Easy Sudoku Puzzles - Free Printable - Free Printable

Let's solve this Sudoku puzzle step by step.

---

🔍 Understanding the Puzzle

This is a standard 9×9 Sudoku grid, divided into 9 smaller 3×3 boxes. The goal is to fill the grid so that:

- Each row contains digits 1–9 with no repetition.
- Each column contains digits 1–9 with no repetition.
- Each 3×3 box contains digits 1–9 with no repetition.

We are given some numbers and need to deduce the missing ones.

---

🧩 Step-by-step Solution

Let me first write down the initial grid for clarity:

```
+-------+-------+-------+
| 7 4 | 3 | 1 |
| 1 9 | 6 8 5 | 2 |
| | 4 3 | |
+-------+-------+-------+
| 5 6 | 3 7 | 1 |
| 1 8 | | 9 5 |
| 9 | 2 | 6 |
+-------+-------+-------+
| 1 3 | 4 | 7 2 |
| 5 2 | | 8 |
| 8 | 1 | 4 7 |
+-------+-------+-------+
```

Now let’s solve it systematically.

---

🔎 Step 1: Fill in obvious cells using elimination

We'll look for cells where only one number can fit based on row, column, and box constraints.

#### Let’s label rows R1 to R9 and columns C1 to C9.

---

✔ Box 1 (Top-left 3x3): R1C1 to R3C3

Current:
```
R1: 7 4 _
R2: 1 9 _
R3: _ _ _
```

Missing: 2,3,5,6,8

But we also have constraints from rows and columns.

Let’s focus on R1C3 — currently empty.

- Row 1 has: 7,4,3,1 → missing: 2,5,6,8,9
- Column 3: R1C3, R2C3, R3C3 → values: ?, ?, 4 → so far: 4
- Box 1: missing 2,3,5,6,8 → but 3 is already in R1C4 → not in box? Wait, R1C4 is outside box 1.

Wait — let's clarify:

Box 1 includes:
- R1C1=7, R1C2=4, R1C3=?
- R2C1=1, R2C2=9, R2C3=?
- R3C1=?, R3C2=?, R3C3=?

So current values in Box 1: 7,4,1,9 → missing: 2,3,5,6,8

Now check R1C3:

- Row 1: has 7,4,3,1 → missing: 2,5,6,8,9
- Column 3: R1C3, R2C3, R3C3, R4C3=3, R5C3=8, R6C3=2, R7C3=4, R8C3=2, R9C3=1 → wait, let's list full column 3:

Column 3:
- R1: ?
- R2: ? (R2C3 = ?)
- R3: ?
- R4: 7
- R5: ?
- R6: 2
- R7: 4
- R8: ?
- R9: 1

Wait — from the grid:

From original:

Row 4: ` 5 6 | 3 7 | 1` → so R4C3 = 3

Row 5: ` 1 8 | | 9 5` → R5C3 = ?

Row 6: ` 9 | 2 | 6 ` → R6C3 = 2

Row 7: `1 3 | 4 | 7 2` → R7C3 = 4

Row 8: `5 2 | | 8` → R8C3 = 2? Wait, no — R8C3 is third cell of row 8.

Wait — let's re-express the full grid clearly.

---

📐 Full Grid with Coordinates

Let me write the grid with positions:

```
Row 1: 7 4 _ | 3 _ _ | _ _ 1
Row 2: 1 9 _ | 6 8 5 | _ _ 2
Row 3: _ _ _ | 4 3 _ | _ _ _
Row 4: _ 5 6 | 3 7 _ | _ _ 1
Row 5: _ 1 8 | _ _ _ | 9 5 _
Row 6: _ 9 _ | 2 _ 6 | _ _ _
Row 7: 1 _ 3 | 4 _ _ | 7 2 _
Row 8: 5 _ 2 | _ _ _ | _ _ 8
Row 9: _ 8 _ | _ 1 4 | 7 _ _
```

Now let’s go step by step.

---

🔎 Look at R1C3 (top-right of box 1)

- Row 1: has 7,4,3,1 → missing: 2,5,6,8,9
- Column 3: values:
- R1: ?
- R2: ?
- R3: ?
- R4: 6 → wait! R4C3 = 6 (from "5 6" in row 4)
- R5: 8
- R6: 2
- R7: 3
- R8: 2 → R8C3 = 2
- R9: ? → R9C3 = ?

Wait — R9C3 is in column 3, row 9.

But R8C3 = 2 (since row 8: 5 _ 2), R9C3 = ? → but R9 has: _ 8 _ | _ 1 4 | 7 _ _ → so R9C3 is third cell of row 9 → yes, unknown.

So column 3:
- R1: ?
- R2: ?
- R3: ?
- R4: 6
- R5: 8
- R6: 2
- R7: 3
- R8: 2 → duplicate? Wait — R6C3 = 2 and R8C3 = 2 → same column → conflict?

No — R6C3 = 2 (row 6: _ 9 _ | 2 _ 6 | _ _ _ → so R6C4 = 2, not R6C3!)

Mistake here!

Let’s fix this.

Row 6: ` 9 | 2 | 6 ` → So:
- R6C1 = ?
- R6C2 = 9
- R6C3 = ?
- R6C4 = 2
- R6C5 = ?
- R6C6 = 6
- R6C7 = ?
- R6C8 = ?
- R6C9 = ?

So R6C3 is not 2 — R6C4 is 2.

Similarly, R8: `5 2 | | 8` → R8C1=5, R8C2=?, R8C3=2, R8C4=?, R8C5=?, R8C6=?, R8C7=?, R8C8=?, R8C9=8

So R8C3 = 2

R7: `1 3 | 4 | 7 2` → R7C1=1, R7C2=?, R7C3=3, R7C4=4, R7C5=?, R7C6=?, R7C7=7, R7C8=2, R7C9=?

So R7C3 = 3

R4: ` 5 6 | 3 7 | 1` → R4C1=?, R4C2=5, R4C3=6, R4C4=3, R4C5=7, R4C6=?, R4C7=?, R4C8=?, R4C9=1

So R4C3 = 6

R5: ` 1 8 | | 9 5` → R5C1=?, R5C2=1, R5C3=8, R5C4=?, R5C5=?, R5C6=?, R5C7=9, R5C8=5, R5C9=?

So R5C3 = 8

R6C3 = ? (unknown)

R9C3 = ? (unknown)

So column 3:
- R1: ?
- R2: ?
- R3: ?
- R4: 6
- R5: 8
- R6: ?
- R7: 3
- R8: 2
- R9: ?

So used in column 3: 6,8,3,2 → missing: 1,4,5,7,9

But R1C3 must be in row 1: missing: 2,5,6,8,9 → but 6 and 8 are already in column 3 → so R1C3 cannot be 6 or 8.

Also, R1C3 is in box 1 → box 1 has: 7,4,1,9 → so missing: 2,3,5,6,8

So possible for R1C3: intersection of:
- Row 1 missing: 2,5,6,8,9
- Col 3 missing: 1,4,5,7,9
- Box 1 missing: 2,3,5,6,8

Intersection: common numbers → 5

Because:
- 5 is in all three sets.

→ So R1C3 = 5

✔ Fill it in.

Now update:

R1: 7 4 5 | 3 _ _ | _ _ 1

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

Row 1: now has 7,4,5,3,1 → missing: 2,6,8,9

---

🔎 Now look at R1C4 = 3 → already filled

R1C5 and R1C6 are empty.

Row 1: missing: 2,6,8,9

Column 5:
- R1: ?
- R2: 8
- R3: 3
- R4: 7
- R5: ?
- R6: ?
- R7: ?
- R8: ?
- R9: 1

So column 5: has 8,3,7,1 → missing: 2,4,5,6,9

Box 2 (middle top): R1C4 to R3C6

Box 2:
- R1C4=3, R1C5=?, R1C6=?
- R2C4=6, R2C5=8, R2C6=5
- R3C4=4, R3C5=3, R3C6=?

So values: 3,6,8,5,4,3 → duplicates? R3C5=3, R1C4=3 → ok, different cells.

Used: 3,4,5,6,8 → missing: 1,2,7,9

So R1C5 and R1C6 must be from {2,6,8,9} (row 1) ∩ {1,2,7,9} (box 2) → intersection: 2,9

So R1C5 and R1C6 ∈ {2,9}

Now column 5: missing 2,4,5,6,9 → so 2 and 9 are allowed.

So possibilities: R1C5=2 or 9, R1C6=9 or 2

But both must be different.

Now look at R1C7, R1C8, R1C9 → row 1 ends with _ _ 1

Row 1 missing: 2,6,8,9 → but we just used 5, so missing: 2,6,8,9

But R1C9 = 1 → so last three cells: R1C7, R1C8, R1C9=1

So R1C7 and R1C8 must be two of 2,6,8,9

But we already assigned R1C3=5, so remaining: R1C4=3, R1C5=?, R1C6=?, R1C7=?, R1C8=?, R1C9=1

So missing: 2,6,8,9 → four numbers for four cells.

But R1C5 and R1C6 must be 2 and 9 (in some order).

So R1C7 and R1C8 must be 6 and 8.

Now look at column 7:

Column 7:
- R1: ?
- R2: ?
- R3: ?
- R4: ?
- R5: 9
- R6: 6
- R7: 7
- R8: ?
- R9: 7 → R9C7 = 7? Wait, R9: _ 8 _ | _ 1 4 | 7 _ _ → so R9C7 = 7

So column 7:
- R1: ?
- R2: ?
- R3: ?
- R4: ?
- R5: 9
- R6: 6
- R7: 7
- R8: ?
- R9: 7 → duplicate? R7C7=7, R9C7=7 → same column → invalid?

Wait — R7C7 = 7 (from "7 2" in row 7), R9C7 = 7 → same column → conflict!

But that can't be — unless I made a mistake.

Wait: R7: `1 3 | 4 | 7 2` → so R7C7 = 7, R7C8 = 2

R9: `_ 8 _ | _ 1 4 | 7 _ _` → so R9C7 = 7

Yes — both R7C7 and R9C7 = 7 → same column → invalid!

But that can’t happen — the puzzle must be valid.

Wait — perhaps I misread.

Let me double-check the original image.

You said:

```
7 4 | 3 | 1
1 9 | 6 8 5 | 2
| 4 3 |
------+-------+------
5 6 | 3 7 | 1
1 8 | | 9 5
9 | 2 | 6
------+-------+------
1 3 | 4 | 7 2
5 2 | | 8
8 | 1 | 4 7
```

So:

- Row 7: 1 _ 3 | 4 _ _ | 7 2 _ → so R7C7 = 7
- Row 9: _ 8 _ | _ 1 4 | 7 _ _ → R9C7 = 7

Yes — both are 7 in column 7 → duplicate → impossible.

Wait — is there a typo?

But maybe I misread the grid.

Wait — the last box in row 9: `| 7 _ _` — that’s R9C7=7, R9C8=?, R9C9=?

And R7C7=7 — yes.

So column 7 has two 7s — invalid.

But that can't be — the puzzle must be solvable.

Wait — perhaps I made a mistake in reading the grid.

Let me recheck the original text:

> ```
> 7 4 | 3 | 1
> 1 9 | 6 8 5 | 2
> | 4 3 |
> ------+-------+------
> 5 6 | 3 7 | 1
> 1 8 | | 9 5
> 9 | 2 | 6
> ------+-------+------
> 1 3 | 4 | 7 2
> 5 2 | | 8
> 8 | 1 | 4 7
> ```

Row 7: `1 3 | 4 | 7 2` → so R7C1=1, R7C2=?, R7C3=3, R7C4=4, R7C5=?, R7C6=?, R7C7=7, R7C8=2, R7C9=?

Row 9: ` 8 | 1 | 4 7` → R9C1=?, R9C2=8, R9C3=?, R9C4=?, R9C5=1, R9C6=4, R9C7=7, R9C8=4? No — wait:

" | 4 7" — so R9C7=4, R9C8=7? No — the format is:

The last box: `| 1 | 4 7` → so:

- R9C7 = 4
- R9C8 = 7
- R9C9 = ?

Ah! Here's the mistake.

The last line: ` 8 | 1 | 4 7` — this means:

- First box: R9C1=?, R9C2=8, R9C3=?
- Second box: R9C4=?, R9C5=1, R9C6=4
- Third box: R9C7=4, R9C8=7, R9C9=?

Wait — no! The third box is `| 4 7` — but that’s only two numbers.

Look: the separator is `|`, so:

The last row: ` 8 | 1 | 4 7`

This means:
- Cells 1-3: _ 8 _
- Cells 4-6: _ 1 4
- Cells 7-9: 4 7 _

Wait — no — the format is:

Each row has three groups separated by `|`.

So:
- First group: three cells → R9C1, R9C2, R9C3 → from ` 8 ` → probably R9C2=8, others blank
- Second group: ` 1 ` → R9C4=?, R9C5=1, R9C6=?
- Third group: `4 7` → R9C7=4, R9C8=7, R9C9=?

But earlier I thought R9C7=7, but it's actually 4.

Similarly, R9C8=7.

So correct:

- R9C7 = 4
- R9C8 = 7
- R9C9 = ?

And R7C7 = 7 → still 7

So column 7:
- R1: ?
- R2: ?
- R3: ?
- R4: ?
- R5: 9
- R6: 6
- R7: 7
- R8: ?
- R9: 4

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

No duplicate — good.

Earlier error was assuming R9C7=7, but it's 4.

So R9C7=4, R9C8=7, R9C9=?

Now back to R1C7 and R1C8.

Row 1: missing: 2,6,8,9

R1C7 and R1C8 must be two of them.

But column 7: R1C7 must be from missing: 1,2,3,5,8 → so possible: 2,8

Column 8: R1C8

Column 8:
- R1: ?
- R2: ?
- R3: ?
- R4: ?
- R5: 5
- R6: ?
- R7: 2
- R8: ?
- R9: 7

So used: 5,2,7 → missing: 1,3,4,6,8,9

So R1C8 can be any of those.

But R1C7 ∈ {2,8}, R1C8 ∈ {6,8,9} (since row 1 missing: 2,6,8,9)

And R1C5 and R1C6 must be 2 and 9 in some order.

Let’s try to find more.

---

🔎 Back to R1C5 and R1C6

We had: must be 2 and 9

Now look at column 5:

- R1: ?
- R2: 8
- R3: 3
- R4: 7
- R5: ?
- R6: ?
- R7: ?
- R8: ?
- R9: 1

Used: 8,3,7,1 → missing: 2,4,5,6,9

So R1C5 ∈ {2,9} → both allowed.

Now column 6:

- R1: ?
- R2: 5
- R3: ?
- R4: ?
- R5: ?
- R6: 6
- R7: ?
- R8: ?
- R9: 4

Used: 5,6,4 → missing: 1,2,3,7,8,9

So R1C6 ∈ {2,9} → both allowed.

Now consider box 2 (R1C4 to R3C6):

Values:
- R1C4=3, R1C5=?, R1C6=?
- R2C4=6, R2C5=8, R2C6=5
- R3C4=4, R3C5=3, R3C6=?

Used: 3,6,8,5,4,3 → so 3 appears twice → okay, different cells.

Used: 3,4,5,6,8 → missing: 1,2,7,9

So R1C5 and R1C6 must be 2 and 9 (as before)

Now R3C6 is also in box 2 → must be from {1,2,7,9}

Now look at R3C6

Row 3: all empty except R3C4=4, R3C5=3

So row 3: missing: 1,2,5,6,7,8,9

Column 6: missing: 1,2,3,7,8,9 → but R3C6 can be 1,2,7,9 (from box)

But R3C6 cannot be 3 (already in row 3, col 6 has 3? R2C6=5, R3C5=3 → no, R3C6 is different)

Wait — R3C5=3, so R3C6 ≠ 3

So R3C6 ∈ {1,2,7,9}

Now back to R1C5 and R1C6: must be 2 and 9

Try to see if we can determine which is which.

Look at R2C1 and R2C3

Row 2: 1 9 _ | 6 8 5 | _ _ 2

So missing: 3,4,7

Column 1:
- R1: 7
- R2: 1
- R3: ?
- R4: ?
- R5: ?
- R6: ?
- R7: 1 → R7C1=1
- R8: 5
- R9: ?

So column 1: 7,1,?, ?, ?, ?, 1,5,? → R2C1=1, R7C1=1 → duplicate! 1 appears twice in column 1 → invalid?

Wait — R2C1 = 1, R7C1 = 1 → same column → duplicate

But that can't be.

Check original:

Row 2: `1 9 | 6 8 5 | 2` → R2C1=1

Row 7: `1 3 | 4 | 7 2` → R7C1=1

Yes — both R2C1 and R7C1 = 1 → same column → invalid

But the puzzle is labeled "EASY" — likely a typo?

Wait — perhaps I misread the grid.

Let me recheck the original image description.

You wrote:

```
7 4 | 3 | 1
1 9 | 6 8 5 | 2
| 4 3 |
------+-------+------
5 6 | 3 7 | 1
1 8 | | 9 5
9 | 2 | 6
------+-------+------
1 3 | 4 | 7 2
5 2 | | 8
8 | 1 | 4 7
```

Row 7: `1 3 | 4 | 7 2` → R7C1=1

Row 2: `1 9 | ...` → R2C1=1

Yes — both are 1 in column 1 → impossible

Unless the puzzle has a typo.

But perhaps the second '1' in row 7 is not in C1.

Wait — the format is:

- First group: `1 3` → so R7C1=1, R7C2=?, R7C3=3

Yes.

So R2C1=1, R7C1=1 → conflict.

Similarly, R7C7=7, R9C7=4 — no conflict.

But R2C1 and R7C1 both 1 → error

Unless the puzzle is incorrect.

But let's assume it's a typo, or perhaps I'm misreading.

Maybe the first number in row 7 is not 1.

But you wrote: `1 3 | 4 | 7 2`

Perhaps it's `1` in C1.

But then it's invalid.

Alternatively, maybe the puzzle is:

Row 7: `1 3 | 4 | 7 2` — R7C1=1

Row 2: `1 9 | ...` — R2C1=1

Yes — both in column 1 → impossible.

So either the puzzle is unsolvable, or I have a mistake.

Wait — perhaps the first row is not `7 4 | 3 | 1` — but let's count the spaces.

Maybe the grid is:

Let me try to reconstruct with proper alignment.

Perhaps it's:

```
Row 1: 7 4 _ | 3 _ _ | _ _ 1
Row 2: 1 9 _ | 6 8 5 | _ _ 2
Row 3: _ _ _ | 4 3 _ | _ _ _
Row 4: _ 5 6 | 3 7 _ | _ _ 1
Row 5: _ 1 8 | _ _ _ | 9 5 _
Row 6: _ 9 _ | 2 _ 6 | _ _ _
Row 7: 1 _ 3 | 4 _ _ | 7 2 _
Row 8: 5 _ 2 | _ _ _ | _ _ 8
Row 9: _ 8 _ | _ 1 4 | 7 _ _
```

Then R7C1=1, R2C1=1 → both in column 1 → conflict.

So unless the puzzle has a typo, it's invalid.

But since you said "I uploaded an image", and I can't see it, perhaps there's a formatting issue.

Perhaps the first number in row 7 is not 1.

But you wrote: `1 3 | 4 | 7 2`

Maybe it's ` _ 3 | 4 | 7 2` — but you have `1`.

Alternatively, perhaps row 2 is not `1 9` — but let's assume it's correct.

Given the time, and since this is a known puzzle type, let's assume that the puzzle is valid, and my interpretation is off.

Perhaps the first number in row 7 is not in C1.

But the format suggests it is.

Another possibility: the puzzle has a typo, and R7C1 should be something else.

But without seeing the image, it's hard.

Perhaps you meant:

Row 7: ` _ 3 | 4 | 7 2` — but you wrote `1`.

Given the complexity, and since this is taking too long, and there's a clear conflict in column 1 (two 1s), I suspect a typo.

But let's assume R7C1 is not 1, but let's say it's blank.

But you wrote `1`.

Perhaps the second row starts with ` _ 9` — but you wrote `1 9`.

I think there might be a formatting issue in your text.

To resolve, let's assume the puzzle is as intended, and proceed with the assumption that R2C1=1, R7C1=1 is a mistake, or perhaps R7C1 is not 1.

But since I can't see the image, I can't know.

Given the time, I'll stop here and suggest that the puzzle may have a typo, or you can provide a clearer version.

But for now, based on your text, there is a duplicate 1 in column 1 (R2C1 and R7C1), which makes the puzzle invalid.

Please double-check the image or provide a corrected version.

If you can confirm the correct numbers, I can continue solving.

Otherwise, the puzzle as described has no solution due to duplicate 1 in column 1.