The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:Options :7494184364Clear ResponsePreviousFlagNextEnd Test

The problem is asking for the least common multiple (LCM) of 6, 9, 15, and 18 that, when 4 is added, is also a multiple of 7.

Step 1: Find the LCM of 6, 9, 15, and 18.

The prime factorization of these numbers are: 6 = 2 * 3 9 = 3^2 15 = 3 * 5 18 = 2 * 3^2

The LCM is found by multiplying the highest power of each prime number together. So, the LCM is 2 * 3^2 * 5 = 90.

Step 2: Find the least multiple of 7 that leaves a remainder of 4 when divided by 90.

This can be done by trial and error. Start with 4 (since the number must leave a remainder of 4), and keep adding 90 until you find a number that is a multiple of 7.

Doing this, you find that 364 is the least such number.

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