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)

Let's solve each part of your problem step by step. ### 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:** \( 1848 \)