1. (USS) Three problems $A, B$, and $C$ were given on a mathematics olympiad. All 25 students solved at least one of these problems. The number of students who solved $B$ and not $A$ is twice the number of students who solved $C$ and not $A$. The number of students who solved only $A$ is greater by 1 than the number of students who along with $A$ solved at least one other problem. Among the students who solved only one problem, half solved $A$. How many students solved only $B$ ?
1. Let $N_{a}, N_{b}, N_{c}, N_{a b}, N_{a c}, N_{b c}, N_{a b c}$ denote the number of students who solved exactly the problems whose letters are stated in the index of the variable. From the conditions of the problem we have
$$
\begin{array}{c}
N_{a}+N_{b}+N_{c}+N_{a b}+N_{b c}+N_{a c}+N_{a b c}=25 \\
N_{b}+N_{b c}=2\left(N_{c}+N_{b c}\right), \quad N_{a}-1=N_{a c}+N_{a b c}+N_{a b}, \quad N_{a}=N_{b}+N_{c} .
\end{array}
$$

From the first and third equations we get $2 N_{a}+N_{b}+N_{0}+N_{6 c}=26$, and from the second and fourth we get $4 N_{b}+N_{c}=26$ and thus $N_{b} \leq 6$. On the other hand, we have from the second equation $N_{b}=2 N_{c}+N_{b c} \Rightarrow$ $N_{c} \leq N_{b} / 2 \Rightarrow 26 \leq 9 N_{b} / 2 \Rightarrow N_{b} \geq 6$; hence $N_{b}=6$.