I multiplied a natural number by 18 and another by 21 and added the products. Which one of the following could be the sum ?

Options

A. 2007
B. 2008
C. 2006
D. 2002

Answer

To solve the problem, we need to find the possible sum of the products of two natural numbers multiplied by 18 and 21 respectively. Let's break it down step by step.

### Step-by-Step Solution:

1. **Define the Natural Numbers:**
Let the natural numbers be \( x \) and \( y \).

2. **Formulate the Expression:**
According to the problem, we multiply one natural number by 18 and the other by 21. Therefore, the expression for the sum of the products can be written as:
\[ S = 18x + 21y \]

3. **Factor the Expression:**
We can factor out the common factor from the expression:
\[ S = 3(6x + 7y) \]
This shows that \( S \) is divisible by 3.

4. **Check Divisibility:**
Since \( S \) is divisible by 3, any possible sum must also be divisible by 3. We need to check the options given in the question to see which one is divisible by 3.

5. **Evaluate Options:**
Let's evaluate the options:
- If the options are, for example, 2004, 2005, 2006, 2007, and 2008:
- 2004: \( 2004 \div 3 = 668 \) (divisible)
- 2005: \( 2005 \div 3 = 668.33 \) (not divisible)
- 2006: \( 2006 \div 3 = 668.67 \) (not divisible)
- 2007: \( 2007 \div 3 = 669 \) (divisible)
- 2008: \( 2008 \div 3 = 669.33 \) (not divisible)

6. **Conclusion:**
From the evaluation, the possible sums that could result from the products \( 18x + 21y \) are those that are divisible by 3. In this case, both 2004 and 2007 are valid sums.

### Final Answer:
Thus, one of the possible sums could be **2007**.