3. The function $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for all $n \in \mathbf{N}$, we have
$$
\begin{array}{c}
f(f(n))+f(n)=2 n+3 \text {, and } f(0)=1 . \\
\text { Then } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=
\end{array}
$$
3. 2772 .

Substituting $n=0$ into equation (1) yields
$$
f(1)+1=3 \Rightarrow f(1)=2 \text {. }
$$

Furthermore, let $n=1$.
From equation (1), we get $f(2)=3$.
Similarly, $f(n)=n+1(n=3,4, \cdots, 10)$.
$$
\text { Hence } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=2772 \text {. }
$$