1) find the haf and LCM berween a) \( 48,56,72 \) b) 198,360

Sure, let's find the **HCF (Highest Common Factor)** and **LCM (Least Common Multiple)** for each set of numbers provided. --- ### **a) For the numbers 48, 56, and 72** #### **Prime Factorization:** 1. **48** - \(48 = 2^4 \times 3^1\) 2. **56** - \(56 = 2^3 \times 7^1\) 3. **72** - \(72 = 2^3 \times 3^2\) #### **Finding the HCF:** - **Identify common prime factors** across all three numbers. - The only common prime factor is **2**. - **Minimum power of 2** among the numbers is \(2^3\). \[ \text{HCF} = 2^3 = 8 \] #### **Finding the LCM:** - **Take the highest power** of all prime factors present. - Prime factors involved: 2, 3, and 7. \[ \begin{align*} \text{LCM} &= 2^{4} \times 3^{2} \times 7^{1} \\ &= 16 \times 9 \times 7 \\ &= 144 \times 7 \\ &= 1008 \end{align*} \] **_Result for part a:_** - **HCF = 8** - **LCM = 1008** --- ### **b) For the numbers 198 and 360** #### **Prime Factorization:** 1. **198** - \(198 = 2^1 \times 3^2 \times 11^1\) 2. **360** - \(360 = 2^3 \times 3^2 \times 5^1\) #### **Finding the HCF:** - **Identify common prime factors** between the two numbers. - Common primes are **2** and **3**. - **Minimum powers**: - \(2^1\) (from 198) - \(3^2\) (common) \[ \text{HCF} = 2^1 \times 3^2 = 2 \times 9 = 18 \] #### **Finding the LCM:** - **Take the highest power** of all prime factors present. - Prime factors involved: 2, 3, 5, and 11. \[ \begin{align*} \text{LCM} &= 2^{3} \times 3^{2} \times 5^{1} \times 11^{1} \\ &= 8 \times 9 \times 5 \times 11 \\ &= 72 \times 5 \times 11 \\ &= 360 \times 11 \\ &= 3960 \end{align*} \] **_Result for part b:_** - **HCF = 18** - **LCM = 3960** --- **Summary:** - **a) 48, 56, 72** - **HCF:** 8 - **LCM:** 1008 - **b) 198, 360** - **HCF:** 18 - **LCM:** 3960