The least number which when divided by 4,5,6,8 and 10 is the LCM of these numbers. But each time to get 3 as remainder, we have to add the remainder 3 to the obtained LCM.LCM of (4, 5, 6, 8, 10)+ 3=120+3=123
Given that, a = 99, b = 147, c = 219 Required number = [H.C.F. of (a-b), (b-c), (c-a)] Now, (a-b) = (99 – 147) = 48 (b-c) = (147 – 219) = 72 (c-a) = (219 – 99) = 120 Also 48 = 24 x 3¹ 72 = 23 x 32 120 = 23 x 3¹ x 5¹ ∴ HCF of 48,72 and 120 = 23 x 3¹ = 8 x 3 = 24
EXAMIANS