6. The unit digit of the number $(5+\sqrt{24})^{2 n}(n \in \mathbf{N})$ is

保留源文本的换行和格式，翻译结果如下：

6. The unit digit of the number $(5+\sqrt{24})^{2 n}(n \in \mathbf{N})$ is
6. When $21 n$, it is 1; when $2 \times n$, it is 7.

Let $x=(5+\sqrt{24})^{2 n}=A+\sqrt{24} B(A, B \in \mathbf{N})$.
By the binomial theorem, we know

Therefore, $x+y=2 A \in \mathbf{N}$. f is
$$
x=(2 A-1)+1-y(\text { where } 0<1-y<1),
$$
the unit digit of $x$ is the unit digit of $(2 A-1)$.
$$
\begin{array}{l}
\text { Hence } 2 A-1=x+y-1=2 \cdot \_{1=0}^{n} \mathrm{C}_{2 n}^{2,} 25^{n-1} \times 24^{\prime}-1 \\
\equiv 2\left(25^{n}+24^{n}\right)-1 \equiv 2 \times 24^{n}-1 \\
\equiv 2(25-1)^{n}-1=2 \times(-1)^{n}-1 \\
\equiv\left\{\begin{array}{l}
1, \text { when } 21 n \text {, }(\bmod 10) \\
7, \text { when } 2 \uparrow n \text { . }
\end{array}\right.
\end{array}
$$