We use ** prime factorization**, that's

This is particularly useful when trying to find the *HCF*, or *LCM*, of large numbers.

To see the *lesson notes*, written in class, click on the button below.

Leaving your answers as *products of prime factors*, find the *Highest Common Factor*, *HCF*, and the *Least Common Multiple*, *LCM*, of each of the following pairs of *whole numbers*:

- \(75\) and \(20\)
- \(90\) and \(84\)
- \(36\) and \(60\)
- \(225\) and \(135\)
- \(100\) and \(150\)
- \(180\) and \(126\)
- \(324\) and \(144\)
- \(495\) and \(525\)

- \(HCF = 5\) and \(LCM = 2^2\times 3 \times 5^2\)
- \(HCF(90,84) = 2\times 3\) and \(LCM(90,84) = 2^2 \times 3^2 \times 5 \times 7\)
- \(HCF(36,60) = 2^2 \times 3\) and \(LCM(36,60) = 2^2 \times 3^2 \times 5\).
- \(HCF(225,135) = 3^2 \times 5\) and \(LCM(225,135) = 3^3 \times 5^2\)
- \(HCF(100,150) = 2\times 5^2\) and \(LCM(100,150) = 2^2 \times 3\times 5^2\)
- \(HCF(180,126) = 2\times 3^2\) and \(LCM(180,126)=2^2 \times 3^2 \times 5 \times 7\)
- \(HCF(324,144) = 2^2\times 3^2\) and \(LCM(324,144) = 2^4\times 3^4\)
- \(HCF(495,525) = 3\times 5\) and \(LCM(495,525) = 3^2\times 5^2 \times 7 \times 11\)

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