Calculate the digits immediately to the left and right of the decimal point in $(\sqrt{2} + \sqrt{3})^{2012}$
We show that $(\sqrt{2}+\sqrt{3})^{2012}=\ldots 7,9 \ldots$. We set $a=(\sqrt{2}+\sqrt{3})^{2}=$ $5+\sqrt{24}$ and $b=(\sqrt{2}-\sqrt{3})^{2}=5-\sqrt{24}$. We need to study $a^{1006}$. The starting point is the observation that $a^{1006}+b^{1006}$ is an integer, while $b^{1006}$ is small. Indeed, $a^{1006}+b^{1006}=\left(S_{1}+\right.$ $\left.\sqrt{24} S_{2}\right)+\left(S_{1}-\sqrt{24} S_{2}\right)$ with $S_{1}=\sum_{k=0}^{503} 5^{2 k} 24^{503-k}\binom{1006}{2 k}$ and $S_{2}=\sum_{k=0}^{502} 5^{2 k+1} 24^{503-k}\binom{1006}{2 k+1}$. Moreover, $ab=25-24=1$, so $0<b<\frac{1}{5}$ and thus $b^{1006}<\frac{1}{5^{1006}}<\frac{1}{10}$. As an integer minus a quantity less than $\frac{1}{10}$, $a^{1006}$ is $\ldots, 9 \ldots$

To evaluate the last digit of $a^{1006}$, it suffices to evaluate the last digit of $a^{1006}+b^{1006}=$ $2 S_{1}$ and subtract 1. Each term of $S_{1}$ divisible by 5 does not contribute to the last digit of $2 S_{1}$. Thus, the last digit is that of $2 \cdot 24^{503}$. Since $24^{503} \equiv 4^{503} \bmod 10$ and a simple induction shows that for all $k$, $4^{k} \equiv 4 \bmod 10$ if and only if $k \equiv 1 \bmod 2$, the last digit of $2 S_{1}$ is 8. We conclude that $a^{1006}=\ldots 7,9 \ldots$