Answer :
To show that \( 15n \) cannot end with the digits 0, 2, 4, 6, or 8 for any natural number \( n \), we need to consider the properties of \( 15n \).
Firstly, let's express \( 15n \):
\[ 15n = 3 \times 5 \times n = 5 \times (3n) \]
This tells us that \( 15n \) is always divisible by 5, regardless of \( n \). For \( 15n \) to end with the digits 0, 2, 4, 6, or 8, it must be divisible by 2 (even number) or end in 0 (multiple of 10).
Now, let's consider the possible cases:
1. **Ending with 0**: If \( 15n \) ends with 0, then \( n \) itself must end with 0 because \( 15n \) is already divisible by 5. However, this contradicts the condition that \( n \) cannot end with 0, 2, 4, 6, or 8.
2. **Ending with 2, 4, 6, or 8**: For \( 15n \) to end with 2, 4, 6, or 8, \( n \) would have to end with 2, 4, 6, or 8, respectively. But again, this contradicts the condition that \( n \) cannot end with these digits.
Therefore, based on these considerations, \( 15n \) cannot end with the digits 0, 2, 4, 6, or 8 for any natural number \( n \). This conclusion follows directly from the fact that \( 15n \) is always divisible by 5 and cannot simultaneously satisfy the condition of ending with these specific digits.