# Using prime factorization, find the HCF and LCM of

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

Solution:

(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 $=2^{7} \times 3^{2} \times 13=14976$

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