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# 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.

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.

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