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

To solve this problem, we need to find the least common multiple (LCM) of 6, 9, 15, and 18 first.

Step 1: Prime factorization of the numbers 6 = 2 * 3 9 = 3^2 15 = 3 * 5 18 = 2 * 3^2

Step 2: Find the LCM The LCM is found by multiplying the highest power of all prime numbers in the factorization. So, LCM = 2 * 3^2 * 5 = 90

Step 3: Find the multiples of 7 that leave a remainder of 4 when divided by 90 We know that the number we are looking for is a multiple of 7 and leaves a remainder of 4 when divided by 90. So, we can write it as 7n = 90k + 4, where n and k are integers.

We can start checking from the smallest multiple of 7 that is greater than 90, which is 98.

98 is not leaving a remainder of 4 when divided by 90.

The next multiple of 7 is 105, which also does not leave a remainder of 4 when divided by 90.

We continue this process until we find a multiple of 7 that leaves a remainder of 4 when divided by 90.

After checking, we find that 364 is the least multiple of 7 that leaves a remainder of 4 when divided by 90.

So, the least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is 364.