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

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 an integer between 100 and 1000, its digits, and the result of subtracting the sum of its digits from the integer itself.

### Step-by-Step Solution:

1. **Define the integer**: Let the integer be represented as \( N \), where \( N \) can be expressed in terms of its digits. For a three-digit number, we can write:
\[ N = 100a + 10b + c \]
Here, \( a \), \( b \), and \( c \) are the digits of the number \( N \), with \( a \) being the hundreds digit, \( b \) the tens digit, and \( c \) the units digit. The digit \( a \) can take values from 1 to 9 (since \( N \) is between 100 and 1000), while \( b \) and \( c \) can take values from 0 to 9.

2. **Calculate the sum of the digits**: The sum of the digits of \( N \) is given by:
\[ S = a + b + c \]

3. **Subtract the sum of the digits from the number**: We need to find the expression for \( N - S \):
\[ N - S = (100a + 10b + c) - (a + b + c) \]
Simplifying this gives:
\[ N - S = 100a + 10b + c - a - b - c \]
\[ N - S = (100a - a) + (10b - b) + (c - c) \]
\[ N - S = 99a + 9b \]

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

5. **Determine divisibility**: Since \( N - S \) is expressed as \( 9(11a + b) \), it is clear that \( N - S \) is divisible by 9.

### Conclusion:
Thus, the result of subtracting the sum of the digits from any integer between 100 and 1000 is always divisible by 9.