The units digit of $33 \cdot 83^{1001} \cdot 7^{1002} \cdot 13^{1003}$ is
(A) 1.
(B) 3.
(C) 5.
(D) 7.
(E) 9.
(34th American High School Mathematics Examination, 1983)
[Solution 1] Consider the units digit of the first $n$ powers of $3, 7, 13$ as shown in the table below:
\begin{tabular}{c|c|c|c}
\hline$a$ & 3 & 7 & 13 \\
\hline$a^{2}$ & 9 & 49 & 169 \\
\hline$a^{3}$ & 27 & 343 & $\cdots 7$ \\
\hline$a^{4}$ & 81 & $\cdots 1$ & $\cdots 1$ \\
\hline$a^{5}$ & 243 & $\cdots 7$ & $\cdots 3$ \\
\hline
\end{tabular}

From this, it is not difficult to see that their units digits follow a cyclic pattern with a period of 4.
Let $\langle n\rangle$ denote the units digit of $n$. Since 4 divides 100, we have
$$
\left\langle 3^{1001}\right\rangle=\left\langle 3^{1}\right\rangle=3, \quad\left\langle 7^{1002}\right\rangle=\left\langle 7^{2}\right\rangle=9 .
$$

Therefore,
$$
\left\langle 13^{1003}\right\rangle=\left\langle 13^{3}\right\rangle=7 \text {. }
$$

Thus,
$$
\left\langle 3^{1001} 7^{1002} 13^{1003}\right\rangle=\langle 3 \cdot 9 \cdot 7\rangle=9 \text {. }
$$

Hence, the answer is $(E)$.
[Solution 2] We can see that $7 \cdot 13=91$ and $3^{4}=81$, so the units digit of any power of these numbers is 1.
Thus,
$$
\begin{aligned}
3^{1001} 7^{1002} 13^{1003} & =3 \cdot 13 \cdot 81^{250} \cdot(7 \cdot 13)^{1002} \\
& =3 \cdot 13 \cdot 81^{250} \cdot 91^{1002},
\end{aligned}
$$

Clearly, its units digit is 9.
Hence, the answer is $(E)$.