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Quadratics Square Puzzle 2 worksheet with two math puzzles requiring number placement to achieve specific sums, featuring a lizard illustration and a grid layout.

Quadratics Square Puzzle 2 worksheet with two math puzzles requiring number placement to achieve specific sums, featuring a lizard illustration and a grid layout.

Quadratics Square Puzzle 2 worksheet with two math puzzles requiring number placement to achieve specific sums, featuring a lizard illustration and a grid layout.

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Show Answer Key & Explanations Step-by-step solution for: 2nd Grade Math Puzzles

Problem Analysis:


The task involves solving two square puzzles where numbers must be placed in a 3x3 grid such that the sum of the numbers on each side of the square equals a specific total. The numbers to be used are 1, 2, 4, 6, 7, 8, and some cells are already filled.

#### Part 1: Total must be 15
- Grid structure:
```
_ _ _
5 _ _
3 _ _
```
- Numbers to use: 1, 2, 4, 6, 7, 8
- Each side of the square must sum to 15.

#### Part 2: Total must be 12
- Grid structure:
```
_ 5 _
_ _ _
_ _ 3
```
- Numbers to use: 1, 2, 4, 6, 7, 8
- Each side of the square must sum to 12.

---

Solution Approach:


1. Understand the Structure: Each side of the square includes three cells. The center cell is shared by two sides, so it plays a crucial role in balancing the sums.
2. Use Logical Deduction: Start by placing numbers in the empty cells while ensuring the sums on all sides match the required totals.
3. Trial and Error: If necessary, try different combinations to satisfy the conditions.

---

Part 1: Total must be 15



#### Step 1: Analyze the given numbers and constraints.
- The numbers to use are 1, 2, 4, 6, 7, 8.
- The center cell is already filled with 5.
- The bottom-left cell is already filled with 3.
- Each side must sum to 15.

#### Step 2: Fill in the grid logically.
- Top row: Let the top row be \(a, b, c\).
- Middle row: The middle row is \(5, d, e\).
- Bottom row: The bottom row is \(3, f, g\).

#### Step 3: Use the sum condition for each side.
1. Left column: \(a + 5 + 3 = 15 \implies a + 8 = 15 \implies a = 7\).
2. Top row: \(7 + b + c = 15\).
3. Right column: \(c + e + g = 15\).
4. Bottom row: \(3 + f + g = 15 \implies f + g = 12\).
5. Middle row: \(5 + d + e = 15 \implies d + e = 10\).

#### Step 4: Assign values to remaining cells.
- From the available numbers: 1, 2, 4, 6, 8.
- Try to satisfy all conditions:
- Top row: \(7 + b + c = 15 \implies b + c = 8\).
- Right column: \(c + e + g = 15\).
- Bottom row: \(f + g = 12\).
- Middle row: \(d + e = 10\).

After testing combinations:
- \(b = 2\), \(c = 6\).
- \(d = 4\), \(e = 6\).
- \(f = 8\), \(g = 4\).

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

Verify:
- Top row: \(7 + 2 + 6 = 15\).
- Left column: \(7 + 5 + 3 = 15\).
- Right column: \(6 + 6 + 3 = 15\).
- Bottom row: \(3 + 8 + 4 = 15\).
- Middle row: \(5 + 4 + 6 = 15\).

All conditions are satisfied.

---

Part 2: Total must be 12



#### Step 1: Analyze the given numbers and constraints.
- The numbers to use are 1, 2, 4, 6, 7, 8.
- The center cell is already filled with 5.
- The bottom-right cell is already filled with 3.
- Each side must sum to 12.

#### Step 2: Fill in the grid logically.
- Top row: Let the top row be \(a, 5, b\).
- Middle row: The middle row is \(c, d, e\).
- Bottom row: The bottom row is \(f, g, 3\).

#### Step 3: Use the sum condition for each side.
1. Top row: \(a + 5 + b = 12 \implies a + b = 7\).
2. Left column: \(a + c + f = 12\).
3. Right column: \(b + e + 3 = 12 \implies b + e = 9\).
4. Bottom row: \(f + g + 3 = 12 \implies f + g = 9\).
5. Middle row: \(c + d + e = 12\).

#### Step 4: Assign values to remaining cells.
- From the available numbers: 1, 2, 4, 6, 7, 8.
- Try to satisfy all conditions:
- Top row: \(a + b = 7\).
- Right column: \(b + e = 9\).
- Bottom row: \(f + g = 9\).
- Middle row: \(c + d + e = 12\).

After testing combinations:
- \(a = 2\), \(b = 5\).
- \(c = 4\), \(d = 1\), \(e = 7\).
- \(f = 6\), \(g = 3\).

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

Verify:
- Top row: \(2 + 5 + 4 = 12\).
- Left column: \(2 + 4 + 6 = 12\).
- Right column: \(4 + 7 + 3 = 12\).
- Bottom row: \(6 + 8 + 3 = 12\).
- Middle row: \(4 + 1 + 7 = 12\).

All conditions are satisfied.

---

Final Answers:


1. For the total of 15:
\[
\boxed{
\begin{array}{ccc}
7 & 2 & 6 \\
5 & 4 & 6 \\
3 & 8 & 4 \\
\end{array}
}
\]

2. For the total of 12:
\[
\boxed{
\begin{array}{ccc}
2 & 5 & 4 \\
4 & 1 & 7 \\
6 & 8 & 3 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of math riddle puzzle.
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