Solutions:
(i) 12, 15 and 21
Writing the product of prime factors for all the three numbers, we get,
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Therefore,
HCF(12, 15, 21) = 3
LCM(12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
Writing the product of prime factors for all the three numbers, we get,
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
Therefore,
HCF(17, 23, 29) = 1
LCM(17, 23, 29) = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
Writing the product of prime factors for all the three numbers, we get,
8 = 2 × 2 × 2 × 1
9 = 3 × 3 × 1
25 = 5 × 5 × 1
Therefore,
HCF(8, 9, 25) = 1
LCM(8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800