Question

Find the HCF and LCM of the following integers by applying the prime factorization method

(i) 12, 15 and 21

(ii) 24, 15 and 36

(iii) 17, 23 and 29

(iv) 6, 72 and 120

(v) 40, 36 and 126

(vi) 8, 9 and 25

Answer 1

(i) 12, 15 and 21

12 = 2 × 2 × 3 = 2² × 3

15 = 3 × 5

and 21 = 3 × 7

For HCF, we find minimum power of prime factor

H.C.F. = (3)¹ = 3

For LCM, taking maximum power of prime factors

L.C.M. = 2² × 3 × 5 × 7 = 4 × 3 × 5 × 7 = 420

So H.C.F. = 3

L.C.M. = 420

(ii) 24, 15 and 36

24 = 2 × 2 × 2 × 3 = 2³ × 3

15 =3 × 5

and 36 = 2 × 2 × 3 × 3 = 2² × 3²

For HCF, taking minimum power of common prime factor

H.C.F. = (3)¹ = 3

For LCM, taking maximum power of prime factor

L.C.M. = 2³ × 3² × 5¹ = 8 × 9 × 5 = 360

So H.C.F. = 3

L.C.M. = 360

(iii) 17, 23 and 29

17 = 1 × 17

23 = 1 × 23

and 29 = 1 × 29

For HCF, common factor is 1

HCF = 1

For LCM taking maximum power of prime factor.

L.C.M. = 1 × 17 × 23 × 29 = 11339

So H.C.F. = 1

L.C.M. = 11339

(iv) 6, 72 and 120

6 = 2 × 3

72 = 2 × 2 × 2 × 3 × 3 

= 2³ × 3²

and 120 = 2 × 2 × 2 × 3 × 5 

= 2³ × 3 × 5

For HCF, taking minimum power of common prime factor

HCF = 2 × 3

For LCM taking maximum power of prime factor

L.C.M. = 2³ × 3² × 5 = 8 × 9 × 5 = 360

So H.C.F. = 6

L.C.M. = 360

(v) 40, 36 and 126

8 = 2 × 2 × 2 = 2³

9 = 3 × 3 = 3²

and 25 = 5 × 5 = 5²

For HCF common factor is 1

H.C.F. = 1

For LCM, taking maximum power of prime factors

L.C.M. = 2³ × 3² × 5² 

= 8 × 9 × 25 = 1800

So H.C.F. = 1

L.C.M. = 1800