Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25.

AcademicMathematicsNCERTClass 10

To find: 

Here we have to find the LCM and HCF of the given integers by applying the prime factorization method.

Solution:

Calculating LCM and HCF using prime factorization method:

Writing the numbers as a product of their prime factors:

(i) Prime factorisation of 12:

  • $2\ \times\ 2\ \times\ 3\ =\ 2^2\ \times\ 3^1$

Prime factorisation of 15:

  • $3\ \times\ 5\ =\ 3^1\ \times\ 5^1$

Prime factorisation of 21:

  • $3\ \times\ 7\ =\ 3^1\ \times\ 7^1$

Multiplying the highest power of each prime number these values together:

$2^2\ \times\ 3^1\ \times\ 5^1\ \times\ 7^1\ =\ 420$

LCM(12, 15, 21)  $=$  420

Multiplying all common prime factors: 

$3^1\ =\ 3$

HCF(12, 15, 21)  $=$  3

So, the LCM and HCF of 12, 15 and 21 are 420 and 3 respectively.

(ii) Prime factorisation of 17:

  • $17\ =\ 17^1$

Prime factorisation of 23:

  • $23\ =\ 23^1$

Prime factorisation of 29:

  • $29\ =\ 29^1$

Multiplying the highest power of each prime number these values together:

$17^1\ \times\ 23^1\ \times\ 29^1\ =\ 11339$

LCM(17, 23, 29)  $=$  11339

Multiplying all common prime factors: 

There is no common prime factor. So,

HCF(17, 23, 29)  $=$  1

So, the LCM and HCF of 17, 23 and 29 are 11339 and 1 respectively.

(iii) Prime factorisation of 8:

  • $2\ \times\ 2\ \times\ 2\ =\ 2^3$

Prime factorisation of 9:

  • $3\ \times\ 3\ =\ 3^2$

Prime factorisation of 25:

  • $5\ \times\ 5\ =\ 5^2$

Multiplying the highest power of each prime number these values together:

$2^3\ \times\ 3^2\ \times\ 5^2\ =\ 1800$

LCM(8, 9, 25)  $=$  1800

Multiplying all common prime factors: 

There is no common prime factor. So,

HCF(8, 9, 25)  $=$  1

So, the LCM and HCF of 8, 9 and 25 are 1800 and 1 respectively.

raja
Updated on 10-Oct-2022 13:19:30

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