Fedot bought a notebook with a volume of 96 sheets. Mathematical Olympiads and Olympiad problems. We wish you success

Sections: Mathematics

Dear Olympiad participant!

The school mathematics Olympiad is held in one round.
There are 5 tasks of varying difficulty levels.
You are not presented with any special requirements regarding the execution of the work. The form of presentation of solutions to problems, as well as methods of solution, can be any. If you have any individual thoughts about a particular task, but you cannot complete the solution, do not hesitate to express all your thoughts. Even partially solved problems will be awarded the appropriate number of points.
Start solving problems that you think are easier, and then move on to the rest. This way you will save work time.

We wish you success!

School stage All-Russian Olympiad schoolchildren in mathematics

5th grade.

Exercise 1. In the expression 1*2*3*4*5, replace “*” with action signs and place the brackets like this. To get an expression whose value is 100.

FIVE - THREE = TWO It is known that instead of a letter A you need to substitute the number 2.

Task 3. How can you use a cup scale without weights to divide 80 kg of nails into two parts - 15 kg and 65 kg?

Task 4. Cut the figure shown in the figure into two equal parts so that each part has one star. You can only cut along the grid lines.

Task 5. A cup and saucer together cost 25 rubles, and 4 cups and 3 saucers cost 88 rubles. Find the price of the cup and the price of the saucer.

6th grade.

Exercise 1. Compare fractions without reducing them to a common denominator.

Task 2. It is required to decipher the notation of an arithmetic equality in which numbers are replaced by letters, and different numbers are replaced by different letters, and identical ones are replaced by identical ones. It is assumed that the original equality is true and written according to the usual rules of arithmetic.

WORK
+WILL
LUCK

Task 3. Three friends came to summer camp to relax: Misha, Volodya and Petya. It is known that each of them has one of the following surnames: Ivanov, Semenov, Gerasimov. Misha is not Gerasimov. Volodya's father is an engineer. Volodya is in 6th grade. Gerasimov studies in the 5th grade. Ivanov's father is a teacher. What is the last name of each of the three friends?

Task 4. Divide the figure along the grid lines into four equal parts so that each part contains one point.

Task 5. The jumping dragonfly slept half the time of every day of the red summer, danced for a third of the time of every day, and sang for a sixth of the time. She decided to devote the rest of her time to preparing for winter. How many hours a day did Dragonfly prepare for winter?

7th grade.

Exercise 1. Solve the puzzle if you know that the largest digit in the number STRONG is 5:

DECIDE
IF
STRONG

Task 2. Solve the equation│7 - x│ = 9.3

Task 3. After seven washes, the length, width and thickness of the soap were halved. How many washes will the remaining soap last?

Task 4 . Divide a rectangle of 4 × 9 cells along the sides of the cells into two equal parts so that you can then make a square from them.

Task 5. The wooden cube was painted white on all sides and then sawn into 64 identical cubes. How many cubes were colored with three sides? On both sides?
On the one side? How many cubes are not colored?

8th grade.

Exercise 1. What two digits does the number 13 end with?

Task 2. Reduce the fraction:

Task 3. The school drama club is preparing to stage an excerpt from A.S.’s fairy tale. Pushkin about Tsar Saltan, decided to distribute the roles between the participants.
“I will be Chernomor,” said Yura.
“No, I will be Chernomor,” said Kolya.
“Okay,” Yura conceded to him, “I can play Guidon.”
“Well, I can become Saltan,” Kolya also showed compliance.
- I agree to be only Guidon! - said Misha.
The boys' wishes were satisfied. How were the roles distributed?

Task 4. In an isosceles triangle ABC with base AB = 8 m, the median AD is drawn. The perimeter of triangle ACD is 2m greater than the perimeter of triangle ABD. Find AC.

Task 5. Nikolai bought a general notebook of 96 sheets and numbered the pages from 1 to 192. Nephew Arthur tore out 35 sheets from this notebook and added up all 70 numbers written on them. Could he have succeeded in 2010?

9th grade.

Exercise 1. Find the last digit of 1989 1989.

Task 2. The sum of the roots of a certain quadratic equation is 1, and the sum of their squares is 2. What is the sum of their cubes?

Task 3. Using three medians m a, m b and m c ∆ ABC, find the length of side AC = b.

Task 4. Reduce the fraction .

Task 5. In how many ways can you choose a vowel and a consonant in the word “kamzol”?

Grade 10.

Exercise 1. Currently there are coins of 1, 2, 5, 10 rubles. List all sums of money that can be paid with both an even and an odd number of coins.

Task 2. Prove that 5 + 5 2 + 5 3 + … + 5 2010 is divisible by 6.

Task 3. In a quadrangle ABCD diagonals intersect at a point M. It is known that AM = 1,
VM = 2, SM = 4. At what values DM quadrilateral ABCD is it a trapezoid?

Task 4. Solve the system of equations

Task 5. Thirty schoolchildren - tenth and eleventh graders - shook hands. It turned out that every tenth grader shook hands with eight eleventh graders, and every eleventh grader shook hands with seven tenth graders. How many tenth graders were there and how many eleventh graders were there?

This work Petya bought a general notebook with a volume of 96 sheets and numbered all its pages in order with numbers from 1 to 192. Vasya tore out (Test) on the subject (ACD and financial analysis), it was custom-made by specialists of our company and passed its successful defense. Work - Petya bought a general notebook with a volume of 96 sheets and numbered all its pages in order with numbers from 1 to 192. Vasya tore out the ACD on the subject and financial analysis reflects its topic and the logical component of its disclosure, the essence of the issue under study is revealed, the main provisions and leading ideas are highlighted this topic.
Work - Petya bought a general notebook with a volume of 96 sheets and numbered all its pages in order with numbers from 1 to 192. Vasya tore it out, contains: tables, drawings, the latest literary sources, the year the work was submitted and defended - 2017. In the work, Petya bought a general notebook volume of 96 sheets and numbered all its pages in order with numbers from 1 to 192. Vasya pulled out (AHD and financial analysis) reveals the relevance of the research topic, reflects the degree of development of the problem, based on a deep assessment and analysis of scientific and methodological literature, in work on the subject of ACD and financial analysis, the object of analysis and its issues are considered comprehensively, both from the theoretical and practical sides, the goal and specific objectives of the topic under consideration are formulated, there is a logic of presentation of the material and its sequence.

Problem 16:

Is it possible to exchange 25 rubles using ten bills in denominations of 1, 3 and 5 rubles? Solution:

Answer: No

Problem 17:

Petya bought a general notebook with a volume of 96 sheets and numbered all its pages in order with numbers from 1 to 192. Vasya tore out 25 sheets from this notebook and added up all 50 numbers written on them. Could he have succeeded in 1990? Solution:

On each sheet, the sum of the page numbers is odd, and the sum of 25 odd numbers is odd.

Problem 18:

The product of 22 integers is 1. Prove that their sum is not zero. Solution:

Among these numbers is an even number of “minus ones”, and in order for the sum to be equal to zero, there must be exactly 11 of them.

Problem 19:

Is it possible to form a magic square from the first 36 prime numbers? Solution:

Among these numbers, one (2) is even and the rest are odd. Therefore, in the line where there is a two, the sum of the numbers is odd, and in others it is even.

Problem 20:

Numbers from 1 to 10 are written in a row. Is it possible to place the “+” and “-” signs between them so that the value of the resulting expression is equal to zero?

Note: Please note that negative numbers are also even and odd. Solution:

In fact, the sum of the numbers from 1 to 10 is 55, and by changing the signs in it, we change the entire expression to an even number.

Problem 21:

The grasshopper jumps in a straight line, and the first time he jumped 1 cm in some direction, the second time - 2 cm, and so on. Prove that after 1985 jumps he cannot end up where he started. Solution:

Note: The sum 1 + 2 + … + 1985 is odd.

Problem 22:

The numbers 1, 2, 3, ..., 1984, 1985 are written on the board. You are allowed to erase any two numbers from the board and write down the modulus of their difference instead. Eventually there will be only one number left on the board. Can it be zero? Solution:

Check that the above operations do not change the parity of the sum of all numbers written on the board.

Problem 23:

Is it possible to cover a chessboard with 1 × 2 dominoes so that only squares a1 and h8 remain free? Solution:

Each domino covers one black and one white square, and when discarding squares a1 and h8, there are 2 fewer black squares than white ones.

Problem 24:

To the 17-digit number we added a number written in the same digits, but in reverse order. Prove that at least one digit of the resulting sum is even. Solution:

Consider two cases: the sum of the first and last digits of a number is less than 10, and the sum of the first and last digits of a number is not less than 10. If we assume that all digits of the sum are odd, then in the first case there should not be a single carryover in the digits (which is obvious , leads to a contradiction), and in the second case, the presence of carry when moving from right to left or left to right alternates with the absence of carry, and as a result we get that the sum digit in the ninth digit is necessarily even.

Problem 25:

There are 100 people in the people's squad, and every evening three of them go on duty. Could it be that after some time it turns out that everyone was on duty with everyone exactly once? Solution:

Since on each duty in which this person participates, he is on duty with two others, then everyone else can be divided into pairs. However, 99 is an odd number.

Problem 26:

There are 45 points on the line that lie outside the segment AB. Prove that the sum of the distances from these points to point A is not equal to the sum of the distances from these points to point B. Solution:

For any point X lying outside AB, we have AX - BX = ± AB. If we assume that the sums of the distances are equal, then we obtain that the expression ± AB ± AB ± … ± AB, which involves 45 terms, is equal to zero. But this is impossible.

Problem 27:

There are 9 numbers arranged in a circle - 4 ones and 5 zeros. Every second the following operation is performed on the numbers: a zero is placed between adjacent numbers if they are different, and a unit if they are equal; after that the old numbers are erased. Can all the numbers become the same after some time? Solution:

It is clear that a combination of nine ones cannot be obtained before nine zeros. If there were nine zeros, then on the previous move the zeros and ones had to alternate, which is impossible, since there are only an odd number of them.

Problem 28:

25 boys and 25 girls are sitting at a round table. Prove that some of the people sitting at the table have both boys as neighbors. Solution:

Let's carry out our proof by contradiction. Let's number everyone sitting at the table in order, starting from some place. If on kth place a boy is sitting, then it is clear that girls are sitting in the (k - 2)th and (k + 2)th places. But since there are equal numbers of boys and girls, then for any girl sitting in the nth place, it is true that there are boys sitting in the (n - 2)th and (n + 2)th places. If we now consider only those 25 people who sit in “even” seats, we find that among them boys and girls alternate if we go around the table in some direction. But 25 is an odd number.

Problem 29:

The snail crawls along the plane at a constant speed, turning at right angles every 15 minutes. Prove that she can return to the starting point only after an integer number of hours. Solution:

It is clear that the number a of areas in which the snail crawled up or down is equal to the number of areas in which it crawled to the right or to the left. It remains only to note that a is even.

Problem 30:

Three grasshoppers play leapfrog on a straight line. Each time one of them jumps over the other (but not both at once!). Can they end up in the same places after the 1991 jump? Solution:

Let's denote the grasshoppers A, B and C. Let's call the arrangement of grasshoppers ABC, BCA and CAB (from left to right) correct, and ACB, BAC and CBA incorrect. It is easy to see that with any jump the type of arrangement changes.

Problem 31:

There are 101 coins, of which 50 are fake, differing in weight by 1 gram from the real ones. Petya took one coin and in one weighing on a scale with an arrow showing the difference in weights on the cups, he wants to determine whether it is counterfeit. Will he be able to do it? Solution:

You need to put this coin aside, and then divide the remaining 100 coins into two piles of 50 coins each, and compare the weights of these piles. If they differ by an even number of grams, then the coin we are interested in is real. If the difference in weights is odd, then the coin is counterfeit.

Problem 32:

Is it possible to write down the numbers from 1 to 9 once in a row so that there is an odd number of digits between one and two, two and three, ..., eight and nine? Solution:

Otherwise, all numbers in a row would be in places of the same parity.