Raskina I.V.

In a singing competition, a Rooster, a Crow, and a Cuckoo participated. Each member of the jury voted for one of the three performers. The Woodpecker calculated that there were 59 judges in total, and that the Rooster and the Crow received a total of 15 votes, the Crow and the Cuckoo received 18 votes, and the Cuckoo and the Rooster received 20 votes. The Woodpecker is a poor counter, but each of the four numbers he mentioned differs from the correct number by no more than 13. How many judges voted for the Crow?
The number of votes cast for the Rooster and the Raven cannot be more than $15+13=28$. Similarly, the total number of votes for the Raven and the Cuckoo cannot exceed $18+13=31$, and for the Cuckoo and the Rooster, it cannot exceed $20+13=33$ votes. By adding these three results, we estimate the upper bound of twice the total number of votes. Thus, the number of jury members is no more than $(28+31+33): 2 = 46$. On the other hand, from the first announcement of the Woodpecker, it is no less than

$59-13=46$. Therefore, the number of jury members is exactly 46, and all inequalities are actually equalities. The number of votes for the Raven can be found as the difference between the total number of jury members and the sum of those who voted for the Cuckoo and the Rooster: 46 - $33=13$ votes.

## [ Chessboard coloring $]$  Problem 116025 Topics: [ Numerical tables and their properties ]  $[\quad$ Mean values $\quad]$

## Author: S. Pricka

In the Pythagorean multiplication table, a rectangular frame one cell thick was highlighted, with each side of the frame consisting of an odd number of cells. The cells of the frame were alternately colored in two colors - black and white. Prove that the sum of the numbers in the black cells is equal to the sum of the numbers in the white cells.

The Pythagorean multiplication table is a grid where the intersection of the $m$-th row and the $n$-th column contains the number $mn$ (for any natural numbers $m$ and $n$).

## Solution

Let the corners of the frame be black. Each white number in the frame is equal to the half-sum of its black neighbors along the frame, and each black number is included in two half-sums. By adding these equalities, we get that the sum of the numbers written in the white cells is equal to the sum of the numbers written in the black cells.

| $1 \cdot 1$ | $1 \cdot 2$ | $1 \cdot 3$ | $1 \cdot 4$ | $1 \cdot 5$ | $1 \cdot 6$ | $1 \cdot 7$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $2 \cdot 1$ | $2 \cdot 1$ | $2 \cdot 1$ | $2 \cdot 1$ | $2 \cdot 1$ | $2 \cdot 1$ | $2 \cdot 1$ |
| $3 \cdot 1$ | $3 \cdot 1$ | $3 \cdot 1$ | $3 \cdot 1$ | $3 \cdot 1$ | $3 \cdot 1$ | $3 \cdot 1$ |
| $4 \cdot 1$ | $4 \cdot 1$ | $4 \cdot 1$ | $4 \cdot 1$ | $4 \cdot 1$ | $4 \cdot 1$ | $4 \cdot 1$ |
| $5 \cdot 1$ | $5 \cdot 1$ | $5 \cdot 1$ | $5 \cdot 1$ | $5 \cdot 1$ | $5 \cdot 1$ | $5 \cdot 1$ |
| $6 \cdot 1$ | $6 \cdot 1$ | $6 \cdot 1$ | $6 \cdot 1$ | $6 \cdot 1$ | $6 \cdot 1$ | $6 \cdot 1$ |
| $7 \cdot 1$ | $7 \cdot 1$ | $7 \cdot 1$ | $7 \cdot 1$ | $7 \cdot 1$ | $7 \cdot 1$ | $7 \cdot 1$ |
| $8 \cdot 1$ | $8 \cdot 1$ | $8 \cdot 1$ | $8 \cdot 1$ | $8 \cdot 1$ | $8 \cdot 1$ | $8 \cdot 1$ |

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