If a is an even positive integer and b is an odd positive integer, then which of the following statement is true?

Options

A. a (b − 1) is even
B. a (b - 1) is odd
C. (a - 1) (0 - 1) is even
D. (a - 1) b is even

Answer

To solve the problem, we need to analyze the statements given the conditions that \( a \) is an even positive integer and \( b \) is an odd positive integer.

### Step-by-Step Solution:

1. **Define the Variables**:
- Let \( a = 2n \) where \( n \) is a positive integer (since \( a \) is an even positive integer).
- Let \( b = 2m + 1 \) where \( m \) is a non-negative integer (since \( b \) is an odd positive integer).

2. **Evaluate the Statements**:
We need to evaluate the truth of the following statements:
- **Statement 1**: \( a(b - 1) \)
- **Statement 2**: \( ab - 1 \)
- **Statement 3**: \( a - 1 \) and \( b - 1 \)
- **Statement 4**: \( (a - 1)b \)

3. **Calculate Each Statement**:
- **Statement 1**:
\[ b - 1 = (2m + 1) - 1 = 2m \quad \text{(which is even)} \]
\[ a(b - 1) = (2n)(2m) = 4nm \quad \text{(which is even)} \]

- **Statement 2**:
\[ ab - 1 = (2n)(2m + 1) - 1 = 4nm + 2n - 1 \]
Since \( 4nm + 2n \) is even, subtracting 1 gives us an odd number.

- **Statement 3**:
\[ a - 1 = 2n - 1 \quad \text{(which is odd)} \]
\[ b - 1 = 2m \quad \text{(which is even)} \]
The product of an odd and an even number is even.

- **Statement 4**:
\[ (a - 1)b = (2n - 1)(2m + 1) \]
The product of two odd numbers is odd, hence this expression is odd.

4. **Conclusion**:
- **Statement 1** is true (even).
- **Statement 2** is false (odd).
- **Statement 3** is false (even).
- **Statement 4** is false (odd).

Thus, the only true statement is **Statement 1**: \( a(b - 1) \) is even.