Problem 877

Consider one of the teams, denoted as A. Over 8 rounds, it played against eight teams and did not play against nine teams. If among these nine teams, there are two teams $B$ and $C$ that did not play against each other, then $A, B$, and $C$ form the desired triplet.

Otherwise, these 9 teams played a full round-robin tournament. This required $9 \cdot 8$ $: 2=36$ matches. However, in each round, they could play no more than four matches, so over 8 rounds, such matches could not exceed 32. This is a contradiction.

## Answer

\section*{[ Numerical Tables and Their Properties ]

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Author: Armunov M.

In the cells of an $8 \times 8$ board, numbers 1 and -1 are placed (one number per cell). Consider all possible placements of a figure $\square$ on the board (the figure can be rotated, but its cells must not go beyond the board's boundaries). We will call such a placement unsuccessful if the sum of the numbers in the four cells of the figure is not equal to 0. Find the smallest possible number of unsuccessful placements.

## Solution

Estimate. We will show that in each "cross" of five cells on the board, there is at least one unsuccessful placement. Let the numbers in the outer cells of the cross be $a, b, c, d$, and in the central cell be $e$; denote the sum of all these five numbers by $S$. Suppose all placements in the cross are successful. Then

$S-a=S-b=S-c=S-d=0$, hence $a=b=c=d$. Therefore, $e + 3a = 0$, which means $e = -3a = \pm 3$, which is impossible. Thus, in each of the 36 "crosses" (with centers in all non-edge cells), there is an unsuccessful placement of the figure. Clearly, each placement is contained in no more than one cross; therefore, there are at least 36 such placements.

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An example of a placement where the number of unsuccessful placements is 36 is shown in the figure (the sign of the corresponding number is indicated in each cell). It is easy to verify that in each cross, there is exactly one unsuccessful placement, and all placements adjacent to the long side of the board are successful.

## Answer

36.

Let's consider one of the teams, denoted as A. Over 8 rounds, it played against eight teams, and did not play against nine teams. If among these nine teams there are two teams $B$ and $C$ that did not play against each other, then $A, B$, and $C$ form the desired triplet.

Otherwise, these 9 teams played a full round-robin tournament among themselves. This required $9 \cdot 8$ $: 2 = 36$ matches. However, in each round, they could play no more than four matches against each other, so in 8 rounds, such matches could not exceed 32. This is a contradiction.

## Answer

Correct.