TE R Find the digit in the decimal representation of $(\sqrt{2}+\sqrt{3})^{1980}$ that is immediately to the right of the decimal point (i.e., the first decimal place) and the digit to the left of the decimal point (i.e., the units digit).
Parse $(\sqrt{3}+\sqrt{2})^{1980}=(5+2 \sqrt{6})^{990},(5+2 \sqrt{6})^{990}+(5-2 \sqrt{6})^{990}$ as a positive even number, and $(5+2 \sqrt{6})^{990}+(5-2 \sqrt{6})^{990} \equiv 2 \cdot(2 \sqrt{6})^{990} \equiv 2^{991} \cdot 6^{995} \equiv 2^{991}(\bmod 10) \equiv$ $2^{4 \times 247+3} \equiv 2^{3} \equiv 3(\bmod 5)$. Therefore, the last digit of $(5+2 \sqrt{6})^{990}+(5-2 \sqrt{6})^{990}$ is 8.

Since $(5-2 \sqrt{6})^{990}=(\sqrt{3}-\sqrt{2})^{1980}=\left(\frac{1}{\sqrt{3}+\sqrt{2}}\right)^{1980}<\frac{1}{2^{1980}}<0.01$, the last digit of $(5+2 \sqrt{6})^{990}$ is 7, and the first decimal place is 9.