Auto: prishn i.f.

There are nine wrestlers of different strengths. In any match between any two of them, the stronger one always wins. Is it possible to divide them into three teams of three wrestlers each so that in matches between the teams on a round-robin system, the first team wins more matches than the second, the second team wins more matches than the third, and the third team wins more matches than the first?
Assign each of the fighters a "rating" from 9 to 1: 9 - the strongest, and so on. Try to form teams so that the sums of the ratings of the fighters in the teams are equal.

## Solution

We will order our fighters by strength and assign each a rating from 9 to 1. Then the sum of the ratings of the fighters is 45. We will try to form teams so that the sums of the ratings of the fighters in the teams are equal. Let's draw a $3 \times 3$ magic square and consider its rows. These are three teams with the same total rating - 15.

| 2 | 7 | 6 |
| :--- | :--- | :--- |
| 9 | 5 | 1 |
| 4 | 3 | 8 |

Let's check. The first with the second - score $5: 4$, the second with the third - score $5: 4$, and the third with the first - score $5: 4$!

## Answer

It is possible.

## [GCD and LCM. Mutual simplicity  Problem 104056 Topics: [Connectivity and decomposition into connected components]  Difficulty: $4-$  [Examples and counterexamples. Constructions] Classes: $7,8,9,10$

The hostess made a pirog (a type of Russian pie) and wants to pre-cut it into such (not necessarily equal) pieces so that the pie can be evenly divided both among five and among seven people. What is the minimum number of pieces she can manage with?

## Solution

This problem is a special case of problem $\underline{98057 .}$

## Answer

11 pieces.