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
(i) 36, 84
Prime factorisation:
$36=2^{2} \times 3^{2}$
$84=2^{2} \times 3 \times 7$
$\mathrm{HCF}=$ product of smallest power of each common prime factor in the numbers $=2^{2} \times 3=12$
$\mathrm{LCM}=$ product of greatest power of each prime factor involved in the numbers $=2^{2} \times 3^{2} \times 7=252$
(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:
$96=2^{5} \times 3$
$404=2^{2} \times 101$
HCF = product of smallest power of each common prime factor in the numbers $=2^{2}=4$
LCM = product of greatest power of each prime factor involved in the numbers $=2^{5} \times 3 \times 101=9696$
(iv) 144, 198
Prime factorisation
$144=2^{4} \times 3^{2}$
$198=2 \times 3^{2} \times 11$
$\mathrm{HCF}=$ 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:
$396=2^{2} \times 3^{2} \times 11$
$1080=2^{3} \times 3^{3} \times 5$
$\mathrm{HCF}=$ 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 $=2^{7} \times 3^{2} \times 13=14976$