Find the LCM and HCF of the following integers by applying the prime factorisation method:
Question:

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

(iv) 40, 36 and 126

(v) 84, 90 and 120

(vi) 24, 15 and 36

Solution:

TO FIND: LCM and HCF of following pairs of integers

(i) 15, 12 and 21

Let us first find the factors of 15, 12 and 21

$12=2^{2} \times 3$

$15=3 \times 5$

$21=3 \times 7$

L.C.M of 12,15 and $21=2^{2} \times 3 \times 5 \times 7$

L.C.M of 12,15 and $21=420$

H.C.F of 12,15 and $21=3$

(ii) 17, 23 and 29

Let us first find the factors of 17, 23 and 29

$17=1 \times 17$

$23=1 \times 23$

$29=1 \times 29$

L.C.M of 17,23 and $29=1 \times 17 \times 23 \times 29$

L.C.M of 17,23 and $29=11339$

H.C.F of 17,23 and $29=1$

(iii) 8, 9 and 25

Let us first find the factors of 8,9 and 25

$8=2^{3}$

$9=3^{2}$

$25=5^{2}$

L.C.M of 8,9 and $25=2^{3} \times 3^{2} \times 5^{2}$

L.C.M of 8,9 and $25=1800$

H.C.F of 8,9 and $25=1$

(iv) 40, 36 and 126

Let us first find the factors of 40, 36 and 126

$40=2^{3} \times 5$

$36=2^{2} \times 3^{2}$

$126=2 \times 3^{2} \times 7$

L.C.M of 40,36 and $126=2^{3} \times 3^{2} \times 5 \times 7$

L.C.M of 40,36 and $126=2520$

H.C.F of 40,36 and $126=2$

(v) 84, 90 and 120

Let us first find the factors of 84, 90 and 120

$84=2^{2} \times 3 \times 7$

$90=2 \times 3^{2} \times 5$

$120=2^{3} \times 3 \times 5$

L.C.M of 84,90 and $120=2^{3} \times 3^{2} \times 5 \times 7$

L.C.M of 84,90 and $120=2520$

H.C.F of 84,90 and $120=6$

(vi) 24, 15 and 36

Let us first find the factors of 24, 15 and 36.

$24=2^{3} \times 3$

$15=3 \times 5$

$36=2^{2} \times 3^{2}$

L.C.M of $24.15$ and $36=2^{3} \times 3^{2} \times 5$

L.C.M of 24,15 and $36=360$

H.C.F of 24,15 and $36=3$