Question -
Answer -
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Q (i) 12,15 and 21
Solutions:
(i) 12, 15 and 21
We will get the product of its prime factor
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
Q (ii) 17,23 and 29
(ii)17, 23 and 29
We will get the product of its prime factor
17=17×1
23=23×1
29=29×1
Therefore,
HCF(17,23,29) = 1
LCM(17,23,29) = 17 × 23 × 29 = 11339
Q (iii) 8, 9 and 25
(iii)8, 9 and 25
We will get the product of its prime factor
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