Let $ f(x)\equal{}5x^{13}\plus{}13x^5\plus{}9ax$. Find the least positive integer $ a$ such that $ 65$ divides $ f(x)$ for every integer $ x$.
1. **Determine the condition modulo 5:**
   \[
   f(x) = 5x^{13} + 13x^5 + 9ax \equiv 13x^5 + 9ax \pmod{5}
   \]
   Since \(13 \equiv 3 \pmod{5}\), we have:
   \[
   f(x) \equiv 3x^5 + 9ax \pmod{5}
   \]
   Since \(9 \equiv 4 \pmod{5}\), we get:
   \[
   f(x) \equiv 3x^5 + 4ax \pmod{5}
   \]
   By Fermat's Little Theorem, \(x^5 \equiv x \pmod{5}\) for any integer \(x\) not divisible by 5. Thus:
   \[
   f(x) \equiv 3x + 4ax \equiv (3 + 4a)x \pmod{5}
   \]
   For \(f(x) \equiv 0 \pmod{5}\) for all \(x\), we need:
   \[
   3 + 4a \equiv 0 \pmod{5}
   \]
   Solving for \(a\):
   \[
   4a \equiv -3 \pmod{5} \implies 4a \equiv 2 \pmod{5} \implies a \equiv 3 \pmod{5}
   \]

2. **Determine the condition modulo 13:**
   \[
   f(x) = 5x^{13} + 13x^5 + 9ax \equiv 5x^{13} + 9ax \pmod{13}
   \]
   Since \(13 \equiv 0 \pmod{13}\), we have:
   \[
   f(x) \equiv 5x^{13} + 9ax \pmod{13}
   \]
   By Fermat's Little Theorem, \(x^{13} \equiv x \pmod{13}\) for any integer \(x\) not divisible by 13. Thus:
   \[
   f(x) \equiv 5x + 9ax \equiv (5 + 9a)x \pmod{13}
   \]
   For \(f(x) \equiv 0 \pmod{13}\) for all \(x\), we need:
   \[
   5 + 9a \equiv 0 \pmod{13}
   \]
   Solving for \(a\):
   \[
   9a \equiv -5 \pmod{13} \implies 9a \equiv 8 \pmod{13}
   \]

3. **Solve the system of congruences:**
   \[
   \begin{cases}
   a \equiv 3 \pmod{5} \\
   9a \equiv 8 \pmod{13}
   \end{cases}
   \]
   First, solve \(9a \equiv 8 \pmod{13}\):
   \[
   9a \equiv 8 \pmod{13}
   \]
   The multiplicative inverse of 9 modulo 13 is 3 (since \(9 \cdot 3 \equiv 27 \equiv 1 \pmod{13}\)). Thus:
   \[
   a \equiv 8 \cdot 3 \pmod{13} \implies a \equiv 24 \pmod{13} \implies a \equiv 11 \pmod{13}
   \]

4. **Combine the results using the Chinese Remainder Theorem:**
   \[
   \begin{cases}
   a \equiv 3 \pmod{5} \\
   a \equiv 11 \pmod{13}
   \end{cases}
   \]
   Let \(a = 5k + 3\). Then:
   \[
   5k + 3 \equiv 11 \pmod{13} \implies 5k \equiv 8 \pmod{13}
   \]
   The multiplicative inverse of 5 modulo 13 is 8 (since \(5 \cdot 8 \equiv 40 \equiv 1 \pmod{13}\)). Thus:
   \[
   k \equiv 8 \cdot 8 \pmod{13} \implies k \equiv 64 \pmod{13} \implies k \equiv 12 \pmod{13}
   \]
   Therefore:
   \[
   a = 5k + 3 = 5(13m + 12) + 3 = 65m + 60 + 3 = 65m + 63
   \]
   The smallest positive integer \(a\) is:
   \[
   a = 63
   \]

The final answer is \(\boxed{63}\).