The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Since
6 = 2 × 3
9 = 3 × 3
15 = 3 × 5
18 = 2 × 3 × 3
L.C.M. of 6, 9, 15 and 18 = 2 × 3 × 3 × 5 = 90.
Let required number be 90k + 4, which is multiple of 7.
Taking k = 1 we get 94, this is not a multiple of 7.
Taking k = 2 we get 184, this is not a multiple of 7.
Taking k = 3 we get 274, this is not a multiple of 7.
Taking k = 4 we get 364, this is a multiple of 7.
So the least value of k for which (90k + 4) is divisible by 7 is k = 4 and the required number is 364.