The All-Russian School Olympiad has become a good tradition. Its main task is to identify gifted children, to motivate schoolchildren to in-depth study of subjects, to develop creative abilities and out-of-the-box thinking in children.

The Olympic movement is gaining more and more popularity among schoolchildren. And there are reasons for that:

• the winners of the all-Russian round are admitted to universities out of competition if the subject is an olympiad subject (diplomas of the winners are valid for 4 years);
• participants and prize-winners receive additional chances upon admission to educational institutions (if the subject is not in the profile of the university, the winner receives an additional 100 points upon admission);
• significant cash reward for prize places (60 thousand, 30 thousand rubles;
• and, of course, fame throughout the country.

Before becoming a winner, you must go through all the stages of the All-Russian Olympiad:

1. The initial school stage, at which worthy representatives for the next stage are determined, is held in September-October 2017. The organization and conduct of the school stage is carried out by specialists of the methodological office.
2. The municipal stage is held between the schools of the city or district. It takes place at the end of December 2017. - early January 2018
3. The third round is more difficult. It is attended by talented students from all over the region. The regional stage takes place in January-February 2018.
4. The final stage determines the winners of the All-Russian Olympiad. In March-April, the best children of the country compete: the winners of the regional stage and the winners of the last year's Olympiad.

The organizers of the final round are representatives of the Ministry of Education and Science of Russia, they are also summing up the results.

You can show your knowledge in any subject: mathematics, physics, geography, even physical education and technology. You can compete in erudition in several subjects at once. There are 24 disciplines in total.

Olympiad subjects are subdivided according to the directions:

№	Direction	Items
1	Exact disciplines	mathematics, computer science
2	Natural science disciplines	geography, biology, physics, chemistry, ecology, astronomy
3	Philological disciplines	literature, Russian language, foreign languages
4	Humanities	economics, social studies, history, law
5	Other	art, technology, physical education, basics of life safety

The peculiarity of the final stage of the Olympiad consists in two types of tasks: theoretical and practical. For example, to get good results in geography, students must complete 6 theoretical tasks, 8 practical tasks, and also answer 30 test questions.

The first stage of the Olympiad begins in September, which means that those wishing to take part in the intellectual marathon should prepare in advance. But above all, they must have a good school-level base, which constantly needs to be replenished with additional knowledge that goes beyond the school curriculum.

The official website of the Olympiad www.rosolymp.ru places tasks from previous years. These materials can be used in preparation for an intellectual marathon. And of course, you cannot do without the help of teachers: additional classes after lessons, classes with tutors.

The winners of the final stage will take part in international Olympiads. They form the Russian national team, which will train at the training camp in 8 subjects.

To provide methodological assistance, the site conducts installation webinars, the Central Organizing Committee of the Olympiad, subject-methodological commissions have been formed.

• Contest
• Olympiad
• Competition-game
• Subject week
• Family competition
• Children with disabilities
• Control test
• Summer camp
• Online tests

Distance Olympiads of the "Snail" Center

Goals and objectives of the distance olympiads of the "Snail" Center:

• checking the level of knowledge of students
• formation of the skill of self-appropriation of knowledge
• formation and development of skills of independent search and analysis of information
• formation and development of skills for using Internet services in education
• increasing motivation to study the subject

Olympiad

They give the participant an opportunity to check and deepen his knowledge of a specific school discipline or even one of its sections. All tasks of distance olympiads are divided by age groups and correspond to school curricula and the requirements of the Federal State Educational Standard.

Competition-game

They give the participant an opportunity to check and deepen his knowledge of a specific school discipline or even one of its sections. All tasks of distance olympiads are divided by age groups and correspond to school curricula and the requirements of the Federal State Educational Standard.

Subject week

They give the participant an opportunity to check and deepen his knowledge of a specific school discipline or even one of its sections. All tasks of distance olympiads are divided by age groups and correspond to school curricula and the requirements of the Federal State Educational Standard.

Family competition

They give the participant an opportunity to check and deepen his knowledge of a specific school discipline or even one of its sections. All tasks of distance olympiads are divided by age groups and correspond to school curricula and the requirements of the Federal State Educational Standard.

Specialist. contests

They give the participant an opportunity to check and deepen his knowledge of a specific school discipline or even one of its sections. All tasks of distance olympiads are divided by age groups and correspond to school curricula and the requirements of the Federal State Educational Standard.

Every year, for schoolchildren of any schools in the Russian Federation, a lot of various Olympiads are held, allowing students to show their knowledge and skills in subjects included in the list of programs of general educational institutions of the country. Participation in such events is considered a very prestigious and responsible task, in which schoolchildren demonstrate the knowledge accumulated over the years of study and defend the honor of their own school. In case of victory, there is an opportunity to earn a certain privilege for further admission to Russian universities and receive a small monetary reward.

Historical summary

For the first time, the educational authorities of Russia provided the opportunity for competition between young students back in the distant 1886. During the prosperity of the Soviet Union, such a movement received an additional impetus for further development. In the 60s of the last century, school Olympiads began to be held in almost every discipline related to the general education program of compulsory education. Initially, such competitions were more of an all-Russian scale, which in the future became all-Union.

To find out which subjects such a competition will consist of in the future period, all school Olympiads for 2017-2018 should be announced.

Present time

In the next academic year, the best students will be able to test their own knowledge in the Olympiads in several categories of disciplines.

1. Natural sciences: geography, physics, biology, chemistry, ecology and astronomy.
2. Humanities: history, social studies, economics and law.
3. Exact sciences: mathematics, computer science.
4. Philology: English, French, Chinese, Italian and Russian, as well as Russian literature.
5. Other disciplines: physical education, life safety, technology and world artistic culture.

In each of the listed disciplines, there are two blocks of tasks: a part aimed at finding practical skills and a part that checks the theoretical base of each participant.

The main stages of the Russian Olympiads

The All-Russian Olympiad consists of the organization and further holding of 4 stages of intellectual competition, held at different levels. Representatives of regional educational institutions and schools determine the final schedule of each Olympiad and its location. Of course, the exact list of each competition for next year has not yet been compiled, but current applicants should be guided by the following dates.

1. School stage. The competition between rivals from one educational application starts almost at the beginning of the academic year - September-October 2017. The Olympiads will concern students of the same parallel, starting from the fifth grade. Members of the city-level methodological commission are responsible for the development of assignments.

2. Municipal stage. The next stage, at which competitions are held between the winners of the previous link of 7-11 grades from the same city. The time of the Olympiad is December 2017-January 2018. The organizers of such an event are representatives of the educational sphere of the regional level, while officials are responsible for the place, time and procedure of the competition.

3. Regional stage. The next level of the All-Russian Olympiad, held in January-February. It is attended by schoolchildren who took leading places in similar competitions at the city level, as well as the winners of the regional selection of the past year.

4. All-Russian stage. The highest level of the subject Olympiad is organized by representatives of the Ministry of Education of the Russian Federation in March-April 2018. The winners of the regional Olympiad and prize-winners of the past year will be able to take part in it. The exceptions are schoolchildren who took 1st place, but lagged behind participants from other cities. The winners of the marked stage get the right to participate in a similar international competition scheduled for next summer.

List of school Olympiads with their main features

Any of the school Olympiads consists of 3 main stages, each of which is characterized by distinctive properties. For example, the winners have a number of privileges over their opponents from the other two groups - the opportunity to enroll in a university, on the basis of which the Olympiad itself was held. In this case, the entrance exams for admission to the first year will be canceled automatically. Winners or prize-winners of the 3rd stage in this sense do not have any indulgences.

Today it is already known that the list of I-level school Olympiads consists of the following directions and disciplines.

1. Olympiad "Lomonosov", consisting of a huge number of different subjects.
2. "Nanotechnology - a breakthrough into the future" - All-Russian Olympiad for every student who wants to.
3. All-Siberian Chemistry Olympiad.
4. "Young talents" - geography.
5. Open Programming Contest.
6. Olympiad in astronomy for schoolchildren from St. Petersburg.
7. Open Olympiad "culture and art".
8. All-Russian economic Olympiad for schoolchildren named after ND Kondratiev in economics.
9. Moscow Olympiad in Physics, Mathematics, Informatics.

The list of the II level Olympiad consists of the following directions.

1. Herzen's Olympiad in a foreign language.
2. South Russian Olympiad for schoolchildren "Architecture and Art" with the following subjects: painting, drawing, composition and drawing.
3. Interregional Olympiad of Moscow State Pedagogical University in Law.
4. All-Siberian Open Olympiad in Informatics, Mathematics, Biology.
5. Interregional Olympiad "Highest standard" in computer science, literature, history of world civilization and oriental studies.
6. Interregional Olympiad "future researchers - the future of science" in biology.
7. City Open Olympiad in Physics.
8. Interdisciplinary Olympiad named after V. I. Vernadsky in social studies and history.
9. Engineering Physics Olympiad.
10. Eurasian linguistic Olympiad in a foreign language of the interregional level.

The 2017-2018 level III Olympiads are represented by the following list of competitions.

1. “Mission is achievable. Your vocation is a financier! " from the economy.
2. Herzen Olympiad in Geography, Biology and Pedagogy.
3. "In the beginning was the Word ..." in history and literature.
4. All-Russian tournament for young physicists.
5. All-Russian Sechenov Olympiad in Chemistry and Biology.
6. All-Russian chemical tournament.
7. "Learn to build the future" with urban planning and architectural graphics.
8. All-Russian Tolstoy Olympiad in history, literature and social studies.
9. All-Russian Olympiad of representatives of musical institutions of the Russian Federation on string instruments, musical pedagogy, folk orchestra instruments, choral conducting and performing.
10. All-Russian competition of scientific works "Junior" in engineering and natural sciences.

The noted list of the most relevant Olympiads in Russia has been operating over the past several years. True, having familiarized yourself with all the competitions, a completely logical question arises: what is the difference between tasks at all levels? First of all, we are talking about the level of preparation of schoolchildren.

To become not only an ordinary representative of the Olympiad, but even to take a prize place, one should have a fairly high level of training. On some Internet portals, you can find olympiad tasks from past years in order to check your own level with the help of ready-made answers, find out the approximate start time of the competition and some organizational points.

All-Russian Olympiads for schoolchildren are held under the auspices of the Ministry of Education and Science of Russia after the official confirmation of the calendar of their dates. Such events cover almost all disciplines and subjects included in the compulsory curriculum of general education schools.

When participating in such competitions, students are given the opportunity to gain experience in answering the questions of intellectual competitions, as well as expand and demonstrate their knowledge. Students begin to respond calmly to various forms of knowledge testing, are responsible for representing and protecting the level of their school or region, which develops a sense of duty and discipline. In addition, a good result can bring a well-deserved cash prize or advantages during admission to the leading universities in the country.

Olympiads for schoolchildren of the 2017-2018 academic year are held in 4 stages, subdivided according to the territorial aspect. These stages in all cities and regions are carried out in the general calendar terms established by the regional leadership of the educational municipal departments.

Schoolchildren taking part in the competition go through four levels of competition in stages:

• Stage 1 (school). In September-October 2017, competitions will be held within each separate school. All parallels of students are tested independently from each other, from the 5th grade to graduates. The assignments for this level are prepared by the methodological commissions of the city level, they also provide assignments for the district and rural secondary schools.
• Stage 2 (regional). In December 2017 - January 2018, the next level will be held, in which the winners of the city and the district will take part - students of grades 7-11. Tests and assignments at this stage are developed by the organizers of the regional (third) stage, and all questions on preparation and locations for conducting are assigned to the local authorities.
• Stage 3 (regional). The time of the event is from January to February 2018. The winners of the Olympiads of the current and completed year of study become participants.
• Stage 4 (All-Russian). Organized by the Ministry of Education and runs from March to April 2018. The winners of the regional stages and the winners of the last year participate in it. However, not all winners of the current year can take part in the All-Russian Olympiads. The exception is children who took 1st place in the region, but significantly lagged behind the other winners in points.

The winners of the All-Russian level, if they wish, can take part in international competitions taking place during the summer holidays.

List of disciplines

In the 2017-2018 academic season, Russian schoolchildren can test their strength in the following areas:

• exact sciences - analytical and physical and mathematical direction;
• natural sciences - biology, ecology, geography, chemistry, etc .;
• philological sector - various foreign languages, native language and literature;
• humanitarian direction - economics, law, historical sciences, etc .;
• other items - art and, BJD.

This year, the Ministry of Education officially announced 97 Olympiads, which will be held in all regions of Russia from 2017 to 2018 (9 more than last year).

Benefits for winners and awardees

Each Olympiad has its own level: I, II or III. The most difficult is the I level, but it gives its diploma winners and prize-winners the most advantages when entering many prestigious universities in the country.

Benefits for winners and runners-up come in two categories:

• admission without exams to the selected university;
• awarding the highest USE score in the discipline in which the student received a prize.

The most famous state competitions of the I level include the following Olympiads:

• St. Petersburg Astronomical;
• "Lomonosov";
• St. Petersburg State Institute;
• "Young talents";
• Moscow school;
• "Highest test";
• "Information Technology";
• "Culture and Art", etc.

Olympiads II level 2017-2018:

• Herzenovskaya;
• Moscow;
• "Eurasian Linguistic";
• "Teacher of the school of the future";
• Lomonosov Tournament;
• TechnoCup, etc.

Competitions of the III level of 2017-2018 include the following:

• "Star";
• "Young talents";
• Scientific works competition "Junior";
• Energy Hope;
• "Step into the Future";
• "Ocean of Knowledge", etc.

According to the Order "On Amendments to the Procedure for Admission to Universities", the winners or awardees of the final stage have the right to enter any university without admission tests in the direction corresponding to the profile of the Olympiad. At the same time, the correlation between the direction of training and the profile of the Olympiad is determined by the university itself and without fail publishes this information on its official website.

The right to use the benefit remains with the winner for 4 years, after which it is canceled and the receipt takes place on a general basis.

Preparation for the Olympiads

The standard structure of Olympiad tasks is divided into 2 types:

• verification of theoretical knowledge;
• the ability to translate theory into practice or demonstration of practical skills.

A decent level of training can be achieved with the help of the official website of the Russian State Olympiads, which contains the tasks of the past rounds. They can be used both to test your knowledge and to identify problem areas in training. On the same website, you can clarify the dates of the tours and get acquainted with the official results.

Video: tasks for the All-Russian Olympiad for schoolchildren appeared on the network

Tasks and keys of the school stage of the All-Russian Olympiad for schoolchildren in mathematics

Download:

Preview:

School stage

4th grade

1. Rectangle area 91

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

Grade 5

The maximum score for each task is 7 points

3. Cut the shape into three identical (overlapping) shapes:

4. Replace the letter A

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

6th grade

The maximum score for each task is 7 points

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

7th grade

The maximum score for each task is 7 points

1. - various numbers.

4. Replace the letters Y, E, A and R with numbers so that you get the correct equality:

YYYY ─ EEE ─ AA + R = 2017.

5. Live on the island the same number of people, and her

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

8th grade

The maximum score for each task is 7 points

ABM, CLD and ADK respectively. Find∠ MKL.

6. Prove that if a, b, c and - whole numbers, then the fractionwill be an integer.

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

Grade 9

The maximum score for each task is 7 points

2. Numbers a and b are such that the equations and also has a solution.

6. What natural x expression

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

Grade 10

The maximum score for each task is 7 points

4 – 5 – 7 – 11 – 19 = 22

3. In the equation

5. In triangle ABC conducted a bisector BL. It turned out that ... Prove that triangle ABL - isosceles.

6. By definition,

Preview:

Problems of the All-Russian Olympiad in Mathematics for Schoolchildren

School stage

Grade 11

The maximum score for each task is 7 points

1. The sum of two numbers is 1. Can their product be greater than 0.3?

2. Sections AM and BH ABC.

It is known that AH = 1 and ... Find side length BC.

3.a inequality true for all values NS ?

Preview:

4th grade

1. Rectangle area 91... The length of one of its sides is 13 cm. What is the sum of all the sides of the rectangle?

Answer. 40

Solution. We find the length of the unknown side of the rectangle from the area and the known side: 91: 13 cm = 7 cm.

The sum of all sides of the rectangle is 13 + 7 + 13 + 7 = 40 cm.

2. Cut the shape into three identical (overlapping) shapes:

Solution.

3. Reconstruct the addition example, where the numbers of the terms are replaced with asterisks: *** + *** = 1997.

Answer. 999 + 998 = 1997.

4 ... The four girls were eating candy. Anya ate more than Yulia, Ira - more than Sveta, but less than Yulia. Arrange the girls' names in ascending order of the candies eaten.

Answer. Sveta, Ira, Julia, Anya.

Preview:

School Olympiad Keys in Mathematics

Grade 5

1. Without changing the order of the digits 1 2 3 4 5, put arithmetic signs and brackets between them so that the result is one. It is impossible to "glue" adjacent digits into one number.

Solution. For example, ((1 + 2): 3 + 4): 5 = 1. Other solutions are possible.

2. Geese and piglets were walking in the barnyard. The boy counted the number of heads, there were 30, and then he counted the number of legs, there were 84. How many geese and how many pigs were there in the schoolyard?

Answer. 12 piglets and 18 geese.

Solution.

Step 1. Imagine that all the piglets are lifted two legs up.

Step 2. There are 30 ∙ 2 = 60 feet left on the ground.

Step 3. Raised up 84 - 60 = 24 feet.

Step 4. Raised 24: 2 = 12 piglets.

Step 5. 30 - 12 = 18 geese.

3. Cut the shape into three identical (overlapping) shapes:

Solution.

4. Replace the letter A to a nonzero digit to get the correct equality. One example is enough.

Answer. A = 3.

Solution. It is easy to show that A = 3 is appropriate, we will prove that there are no other solutions. Let us cancel the equality by A . We will receive.
If A ,
if A> 3, then.

5. Girls and boys went to the store on their way to school. Each student bought 5 thin notebooks. In addition, each girl bought 5 pens and 2 pencils, and each boy bought 3 pencils and 4 pens. How many notebooks were bought if the children bought 196 pens and pencils in total?

Answer. 140 notebooks.

Solution. Each of the students bought 7 pens and pencils. A total of 196 pens and pencils were purchased.

196: 7 = 28 students.

Each of the students bought 5 notebooks, which means that only bought
28 ⋅ 5 = 140 notebooks.

Preview:

School Olympiad Keys in Mathematics

6th grade

1. There are 30 points on a straight line, the distance between any two adjacent points is 2 cm. What is the distance between two extreme points?

Answer. 58 cm.

Solution. Between the extreme points, 29 pieces of 2 cm are placed.

2 cm * 29 = 58 cm.

2. Will the sum of the numbers 1 + 2 + 3 + ...... + 2005 + 2006 + 2007 be divisible by 2007? Justify the answer.

Answer. Will.

Solution. We represent this amount in the form of the following terms:
(1 + 2006) + (2 + 2005) + …..+ (1003 + 1004) + 2007.

Since each term is divisible by 2007, the entire sum will be divisible by 2007.

3. Cut the figurine into 6 equal checkered figurines.

Solution. The figurine can only be cut this way

4. Nastya places the numbers 1, 3, 5, 7, 9 in squares 3 by 3. She wants the sum of the numbers on all horizontals, verticals and diagonals to be divisible by 5. Give an example of such an arrangement, provided that Nastya is going to use each number no more than two times.

Solution. Below is one of the constellations. There are other solutions as well.

5. Dad usually comes to pick Pavlik after school by car. Once the lessons ended earlier than usual and Pavlik went home on foot. After 20 minutes, he met Dad, got into the car and arrived home 10 minutes early. How many minutes earlier did lessons end that day?

Answer. 25 minutes early.

Solution. The car arrived home earlier, because it did not have to get from the meeting point to the school and back, which means that a doubled car travels this way in 10 minutes, and in one direction - in 5 minutes. So, the car met Pavlik 5 minutes before the usual end of lessons. By this time Pavlik had already walked for 20 minutes. Thus, the lessons ended 25 minutes early.

Preview:

School Olympiad Keys in Mathematics

7th grade

1. Find the solution to the number puzzle a, bb + bb, ab = 60 where a and b - various numbers.

Answer. 4.55 + 55.45 = 60

2. After Natasha ate half of the peaches from the jar, the compote level dropped by one third. How much (of the received level) will the compote level decrease if you eat half of the remaining peaches?

Answer. One quarter.

Solution. It is clear from the condition that half of the peaches take up a third of the can. This means that after Natasha has eaten half of the peaches, there are equal parts of the peaches and compote left in the jar (one third each). This means that half of the number of remaining peaches is a quarter of the total volume of contents.

banks. If you eat that half of the leftover peaches, the compote level will drop by a quarter.

3. Cut the rectangle shown in the figure along the grid lines into five rectangles of different sizes.

Solution. For example, so

4. Replace the letters Y, E, A and R with numbers so that you get the correct equality: YYYY ─ EEE ─ AA + R = 2017.

Answer. With Y = 2, E = 1, A = 9, R = 5 we get 2222 ─ 111 ─ 99 + 5 = 2017.

5. Live on the island the same number of people, and e m each of them is either a knight who always speaks the truth, or a liar who always lies e t. Once all the knights said: - "I am friends with only 1 liar", and all liars: - "I am not friends with the knights." Who is more on the island, knights or liars?

Answer. More knights

Solution. Each liar is friends with at least one knight. But since each knight is friends with exactly one liar, two liars cannot have a common knight friend. Then each liar can be associated with his friend a knight, from which it turns out that there are at least as many knights as there are liars. Since there are no inhabitants on the island e number, then equality is impossible. This means there are more knights.

Preview:

School Olympiad Keys in Mathematics

8th grade

1. There are 4 people in the family. If Masha is doubled the scholarship, the total income of the whole family will increase by 5%, if, instead, the mother's salary is doubled - by 15%, if the salary is doubled to the father - by 25%. By what percentage will the income of the whole family increase if grandfather's pension is doubled?

Answer. 55%.

Solution ... When Masha's scholarship is doubled, the total family income increases by exactly the amount of this scholarship, so it is 5% of income. Likewise, mom and dad salaries are 15% and 25%. This means that the grandfather's pension is 100 - 5 - 15 - 25 = 55%, and if e e double, the family income will grow by 55%.

2. On sides AB, CD and AD of square ABCD outside equilateral triangles are built ABM, CLD and ADK respectively. Find∠ MKL.

Answer. 90 °.

Solution. Consider a triangle MAK: angle MAK equal to 360 ° - 90 ° - 60 ° - 60 ° = 150 °. MA = AK by hypothesis, therefore, the triangle MAK isosceles,∠ AMK = ∠ AKM = (180 ° - 150 °): 2 = 15 °.

Similarly, we find that the angle DKL is equal to 15 °. Then the required angle MKL is equal to the sum ∠ MKA + ∠ AKD + ​​∠ DKL = 15 ° + 60 ° + 15 ° = 90 °.

3. Nif-Nif, Naf-Naf and Nuf-Nuf shared three pieces of truffle in masses of 4, 7 and 10 g. The wolf decided to help them. He can cut off any two pieces at the same time and eat 1 g of truffle. Can the wolf leave equal pieces of truffle to the piglets? If so, how?

Answer. Yes.

Solution. The wolf can first cut off 1 g three times from pieces of 4 g and 10 g. You will get one piece in 1 g and two pieces of 7 g each. Now it remains to cut off six times and eat 1 g from pieces in 7 g, then the piglets will get 1 g of truffle.

4. How many four-digit numbers are there that are divisible by 19 and end in 19?

Answer. 5 .

Solution. Let be Is such a number. Thenis also a multiple of 19. But
Since 100 and 19 are coprime, the two-digit number is divisible by 19. And there are only five of them: 19, 38, 57, 76 and 95.

It is easy to verify that all numbers 1919, 3819, 5719, 7619 and 9519 are suitable for us.

5. A team of Petit, Vasya and a single scooter is taking part in the race. The distance is divided into sections of the same length, their number is 42, at the beginning of each there is a checkpoint. Petya runs the section in 9 minutes, Vasya - in 11 minutes, and on a scooter, any of them passes the section in 3 minutes. They start at the same time, and at the finish line the time of the one who came last is taken into account. The guys agreed that one rides the first part of the way on a scooter, the rest is running, and the other is the other way around (the scooter can be left at any checkpoint). How many sections does Petya have to ride on a scooter for the team to show the best time?

Answer. eighteen

Solution. If the time of one becomes less than the time of the other of the guys, then the time of the other and, consequently, the time of the team will increase. This means that the guys' time must coincide. Designating the number of sections Petya passes through x and solving the equation, we get x = 18.

6. Prove that if a, b, c and - whole numbers, then the fractionwill be an integer.

Solution.

Consider , by condition this number is an integer.

Then and will also be an integer as a difference N and doubled integer.

Preview:

School Olympiad Keys in Mathematics

Grade 9

1. Sasha and Yura have been together for 35 years now. Sasha is now twice as old as Yura was when Sasha was as old as Yura is now. How old is Sasha now and how old is Yura?

Answer. Sasha is 20 years old, Yura is 15 years old.

Solution. Let Sasha now x years, then Yura , and when Sasha wasyears, then Yura, by condition,... But equal time has passed for both Sasha and Yura, so we get the equation

from which .

2. Numbers a and b are such that the equations and have solutions. Prove that the equationalso has a solution.

Solution. If the first equations have solutions, then their discriminants are nonnegative, whence and ... Multiplying these inequalities, we obtain or , whence it follows that the discriminant of the last equation is also non-negative and the equation has a solution.

3. The fisherman caught a large number of fish weighing 3.5 kg. and 4.5 kg. His backpack holds no more than 20 kg. What is the maximum weight of fish he can take with him? Justify the answer.

Answer. 19.5 kg

Solution. The backpack can hold 0, 1, 2, 3 or 4 fish weighing 4.5 kg.
(no more since). For each of these options, the remainder of the backpack capacity is not divisible entirely by 3.5 and, at best, can be packed kg. fishes.

4. The shooter shot ten times at a standard target and knocked out 90 points.

How many hits were there in the seven, eight and nine, if there were four ten, and there were no other hits or misses?

Answer. In the seven - 1 hit, in the eight - 2 hits, in the nine - 3 hits.

Solution. Since the shooter hit only seven, eight and nine in the other six shots, then in three shots (since at least once the shooter hit seven, eight and nine) he will dialpoints. Then for the remaining 3 shots you need to score 26 points. What is possible with the only combination 8 + 9 + 9 = 26. So, the arrow hit the seven 1 time, the eight - 2 times, the nine - 3 times.

5 ... The midpoints of adjacent sides in a convex quadrilateral are connected by segments. Prove that the area of ​​the resulting quadrangle is half the area of ​​the original.

Solution. We denote the quadrilateral by ABCD , and the middle of the sides AB, BC, CD, DA for P, Q, S, T respectively. Note that in the triangle ABC segment PQ is the middle line, which means that it cuts off the triangle from it PBQ four times less area than area ABC. Similarly, ... But triangles ABC and CDA add up to the entire quadrilateral ABCD means Similarly, we find thatThen the total area of ​​these four triangles is half the area of ​​the quadrilateral ABCD and the area of ​​the remaining quadrilateral PQST is also equal to half the area ABCD.

6. What natural x expression is the square of a natural number?

Answer. For x = 5.

Solution. Let be . Note that - also the square of some integer less than t. We get that. Numbers and - natural and the first is greater than the second. Means, a ... Having solved this system, we get, , what gives .

Preview:

School Olympiad Keys in Mathematics

Grade 10

1. Arrange the module signs so that you get the correct equality

4 – 5 – 7 – 11 – 19 = 22

Solution. For example,

2. When Winnie the Pooh came to visit the Rabbit, he ate 3 plates of honey, 4 plates of condensed milk and 2 plates of jam, and after that he could not go outside due to the fact that he was very fat from such food. But it is known that if he ate 2 bowls of honey, 3 bowls of condensed milk and 4 bowls of jam or 4 bowls of honey, 2 bowls of condensed milk and 3 bowls of jam, then he could calmly leave the burrow of the hospitable Rabbit. What makes you fat more: from jam or from condensed milk?

Answer. From condensed milk.

Solution. Let us denote through M - the nutritional value of honey, through C - the nutritional value of condensed milk, through B - the nutritional value of jam.

By the condition 3M + 4C + 2B> 2M + 3C + 4B, whence M + C> 2B. (*)

According to the condition, 3M + 4C + 2B> 4M + 2C + 3B, whence 2C> M + B (**).

Adding inequality (**) with inequality (*), we obtain M + 3C> M + 3B, whence C> B.

3. In the equation one of the numbers is replaced by dots. Find this number if it is known that one of the roots is 2.

Answer. 2.

Solution. Since 2 is the root of the equation, we have:

whence we get that, which means that the number 2 was written instead of an ellipsis.

4. Marya Ivanovna went out from town to village, and Katerina Mikhailovna went out to meet her from village to town at the same time. Find the distance between the village and the city if it is known that the distance between pedestrians was 2 km twice: first, when Marya Ivanovna passed half the way to the village, and then, when Katerina Mikhailovna passed a third of the way to the city.

Answer. 6 km.

Solution. Let us denote the distance between the village and the city for S km, the speed of Marya Ivanovna and Katerina Mikhailovna for x and y , and count the time spent by pedestrians in the first and second cases. In the first case, we get

In the second. Hence, excluding x and y, we have
, whence S = 6 km.

5. In triangle ABC conducted a bisector BL. It turned out that ... Prove that triangle ABL - isosceles.

Solution. By the property of the bisector, we have BC: AB = CL: AL. Multiplying this equality by, we get, whence BC: CL = AC: BC ... The last equality implies the similarity of triangles ABC and BLC by corner C and adjacent sides. From the equality of the corresponding angles in similar triangles, we obtain from where in

triangle ABL vertex angles A and B are equal, i.e. he is isosceles: AL = BL.

6. By definition, ... What factor should be deleted from the workso that the remaining product becomes the square of some natural number?

Answer. ten!

Solution. notice, that

x = 0.5 and is 0.25.

2. Lines AM and BH - respectively, the median and the height of the triangle ABC.

It is known that AH = 1 and ... Find side length BC.

Answer. 2 cm.

Solution. Let's draw a segment MN, it will be the median of a right triangle BHC drawn to the hypotenuse BC and is equal to half of it. Then- isosceles, therefore, hence, therefore, AH = HM = MC = 1 and BC = 2MC = 2 cm.

3. At what values ​​of the numeric parameter and inequality true for all values NS ?

Answer . ...

Solution . For, we have, which is not true.

At 1 we can reduce the inequality by keeping the sign:

This inequality is true for all x only for.

At we can reduce the inequality by, changing the sign to the opposite:... But the square of a number is never negative.

4. There is one kilogram of 20% saline solution. The laboratory assistant placed a flask with this solution in an apparatus in which water is evaporated from the solution and at the same time a 30% solution of the same salt is poured into it at a constant rate of 300 g / h. The evaporation rate is also constant at 200 g / h. The process stops as soon as a 40% solution is in the flask. What will be the mass of the resulting solution?

Answer. 1.4 kilograms.

Solution. Let t be the time during which the apparatus worked. Then, at the end of the work in the flask, it turned out 1 + (0.3 - 0.2) t = 1 + 0.1t kg. solution. In this case, the mass of salt in this solution is 1 · 0.2 + 0.3 · 0.3 · t = 0.2 + 0.09t. Since the resulting solution contains 40% salt, we get
0.2 + 0.09t = 0.4 (1 + 0.1t), that is, 0.2 + 0.09t = 0.4 + 0.04t, hence t = 4 hours. Therefore, the mass of the resulting solution is 1 + 0.1 4 = 1.4 kg.

5. In how many ways can one choose 13 different numbers among all natural numbers from 1 to 25 so that the sum of any two chosen numbers does not equal 25 or 26?

Answer. The only one.

Solution. Let's write all our numbers in the following order: 25,1,24,2,23,3, ..., 14,12,13. It is clear that any two of them are equal to the sum of 25 or 26 if and only if they are adjacent in this sequence. Thus, among the thirteen numbers we have chosen, there should be no adjacent ones, from which we immediately get that these should be all members of this sequence with odd numbers - there is only one choice.

6. Let k be a natural number. It is known that among 29 consecutive numbers 30k + 1, 30k + 2, ..., 30k + 29, there are 7 primes. Prove that the first and the last are simple.

Solution. We delete from this series numbers that are multiples of 2, 3, or 5. There remain 8 numbers: 30k + 1, 30k + 7, 30k + 11, 30k + 13, 30k + 17, 30k + 19, 30k + 23, 30k + 29. Let's say there is a composite number among them. Let us prove that this number is a multiple of 7. The first seven of these numbers give different remainders when divided by 7, since the numbers 1, 7, 11, 13, 17, 19, 23 give different remainders when divided by 7. So, one of of these numbers is a multiple of 7. Note that the number 30k + 1 is not a multiple of 7, otherwise 30k + 29 will also be a multiple of 7, and the composite number must be exactly one. Hence, the numbers 30k + 1 and 30k + 29 are prime.