If the sum of the digits of any integer lying between 100 and 1000 is subtracted from the number, the result always is

Options

A. divisible by 6
B. divisible by 2
C. divisible by 9
D. divisible by 5

Answer

To solve the problem, we need to analyze the relationship between a three-digit number and the sum of its digits. Let's break it down step by step.

### Step-by-Step Solution:

1. **Define the Three-Digit Number**:
Let the three-digit number be represented as \( xyz \), where \( x \), \( y \), and \( z \) are the digits in the hundreds, tens, and units places respectively. Therefore, the value of the number can be expressed as:
\[ \text{Number} = 100x + 10y + z \]

2. **Calculate the Sum of the Digits**:
The sum of the digits of the number \( xyz \) is:
\[ \text{Sum of digits} = x + y + z \]

3. **Subtract the Sum of the Digits from the Number**:
We need to find the result of subtracting the sum of the digits from the number:
\[ \text{Result} = (100x + 10y + z) - (x + y + z) \]

4. **Simplify the Expression**:
Simplifying the result gives:
\[ \text{Result} = 100x + 10y + z - x - y - z \]
\[ = (100x - x) + (10y - y) + (z - z) \]
\[ = 99x + 9y \]

5. **Factor the Result**:
We can factor out the common term from the result:
\[ \text{Result} = 9(11x + y) \]

6. **Conclusion**:
Since \( 9(11x + y) \) is clearly divisible by 9, we conclude that the result of subtracting the sum of the digits from the number is always divisible by 9.

### Final Answer:
The result is always divisible by **9**.