A number 'x' is added to 2000. The resultant sum is completely divisible by 12,16,18 & 21. Find the sum of digits of 'x'?

Options

A. 7
B. 5
C. 6
D. 4

Answer

To solve the problem step by step, we need to find a number 'x' such that when added to 2000, the result is divisible by 12, 16, 18, and 21.

### Step 1: Find the LCM of 12, 16, 18, and 21
To determine the least common multiple (LCM), we first find the prime factorization of each number:

- **12** = 2^2 * 3^1
- **16** = 2^4
- **18** = 2^1 * 3^2
- **21** = 3^1 * 7^1

Next, we take the highest power of each prime number that appears in these factorizations:

- 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).

Now, we can calculate the LCM:

\[ \text{LCM} = 2^4 \times 3^2 \times 7^1 = 16 \times 9 \times 7 \]

Calculating this step by step:

- First, calculate \(16 \times 9 = 144\).
- Then, calculate \(144 \times 7 = 1008\).

Thus, the LCM of 12, 16, 18, and 21 is **1008**.

### Step 2: Determine the value of 'x'
We need \(2000 + x\) to be a multiple of 1008. This means we can express it as:

\[ 2000 + x = k \times 1008 \]

for some integer \(k\). Rearranging gives us:

\[ x = k \times 1008 - 2000 \]

### Step 3: Find the smallest \(k\) such that \(x\) is non-negative
We need \(k \times 1008\) to be greater than 2000.

Calculating the smallest \(k\):

\[ k \geq \frac{2000}{1008} \approx 1.976 \]

Thus, the smallest integer \(k\) is 2.

### Step 4: Calculate \(x\) using \(k = 2\)
Now substituting \(k = 2\):

\[ x = 2 \times 1008 - 2000 = 2016 - 2000 = 16 \]

### Step 5: Find the sum of the digits of \(x\)
Now we need to find the sum of the digits of \(x = 16\):

\[ 1 + 6 = 7 \]

### Final Answer
The sum of the digits of \(x\) is **7**.

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