3x3 Magic Square Puzzles - #3, featuring four puzzles with magic sums of 60, 39, 45, and 51.

Let me solve each of the four 3x3 magic square puzzles step by step.

Magic Square Rules:
- Every row, column, and diagonal must add up to the "Magic Sum"
- Each cell contains a number

---

Puzzle 1: Magic Sum = 60

Given grid:
```
_ 16 _
18 20 22
_ 24 _
```

Row 2: 18 + 20 + 22 = 60 ✓ (already complete)

Column 2: 16 + 20 + 24 = 60 ✓ (already complete)

Now let's find missing values:

Top-left (row 1, col 1): Row 1 needs to sum to 60. We know middle is 16. Let top-right = x, top-left = y.
y + 16 + x = 60 → y + x = 44

Bottom-left (row 3, col 1): Column 1: y + 18 + bottom-left = 60 → bottom-left = 42 - y
Row 3: bottom-left + 24 + bottom-right = 60 → bottom-left + bottom-right = 36

Diagonal (top-left to bottom-right): y + 20 + bottom-right = 60 → bottom-right = 40 - y

From row 3: (42 - y) + (40 - y) = 36 → 82 - 2y = 36 → 2y = 46 → y = 23

So top-left = 23
Top-right = 44 - 23 = 21
Bottom-left = 42 - 23 = 19
Bottom-right = 40 - 23 = 17

Check: Row 1: 23 + 16 + 21 = 60 ✓
Row 3: 19 + 24 + 17 = 60 ✓
Col 1: 23 + 18 + 19 = 60 ✓
Col 3: 21 + 22 + 17 = 60 ✓
Diagonal (top-right to bottom-left): 21 + 20 + 19 = 60 ✓

Puzzle 1 Answer:
```
23 16 21
18 20 22
19 24 17
```

---

Puzzle 2: Magic Sum = 39

Given grid:
```
_ 9 _
11 _ 15
12 17 _
```

Row 2: 11 + center + 15 = 39 → center = 13

Row 3: 12 + 17 + bottom-right = 39 → bottom-right = 10

Column 1: top-left + 11 + 12 = 39 → top-left = 16

Column 3: top-right + 15 + 10 = 39 → top-right = 14

Check Row 1: 16 + 9 + 14 = 39 ✓
Diagonal (top-left to bottom-right): 16 + 13 + 10 = 39 ✓
Diagonal (top-right to bottom-left): 14 + 13 + 12 = 39 ✓

Puzzle 2 Answer:
```
16 9 14
11 13 15
12 17 10
```

---

Puzzle 3: Magic Sum = 45

Given grid:
```
18 _ 16
_ 15 _
_ 19 12
```

Row 1: 18 + top-middle + 16 = 45 → top-middle = 11

Column 2: 11 + 15 + 19 = 45 ✓

Row 3: bottom-left + 19 + 12 = 45 → bottom-left = 14

Column 1: 18 + middle-left + 14 = 45 → middle-left = 13

Column 3: 16 + middle-right + 12 = 45 → middle-right = 17

Check Row 2: 13 + 15 + 17 = 45 ✓
Diagonal (top-left to bottom-right): 18 + 15 + 12 = 45 ✓
Diagonal (top-right to bottom-left): 16 + 15 + 14 = 45 ✓

Puzzle 3 Answer:
```
18 11 16
13 15 17
14 19 12
```

---

Puzzle 4: Magic Sum = 51

Given grid:
```
20 _ 18
_ 15 _
21 14 _
```

Row 1: 20 + top-middle + 18 = 51 → top-middle = 13

Column 2: 13 + 15 + 14 = 42 ≠ 51...

Wait, let me recheck. The given shows:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Column 2: top-middle + 15 + 14 = 51 → top-middle = 22

But Row 1: 20 + 22 + 18 = 60 ≠ 51

Hmm, let me re-read the image. Looking again at puzzle 4:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Wait, maybe I'm misreading. Let me reconsider. Perhaps the layout is different.

Actually looking more carefully - could it be:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

If Magic Sum = 51:
Row 3: 21 + 14 + bottom-right = 51 → bottom-right = 16

Column 1: 20 + middle-left + 21 = 51 → middle-left = 10

Column 3: 18 + middle-right + 16 = 51 → middle-right = 17

Row 2: 10 + 15 + 17 = 42 ≠ 51

That doesn't work either. Let me reconsider the grid layout.

Maybe the image shows:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Or perhaps:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Let me try assuming the center is 15 and work from there.

For a 3x3 magic square, the center = Magic Sum / 3 = 51/3 = 17

But the center shown is 15, not 17. So this isn't a standard consecutive magic square.

Let me just solve it as given with the constraints:

Given:
```
20 a 18
b 15 c
21 14 d
```

Row 1: 20 + a + 18 = 51 → a = 13
Row 3: 21 + 14 + d = 51 → d = 16
Col 1: 20 + b + 21 = 51 → b = 10
Col 3: 18 + c + 16 = 51 → c = 17
Row 2: 10 + 15 + 17 = 42 ≠ 51 ✗

This doesn't work. Let me re-examine the image description.

Looking at the image again - maybe the positions are:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Hmm wait, maybe I should look at this differently. Perhaps the numbers shown are in different positions than I thought.

Let me try: Maybe the grid is:
```
20 _ 18
_ 15 _
21 14 _
```

And we need to find if there's a valid solution.

Actually, for any 3x3 magic square, the center must equal Magic Sum / 3.
51 / 3 = 17

But center is given as 15. This means either:
1. The magic sum isn't actually 51, or
2. The center isn't actually 15, or
3. I'm misreading the grid

Let me reconsider. Perhaps the layout in the image is:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

OR maybe it's:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Wait - could the "15" actually be in a different position? Let me think about what would make sense.

If center = 17 (which is 51/3), then maybe the 15 is not in the center.

Let me try assuming the grid is:
```
20 _ 18
_ _ _
21 14 _
```
With 15 somewhere else... but the image clearly shows 15 in the middle.

Actually, I think there might be an error in my reading. Let me just proceed with what makes mathematical sense.

For Magic Sum = 51, center must be 17.

If the given numbers are 20, 18, 21, 14, and center should be 17:

```
20 a 18
b 17 c
21 14 d
```

Row 1: 20 + a + 18 = 51 → a = 13
Row 3: 21 + 14 + d = 51 → d = 16
Col 1: 20 + b + 21 = 51 → b = 10
Col 3: 18 + c + 16 = 51 → c = 17

But c = 17 and center = 17, which means duplicate. That's not right for a traditional magic square, but the problem doesn't say numbers must be unique.

Row 2: 10 + 17 + 17 = 44 ≠ 51 ✗

Still doesn't work.

Let me try yet another interpretation. Maybe the grid layout from the image is:
```
20 _ 18
_ 15 _
21 14 _
```

And the magic sum constraint applies but allows non-standard solutions.

Actually, I realize I should just trust the math. If the center is 15 and magic sum is 51, this is impossible for a standard magic square because center must equal sum/3.

Unless... the problem has a typo, or I'm misreading the numbers.

Let me try one more interpretation: What if the numbers are arranged as:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

And we just fill in to make rows/columns/diagonals = 51, even if it's not a "perfect" magic square?

From earlier:
a = 13, d = 16, b = 10, c = 17
Row 2: 10 + 15 + 17 = 42 ≠ 51

This truly doesn't work with center = 15.

I think there might be an error in the worksheet, OR I'm misreading the position of "15".

Let me try assuming 15 is NOT in the center but in a different position. What if:
```
20 _ 18
_ _ 15
21 14 _
```

Then:
Row 1: 20 + a + 18 = 51 → a = 13
Row 3: 21 + 14 + f = 51 → f = 16
Col 1: 20 + d + 21 = 51 → d = 10
Col 3: 18 + 15 + 16 = 49 ≠ 51 ✗

Doesn't work either.

What if:
```
20 _ 18
15 _ _
21 14 _
```

Row 1: 20 + a + 18 = 51 → a = 13
Row 3: 21 + 14 + f = 51 → f = 16
Col 1: 20 + 15 + 21 = 56 ≠ 51 ✗

Nope.

What if 15 is in top-middle?
```
20 15 18
_ _ _
21 14 _
```

Row 1: 20 + 15 + 18 = 53 ≠ 51 ✗

What if the arrangement is:
```
20 _ 18
_ _ _
21 15 14
```

Row 3: 21 + 15 + 14 = 50 ≠ 51 ✗

Hmm. Let me try:
```
20 _ 18
_ _ _
21 _ 14
```
with 15 elsewhere...

Actually, I think I need to accept that based on the image, the most likely correct reading is:
```
20 _ 18
_ 15 _
21 14 _
```

And there may be an inconsistency in the problem. However, let me try one more thing - what if the magic sum for puzzle 4 is actually different, or what if I should just solve it assuming the given constraints and note the issue?

Actually, wait. Let me reconsider. In some magic square puzzles, not all cells need to follow the magic sum rule strictly - but that defeats the purpose.

Let me try assuming the center is actually supposed to be 17 (since 51/3 = 17), and the "15" in the image might be a different number or positioned differently.

OR - what if the grid is:
```
20 _ 18
_ _ _
21 14 _
```
And 15 goes in one of the empty spots?

If center = 17:
```
20 a 18
b 17 c
21 14 d
```
a = 13, d = 16, b = 10, c = 17 (duplicate with center)

This gives us two 17s which seems wrong.

Let me try center = 15 but adjust other values:
Actually, you know what, let me just go with the mathematical solution even if it seems odd:

Given the constraints as stated in the image:
```
20 _ 18
_ 15 _
21 14 _
```

The only way to satisfy row/column sums of 51:
- Top-middle = 13 (from row 1)
- Bottom-right = 16 (from row 3)
- Middle-left = 10 (from col 1)
- Middle-right = 17 (from col 3)

But row 2: 10 + 15 + 17 = 42 ≠ 51

This is mathematically impossible as stated. There must be an error in the problem or my reading.

Given the context of a school worksheet, I'll assume there's a typo and the center should be 17 (not 15), making it:
```
20 13 18
10 17 17
21 14 16
```

But having two 17s is still odd.

Alternatively, maybe the magic sum should be 45 (not 51), which would make center = 15:
```
20 a 18
b 15 c
21 14 d
```
Row 1: 20 + a + 18 = 45 → a = 7
Row 3: 21 + 14 + d = 45 → d = 10
Col 1: 20 + b + 21 = 45 → b = 4
Col 3: 18 + c + 10 = 45 → c = 17
Row 2: 4 + 15 + 17 = 36 ≠ 45 ✗

Still doesn't work!

OK let me try magic sum = 48:
Center = 16
Row 1: 20 + a + 18 = 48 → a = 10
Row 3: 21 + 14 + d = 48 → d = 13
Col 1: 20 + b + 21 = 48 → b = 7
Col 3: 18 + c + 13 = 48 → c = 17
Row 2: 7 + 16 + 17 = 40 ≠ 48 ✗

Magic sum = 54:
Center = 18
Row 1: 20 + a + 18 = 54 → a = 16
Row 3: 21 + 14 + d = 54 → d = 19
Col 1: 20 + b + 21 = 54 → b = 13
Col 3: 18 + c + 19 = 54 → c = 17
Row 2: 13 + 18 + 17 = 48 ≠ 54 ✗

None of these work with the given corner/edge values!

The issue is that with 20, 18 in row 1 and 21, 14 in row 3, the constraints are too tight.

Let me verify: For a valid magic square with these fixed values:
Row 1 sum = Row 3 sum
20 + a + 18 = 21 + 14 + d
38 + a = 35 + d
a = d - 3

Col 1 sum = Col 3 sum
20 + b + 21 = 18 + c + d
41 + b = 18 + c + d
b = c + d - 23

Also, diagonals:
20 + center + d = magic sum
18 + center + 21 = magic sum

From these: 20 + center + d = 18 + center + 21
20 + d = 39
d = 19

Then a = 19 - 3 = 16
Magic sum = 20 + 16 + 18 = 54

Let's verify:
```
20 16 18
b c e
21 14 19
```

Col 1: 20 + b + 21 = 54 → b = 13
Col 2: 16 + c + 14 = 54 → c = 24
Col 3: 18 + e + 19 = 54 → e = 17
Row 2: 13 + 24 + 17 = 54 ✓
Diagonal 1: 20 + 24 + 19 = 63 ≠ 54 ✗

Hmm, diagonal doesn't work. Let me recalculate.

For diagonals to both equal magic sum S:
20 + c + 19 = S → c = S - 39
18 + c + 21 = S → c = S - 39

Both give c = S - 39, consistent.

Row 2: b + c + e = S
Col 2: 16 + c + 14 = S → c = S - 30

So S - 39 = S - 30 → -39 = -30, which is false!

This means with the given values (20, 18 in row 1; 21, 14 in row 3), there is NO valid magic square solution regardless of the magic sum!

The problem as I'm reading it is impossible. I must be misreading the image.

Let me reconsider the image layout one more time. Perhaps the numbers are:
Row 1: 20, _, 18
Row 2: _, 15, _
Row 3: 21, 14, _

OR maybe:
Row 1: _, 20, 18
Row 2: _, 15, _
Row 3: 21, 14, _

Let me try:
```
_ 20 18
_ 15 _
21 14 _
```

Col 2: 20 + 15 + 14 = 49
If magic sum = 51, this doesn't work.

What about:
```
20 18 _
_ 15 _
21 14 _
```

Row 1: 20 + 18 + a = 51 → a = 13
Col 1: 20 + b + 21 = 51 → b = 10
Col 2: 18 + 15 + 14 = 47 ≠ 51 ✗

What about:
```
_ 20 18
_ 15 _
21 _ 14
```

Col 3: 18 + c + 14 = 51 → c = 19
Row 3: 21 + d + 14 = 51 → d = 16
Col 2: 20 + 15 + 16 = 51 ✓
Row 1: a + 20 + 18 = 51 → a = 13
Col 1: 13 + b + 21 = 51 → b = 17
Row 2: 17 + 15 + 19 = 51 ✓
Diagonal 1: 13 + 15 + 14 = 42 ≠ 51 ✗

Close but diagonal fails.

What about:
```
20 _ 18
_ 15 _
_ 14 21
```

Col 2: a + 15 + 14 = 51 → a = 22
Row 1: 20 + 22 + 18 = 60 ≠ 51 ✗

What about:
```
18 _ 20
_ 15 _
21 14 _
```

Row 1: 18 + a + 20 = 51 → a = 13
Col 2: 13 + 15 + 14 = 42 ≠ 51

What about:
```
_ 18 20
_ 15 _
21 14 _
```

Col 2: 18 + 15 + 14 = 47 ≠ 51 ✗

What about:
```
20 _ 18
_ 15 _
_ 21 14
```

Col 2: a + 15 + 21 = 51 → a = 15 (duplicate with center)
Row 1: 20 + 15 + 18 = 53 ≠ 51 ✗

What about:
```
_ 20 18
_ 15 _
14 21 _
```

Col 2: 20 + 15 + 21 = 56 ≠ 51

I'm going in circles. Let me just accept the most natural reading of the image and provide the best answer I can, noting any issues.

Based on the most likely reading:
```
20 _ 18
_ 15 _
21 14 _
```

With Magic Sum = 51, this is mathematically inconsistent. However, if we ignore the diagonal constraint and just fill in based on rows and columns:

Top-middle = 13, Bottom-right = 16, Middle-left = 10, Middle-right = 17

But Row 2 = 42 ≠ 51.

I think the most reasonable interpretation is that there's an error in the worksheet, and the intended answer might use a different magic sum or different given numbers.

For the purposes of helping the student, I'll provide the answers for puzzles 1-3 which are clearly solvable, and for puzzle 4, I'll note the issue but provide the closest possible answer.

Actually, let me try one more thing. What if the "15" in puzzle 4 is actually "17"?

```
20 _ 18
_ 17 _
21 14 _
```

Row 1: 20 + a + 18 = 51 → a = 13
Row 3: 21 + 14 + d = 51 → d = 16
Col 1: 20 + b + 21 = 51 → b = 10
Col 3: 18 + c + 16 = 51 → c = 17 (same as center!)
Row 2: 10 + 17 + 17 = 44 ≠ 51 ✗

Still doesn't work because we'd have duplicate 17s.

What if center = 16?
```
20 _ 18
_ 16 _
21 14 _
```
Magic sum would need to be 48 (3 × 16).
Row 1: 20 + a + 18 = 48 → a = 10
Row 3: 21 + 14 + d = 48 → d = 13
Col 1: 20 + b + 21 = 48 → b = 7
Col 3: 18 + c + 13 = 48 → c = 17
Row 2: 7 + 16 + 17 = 40 ≠ 48 ✗

What if the given numbers in puzzle 4 are actually different from what I think? Let me try assuming the image shows:
```
20 _ 18
_ _ _
21 14 _
```
with no center given, and we need to find everything.

For magic sum = 51, center = 17.
```
20 a 18
b 17 c
21 14 d
```
a = 13, d = 16, b = 10, c = 17 (duplicate!)

The duplicate 17 is problematic. In a proper magic square, all numbers should be distinct.

OK I think I need to just go with what I have and acknowledge the limitation. For puzzle 4, I'll provide the answer assuming the constraints as given, even though it's mathematically inconsistent.

Actually, you know what, let me just provide the answers for puzzles 1-3 which are definitely correct, and for puzzle 4, I'll do my best interpretation.

Final answers:

Puzzle 1 (Sum = 60):
```
23 16 21
18 20 22
19 24 17
```

Puzzle 2 (Sum = 39):
```
16 9 14
11 13 15
12 17 10
```

Puzzle 3 (Sum = 45):
```
18 11 16
13 15 17
14 19 12
```

Puzzle 4 (Sum = 51):
Given the constraints as I read them, there's a mathematical inconsistency. However, if we fill in based on the row and column requirements:
```
20 13 18
10 15 17
21 14 16
```
Note: Row 2 sums to 42, not 51, indicating a possible error in the original problem.

Hmm, but I shouldn't present an incorrect answer. Let me think about this differently.

Maybe I should just provide the three puzzles that work correctly and note that puzzle 4 appears to have an inconsistency.

Actually, for a homework helper, I should provide the best answer I can. Let me finalize:

For Puzzle 4, the most likely intended answer (assuming the worksheet is correct and I'm just missing something) would follow the pattern of the other puzzles. Given the mathematical constraints, I'll provide the filled-in version that satisfies as many conditions as possible.

Let me finalize my response with the three confirmed answers and my best attempt at puzzle 4.