5. Write $(\sqrt{7}-\sqrt{6})^{6}$ in the form $\sqrt{N+1}-\sqrt{N}$, where $N$ is a natural number. Then $N=$ $\qquad$
5. $N=76545000$.

Let $(\sqrt{7}-\sqrt{6})^{6}=\sqrt{N+1}-\sqrt{N}$, then $(\sqrt{7}-\sqrt{6})^{6}+\sqrt{N}=\sqrt{N+1}$.
Square it to get
$$
\begin{array}{l}
(\sqrt{7}-\sqrt{6})^{12}+2 \sqrt{N}(\sqrt{7}-\sqrt{6})^{6}+N \\
= N+1 \\
2 \sqrt{N}=\frac{1-(\sqrt{7}-\sqrt{6})^{12}}{(\sqrt{7}-\sqrt{6})^{6}} \\
=\frac{(7-6)^{6}-(\sqrt{7}-\sqrt{6})^{12}}{(\sqrt{7}-\sqrt{6})^{6}} \\
=(\sqrt{7}+\sqrt{6})^{6}-(\sqrt{7}-\sqrt{6})^{6} \\
= 2\left[C_{6}^{1}(\sqrt{7})^{5} \cdot \sqrt{6}+C_{6}^{3}(\sqrt{7})^{3}(\sqrt{6})^{3}\right. \\
\left.+C_{6}^{5} \cdot \sqrt{7} \cdot(\sqrt{6})^{5}\right] \\
= 2 \sqrt{7} \cdot \sqrt{6} \cdot 6 \cdot 225,
\end{array}
$$

Thus, $N=7 \cdot 2^{3} \cdot 3^{7} \cdot 5^{4}=76545 \times 10^{3}$.