Find the largest four digit number which is divisible by 12, 15, 18 and 27 ?

Options

A. 9690
B. 9720
C. 9980
D. 9960

Answer

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

### Step 1: Find the LCM of the numbers
To find the largest four-digit number divisible by 12, 15, 18, and 27, we first need to calculate the Least Common Multiple (LCM) of these numbers.

**Prime Factorization:**
- 12 = 2² × 3¹
- 15 = 3¹ × 5¹
- 18 = 2¹ × 3²
- 27 = 3³

**Finding LCM:**
- Take the highest power of each prime factor:
- For 2: highest power is 2² (from 12)
- For 3: highest power is 3³ (from 27)
- For 5: highest power is 5¹ (from 15)

Thus, the LCM is:
\[ \text{LCM} = 2^2 \times 3^3 \times 5^1 = 4 \times 27 \times 5 \]

Calculating this step-by-step:
- \( 4 \times 27 = 108 \)
- \( 108 \times 5 = 540 \)

So, the LCM of 12, 15, 18, and 27 is **540**.

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

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

\[ 9999 \div 540 \approx 18.5148 \]

Taking the integer part, we find that 540 fits into 9999 a total of **18** times.

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

\[ 18 \times 540 = 9720 \]

### Step 5: Conclusion
Thus, the largest four-digit number that is divisible by 12, 15, 18, and 27 is **9720**.

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