All-Russian Olympiads. International distance competitions and olympiads. Goals and objectives of the distance Olympiads of the Snail Center

Federal State Unitary Enterprise "Central Aerohydrodynamic Institute named after Professor N.E. Zhukovsky", Ministry of Education and Science of the Chelyabinsk Region, Department of Education of the Yamalo-Nenets Autonomous District, Department of Education, Youth Policy and Sports of the Administration of the Shelekhov Municipal District of the Irkutsk Region, Federal State Autonomous Educational institution higher education"South Ural State University (national research university )", Budgetary Institution of Higher Education of the Khanty-Mansiysk Autonomous Okrug - Ugra "Surgut State University", State Budgetary Educational Institution of Higher Education of the Moscow Region "Dubna University", Federal State Budgetary Educational Institution of Higher Education "Togliatti State University", Federal State Autonomous educational institution of higher education "North-Eastern Federal University named after M.K. Ammosov", Federal State Autonomous Educational Institution of Higher Education "Far Eastern Federal University", Federal State Autonomous Educational Institution of Higher Education "Samara National Research University named after Academician S.P. Korolev ", Federal State Autonomous Educational Institution of Higher Education "Sevastopol State University", Federal State Autonomous Educational Institution of Higher Education "National Research Technological University "MISiS", Federal State Autonomous Educational Institution of Higher Education "St. IN AND. Ulyanov (Lenin)", Federal State Autonomous Educational Institution of Higher Education "National Research Tomsk Polytechnic University", Federal State Autonomous Educational Institution of Higher Education "Southern Federal University", Federal State Autonomous Educational Institution of Higher Education "Northern (Arctic) Federal University named after M V. Lomonosov", Federal State Autonomous Educational Institution of Higher Education "National Research Nuclear University "MEPhI", Federal State Budgetary Educational Institution of Higher Education "Altai State University", Federal State Budgetary Educational Institution of Higher Education "Amur State University", Federal State budgetary educational institution of higher education "Volgograd State Technical University", Federal State Budgetary Educational Institution of Higher Education "Voronezh State University", Federal State Budgetary Educational Institution of Higher Education "Don State Technical University", Federal State Budgetary Educational Institution of Higher Education "Izhevsk State Technical University" university named after M. T. Kalashnikova", Federal State Budgetary Educational Institution of Higher Education "Kovrov State Technological Academy named after V.A. Degtyarev", Federal State Budgetary Educational Institution of Higher Education "Kuban State Technological University", Federal State Budgetary Educational Institution of Higher Education "Moscow State Technological University "STANKIN", Federal State Budgetary Educational Institution of Higher Education "Moscow Technological University", Federal State Budgetary Educational institution of higher education "Nizhny Novgorod State Technical University named after R.E. Alekseev", Federal State Budgetary Educational Institution of Higher Education "Novosibirsk State Technical University", Federal State Budgetary Educational Institution of Higher Education "Oryol State University named after I.S. Turgenev", Federal State Budgetary Educational Institution of Higher Education "Perm National Research Polytechnic University", Federal State Budgetary Educational Institution of Higher Education "Russian State University of Oil and Gas (National Research University) named after I.M. Gubkin", Federal State Budgetary Educational Institution of Higher Education "Samara State Technical University", Federal State Budgetary Educational Institution of Higher Education "Saint Petersburg Mining University", Federal State Budgetary Educational Institution of Higher Education "St. . Kirov", Federal State Budgetary Educational Institution of Higher Education "Saratov State Technical University named after Yuri Gagarin", Federal State Budgetary Educational Institution of Higher Education "North Caucasus Mining and Metallurgical Institute (State Technological University)", Federal State Budgetary Educational Institution higher education "Siberian State University of Science and Technology named after Academician M.F. Reshetnev, Federal State Budgetary Educational Institution of Higher Education "Sochi State University", Federal State Budgetary Educational Institution of Higher Education "Pacific State University", Federal State Budgetary Educational Institution of Higher Education "Ural State Transport University", Federal State Budgetary Educational Institution of Higher Education education "South-Western State University", Federal State Budgetary Educational Institution of Higher Education "South-Russian State Polytechnic University (NPI) named after M. I. Platova", Federal State Budgetary Educational Institution of Higher Education "Yaroslavl State Technical University", Federal State Budgetary Educational Institution of Higher Education "Transbaikal State University", Federal State Budgetary Educational Institution of Higher Education "Omsk State Technical University", Federal State Budgetary Educational institution of higher education "Ulyanovsk State University", Federal State Budgetary Educational Institution of Higher Education "Moscow State University of Technology and Management named after K.G. Razumovsky (First Cossack University)", Federal State Budgetary Educational Institution of Higher Education "Belgorod State Technological University. V.G. Shukhov", Federal State Budgetary Educational Institution of Higher Education "Penza State Technological University", Federal State Budgetary Educational Institution of Higher Education "Tver State University", Federal State Budgetary Educational Institution of Higher Education "Tula State University", Federal State Budgetary Educational Institution of Higher Education "Ufa State Aviation Technical University", Federal State Budgetary Educational Institution of Higher Education "Moscow Aviation Institute (National Research University)", Federal State Budgetary Educational Institution of Higher Education "Irkutsk National Research Technical University", Federal State Budgetary Educational Institution of Higher Education "Yuzhno -Ural State Agrarian University"

Contest
Olympics
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 independent appropriation of knowledge
formation and development of skills for independent search and analysis of information
formation and development of skills for using Internet services in education
increasing motivation to study the subject

Olympics

They give the participant the opportunity to test and deepen knowledge in a particular school discipline or even in one of its sections. All tasks of distance Olympiads are divided by age groups and comply with school programs and the requirements of the Federal State Educational Standard.

Competition-game

They give the participant the opportunity to test and deepen knowledge in a particular school discipline or even in one of its sections. All tasks of remote Olympiads are divided into age groups and correspond to school programs and the requirements of the Federal State Educational Standard.

subject week

family competition

Specialist. contests

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

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school stage

4th grade

1. Rectangle area 91

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

school stage

5th grade

The maximum score for each task is 7 points

3. Cut the figure into three identical (coinciding when superimposed) figures:

4. Replace the letter A

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

school stage

6th grade

The maximum score for each task is 7 points

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

school stage

7th grade

The maximum score for each task is 7 points

1. - different 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. There is something alive on the island th number of people, with her

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

school stage

8th grade

The maximum score for each task is 7 points

AVM, CLD and ADK respectively. Find∠ MKL .

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

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

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. At what natural x expression

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

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 held a bisector B.L. It turned out that . Prove that the triangle ABL - isosceles.

6. By definition,

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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics

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. Segments AM and BH ABC.

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

3. a inequality true for all values X ?

Preview:

4th grade

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

Answer. 40

Solution. The length of the unknown side of the rectangle is found from the area and the known side: 91:13 cm = 7 cm.

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

2. Cut the figure into three identical (coinciding when superimposed) figures:

Solution.

3. Restore the addition example, where the digits of the terms are replaced by asterisks: *** + *** = 1997.

Answer. 999 + 998 = 1997.

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

Answer. Sveta, Ira, Julia, Anya.

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Keys school olympiad mathematics

5th grade

1. Without changing the order of the numbers 1 2 3 4 5, put signs of arithmetic operations and brackets between them so that the result is one. It is impossible to “glue” adjacent numbers 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 school yard?

Answer. 12 piglets and 18 geese.

Solution.

1 step. Imagine that all the pigs raised two legs up.

2 step. There are 30 ∙ 2 = 60 legs left to stand on the ground.

3 step. Raised up 84 - 60 \u003d 24 legs.

4 step. Raised 24: 2 = 12 piglets.

5 step. 30 - 12 = 18 geese.

3. Cut the figure into three identical (coinciding when superimposed) figures:

Solution.

4. Replace the letter A to a non-zero digit to get the correct equality. It suffices to give one example.

Answer. A = 3.

Solution. It is easy to show that A = 3 is suitable, we prove that there are no other solutions. Reduce equality by A . We get .
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 pieces of pens and pencils in total?

Answer. 140 notebooks.

Solution. Each student 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 everything was bought
28 ⋅ 5=140 notebooks.

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Keys of the School Olympiad in Mathematics

6th grade

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

Answer. 58 cm

Solution. 29 parts of 2 cm are placed between the extreme points.

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 sum in the form of the following terms:
(1 + 2006) + (2 + 2005) + …..+ (1003 + 1004) + 2007.

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

3. Cut the figurine into 6 equal checkered figurines.

Solution. The figurine can only be cut

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

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

5. Usually dad comes to pick up 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 early did class end that day?

Answer. 25 minutes early.

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

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Keys of the School Olympiad in Mathematics

7th grade

1. Find the solution to the numerical puzzle a,bb + bb,ab = 60 , where a and b - different 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. By what part (from the received level) will the compote level decrease if you eat half of the remaining peaches?

Answer. For one quarter.

Solution. It is clear from the condition that half of the peaches take up a third of the jar. So, after Natasha ate half of the peaches, the jar of peaches and compote remained equally (one third each). So half of the number of remaining peaches is a quarter of the total content

banks. If you eat this half of the remaining 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. There is something alive on the island th number of people, with yo m each of them is either a knight who always tells the truth, or a liar who always lies yo m. Once all the knights said: - "I am friends with only 1 liar", and all the liars: - "I am not friends with the knights." Who is more on the island, knights or knaves?

Answer. more knights

Solution. Every knave is friends with at least one knight. But since each knight is friends with exactly one knave, two knaves cannot have a common knight friend. Then each knave can be associated with his friend a knight, whence it turns out that there are at least as many knights as there are knaves. Since there are no inhabitants on the island yo number, then equality is impossible. So more knights.

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Keys of the School Olympiad in Mathematics

8th grade

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

Answer. By 55%.

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

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

Answer. 90°.

Solution. Consider a triangle MAK : angle MAK equals 360° - 90° - 60° - 60° = 150°. MA=AK by condition, then a triangle MAC isosceles,∠AMK = ∠AKM = (180° - 150°) : 2 = 15°.

Similarly, we get that the angle DKL equals 15°. Then the required angle MKL is the sum of ∠MKA + ∠AKD + ∠DKL = 15° + 60° + 15° = 90°.

3. Nif-Nif, Naf-Naf and Nuf-Nuf shared three pieces of truffle with masses of 4 g, 7 g and 10 g. The wolf decided to help them. He can cut off and eat 1 g of truffle from any two pieces at the same time. Can the wolf leave the piglets equal pieces of truffle? 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 of 1 g and two pieces of 7 g. Now it remains to cut and eat 1 g six times from pieces of 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 - such a number. Thenis also a multiple of 19. But
Since 100 and 19 are coprime, a two-digit number is divisible by 19. And there are only five of them: 19, 38, 57, 76 and 95.

It is easy to make sure that all numbers 1919, 3819, 5719, 7619 and 9519 suit us.

5. A team of Petit, Vasya and a single scooter is participating 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 of them rides the first part of the way on a scooter, the rest is running, and the other - vice versa (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. 18

Solution. If the time of one becomes less than the time of the other of the guys, then the time of the other will increase and, consequently, the time of the team. So, the time of the guys should coincide. Denoting 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 a fractionwill be an integer.

Solution.

Consider , by the condition this number is an integer.

Then and will also be an integer as the difference N and double integer.

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Keys of the School Olympiad in Mathematics

Grade 9

1. Sasha and Yura are now together for 35 years. 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, according to the condition,. But the time for both Sasha and Yura has passed equally, 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 get 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 can hold no more than 20 kg. Which Weight Limit Can he take fish 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 because). For each of these options, the remaining capacity of the backpack is not divisible by 3.5 and at best it will be possible to pack kg. fish.

4. The shooter fired ten times at the standard target and hit 90 points.

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

Answer. Seven - 1 hit, eight - 2 hits, nine - 3 hits.

Solution. Since the shooter hit only the seven, eight and nine in the remaining six shots, then for three shots (since the shooter hit the seven, eight and nine at least once) he will scorepoints. Then for the remaining 3 shots you need to score 26 points. What is possible with a single combination of 8 + 9 + 9 = 26. So, the shooter 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 quadrilateral is half the area of the original.

Solution. Let's denote the quadrilateral by ABCD , and the midpoints of the sides AB , BC , CD , DA for P , Q , S , T respectively. Note that in the triangle ABC segment PQ is the median line, which means that it cuts off the triangle from it PBQ four times less area than area ABC. Likewise, . But triangles ABC and CDA add up to the whole quadrilateral ABCD means Similarly, we get 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 half the area ABCD.

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

Answer. For x = 5.

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

Preview:

Keys of the School Olympiad in Mathematics

Grade 10

1. Arrange the signs of the module so that the correct equality is obtained

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 because he was very fat from such food. But it is known that if he ate 2 plates of honey, 3 plates of condensed milk and 4 plates of jam or 4 plates of honey, 2 plates of condensed milk and 3 plates of jam, he could easily leave the hole of the hospitable Rabbit. What makes them fatter 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 condition 3M + 4C + 2B > 2M + 3C + 4B, whence M + C > 2B. (*)

By 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 one of the roots is known to be 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 the ellipsis.

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

Answer. 6 km.

Solution. Let us denote the distance between the village and the city as S km, the speeds of Marya Ivanovna and Katerina Mikhailovna as x and y , and calculate the time spent by pedestrians in the first and second cases. We get in the first case

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

5. In triangle ABC held a bisector B.L. It turned out that . Prove that the triangle ABL - isosceles.

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

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

6. By definition, . Which factor should be removed from the productso that the remaining product becomes the square of some natural number?

Answer. 10!

Solution. notice, that

x = 0.5 and is 0.25.

2. Segments AM and BH are the median and height of the triangle, respectively ABC.

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

Answer. 2 cm

Solution. Let's spend a segment MN, it will be the median of a right triangle BHC drawn to the hypotenuse BC and equal to half of it. Thenisosceles, therefore, so, therefore, AH = HM = MC = 1 and BC = 2MC = 2 cm.

3. At what values of the numerical parameter and inequality true for all values X ?

Answer . .

Solution . When we have , which is not true.

At 1 reduce the inequality by, keeping the sign:

This inequality is true for all x only for .

At reduce 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 the flask with this solution into 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 h. Therefore, the mass of the resulting solution is 1 + 0.1 4 = 1.4 kg.

5. In how many ways can 13 different numbers be chosen 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 add up to 25 or 26 if and only if they are adjacent in this sequence. Thus, among the thirteen numbers we have chosen, there should not be neighboring ones, from which we immediately get that these must be all members of this sequence with odd numbers - the only 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 last of them are simple.

Solution. Let's cross out the numbers that are multiples of 2, 3 or 5 from this row. There will be 8 numbers left: 30k+1, 30k+7, 30k+11, 30k+13, 30k+17, 30k+19, 30k+23, 30k+29. Let's assume that among them there is a composite number. 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. Hence, 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.

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

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

winners of the All-Russian round are accepted to universities without competition if the profile 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 establishments(if the subject is not in the profile of the university, the winner receives an additional 100 points upon admission);
significant cash reward for prizes (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:

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

Organizers final round are representatives of the Ministry of Education and Science of Russia, they also sum 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 divided into areas:

№	Direction	Items
1	Exact disciplines	mathematics, computer science
2	Natural sciences	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 Culture, basics of life safety

Peculiarity final stage Olympiad consists of 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 must be constantly replenished with additional knowledge that goes beyond 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 can’t do without the help of teachers: additional classes after school, classes with tutors.

The winners of the final stage will take part in international olympiads. They form the national team of Russia, which will be trained at training camps in 8 subjects.

To provide methodological assistance the site hosts introductory webinars, the Central Organizing Committee of the Olympiad, subject and methodological commissions have been formed.