If the sum of the digits of a three digit number is subtracted from that number, then it will always be divisible by

Options

A. a) 3 only
B. b) 9 only
C. c) Both 3 and 9
D. d) All of 3,6 and 9

Answer

To solve the problem, we need to analyze the three-digit number and its properties. Let's break it down step by step.

### Step 1: Define the Three-Digit Number
Let the three-digit number be represented as \( abc \), where:
- \( a \) is the hundreds digit,
- \( b \) is the tens digit,
- \( c \) is the units digit.

### Step 2: Express the Number in Terms of Digits
The three-digit number can be expressed mathematically as:
\[ N = 100a + 10b + c \]

### Step 3: Calculate the Sum of the Digits
The sum of the digits of the number \( abc \) is:
\[ S = a + b + c \]

### Step 4: Subtract the Sum from the Number
According to the problem, we need to subtract the sum of the digits from the number:
\[ D = N - S = (100a + 10b + c) - (a + b + c) \]

### Step 5: Simplify the Expression
Now, let's simplify the expression for \( D \):
\[ D = (100a + 10b + c) - (a + b + c) = 100a + 10b + c - a - b - c \]
This simplifies to:
\[ D = (100a - a) + (10b - b) + (c - c) = 99a + 9b \]

### Step 6: Factor Out Common Terms
We can factor out the common term from \( D \):
\[ D = 9(11a + b) \]

### Step 7: Conclusion about Divisibility
Since \( D \) is expressed as \( 9(11a + b) \), it is clear that \( D \) is divisible by \( 9 \). Additionally, since \( 9 \) is a multiple of \( 3 \), \( D \) is also divisible by \( 3 \).

### Final Answer
Thus, if the sum of the digits of a three-digit number is subtracted from that number, the result will always be divisible by **9** (and also by 3).

---