If the sum of the digits of any Integer between 100 and 1000 is subtracted from the same integer, the resulting number is always divisible by

Options

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

Answer

To solve the problem, we need to analyze the expression formed when we subtract the sum of the digits of a three-digit integer from the integer itself. Let's break it down step by step.

### Step-by-Step Solution:

1. **Define the three-digit integer**:
Let the three-digit integer be represented as \( N \). We can express \( N \) in terms of its digits:
\[ N = 100x + 10y + z \]
where \( x \), \( y \), and \( z \) are the digits in the hundreds, tens, and units place, respectively. Here, \( x \) can range from 1 to 9 (since \( N \) is a three-digit number), and \( y \) and \( z \) can range from 0 to 9.

2. **Calculate the sum of the digits**:
The sum of the digits of \( N \) is:
\[ S = x + y + z \]

3. **Subtract the sum of the digits from the integer**:
We need to find \( N - S \):
\[ N - S = (100x + 10y + z) - (x + y + z) \]

4. **Simplify the expression**:
Now, simplifying the expression:
\[ N - S = 100x + 10y + z - x - y - z \]
\[ = (100x - x) + (10y - y) + (z - z) \]
\[ = 99x + 9y \]

5. **Factor out common terms**:
We can factor out 9 from the expression:
\[ N - S = 9(11x + y) \]

6. **Conclusion**:
Since \( N - S \) can be expressed as \( 9(11x + y) \), it is clear that \( N - S \) is always divisible by 9.

### Final Answer:
The resulting number \( N - S \) is always divisible by **9**.

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