`(a + 1)(b-1) = 625, (a != b) in I^(+)` then the value of `(a +b)` is

Options

A. a.`a +b ge 25`
B. b. `a +b ge 50`
C. c. `a + b = 24`
D. d. `a + b = 26`

Answer

To solve the equation \((a + 1)(b - 1) = 625\) with the condition \(a \neq b\), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ (a + 1)(b - 1) = 625 \]

### Step 2: Factor 625
Next, we need to find pairs of factors of 625. The factors of 625 are:
\[ 1 \times 625, \quad 5 \times 125, \quad 25 \times 25 \]
However, since \(a\) must not equal \(b\), we will only consider the pairs where \(a + 1\) and \(b - 1\) are different.

### Step 3: Test Factor Pairs
We can express \(a + 1\) and \(b - 1\) using the factor pairs:
1. If \(a + 1 = 25\) and \(b - 1 = 25\), then \(a = 24\) and \(b = 26\) (valid since \(a \neq b\)).
2. If \(a + 1 = 5\) and \(b - 1 = 125\), then \(a = 4\) and \(b = 126\) (valid since \(a \neq b\)).
3. If \(a + 1 = 125\) and \(b - 1 = 5\), then \(a = 124\) and \(b = 6\) (valid since \(a \neq b\)).
4. If \(a + 1 = 1\) and \(b - 1 = 625\), then \(a = 0\) and \(b = 626\) (valid since \(a \neq b\)).
5. If \(a + 1 = 625\) and \(b - 1 = 1\), then \(a = 624\) and \(b = 2\) (valid since \(a \neq b\)).

### Step 4: Calculate \(a + b\)
Now we calculate \(a + b\) for the valid pairs:
1. For \(a = 24\) and \(b = 26\):
\[ a + b = 24 + 26 = 50 \]
2. For \(a = 4\) and \(b = 126\):
\[ a + b = 4 + 126 = 130 \]
3. For \(a = 124\) and \(b = 6\):
\[ a + b = 124 + 6 = 130 \]
4. For \(a = 0\) and \(b = 626\):
\[ a + b = 0 + 626 = 626 \]
5. For \(a = 624\) and \(b = 2\):
\[ a + b = 624 + 2 = 626 \]

### Step 5: Conclusion
The minimum value of \(a + b\) occurs when \(a = 24\) and \(b = 26\), giving us:
\[ a + b = 50 \]

Thus, the final answer is:
\[ \boxed{50} \]