Problems on H.C.F and L.C.M
Find the least number which when divide by 2, 3, 4, 5 and 6 leaves 1, 2, 3, 4 and 5 as remainders respectively, 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
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.
LCM of 5,6,8,10 and 12 = 120But in each case , we have a remainder 2 .So the required number is 120n+2 which is exactly divisible by 13. 120 n + 2 = 13 x 9 n + 2 So clearly 3n+2 must be divisible by 13. For n=8, 3n +2 is exactly divisible by 13. Therefore , the required number = 120n + 2 = 120 x 8 + 2 = 962
EXAMIANS