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Problem 3. During the physical education class, the entire class lined up by height (all children have different heights). Dima noticed that the number of people taller than him is four times the number of people shorter than him. And Lёnya noticed that the number of people taller than him is three times less than the ... | Answer: 21.
Solution. Let $x$ be the number of people who are shorter than Dima. Then, the total number of students in the class is $x$ (people who are shorter than Dima) $+4 x$ (people who are taller than Dima) +1 (Dima) $=5 x+1$ (total number of people in the class).
Let $y$ be the number of people who are taller t... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Koschei the Deathless has 11 large chests. In some of them lie 8 medium chests. And in some of the medium chests lie 8 small chests. The chests contain nothing else. In total, Koschei has 102 empty chests. How many chests does Koschei have in total? | Answer: 115.
Solution. Let $x$ be the number of non-empty chests.
Consider the process when Koschei just started placing chests inside each other. Initially, he had 11 empty large chests. Each time he placed 8 smaller chests into a single empty chest, the total number of empty chests increased by 7 ( -1 old empty che... | 115 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. The dragon has 40 piles of gold coins, and the number of coins in any two of them differs. After the dragon plundered a neighboring city and brought back more gold, the number of coins in each pile increased by either 2, 3, or 4 times. What is the smallest number of different piles of coins that could result... | Answer: 14.
Solution. Evaluation. Suppose that no more than 13 piles increased by the same factor. Then no more than 13 piles increased by a factor of 2, no more than 13 by a factor of 3, and no more than 13 by a factor of 4. Thus, the dragon has no more than 39 piles in total. Contradiction.
Therefore, there will be... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. On one face of a die, one dot is drawn, on another face - two, on the third - three, and so on. Four identical dice are stacked as shown in the figure. How many dots in total are on the 6 faces where the dice touch?
 writing 10 digits on the board from left to right to form a ten-digit number. Moreover, it is not allowed to write two consecutive identical digits. If the resulting number is divisible by 9, then Sasha wins; otherwise, Masha wins. Who will win with correct play fro... | Answer: Sasha.
Solution. A number is divisible by 9 if the sum of its digits is divisible by 9. Therefore, one of Sasha's possible strategies is to complement each of Masha's digits to 9. That is, if Masha writes 0, then Sasha writes 9; if Masha writes 1, then Sasha writes 8, and so on. Thus, after each pair of moves,... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1 Karlson eats three jars of jam and one jar of honey in 25 minutes, while Little Man takes 55 minutes. One jar of jam and three jars of honey Karlson eats in 35 minutes, while Little Man takes 1 hour 25 minutes. How long will it take them to eat six jars of jam together? | 9.1 20 minutes.
From the condition, it follows that if Karlson eats three jars of jam and one jar of honey, and then immediately eats one jar of jam and three jars of honey, he will spend $25+35=60$ minutes. For Little Man, this time will be 140 minutes. Therefore, Karlson will spend 15 minutes on one jar of jam and o... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. At the first stop, 18 passengers entered an empty bus. Then at each stop, 4 people got off and 6 people got on. How many passengers were in the bus between the fourth and fifth stops? | Answer: 24 people.
## Solution.
## Method 1.
After each stop, except the first one, the number of passengers in the bus increases by 2 people. Therefore, from the second to the fourth stop, the number of people increased by 6 people. That is, it became $18+6=24$ people.
Method 2.
From the second to the fourth stop... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. A sign engraver makes signs with letters. He engraves identical letters in the same amount of time, and different letters possibly in different times. For two signs “ДОМ МОДЫ” and “ВХОД” together, he spent 50 minutes, and one sign “В ДЫМОХОД” he made in 35 minutes. How long will it take him to make the sign “ВЫХОД”? | Answer: 20 minutes.
## Solution.
In the signs FASHION HOUSE ENTRANCE and IN CHIMNEY, we separate the letters that form the word EXIT, then from the first sign, D, O, M, M, O, D will remain, and from the second - D, M, O. Note that FASHION HOUSE ENTRANCE differs from IN CHIMNEY by the letters D, O, M, and in time - by... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. The younger brother takes 25 minutes to reach school, while the older brother takes 15 minutes to walk the same route. How many minutes after the younger brother leaves home will the older brother catch up to him if he leaves 8 minutes later? | Answer: in 17 minutes. Solution: Let $S$ be the distance from home to school. Since the younger brother covers this distance in 25 minutes, in 8 minutes he will cover the distance $\frac{8 S}{25}$. After the older brother leaves the house, the rate at which they are closing the distance between them will be $\frac{S}{1... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Petya tells his neighbor Vova: «In our class, there are 30 people, and there is an interesting situation: any two boys have a different number of girlfriends in the class, and any two girls have a different number of friends among the boys in the class. Can you determine how many boys and how many girls we have in... | Answer. a) Vova is wrong; b) 15 boys and 15 girls. Solution. a) To show that Vova is wrong, we will provide an example of a class that meets the conditions of the problem. For a class with 15 girls and 15 boys, we will number them and present a "friendship" table. In the cells of the table, there is a "+", if the corre... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the USA, the date is typically written as the month number, followed by the day number, and then the year. In Europe, however, the day comes first, followed by the month and the year. How many days in a year cannot be read unambiguously without knowing which format it is written in? | 2. It is clear that these are the days whose date can be the number of the month, i.e., takes values from 1 to 12. There are such days $12 \cdot 12=144$. However, the days where the number matches the month are unambiguous. There are 12 such days. Therefore, the number of days sought is $144-12=132$. | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A traveler arrived on an island inhabited by liars (L) and truth-tellers (P). Each L, when asked a question "How many..?", gives a number that is 2 more or 2 less than the correct answer, while each P answers correctly. The traveler met two residents of the island and asked each how many L and P live on the island. ... | 5. I - L, II - P. On the island, there are 1000 L and 1000 P. The answers of the first and second are different, so the option P and P is impossible. The option L and L is also impossible, because the numbers 1001 and 1000 differ by 1, while the answers of the liars regarding the number of L should differ by 4, or coin... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A mathematician left point A for point B. After some time, a physicist also left point A for point B. Catching up with the mathematician after 20 km, the physicist, without stopping, continued to point B and turned back. They met again 20 km from B. Then each continued in their respective directions. Upon reaching p... | # Answer: 45.
Solution. From the first meeting to the second meeting, the mathematician walked a total of 100 - 20 - 20 = 60 km, while the physicist walked -100 - 20 + 20 = 100 km. It is clear from this that the ratio of their speeds is 6:10 or 3:5. From the second to the third meeting, they will walk together 100 + 1... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.4. Three pirates were dividing a bag of coins. The first took 3/7 of all the coins; the second took 51 percent of the remainder. After this, the third received 8 fewer coins than the second. How many coins were in the bag? Justify your answer. | Solution: The third pirate received $49\%$ of the remainder, which is $2\%$ less than the third. These two percent of the remainder amount to 8 coins, so one percent is
4 coins, and the entire remainder is 400 coins. These 400 coins make up $1-\frac{3}{7}=\frac{4}{7}$ of the total. Therefore, the total number of coins... | 700 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.5. On cards, all two-digit numbers from 10 to 99 are written (one number per card). All these cards are lying on the table face down. What is the minimum number of cards that need to be flipped to guarantee that at least one of the revealed numbers is divisible by 7? Justify your answer. | Solution: There are exactly 13 two-digit numbers divisible by 7 (14 = 7 * 2, 21 = 7 * 3, ..., 98 = 7 * 14). If at least 13 cards are not flipped, it is impossible to exclude the situation where all these numbers remain covered. Therefore, no more than 12 cards can be left in their original position, and the rest must b... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. January first of a certain non-leap year fell on a Saturday. And how many Fridays are there in this year? | Answer: 52
Solution. In a non-leap year, there are 365 days. The first two days fell on Saturday and Sunday, followed by 51 full weeks ( $51 \cdot 7=357$ days) and 6 more days. In total, there are 52 Fridays. | 52 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1. Lisa drew graphs of all functions of the form $y=a x+b$, where $a$ and $b$ take all natural values from 1 to 100. How many of these graphs pass through the point $(3,333)$? | Answer: 23
Solution. We are looking for such $a$ and $b$ that $333=3a+b$. That is, $b=3(111-a)$. Therefore, $0<111-a \leqslant 33$, $78 \leqslant a$. There are exactly 23 such $a$ and for each of them, $b$ can be uniquely determined. | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. The sea includes a bay with more saline water. The salinity of the water in the sea is 120 per mille, in the bay 240 per mille, in the part of the sea not including the bay - 110 per mille. How many times is the volume of water in the sea larger than the volume of water in the bay? The volume of water is considere... | Answer: 13
Solution. Let the volume of salt in the bay be $s_{1}$, and the volume of water $v_{1}$; in the part of the sea not including the bay, the volume of salt is $s_{2}$, and the total volume is $v_{2}$. We have the equations
$$
\frac{s_{1}}{v_{1}}=\frac{240}{1000} ; \quad \frac{s_{2}}{v_{2}}=\frac{110}{1000} ;... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. In the club, there are 9 people. Every day, some three of them went to the cafe together, while the others did not go to the cafe. After 360 days, it turned out that any two people from the club had been to the cafe together the same number of times. How many times? | Answer: 30
Solution. The total number of pairs of people in the circle is $9 \cdot 8 / 2=36$. Over 360 days, the cafe was visited by $360 \cdot 3$ pairs (since three new pairs are added each day). Since all pairs visited the cafe an equal number of times, this number is $360 \cdot 3 / 36=30$. | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Of all numbers with the sum of digits equal to 25, find the one whose product of digits is maximal. If there are several such numbers, write the smallest of them in the answer. | Answer: 33333334
Solution. Obviously, there is no 0 in the number. If the number contains the digit 1, then it can be removed and one of the remaining digits can be increased by 1, which does not change the sum, but increases the product. If the number contains a digit $x \geqslant 5$, then it can be replaced by the d... | 33333334 | Other | math-word-problem | Yes | Yes | olympiads | false |
11.1 The number 890 has the property that by changing any of its digits to 1 (increasing or decreasing), you can obtain a number that is a multiple of 11. Find the smallest three-digit number with the same property. | Answer: 120.
Solution. Since the last digit of the number must be changed to obtain a number divisible by 11, the required number should differ from it by 1. The smallest three-digit number divisible by 11 is 110. However, the numbers adjacent to it, 109 and 111, do not have the required property. Indeed, if the secon... | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. All graduates of the mathematics school took the Unified State Exam (USE) in mathematics and physical education. Each student's result in mathematics turned out to be equal to the sum of the results of all other students in physical education. How many graduates are there in the school if the total number of points ... | # Answer: 51.
## Solution:
Let the total number of graduates be $n$. Denote the total score in mathematics by $M$. This sum is equal to the sum of the physical education scores of all students, counted $(n-1)$ times.
Thus, we have the equation: $M=50 \frac{M}{n-1}$.
From this, the answer is
Instructions for checki... | 51 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The letters А, Б, К, М, П, У, Ш were encoded with sequences of zeros and ones (each with its own). Then, in the word ПАПАМАМАБАБУШКА, the letters were replaced with their codes. Could the length of the resulting sequence be shorter than 40 characters, if the sequence can be uniquely decoded? | Answer: She could.
## Solution:
Here is an example of a code table:
| A | B | K | M | P | U | Sh |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 110 | 1111 | 100 | 101 | 11100 | 11101 |
The word will look like this: 10101010100010001100110111001110111110 - a total of 38 characters. Decoding is una... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. A travel agency ran a promotion: "Buy a trip to Egypt, bring four friends who also buy a trip, and get the cost of your trip back." During the promotion, 13 customers came on their own, and the rest were brought by friends. Some of them brought exactly four new customers, while the other 100 did not bring anyone.... | Answer: 29.
Let each of the $x$ potential "lucky ones" bring 4 friends. Then the number of "brought" customers is $4 x$, and 13 came on their own, so the total number of tourists was $13+4 x$.
On the other hand, $x$ people brought new customers, while 100 people did not, so the total number of tourists was $x+100$.
... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Papa always picked up Misha from kindergarten at the same time and in the same car. But today, after firmly establishing that there would be pea porridge for dinner, Misha resolutely left the kindergarten and set off down the road to meet Papa. Having walked 5 km, Misha met Papa, got into the car, and they arrived h... | 3. Speed: 60 km/h (or 1 km/min). Today, Misha and his dad saved 10 minutes that were previously spent driving 10 km (from the meeting point to the kindergarten and back). This means that in 1 minute, Misha's dad drives 1 km, i.e., his speed is 60 km/h.
Criteria: Correct reasoning and correct answer - 7 points; only an... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. How many natural numbers greater than one exist, the product of which with their smallest prime divisor does not exceed 100? | Answer: 33.
Sketch of the solution. The smallest prime divisor can be 2. These are the numbers: $2, 4, \ldots, 50$ (25 numbers). The smallest prime divisor can be 3. These are the numbers: 3, 9, 15, 21, 27, 33 (6 numbers). The smallest prime divisor can be 5. This number: 5 (1 number). The smallest prime divisor can b... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Every week, listeners choose the ten most popular songs via SMS. It is known that 1) the same set of songs in the same order is never chosen twice in a row; 2) a song that has once dropped in the ranking will not rise again in the future. What is the maximum number of weeks that the same 10 songs can remain in the r... | Answer: 46 weeks.
Sketch of the solution. Suppose that for two consecutive weeks, the same set of 10 songs was on the list. Since the order of the songs is different, at least one song in the list has moved up in the ranking, and at least one has moved down. Since a song, once it has moved down, does not move up in th... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A student's income comes from three sources: scholarship, temporary part-time work, and parental support. If the government doubles the scholarship, the income will increase by $5 \%$. If the time spent on part-time work is doubled, the income will increase by $15 \%$. By what percentage will the student's income in... | Solution. Let $\mathrm{S}$ be the student's monthly income, $a, b$ and $c$ be the amounts of the scholarship, part-time job, and parental support, respectively (expressed, for example, in rubles). Clearly, $S=a+b+c$. Then, according to the conditions, $2 a+b+c=1.05 S$ and $a+2 b+c=1.15 S$. From the first equation, $a=0... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. In each cell of a table consisting of 10 columns and $n$ rows, a digit is written. It is known that for any row $A$ and any two columns, there exists a row that differs from $A$ exactly in these two columns. Prove that $n \geqslant 512$.
(P. Karasev) | 10.1. Let $R_{0}$ be the first row of the table. Consider any set of an even number of columns and number them from left to right: $C_{1}, \ldots, C_{2 m}$. Then there is a row $R_{1}$ that differs from $R_{0}$ exactly in columns $C_{1}$ and $C_{2}$; further, there is a row $R_{2}$ that differs from $R_{1}$ exactly in ... | 512 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Write the number 2021 using each of the digits from 0 to 9 exactly once, along with parentheses and arithmetic operations. (Parentheses and arithmetic operations can be used in any quantity. "Sticking" digits together to form a single number is allowed) | For example, $43 \cdot (8 \cdot 5 + 7) + 0 \cdot 1 \cdot 2 \cdot 6 \cdot 9 = 2021$. | 2021 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. A mathematician and a physicist started running on a running track towards the finish line at the same time. After finishing, the mathematician said: "If I had run twice as fast, I would have beaten the physicist by 12 seconds." And after finishing, the physicist said: "If I had run twice as fast, I would have beate... | # Answer: 16.
Solution: Let the time it took the mathematician to run the entire distance be $2 x$ seconds, and the time it took the physicist be $2 y$ seconds. Then, under the condition that the mathematician ran twice as fast, we get $2 y-x=12$. And from the condition that the physicist ran twice as fast, we get $2 ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On the island, there live 7 natives who know mathematics and physics, 6 natives who know physics and chemistry, 3 natives who know chemistry and mathematics, and 4 natives who know physics and biology. In how many ways can a team of three people be formed who together know at least three subjects out of the four? Th... | # Answer: 1080.
## Solution.
Let's choose three people. If this trio includes representatives of at least two different groups of aborigines, then together they know no fewer than three subjects.
If this trio includes representatives of only one group, then together they know only two subjects.
Therefore, we will f... | 1080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Islandland consists of ten islands, some of which are connected by two-way air routes. If you choose any 9 islands, you can fly around them one by one and return to the starting island at the end. Find the minimum number of air routes that can exist in this country. | Answer: 15 airlines.
Solution. Evaluation. Let islands $A$ and $B$ be connected by an airline. If we choose 9 islands for the tour, excluding $A$, then a circular path passes through $B$, meaning $B$ is connected by airlines to at least two other islands. Thus, at least three airlines depart from $B$, and a similar st... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will be t... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we get $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
$. It turns out that $\angle A B D=\angle B C D$. Find the length of segment $B D$, if $B C=36$ and $A D=64$.
 | Answer: 48.
Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4.1. A rectangle was cut into six smaller rectangles, the areas of five of them are marked on the diagram. Find the area of the remaining rectangle.
 | Answer: 101.
Solution. Let's introduce the notation as shown in the figure.
Notice that
$$
2=\frac{40}{20}=\frac{S_{H I L K}}{S_{D E I H}}=\frac{H K \cdot H I}{H D \cdot H I}=\frac{H K}{H ... | 101 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2. Toshа travels from point A to point B via point C. From A to C, Toshа travels at an average speed of 75 km/h, and from C to B, Toshа travels at an average speed of 145 km/h. The entire journey from A to B took Toshа 4 hours and 48 minutes. The next day, Toshа travels back at an average speed of 100 km/h. The journ... | Answer: 290 km.
Solution. Let $\mathrm{X}$ km be the distance between B and C, and $y$ km be the distance between A and C. From the system of equations $x / 145 + y / 75 = 24 / 5$ and $(x + y) / 100 = 2 + y / 70$, we find: $x = 290$ and $y = 210$.
. Two chips are called "connected" if they are in the same column or row, and there are no other chips between them. What is the maximum number of chips that can be placed on the board so that each has no more than two "con... | Answer: 35.
Solution. Let $x_{1}, x_{2}, \ldots, x_{20}$ be the number of chips in rows $1,2, \ldots, 20$, and $y_{1}, y_{2}, \ldots, y_{15}$ be the number of chips in columns $1,2, \ldots, 15$. Then the total number of chips $S=x_{1}+x_{2}+\ldots+x_{20}=y_{1}+y_{2}+\ldots+y_{15}$. The number of "connectivities" $k \g... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1. Petya thought of two numbers and wrote down their product. After that, he decreased the first of the thought numbers by 3, and increased the other by 3. It turned out that the product increased by 900. By how much would the product have decreased if Petya had done the opposite: increased the first number ... | Answer: 918.
Solution: Let these numbers be $a$ and $b$. Then, according to the condition, $(a-3)(b+3)-ab=600$. Expanding the brackets: $ab+3a-3b-9-ab=900$, so $a-b=303$. We need to find the difference $ab-(a+3)(b-3)=$ $ab-ab+3a-3b+9=3(a-b)+9=3 \cdot 303+9=918$. | 918 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Variant 1. At the intersection of perpendicular roads, a highway from Moscow to Kazan and a road from Vladimir to Ryazan intersect. Dima and Tolya set out with constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively. When Dima passed the intersection, Tolya had 3500 meters left to reach it. Wh... | Answer: 9100.
Solution. When Tolya has traveled 3500 meters, Dima will have traveled 4200 meters, so at the moment when Dima is 8400 meters past the intersection, Tolya will be 3500 meters past the intersection. By the Pythagorean theorem, the distance between the boys is 9100 meters. | 9100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Variant 1. Find the largest root of the equation
$(x+1)(x+2)-(x+2)(x+3)+(x+3)(x+4)-(x+4)(x+5)+\cdots-(x+1000)(x+1001)=0$. | Answer: -501.
Solution. Let's break the terms into pairs of adjacent terms and factor out the common factor, we get $(x+2)(x+1-x-3)+\cdots+(x+1000)(x+999-x-1001)=-2 \cdot(x+2+x+4+\cdots+x+1000)=$ $-2 \cdot\left(500 x+\frac{2+1000}{2} \cdot 500\right)=-1000 \cdot(x+501)=0$. From this, $x=-501-$ the only root. | -501 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1
Unit cubes were used to assemble a large parallelepiped with sides greater than 4. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 836. Find the number of cubes that have no more th... | Answer: 144.
Solution. Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$ and $c-2$ is greater than 2 and their product equals the product of four prime numbers $2,2,11$ and 19, we ha... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.5. Three hundred non-zero integers are written in a circle such that each number is greater than the product of the next three numbers in the clockwise direction. What is the maximum number of positive numbers that can be among these 300 written numbers? | Answer: 200.
Solution: Note that three consecutive numbers cannot all be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number preceding them (counterclockwise) is also a natural number. Since it is greater than the product of these three natural numbers, it is greater ... | 200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. Misha and Grisha are writing numbers on the board. Misha writes threes, and Grisha writes fives. In total, they wrote 20 numbers. How many fives were written if the sum of all the numbers is 94? | Answer: 17
Solution. If all 20 numbers were "5", their sum would be 100. We will replace "5" with "3". With each replacement, the sum decreases by 2. Since $100-94=6, 6: 2=3$, we need 3 replacements. Therefore, the number of fives is $20-3=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Five identical squares, standing in a row, were cut by two horizontal lines. The sum of the perimeters of the resulting 15 rectangles is 800 cm. Indicate in centimeters the length of the original squares.
| | | | | |
| :--- | :--- | :--- | :--- | :--- |
| | | | | | | # Answer: 20
Solution. Let's calculate how many times the side of the original square is repeated in the sum of all perimeters. The sides of the rectangle are counted once (a total of 12), and the crossbars are counted twice ($4 \cdot 2 + 10 \cdot 2 = 28$). In total, $40.800: 40=20$ cm - the side of the square. | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. In a row, 64 people are standing - each one is either a knight, who always tells the truth, or a liar, who always lies. One of the standing knights said that he stands next to a knight and a liar, and all the other 63 people repeated his phrase. Indicate how many of them were knights. | Answer: 42
Solution. The people at the ends must be liars, as they do not have a second neighbor, and by saying this phrase, those standing there lied. All others can be liars, in which case there are no knights at all, but this option contradicts the condition that at least one knight is present. If there is a knight... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.10. In the company, there are 100 children, some of whom are friends (friendship is always mutual). It is known that by selecting any child, the remaining 99 children can be divided into 33 groups of three such that in each group, all three are pairwise friends. Find the minimum possible number of pairs of friends.
... | Answer: 198.
Solution: Let's translate the problem into the language of graphs, associating each child with a vertex and each friendship with an edge. Then we know that in this graph with 100 vertices, after removing any vertex, the remaining vertices can be divided into 33 triples such that the vertices in each tripl... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
 are there on the parabola $y=x^{2}$, such that the tangent at these points intersects both coordinate axes at points with integer coordinates not exceeding 2020 in absolute value? | Answer: 44. Solution. The equation of the tangent to the parabola $y=x^{2}$ at the point $\left(x_{0}, y_{0}\right)$, where $y_{0}=x_{0}^{2}$, is $y-y_{0}=2 x_{0}\left(x-x_{0}\right)$. From this, we find the coordinates of the points of intersection of the tangent with the axes, namely $x_{1}=\frac{x_{0}}{2}, y_{1}=-x_... | 44 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain company, there are 100 shareholders, and any 66 of them own no less than $50 \%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? | 4. Let $M$ be the shareholder owning the largest percentage of shares - $\mathrm{x} \%$ of shares. Divide the other 99 shareholders into three groups A, B, C, each with 33 shareholders. Let them own a, b, c percentages of shares, respectively. Then $2(100-x)=2(a+b+c)=(a+b)+(b+c)+(c+a) \geq 50+50+50$, i.e., $x \leq 25$.... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. The age of a certain person in 1988 was equal to the sum of the digits of their birth year. How old was he
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9.2. The age of a certain person in 1988 was equal to the sum of the digits of their birth y... | Answer: 22.
Solution. The number of years a person has lived is equal to the sum of the digits of a four-digit number, each of which is no more than 9. Therefore, he is no more than 36 years old, and he was born in the 20th century. Let $x$ be the number of tens, $y$ be the number of units in his birth year.
Then, ac... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4 For each integer from 10 to 2021, we found the product of the digits, and then added all the obtained results. What is the sum that was obtained? | Solution: Consider the product $(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9)$. If we expand the brackets, we get the products of pairs of digits that form all two-digit numbers. Similarly, in the product
$$
(1+2+3+\cdots+9) \cdot(0+1+2+\cdots+9) \cdot(0+1+2+\cdots+9)
$$
after expanding the brackets, we get all combinations... | 184275 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3 There are 12 natural numbers. It is known that the sum of any three of them is not less than 100. Prove that the sum of all the numbers is not less than 406. | Solution. Arrange these numbers in non-decreasing order: $\mathrm{a}_{1} \leq \mathrm{a}_{2} \leq \mathrm{a}_{3} \leq \mathrm{a}_{4} \leq \ldots \leq \mathrm{a}_{12}$. By the condition $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3} \geq 100$, therefore $3 \mathrm{a}_{3} \geq 100, \mathrm{a}_{3} \geq \frac{100}{3}>33$. S... | 406 | Inequalities | proof | Yes | Yes | olympiads | false |
5. A semicircle with diameter $A B$ and center at point $O$ is divided into three parts by points $C$ and $D$ such that point $C$ lies on the arc $A D$. Perpendiculars $D E$ and $D F$ are dropped from point $D$ to segments $O C$ and $A B$ respectively. It turns out that $D E$ is the angle bisector of triangle $A D C$, ... | Answer: $20^{\circ}$.
Solution. Triangle $A O D$ is isosceles ($O D=O A$, as radii), hence, $\angle O A D=\angle O D A$. Since $D O$ is the bisector of angle $A D F$, then $\angle O A D=$ $\angle O D F$. Calculation of angles in the right triangle $A F D$ shows that $\angle O A D=30^{\circ}$. Let $G$ be the point of i... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. Vlad and Dima decided to earn some money. Each of them decided to deposit 3000 rubles in the bank and withdraw all the money after a year.
Vlad chose the deposit "Confidence": the amount increases by $20\%$ over the year, but the bank charges a $10\%$ fee upon withdrawal.
Dima chose the deposit "Reliabil... | Answer: Dima will earn 120 rubles more.
Solution. Vlad's deposit amount will increase to $3000 \cdot 1.2$ rubles in a year, and after withdrawal, it will decrease to $3000 \cdot 1.2 \cdot 0.9=3240$ rubles.
Dima's deposit amount will increase to $3000 \cdot 1.4$ rubles in a year, and after withdrawal, it will decrease... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. The Smeshariki Kros, Yozhik, Nyusha, and Barash ate a total of 86 candies, and each of them ate no fewer than 5 candies. It is known that:
- Nyusha ate more candies than each of the other Smeshariki;
- Kros and Yozhik together ate 53 candies.
How many candies did Nyusha eat | Answer: 28.
Solution. Krosh or Yozhik ate at least 27 candies (otherwise, they would have eaten no more than $26+26=52$ candies in total), so Nusha ate at least 28 candies. Considering that Barash ate at least 5 candies, we get that all of them together ate at least $53+28+5=86$ candies. Therefore, this is only possib... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. A store sells four types of nuts: hazelnuts, almonds, cashews, and pistachios. Stepan wants to buy 1 kilogram of nuts of one type and another 1 kilogram of nuts of a different type. He calculated how much such a purchase would cost him depending on which two types of nuts he chooses. Five out of six possib... | Answer: 2290.
Solution. Let $a, b, c, d$ be the cost of 1 kilogram of hazelnuts, almonds, cashews, and pistachios, respectively. From the condition, it follows that the set $A=\{1900,2070,2110,2330,2500\}$ is contained in the set $B=\{a+b, b+c, c+d, d+a, a+c, b+d\}$.
Note that the 6 elements of set $B$ can be divided... | 2290 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.6. A magic square is a $3 \times 3$ table in which numbers are arranged so that the sums of all rows, columns, and the two main diagonals are the same. The figure shows a magic square in which all numbers except three have been erased. Find what the number in the upper left corner of the square is.
| $?$ | 3... | Answer: 14.
Solution. Let the unknown number be $x$, then the sums in all rows, columns, and on the main diagonals are $9+31+x=40+x$.
1) Considering the left column, we get that the number in the lower left corner of the square is 27.
2) Considering the main diagonal going up to the right, we get that the number in t... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
84. a) Prove that there exists a pair of two-digit numbers such that if 20 is added to the first number and 15 is subtracted from the second, the resulting numbers will remain two-digit, and their product will be equal to the product of the original numbers? b) How many such pairs are there? | Answer: b) 16 pairs. Hint For part a), it is sufficient to provide a specific example (see problem 7.4). b) Let $a, b$ be the desired pair of numbers. Then $(a+20)(b-15)=a b$. From this, $20 b-15 a=20 \cdot 15 \Leftrightarrow 4 b-3 a=60$.
Since 60 and $3a$ are divisible by 3, $b$ must also be divisible by 3, i.e., $b=... | 16 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation
$$
y=\frac{1}{5} x^{2}+a x+b
$$
passes through points $B$ and $C$. Additionally, the vertex of this parabola (po... | Answer: 20.
Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, throu... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is such that $\angle ADC = 2 \angle CAD = 82^{\circ}$. Inside the trapezoid, a point $T$ is chosen such that $CT = CD, AT = TD$. Find $\angle TCD$. Give your answer in degrees.
.
. From this similarity and the cyclic nature of the pentagon
. It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
$ of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will tell them $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$, and separately inform them o... | Answer. For $k=2017$.
Solution. First, we prove that $k>2016$. Suppose the teacher used some $k \leqslant 2016$, thinking of the polynomial $P(x)$. Consider the polynomial $Q(x)=P(x)+(x-n_{1})(x-n_{2}) \ldots(x-n_{k})$. Note that the degree of the polynomial $Q(x)$ is also 2017, and its leading coefficient is also 1. ... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. From Zlatoust to Miass, "GAZ", "MAZ", and "KamAZ" set off simultaneously. "KamAZ", having reached Miass, immediately turned back and met "MAZ" 18 km from Miass, and "GAZ" - 25 km from Miass. "MAZ", having reached Miass, also immediately turned back and met "GAZ" 8 km from Miass. What is the distance from Zlatoust to... | Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" - $k$ km/h. For each pair of vehicles, we equate their travel time until they meet.
We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Two types of tiles were laid on the wall in a checkerboard pattern. Several tiles fell off the wall. The remaining tiles are shown in the picture. How many striped tiles fell off? Be sure to explain your answer.
Method 2. Look at the number of fallen cells by rows: in the second row from the top, 2 tiles fell out, of which 1 ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Mowgli asked five monkeys to bring him some nuts. The monkeys collected an equal number of nuts and carried them to Mowgli. On the way, they quarreled, and each monkey threw one nut at each of the others. As a result, they brought Mowgli half as many nuts as they had collected. How many nuts did Mowgli receive? Be s... | Answer: 20 nuts.
## Solution.
Each monkey threw 4 nuts, so the monkeys threw a total of $5 \cdot 4=20$ nuts together.
If half of the nuts remained, it means that Mowgli brought as many nuts as were thrown, which is 20 nuts.
## Grading Criteria.
- Correct solution - 7 points.
- Very brief solution (such as "5 - $4=... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. The distance from the home of Vintik and Shpuntik to school is 6 kilometers. Vintik and Shpuntik left for school at the same time, with Vintik spending half of the time riding a scooter at a speed of 10 km/h and then walking, while Shpuntik traveled half the distance by bicycle and then walked. They arrived at schoo... | Answer: 15 km/h.
## Solution:
First method.
Since Shtyubik rides twice as fast as he walks, he also covers twice the distance he walks (since he spends the same amount of time on both), which is 4 km. Since Shtyubik and Shpuntyk walk at the same speed, the last 2 km they walked together. Therefore, while Shtyubik wa... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In triangle $A B C$, the median $B M$ was drawn. It turned out that $A B=2 B M$ and $\angle M B A=40^{\circ}$. Find $\angle C B A$.
# | # Answer: $110^{\circ}$.
## Solution:
Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a p... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. Five athletes came to training with their balls, and when leaving, each took someone else's. In how many ways is this possible. | Answer: 44.
Solution. First, let's assume that no two athletes have exchanged balls. Imagine them sitting around a round table. Then, depending on their different relative positions, they would choose the ball of the next (for example, clockwise) athlete. However, this is only possible if 5 are seated at one table, wh... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2 The sequence of numbers $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots$ is formed according to the rule: $\mathrm{x}_{1}=1, \mathrm{x}_{\mathrm{n}+1}=1+\frac{\mathrm{x}_{\mathrm{n}}^{2}}{\mathrm{n}}$ for $\mathrm{n}=1,2,3, \ldots$ Find $\mathrm{x}_{2019}$. | Answer: 2019
Reasoning. $x_{2}=1+\frac{1^{2}}{1}=2, x_{3}=1+\frac{2^{2}}{2}=3$. Hypothesis: $x_{n}=n$. We proceed by induction. If $x_{n}=n$, then $x_{n+1}=1+\frac{n^{2}}{n}=1+n$. The hypothesis is confirmed. | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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