When the sum of n digits of an n digit number is subtracted from the number itself, where the number must be atleast two digit number, then the correct statement is :

Options

A. the difference is a prime number
B. the resultant value is a perfect square
C. the resultant value is an odd number
D. the resultant value is a multiple of 9

Answer

To solve the problem, we need to analyze the situation where we subtract the sum of the digits of an n-digit number from the number itself. Let's break it down step by step.

### Step 1: Define the n-digit number
Let’s denote the n-digit number as \( N \). For an n-digit number, we can express it in terms of its digits. For example, if \( n = 2 \), the number can be represented as:
\[ N = 10a + b \]
where \( a \) is the digit in the tens place and \( b \) is the digit in the units place.

For a general n-digit number:
\[ N = d_1 \cdot 10^{n-1} + d_2 \cdot 10^{n-2} + \ldots + d_{n-1} \cdot 10^1 + d_n \cdot 10^0 \]
where \( d_1, d_2, \ldots, d_n \) are the digits of the number.

### Step 2: Calculate the sum of the digits
The sum of the digits \( S \) of the number \( N \) is:
\[ S = d_1 + d_2 + \ldots + d_n \]

### Step 3: Subtract the sum from the number
Now, we subtract the sum of the digits from the number:
\[ D = N - S \]

### Step 4: Analyze the result
From the representation of \( N \), we can see that when we subtract \( S \) from \( N \), the resulting number \( D \) can be simplified.

For example, if we consider the case of a two-digit number:
\[ D = (10a + b) - (a + b) = 9a \]
This shows that \( D \) is a multiple of 9.

For a three-digit number:
\[ N = 100a + 10b + c \]
\[ S = a + b + c \]
\[ D = (100a + 10b + c) - (a + b + c) = 99a + 9b = 9(11a + b) \]
Again, \( D \) is a multiple of 9.

### Step 5: Generalize the result
Continuing this pattern for an n-digit number, we can conclude that:
\[ D = N - S = 9 \times \text{(some integer)} \]
Thus, \( D \) is always a multiple of 9.

### Conclusion
The statement we can conclude from this analysis is that when the sum of the digits of an n-digit number is subtracted from the number itself, the result is always a multiple of 9.