Question : Find the H.C.F and L.C.M of the following integers by applying prime factorisation method.
(i) 12, 15, 21
(ii) 17, 23, 29
(iii) 8, 9 and 25
Solution :
(i) 12, 15, 21
12 = \(2 \times 2 \times 3\)
15 = \(3 \times 5\)
21 = \(3 \times 7\)
Here 3 is a common prime factor of the given numbers.
Hence, HCF = 3
H.C.F. (12, 15, 21) = 3
L.C.M is the product of the prime factors \(2 \times 2 \times 3 \times 7 \times 5\)
The common factor 3 is repeated three times, so we write 3 only one times in multiplication..
L.C.M. (12, 15, 21) = 420
(ii) 17, 23, 29
H.C.F. There is no common factor as 17, 23, 29 they are prime. Hence, H.C.F is 1.
L.C.M. is the product of all prime factors 17, 23 and 29.
\(17 \times 23 \times 29\) = 11339
Hence, H.C.F. (17, 23, 29) = 1.
L.C.M (17, 23, 29) = 11339
(iii) 8, 9 and 25
First we write the prime factorisation of each of the given numbers.
8 = \(2 \times 2 \times\) = \(2^3\),
9 = \(3 \times 3\) = \(3^2\)
25 = \(5 \times 5\) = \(5^2\)
\(\therefore\) L.C.M = \(2^2 \times 3^2 \times 5^2\) = \(8 \times 9 \times 25\) = 1800
and H.C.F = 1.
