Question

Using prime factorisation, find the HCF of 144, 180 and 192.

Answer 1

We need to find the HCF (Highest Common Factor) of 144, 180, and 192 using prime factorisation. First, find the prime factorisation of each number: For 144: \(144 = 2 \times 72\) \(72 = 2 \times 36\) \(36 = 2 \times 18\) \(18 = 2 \times 9\) \(9 = 3 \times 3\) So, \(144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^4 \times 3^2\). For 180: \(180 = 10 \times 18 = (2 \times 5) \times (2 \times 9) = (2 \times 5) \times (2 \times 3 \times 3)\) So, \(180 = 2^2 \times 3^2 \times 5^1\). For 192: \(192 = 2 \times 96\) \(96 = 2 \times 48\) \(48 = 2 \times 24\) \(24 = 2 \times 12\) \(12 = 2 \times 6\) \(6 = 2 \times 3\) So, \(192 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^6 \times 3^1\). Now, identify the common prime factors and their lowest powers: Prime factorisations: \(144 = 2^4 \times 3^2\) \(180 = 2^2 \times 3^2 \times 5^1\) \(192 = 2^6 \times 3^1\) The common prime factors are 2 and 3. The lowest power of 2 is \(2^{\min(4, 2, 6)} = 2^2\). The lowest power of 3 is \(3^{\min(2, 2, 1)} = 3^1\). The prime factor 5 is not common to all three numbers. HCF = Product of the lowest powers of common prime factors. HCF = \(2^2 \times 3^1 = 4 \times 3 = 12\). \[ \boxed{12} \]