If the sum of digits of the number `N = 2000^11 -2011` is `S,` then

Options

A. S is a prime numbers
B. Sum of digits of S is 10
C. `(S+1)` is divisible by exactly 3 prime number
D. S is a composite numbers

Answer

To solve the problem, we need to find the sum of the digits of the number \( N = 2000^{11} - 2011 \).

### Step 1: Rewrite the expression for \( N \)
We can express \( 2000 \) as \( 2 \times 10^3 \). Therefore:
\[ N = (2 \times 10^3)^{11} - 2011 = 2^{11} \times 10^{33} - 2011 \]

### Step 2: Calculate \( 2^{11} \)
Calculating \( 2^{11} \):
\[ 2^{11} = 2048 \]
Thus, we have:
\[ N = 2048 \times 10^{33} - 2011 \]

### Step 3: Understand the structure of \( N \)
The term \( 2048 \times 10^{33} \) represents the number \( 2048 \) followed by \( 33 \) zeros:
\[ N = 204800000000000000000000000000000000000 - 2011 \]

### Step 4: Perform the subtraction
Now, we subtract \( 2011 \) from \( 204800000000000000000000000000000000000 \):
\[ N = 204799999999999999999999999999999999789 \]
This means \( N \) will have \( 33 \) nines followed by \( 789 \) at the end.

### Step 5: Identify the digits of \( N \)
The digits of \( N \) are:
- 2 (from 2048)
- 0 (from 2048)
- 4 (from 2048)
- 7 (from the subtraction)
- 9 (thirty times)
- 8 (from the subtraction)
- 9 (from the subtraction)

### Step 6: Calculate the sum of the digits
Now, we sum these digits:
\[ S = 2 + 0 + 4 + 7 + (9 \times 30) + 8 + 9 \]
Calculating this step-by-step:
1. \( 2 + 0 = 2 \)
2. \( 2 + 4 = 6 \)
3. \( 6 + 7 = 13 \)
4. \( 13 + (9 \times 30) = 13 + 270 = 283 \)
5. \( 283 + 8 = 291 \)
6. \( 291 + 9 = 300 \)

Thus, the sum of the digits \( S \) is:
\[ S = 300 \]

### Step 7: Final answer
The sum of the digits of the number \( N \) is \( S = 300 \).