What's the Unit Digit #2

Note that $0 \; < \; 15 - \sqrt{220} \; = \; \frac{5}{15 + \sqrt{220}} \; < \; \frac13$ and so certainly $0 \; < \; (15 - \sqrt{220})^{82} \; < \; (15 - \sqrt{220})^{19} \; < \; \tfrac13$ so that $0 \; < \; B \; = \; (15 - \sqrt{220})^{19} + (15 - \sqrt{220})^{82} \; < \; \tfrac23 \; < \; 1$ In fact, $B$ is very close to $0$ , but the above is enough. Now $\begin{aligned} (15 + \sqrt{220})^{19} + (15 - \sqrt{220})^{19} & = 2\sum_{j=0}^9 {19 \choose 2j}15^{19-2j} 220^j \\ (15 + \sqrt{220})^{82} + (15 - \sqrt{220})^{82} & = 2\sum_{j=0}^{41} {82 \choose 2j}15^{82-2j} 220^j \end{aligned}$ and so, since $220 \equiv 0 \pmod{10}$ , we see that $\begin{aligned} (15 + \sqrt{220})^{19} + (15 - \sqrt{220})^{19} & \equiv \; 2\times 15^{19} \; \equiv \; 0 \\ (15 + \sqrt{220})^{82} + (15 - \sqrt{220})^{82} & \equiv \; 2\times 15^{82} \; \equiv \; 0 \end{aligned}$ modulo $10$ . Thus, if we define $A \; = \; (15 + \sqrt{220})^{19} + (15 + \sqrt{220})^{82}$ then we have shown that $A + B \equiv 0 \pmod{10}$ , and so (since $0 < B < 1$ ) the last digit of $A$ before the decimal point is $\boxed{9}$