1. Find the HCF of a) 21 and 8 \[ \begin{array}{l} 21- \\ 8- \end{array} \] b) 108 and 270 \[ \begin{array}{l} 108- \\ 270- \end{array} \] 416 and 8 6- 8- d) \( 12-9 \) \( 12- \) \( 4- \)

Sure, let's find the Highest Common Factor (HCF) of each pair of numbers using the **Euclidean Algorithm**. The Euclidean Algorithm is an efficient method for computing the HCF of two numbers based on the principle that the HCF of two numbers also divides their difference. ### a) HCF of 21 and 8 **Step-by-Step Process:** 1. **Divide 21 by 8:** \[ 21 \div 8 = 2 \quad \text{with a remainder of} \quad 21 - (8 \times 2) = 5 \] \[ 21 = 8 \times 2 + 5 \] 2. **Now, divide 8 by the remainder 5:** \[ 8 \div 5 = 1 \quad \text{with a remainder of} \quad 8 - (5 \times 1) = 3 \] \[ 8 = 5 \times 1 + 3 \] 3. **Next, divide 5 by the remainder 3:** \[ 5 \div 3 = 1 \quad \text{with a remainder of} \quad 5 - (3 \times 1) = 2 \] \[ 5 = 3 \times 1 + 2 \] 4. **Then, divide 3 by the remainder 2:** \[ 3 \div 2 = 1 \quad \text{with a remainder of} \quad 3 - (2 \times 1) = 1 \] \[ 3 = 2 \times 1 + 1 \] 5. **Finally, divide 2 by the remainder 1:** \[ 2 \div 1 = 2 \quad \text{with a remainder of} \quad 2 - (1 \times 2) = 0 \] \[ 2 = 1 \times 2 + 0 \] **Conclusion:** When the remainder becomes **0**, the last non-zero remainder is the HCF. \[ \boxed{1} \] --- ### b) HCF of 108 and 270 **Step-by-Step Process:** 1. **Divide 270 by 108:** \[ 270 \div 108 = 2 \quad \text{with a remainder of} \quad 270 - (108 \times 2) = 54 \] \[ 270 = 108 \times 2 + 54 \] 2. **Now, divide 108 by the remainder 54:** \[ 108 \div 54 = 2 \quad \text{with a remainder of} \quad 108 - (54 \times 2) = 0 \] \[ 108 = 54 \times 2 + 0 \] **Conclusion:** When the remainder becomes **0**, the last non-zero remainder is the HCF. \[ \boxed{54} \] --- ### c) HCF of 416 and 8 **Step-by-Step Process:** 1. **Divide 416 by 8:** \[ 416 \div 8 = 52 \quad \text{with a remainder of} \quad 416 - (8 \times 52) = 0 \] \[ 416 = 8 \times 52 + 0 \] **Conclusion:** Since the remainder is already **0**, the divisor at this step is the HCF. \[ \boxed{8} \] --- ### d) HCF of 12 and 9 **Step-by-Step Process:** 1. **Divide 12 by 9:** \[ 12 \div 9 = 1 \quad \text{with a remainder of} \quad 12 - (9 \times 1) = 3 \] \[ 12 = 9 \times 1 + 3 \] 2. **Now, divide 9 by the remainder 3:** \[ 9 \div 3 = 3 \quad \text{with a remainder of} \quad 9 - (3 \times 3) = 0 \] \[ 9 = 3 \times 3 + 0 \] **Conclusion:** When the remainder becomes **0**, the last non-zero remainder is the HCF. \[ \boxed{3} \] --- **Summary of HCFs:** - **a) HCF of 21 and 8:** 1 - **b) HCF of 108 and 270:** 54 - **c) HCF of 416 and 8:** 8 - **d) HCF of 12 and 9:** 3