Consider the nine digit number n = 7 3 `alpha` 4 9 6 1 `beta` 0. The number of ordered pairs `(alpha,beta)` for which the given number is divisible by 88, is

Options

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

Answer

To determine the number of ordered pairs \((\alpha, \beta)\) for which the nine-digit number \(n = 73\alpha4961\beta0\) is divisible by 88, we need to check the divisibility by both 8 and 11, since \(88 = 8 \times 11\).

### Step 1: Check Divisibility by 8
A number is divisible by 8 if its last three digits form a number that is divisible by 8. The last three digits of our number are \(1\beta0\).

We can express the last three digits as \(100 + 10\beta + 1\). Thus, we need to check the values of \(\beta\) such that \(100 + 10\beta + 1\) is divisible by 8.

Calculating the last three digits:
\[ 1\beta0 = 100 + 10\beta + 0 = 100 + 10\beta \]

Now, we will check the possible values of \(\beta\) (which can be from 0 to 9):
- For \(\beta = 0\): \(100\) (divisible by 8)
- For \(\beta = 1\): \(110\) (not divisible by 8)
- For \(\beta = 2\): \(120\) (divisible by 8)
- For \(\beta = 3\): \(130\) (not divisible by 8)
- For \(\beta = 4\): \(140\) (not divisible by 8)
- For \(\beta = 5\): \(150\) (not divisible by 8)
- For \(\beta = 6\): \(160\) (divisible by 8)
- For \(\beta = 7\): \(170\) (not divisible by 8)
- For \(\beta = 8\): \(180\) (not divisible by 8)
- For \(\beta = 9\): \(190\) (not divisible by 8)

The valid values of \(\beta\) that make \(1\beta0\) divisible by 8 are \(\beta = 0, 2, 6\).

### Step 2: Check Divisibility by 11
Next, we need to check the divisibility by 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 a multiple of 11.

The digits in \(n\) are:
- Odd positions: \(7, \alpha, 9, 1, 0\) (sum = \(7 + \alpha + 9 + 1 + 0 = \alpha + 17\))
- Even positions: \(3, 4, 6, \beta\) (sum = \(3 + 4 + 6 + \beta = 13 + \beta\))

Now, we need to find:
\[ (\alpha + 17) - (13 + \beta) = \alpha - \beta + 4 \]
This expression must be a multiple of 11.

### Step 3: Analyze Cases for \(\beta\)
1. **Case \(\beta = 0\)**:
\[ \alpha - 0 + 4 = \alpha + 4 \equiv 0 \mod 11 \]
This gives \(\alpha \equiv -4 \equiv 7 \mod 11\). Thus, \(\alpha = 7\) is valid.

2. **Case \(\beta = 2\)**:
\[ \alpha - 2 + 4 = \alpha + 2 \equiv 0 \mod 11 \]
This gives \(\alpha \equiv -2 \equiv 9 \mod 11\). Thus, \(\alpha = 9\) is valid.

3. **Case \(\beta = 6\)**:
\[ \alpha - 6 + 4 = \alpha - 2 \equiv 0 \mod 11 \]
This gives \(\alpha \equiv 2 \mod 11\). Thus, \(\alpha = 2\) is valid.

### Step 4: Summary of Valid Pairs
The valid pairs \((\alpha, \beta)\) are:
- For \(\beta = 0\): \((7, 0)\)
- For \(\beta = 2\): \((9, 2)\)
- For \(\beta = 6\): \((2, 6)\)

### Conclusion
The number of ordered pairs \((\alpha, \beta)\) for which the number \(n\) is divisible by 88 is **3**.