Negative Numbers - Free Printable

Problem Analysis

The task consists of two parts:
1. Warm-up Exercises: These involve basic arithmetic operations with negative numbers, including multiplication and division.
2. Multiplication Arithmagon Puzzles: These puzzles require filling in missing numbers in a triangular structure where the product of the numbers in the circles equals the number in the square between them.

Let's solve each part step by step.

---

Part 1: Warm-up Exercises

We need to compute the results of the given arithmetic operations involving negative numbers.

#### a) \( 4 \times -7 \)
- Multiplying a positive number by a negative number results in a negative number.
- \( 4 \times -7 = -28 \)

#### b) \( -3 \times -1 \)
- Multiplying two negative numbers results in a positive number.
- \( -3 \times -1 = 3 \)

#### c) \( -20 \div 4 \)
- Dividing a negative number by a positive number results in a negative number.
- \( -20 \div 4 = -5 \)

#### d) \( 15 \div -5 \)
- Dividing a positive number by a negative number results in a negative number.
- \( 15 \div -5 = -3 \)

#### e) \( -9 \times -9 \)
- Multiplying two negative numbers results in a positive number.
- \( -9 \times -9 = 81 \)

#### f) \( 42 \div -6 \)
- Dividing a positive number by a negative number results in a negative number.
- \( 42 \div -6 = -7 \)

#### g) \( -12 \times 4 \)
- Multiplying a negative number by a positive number results in a negative number.
- \( -12 \times 4 = -48 \)

#### h) \( -4 \times 0 \)
- Any number multiplied by zero is zero.
- \( -4 \times 0 = 0 \)

#### i) \( -8 \div -4 \)
- Dividing two negative numbers results in a positive number.
- \( -8 \div -4 = 2 \)

Final Answers for Warm-up Exercises

\[
\begin{aligned}
&\text{a) } -28, \quad \text{b) } 3, \quad \text{c) } -5, \\
&\text{d) } -3, \quad \text{e) } 81, \quad \text{f) } -7, \\
&\text{g) } -48, \quad \text{h) } 0, \quad \text{i) } 2.
\end{aligned}
\]

---

Part 2: Multiplication Arithmagon Puzzles

In these puzzles, the numbers in the circles multiply to give the numbers in the squares between them. We need to fill in the missing numbers.

#### a)
```
-3
/ \
? ?
/ \
6 -2
```
- The top circle is \(-3\).
- The left square is \(6\), so the product of \(-3\) and the left circle is \(6\):
\[
-3 \times \text{(left circle)} = 6 \implies \text{(left circle)} = 6 \div -3 = -2
\]
- The right square is \(-2\), so the product of \(-3\) and the right circle is \(-2\):
\[
-3 \times \text{(right circle)} = -2 \implies \text{(right circle)} = -2 \div -3 = \frac{2}{3}
\]

#### b)
```
8
/ \
? -40
/ \
-4 ?
```
- The top circle is \(8\).
- The left square is \(-4\), so the product of \(8\) and the left circle is \(-4\):
\[
8 \times \text{(left circle)} = -4 \implies \text{(left circle)} = -4 \div 8 = -\frac{1}{2}
\]
- The right square is \(-40\), so the product of \(8\) and the right circle is \(-40\):
\[
8 \times \text{(right circle)} = -40 \implies \text{(right circle)} = -40 \div 8 = -5
\]

#### c)
```
?
/ \
48 -42
/ \
? -56
```
- The bottom left square is \(48\), and the bottom right square is \(-42\).
- Let the top circle be \(x\), the left circle be \(y\), and the right circle be \(z\).
- From the bottom left square:
\[
y \times x = 48
\]
- From the bottom right square:
\[
z \times x = -42
\]
- From the bottom edge:
\[
y \times z = -56
\]

We solve these equations step by step:
1. From \(y \times x = 48\), we have \(y = \frac{48}{x}\).
2. From \(z \times x = -42\), we have \(z = \frac{-42}{x}\).
3. Substitute \(y\) and \(z\) into \(y \times z = -56\):
\[
\left(\frac{48}{x}\right) \times \left(\frac{-42}{x}\right) = -56
\]
\[
\frac{48 \times -42}{x^2} = -56
\]
\[
\frac{-2016}{x^2} = -56
\]
\[
x^2 = \frac{2016}{56} = 36
\]
\[
x = \pm 6
\]

- If \(x = 6\):
\[
y = \frac{48}{6} = 8, \quad z = \frac{-42}{6} = -7
\]

- If \(x = -6\):
\[
y = \frac{48}{-6} = -8, \quad z = \frac{-42}{-6} = 7
\]

Both solutions are valid, but typically, the positive value is preferred unless specified otherwise. Thus, we use \(x = 6\).

Final Answers for Multiplication Arithmagon Puzzles

\[
\begin{aligned}
&\text{a) Left circle: } -2, \text{ Right circle: } \frac{2}{3}, \\
&\text{b) Left circle: } -\frac{1}{2}, \text{ Right circle: } -5, \\
&\text{c) Top circle: } 6, \text{ Left circle: } 8, \text{ Right circle: } -7.
\end{aligned}
\]

---

Final Answer

\[
\boxed{
\begin{aligned}
&\text{Warm-up: } -28, 3, -5, -3, 81, -7, -48, 0, 2, \\
&\text{Arithmagon: }
\begin{aligned}
&\text{a) } -2, \frac{2}{3}, \\
&\text{b) } -\frac{1}{2}, -5, \\
&\text{c) } 6, 8, -7.
\end{aligned}
\end{aligned}
}
\]