EXERCISE 1.6 Use prime factorisation to determine the HCF of the following numbers: \( \begin{array}{llll}\text { (1) } 8 \text { and } 16 & \text { (2) } 9 \text { and } 27 & \text { (3) } 18 \text { and } 42 \\ \text { (4) } 54 \text { and } 98 & \text { (5) } 90 \text { and } 75 & \text { (6) } 112 \text { and } 160 \\ \text { (7) } 84 \text { and } 126 & \text { (8) } 300 \text { and } 250 & \text { (9) } 396 \text { and } 693 \\ \text { (10)* } 2250,7560 \text { and } 10500 & \end{array} \)

To find the highest common factor (HCF) using prime factorization, we need to break down each number into its prime factors, then identify the common factors among them. Let’s see how this works out for a couple of pairs: 1. **For 8 and 16**: - 8 = \(2^3\) - 16 = \(2^4\) - The common prime factor is \(2\), and the lowest power is \(2^3\), so the HCF is **8**. 2. **For 9 and 27**: - 9 = \(3^2\) - 27 = \(3^3\) - The common prime factor is \(3\), and the lowest power is \(3^2\), so the HCF is **9**. Now, let’s solve the HCF for all the pairs as per your list: **(3) For 18 and 42**: - 18 = \(2 \times 3^2\) - 42 = \(2 \times 3 \times 7\) - The common prime factors are \(2\) and \(3\), so the HCF is \(2^1 \times 3^1 = **6**\). **(4) For 54 and 98**: - 54 = \(2 \times 3^3\) - 98 = \(2 \times 7^2\) - The HCF is \(2^1 = **2**\). **(5) For 90 and 75**: - 90 = \(2 \times 3^2 \times 5\) - 75 = \(3 \times 5^2\) - The common prime factors are \(3\) and \(5\), so the HCF is \(3^1 \times 5^1 = **15**\). **(6) For 112 and 160**: - 112 = \(2^4 \times 7\) - 160 = \(2^5 \times 5\) - The common prime factor is \(2\), so the HCF is \(2^4 = **16**\). **(7) For 84 and 126**: - 84 = \(2^2 \times 3 \times 7\) - 126 = \(2 \times 3^2 \times 7\) - The HCF is \(2^1 \times 3^1 \times 7^1 = **42**\). **(8) For 300 and 250**: - 300 = \(2^2 \times 3 \times 5^2\) - 250 = \(2 \times 5^3\) - The HCF is \(2^1 \times 5^2 = **50**\). **(9) For 396 and 693**: - 396 = \(2^2 \times 3^2 \times 11\) - 693 = \(3 \times 7 \times 11\) - The common prime factors are \(3\) and \(11\), so the HCF is \(3^1 \times 11^1 = **33**\). **(10) For 2250, 7560 and 10500**: - 2250 = \(2 \times 3^2 \times 5^2\) - 7560 = \(2^3 \times 3^3 \times 5 \times 7\) - 10500 = \(2^2 \times 3 \times 5^2 \times 7\) - The common prime factors are \(2^1\), \(3^1\), and \(5^2\), resulting in the HCF of \(2^1 \times 3^1 \times 5^2 = **450**\). Happy calculating!