The greatest four digit numbers which is exactly divisible by each one of the numbers 12, 18, 21 and 28 is

Options

A. 9828
B. 9288
C. 9882
D. 9928

Answer

To find the greatest four-digit number that is exactly divisible by 12, 18, 21, and 28, we will follow these steps:

### Step 1: Find the LCM of the numbers
To determine the greatest four-digit number that is divisible by 12, 18, 21, and 28, we first need to find the least common multiple (LCM) of these numbers.

1. **Prime Factorization**:
- 12 = 2^2 × 3^1
- 18 = 2^1 × 3^2
- 21 = 3^1 × 7^1
- 28 = 2^2 × 7^1

2. **Determine the LCM**:
- Take the highest power of each prime factor:
- 2^2 (from 12 and 28)
- 3^2 (from 18)
- 7^1 (from 21 and 28)
- Therefore, LCM = 2^2 × 3^2 × 7^1 = 4 × 9 × 7 = 252.

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

### Step 3: Divide the greatest four-digit number by the LCM
Next, we will divide 9999 by 252 to find the largest multiple of 252 that is less than or equal to 9999.

1. **Perform the Division**:
- 9999 ÷ 252 ≈ 39.6825 (we take the integer part, which is 39).

### Step 4: Calculate the largest multiple of LCM
Now, we multiply the integer part by the LCM to find the largest multiple of 252 that is a four-digit number.

1. **Multiply**:
- 39 × 252 = 9828.

### Step 5: Conclusion
Thus, the greatest four-digit number that is exactly divisible by 12, 18, 21, and 28 is **9828**.

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