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?
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