Problems on H.C.F and L.C.M
In finding the HCF of two numbers, the last divisor was 41 and the successive quotients, starting from the first, where 2, 4 and 2. The numbers are?
Let the numbers be 3x, 4x and 5x.Then, their L.C.M. = 60x.So, 60x = 2400 or x = 40. The numbers are (3 x 40), (4 x 40) and (5 x 40).Hence, required H.C.F. = 40.
EXAMIANS