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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.