For the expression $\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}$, when written as a decimal, find the digit before the decimal point.
Let $a_{n}=\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{n}-\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
Then $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
The last digits are
$$
\begin{array}{l}
1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7 \\
7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9 \\
8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6 \\
9,5,4,9,3,2,5,7,2,9,1,0,1,1, \cdots .
\end{array}
$$

It is not hard to see that the last digits form a periodic sequence with a period of 60.
Thus, the last digit of $a_{2013}$ is 8.
$$
\text { Also, }-1<\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{2013}<0 \text {, so } \frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}
$$

has an integer part of $a_{2013}-1$.
Therefore, the digits before the decimal point are 7.
(Jin Lei, Xi'an Jiaotong University Affiliated High School,
710054)