The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9,…

To find the least multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18, we follow a systematic process.

Step-by-Step Calculation
1. Find the Least Common Multiple (LCM) of 6, 9, 15, and 18:

Prime factorization:

• 6 = 2 × 3

• 9 = 3²

• 15 = 3 × 5

• 18 = 2 × 3²

The LCM uses the highest power of each prime factor present:

• Factors: 2¹, 3², 5¹

• LCM = 2 × 9 × 5 = 90.

2. Formulate the expression for the number:
The number must leave a remainder of 4 when divided by the LCM (90). We can write this as:
Number = 90k + 4 (where k is a positive integer 0, 1, 2, 3...).

3. Apply the divisibility condition:
We need this number (90k + 4) to be exactly divisible by 7.

• Divide 90 by 7: 90 = (7 × 12) + 6.

• The expression becomes: (7 × 12k + 6k + 4).

• For divisibility by 7, (6k + 4) must be divisible by 7.

Test values for k:

• If k = 1: 6(1) + 4 = 10 (not divisible by 7)

• If k = 2: 6(2) + 4 = 16 (not divisible by 7)

• If k = 3: 6(3) + 4 = 22 (not divisible by 7)

• If k = 4: 6(4) + 4 = 28 (divisible by 7: 28 / 7 = 4)

4. Calculate the final number:
Using k = 4:
Number = 90(4) + 4 = 360 + 4 = 364.