Which one statisfies $n^{29} \equiv 7 \pmod {65}$?

$ \textbf{(A)}\ 37 \qquad \textbf{(B)}\ 39 \qquad \textbf{(C)}\ 43 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 55$
1. **Apply the Chinese Remainder Theorem**: We need to solve \( n^{29} \equiv 7 \pmod{65} \). Since \( 65 = 5 \times 13 \), we can use the Chinese Remainder Theorem to break this into two congruences:
   \[
   n^{29} \equiv 7 \pmod{5} \quad \text{and} \quad n^{29} \equiv 7 \pmod{13}
   \]

2. **Check modulo 5**: We need to find \( n \) such that \( n^{29} \equiv 2 \pmod{5} \) because \( 7 \equiv 2 \pmod{5} \).
   - For \( n = 37 \):
     \[
     37 \equiv 2 \pmod{5} \quad \Rightarrow \quad 37^{29} \equiv 2^{29} \pmod{5}
     \]
     Since \( 2^{4} \equiv 1 \pmod{5} \), we have:
     \[
     2^{29} = 2^{4 \cdot 7 + 1} = (2^4)^7 \cdot 2 \equiv 1^7 \cdot 2 \equiv 2 \pmod{5}
     \]
     Thus, \( 37^{29} \equiv 2 \pmod{5} \).

3. **Check modulo 13**: We need to find \( n \) such that \( n^{29} \equiv 7 \pmod{13} \).
   - For \( n = 37 \):
     \[
     37 \equiv 11 \pmod{13} \quad \Rightarrow \quad 37^{29} \equiv 11^{29} \pmod{13}
     \]
     Since \( 11 \equiv -2 \pmod{13} \), we have:
     \[
     11^{29} \equiv (-2)^{29} \equiv -2^{29} \pmod{13}
     \]
     We need to compute \( 2^{29} \pmod{13} \). Using the fact that \( 2^{12} \equiv 1 \pmod{13} \) (by Fermat's Little Theorem), we get:
     \[
     2^{29} = 2^{24 + 5} = (2^{12})^2 \cdot 2^5 \equiv 1^2 \cdot 32 \equiv 32 \pmod{13}
     \]
     Since \( 32 \equiv 6 \pmod{13} \), we have:
     \[
     -2^{29} \equiv -6 \equiv 7 \pmod{13}
     \]
     Thus, \( 37^{29} \equiv 7 \pmod{13} \).

4. **Conclusion**: Since \( 37 \) satisfies both congruences \( n^{29} \equiv 2 \pmod{5} \) and \( n^{29} \equiv 7 \pmod{13} \), it is the correct answer.

The final answer is \( \boxed{37} \).