Write each number as the product of prime factors using LADDER method AND determine the LCM and HCF of each set of numbers \( 54: 36: 60 \)

1. **Ladder Method Factorization** Write the numbers \( 54 \), \( 36 \), and \( 60 \) in a row and divide by common prime factors step‐by‐step. - **Step 1:** Divide by \(2\) (since all are even). \[ \begin{array}{rcccc} 2 & | & 54 & 36 & 60 \\ & & \frac{54}{2}=27 & \frac{36}{2}=18 & \frac{60}{2}=30 \\ \end{array} \] Now, the numbers are: \(27\), \(18\), \(30\). - **Step 2:** Divide by \(3\) (all numbers are divisible by \(3\)). \[ \begin{array}{rcccc} 3 & | & 27 & 18 & 30 \\ & & \frac{27}{3}=9 & \frac{18}{3}=6 & \frac{30}{3}=10 \\ \end{array} \] Now, the numbers are: \(9\), \(6\), \(10\). - **Step 3:** At this point, there is no prime that divides all three numbers. So, factor each remaining number individually: - \(9 = 3 \times 3 = 3^2\) - \(6 = 2 \times 3\) - \(10 = 2 \times 5\) Using the ladder factors from the left-hand side: - The common factors that went outside the ladder are \(2 \times 3\). 2. **Express Each Number as the Product of Prime Factors** - For \(54\): After the ladder, we had \(27\) remaining from the division by \(2\) and \(3\). But note, \(27 = 3^3\). Thus, \[ 54 = 2 \times 3 \times 3^3 = 2 \times 3^3. \] - For \(36\): Initially, \(36\) was divided by \(2\) to get \(18\), and by \(3\) to get \(6\). Factorizing \(6\) gives \(2 \times 3\). Thus, the factors are: one \(2\) (from ladder) times another \(2\) (from \(6\)) and one \(3\) (from ladder) times another \(3\) (from \(6\)). In total, \[ 36 = 2^2 \times 3^2. \] - For \(60\): Initially, \(60\) was divided by \(2\) to get \(30\) and then by \(3\) to get \(10\). Factorizing \(10\) gives \(2 \times 5\). Thus, including the common factors, \[ 60 = 2^2 \times 3 \times 5. \] 3. **Determine the Highest Common Factor (HCF)** The HCF is found by taking the lowest power of each prime common to all three numbers. - Prime \(2\): Exponents: In \(54\) it is \(2^1\), in \(36\) it is \(2^2\), in \(60\) it is \(2^2\). Minimum exponent is \(1\). - Prime \(3\): Exponents: In \(54\) it is \(3^3\), in \(36\) it is \(3^2\), in \(60\) it is \(3^1\). Minimum exponent is \(1\). - Prime \(5\) is not common to all numbers. Therefore, \[ \text{HCF} = 2^1 \times 3^1 = 2 \times 3 = 6. \] 4. **Determine the Least Common Multiple (LCM)** The LCM is found by taking the highest exponent of each prime that appears in the factorization. - Prime \(2\): Highest exponent is \(2\) (from \(36\) and \(60\)). - Prime \(3\): Highest exponent is \(3\) (from \(54\)). - Prime \(5\): Highest exponent is \(1\) (from \(60\)). Hence, \[ \text{LCM} = 2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540. \] 5. **Final Answers** - \(54 = 2 \times 3^3\) - \(36 = 2^2 \times 3^2\) - \(60 = 2^2 \times 3 \times 5\) \[ \text{HCF} = 6 \quad \text{and} \quad \text{LCM} = 540. \]