1.64 Given that for any integer $k \neq 27$, the number $a-k^{1964}$ is divisible by $27-k$. Find $a$.
(27th Moscow Mathematical Olympiad, 1964)
［Solution］ Let $27-k=m$, then $k=27-m$.
$$
\begin{aligned}
& \frac{a-k^{1964}}{27-k} \\
= & \frac{a-(27-m)^{1964}}{m} \\
= & \frac{a-27^{1964}}{m}+C_{1964}^{1} 27^{1963}-C_{1964}^{2} 27^{1962} m+\cdots-m^{1963} .
\end{aligned}
$$

Since all terms from the second term onward in the above expression are integers, it must be that for any integer $m$, $\frac{a-27^{1964}}{m}$ is an integer.
Therefore,
$$
a=27^{1964} \text {. }
$$