Question

1. Find the HCF of the following numbers by prime factorisation and continued division method? i 18, 27, 36 ii. 106, 159, 265 10, 35, 40 iv. 32, 64,96, 128
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Answer 1

Step-by-step explanation:

The Solution of the following questions are as follows: –

(i) 18 , 27 , 36

Solution: – Prime factors of 18 = 3*6=3*3*2

Prime factors of 27 =3*9=3*3*3

Prime factors of 36=4*9=2*2*3*3

Thus, HCF (18, 24,36) = 3*3=9

(ii) 106, 159, 265

Solution: – prime factors of 106 = 2*53

Prime factors of 159 = 3*53

Prime factors of 265 = 5* 53

Thus, HCF ( 106, 159, 265) = 53

(iii) 10, 35, 40

Solution: – prime factors of 10= 2*5

Prime factors of 35 = 5*7

Prime factors of 40 = 5*8=5*2*2*2

Thus, HCF ( 10, 35, 40 ) = 5

(iv) 32, 64, 96, 128

Solution: – Prime factors of 32 = 2*2*2*2*2

Prime factors of 64 = 2*32 = 2*2*2*2*2*2

Prime factors of 96 = 3*32 =3*2*2*2*2*2

Prime factors of 128 = 4* 32 = 2*2*2*2*2*2*2

Thus, HCF ( 32, 64, 96, 128 ) =2*2*2*2*2 =32

Answer 2

Step-by-step explanation:

(i) 18 , 27 , 36

Solution: – Prime factors of 18 = 3*6=3*3*2

Prime factors of 27 =3*9=3*3*3

Prime factors of 36=4*9=2*2*3*3

Thus, HCF (18, 24,36) = 3*3=9

(ii) 106, 159, 265

Solution: – prime factors of 106 = 2*53

Prime factors of 159 = 3*53

Prime factors of 265 = 5* 53

Thus, HCF ( 106, 159, 265) = 53

(iii) 10, 35, 40

Solution: – prime factors of 10= 2*5

Prime factors of 35 = 5*7

Prime factors of 40 = 5*8=5*2*2*2

Thus, HCF ( 10, 35, 40 ) = 5

(iv) 32, 64, 96, 128

Solution: – Prime factors of 32 = 2*2*2*2*2

Prime factors of 64 = 2*32 = 2*2*2*2*2*2

Prime factors of 96 = 3*32 =3*2*2*2*2*2

Prime factors of 128 = 4* 32 = 2*2*2*2*2*2*2

Thus, HCF ( 32, 64, 96, 128 ) =2*2*2*2*2 =32

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