Question

Using prime factorization, find the HCF and LCM of

(i) 8, 9, 25
(ii) 12, 15, 21
(iii) 17, 23, 29
(iv) 24, 36, 40
(v) 30, 72, 432
(vi) 21, 28, 36, 45

Answer 1

(i) 8 = 2 ⨯ 2 ⨯ 2 = 2³
9 = 3 ⨯ 3 = 3²
25 = 5 ⨯ 5 = 5²
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 = 2³ ⨯ 3² ⨯ 5² = 1800

(ii) 12 = 2 ⨯ 2 ⨯ 3 = 2² ⨯ 3
15 = 3 ⨯ 5
21 = 3 ⨯ 7
HCF = Product of smallest power of each common prime factor in the numbers = 3
LCM = Product of the greatest power of each prime factor involved in the numbers = 2² ⨯ 3 ⨯ 5 ⨯ 7 = 420

(iii) 17 = 17
23 = 23
29 = 29
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 = 17 ⨯ 23 ⨯ 29 = 11339

(iv) 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 = $2^3\times 3^2\times 5=360$

(v) $30=2\times 3\times 5$
$72=2\times 2\times 2\times 3\times 3$ = $2^3\times 3^2$
$432=2\times 2\times 2\times 2\times 3\times 3\times 3$ = $2^4\times 3^3$
∴ 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 = $2^4\times 3^3\times 5=2160$

(vi) $21=3\times 7$
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 = $2^2\times 3^2\times 5\times 7=1260$