The greatest number which always divides the product of the predecessor and successor of an odd natural number other than 1, is

Options

A. 6
B. 4
C. 16
D. 8

Answer

To solve the question, we need to find the greatest number that always divides the product of the predecessor and successor of any odd natural number greater than 1.

Let's break this down step by step:

### Step 1: Define the Odd Natural Number
Let’s denote an odd natural number as \( n \). The odd natural numbers greater than 1 are 3, 5, 7, 9, 11, etc.

### Step 2: Identify the Predecessor and Successor
For any odd natural number \( n \):
- The predecessor (the number before \( n \)) is \( n - 1 \).
- The successor (the number after \( n \)) is \( n + 1 \).

### Step 3: Calculate the Product
The product of the predecessor and successor can be expressed as:
\[ (n - 1)(n + 1) \]
This can be simplified using the difference of squares:
\[ (n - 1)(n + 1) = n^2 - 1 \]

### Step 4: Analyze the Expression
Now we need to determine what number always divides \( n^2 - 1 \) for any odd natural number \( n \):
- Since \( n \) is odd, \( n^2 \) will also be odd (as the square of an odd number is odd).
- Therefore, \( n^2 - 1 \) will be even (since an odd number minus one is even).

### Step 5: Factor \( n^2 - 1 \)
The expression \( n^2 - 1 \) can be factored as:
\[ n^2 - 1 = (n - 1)(n + 1) \]
Both \( n - 1 \) and \( n + 1 \) are consecutive even numbers, meaning one of them is divisible by 2 and the other is divisible by 4.

### Step 6: Find the Greatest Common Divisor
Since one of the two consecutive even numbers is divisible by 4, the product \( (n - 1)(n + 1) \) is always divisible by 8.

### Conclusion
Thus, the greatest number that always divides the product of the predecessor and successor of any odd natural number greater than 1 is:
\[ \boxed{8} \]