School stage. School stage Wash tasks
A good tradition was the holding of the All-Russian School Olympiad. Its main task is to identify gifted children, the motivation of schoolchildren to an in-depth study of objects, the development of creative abilities and non-standard thinking in children.
Olympiad motion is becoming increasingly popular with schoolchildren. And there are reasons:
- the winners of the All-Russian Tour are accepted into universities outside the competition, if the profile subject is the Olympic Item (the diplomas of the winners are valid for 4 years);
- participants and winners receive additional chances when entering educational institutions (if the subject is not at the university's profile, the winner receives additional 100 points during admission);
- significant monetary remuneration for prizes (60 thousand, 30 thousand rubles;
- and, of course, fame to the whole country.
Before becoming the winner, you must go through all the stages of the All-Russian Olympiad:
- The initial school stage, which is determined by decent representatives to the next step, to be carried out in September-October 2017. The organization and holding of the school stage are carried out by specialists of a methodical office.
- The municipal stage is carried out between the schools of the city or district. It takes place at the end of December 2017. - early January 2018
- The third round is more complicated. It takes the participation of talented students from the whole area. Regional Stage takes place in January-February 2018
- The final stage defines the winners of the All-Russian Olympiad. In March - April, the best children of the country are competing: the winners of the regional stage and the winners of last year's Olympiad.
The organizers of the final tour are representatives of the Ministry of Education and Science of Russia, they also sum up.
You can show your knowledge by any subject: in mathematics, in physics, geography, even on physical education and technology. You can compete in eruditions in several subjects at once. Total 24 disciplines.
Olympiad objects are divided into directions:
| № | Direction | Items |
| 1 | Accurate disciplines | mathematics, informatics |
| 2 | Natural science disciplines | geography, Biology, Physics, Chemistry, Ecology, Astronomy |
| 3 | Philological disciplines | literature, Russian, Foreign Languages |
| 4 | Humanitarian disciplines | economy, social studies, history, right |
| 5 | Others | art, Technology, Physical Culture, Life Safety Basics |
The feature of the final stage of the Olympiad consists of two types of tasks: theoretical and practical. For example, to get good geography results. Students must perform 6 theoretical tasks, 8 practical tasks, as well as to respond to 30 test questions.
The first stage of the Olympiad begins in September, which means that you want to take part in the intellectual marathon should prepare in advance. But first of all, should have a good school-based base, which you constantly need to replenish additional knowledge that go beyond the school program.
The official website of the Olympiad www.rosolymp.ru places the tasks of past years. These materials can be used in preparation for an intelligent marathon. And of course not to do without the help of teachers: additional classes after lessons, classes with tutoring.
The winners of the final stage will take part in international olympiads. They form the national team of Russia, which will be prepared at training campaigns of 8 subjects.
To provide methodological assistance on the site, installation webinars are carried out, the Central Organizing Committee of the Olympics, the subject-methodical commissions are formed.
It is a whole system of Olympics on subjects, which is included in the obligatory program of general educational institutions of the country. Participation in such an Olympiad is an honorary and responsible mission, because it is a chance of a schoolboy to show accumulated luggage knowledge, protect the honor of his educational institution, and in case of victory - also the opportunity to get monetary stimulation and earn a privilege when entering the best universities in Russia.
The practice of conducting subject Olympiad exists in the country for more than a hundred years - back in 1886, representatives of education authorities initiated competitions between young talents. During the Soviet Union, this movement did not simply stop existence, but also received an additional impetus to development. Since the 60s of the last century, in almost all major school disciplines, intelligent competitions of the All-Union, and then All-Russian scale began to be carried out.
What subjects come to the Olympics?
In the 2017-2018 school year, schoolchildren of the countries will be able to fight for prizes in several categories of disciplines:
- in the exact sciences, which includes computer science and mathematical block;
- in the natural sciences, which include geography, biology, astronomy, physics, chemistry and ecology;
- in the field of philology, including the Olympiad in German, English, Chinese, French, Italian, as well as the Russian language and literature;
- in the field of humanitarian sciences consisting of history, social studies, rights and economics;
- according to other disciplines, which include physical education, world artistic culture, technology and safety of vital activity.
In the Olympics tasks for each of the listed disciplines, two tasks block are usually allocated: a part that checks theoretical preparation, and a part aimed at identifying practical skills.
The main stages of the Olympics 2017-2018
The All-Russian School Olympiad includes the organization of the four stages of competitions held at various levels. The final schedule of intellectual battles between schoolchildren is determined by representatives of schools and regional educational authorities, however, you can navigate for such periods of time.
Schoolchildren expect 4 stages of competitions of different levels of complexity - Stage 1. School. Competitions between representatives of one school will be held in September-October 2017. The Olympiad is held between students parallel, starting from the fifth grade. Development of tasks for subject Olympiads in this case is assigned to members of the method of urban level.
- Stage 2. Municipal. The stage on which competitions between the winners of the schools of one city, representing the 7-11 classes, will be held from December 2017 to January 2018. The mission of the compilation of olympics assigned to the organizers of the regional level, and local officials respond to issues related to the provision of place and provision of the Olympics procedure.
- Stage 3. Regional. The third level of the Olympics, which will be held in January-February 2018. At this stage, schoolchildren who received prizes in the city Olympiad, and those who won the regional selections of last year take part in the competition.
- Stage 4. All-Russian. The highest level of subject Olympiad will be organized by representatives of the Ministry of Education of the Russian Federation in March-April 2018. It is invited to winners of the regional level and the guys who won last year. However, not every winner of regional selection can be a member of this stage. The exception is the schoolchildren who have received 1 place in the region, but lagging behind the glasses from the winners at the level of other cities. The prize-winners of the All-Russian stage can then go to the international competition competitions that take place in the summer.
Where to find typical tasks in the Olympics?
Of course, to adequately perform in this event, you need to have a high level of preparation. The All-Russian Olympiad is represented in the network Own Site - Rosolymp.ru, - on which students can familiarize themselves with the tasks of past years, check their level with the help of answers to them, find out specific dates and requirements for organizational moments.
Tasks and keys of the school stage of the All-Russian Olympiad of schoolchildren in mathematics
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Preview:
School stage
4th grade
1. Rectangle area 91.
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
Grade 5.
Maximum evaluation of each task - 7 points
3. Cut the figure to three the same (matching when applied) figures:
4. Replace the letter A
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
6th grade
Maximum evaluation of each task - 7 points
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
7th grade
Maximum evaluation of each task - 7 points
1. - Various numbers.
4. Replace the letters Y, E, A and R numbers so that the right equality is:
Yyyy ─ Eee ─ AA + R \u003d 2017.
5. On the island alive, it is not number of peopleher
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
8th grade
Maximum evaluation of each task - 7 points
AVM, CLD and ADK respectively. Find∠ mkl.
6. Prove that ifa, B, C and - integers, and fraction It will be an integer.
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
Grade 9.
Maximum evaluation of each task - 7 points
2. Numbers a and b Such that equations and also has a solution.
6. At what naturalx Expression
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
Grade 10
Maximum evaluation of each task - 7 points
4 – 5 – 7 – 11 – 19 = 22
3. In equation
5. In the ABC triangle Wired bisectorBl. It turned out that . Prove that triangleABL - Isol.
6. By definition,
Preview:
Objectives of the All-Russian Olympiad of schoolchildren in mathematics
School stage
Grade 11
Maximum evaluation of each task - 7 points
1. The sum of the two numbers is 1. Can their work be greater than 0.3?
2. Segments AM and BH ABC.
It is known that AH \u003d 1 and . Find side lengthBC.
3. And inequality right with all valuesx?
Preview:
4th grade
1. Rectangle area 91.. The length of one of his sides is 13 cm. What is the sum of all sides of the rectangle?
Answer. 40.
Decision. The length of the unknown side of the rectangle is found from the area and the famous side: 91: 13 cm \u003d 7 cm.
The sum of all sides of the rectangle is equal to 13 + 7 + 13 + 7 \u003d 40 cm.
2. Cut the figure to three the same (matching when applied) figures:
Decision.
3. Restore an example of addition, where the digits of the components are replaced with asterisks: *** + *** \u003d 1997.
Answer. 999 + 998 \u003d 1997.
4 . Four girls ate candy. Anya ate more than Julia, Ira - more than light, but less than Julia. Arrange the names of girls in order of increasing edible candy.
Answer. Light, Ira, Julia, Anya.
Preview:
Keys of the School Olympiad in Mathematics
Grade 5.
1. Not changing the order of the arrangement of numbers 1 2 3 4 5, put the signs of arithmetic actions between them and the brackets so that the result is one. "Glue" adjacent numbers to one number cannot be.
Decision. For example, ((1 + 2): 3 + 4): 5 \u003d 1. Other solutions are possible.
2. Geese and piglets were walking in the livestock. The boy counted the number of heads, they turned out to be 30, and then he counted the number of legs, they turned out to be 84. How many geese and how many piglets were on the school yard?
Answer. 12 piglets and 18 geese.
Decision.
1 step. Imagine that all piglets raised two legs up.
2 step. On Earth, it remains 30 ∙ 2 \u003d 60 legs.
3 step. Raised up 84 - 60 \u003d 24 legs.
4 step. Raised 24: 2 \u003d 12 piglets.
5 step. 30 - 12 \u003d 18 geese.
3. Cut the figure to three the same (matching when applied) figures:
Decision.
4. Replace the letter A On the nonzero digit to make faithful equality. It is enough to bring one example.
Answer. A \u003d 3.
Decision. It is easy to show thatBUT \u003d 3 Suitable, we prove that there are no other solutions. Reduce equality byBUT . We get.
If A. ,
if a\u003e 3, then.
5. Girls and boys on the way to school went to the store. Each student bought 5 thin notebooks. In addition, every girl bought 5 handles and 2 pencils, and every boy bought 3 pencils and 4 handles. How much did Notebooks bought, if the total handles and pencils of children bought 196 pieces?
Answer. 140 notebooks.
Decision. Each of the students bought 7 pens and pencils. In total, 196 pens and pencils were bought.
196: 7 \u003d 28 students.
Each of the students bought at 5 notebooks, it means that everything was bought.
28
⋅ 5 \u003d 140 notebooks.
Preview:
Keys of the School Olympiad in Mathematics
6th grade
1. On a straight line 30 points, the distance between any two adjacent is 2 cm. What is the distance between two extreme dots?
Answer. 58 cm.
Decision. There are 29 parts of 2 cm between extreme dots.
2 cm * 29 \u003d 58 cm.
2. Will the amount of numbers 1 + 2 + 3 + ...... + 2005 + 2006 + 2007 sharing for 2007? Justify the answer.
Answer. Will be.
Decision. Imagine this amount in the form of the following terms:
(1 + 2006) + (2 + 2005) + …..+ (1003 + 1004) + 2007.
Since each group is divided into 2007, then the whole amount will be divided into 2007.
3. Cut the figure on 6 equal checkered figures.
Decision. The figure can only be cut
4. Nastya places in square 3 to 3 numbers 1, 3, 5, 7, 9. It wants the amount of numbers on all horizontals, verticals and diagonals to 5. Give an example of such an arrangement, provided that each Nastya is going to use no more than two times.
Decision. Below is one of the alignments. There are other solutions.
5. Usually, dad arrives at the car after lessons. 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 earlier. How many minutes did the lessons ended on this day?
Answer. 25 minutes earlier.
Decision. The car came home earlier, because she did not have to reach the meeting from the meeting to school and back, it means that the car doubted this path passing for 10 minutes, and one way in 5 minutes. So, the car met with Pavlik 5 minutes before the usual graduation. By this point, Pavlik was already 20 minutes. Thus, the lessons ended 25 minutes earlier.
Preview:
Keys of the School Olympiad in Mathematics
7th grade
1. Find a numerical rebus solutiona, bb + bb, ab \u003d 60, where a and b - Various numbers.
Answer. 4.55 + 55.45 \u003d 60
2. After Natasha ate half peaches from the jar, the level of the compote has dropped one-third. What part (from the level received) drops the level of the compote, if you eat half of the remaining peaches?
Answer. One quarter.
Decision. It is clear from the condition that half of the peaches occupies a third of banks. So, after Natasha ate half of the peaches, the joke of peaches and the compote remains equally (one third). So half from the number of remaining peaches is a quarter of the total content
banks. If you eat this half of the remaining peaches, the level of the compote is dropped by a quarter.
3. Cut on the grid lines the rectangle depicted in the figure, on five rectangles of various areas.
Decision. For example, so
4. Replace the letters y, e, a and r numbers so that the right equality is: yyyy ─ Eee ─ AA + R \u003d 2017.
Answer. At y \u003d 2, e \u003d 1, a \u003d 9, r \u003d 5, we obtain 2222 ─ 111 ─ 99 + 5 \u003d 2017.
5. On the island alive, it is not number of peoplee. m each of them is either a knight who always says the truth or a liar who always lze. since once, all the knights stated: - "I am friends only with 1 liar", and all the liars: - "I am not friends with knights." Who is on the island more, knights or liars?
Answer. Knights bigger
Decision. Each liar is friendly at least with one knight. But since each knight is friendly smoothly with one liar, two liars can not be a common friend-knight. Then every liar can be put in line with the knight's friend, from where it turns out that the knights, at least as much as liars. Since there is no entire inhabitants on the islande. number, then equality is impossible. So, knights are more.
Preview:
Keys of the School Olympiad in Mathematics
8th grade
1. In the family of 4 people. If Masha double the scholarship, the total income of the whole family will increase by 5%, if, instead, the mother will double the salary - by 15%, if the salary double dad - by 25%. How much percentage will increase the income of the whole family, if the grandfather give a pension?
Answer. By 55%.
Decision . When doubling the scholarship of Machine, the total family income increases exactly by the magnitude of this scholarship, so it is 5% of income. Similarly, the salaries of mom and dad are 15% and 25%. It means that the grandfather's pension is 100 - 5 - 15 - 25 \u003d 55%, and if ee. Frequently, the family's income will grow by 55%.
2. On the sides of AV, CD and AD Square ABSD Equilaf triangles are built outsideAVM, CLD and ADK respectively. Find∠ mkl.
Answer. 90 °.
Decision. Consider a triangleMAK: Mak angle equal to 360 ° - 90 ° - 60 ° - 60 ° \u003d 150 °.Ma \u003d AK. By condition, it means, a triangleMak. isosceles,∠ amk \u003d ∠ AKM \u003d (180 ° - 150 °): 2 \u003d 15 °.
Similarly, we get that the angleDKL equal to 15 °. Then the desired cornerMKL is equal to the sum ∠ MKA + ∠ AKD + \u200b\u200b∠ DKL \u003d 15 ° + 60 ° + 15 ° \u003d 90 °.
3. NIF-NIF, NAF-NAF and NUF-NUF divided three pieces of truffle by the masses of 4 g., 7 G. and 10. The wolf decided to help them. It can from any two pieces to simultaneously cut off and eat for 1 g. Truffle. Can the wolf leave the piglets equal pieces of truffle? If so, how?
Answer. Yes.
Decision. The wolf may first cut off three times from pieces in 4 g and 10 g. It turns out one piece in 1 g and two pieces of 7 g. Now it remains six times to cut off and eat 1 g from pieces in 7 g., Then piglets It will get to 1 g. Truffle.
4. How much is the four-digit numbers that are divided into 19 and ends at 19?
Answer. five .
Decision. Let be - such a number. Then also multiple 19. But
Since 100 and 19 are mutually simple, then the two-digit number is divided into 19. And there are only five of these: 19, 38, 57, 76 and 95.
It is easy to make sure that all numbers are 1919, 3819, 5719, 7619 and 9519 we are suitable for us.
5. The team from Petit, Vasi and a single scooter participates in the race. The distance is divided into sections of the same length, their quantity is 42, at the beginning of each test. Petya runs through a plot for 9 minutes, Vasya - for 11 minutes, and on the scooter, any of them drives a plot for 3 minutes. They start at the same time, and the finisher takes into account the time who came the last. The guys agreed that one drives the first part of the path on the scooter, the rest of the junction, and the other - on the contrary (the scooter can be left at any control point). How many areas of Petya should drive on the scooter so that the team show the best time?
Answer. eighteen
Decision. If one time becomes less than the time of another of the guys, then the time of the other and, therefore, the time of the team will increase. So, the time of the guys must coincide. Denote by the number of people passing areas throughx. And deciding the equation, I get x \u003d 18.
6. Prove that ifa, B, C and - integers, and fraction It will be an integer.
Decision.
Consider , By condition, this is a number of integer.
Then I. will also be an integer as a differenceN. and doubled integer.
Preview:
Keys of the School Olympiad in Mathematics
Grade 9.
1. Sasha and Yura now together 35 years old. Sasha is now twice as old than Yura then when Sasha was so many years as Yura now. How many years now Sasha and how much - Yura?
Answer. Sasha 20 years, Yura 15 years.
Decision. Let Sasha nowx years, then jure , and when Sasha wasyears, then Jura, under the condition . But the time and for Sasha and for Yura passed equally, so we obtain the equation
from which .
2. Numbers a and b Such that equations and have solutions. Prove that equation also has a solution.
Decision. If the first equations have solutions, their discriminants are non-negative, from whereand . Multiplying these inequalities, get or From where it follows that the discriminant of the last equation is also non-negative and the equation has a solution.
3. Fisherman caught a large number of fish weighing 3.5 kg. and 4.5 kg. His backpack accommodates no more than 20 kg. What is the maximum weight of the fish he can take with him? Justify the answer.
Answer. 19.5 kg.
Decision. In the backpack you can put 0, 1, 2, 3 or 4 fish weighing 4.5 kg.
(no more because). For each of these options, the residue of the backpack is not divided by 3.5 and at best will be able to pack kg. fish.
4. The shooter tent times shot on a standard target and knocked out 90 points.
How many hits were in the seven, eight and nine, if there were four dozen, and there were no other hits and misses?
Answer. In the seven - 1 hit, in the eight - 2 hitting, in the nine - 3 hit.
Decision. Since the shooter got only in the seven, eight and nine to the remaining six shots, then for three shots (as at least one time in the seven, the eight and nine arrows hit) he will pick up Point. Then for the remaining 3 shots need to dial 26 points. What is possible with the only combination of 8 + 9 + 9 \u003d 26. So, in the seven arrows got 1 time, 2 times, 2 times, in the nine - 3 times.
5 . The middle of the neighboring sides in the convex of the quadrangle are connected by segments. Prove that the area of \u200b\u200bthe resulting quadrilateer is twice as fewer areas of the initial.
Decision. Denote by a four-triggerAbcd. , and the middle of the partiesAB, BC, CD, DA for P, Q, S, T respectively. Note that in the triangleABC Cut PQ. It is the middle line, it means that it cuts off from him a trianglePBQ. four times less square than the areaABC. Similarly, . But trianglesABC and CDA. in total constitute the entire quadricleABCD, meaning Similarly, we get that Then the total area of \u200b\u200bthese four triangles is half the square of the quadricleAbcd. and the area of \u200b\u200bthe remaining quadriclePQST. Also equal half of the squareABCD.
6. At what naturalx Expression Is a square of a natural number?
Answer. At x \u003d 5.
Decision. Let be . Note that - Also square some integersmaller t. We get that. Numbers I. - Natural and first larger. So, but . Deciding 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 faithful equality
4 – 5 – 7 – 11 – 19 = 22
Decision. For example,
2. When Winnie Pooh came to visit the rabbit, he ate 3 honey plates, 4 sackcloth plates and 2 dishes of jams, and after that I could not get out out due to the fact that he was very fat from such a meal. But it is known that if he ate 2 honey plates, 3 sackcloth plates and 4 dishes of jam or 4 honey plates, 2 sackcloth plates and 3 savage plates, I could calmly leave the hole of the hospitable rabbit. What makes more fat: from jam or from condensed milk?
Answer. From condensed milk.
Decision. Denote by M - Honey nutritionality, through C - the feederness of the condensed milk, through B is the feederness of the jam.
Under the condition 3M + 4C + 2V\u003e 2M + 3C + 4V, where M + C\u003e 2B. (*)
By the condition of 3m + 4C + 2V\u003e 4M + 2C + 3V, from where 2c\u003e M + in (**).
Folding inequality (**) with inequality (*), we get M + 3C\u003e M + 3V, from where with\u003e B.
3. In equation One of the numbers is replaced by points. Find this number if it is known that one of the roots is 2.
Answer. 2.
Decision. Since 2 is the root of the equation, we have:
where do we get thatSo, instead of dall, the number 2 was recorded.
4. Marya Ivanovna came from the city to the village, and Katerina Mikhailovna came to the city to the city at the same time. 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 passed half the way to the village, and then, when Katerina Mikhailovna passed a third of the road to the city.
Answer. 6 km.
Decision. Denote the distance between the village and the city for s km, the speed of Maryia Ivanovna and Katerina Mikhailovna forx and Y. , and consider the time spent by pedestrians in the first and second cases. We get in the first case
In the second From here, excludingx and Y, we have
Where S \u003d 6 km.
5. In the ABC triangle Wired bisectorBl. It turned out that . Prove that triangleABL - Isol.
Decision. By the property of bisector, we have BC: AB \u003d CL: AL. Multiplying is equality on, Get where BC: CL \u003d AC: BC . The latter equality entails the similarity of trianglesABC and BLC on the corner C and adjacent to the parties. From the equality of the corresponding angles in such triangles we getFrom where in
aBL triangle Corners at the verticesA and B. equal, i.e. He is a chained:Al \u003d bl.
6. By definition, . What a doubt need to strike out So that the remaining work has become a square of some natural number?
Answer. 10!
Decision. notice, that
x. \u003d 0.5 and is 0.25.2. Segments AM and BH - Accordingly, the median and the height of the triangleABC.
It is known that AH \u003d 1 and . Find side lengthBC.
Answer. 2 cm.
Decision. Cut the cutMN, it will be a median rectangular triangleBHC. conducted to hypotenuseBC. And equal to her half. Then - Isoic, so, Therefore, therefore, AH \u003d Hm \u003d Mc \u003d 1 and Bc \u003d 2mc \u003d 2 cm.
3. At what values \u200b\u200bof the numerical parameterand inequality right with all valuesx?
Answer. .
Decision . When we have something wrong.
For 1 will reduce the inequality on, Saving a sign:
Such inequality is true for allx only at.
For Reduction inequality onBy changing the sign to the opposite:. But the square of the number is never negative.
4. There is one kilogram of a 20% hydrochlorian solution. The laboratory assistant placed the flask with this solution into the apparatus, in which water was evaporated from the solution and at the same time, with a constant speed of 300 g / h, a 30% solution of the same salt is added. The evaporation rate is also constant and is 200 g / h. The process stops as soon as the flask will be 40% solution. What will be the mass of the resulting solution?
Answer. 1.4 kilograms.
Decision. Let T - the time during which the device worked. Then at the end of work in the flask it turned out 1 + (0.3 - 0.2) t \u003d 1 + 0,1t kg. Solid. In this case, the mass of salt in this solution is 1 · 0.2 + 0.3 · 0.3 · t \u003d 0.2 + 0.09t. Since the resulting solution contains 40% of the salt, we get
0.2 + 0.09t \u003d 0.4 (1 + 0,1t), that is, 0.2 + 0.09t \u003d 0.4 + 0.04t, hence T \u003d 4 hours. Consequently, the mass of the resulting solution is equal to 1 + 0.1 · 4 \u003d 1.4 kg.
5. In some ways, among all natural numbers from 1 to 25, you can select 13 different so that the sum of any two selected numbers is not equal to 25 or 26?
Answer. The only one.
Decision. We 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 in the amount of 25 or 26 if and only when they are in this sequence adjacent. Thus, among the thirteen numbers selected by us, there should be no neighboring, where we immediately get that it should be all members of this sequence with odd numbers - the only one.
6. Let K be a natural number. It is known that among 29 consecutive numbers 30k + 1, 30k + 2, ..., 30k + 29 there are 7 simple. Prove that the first and last of them are simple.
Decision. I cross out this number of numbers, multiple 2, 3 or 5. 8 numbers will remain: 30k + 1, 30k + 7, 30k + 11, 30k + 13, 30k + 17, 30k + 19, 30k + 23, 30k + 29. Suppose that among them there is a composite number. We prove that this is the number of multiple 7. The first seven of these numbers give different remnants during division by 7, t. C. Numbers 1, 7, 11, 13, 17, 19, 23 give different remains during division by 7. So one of These numbers are multiple 7. Note that the number 30k + 1 is not multiple 7, otherwise 30k + 29 will also be more than 7, and the composite number should be exactly one. So, the numbers 30k + 1 and 30k + 29 are simple.
2019-2020 academic year
ORDER №336 dated 05.06.2019 "On the holding of the school stage of the All-Russian School Olympiad in the 2019-2020 school year."
The consent of the parents (legal representatives) for personal data processing (blank).
Analytical report template.
ATTENTION!!!Protocols for the results of Vosh grades 4-11 are accepted only in the program "Exsel" (Archived documents in programs Zip and Rar., except 7z.).
Data for 2019-2020 academic year
- Guidelines By holding the school stage of the Vosh 2018-2019 academic year on the subjects can download on the site.
- Presentation Meetings on the All-Russian Olympiad Schoolchildren 2019-2020 academic year.
- Presentation "Features of the organization and holding school stage Vosh for students with disabilities" by
- Presentation "Regional Center for Work with Gifted Children".
- Diploma Winner / winner of the school stage Vosh.
- Regulations The fulfillment of the olympics tasks of the school stage of the All-Russian Olympiad of schoolchildren.
- Schedule The school stage of the All-Russian School Olympiad in 2018-2019 academic year.
Explained in the order of the All-Russian Schoolchildren Olympiad - School Stage for 4 Classes
According to the order of the Ministry of Education and Science of the Russian Federation of December 17, 2015 No. 1488 All-Russian School Olympiad since September 2016 is held for students of 4 classes only in the Russian language and mathematics. According to schedule 09/21/2018 - in the Russian language; 09/26/2018 - in mathematics. A detailed schedule for holding a school stage Vosh for all parallel students is placed in the Plan of MBU "Center for Educational Innovations" for September 2018
Time to perform work in the Russian language 60 minutes, in mathematics - 9 0 minutes.
Attention responsible for the holding of the Olympiad
in educational organizations!
The tasks of the school stage of the All-Russian Schoolchildren Olympiad 2018-2019 uch. year. For 4-11 classes will be sent to educational organizations by email, starting from September 10, 2018, all changes and clarifications associated with E-mail addresses are pleased to send by email: [Email Protected]no later than 09/09/2018
Olympiad tasks (at 08.00 h.) And solutions (at 15.00) will be sent at the electronic address of the school. And the answers will be duplicated the next day on the site www.Syt
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School Stage Answers. Download
School Stage Answers (Girls) for 5 CL. Download
School Stage Answers (Girls) for 6 CL. c.
School Stage Answers for Technology (Youth) for 5-6 CB. Download
Answers to the school stage in literature.
School Stage Answers.
School Stage Answers.
School Stage Answers for 5 CL.
School Stage Answers for 6 CL.
School responses in geography for 5-6 cl.
School Stage Says for Biology for 5-6 CB.
School Stage Answers for 5-6 CL.
School Stage Answers.
School Stage Answers.
School Stage Answers.
School stages in Spanish.
Answers to the school stage in astronomy.
Schools of the school stage in the Russian language for the 4-CL.
Schools of the school stage in the Russian language for 5-6 cells.
School Stage Says for Mathematics for 4 CB.
School Stage Says for Mathematics for 5 CL.
Schools of the school stage in mathematics for 6 cl.
School responses in physical culture.
7-11 classes
Answers to the school stage in literature 7-8 cl.
Answers of the school stage in literature 9 CL.
Answers of the school stage in literature 10 cl.
School responses in literature 11 CL.
Schools School Stage 7-9 CL.
Schools of the school stage in geography 10-11 CL.
Schools School Stage (Girl) 7 CL.
School Stage Answers (Girls) 8-9 CL.
School Stage Answers (Girls) 10-11 CL.
School Stage Answers (young men).
Essay evaluation criteria for creative project.
Criteria for estimating practical work.
Answers to the school stage for astronomy 7-8 cl.
School Stage Answers 9 CL.
Answers of the school stage for astronomy 10 cl.
Answers of the school stage on astronomy 11 cl.
School Stage Answers on MHC 7-8 CL.
School Stage Answers 9 CL.
School Stage Answers on MHC 10 CL.
School Stage Answers on MHC 11 CL.
School Stage Summary School Stage for 8 CL.
School Stage Schools for 9 CL.
School Stage Summary Answers for 10 CL.
School Stage Answers for 11 CL.
School Stage Answers for 7-8 CB.
School Stage Answers for 9 CL.
School Stage Answers for 10-11 CB.
School Stage Answers.
School Stage Answers 7 CL.
School Stage Answers 8 CL.
School Stage Answers 9 CL History.
Schools of the school stage on the history of 10-11 CL.
Answers of the school stage in physical culture (7-8 classes).
School stage answers in physical culture (9-11 classes).
School Stage Answers in German 7-8 CB.
