C5 (8-1, USSR) In a middle school mathematics competition, three problems $A, B, C$ were given, with 25 students participating. Each student solved at least one of the problems. Among the students who did not solve problem $A$, the number of students who solved problem $B$ is twice the number of students who solved problem $C$. Among the students who solved problem $A$, the number of students who only solved $A$ is one more than the number of students who solved $A$ and at least one other problem. Additionally, among the students who only solved one problem, half did not solve problem $A$. How many students only solved problem $B$?
Let $x_{A}, x_{B}, x_{C}, x_{A B}, x_{B C}, x_{C A}, x_{A B C}$ represent the number of students who solved only the problems indicated by the subscripts. According to the problem, we have
$$
x_{A}+x_{B}+x_{C}+x_{A B}+x_{B C}+x_{C A}+x_{A B C}=25 .
$$

Among the students who did not solve problem $A$, $x_{B}+x_{C}+x_{\mathrm{BC}}$, $x_{B}+x_{\mathrm{BC}}$ represents the number of students who solved problem $B$, and $x_{C}+x_{B}$ represents the number of students who solved problem $C$. According to the problem, we have
$$
\begin{array}{c}
x_{B}+x_{B C}=2\left(x_{C}+x_{B C}\right), \\
x_{B}-2 x_{C}-x_{B C}=0 .
\end{array}
$$

The number of students who only solved problem $A$ is $x_{1}$, and the number of students who solved problem $A$ and at least one other problem is $x_{A B}+x_{A}+x_{A B C}$. Therefore,
$$
x_{A}=x_{A B}+x_{A C}+x_{A B C}+1 .
$$

Among the students who only solved one problem, half did not solve problem $A$, so
$$
x_{A}=x_{B}+x_{C} \text {. }
$$
(1) + (3) + (4), we get
$$
3 x_{A}+x_{I C}=26,
$$

Therefore, $x_{A} \leqslant 8$.
From (2), (4), and (5), we get
$$
\begin{array}{l}
x_{1}+3 x_{B}=26, \\
x_{B}=\frac{26-x_{A}}{3},
\end{array}
$$

Therefore, $x_{A}=2,5,8$.
When $x_{A}=2$, $x_{B}=8$, which contradicts (4).
When $x_{A}=5$, $x_{B}=7$, which contradicts (4).
When $x_{A}=8$, $x_{B}=6$, which satisfies the problem.
Therefore, the number of students who only solved problem $B$ is 6.