The sum of the digits of a 3-digit number is subtracted from the number. The resulting number is always divisible by:

Options

A. 2
B. 5
C. 8
D. 9

Answer

To solve the problem, we need to analyze the situation mathematically. Let's break it down step by step.

### Step 1: Define the 3-digit number
Let the 3-digit number be represented as \( xyz \), where \( x \), \( y \), and \( z \) are its digits. We can express this number mathematically as:
\[ N = 100x + 10y + z \]

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

### Step 3: Subtract the sum of the digits from the number
Now, we subtract the sum of the digits from the number:
\[ R = N - S = (100x + 10y + z) - (x + y + z) \]

### Step 4: Simplify the expression
Now, we simplify the expression for \( R \):
\[ R = 100x + 10y + z - x - y - z \]
\[ R = (100x - x) + (10y - y) + (z - z) \]
\[ R = 99x + 9y \]

### Step 5: Factor out the common terms
We can factor out 9 from the expression:
\[ R = 9(11x + y) \]

### Step 6: Conclusion about divisibility
Since \( R \) can be expressed as \( 9(11x + y) \), it is clear that \( R \) is always divisible by 9, regardless of the values of \( x \) and \( y \) (as long as they are digits).

### Final Answer
Thus, the resulting number \( R \) is always divisible by **9**.

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