1.2. Write down the additive inverse of 16 - 1.3. determine the sum of the prime factors of 180 1.4. Use prime factorisation to find the HCF and LCM of $$ 1848 ; 132 $$ and 462 QUESTION 2 (SOLVING PROBLEMS)

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### 1.2. Additive Inverse of 16
The additive inverse of a number is the value that, when added to the original number, results in zero. For any number $$ a $$, the additive inverse is $$ -a $$.

For $$ 16 $$:
$$
\text{Additive Inverse} = -16
$$

### 1.3. Sum of the Prime Factors of 180
First, we need to find the prime factorization of $$ 180 $$.

1. Start dividing by the smallest prime number, which is $$ 2 $$:
   $$
   180 \div 2 = 90
   $$
2. Divide $$ 90 $$ by $$ 2 $$:
   $$
   90 \div 2 = 45
   $$
3. Now, divide $$ 45 $$ by the next smallest prime number, which is $$ 3 $$:
   $$
   45 \div 3 = 15
   $$
4. Divide $$ 15 $$ by $$ 3 $$:
   $$
   15 \div 3 = 5
   $$
5. Finally, $$ 5 $$ is a prime number.

So, the prime factorization of $$ 180 $$ is:
$$
180 = 2^2 \times 3^2 \times 5^1
$$

The prime factors are $$ 2, 3, $$ and $$ 5 $$. Now, we find the sum of these prime factors:
$$
\text{Sum} = 2 + 3 + 5 = 10
$$

### 1.4. HCF and LCM of $$ 1848, 132, $$ and $$ 462 $$ using Prime Factorization
First, we will find the prime factorization of each number.

**Prime Factorization of 1848:**
1. Divide by $$ 2 $$:
   $$
   1848 \div 2 = 924
   $$
2. Divide $$ 924 $$ by $$ 2 $$:
   $$
   924 \div 2 = 462
   $$
3. Divide $$ 462 $$ by $$ 2 $$:
   $$
   462 \div 2 = 231
   $$
4. Divide $$ 231 $$ by $$ 3 $$:
   $$
   231 \div 3 = 77
   $$
5. Divide $$ 77 $$ by $$ 7 $$:
   $$
   77 \div 7 = 11
   $$
6. Finally, $$ 11 $$ is a prime number.

So, the prime factorization of $$ 1848 $$ is:
$$
1848 = 2^3 \times 3^1 \times 7^1 \times 11^1
$$

**Prime Factorization of 132:**
1. Divide by $$ 2 $$:
   $$
   132 \div 2 = 66
   $$
2. Divide $$ 66 $$ by $$ 2 $$:
   $$
   66 \div 2 = 33
   $$
3. Divide $$ 33 $$ by $$ 3 $$:
   $$
   33 \div 3 = 11
   $$
4. Finally, $$ 11 $$ is a prime number.

So, the prime factorization of $$ 132 $$ is:
$$
132 = 2^2 \times 3^1 \times 11^1
$$

**Prime Factorization of 462:**
1. Divide by $$ 2 $$:
   $$
   462 \div 2 = 231
   $$
2. Divide $$ 231 $$ by $$ 3 $$:
   $$
   231 \div 3 = 77
   $$
3. Divide $$ 77 $$ by $$ 7 $$:
   $$
   77 \div 7 = 11
   $$
4. Finally, $$ 11 $$ is a prime number.

So, the prime factorization of $$ 462 $$ is:
$$
462 = 2^1 \times 3^1 \times 7^1 \times 11^1
$$

### Finding HCF and LCM
**HCF (Highest Common Factor):**
To find the HCF, we take the lowest power of all common prime factors:
- For $$ 2 $$: $$ \min(3, 2, 1) = 1 $$
- For $$ 3 $$: $$ \min(1, 1, 1) = 1 $$
- For $$ 7 $$: $$ \min(1, 0, 1) = 0 $$ (not common)
- For $$ 11 $$: $$ \min(1, 1, 1) = 1 $$

Thus, the HCF is:
$$
\text{HCF} = 2^1 \times 3^1 \times 11^1 = 66
$$

**LCM (Lowest Common Multiple):**
To find the LCM, we take the highest power of all prime factors:
- For $$ 2 $$: $$ \max(3, 2, 1) = 3 $$
- For $$ 3 $$: $$ \max(1, 1, 1) = 1 $$
- For $$ 7 $$: $$ \max(1, 0, 1) = 1 $$
- For $$ 11 $$: $$ \max(1, 1, 1) = 1 $$

Thus, the LCM is:
$$
\text{LCM} = 2^3 \times 3^1 \times 7^1 \times 11^1
$$

Calculating the LCM:
$$
\text{LCM} = 8 \times 3 \times 7 \times 11
$$

Let's calculate this step by step:
1. $$ 8 \times 3 = 24 $$
2. $$ 24 \times 7 = 168 $$
3. $$ 168 \times 11 = 1848 $$

So, the LCM is:
$$
\text{LCM} = 1848
$$

### Summary of Results
- **Additive Inverse of 16:** $$ -16 $$
- **Sum of Prime Factors of 180:** $$ 10 $$
- **HCF of 1848, 132, and 462:** $$ 66 $$
- **LCM of 1848, 132, and 462:** $$ 1848 $$
- **1.2.** The additive inverse of 16 is -16. - **1.3.** The sum of the prime factors of 180 is 10. - **1.4.** The HCF of 1848, 132, and 462 is 66, and the LCM is 1848.

1.2. Write down the additive inverse of 16 - 1.3. determine the sum of the prime factors of 180 1.4. Use prime factorisation to find the HCF and LCM of \( 1848 ; 132 \) and 462 QUESTION 2 (SOLVING PROBLEMS)

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