Both the end digits of a 99 digit number N are 2. N is divisible by 11, then all the middle digits are :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the middle digits of a 99-digit number \( N \) that starts and ends with the digit 2 and is divisible by 11.

### Step-by-Step Solution:

1. **Understanding the Structure of the Number**:
The number \( N \) has 99 digits, and both the first and last digits are 2. Thus, we can represent \( N \) as:
\[ N = 2, x_1, x_2, x_3, \ldots, x_{97}, 2 \]
where \( x_1, x_2, \ldots, x_{97} \) are the middle digits.

2. **Applying the Divisibility Rule for 11**:
A number is divisible by 11 if the difference between the sum of the digits at odd positions and the sum of the digits at even positions is either 0 or divisible by 11.
- Odd-positioned digits: \( 2, x_1, x_3, \ldots, x_{97} \)
- Even-positioned digits: \( x_2, x_4, \ldots, 2 \)

3. **Counting the Positions**:
- There are 50 odd positions (including the first and last digit).
- There are 49 even positions (from the second digit to the second last digit).

4. **Calculating the Sums**:
- Let \( S_{odd} \) be the sum of the digits at odd positions:
\[ S_{odd} = 2 + x_1 + x_3 + \ldots + x_{97} \]
- Let \( S_{even} \) be the sum of the digits at even positions:
\[ S_{even} = x_2 + x_4 + \ldots + 2 \]

5. **Setting Up the Equation**:
The condition for divisibility by 11 gives us:
\[ S_{odd} - S_{even} = (2 + x_1 + x_3 + \ldots + x_{97}) - (x_2 + x_4 + \ldots + 2) \]
Simplifying this, we get:
\[ S_{odd} - S_{even} = (x_1 + x_3 + \ldots + x_{97}) - (x_2 + x_4 + \ldots) \]

6. **Finding the Total Number of Digits**:
Since there are 99 digits in total, we can express the total sum of all digits as:
\[ S_{total} = 2 + (x_1 + x_2 + x_3 + \ldots + x_{97}) + 2 = 4 + (x_1 + x_2 + \ldots + x_{97}) \]

7. **Conclusion**:
To satisfy the divisibility condition, we need to ensure that the sums of the odd and even positioned digits balance out correctly. Given that both ends are 2, the middle digits must be arranged such that the overall difference is 0 or divisible by 11.

After analyzing the structure and the requirement for divisibility, we can conclude that all middle digits \( x_1, x_2, \ldots, x_{97} \) must be 2 to maintain the balance:
\[ x_1 = x_2 = x_3 = \ldots = x_{97} = 2 \]

### Final Answer:
All the middle digits are 2.