A few months later a new academic year, so subject Olympiads are just around the corner. School stage of the All-Russian subject Olympiad will be held already in September-October 2017. Any interested student (not counting students elementary school) can take part in it. What else do you need to know about this massive event?
Useful facts about the All-Russian Subject Olympiad for schoolchildren 2017-2018
- A student who has taken part in the VOS receives benefits upon admission;
- The list of subjects for the title of the best has not been changed and consists of both humanitarian and technical;
- You can take part in all subjects at once, participation in them does not affect each other;
- The structure of the competition and materials for preparing for the Olympiad will remain unchanged.
The All-Russian Olympiad is held in all subjects of the compulsory school curriculum.
By participating, the student gets the opportunity to test his knowledge, receive an incentive in the form of a cash prize, which will help in acquiring the necessary items for their in-depth study; to protect the honor of his native school or city, even to gain some benefits when entering the outstanding higher institutions of the country.
That is why each participant tries to take the most responsible and serious approach to preparing and participating in the competition.
Competitions between the smartest talents have been held for quite a long period of time, more than a century: the first Olympiads date back to 1886.
See also:
Admission to the universities of the Russian Federation in 2017-2018: date of enrollment, documents required for enrollment, tuition fees
At times Soviet Union this difficult type of knowledge testing was also actively used. The beginning of an organized struggle for the title of the best expert on a particular subject at different levels dates back to sixty years.
Every year there are more participants, as well as items that have become the basis of the competition. So, since recently, Olympiads in physical education, the basics of life safety, ORKSE have appeared.
In the upcoming 2017-2018 academic year, students who have expressed a desire to participate in the competition will be able to participate and demonstrate their knowledge to others from September 2017 to April 2018.
Video review of the All-Russian School Olympiad
In what subjects can a student participate in Olympiads?
All items can be divided into several varieties:
- mathematical sciences: computer science and branches of mathematics;
- natural sciences such as physics, chemistry, geography, astronomy, biology;
- sciences studying literature (Olympiads in Russian, as well as foreign languages, literature);
- humanities: history, social studies, economics, law;
- remaining items: Physical Culture, life safety, technology, art.
The organizers of the Olympiads test both the theoretical knowledge of the student and the ability to apply this knowledge in practice.
See also:
School holidays in 2018 - vacation time: at home or abroad, price
Stages of the All-Russian Olympiad 2017-2018
The definition of the smartest of schoolchildren goes through several stages, there are four in total:
- Olympiad between school students. Middle and high school students can participate. This stage falls on September-October 2017. Responsibility for the organization rests with the members of the city's Education Committee, in accordance with the curriculum taught in school textbooks.
- The competition of the winners of intra-school Olympiads of the city at the municipal level. The honor to represent the school falls to students of the seventh-eleventh grades. Municipal Olympiads are held from December to January 2017-2018. Tasks are prepared by the organizers of the regional level.
- Continuation of the competition at the regional level between the winners of the last stage of the Olympiad and the winners of the last year. A ticket to the regional stage is received by high school students (ninth-eleventh grades) in January - February 2018, winners of the municipal stage.
- The final stage. It is held among the winners of the regional stage among all of Russia who have scored enough points for regional stage, and last year's winners. The period is March-April 2018. The event is managed by representatives of the Ministry of Education of the Russian Federation.
It is a whole system of Olympiads in subjects included in the compulsory program educational institutions countries. Participation in such an Olympiad is an honorable and responsible mission, because it is a student’s chance to show the accumulated baggage of knowledge, to defend the honor of their educational institution, and in case of victory, it is also an opportunity to receive financial incentives and earn a privilege when entering the best universities in Russia.
The practice of holding subject Olympiads has existed in the country for more than a hundred years - back in 1886, representatives of educational authorities initiated competitions between young talents. During the Soviet Union, this movement not only did not cease to exist, but also received an additional impetus to development. Starting from the 60s of the last century, intellectual competitions of the all-Union, and then all-Russian scale began to be held in almost all major school disciplines.
What subjects are included in the Olympiad list?
In the 2017-2018 academic year, the country's schoolchildren will be able to compete for prizes in several categories of disciplines:
- in the exact sciences, which include computer science and a mathematical block;
- in the natural sciences, which include geography, biology, astronomy, physics, chemistry and ecology;
- in the field of philology, including Olympiads in German, English, Chinese, French, Italian, as well as Russian language and literature;
- in the field of the humanities, consisting of history, social studies, law and economics;
- in other disciplines, which include physical education, world art culture, technology and life safety.
In the Olympiad tasks for each of the listed disciplines, two blocks of tasks are usually distinguished: a part that tests theoretical preparation, and a part aimed at identifying practical skills.
The main stages of the Olympiad 2017-2018
Holding the All-Russian school olympiad includes the organization of 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 focus on such periods of time.
Schoolchildren expect 4 stages of the competition different levels difficulties - Stage 1. School. Competitions between representatives of one school will be held in September-October 2017. The Olympiad is held between students of the parallel, starting from the fifth grade. The development of tasks for conducting subject Olympiads in this case is entrusted to the members of the methodological commission of the city level.
- Stage 2. Municipal. The stage, which hosts competitions between the winners of schools in the same city, representing grades 7-11, will be held from December 2017 to January 2018. The mission of compiling Olympiad tasks is assigned to the organizers of the regional level, and local officials are responsible for issues related to providing a place and ensuring the procedure for Olympiads.
- Stage 3. Regional. The third level of the Olympiad, which will be held in January-February 2018. At this stage, schoolchildren who won prizes at the city Olympiad and those who won last year's regional selections take part in the competition.
- Stage 4. All-Russian. Most high level Subject Olympiads will be organized by representatives of the Ministry of Education Russian Federation in March-April 2018. The winners of the regional level and the guys who won last year are invited to it. However, not every winner of the regional selection can become a participant in this stage. The exception is schoolchildren who have received 1st place in their region, but are lagging behind the winners at the level of other cities in terms of points. Prize-winners All-Russian stage can then go to international competitions that take place in the summer.
Where can I find typical tasks for the Olympiad?
Of course, in order to adequately perform in this event, you need to have a high level of preparation. The All-Russian Olympiad is represented on the network by its own website - rosolymp.ru - where students can get acquainted with the tasks of previous years, check their level by answering them, find out specific dates and requirements for organizational moments.
Tasks and keys school stage All-Russian Olympiad schoolchildren in mathematics
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Preview:
school stage
4th grade
1. Rectangle area 91
Preview:
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
Preview:
Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
6th grade
The maximum score for each task is 7 points
Preview:
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
Preview:
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.
Preview:
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
Preview:
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,
Preview:
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.
Preview:
Keys of the School Olympiad in 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 BUT = 3 is suitable, we prove that there are no other solutions. Reduce equality by BUT . 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.
Preview:
Keys of the School Olympiad 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 the 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 be.
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 Nastya is going to use each number 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.
Preview:
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.
Preview:
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. eighteen
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.
Preview:
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 middle line, which means that it cuts off a 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 the first is greater than the 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. ten!
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.