Consider the number `N = 8 7 a 2 7 9 3 1 b` , where a , b are single digit whole numbers.
Number of values of a + b for which N is divisible by 11 is

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To determine the number of values of \( a + b \) for which the number \( N = 87a27931b \) is divisible by 11, we will use the divisibility rule for 11. According to this rule, a number is divisible by 11 if the difference between the sum of its digits in odd positions and the sum of its digits in even positions is either 0 or divisible by 11.

### Step-by-Step Solution:

1. **Identify the positions of the digits:**
- The digits in \( N \) are positioned as follows:
- Odd positions: 8, a, 2, 9, b (1st, 3rd, 5th, 7th, 9th)
- Even positions: 7, 2, 7, 3, 1 (2nd, 4th, 6th, 8th)

2. **Calculate the sum of the digits in odd positions:**
\[ \text{Sum of odd positions} = 8 + a + 2 + 9 + b = 19 + a + b \]

3. **Calculate the sum of the digits in even positions:**
\[ \text{Sum of even positions} = 7 + 2 + 7 + 3 + 1 = 20 \]

4. **Set up the divisibility condition:**
The difference between the sums must satisfy:
\[ |(19 + a + b) - 20| = |a + b - 1| \]
This difference must be divisible by 11. Therefore, we have:
\[ |a + b - 1| \equiv 0 \mod{11} \]

5. **Solve the equation:**
This gives us two cases:
- Case 1: \( a + b - 1 = 0 \) → \( a + b = 1 \)
- Case 2: \( a + b - 1 = 11 \) → \( a + b = 12 \)
- Case 3: \( a + b - 1 = -11 \) → \( a + b = -10 \) (not possible since \( a \) and \( b \) are non-negative)

6. **Determine valid values for \( a + b \):**
- For \( a + b = 1 \):
- Possible pairs: \( (0, 1), (1, 0) \) → 2 valid combinations.
- For \( a + b = 12 \):
- Possible pairs: \( (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3) \) → 7 valid combinations.

7. **Count the total valid combinations:**
- Total combinations = 2 (from \( a + b = 1 \)) + 7 (from \( a + b = 12 \)) = 9.

Thus, the number of values of \( a + b \) for which \( N \) is divisible by 11 is **9**.