Let, x be the smallest number, which when added to 2000 makes the resulting number divisible by 12, 16, 18 and 21. The sum of the digits of x is:

Options

A. 6
B. 5
C. 7
D. 8

Answer

To solve the problem step by step, we need to find the smallest number \( x \) such that when added to 2000, the resulting number is divisible by 12, 16, 18, and 21.

### Step 1: Find the LCM of the numbers
We start by finding the least common multiple (LCM) of the numbers 12, 16, 18, and 21.

- **Prime Factorization:**
- \( 12 = 2^2 \times 3^1 \)
- \( 16 = 2^4 \)
- \( 18 = 2^1 \times 3^2 \)
- \( 21 = 3^1 \times 7^1 \)

- **Taking the highest power of each prime:**
- For \( 2 \): the highest power is \( 2^4 \) (from 16)
- For \( 3 \): the highest power is \( 3^2 \) (from 18)
- For \( 7 \): the highest power is \( 7^1 \) (from 21)

Thus, the LCM is:
\[ LCM = 2^4 \times 3^2 \times 7^1 = 16 \times 9 \times 7 \]

Calculating this step by step:
- \( 16 \times 9 = 144 \)
- \( 144 \times 7 = 1008 \)

So, \( LCM(12, 16, 18, 21) = 1008 \).

### Step 2: Find the smallest multiple of LCM greater than 2000
Next, we need to find the smallest multiple of 1008 that is greater than 2000.

- The multiples of 1008 are:
- \( 1008 \times 1 = 1008 \)
- \( 1008 \times 2 = 2016 \) (this is the first multiple greater than 2000)

### Step 3: Set up the equation
Now, we set up the equation based on the problem statement:
\[ 2000 + x = 2016 \]

### Step 4: Solve for \( x \)
Now, we solve for \( x \):
\[ x = 2016 - 2000 = 16 \]

### Step 5: Find the sum of the digits of \( x \)
Finally, we need to find the sum of the digits of \( x \):
- The digits of \( 16 \) are \( 1 \) and \( 6 \).
- The sum is \( 1 + 6 = 7 \).

### Conclusion
Thus, the sum of the digits of \( x \) is \( 7 \).