Impressions of an Olympiad participant

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student championship. However, the unofficial results of the team championship based on the total points scored by teams from different countries are always of great interest.

The best result was shown by the Vengrin team, which scored 255 points out of 336 possible; the USSR team had 205 points. This is followed by the teams of the GDR - 142 points, Poland - 118 points, Great Britain - and Romania - 110 points each. Each of the other teams scored less than 100 points.

The ceremonial closing of the Olympiad took place at the Jan Amos Kamensky Brotherhood University, where the winners of the Olympiad were awarded diplomas and prizes. There is no doubt that for many young Olympiad participants, this international mathematical meeting is just the beginning of a great, interesting life in modern mathematics.

At the final meeting, the head of the Polish delegation began to merge the countries participating in the XIII Olympid into the XIV Olympid in the Polish People’s Republic.

Objectives of the XIII International Olympiad

First day of perfection

1. Prove that the following statement:
   “For any valid

On July 10, our team arrived in Bratcheskaya. We were met at the station by the organizers, who immediately managed to create a telluo and relaxed atmosphere at the Olympiad. At the hotel we were able to meet our opponents. Members of the other teams, as soon as they met us, began to give their tasks. Among the problems whose conditions we were able to analyze, one proposed by the Yugoslavian ribs stood out:
“Users. The first one says: “Two turtles are crawling behind me,” the second one says: “One turtle is crawling in front of me and one turtle is crawling behind me,” and the third is: “Two turtles are crawling in front of me and one turtle is crawling behind me.” How can this be? Its solution generates its own strength and persuasiveness*).

The next morning the students were taken to Zilini

*) Answer: the third turtle is lying.

nau, where the Olympics were to take place. Along the way, everyone looked with satisfaction at the modern resort of Pstyatyny and the ancient town of Trencin. The cameras worked without interruption. In Zilina, we first examined the significance of the construction technical, where the condition of the northern loophole should be. On the way back, on one of the mountain roads, the bus with the teams of Austria, Bulgaria, Cuba and the GDR “lay down” on its side. Everyone rushed

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Members of the Soviet commission. Glory to the right: V. Erokhin, D. Logachev, O. Lyashko, I. S. Petrakov, A. Aleksadrov; second row: S. Gashkov, E. Sallinen, A. Chistov, M. Tsfasman.

to help, but when they found out that he was not injured in any way, the cameras started rolling again. The incident distracted us from the thought of the upcoming event and convinced us that JECTS, the familiarization of “the bus overturned,” “sound” approximately the same in all languages.

The main days have arrived—the days of accomplishment. On the 13th the official opening of the Olympids took place. At the Reznidum are the organizers and leaders of the commission and the leaders of our team, V. A. Svortsov and I. S. Petrakov. They wished us success. Even before this, the serial numbers of the participants were reported (according to the Slovak alphabet):

1. Aleksadrov Alexey.
2. Tsfasman Mikhail.
3. Chistov Alexander.
4. Gashkov Sergey.
5. Erokhin Vasily.
6. Logachev Dmitry.
7. Lyashko Oleg.
8. Sallinen Evgeniy. The guys from other commissions received similar numbers. All participants with number 1 were placed in room 1, number 2 in room 2, and so on. The envelopes lying on the tables contained information about the first day of the commission.

The first task is quite simple. Our team received 34 okans for it. The case n=3 can be proved algebraically. For n=5, we note that for $a_1 > a_2 > ... > a_5$ (for definition), the first, third and fifth terms of the sum are positive, the second and fourth are negative, and in absolute value the first is greater than the second, and the fifth is greater than the fourth. For other n, contradictory examples are constructed.

The second and third tasks turned out to be the most difficult for our commission. The second problem has a non-trivial idea: all polytopes $P_1, P_2, ..., P_9$ are contained in the polytopes, we obtain from $P_1$ is prepared for the coefficient 2 of the relative point $A_1$. For this task the team received a total of 21 okans. In the third problem we prove that from the first k terms of such a sequence we can construct the $k+1$th term. Let’s take all prime divisors of these k terms of the sequence: $p_1, p_2, ..., p_m$; number

$$2^{(p_1-1) (p_2-1) ... (p_m-1)} - 1$$

is divisible by $p_1, p_2, ..., p_m$. But since all these primes are odd, the number $2^{(p_1-1) (p_2-1) ... (p_m-1)} - 3$ is not divisible by any of them, therefore, it is coprime with all terms of the sequence. For this task we also received little, only 27 okans. In the evening of the same day, some kind of excursion took place, there was no memory of it, since all thoughts were absorbed in tasks.

The tasks of the second day, according to the Soviet commission, were easier than the first. The fourth problem would probably be more difficult if the question did not contain a hint: the fact that the angle $\alpha$ appears in the expression for the long one causes a natural desire to “create” it on the plane. It is solved by “unfolding” the notebook onto a plane. This task gave the team 34 okans.

Please, the easiest of all the problems was the fifth one. Further, I can’t believe that 7 okans were given for her solution, the same as for the second one. It is solved by the method of mathematical induction. It is only necessary to prove that for the system $S$, which satisfies the conditions of the problem for $k$ points, there is a parallel transfer that forms (together with $S$) the system $S_1$, for which the conditions of the problem for $k+1$ points are satisfied. For this task our team received 51 okans.

In Problem 6, the easiest way is to assume the opposite and write down all possible inequalities for the row with the minimum sum of the numbers in it and, similarly, for the column. Then you need to add up all the inequalities and take into account that the sum in the selected row (column) does not exceed $\frac{n}{2}$. For the sixth task, our team received 38 okans.

We learned these results, but immediately after the performance it seemed to the whole team that the weights had performed poorly. The only thing that consoled us a little was a conversation with