Problems on H.C.F and L.C.M
An officer was appointed on maximum daily wages on contract money of Rs. 4956. But on being absent for some days, he was paid only Rs. 3894. For how many days was he absent?
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
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
EXAMIANS