Find the four digit number which is completely divisible by 12, 15, 18 & 27?

Options

A. 9690
B. 9720
C. 9930
D. 9960

Answer

To find a four-digit number that is completely divisible by 12, 15, 18, and 27, we will follow these steps:

### Step 1: Find the Least Common Multiple (LCM)
To find a number that is divisible by all the given numbers, we need to calculate the LCM of 12, 15, 18, and 27.

**Prime Factorization:**
- 12 = 2^2 × 3^1
- 15 = 3^1 × 5^1
- 18 = 2^1 × 3^2
- 27 = 3^3

**Taking the highest power of each prime factor:**
- For 2: The highest power is 2^2 (from 12).
- For 3: The highest power is 3^3 (from 27).
- For 5: The highest power is 5^1 (from 15).

**Calculating LCM:**
LCM = 2^2 × 3^3 × 5^1
= 4 × 27 × 5
= 108 × 5
= 540

### Step 2: Identify the largest four-digit number
The largest four-digit number is 9999.

### Step 3: Check divisibility by LCM
To find the largest four-digit number that is divisible by 540, we need to divide 9999 by 540 and find the largest integer quotient.

**Calculating:**
9999 ÷ 540 ≈ 18.5148

The largest integer less than or equal to 18.5148 is 18.

### Step 4: Multiply back to find the largest four-digit number
Now, we multiply 540 by 18 to find the largest four-digit number that is divisible by 540.

540 × 18 = 9720

### Conclusion
The four-digit number that is completely divisible by 12, 15, 18, and 27 is **9720**.

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