Can the number `6^(n),n` being a natural number, end with the digit 5? Give reason.

Answer

To determine if the number \( 6^n \) (where \( n \) is a natural number) can end with the digit 5, we will analyze the prime factorization of \( 6^n \) and the conditions required for a number to end with the digit 5.

### Step-by-Step Solution:

1. **Understanding the Prime Factorization of 6:**
\[ 6 = 2 \times 3 \]
Therefore, we can express \( 6^n \) as:
\[ 6^n = (2 \times 3)^n = 2^n \times 3^n \]

2. **Identifying the Last Digit of a Number:**
For a number to end with the digit 5, it must have at least one factor of 5 in its prime factorization. Additionally, it should not have any factor of 2, because if a number has both 2 and 5 as factors, it will end with 0 (since \( 2 \times 5 = 10 \)).

3. **Analyzing the Factors of \( 6^n \):**
From the expression \( 6^n = 2^n \times 3^n \):
- We see that \( 6^n \) contains \( 2^n \), which means it always has a factor of 2.
- There is no factor of 5 in \( 6^n \).

4. **Conclusion:**
Since \( 6^n \) does not contain a factor of 5 and contains a factor of 2, it is impossible for \( 6^n \) to end with the digit 5.

Thus, we conclude that **the number \( 6^n \) cannot end with the digit 5**.