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 step by step, we will analyze the integer lying between 100 and 1000, subtract the sum of its digits, and determine the result.

### Step 1: Define the three-digit number
Let the three-digit number be represented as:
\[ N = 100x + 10y + z \]
where \(x\), \(y\), and \(z\) are the digits of the number. Here, \(x\) is the digit in the hundreds place, \(y\) is the digit in the tens place, and \(z\) is the digit in the units place. Since the number is between 100 and 1000, \(x\) can range from 1 to 9, while \(y\) and \(z\) can range from 0 to 9.

### 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 need to subtract the sum of the digits from the number:
\[ R = N - S = (100x + 10y + z) - (x + y + z) \]

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

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

### Step 6: Conclusion
Since \(R\) is expressed as \(9(11x + y)\), it is clear that \(R\) is divisible by 9. Therefore, the result of subtracting the sum of the digits from any three-digit number between 100 and 1000 is always divisible by 9.

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

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