If n is an odd natural number, then `n(n^(2)-1)` is divisible by:
(i) 24
(ii) 48
(iii) 120
(iv) 64

Options

A. 24
B. 48
C. 120
D. 64

Answer

To determine if \( n(n^2 - 1) \) is divisible by 24, 48, 120, or 64 when \( n \) is an odd natural number, we can follow these steps:

### Step 1: Express \( n \) as an odd number
An odd natural number can be expressed in the form:
\[ n = 2k + 1 \quad \text{for some integer } k \]

### Step 2: Substitute \( n \) into the expression
We need to evaluate:
\[ n(n^2 - 1) = n(n - 1)(n + 1) \]
Since \( n \) is odd, both \( n - 1 \) and \( n + 1 \) are even numbers. Specifically:
- \( n - 1 = 2k \) (even)
- \( n + 1 = 2k + 2 = 2(k + 1) \) (even)

Thus, we can write:
\[ n(n^2 - 1) = n \cdot (n - 1) \cdot (n + 1) = (2k + 1) \cdot (2k) \cdot (2(k + 1)) \]

### Step 3: Factor out the even components
We can factor out the 2's:
\[ = (2k + 1) \cdot 2k \cdot 2(k + 1) = 4k(2k + 1)(k + 1) \]

### Step 4: Analyze divisibility by 24
To check if this expression is divisible by 24, we note that:
- \( 4k \) contributes a factor of 4.
- \( 2k + 1 \) is odd and does not contribute any additional factors of 2.
- \( k + 1 \) can be either odd or even.

For \( n(n^2 - 1) \) to be divisible by 24, we need at least:
- \( 3 \) factors of \( 2 \) (which gives us \( 8 \))
- \( 3 \) factors of \( 3 \)

### Step 5: Check for factors of 3
Among \( n - 1 \), \( n \), and \( n + 1 \):
- One of these three consecutive integers must be divisible by 3.

### Step 6: Conclusion
Since we have established that:
- \( n(n^2 - 1) \) is divisible by \( 8 \) (from the \( 4k \) and \( 2(k + 1) \))
- One of \( n - 1, n, n + 1 \) is divisible by \( 3 \)

Thus, \( n(n^2 - 1) \) is divisible by:
\[ 8 \times 3 = 24 \]

### Final Answer
The expression \( n(n^2 - 1) \) is divisible by \( \boxed{24} \).