Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]
$\textbf{(A)}\ 0\qquad  \textbf{(B)}\ 2\qquad  \textbf{(C)}\ 5\qquad  \textbf{(D)}\ 9\qquad  \textbf{(E)}\ \text{none of these}$
Let $A=15+\sqrt{220}$ and $B=15-\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.
We have
\begin{align*} A^{19}+B^{19} &= \left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{1}15^{18}\sqrt{220}^1+\cdots+\binom{19}{19}15^0\sqrt{220}^{19}\right] + \left[\binom{19}{0}15^{19}\sqrt{220}^0-\binom{19}{1}15^{18}\sqrt{220}^1+\cdots-\binom{19}{19}15^0\sqrt{220}^{19}\right] \\ &= 2\left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{2}15^{17}\sqrt{220}^2+\cdots+\binom{19}{18}15^1\sqrt{220}^{18}\right] \\ &= 2\left[\binom{19}{0}15^{19}+\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9\right]. \end{align*}
Similarly, we have \[A^{82}+B^{82}=2\left[\binom{82}{0}15^{82}+\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}\right].\]
We add the two equations and take the sum modulo $10:$
\begin{align*} \left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right) &= 2\Biggl[\binom{19}{0}15^{19}+\phantom{ }\underbrace{\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9}_{0\pmod{10}}\phantom{ }\Biggr]+2\Biggl[\binom{82}{0}15^{82}+\phantom{ }\underbrace{\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}}_{0\pmod{10}}\phantom{ }\Biggr] \\ &\equiv 2\left[\binom{19}{0}15^{19}\right]+2\left[\binom{82}{0}15^{82}\right] \\ &\equiv 2\left[5\right]+2\left[5\right] \\ &\equiv 0\pmod{10}. \end{align*}
It is clear that $0<B^{82}<B^{19}<B<0.5,$ from which $0<B^{19}+B^{82}<1.$ We conclude that the units digit of the decimal expansion of $B^{19}+B^{82}$ is $0.$ Since the units digit of the decimal expansion of $\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)$ is $0,$ the units digit of the decimal expansion of $A^{19}+A^{82}$ is $\boxed{\textbf{(D)}\ 9}.$
~MRENTHUSIASM