$11 \cdot 19$ Find the units digit of the decimal expansion of $(15+\sqrt{220})^{19}+(15+\sqrt{220})^{82}$
(A) 0 .
(B) 2 .
(C) 5 .
(D) 9 .
(E) None of the above values.
(33rd American High School Mathematics Examination, 1982)
[Solution] Let $d_{1}=a+\sqrt{b}, d_{2}=a-\sqrt{b}$, where $a=15, b=220$. By the binomial theorem, we have
$$
\begin{aligned}
d_{1}{ }^{n}= & a^{n}+C_{n}^{1} a^{n-1} \sqrt{b}+C_{n}^{2} a^{n-2} b+C_{n}^{3} a^{n-3} b \sqrt{b}+C_{n}^{4} a^{n-4} b^{2} \\
& +\cdots+(\sqrt{b})^{n}, \\
d_{2}{ }^{n}= & a^{n}-C_{n}^{1} a^{n-1} \sqrt{b}+C_{n}^{2} a^{n-2} b-C_{n}^{3} a^{n-3} b \sqrt{b}+C_{n}^{4} a^{n-4} b^{2}+\cdots, \\
\therefore & d_{1}{ }^{n}+d_{2}{ }^{n}=2\left(a^{n}+C_{n}^{2} a^{n-2} b+C_{n}^{4} a^{n-4} b^{2}+\cdots\right) .
\end{aligned}
$$

Here, $n$ is a positive integer. Since $a$ and $b$ are both divisible by 5, we conclude that $d_{1}{ }^{n}+d_{2}{ }^{n}$ is divisible by 10.
Taking $n=19$ and $n=82$, we get
$$
d_{1}{ }^{19}+d_{2}{ }^{19}=10 k_{1}, \quad d_{1}{ }^{82}+d_{2}{ }^{82}=10 k_{2},
$$

where $k_{1}$ and $k_{2}$ are positive integers. Adding the two equations, we have
$d_{1}{ }^{19}+d_{1}{ }^{82}=10 k-\left(d_{2}{ }^{19}+d_{2}{ }^{82}\right)$, where $k=k_{1}+k_{2}$.
However, $d_{2}=15-\sqrt{220}=\frac{5}{15+\sqrt{220}}<\frac{1}{3}$,

so
$$
d_{2}{ }^{19}+d_{2}{ }^{82}<1 \text {. }
$$

Therefore, the units digit of $d_{1}{ }^{19}+d_{1}{ }^{82}=10 k-\left(d_{2}{ }^{19}+d_{2}{ }^{82}\right)$ should be 9. Hence, the answer is $(D)$.