arithmetic - find the least number which when divided by 6 8 10 and 15 leaves 3 as remainder in each case but when divided by 9 leaves no remainder - skillgun

L.C.M. of 6, 8, 10, 15 = 120.
Required number is of the form 120k + 3
Least value of k for which (120k + 3) is divisible by 9 is k = 2.
Required number = (120 x 2 + 3) = 243