Magic Square Math Worksheet with Two Puzzles for Practicing Addition and Logical Thinking

Problem Overview:

We are tasked with solving two magic square problems. A magic square is a square grid filled with distinct numbers such that the sum of the numbers in each row, each column, and both main diagonals is the same.

#### Part 1:
- Use the numbers 1, 2, 3, 4, 6, 7, 8, 9.
- Each line (row, column, diagonal) must add up to 15.
- The partially filled grid is given as:
```
_ | _ | _
9 | 5 | _
_ | _ | _
```

#### Part 2:
- Use the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8.
- Each line (row, column, diagonal) must add up to 12.
- The partially filled grid is given as:
```
_ | 8 | _
_ | _ | _
5 | _ | _
```

---

Solution for Part 1:

#### Step 1: Understand the properties of a 3x3 magic square
For a 3x3 magic square:
- The sum of all numbers from 1 to 9 is \(1 + 2 + 3 + 4 + 6 + 7 + 8 + 9 = 45\).
- Since there are 3 rows, the sum of each row (and thus each column and diagonal) is \( \frac{45}{3} = 15 \).

#### Step 2: Analyze the given grid
The grid is:
```
_ | _ | _
9 | 5 | _
_ | _ | _
```
We know:
- The center number is 5.
- The bottom-left number is 9.

#### Step 3: Fill in the missing numbers
To solve this, we use the property that the sum of each row, column, and diagonal must be 15.

1. First Row:
- Let the first row be \(a, b, c\). Then \(a + b + c = 15\).

2. Second Row:
- The second row is \(9, 5, d\). So, \(9 + 5 + d = 15 \implies d = 1\).

3. Third Row:
- Let the third row be \(e, f, g\). Then \(e + f + g = 15\).

4. First Column:
- The first column is \(a, 9, e\). So, \(a + 9 + e = 15 \implies a + e = 6\).

5. Second Column:
- The second column is \(b, 5, f\). So, \(b + 5 + f = 15 \implies b + f = 10\).

6. Third Column:
- The third column is \(c, 1, g\). So, \(c + 1 + g = 15 \implies c + g = 14\).

7. Main Diagonal (Top-Left to Bottom-Right):
- The diagonal is \(a, 5, g\). So, \(a + 5 + g = 15 \implies a + g = 10\).

8. Other Diagonal (Top-Right to Bottom-Left):
- The diagonal is \(c, 5, e\). So, \(c + 5 + e = 15 \implies c + e = 10\).

#### Step 4: Solve the system of equations
From the equations:
- \(a + e = 6\)
- \(a + g = 10\)
- \(c + e = 10\)
- \(c + g = 14\)

We can solve these step by step:
- From \(a + g = 10\) and \(a + e = 6\), we get \(g - e = 4\).
- From \(c + g = 14\) and \(c + e = 10\), we get \(g - e = 4\).

This confirms consistency. Solving further:
- Assume \(a = 2\), then \(e = 4\) (from \(a + e = 6\)).
- Then \(g = 8\) (from \(a + g = 10\)).
- Then \(c = 6\) (from \(c + g = 14\)).

Now, fill in the remaining values:
- \(b + f = 10\). Since \(b\) and \(f\) must be distinct and fit the remaining numbers, let \(b = 7\) and \(f = 3\).

#### Final Grid for Part 1:
```
2 | 7 | 6
9 | 5 | 1
4 | 3 | 8
```

Verify:
- Rows: \(2 + 7 + 6 = 15\), \(9 + 5 + 1 = 15\), \(4 + 3 + 8 = 15\).
- Columns: \(2 + 9 + 4 = 15\), \(7 + 5 + 3 = 15\), \(6 + 1 + 8 = 15\).
- Diagonals: \(2 + 5 + 8 = 15\), \(6 + 5 + 4 = 15\).

All conditions are satisfied.

---

Solution for Part 2:

#### Step 1: Understand the properties of a 3x3 magic square
For a 3x3 magic square:
- The sum of all numbers from 0 to 8 is \(0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36\).
- Since there are 3 rows, the sum of each row (and thus each column and diagonal) is \( \frac{36}{3} = 12 \).

#### Step 2: Analyze the given grid
The grid is:
```
_ | 8 | _
_ | _ | _
5 | _ | _
```
We know:
- The center number is 8.
- The bottom-left number is 5.

#### Step 3: Fill in the missing numbers
To solve this, we use the property that the sum of each row, column, and diagonal must be 12.

1. First Row:
- Let the first row be \(a, 8, b\). Then \(a + 8 + b = 12 \implies a + b = 4\).

2. Second Row:
- Let the second row be \(c, d, e\). Then \(c + d + e = 12\).

3. Third Row:
- The third row is \(5, f, g\). So, \(5 + f + g = 12 \implies f + g = 7\).

4. First Column:
- The first column is \(a, c, 5\). So, \(a + c + 5 = 12 \implies a + c = 7\).

5. Second Column:
- The second column is \(8, d, f\). So, \(8 + d + f = 12 \implies d + f = 4\).

6. Third Column:
- The third column is \(b, e, g\). So, \(b + e + g = 12\).

7. Main Diagonal (Top-Left to Bottom-Right):
- The diagonal is \(a, d, g\). So, \(a + d + g = 12\).

8. Other Diagonal (Top-Right to Bottom-Left):
- The diagonal is \(b, d, 5\). So, \(b + d + 5 = 12 \implies b + d = 7\).

#### Step 4: Solve the system of equations
From the equations:
- \(a + b = 4\)
- \(a + c = 7\)
- \(d + f = 4\)
- \(b + d = 7\)
- \(f + g = 7\)

We can solve these step by step:
- Assume \(a = 3\), then \(b = 1\) (from \(a + b = 4\)).
- Then \(c = 4\) (from \(a + c = 7\)).
- Then \(d = 6\) (from \(b + d = 7\)).
- Then \(f = -2\) (from \(d + f = 4\)), but this is not valid since we need non-negative integers.

Re-evaluate:
- Assume \(a = 1\), then \(b = 3\) (from \(a + b = 4\)).
- Then \(c = 6\) (from \(a + c = 7\)).
- Then \(d = 4\) (from \(b + d = 7\)).
- Then \(f = 0\) (from \(d + f = 4\)).
- Then \(g = 7\) (from \(f + g = 7\)).

#### Final Grid for Part 2:
```
1 | 8 | 3
6 | 4 | 2
5 | 0 | 7
```

Verify:
- Rows: \(1 + 8 + 3 = 12\), \(6 + 4 + 2 = 12\), \(5 + 0 + 7 = 12\).
- Columns: \(1 + 6 + 5 = 12\), \(8 + 4 + 0 = 12\), \(3 + 2 + 7 = 12\).
- Diagonals: \(1 + 4 + 7 = 12\), \(3 + 4 + 5 = 12\).

All conditions are satisfied.

---

Final Answers:

1. For Part 1:
\[
\boxed{
\begin{array}{|c|c|c|}
\hline
2 & 7 & 6 \\
\hline
9 & 5 & 1 \\
\hline
4 & 3 & 8 \\
\hline
\end{array}
}
\]

2. For Part 2:
\[
\boxed{
\begin{array}{|c|c|c|}
\hline
1 & 8 & 3 \\
\hline
6 & 4 & 2 \\
\hline
5 & 0 & 7 \\
\hline
\end{array}
}
\]