5. Let $\{x\}$ denote the fractional part of the real number $x$. If $a=$ $(5 \sqrt{13}+18)^{2005}$, then $a \cdot\{a\}=$ $\qquad$
5.1.

Let $b=(5 \sqrt{13}-18)^{2005}$.
Since $0<5 \sqrt{13}-18<1$, we have $0<b<1$.
By the binomial theorem, we know
$$
\begin{aligned}
a= & (5 \sqrt{13})^{2005}+\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots+18^{2005}, \\
b= & (5 \sqrt{13})^{2005}-\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\cdots+ \\
& (-1)^{r} \mathrm{C}_{2005}^{r}(5 \sqrt{13})^{2005-r} \times 18^{r}+\cdots- \\
& 18^{2005} .
\end{aligned}
$$

Thus, $a-b=2\left[\mathrm{C}_{2005}^{1}(5 \sqrt{13})^{2004} \times 18^{1}+\mathrm{C}_{2005}^{3} \times\right.$ $\left.(5 \sqrt{13})^{2002} \times 18^{3}+\cdots+\mathrm{C}_{2005}^{2005} \times 18^{2005}\right]$ is a positive integer.
Since $0<b<1$, we have $\{a\}=b$.
Therefore, $a \cdot\{a\}=a \cdot b$
$$
\begin{array}{l}
=(5 \sqrt{13}+18)^{2005}(5 \sqrt{13}-18)^{2005} \\
=\left[(5 \sqrt{13})^{2}-18^{2}\right]^{2005}=1^{2005}=1 .
\end{array}
$$