$ 1 \le n \le 455$ and $ n^3 \equiv 1 \pmod {455}$. The number of solutions is ?

$\textbf{(A)}\ 9  \qquad\textbf{(B)}\ 6  \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 1  \qquad\textbf{(E)}\ \text{None}$
1. We start by noting that \( 455 = 5 \times 7 \times 13 \). We need to solve \( n^3 \equiv 1 \pmod{455} \). This can be broken down into solving the congruences modulo 5, 7, and 13 separately.

2. First, consider the congruence modulo 5:
   \[
   n^3 \equiv 1 \pmod{5}
   \]
   The possible values of \( n \) modulo 5 are 0, 1, 2, 3, and 4. We check each:
   \[
   \begin{aligned}
   0^3 &\equiv 0 \pmod{5}, \\
   1^3 &\equiv 1 \pmod{5}, \\
   2^3 &\equiv 8 \equiv 3 \pmod{5}, \\
   3^3 &\equiv 27 \equiv 2 \pmod{5}, \\
   4^3 &\equiv 64 \equiv 4 \pmod{5}.
   \end{aligned}
   \]
   Only \( n \equiv 1 \pmod{5} \) satisfies the congruence.

3. Next, consider the congruence modulo 7:
   \[
   n^3 \equiv 1 \pmod{7}
   \]
   The possible values of \( n \) modulo 7 are 0, 1, 2, 3, 4, 5, and 6. We check each:
   \[
   \begin{aligned}
   0^3 &\equiv 0 \pmod{7}, \\
   1^3 &\equiv 1 \pmod{7}, \\
   2^3 &\equiv 8 \equiv 1 \pmod{7}, \\
   3^3 &\equiv 27 \equiv 6 \pmod{7}, \\
   4^3 &\equiv 64 \equiv 1 \pmod{7}, \\
   5^3 &\equiv 125 \equiv 6 \pmod{7}, \\
   6^3 &\equiv 216 \equiv 6 \pmod{7}.
   \end{aligned}
   \]
   The values \( n \equiv 1, 2, 4 \pmod{7} \) satisfy the congruence.

4. Finally, consider the congruence modulo 13:
   \[
   n^3 \equiv 1 \pmod{13}
   \]
   The possible values of \( n \) modulo 13 are 0 through 12. We check each:
   \[
   \begin{aligned}
   0^3 &\equiv 0 \pmod{13}, \\
   1^3 &\equiv 1 \pmod{13}, \\
   2^3 &\equiv 8 \pmod{13}, \\
   3^3 &\equiv 27 \equiv 1 \pmod{13}, \\
   4^3 &\equiv 64 \equiv 12 \pmod{13}, \\
   5^3 &\equiv 125 \equiv 8 \pmod{13}, \\
   6^3 &\equiv 216 \equiv 8 \pmod{13}, \\
   7^3 &\equiv 343 \equiv 5 \pmod{13}, \\
   8^3 &\equiv 512 \equiv 5 \pmod{13}, \\
   9^3 &\equiv 729 \equiv 1 \pmod{13}, \\
   10^3 &\equiv 1000 \equiv 12 \pmod{13}, \\
   11^3 &\equiv 1331 \equiv 8 \pmod{13}, \\
   12^3 &\equiv 1728 \equiv 12 \pmod{13}.
   \end{aligned}
   \]
   The values \( n \equiv 1, 3, 9 \pmod{13} \) satisfy the congruence.

5. Combining the results using the Chinese Remainder Theorem, we have:
   \[
   n \equiv 1 \pmod{5}
   \]
   \[
   n \equiv 1, 2, 4 \pmod{7}
   \]
   \[
   n \equiv 1, 3, 9 \pmod{13}
   \]
   The total number of solutions is the product of the number of solutions for each modulus:
   \[
   1 \cdot 3 \cdot 3 = 9
   \]

The final answer is \(\boxed{9}\).