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

1. **Prime Factorization of Each Number** We start by writing the numbers \(54\), \(36\), and \(60\) side by side and then “laddering” by dividing with primes. **Step 1: Divide by \(2\)** Since all three numbers are even, we can divide each by \(2\): \[ \begin{array}{cccc} 2 & | & 54 & 36 \quad 60 \\ & & \downarrow & \downarrow \quad \downarrow \\ & & 27 & 18 \quad 30 \\ \end{array} \] **Step 2: Divide by \(3\)** Now, look at the new row: \(27\), \(18\), and \(30\). All are divisible by \(3\): \[ \begin{array}{cccc} 2 & 3 & | & 54 \quad 36 \quad 60 \\ & & & \downarrow \quad \downarrow \quad \downarrow \\ & & & 27 \quad 18 \quad 30 \\ & & 3 & \downarrow \quad \downarrow \quad \downarrow \\ & & & 9 \quad 6 \quad 10 \\ \end{array} \] **Step 3: Divide by \(2\)** In the row \(9\), \(6\), \(10\), both \(6\) and \(10\) are divisible by \(2\) (while \(9\) is not), so we divide those numbers: \[ \begin{array}{cccc} 2 & 3 & 2 & | \; 54 \quad 36 \quad 60 \\ & & & \downarrow \quad \downarrow \quad \downarrow \\ & & & 27 \quad 18 \quad 30 \\ & & 3 & \downarrow \quad \downarrow \quad \downarrow \\ & & & 9 \quad 6 \quad 10 \\ & & & \quad (\text{divide }6\text{ and }10\text{ by }2)\\ & & & 9 \quad 3 \quad 5\\ \end{array} \] **Step 4: Divide by \(3\)** Now, in the row \(9\), \(3\), \(5\), the first two numbers are divisible by \(3\): \[ \begin{array}{cccc} 2 & 3 & 2 & 3\, | \; 54 \quad 36 \quad 60 \\ & & & & \downarrow \quad \downarrow \quad \downarrow \\ & & & & 27 \quad 18 \quad 30 \\ & & 3 & & \downarrow \quad \downarrow \quad \downarrow \\ & & & & 9 \quad 6 \quad 10 \\ & & & & \quad (\text{divide }6\text{ and }10\text{ by }2)\\ & & & & 9 \quad 3 \quad 5\\ & & & & (\text{divide }9\text{ and }3\text{ by }3)\\ & & & & 3 \quad 1 \quad 5\\ \end{array} \] **Step 5: Divide by \(3\)** In the row \(3\), \(1\), \(5\), only the first number is divisible by \(3\): \[ \begin{array}{cccc} 2 & 3 & 2 & 3\, 3\, | \; 54 \quad 36 \quad 60 \\ & & & & \downarrow \quad \downarrow \quad \downarrow \\ & & & & 27 \quad 18 \quad 30 \\ & & 3 & & \downarrow \quad \downarrow \quad \downarrow \\ & & & & 9 \quad 6 \quad 10 \\ & & & & (\text{after dividing by }2)\\ & & & & 9 \quad 3 \quad 5\\ & & & & (\text{after dividing by }3)\\ & & & & 3 \quad 1 \quad 5\\ & & & & (\text{divide }3\text{ by }3)\\ & & & & 1 \quad 1 \quad 5\\ \end{array} \] **Step 6: Divide by \(5\)** Finally, in the row \(1\), \(1\), \(5\), the third number is divisible by \(5\): \[ \begin{array}{cccc} 2 & 3 & 2 & 3\, 3\, 5\, | \; 54 \quad 36 \quad 60 \\ & & & & \downarrow \quad \downarrow \quad \downarrow \\ & & & & 27 \quad 18 \quad 30 \\ & & 3 & & \downarrow \quad \downarrow \quad \downarrow \\ & & & & 9 \quad 6 \quad 10 \\ & & & & (\text{after dividing by }2)\\ & & & & 9 \quad 3 \quad 5\\ & & & & (\text{after dividing by }3)\\ & & & & 3 \quad 1 \quad 5\\ & & & & (\text{after dividing }3\text{ by }3)\\ & & & & 1 \quad 1 \quad 5\\ & & & & (\text{divide }5\text{ by }5)\\ & & & & 1 \quad 1 \quad 1\\ \end{array} \] 2. **Expressing the Prime Factorizations** The primes on the left represent the factors extracted in order. For each original number, the prime factors are: - For \(54\): \[ 54 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3 \] - For \(36\): \[ 36 = 2 \times 3 \times 2 \times 3 = 2^2 \times 3^2 \] - For \(60\): \[ 60 = 2 \times 3 \times 2 \times 5 = 2^2 \times 3 \times 5 \] 3. **Finding the HCF (Highest Common Factor)** To find the HCF, we take the common prime factors with the smallest exponents in all three numbers. - For prime \(2\): Exponents are \(1\) (in \(54\)), \(2\) (in \(36\)), and \(2\) (in \(60\)). The smallest exponent is \(1\). - For prime \(3\): Exponents are \(3\) (in \(54\)), \(2\) (in \(36\)), and \(1\) (in \(60\)). The smallest exponent is \(1\). - The prime \(5\) does not appear in all three. Hence, the HCF is: \[ \text{HCF} = 2^1 \times 3^1 = 6 \] 4. **Finding the LCM (Least Common Multiple)** To find the LCM, we take each prime factor with the highest exponent present in any of the numbers. - For prime \(2\): The highest exponent among the numbers is \(2\). - For prime \(3\): The highest exponent is \(3\) (from \(54\)). - For prime \(5\): The highest exponent is \(1\) (from \(60\)). Hence, the LCM is: \[ \text{LCM} = 2^2 \times 3^3 \times 5 = 4 \times 27 \times 5 = 540 \] 5. **Final Answer** \[ \begin{aligned} 54 &= 2 \times 3^3,\\[1mm] 36 &= 2^2 \times 3^2,\\[1mm] 60 &= 2^2 \times 3 \times 5,\\[2mm] \text{HCF} &= 6,\\[1mm] \text{LCM} &= 540. \end{aligned} \]