Let $ t_{k} \left(n\right)$ show the sum of $ k^{th}$ power of digits of positive number $ n$. For which $ k$, the condition that $ t_{k} \left(n\right)$ is a multiple of 3 does not imply the condition that $ n$ is a multiple of 3?

$\textbf{(A)}\ 3  \qquad\textbf{(B)}\ 6  \qquad\textbf{(C)}\ 9  \qquad\textbf{(D)}\ 15  \qquad\textbf{(E)}\ \text{None}$
1. We need to determine for which \( k \) the condition that \( t_k(n) \) is a multiple of 3 does not imply that \( n \) is a multiple of 3. Here, \( t_k(n) \) represents the sum of the \( k \)-th powers of the digits of \( n \).

2. First, observe the behavior of \( a^k \mod 3 \) for different values of \( k \). We know that:
   - \( a \equiv 0 \mod 3 \implies a^k \equiv 0 \mod 3 \)
   - \( a \equiv 1 \mod 3 \implies a^k \equiv 1 \mod 3 \)
   - \( a \equiv 2 \mod 3 \implies a^k \equiv 2^k \mod 3 \)

3. For \( k = 3 \):
   - \( 2^3 = 8 \equiv 2 \mod 3 \)
   - Therefore, \( a^3 \equiv a \mod 3 \) for any digit \( a \).

4. In general, for any odd \( k \), \( a^{2k+1} \equiv a \mod 3 \). This is because:
   - \( 2^{2k+1} \equiv 2 \mod 3 \)
   - Hence, \( a^{2k+1} \equiv a \mod 3 \)

5. Now, consider even \( k \). For \( k = 6 \):
   - \( 2^6 = 64 \equiv 1 \mod 3 \)
   - Therefore, \( a^6 \equiv 1 \mod 3 \) if \( a \equiv 2 \mod 3 \)

6. To verify, let's take an example:
   - Consider \( n = 221 \)
   - The digits are 2, 2, and 1.
   - Calculate \( t_6(221) \):
     \[
     t_6(221) = 2^6 + 2^6 + 1^6 = 64 + 64 + 1 = 129
     \]
   - Check \( 129 \mod 3 \):
     \[
     129 \equiv 0 \mod 3
     \]
   - However, \( 221 \mod 3 \):
     \[
     221 \equiv 2 \mod 3
     \]

7. This shows that \( t_6(n) \equiv 0 \mod 3 \) does not imply \( n \equiv 0 \mod 3 \).

Conclusion:
The value of \( k \) for which \( t_k(n) \) being a multiple of 3 does not imply \( n \) is a multiple of 3 is \( k = 6 \).

The final answer is \(\boxed{6}\)