The largest 4-digit number exactly divisible by each of 12, 15, 18 and 27 is:

Options

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

Answer

To find the largest 4-digit number that is exactly divisible by 12, 15, 18, and 27, we will follow these steps:

### Step 1: Find the LCM of the given numbers
To find the largest 4-digit number divisible by 12, 15, 18, and 27, we first need to calculate the least common multiple (LCM) of these numbers.

**Finding the 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:**
- 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 the LCM:**
LCM = 2^2 × 3^3 × 5^1
= 4 × 27 × 5
= 108 × 5
= 540

### Step 2: Find the largest 4-digit number
The largest 4-digit number is 9999.

### Step 3: Divide the largest 4-digit number by the LCM
Now, we need to find how many times 540 fits into 9999.

9999 ÷ 540 ≈ 18.51

### Step 4: Multiply the LCM by the largest integer less than or equal to the result
We take the integer part, which is 18, and multiply it by the LCM.

540 × 18 = 9720

### Step 5: Verify if 9720 is divisible by all the numbers
To ensure that 9720 is divisible by 12, 15, 18, and 27:
- 9720 ÷ 12 = 810 (divisible)
- 9720 ÷ 15 = 648 (divisible)
- 9720 ÷ 18 = 540 (divisible)
- 9720 ÷ 27 = 360 (divisible)

Since 9720 is divisible by all four numbers, it is confirmed as the answer.

### Final Answer
The largest 4-digit number exactly divisible by 12, 15, 18, and 27 is **9720**.

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