Problems on H.C.F and L.C.M
Find the least which when divided y 16,18,20, and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
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
L.C.M. of 5, 6, 7, 8 = 840.Required number is of the form 840k + 3Least value of k for which (840k + 3) is divisible by 9 is k = 2.Required number = (840 x 2 + 3) = 1683.
EXAMIANS