Question

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

Answer 1

Answer:

HERE IS UR ANSWER DEAR

Explanation:

(i) 18,27,36  

The three numbers can be represented as the product of their prime factors, as 3  

2

×2,  3  

3

 and 3  

2

×2  

2

 respectively.

The product of the lowest powers of the common factors among the three numbers is 2  

1

×3  

2

 

∴ HCF=2×3  

2

=18

(ii)106,159,165

The three numbers can be represented as the product of their prime factors, as 2×53,3×53 and 5×53 respectively.

The product of the lowest powers of the common factors among the three numbers is 53  

1

 

∴ HCF=53

(iii)10,35,40

The three numbers can be represented as the product of their prime factors, as 5×2,5×7, and 2  

3

×5 respectively.

The product of the lowest powers of the common factors among the three numbers is 5  

1

 

∴ HCF=5  

(iv)32,64,96,128  

The three numbers can be represented as the product of their prime factors, as 2  

5

,2  

6

,2  

5

×3 and 2  

7

 respectively.

The product of the lowest powers of the common factors among the three numbers is 2  

5

 

∴ HCF=2  

5

=32

HOPE IT HELPS U !!