`19^(n) - 1` is

Options

A. always divisible by 9
B. always divisible by 20
C. is never divisible by 19
D. only (a) and (c ) are true

Answer

To solve the question `19^(n) - 1`, we need to determine the divisibility of this expression by the given options.

### Step-by-Step Solution:

1. **Understanding the Expression**: The expression is `19^n - 1`, where `n` is a natural number. We need to check if this expression is divisible by 9, 20, and 3.

2. **Testing for Divisibility by 9**:
- Let's start with `n = 1`:
\[ 19^1 - 1 = 19 - 1 = 18 \]
- Since \(18\) is divisible by \(9\), we can say that for \(n = 1\), the expression is divisible by \(9\).
- Now, let’s try `n = 2`:
\[ 19^2 - 1 = 361 - 1 = 360 \]
- \(360\) is also divisible by \(9\) (since \(3 + 6 + 0 = 9\)).
- We can generalize this:
- By the property of numbers, \(19^n - 1\) can be expressed as a difference of squares:
\[ 19^n - 1 = (19^{n/2} - 1)(19^{n/2} + 1) \quad \text{(if n is even)} \]
- For any natural number \(n\), the expression will always yield a result that is divisible by \(9\).

3. **Testing for Divisibility by 20**:
- Let’s check if `19^n - 1` is divisible by \(20\):
- For \(n = 1\):
\[ 19^1 - 1 = 18 \]
- \(18\) is not divisible by \(20\).
- Therefore, `19^n - 1` is not always divisible by \(20\).

4. **Testing for Divisibility by 3**:
- For \(n = 1\):
\[ 19^1 - 1 = 18 \]
- \(18\) is divisible by \(3\).
- For \(n = 2\):
\[ 19^2 - 1 = 360 \]
- \(360\) is also divisible by \(3\).
- In general, since \(19 \equiv 1 \mod 3\), we have:
\[ 19^n \equiv 1^n \equiv 1 \mod 3 \implies 19^n - 1 \equiv 0 \mod 3 \]
- Thus, `19^n - 1` is always divisible by \(3\).

5. **Conclusion**:
- From the above checks, we conclude:
- `19^n - 1` is always divisible by \(9\) and \(3\).
- It is not divisible by \(20\).
- Therefore, the correct options are **A (always divisible by 9)** and **C (always divisible by 3)**.

### Final Answer:
The expression `19^n - 1` is always divisible by \(9\) and \(3\).