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. 7
B. 5
C. 6
D. 4

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 12, 16, 18, and 21.

To find the least common multiple (LCM), we can use the prime factorization method:

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

Now, we take the highest power of each prime factor:

- 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:

1. \( 16 \times 9 = 144 \)
2. \( 144 \times 7 = 1008 \)

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

### Step 2: Find the smallest number \( x \) such that \( 2000 + x \) is divisible by 1008.

We need to find the smallest \( x \) such that:

\[ 2000 + x \equiv 0 \mod 1008 \]

This means we need to find \( x \) such that:

\[ x \equiv -2000 \mod 1008 \]

First, we calculate \( 2000 \mod 1008 \):

\[ 2000 \div 1008 \approx 1.976 \quad \text{(take the integer part, which is 1)} \]
\[ 1 \times 1008 = 1008 \]
\[ 2000 - 1008 = 992 \]

So, \( 2000 \mod 1008 = 992 \).

Now we need to find \( x \):

\[ x \equiv -992 \mod 1008 \]

To find a positive equivalent, we can add 1008 to -992:

\[ x = 1008 - 992 = 16 \]

### Step 3: Calculate 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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