Sudoku Puzzle Worksheet Set Instant Download Kid-friendly Sudoku ... - Free Printable

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 singles in Rows and Columns
* Row 6: The numbers present are 6, 9, 7, 3, 1. The missing numbers are 2, 4, 5, 8. Looking at Column 2, it already has a 9, 1, 6. Let's look closer. In Row 6, the cell R6C2 (Row 6, Column 2) must be 4 because Column 2 already has 1, 9, 6, and Row 6 needs a 4 to complete the set with the other visible numbers? Let's re-evaluate systematically.

Let's use a more rigorous method by looking at specific boxes and intersections.

Step 2: Analyze Box 5 (Center Box)
* Current numbers in Box 5: 1, 9, 7, 3.
* Missing numbers: 2, 4, 5, 6, 8.
* Look at Row 5: It has 7, 1, 2, 6. Missing: 3, 4, 5, 8, 9.
* Look at Column 5: It has 2, 3, 1, 7, 9, 8. Missing: 4, 5, 6.
* Cell R4C5 is in Box 5. Column 5 needs 4, 5, 6. Row 4 has 1, 6, 9. So R4C5 cannot be 6.
* Let's look at Row 4: Numbers are 1, 6, 9. Missing: 2, 3, 4, 5, 7, 8.
* Let's look at Column 6: Numbers are 9, 8, 5, 6, 3. Missing: 1, 2, 4, 7.
* Cell R4C6 is in Box 5. Row 4 has 1, 6, 9. Col 6 has 9,8,5,6,3.
* This is getting complex. Let's find the "low hanging fruit" (cells with only one possibility).

Revised Step-by-Step Solution:

1. Row 1: Missing numbers are 1, 3, 5, 6.
* R1C1: Col 1 has 2,8,7,6,5,4. Missing in Col 1: 1,3,9. Row 1 needs 1,3,5,6. Intersection: 1,3. Box 1 has 2,8,9,6. So R1C1 can be 1 or 3.
* Let's look at Box 1 (Top Left). Numbers: 2, 8, 9, 6. Missing: 1, 3, 4, 5, 7.
* R1C3 is 7. So 7 is placed.
* R2C1 is 2. R3C1 is 8, R3C2 is 9, R3C3 is 6.
* Cells remaining in Box 1: R1C1, R1C2, R2C2, R2C3.
* Missing in Box 1: 1, 3, 4, 5.
* R1C2: Row 1 has 7,4,2,9,8. Wait, R1C4=4, R1C5=2, R1C6=9, R1C8=8.
* Row 1 so far: [?, ?, 7, 4, 2, 9, ?, 8, ?]. Missing: 1, 3, 5, 6.
* Col 2 has 9, 1, 6. So R1C2 cannot be 1 or 6. Must be 3 or 5.
* Col 1 has 2, 8, 7, 6, 5, 4. So R1C1 cannot be 5 or 6. Must be 1 or 3.

Let's solve Column 5 first.
* Col 5 values: 2, ?, 3, ?, 1, 7, ?, 9, 8.
* Missing: 4, 5, 6.
* R2C5: Row 2 has 2, 8, 3. Box 2 has 4,2,9,8,3,5. Wait, Box 2 (Top Middle) has 4,2,9,8,3,5. Missing: 1,6,7.
* Cells in Box 2: R1C4(4), R1C5(2), R1C6(9), R2C4(?), R2C5(?), R2C6(8), R3C4(?), R3C5(3), R3C6(5).
* Missing in Box 2: 1, 6, 7.
* R2C4, R2C5, R3C4 are empty.
* Row 2 has 2, 8, 3. So R2C4, R2C5 cannot be 2,3,8.
* Col 4 has 4, 9, 5. Col 5 has 2, 3, 1, 7, 9, 8.
* Let's look at R2C5. Col 5 needs 4,5,6. Box 2 needs 1,6,7. The common number is 6. So R2C5 = 6.
* Now Box 2 missing: 1, 7. Cells: R2C4, R3C4.
* Row 2 has 2, 6, 8, 3. Missing 1,4,5,7,9.
* Col 4 has 4, 9, 5. So R2C4 cannot be 4,5,9. It must be 1 or 7.
* Row 3 has 8,9,6,3,5,1,4. Missing 2,7.
* R3C4 is in Row 3. Row 3 missing 2,7. Box 2 needs 1,7. So R3C4 must be 7.
* Therefore, R2C4 = 1.
* Now Row 3 missing is just 2. So R3C8 = 2? No, R3C8 is in Box 3. Let's check Row 3: 8,9,6,7,3,5,1,?,4. Missing is 2. So R3C8 = 2.

2. Continue with Box 2 and Row 1/2
* We found R2C5=6, R3C4=7, R2C4=1.
* Row 2 now: 2, ?, ?, 1, 6, 8, 3, ?, ?. Missing: 4, 5, 7, 9.
* Box 1 (Top Left) missing: 1, 3, 4, 5. We know R1C1, R1C2, R2C2, R2C3 are empty.
* R2C2 and R2C3 are in Row 2. Row 2 missing 4,5,7,9. Box 1 missing 1,3,4,5. Intersection for R2C2, R2C3 is 4,5.
* Col 2 has 9,1,6. Col 3 has 7,6,8.
* Let's look at Col 3. Values: 7, ?, 6, ?, ?, ?, ?, 8, ?.
* Let's go back to Row 1. Missing 1,3,5,6.
* R1C1, R1C2, R1C7, R1C9.
* Col 9 has 8,4,6,1. Missing 2,3,5,7,9.
* R1C9 is in Box 3. Box 3 has 8,3,2,4,1. (From R1C8=8, R2C7=3, R3C7=1, R3C8=2, R3C9=4).
* Wait, let's list Box 3 knowns:
* R1C8=8.
* R2C7=3.
* R3C7=1, R3C8=2, R3C9=4.
* So Box 3 has 1,2,3,4,8. Missing: 5,6,7,9.
* Empty cells in Box 3: R1C7, R1C9, R2C8, R2C9.
* Row 1 missing: 1,3,5,6. But 1,3 are not in Box 3 missing list. So R1C7 and R1C9 must be from {5,6}.
* Since R1C7 and R1C9 are 5 or 6, the remaining Row 1 cells R1C1 and R1C2 must be 1 and 3.
* Col 1 has 2,8,7,6,5,4. Missing 1,3,9. R1C1 is 1 or 3.
* Col 2 has 9,1,6. R1C2 is 1 or 3. But Col 2 already has 1 (in R4C2). So R1C2 cannot be 1. Thus R1C2 = 3 and R1C1 = 1.
* Now Row 1 is: 1, 3, 7, 4, 2, 9, ?, 8, ?. Missing 5,6.
* Col 7 has 3,1,2. Col 9 has 8,4,6,1.
* Check Col 7: R1C7. If R1C7=5, R1C9=6. If R1C7=6, R1C9=5.
* Look at Col 7. It has 3,1,2.
* Look at Col 9. It has 8,4,6,1. It already has a 6 (R5C9=6). So R1C9 cannot be 6.
* Therefore, R1C9 = 5 and R1C7 = 6.
* Row 1 Complete: 1, 3, 7, 4, 2, 9, 6, 8, 5.

3. Finish Box 1 and Row 2
* Box 1 missing: 4, 5. Cells: R2C2, R2C3.
* Row 2 missing: 4, 5, 7, 9. (We have 2,1,6,8,3).
* R2C2 and R2C3 are 4 or 5.
* Col 2 has 3,9,1,6. Missing 2,4,5,7,8.
* Col 3 has 7,6,8. Missing 1,2,3,4,5,9.
* Let's look at Col 2. R2C2 is 4 or 5.
* Let's look at Row 2 remaining cells: R2C8, R2C9. They must be 7, 9.
* Box 3 missing: 7, 9. (We have 6,8,5,3,1,2,4). Cells R2C8, R2C9.
* Col 8 has 8,2,9,6,1. Missing 3,4,5,7. R2C8 is 7 or 9. Col 8 has 9 (R4C8=9). So R2C8 cannot be 9. Thus R2C8 = 7 and R2C9 = 9.
* Now back to R2C2, R2C3. They are 4,5.
* Col 2 has 1,3,9,6.
* Col 3 has 7,6,8.
* Let's hold this. Look at Col 3.
* Row 3 is complete: 8,9,6,7,3,5,1,2,4.
* Row 2 so far: 2, [4/5], [5/4], 1, 6, 8, 3, 7, 9.

4. Solve Middle Band (Rows 4,5,6)
* Row 4: ?, 1, ?, ?, ?, 6, ?, 9, ?.
* Knowns in Row 4: 1, 6, 9.
* Box 4 (Middle Left) cells: R4C1, R4C2(1), R4C3, R5C1(7), R5C2, R5C3, R6C1(6), R6C2, R6C3.
* Box 4 numbers present: 1, 7, 6. Missing: 2,3,4,5,8,9.
* Col 1: 1,2,8,?,7,6,?,5,4. Missing: 3,9.
* R4C1 and R7C1 are empty in Col 1.
* R4C1 is in Row 4. Row 4 missing 2,3,4,5,7,8.
* Col 1 missing 3,9. So R4C1 is 3 or 9.
* Box 4 needs 2,3,4,5,8,9.
* Let's look at Col 1. R7C1 is also empty.
* Row 7: ?, ?, ?, ?, ?, ?, ?, 6, ?.
* Let's determine R4C1.
* Look at Box 7 (Bottom Left). Cells R7C1-3, R8C1-3, R9C1-3.
* Col 1 bottom: R7C1, R8C1(5), R9C1(4).
* Col 1 missing 3,9. So R7C1 is 3 or 9. R4C1 is 3 or 9.

Let's look at Row 5: 7, ?, ?, ?, 1, ?, 2, ?, 6.
* Missing: 3,4,5,8,9.
* Box 5 (Center) missing: 2,4,5,8. (Has 1,9,7,3,6 from previous steps? No. Box 5 has R4C4-6, R5C4-6, R6C4-6).
* Let's restart Box 5 content.
* R4C4=?, R4C5=?, R4C6=6.
* R5C4=?, R5C5=1, R5C6=?.
* R6C4=9, R6C5=7, R6C6=3.
* So Box 5 has: 6,1,9,7,3. Missing: 2,4,5,8.
* Cells: R4C4, R4C5, R5C4, R5C6.
* Col 5 missing: 4,5. (Has 2,6,3,1,7,9,8). Wait, Col 5 has R1=2, R2=6, R3=3, R5=1, R6=7, R8=9, R9=8. Missing 4,5.
* Cells in Col 5: R4C5, R7C5.
* So R4C5 and R7C5 are 4 and 5.
* Row 4 missing 2,3,4,5,7,8.
* Box 5 missing 2,4,5,8.
* R4C5 is 4 or 5.
* R4C4 is in Box 5. Col 4 has 4,1,7,9,5. (R1=4, R2=1, R3=7, R6=9, R9=5). Missing 2,3,6,8.
* R4C4 cannot be 6 (Row 4 has 6). Cannot be 3?
* Let's look at R5C6. Col 6 has 9,8,5,6,3,1. Missing 2,4,7.
* Box 5 missing 2,4,5,8. R5C6 must be 2,4,8. Intersection with Col 6 (2,4,7) is 2,4.
* Row 5 missing 3,4,5,8,9.

This is tricky without a grid. Let's deduce Col 4.
* Col 4: 4, 1, 7, ?, ?, 9, ?, ?, 5.
* Missing: 2,3,6,8.
* R4C4, R5C4, R7C4, R8C4.
* Row 4 has 1,6,9. So R4C4 != 6.
* Row 5 has 7,1,2,6. So R5C4 != 2,6.
* Row 8 has 5,8,9. So R8C4 != 8.
* Row 7 has 6.

Let's solve Box 6 (Middle Right).
* Cells: R4C7-9, R5C7-9, R6C7-9.
* Knowns: R4C8=9. R5C7=2, R5C9=6. R6C7=?, R6C8=?, R6C9=1.
* Row 6: 6, ?, ?, 9, 7, 3, ?, ?, 1. Missing: 2,4,5,8.
* Box 6 has 9,2,6,1. Missing: 3,4,5,7,8.
* R4C7, R4C9, R6C7, R6C8 are empty in Box 6? No, R5C8 is also empty.
* Row 5: 7, ?, ?, ?, 1, ?, 2, ?, 6.
* R5C8 is in Box 6.

Let's look at Col 9.
* Values: 5, 9, 4, ?, 6, 1, ?, ?, ?.
* Missing: 2,3,7,8.
* R4C9, R7C9, R8C9, R9C9.
* Row 4 missing 2,3,4,5,7,8.
* Row 9: 4,6,?,5,8,?,?,1,?.

Let's try filling Row 6.
* Missing: 2,4,5,8.
* Cells: R6C2, R6C3, R6C7, R6C8.
* Col 2 has 3,9,1,6. Missing 2,4,5,7,8.
* Col 3 has 7,6,8. Missing 1,2,3,4,5,9.
* Col 7 has 6,3,1,2. Missing 4,5,7,8,9.
* Col 8 has 8,7,2,9,6,1. Missing 3,4,5.
* R6C8 is in Col 8. Col 8 missing 3,4,5. Row 6 missing 2,4,5,8. Intersection: 4,5.
* R6C7 is in Col 7. Col 7 missing 4,5,7,8,9. Row 6 missing 2,4,5,8. Intersection: 4,5,8.

Let's look at Box 4 again.
* Missing: 2,3,4,5,8,9.
* R4C1, R4C3, R5C2, R5C3, R6C2, R6C3.
* Col 1 missing 3,9. R4C1 is 3 or 9.
* If R4C1=3, R7C1=9. If R4C1=9, R7C1=3.
* Row 4 starts with R4C1.

Let's look at Row 9.
* 4, 6, ?, 5, 8, ?, ?, 1, ?.
* Missing: 2,3,7,9.
* Col 3 missing 1,2,3,4,5,9. R9C3 is 2,3,9.
* Col 6 missing 2,4,7. R9C6 is 2,7.
* Col 7 missing 4,5,7,8,9. R9C7 is 7,9.
* Col 9 missing 2,3,7,8. R9C9 is 2,3,7.

Let's solve Col 6.
* Values: 9,8,5,6,?,3,?,?,?.
* Missing: 1,2,4,7.
* R5C6, R7C6, R8C6, R9C6.
* Row 5 missing 3,4,5,8,9. R5C6 must be 4. (Since 1,2,7 not in Row 5 missing? No, Row 5 has 1,2,6,7,9? No. Row 5 has 7,1,2,6. Missing 3,4,5,8,9. So R5C6 can be 4. Can it be others? Col 6 needs 1,2,4,7. Intersection is 4. So R5C6 = 4.
* Now Col 6 missing: 1,2,7.
* Row 7 missing: 1,2,3,4,5,7,8,9? No.
* Row 8: 5,?,8,?,9,?,?,?,?.
* Row 9: 4,6,?,5,8,?,?,1,?.

Since R5C6=4:
* Box 5 missing now: 2,5,8. Cells: R4C4, R4C5, R5C4.
* Col 5 missing: 4,5. We established R4C5, R7C5 are 4,5.
* Since R5C6=4, Row 5 has 4.
* In Box 5, R4C5 is 4 or 5.
* If R4C5=4, then R7C5=5.
* If R4C5=5, then R7C5=4.
* Box 5 missing 2,5,8. R4C4, R4C5, R5C4.
* If R4C5=4, then Box 5 missing 2,5,8 for R4C4, R5C4? No, R4C5 is filled.
* Wait, Box 5 cells are R4C4, R4C5, R4C6(6), R5C4, R5C5(1), R5C6(4), R6C4(9), R6C5(7), R6C6(3).
* So Box 5 missing: 2,5,8.
* Empty cells: R4C4, R4C5, R5C4.
* Col 5 missing 4,5. So R4C5 is 4 or 5. But Box 5 needs 2,5,8. So R4C5 MUST be 5.
* Therefore, R7C5 = 4 (last spot in Col 5).
* Now Box 5 missing: 2,8. Cells: R4C4, R5C4.
* Col 4 missing: 2,3,6,8.
* Row 4 missing: 2,3,4,7,8. (Has 1,5,6,9).
* Row 5 missing: 3,5,8,9. (Has 7,1,4,2,6). Wait, R5C6=4. Row 5: 7,?,?,?,1,4,2,?,6. Missing 3,5,8,9.
* R5C4 is in Box 5. Must be 2 or 8. But Row 5 missing doesn't have 2. Contradiction?
* Let's re-check Row 5.
* Row 5: 7, R5C2, R5C3, R5C4, 1, 4, 2, R5C8, 6.
* Missing: 3,5,8,9.
* Box 5 missing was 2,8. R5C4 must be 8? Because 2 is already in Row 5 (R5C7=2).
* So R5C4 = 8.
* Then R4C4 = 2.
* Check Box 5: 2,5,6 / 8,1,4 / 9,7,3. All good.

Now we have:
* R4C4=2, R4C5=5, R4C6=6.
* R5C4=8, R5C5=1, R5C6=4.
* R6C4=9, R6C5=7, R6C6=3.

Back to Row 4:
* Values: R4C1=?, R4C2=1, R4C3=?, R4C4=2, R4C5=5, R4C6=6, R4C7=?, R4C8=9, R4C9=?.
* Missing: 3,4,7,8.
* Col 1 missing 3,9. R4C1 is 3 or 9. But Row 4 has 9. So R4C1 = 3.
* Therefore R7C1 = 9 (last in Col 1).
* Row 4 missing now: 4,7,8. Cells: R4C3, R4C7, R4C9.
* Col 3 missing 1,2,3,4,5,9. R4C3 is 4,7,8. Intersection: 4. So R4C3 = 4?
* Check Col 3. Has 7,6,8,4? No, 4 is not in Col 3 yet.
* Is 7 in Col 3? Yes (R1C3=7). So R4C3 != 7.
* Is 8 in Col 3? Yes (R8C3=8). So R4C3 != 8.
* Therefore R4C3 = 4.
* Row 4 missing: 7,8. Cells: R4C7, R4C9.
* Col 7 missing 4,5,7,8,9.
* Col 9 missing 2,3,7,8.
* Let's check Box 6.
* Box 6 missing: 3,5,7,8. (Has 2,4,6,1,9).
* Cells: R4C7, R4C9, R5C8, R6C7, R6C8.
* R4C7, R4C9 are 7,8.
* So R5C8, R6C7, R6C8 are 3,5. And one more?
* Box 6 has 5 cells empty? No.
* R4C7, R4C9 (2 cells).
* R5C8 (1 cell).
* R6C7, R6C8 (2 cells).
* Total 5 cells. Missing 3,5,7,8? No, Box 6 has 1,2,4,6,9. Missing 3,5,7,8. That's 4 numbers. But 5 cells?
* Ah, R5C7=2, R5C9=6. R6C9=1. R4C8=9.
* Box 6 cells:
* R4C7, R4C8(9), R4C9
* R5C7(2), R5C8, R5C9(6)
* R6C7, R6C8, R6C9(1)
* Present: 1,2,6,9. Missing: 3,4,5,7,8.
* We determined R4C7, R4C9 are 7,8.
* So remaining cells R5C8, R6C7, R6C8 must be 3,4,5.
* Row 5 missing: 3,5,9. (Has 7,4,8,1,4? No. R5: 7,?,?,8,1,4,2,?,6. Missing 3,5,9).
* R5C8 is in Row 5. Must be 3,5,9. Intersection with Box 6 remainder (3,4,5) is 3,5.
* Row 6 missing: 2,4,5,8. (Has 6,?, ?, 9,7,3,?,?,1).
* R6C7, R6C8 are in Row 6. Must be from 2,4,5,8. Intersection with Box 6 remainder (3,4,5) is 4,5.
* So R6C7, R6C8 are 4,5.
* Therefore R5C8 must be 3 (since 4,5 are taken by Row 6 in Box 6, and 3 is left for Box 6? No. Box 6 missing 3,4,5,7,8. R4 takes 7,8. Remaining 3,4,5. R6 takes 4,5. So R5C8=3).
* So R5C8 = 3.
* Now Row 5 missing: 5,9. Cells: R5C2, R5C3.
* Col 2 missing 2,4,5,7,8. R5C2 is 5 or 9. Col 2 has no 9? Wait, Col 2 has 3,9,1,6. Yes, 9 is there (R3C2=9). So R5C2 cannot be 9. Thus R5C2 = 5 and R5C3 = 9.
* Row 5 Complete: 7, 5, 9, 8, 1, 4, 2, 3, 6.

Back to Row 6:
* Missing: 2,4,8. (Has 6,5,9,9? No. R6: 6, R6C2, R6C3, 9,7,3, R6C7, R6C8, 1).
* We said R6C7, R6C8 are 4,5. But R5C2=5, so Col 2 has 5.
* Wait, R6C2, R6C3 are in Box 4.
* Row 6 missing: 2,4,5,8.
* R6C7, R6C8 are 4,5.
* So R6C2, R6C3 are 2,8.
* Col 2 has 3,9,1,5,6. Missing 2,4,7,8.
* Col 3 has 7,6,4,9,8. Missing 1,2,3,5.
* R6C3 is 2 or 8. Col 3 missing 1,2,3,5. So R6C3 cannot be 8. Thus R6C3 = 2 and R6C2 = 8.
* Now R6C7, R6C8 are 4,5.
* Col 7 missing 4,5,7,8,9.
* Col 8 missing 4,5. (Has 8,7,2,9,3,6,1). Wait, Col 8 has R1=8, R2=7, R3=2, R4=9, R5=3, R6=?, R7=6, R8=?, R9=1.
* Col 8 missing 4,5.
* So R6C8 is 4 or 5. R8C8 is 4 or 5.
* Row 6: R6C7, R6C8 are 4,5.
* Check Col 7. R6C7 is 4 or 5.
* Let's check R4C7, R4C9. They were 7,8.
* Col 7 has 6,3,1,2. Missing 4,5,7,8,9.
* Col 9 has 5,9,4,6,1. Missing 2,3,7,8.
* R4C9 is 7 or 8.
* R4C7 is 7 or 8.

Let's solve Box 4 (Middle Left).
* Cells: R4C1(3), R4C2(1), R4C3(4), R5C1(7), R5C2(5), R5C3(9), R6C1(6), R6C2(8), R6C3(2).
* Box 4 Complete: 3,1,4 / 7,5,9 / 6,8,2.

Now Row 4:
* 3, 1, 4, 2, 5, 6, ?, 9, ?.
* Missing: 7,8.
* Col 7 missing 4,5,7,8,9.
* Col 9 missing 2,3,7,8.
* Look at Col 7. R4C7 is 7 or 8.
* Look at Col 9. R4C9 is 7 or 8.

Let's look at Row 7.
* 9, ?, ?, ?, 4, ?, ?, 6, ?.
* Missing: 1,2,3,5,7,8.
* Col 2 missing 2,4,7. (Has 3,9,1,5,8,6). Wait, Col 2: R1=3, R2=?, R3=9, R4=1, R5=5, R6=8, R7=?, R8=?, R9=6.
* Col 2 missing: 2,4,7.
* R2C2 is 4 or 5? No, R2C2 was 4 or 5 in Box 1.
* Let's finish Box 1.
* R2C2, R2C3 were 4,5.
* Col 2 missing 2,4,7. R2C2 is 4 or 5. So R2C2 = 4.
* Therefore R2C3 = 5.
* Then Col 2 missing 2,7. Cells R7C2, R8C2.
* Row 7 missing 1,2,3,5,7,8. R7C2 is 2 or 7.
* Row 8: 5, ?, 8, ?, 9, ?, ?, ?, ?.
* R8C2 is 2 or 7.

Back to Col 3.
* Values: 7,5,6,4,9,2,?,8,?.
* Missing: 1,3.
* Cells: R7C3, R9C3.
* Row 7 missing 1,2,3,5,7,8. R7C3 is 1 or 3.
* Row 9 missing 2,3,7,9. R9C3 is 1 or 3. But Row 9 has no 1? Wait, R9C8=1. So Row 9 has 1. Thus R9C3 cannot be 1. So R9C3 = 3 and R7C3 = 1.
* Col 3 Complete: 7,5,6,4,9,2,1,8,3.

Now Row 9:
* 4, 6, 3, 5, 8, ?, ?, 1, ?.
* Missing: 2,7,9.
* Col 6 missing 1,2,7. (Has 9,8,5,6,4,3). Wait, Col 6: R1=9, R2=8, R3=5, R4=6, R5=4, R6=3, R7=?, R8=?, R9=?.
* Missing: 1,2,7.
* R9C6 is 2,7,9. Intersection with Col 6 (1,2,7) is 2,7.
* Col 7 missing 4,5,7,8,9. R9C7 is 2,7,9. Intersection: 7,9.
* Col 9 missing 2,3,7,8. R9C9 is 2,7,9. Intersection: 2,7.

Let's look at Row 7.
* 9, R7C2, 1, R7C4, 4, R7C6, R7C7, 6, R7C9.
* Missing: 2,3,5,7,8.
* Col 2 missing 2,7. R7C2 is 2 or 7.
* Col 4 missing 3,6. (Has 4,1,7,2,8,9,5). Wait, Col 4: R1