13. In a math competition, there were three problems: A, B, and C. Among the 25 participating students, each student solved at least one problem. 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; the number of students who only solved problem A is one more than the number of students who solved problem A among the remaining students; among the students who only solved one problem, half did not solve problem A. How many students only solved problem B?
13. Solution Let the sets of students who solved questions A, B, and C be $A, B, C$, respectively, and represent them with three circles. The overlapping parts represent the sets of students who solved two or three questions, with the number of students in each part denoted by $a, b, c, d, e, f, g$.
Since each student solved at least one question, we have
$$
a+b+c+d+e+f+g=25
$$

Since among the students who did not solve question A, the number of students who solved question B is twice the number of students who solved question C, we have
$$
b+f=2(c+f)
$$

Since the number of students who only solved question A is one more than the number of students who solved question A among the remaining students, we have
$$
a=d+e+g+1
$$

Since among the students who only solved one question, half did not solve question A, we have
$$
a=b+c
$$

From (2), we get:
$$
b=2 c+f, f=b-2 c
$$

Substituting (5) into (1) to eliminate $f$, we get
$$
a+2 b-c+d+e+g=25
$$

Substituting (3) and (4) into (6), we get:
$$
\begin{array}{c}
2 b-c+2 d+2 e+2 g=24 \\
3 b+d+e+g=25
\end{array}
$$

Multiplying (8) by 2 and subtracting (7), we get:
$$
4 b+c=26
$$

Since $c \geqslant 0$, we have $4 b \leqslant 26, b \leqslant 6 \frac{1}{2}$.
Using (5) and (9) to eliminate $c$, we get $f=b-2(26-4 b)=9 b-52$
Since $f \geqslant 0$, we have $9 b \geqslant 52, b \geqslant \frac{52}{9}$. Since $b \in \mathbf{Z}$, we have $b=6$. That is, the number of students who only solved question B is 6.