If `x^(n)-1` is divisible by x-k, then the least positive integral value of k is:
(i) 1
(ii) 2
(iii) 3
(iv) 4

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the least positive integral value of \( k \) such that \( x^n - 1 \) is divisible by \( x - k \).

### Step-by-Step Solution:

1. **Understanding the Divisibility Condition**:
We know that if \( x^n - 1 \) is divisible by \( x - k \), then \( k \) must be a root of the polynomial \( x^n - 1 \). This means that substituting \( x = k \) into the polynomial should yield zero:
\[ k^n - 1 = 0 \implies k^n = 1 \]

2. **Finding the Roots**:
The equation \( k^n = 1 \) implies that \( k \) must be a root of unity. The roots of unity are the complex numbers that satisfy this equation. However, we are interested in the positive integral values of \( k \).

3. **Identifying Positive Integral Roots**:
The positive integral solutions to \( k^n = 1 \) are limited to \( k = 1 \). This is because:
- For \( n = 1 \), \( k^1 = 1 \) gives \( k = 1 \).
- For \( n = 2 \), \( k^2 = 1 \) gives \( k = 1 \) or \( k = -1 \) (but we only consider positive integers).
- For any \( n > 1 \), \( k^n = 1 \) still only yields \( k = 1 \) as the positive integer solution.

4. **Conclusion**:
Since the only positive integral value of \( k \) that satisfies \( k^n = 1 \) for all \( n \) is \( k = 1 \), we conclude that the least positive integral value of \( k \) is:
\[ \boxed{1} \]