7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of

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7. 6 .

Solution: Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$, then $-1<\beta<0,1<\alpha<2$, and $\alpha+\beta=1, \alpha \beta=-1$,

Therefore, $\alpha, \beta$ are the two distinct real roots of the equation $x^{2}-x-1=0$. Let $A_{n}=\alpha^{n}+\beta^{n}$, then $A_{n+2}=A_{n+1}+A_{n}$,

Given $A_{1}=\alpha+\beta=1, A_{2}=\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3$, 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, ..., which is a periodic sequence with a period of 16. Since $2009=16 \times 125+9$, the remainder when $A_{2009}$ is divided by 7 is 6.

Since $-1<\left(\frac{1-\sqrt{5}}{2}\right)^{2009}$, the remainder when $\left(\frac{1+\sqrt{5}}{2}\right)^{2009}$ is divided by 7 is 6.