10. Tank A contains 4 liters of liquid A, Tank B contains 2 kilograms of liquid B, and Tank C contains 2 kilograms of liquid C. These liquids can all be mixed. First, 1 liter of liquid from Tank A is poured into Tank B, and then 1 liter of the mixed liquid from Tank B is poured into Tank C. Finally, 1 liter of the mixed liquid from Tank C is poured back into Tank A. This operation is called one mixing. After $n$ mixings, the amount of liquid A in Tank B is denoted as $f(n)$. If $f(n)>0.9999$, what is the minimum value of $n$? $\qquad$ .
Let $a_{n}, b_{n}, c_{n}$ be the amounts of liquid A in containers 甲, 乙, and 丙, respectively, after $n$ mixings. Then,
$$
a_{n}+b_{n}+c_{n}=4 \text{. }
$$

By symmetry, we have
$$
b_{n}=c_{n} \text{. }
$$

Now, let's examine the amount of liquid A in container 甲 after the $(n+1)$-th mixing. According to the problem,
$$
\begin{aligned}
a_{n+1} & =\frac{1}{2} a_{n}+\frac{1}{3}\left(\frac{1}{4} a_{n}+b_{n}\right) \\
+ & \left.\frac{1}{3} \div \frac{1}{4} a_{n}+c_{5}\right) \\
& =\frac{1}{2} a_{n}+\frac{1}{6} a_{n}+\frac{2}{3} b_{n} \\
& =\frac{2}{3} a_{n}+\frac{1}{3}\left(4-a_{n}\right) \\
& =\frac{1}{3} a_{n}+\frac{4}{3}
\end{aligned}
$$

Thus, $a_{n+1}-2=\frac{1}{3}\left(a_{n}-2\right)$.
By induction, we get
$$
\begin{array}{l}
a_{n+1}-2=\frac{1}{3}\left(a_{n}-2\right)=\frac{1}{3^{2}}\left(a_{n-1}-2\right) \\
=\cdots=\left(\frac{1}{3}\right)^{n+1}\left(a_{0}-2\right) .
\end{array}
$$

Substituting $a_{0}=4$ yields
$$
\begin{array}{l}
a_{n}=2\left[1+\left(\frac{1}{3}\right)^{n}\right], \\
b_{n}=\frac{1}{2}\left(4-a_{n}\right)=1-\left(\frac{1}{3}\right)^{n} .
\end{array}
$$

Given $f(n)>0.9999$, i.e.,
$1-\left(\frac{1}{3}\right)^{n}>0.9999$ or $3^{n}>10000$. The smallest value of $n$ is 9.
(Li Yongle, Yu Rongpei)