[Mathematical logic (miscellaneous).] Case analysis $\quad]$

After a hockey game, Anton said that he scored 3 goals, and Ilya only one. Ilya said that he scored 4 goals, and Seryozha as many as 5. Seryozha said that he scored 6 goals, and Anton only two. Could it be that together they scored 10 goals, given that each of them told the truth once and lied once?
There are two cases.

1) Anton told the truth about himself, that is, he scored 3 goals. Then Seryozha lied about Anton, so he told the truth about himself, that is, he scored 6 goals. Therefore, Ilya lied about Seryozha and told the truth about himself, that is, he scored 4 goals. In this case, the boys scored a total of $3+6+4=13$ goals.
2) Anton lied about himself. Then he told the truth about Ilya, that is, Ilya scored 1 goal. Then Ilya also lied about himself and told the truth about Seryozha: Seryozha scored 5 goals. Therefore, Seryozha lied about himself and told the truth about Anton: the latter scored 2 goals. In this case, the boys scored a total of $1+5+2$ = 8 goals.

## Problem 64941 Topics: [ \[\begin{array}{l}\text { Equations in integers } \\ \text { [Proof by contradiction }]\end{array}\] \]

Can different natural numbers be placed in the circles (see figure) such that the sums of three numbers along each segment are equal?

![](https://cdn.mathpix.com/cropped/2024_05_06_e123f5c8ab3546621c69g-05.jpg?height=395&width=458&top_left_y=474&top_left_x=800)

## Solution

Suppose the required arrangement exists, $S$ - the sum of all the numbers placed, $a$ and $b$ - the numbers in the circles located at any two vertices of the triangle. Then for the vertex where the number $a$ is located, the sum of the numbers along the three segments containing this vertex is $S+2 a$. Similarly, for the vertex where the number $b$ is located, this sum is $S+2 b$. Since the sums of the numbers along each segment are equal, then $S+2 a=$ $S+2 b$, that is, $a=b$. But this contradicts the condition.

## Answer

No.