Show that any number of the form 4^(n),n ne N` can never end with the digit 0.

Answer

To show that any number of the form \( 4^n \) (where \( n \) is a natural number) can never end with the digit 0, we will analyze the last digits of \( 4^n \) for various values of \( n \).

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to prove that \( 4^n \) does not end with the digit 0 for any natural number \( n \).

2. **Calculating \( 4^n \) for Small Values of \( n \)**:
- For \( n = 1 \):
\[ 4^1 = 4 \]
The last digit is 4.
- For \( n = 2 \):
\[ 4^2 = 16 \]
The last digit is 6.
- For \( n = 3 \):
\[ 4^3 = 64 \]
The last digit is 4.
- For \( n = 4 \):
\[ 4^4 = 256 \]
The last digit is 6.

3. **Identifying a Pattern**: From the calculations above, we can observe:
- When \( n \) is odd (1, 3, ...), \( 4^n \) ends with the digit 4.
- When \( n \) is even (2, 4, ...), \( 4^n \) ends with the digit 6.

4. **Generalizing the Pattern**:
- The last digit of \( 4^n \) alternates based on whether \( n \) is odd or even:
- If \( n \) is odd: Last digit is 4.
- If \( n \) is even: Last digit is 6.

5. **Conclusion**: Since the last digit of \( 4^n \) is either 4 or 6 for all natural numbers \( n \), it can never be 0.

### Final Statement:
Thus, we conclude that any number of the form \( 4^n \) (where \( n \) is a natural number) can never end with the digit 0.

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