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

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

Solution:
L.C.M. of 6, 9, 15 and 18 is 90.
Let the required number be 90k + 4, which is a multiple of 7

The least value of k for which (90k + 4) is divisible by 7 is k = 4

Hence, the equired number = (90 × 4) + 4 = 364