$ a=\text{the number of participants who solved only problem A},\\b=\text{the number of participants who solved only problem B},\\c=\text{the number of participants who solved only problem C},\\d=\text{the number of participants who solved problem B and problem C but not problem A}$ DPopov wrote: Of all the contestants who did not solve problem $ A$, the number who solved $ B$ was twice the number who solved $ C$. $ b+d=2(c+d)\quad\ldots(1)$ DPopov wrote: The number of students who solved only problem $ A$ was one more than the number of students who solved $ A$ and at least one other problem. $ a=1+(25-a-b-c-d)\quad\ldots(2)$ DPopov wrote: Of all students who solved just one problem, half did not solve problem $ A$. $ a=b+c\quad\ldots(3)$ (2) and (3): $ 3b+3c=26-d\quad\ldots(4)$. (1): $ b-2c=d\quad\ldots(5)$. (4) and (5): $ 4b+c=26\quad\ldots(6)$ (5): $ b-2c=d\ge0\implies b\ge 2c\quad\ldots(7)$ (i) $ b\le 5$ (7): $ c\le 2$ (6): $ c\ge 6$ Contradiction (ii) $ b\ge 7$ (6): $ c\le -2$ Contradiction $ \boxed{b=6}$

i didnt get how did you get B = 6 ... can you please explain how did you compare at the end of your solution....

I did some algebraic manipulations, but without inequalities. Here's what I did: Define the following variables: $ a$: Number of students who only solved A $ b$: " " B $ c$: " " C $ w$: Number of students who solved A and B but not C $ x$: " " A and C but not B $ z$: " " B and C but not A $ y$: Number of students who solved all 3 problems Then, based on the given information, we can get the following equations: $ a + b + c + w + x + y + z = 25$ $ b + z = 2(z + c)$ $ \Rightarrow b + z = 2z + 2c$ $ \Rightarrow z = b - 2c$ $ a = 1 + w + x + y$ $ a = b + c$ Using substitution, the first equation becomes $ 2b + 2c + a - 1 + b - 2c = 25$ $ 3b + 3c - 1 + b - 2c = 25$ $ 4b + c = 26$ Now the possible ordered pairs $ (b,c)$ of positive integers $ b$, $ c$ that satisfy this equation are: $ b = 1, c = 22$ $ b = 2, c = 18$ $ b = 3, c = 14$ $ b = 4, c = 10$ $ b = 5, c = 6$ $ b = 6, c = 2$ By doing some calculations, we see that the first four ordered pairs listed above are not possible because that will resut in $ a + b + c = 2(b + c) > 25$, which is not possible. Also, $ b = 5$ and $ c = 6$ won't work because that will result in $ z = 5 - 12 = - 7$, which isn't possible. Now let's try $ b = 6$ and $ c = 2$. Substituting and calculating gives us $ a = 8$, $ z = 2$, and $ w + x + y + z = 7$, all of which are possible. Therefore $ \boxed{6}$ students in the olympiad solved only problem $ B$.