In a mathematical student competition, three problems were posed: $A$, $B$, and $C$. 25 students in total solved at least one of the problems. Among the students who did not solve $A$, there were twice as many who solved $B$ as those who solved $C$. One more student solved only $A$ than those who solved $A$ along with other problems. Half of the students who solved only one problem did not solve $A$. - How many students solved only $B$?
We categorize the contestants who solved at least one problem into the following 5 groups:

I. those who only solved $A$,

II. those who solved $A$ and at least one other problem,

III. those who only solved $B$,

IV. those who only solved $C$,

V. those who solved both $B$ and $C$.

Thus, each of them falls into one and only one group.

Let the number of members in groups III, IV, and V be $b, c, d$ respectively. Only these did not solve $A$. Among them, $c+d$ solved $C$, and $b+d$ solved $B$. The latter number is twice the former:

\[
b+d=2(c+d)
\]

From this,

\[
d=b-2c
\]

and since this number cannot be negative,

\[
b \geq 2c
\]

Those who solved only one problem are members of groups I, III, and IV. Among them, the members of groups III and IV are those who did not solve $A$. Their number is $b+c$, so the number of those who solved only one problem is $2(b+c)$, and thus the number of those who only solved $A$ is also $b+c$, and therefore the number of members in group II is $b+c-1$.

Now, from the total number of members in the 5 groups, taking (1) into account,

\[
\begin{gathered}
(b+c)+(b+c-1)+b+c+(b-2c)=4b+c-1=25 \\
4b+c=26, \quad \text{or} \quad c=4(6-b)+2
\end{gathered}
\]

Since $b$ and $c$ are non-negative integers, from the latter form, $b \leq 6$ and $c \geq 2$. On the other hand, from (2) and the first form of (3), $8c+c \leq 26 < 27, c < 3$, so $c=2$, and $b=6$. Thus, there were 6 students who only solved the $B$ problem. (The number of members in groups I, II, and V are 8, 7, and 2, respectively.)

Simon Csaba (Szombathely, Nagy Lajos Gymnasium IV. grade)

Note. In the last step, it is enough to read from (1) in the form $b=2c+d$: $b \geq c$. Thus, from (3), $5b \geq 26, b \geq 6$, so $b=6$.