$33 \cdot 1$. The units digit of the number $2137^{753}$ is
(A) 1.
(B) 3.
(C) 5.
(D) 7.
(E) 9.
(12th American High School Mathematics Examination, 1961)
[Solution] Let $\langle a\rangle$ denote the unit digit of the integer $a$. Clearly, if the unit digit of $a$ is $b$, then $\langle a\rangle=\langle b\rangle$, and $\left\langle a^{n}\right\rangle=\left\langle b^{n}\right\rangle$ (where $n$ is a natural number).
Also, $\left\langle 7^{0}\right\rangle=1,\left\langle 7^{1}\right\rangle=7,\left\langle 7^{2}\right\rangle=9,\left\langle 7^{3}\right\rangle=3$.
It is easy to prove that $\left\langle 7^{4 n+r}\right\rangle=\left\langle 7^{r}\right\rangle$, meaning the unit digit of the powers of 7 cycles with a period of 4.
Thus, $\left\langle 2137^{753}\right\rangle=\left\langle 2137^{4 \times 188+1}\right\rangle=\left\langle 2137^{1}\right\rangle=7$.
Therefore, the answer is $(D)$.
Note: Regarding the unit digit (last digit) of $a^{n}$ for natural numbers, the following conclusions can be drawn:
(1) If the unit digit of $a$ is $b$, then $\left\langle a^{n}\right\rangle=\left\langle b^{n}\right\rangle$;
(2) The unit digit of any power of a natural number ending in $0,1,5,6$ remains $0,1,5,6$;
(3) The unit digit of the $n$-th power of $2,3,4,7,8,9$ is as follows:
Table of unit digits of $a$:
\begin{tabular}{c|c|c|c|c|c|c}
\hline$n-a$ & 2 & 3 & 4 & 7 & 8 & 9 \\
\hline $4 k+1$ & 2 & 3 & 4 & 7 & 8 & 9 \\
$4 k+2$ & 4 & 9 & 6 & 9 & 4 & 1 \\
$4 k+3$ & 8 & 7 & 4 & 3 & 2 & 9 \\
$4 k$ & 6 & 1 & 6 & 1 & 6 & 1 \\
\hline
\end{tabular}