The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
364
L.C.M. of 6, 9, 15 and 18 is 90. Let required number be 90k + 4, which is a 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.