Question

find the hcf and lcm using the prime factorization method (1) 15 and 12. (2) 18 and 45.​

Answer 1

$i) Given \: numbers \: 15\:and \: 12$

We have 15 = 3¹ × 5¹

12 = 2×2×3 = 2² × 3¹

$HCF( 15,12) = 3$

$\blue{( Product \:of \: the \: smallest\:power}$

$\blue{ of \:each \: Common \:prime \:factors}$

$\blue{ of \:the \: numbers .)}$

$LCM ( 15,12) = 3\times 2^{2} \times 5$

$= 60$

$\pink{ ( Product \:of \: greatest \:power \:of }$

$\pink{ each \:prime \:factors \:of \:the \: numbers )}$

$ii) Given \: numbers \: 18\:and \: 45$

We have 18 = 2 × 3 × 3 = 2 × 3²

45 = 3 × 3 × 5 = 3² × 5¹

$HCF( 18,45) = 3^{2} = 9$

$\blue{( Product \:of \: the \: smallest\:power}$

$\blue{ of \:each \: Common \:prime \:factors}$

$\blue{ of \:the \: numbers .)}$

$LCM ( 18,45) = 2 \times 3^{2}\times 5$

$= 90$

$\pink{ ( Product \:of \: greatest \:power \:of }$

$\pink{ each \:prime \:factors \:of \:the \: numbers )}$

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