## Task B-2.5.

The math test consisted of three problems. Professor Max informed the students of their test results in a somewhat vague manner, setting a new problem:

The first and second problems were correctly solved by 7 students, the second and third problems by 9 students, and the first and third problems by as many as 13 students. What is the minimum number of students who correctly solved all three problems?

It is known that there are 24 students in the class and that no student failed to solve any of the problems.
## Solution.

Let the number of students who solved only the first and second problems be denoted by $a$, only the first and third by $b$, only the second and third by $c$. Let $d$ be the number of students who correctly solved all three problems, $e$ the number who correctly solved only the first problem, $f$ only the second, and $g$ only the third.

![](https://cdn.mathpix.com/cropped/2024_05_30_1c7eac828d9f4d45279fg-10.jpg?height=599&width=651&top_left_y=1408&top_left_x=722)

Then we have $a+d=7, b+d=13, c+d=9, a+b+c+d+e+f+g=24$.

Adding the first three equations: $a+b+c+3d=29$. Since $a+b+c+d+e+f+g=24$, it follows that $a+b+c+3d+e+f+g=24+2d, 29+e+f+g=24+2d, e+f+g+5=2d$.

According to the problem's condition, $e+f+g$ is a non-negative number.

If $e+f+g=0$, then $2d=5$, which is impossible.

When $e+f+g=1$, we have $d=3$. This is the smallest possible value for $d$.

The smallest number of students who correctly solved all three problems is three.

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B variant

## October 26, 2020.