Question

Using prime factorisation, find the HCF and LCM of

(i) 24, 36, 40
(ii) 30, 72, 432
(iii) 21, 28, 36, 45

Answer 1

(i) 24 = 2$\times 2\times 2\times 3$ = $2^3\times 3$
36 = $2\times 2\times 3\times 3$ = $2^2\times$$3^2$
40 = 2 $\times 2\times 2\times 5$ = $2^3$ $\times 5$
∴ ​HCF = Product of smallest power of each common prime factor in the numbers = $2^2$ = 4
∴​ LCM = Product of the greatest power of each prime factor involved in the numbers = $2^3\times 3^2\times 5=360$

(ii) $30=2\times 3\times 5$
$72=2\times 2\times 2\times 3\times 3$ = $2^3\times 3^2$
$432=2\times 2\times 2\times 2\times 3\times 3\times 3$ = $2^4\times 3^3$
∴ HCF = Product of smallest power of each common prime factor in the numbers = $2\times 3=6$
∴ ​LCM = Product of the greatest power of each prime factor involved in the numbers = $2^4\times 3^3\times 5=2160$

(iii) $21=3\times 7$
28 = $28=2\times 2\times 7$ = $2^2\times 7$
$36=2\times 2\times 3\times 3$ = $2^2\times 3^2$
$45=5\times 3\times 3$ = $5\times 3^2$
∴​ HCF = Product of smallest power of each common prime factor in the numbers = 1
∴​ LCM = Product of the greatest power of each prime factor involved in the numbers = $2^2\times 3^2\times 5\times 7=1260$