Question

Using prime factorization, find the HCF and LCM of

(i) 36, 84
(ii) 23, 31
(iii) 96, 404
(iv) 144, 198
(v) 396, 1080
(vi) 1152, 1664

Answer 1

(i) 36, 84
Prime factorisation:
36 = 2² ⨯ 3²
84 = 2² ⨯ 3 ⨯ 7
​ HCF = product of smallest power of each common prime factor in the numbers = 2² ⨯ 3 = 12
LCM = product of greatest power of each prime factor involved in the numbers = 2² ⨯ 3² ⨯ 7 = 252

(ii) 23, 31
Prime factorisation:
23 = 23
31 = 31
​ HCF = product of smallest power of each common prime factor in the numbers = 1
LCM = product of greatest power of each prime factor involved in the numbers = 23 ⨯ 31 = 713

(iii) 96, 404
Prime factorisation:
96 = 2⁵ ⨯ 3
404 = 2² ⨯ 101
​ HCF = product of smallest power of each common prime factor in the numbers = 2² = 4
LCM = product of greatest power of each prime factor involved in the numbers = 2⁵ ⨯ 3 ⨯ 101 = 9696

(iv) 144, 198
Prime factorisation:
144 = $2^4\times 3^2$
198 = $2\times 3^2\times 11$
​ HCF = product of smallest power of each common prime factor in the numbers = $2\times 3^2$ = 18
LCM = product of greatest power of each prime factor involved in the numbers = $2^4\times 3^2\times 11$ = 1584

(v) 396, 1080
Prime factorisation:
396 = $2^2\times 3^2\times 11$
1080 = $2^3\times 3^3\times 5$
​ HCF = product of smallest power of each common prime factor in the numbers = $2^2\times 3^2$ = 36
LCM = product of greatest power of each prime factor involved in the numbers = $2^3\times 3^3\times 5\times 11$ = 11880

(vi) 1152 , 1664
Prime factorisation:
1152 = $2^7\times 3^2$
1664 = $2^7\times 13$
HCF = product of smallest power of each common prime factor involved in the numbers = $2^7$ = 128
LCM = product of greatest power of each prime factor involved in the numbers = $2^7\times 3^2\times 13$ = 14976