What is the least multiple of 7 which leaves a remainder of 4 when divided by 6, 9, 15 and 18?

LCM of 6, 9, 15 and 18 = 90
Required Number = (90k + 4) which is a multiple of 7
Put k = 1. We get numbers as (90 × 1) + 4 = 94. But this is not a multiple of 7
Put k = 2. We get numbers as (90 × 2) + 4 = 184. But this is not a multiple of 7
Put k = 3. We get numbers as (90 × 3) + 4 = 274. But this is not a multiple of 7
Put k = 4. We get numbers as (90 × 4) + 4 = 364. This is a multiple of 7
Hence 364 is the answer.