BE 6.
So the chain holds. R1C6=1.
Is R1C7 actually 1?
Looking at the image crop 1: `9 6 . | . 4 . | 1 . .`
Yes.
Is there an error in my Col 6 filling?
R7C6=4.
Image Row 7: `8 . . | . . 4 | . 7 .`
Yes, R7C6=4.
R8C6=8.
Image Row 8: `. . . | . 3 8 | . . .`
Yes, R8C6=8.
R9C6=7.
Derived.
R6C6=2.
Image Row 6: `3 . 5 | . . 2 | . 6 4`
Yes.
R5C6=3.
Derived from Box 5.
R4C6=6.
Derived.
R2C6=9, R3C6=5.
Derived.
So R1C6=1.
Conclusion: The puzzle as interpreted has a contradiction in Row 1 (1 at C6 and C7).
However, often in these puzzles, I might have misidentified a number.
Let's check R1C5. Image: `4`.
Let's check R1C7. Image: `1`.
Wait, look at Box 3 (Top Right).
Cells:
R1C7=1, R1C8, R1C9
R2C7, R2C8, R2C9
R3C7, R3C8, R3C9=9
If R1C6=1, then R1C7 cannot be 1.
Is it possible R1C7 is NOT 1?
Image: `1` is clearly in R1C7.
Is it possible R1C6 is NOT 1?
This requires Col 6 to have a 1 elsewhere.
R2-R9 Col 6: `9, 5, 6, 3, 2, 4, 8, 7`.
None of these are 1.
Did I place a number wrong in Col 6?
R9C6=7.
If R9C6 is not 7, what can it be?
Row 9: `_ _ _ 9 2 _ _ 8 5`.
Missing `1, 3, 4, 6, 7`.
Col 6 missing `1, 5, 7, 9` (initially).
If R9C6=1?
Then Col 6 has 1. R1C6 is not 1.
If R9C6=1, then Row 9 has 1.
Then Col 6 missing `5, 7, 9`.
R2C6, R3C6, R1C6.
R3C6 cannot be 9.
Why did I say R9C6=7?
Intersection of Row 9 missing `1,3,4,6,7` and Col 6 missing `1,5,7,9`.
Common: `1, 7`.
I eliminated 1 because... I didn't.
I said "Intersection: 7". That was a mistake. Intersection is `1, 7`.
If R9C6=1:
Then R1C6 is `5, 7, 9`.
Col 6 missing `5, 7, 9`.
R1C6, R2C6, R3C6.
Parent Tip: Review the logic above to help your child master the concept of printable games sudoku.
