Competition problem — Combinatorics

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Combinatorics Difficulty 6.5 National Olympiad Prove it Mongolia

Three problems $A$ , $B$ and $C$ were given in a mathematical olympiad and there were $25$ participants. Every participant solved at least $1$ problem. Among the participants who didn't solve problem $A$ , the number of participants who solved $B$ is twice as much as the number of participants who solved problem $C$ . The number of participants who solved only problem $A$ is $1$ more than the number of remaining participants who solved problem $A$ . Half of the participants who solved just one problem didn't solve problem $A$ . How many participants solved just one problem $B$ ?

Solution

Let $A$ , $B$ and $C$ be the numbers of participants who solved just one problem $A$ , $B$ , $C$ respectively. And let $AB$ be the number of participants who solved only $A$ and $B$ and etc. Then
$\left\{ \begin{array}{l} A + B + C + AB + AC + BC + ABC = 25 \quad (1) \\ B + BC = 2(C + BC) \quad (2) \\ A - 1 = AB + AC + ABC \quad (3) \\ A + B + C = 2(B + C) \quad (4) \end{array} \right.$
From (1) and (3) implies $2A + B + C + BC = 26$ (5). (4) is equivalent to $A = B + C$ and (2) is equivalent to $BC = B - 2C$ . From (5), (2) and (4), $4B + C = 26$ implies $C = 26 - 4B$ (6) and $BC = B - 2C = B - 2(26 - 4B) = 9B - 52$ (7). Considering $B, C, BC$ are non-negative integers, $B = 6$ will imply from (6) and (7).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.