L.C.M. of 6, 9, 15 and 18 is 90.Let required number be 90k + 4, which is multiple of 7.Least value of k for which (90k + 4) is divisible by 7 is k = 4.Required number = (90 x 4) + 4 = 364.
The greatest number that will divide 172,205 and 304 leaving the same remainder in each case is HCF ((205-172), (304-205), (304-172)) = HCF of 33, 99, and 132 33=31x11 99=3 x 3 x 11 132=3 x 2 x 2 x11 Required number =3 x 11=33