Question

Q3 Use prime factors to find (i) the HCF and (ii) the LCM of each of the following pairs of numbers. a) 60 and 75 HCF-is b) 54 and 96 c) 108 and 144 d) 200 and 240 e) 168 and 196 f) 150 and 180

Ask by Wright Hampton. in the United Kingdom
Mar 13,2025

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Answer

**a) 60 and 75** - **HCF:** 15 - **LCM:** 300 **b) 54 and 96** - **HCF:** 6 - **LCM:** 864 **c) 108 and 144** - **HCF:** 36 - **LCM:** 432 **d) 200 and 240** - **HCF:** 40 - **LCM:** 1200 **e) 168 and 196** - **HCF:** 28 - **LCM:** 1176 **f) 150 and 180** - **HCF:** 30 - **LCM:** 900

Solution

**a) For 60 and 75** 1. Write the prime factorizations: \[ 60 = 2^2 \times 3 \times 5, \quad 75 = 3 \times 5^2 \] 2. **HCF** is found by taking the lowest power of the common primes: \[ \text{HCF} = 3^1 \times 5^1 = 15 \] 3. **LCM** is found by taking the highest power of each prime that appears: \[ \text{LCM} = 2^2 \times 3^1 \times 5^2 = 4 \times 3 \times 25 = 300 \] --- **b) For 54 and 96** 1. Write the prime factorizations: \[ 54 = 2 \times 3^3, \quad 96 = 2^5 \times 3 \] 2. **HCF** (lowest power of common factors): \[ \text{HCF} = 2^1 \times 3^1 = 6 \] 3. **LCM** (highest power of each prime): \[ \text{LCM} = 2^5 \times 3^3 = 32 \times 27 = 864 \] --- **c) For 108 and 144** 1. Write the prime factorizations: \[ 108 = 2^2 \times 3^3, \quad 144 = 2^4 \times 3^2 \] 2. **HCF**: \[ \text{HCF} = 2^2 \times 3^2 = 4 \times 9 = 36 \] 3. **LCM**: \[ \text{LCM} = 2^4 \times 3^3 = 16 \times 27 = 432 \] --- **d) For 200 and 240** 1. Write the prime factorizations: \[ 200 = 2^3 \times 5^2, \quad 240 = 2^4 \times 3 \times 5 \] 2. **HCF**: \[ \text{HCF} = 2^3 \times 5^1 = 8 \times 5 = 40 \] 3. **LCM**: \[ \text{LCM} = 2^4 \times 3^1 \times 5^2 = 16 \times 3 \times 25 = 1200 \] --- **e) For 168 and 196** 1. Write the prime factorizations: \[ 168 = 2^3 \times 3 \times 7, \quad 196 = 2^2 \times 7^2 \] 2. **HCF**: \[ \text{HCF} = 2^2 \times 7^1 = 4 \times 7 = 28 \] 3. **LCM**: \[ \text{LCM} = 2^3 \times 3^1 \times 7^2 = 8 \times 3 \times 49 = 1176 \] --- **f) For 150 and 180** 1. Write the prime factorizations: \[ 150 = 2 \times 3 \times 5^2, \quad 180 = 2^2 \times 3^2 \times 5 \] 2. **HCF**: \[ \text{HCF} = 2^1 \times 3^1 \times 5^1 = 30 \] 3. **LCM**: \[ \text{LCM} = 2^2 \times 3^2 \times 5^2 = 4 \times 9 \times 25 = 900 \]

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To find the Highest Common Factor (HCF) and the Lowest Common Multiple (LCM) using prime factorization, start by breaking each number down into its prime factors. For example, let’s look at the first pair: **a) 60 and 75** - The prime factors of 60 are: 2 × 2 × 3 × 5 (or \(2^2 × 3^1 × 5^1\)). - The prime factors of 75 are: 3 × 5 × 5 (or \(3^1 × 5^2\)). - **HCF**: The common prime factors are 3 and 5, so HCF = 3 × 5 = 15. - **LCM**: Take the highest power of each prime: 2, 3, \(5^2\). So, LCM = 2^2 × 3^1 × 5^2 = 300. Repeat this process for the other pairs: **b) 54 and 96** - 54: \(2^1 × 3^3\) - 96: \(2^5 × 3^1\) - HCF = \(2^1 × 3^1 = 6\) - LCM = \(2^5 × 3^3 = 486\) **c) 108 and 144** - 108: \(2^2 × 3^3\) - 144: \(2^4 × 3^2\) - HCF = \(2^2 × 3^2 = 36\) - LCM = \(2^4 × 3^3 = 432\) **d) 200 and 240** - 200: \(2^3 × 5^2\) - 240: \(2^4 × 3^1 × 5^1\) - HCF = \(2^3 × 5^1 = 40\) - LCM = \(2^4 × 3^1 × 5^2 = 1200\) **e) 168 and 196** - 168: \(2^3 × 3^1 × 7^1\) - 196: \(2^2 × 7^2\) - HCF = \(2^2 × 7^1 = 28\) - LCM = \(2^3 × 3^1 × 7^2 = 1176\) **f) 150 and 180** - 150: \(2^1 × 3^1 × 5^2\) - 180: \(2^2 × 3^2 × 5^1\) - HCF = \(2^1 × 3^1 × 5^1 = 30\) - LCM = \(2^2 × 3^2 × 5^2 = 900\) Now you’ve got the HCF and LCM for each pair of numbers! Prime factorization is like a treasure hunt for numbers, revealing their hidden relationships! Happy calculating!

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