7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
7.6.

Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
-1<\beta<0,1<\alpha<2 \text {, }
$$

and $\alpha+\beta=1, \alpha \beta=-1$.
Therefore, $\alpha, \beta$ are the two distinct real roots of $x^{2}-x-1=0$.
Let $A_{n}=\alpha^{n}+\beta^{n}$. Then $A_{n+2}=A_{n+1}+A_{n}$.
By $A_{1}=\alpha+\beta=1$,
$$
A_{2}=\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3 \text {, }
$$

we know that $\left\{A_{n}\right\}$ is an integer sequence, and the sequence of remainders when divided by 7 is 1, $3,4,0,4,4,1,5,6,4,3,0,3,3,6,2,1,3,4, \cdots$, which is a periodic sequence with a period of 16.

Since $2009=16 \times 125+9$, we know that the remainder when $A_{2000}$ is divided by 7 is 6.
$$
\text { Since }-1<\left(\frac{1-\sqrt{5}}{2}\right)^{200}<0 \text {, hence }\left[\left(\frac{1+\sqrt{5}}{2}\right)^{209}\right]
$$

is 6 when divided by 7.