The greatest number which always divides the product of the predecessor and successor of an odd natural number \((\neq 1)\) is
6
4
16
8
Step 1: Write the odd number.
Let the odd number be \(2m+1\) where \(m \ge 1\).
Step 2: Find predecessor and successor.
Predecessor: \(2m+1 - 1 = 2m\).
Successor: \(2m+1 + 1 = 2m+2\).
Step 3: Multiply them.
Product \(= (2m)\times(2m+2)\).
Step 4: Factor the product.
\((2m)\times(2m+2) = 2\times 2 \times m \times (m+1)\).
So, \(\;\;= 4\,m\,(m+1)\).
Step 5: Use the idea of consecutive numbers.
Among \(m\) and \(m+1\), one is even.
Therefore, \(m\,(m+1)\) has at least one extra factor \(2\).
Step 6: Count the guaranteed factors of 2.
From \(4\) we already have two 2's: \(2\times 2\).
From \(m\,(m+1)\) we get at least one more \(2\).
So the product has at least \(2\times 2\times 2 = 8\) as a factor.
Step 7: Check that 16 is not always a factor.
Take \(2m+1=3\) (odd). Then predecessor \(=2\), successor \(=4\).
Product \(=2\times 4=8\), which is not divisible by \(16\).
Conclusion: The greatest number that always divides the product is \(\boxed{8}\).
Correct option: D