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Problem 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:
a) in 5 or fewer;
b) in 4 or fewer;
.
Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)
Comment. It is impossible to paint all cells ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,877 |
Problem 3. A bag of sunflower seeds was passed around a table. The first person took 1 seed, the second took 2, the third took 3, and so on: each subsequent person took one more seed than the previous one. It is known that in the second round, the total number of seeds taken was 100 more than in the first round. How ma... | Answer: 10 people.
Solution: Let there be $n$ people sitting at the table. Then on the second round, the first person took the $n+1$-th sunflower seed, the second person took the $n+2$-th - and generally, each person took $n$ more seeds than on the first round. Altogether, on the second round, they took $n \cdot n=n^{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,878 |
Problem 4. Dima saw strange clocks in a museum (see figure). They differ from ordinary clocks in that their dial has no numbers and it is generally unclear where the top of the clock is; moreover, the second, minute, and hour hands are of the same length. What time did the clocks show?
![](https://cdn.mathpix.com/crop... | Answer: Ten to five.
Solution: Let's think about which of the hands is the hour hand. If the hour hand were pointing exactly at an hour mark, the minute and second hands would be pointing exactly at the "12" mark - but there are no matching hands in the picture. Therefore, the hour hand must be hand B.
The remaining ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,879 |
Problem 5. Three square paths with a common center are 1 m apart from each other (see figure). Three ants start simultaneously from the lower left corners of the paths and run at the same speed: Mu and Ra counterclockwise, and Wei clockwise. When Mu reaches the lower right corner of the largest path, the other two, who... | Answer: 4 m, 6 m, 8 m.
Solution: The lengths of the sides of two adjacent paths differ by 2 m (Fig. 3). Therefore, at the moment when Mu reached the corner, Ra had run 2 m along the right side of the path and was at a distance of $2+1=3$ m from the "lower" side of the outer path. Since $\mathrm{Pa}$ is halfway between... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,880 |
Problem 6. Fox Alice and Cat Basil have grown 20 fake banknotes on a tree and are now filling in seven-digit numbers on them. Each banknote has 7 empty cells for digits. Basil calls out one digit at a time, either "1" or "2" (he doesn't know any others), and Alice writes the called digit in any free cell of any banknot... | Answer: 2.
Solution: Basil can always get two banknotes: he knows the place where the last digit should be written and names it so that it differs from the digit in the same place on some other banknote. Then the numbers on these two banknotes will be different, and the cat can take them.
We will show how Alice can e... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,881 |
Problem 2. Two identical rectangular triangles made of paper were placed one on top of the other as shown in the figure (with the vertex of the right angle of one landing on the side of the other). Prove that the shaded triangle
; therefore, its sides $A B$ and $B C$ are equal. But its sides $A B$ and $A C$ are also equal (as corresponding sides of equal paper triangles); therefore, triangle $A B C$ is equilateral.
Comments. 1. The so... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,884 |
Problem 2. a) Fill in each circle with a non-zero digit so that the sum of the digits in the two top circles is 7 times less than the sum of the other digits, and the sum of the digits in the two left circles is 5 times
 See the diagram.
Solution. b) If the sum of the digits in the two upper circles is 7 times less than the sum of the remaining digits, then it is 8 times less than the sum of all five digits. Reasoning similarly, we get that
 on graph paper and painted a picture on it. After that, he drew a frame one cell wide around the picture (see figure). It turned out that the area of the picture is equal to the area of the frame.
. Then, in the small frame, as in the large one, there will be four corner cells (they are shaded), and each sid... | 3\times10or4\times6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,887 |
Problem 4. In a singing competition, a Rooster, a Crow, and a Cuckoo participated. Each member of the jury voted for one of the three performers. The Woodpecker calculated that there were 59 judges in total, and that the Rooster and the Crow received a total of 15 votes, the Crow and the Cuckoo received 18 votes, and t... | Answer: 13 judges.
Solution. The number of votes for the Rooster and the Raven cannot be more than $15+13=28$. Similarly, the number of votes for the Raven and the Cuckoo cannot exceed $18+13=31$, and the number of votes for the Cuckoo and the Rooster cannot exceed $20+13=33$. Adding these three quantities of votes, w... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,890 |
Problem 3. An equilateral triangle with a side length of 8 was divided into smaller equilateral triangles with a side length of 1 (see figure). What is the minimum number of small triangles that need to be shaded so that all intersection points of the lines (including those on the edges) are vertices of at least one sh... | Answer: 15 small triangles. See the example in the figure.
Solution: The total number of intersection points of the lines is $1+2+3+\ldots+9=45$. Since a triangle has three vertices, at leas... | 15 | Combinatorics | proof | Yes | Yes | olympiads | false | 22,891 |
Problem 1. On the surface of a planet shaped like a donut, two snails crawled, leaving trails behind: one along the outer equator, and the other along a spiral line (see figure). Into how many parts did the snails' trails divide the surface of the planet? (It is sufficient to write the answer.)
 can be found as follows. The creatures weighing 10, 9, and 8 kg must be placed in different suitcases (otherwise, one suitcase would be too h... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,893 |
Problem 2. For breakfast, a group of 5 elephants and 7 hippos ate 11 round and 20 cubic watermelons, while a group of 8 elephants and 4 hippos ate 20 round and 8 cubic watermelons.
All elephants ate the same whole number of watermelons. And all hippos ate the same number of watermelons. However, one type of animal eat... | Answer. Elephants eat only round watermelons.
Solution. First, let's find out how many watermelons each animal eats for breakfast. According to the problem, 5 elephants and 7 hippos eat 31 watermelons, while 8 elephants and 4 hippos eat 28 watermelons. We see that if we replace three hippos with three elephants, we ne... | Elephantseatonlyroundwatermelons | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,894 |
Problem 4. There are three piles of 40 stones each. Petya and Vasya take turns, Petya starts. On a turn, one must combine two piles, then divide these stones into four piles. The player who cannot make a move loses. Which of the players (Petya or Vasya) can win, no matter how the opponent plays?
$[6$ points] (A.V. Sha... | Answer: Vasya.
Comment: In fact, this is a joke game: Vasya wins regardless of the players' actions.
Solution: In one move, two piles are replaced by four, i.e., the number of piles increases by 2. So, how many piles will there be when the game ends? Initially, the number of piles was odd, so, increasing by two, it w... | Vasya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,896 |
Problem 5. Maxim laid out a polygon on the table using 9 squares and 19 equilateral triangles (without overlapping them) from a set where the sides of all squares and triangles are 1 cm. Could the perimeter of this polygon be 15 cm? [9 points] (M.A. Volchkevich) | Answer. Yes, it could (see fig.).
Comments. 1. The difficulty here is that the figure is proposed to be constructed with a rather small perimeter: even if we were to construct a polygon only f... | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,897 |
Problem 6. A row of 100 coins lies on a table, some showing heads and the rest showing tails. In one operation, it is allowed to select seven coins lying at equal intervals (i.e., seven consecutive coins, or seven coins lying every other coin, etc.), and flip all seven coins. Prove that using such operations, it is pos... | Solution. It is clear that it is enough to learn how to flip each of the coins (while keeping the position of the others).
First, let's show how to flip a pair of coins, with exactly 6 coins lying between them. If we mentally combine this pair with the coins between them, we get a group of 8 coins in a row. We will fl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,898 |
1. In Fedia the gardener's garden, there is a wonder-tree with seven branches. On each branch, either 6 apples, 5 pears, or 3 oranges can grow. Fedia noticed that there are fruits of all types on the tree, and the most pears have grown, while the fewest apples have grown.
How many fruits have grown on the wonder-tree?... | # Problem 1.
Answer: 30.
Since the tree ended up with fruits of all kinds, there must be at least one branch with apples, meaning there are no fewer than 6 apples. Since apples are the fewest, oranges must have grown on at least three branches, meaning there are at least 9.
There are three branches left. Suppose tha... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,899 |
2. The magician thought of two natural numbers and told Sime their sum, and told Provo their product. Knowing that the product is 2280, Provo was able to guess the numbers only after Sime mentioned that the sum is odd and two-digit. So what numbers did the magician think of?
[4 points] | # Problem 2.
Answer: 40 and 57.
Let the numbers guessed by the magician be denoted as $a$ and $b$. Then, according to the problem, $a \cdot b = 2280$ and the number $a + b$ is a two-digit odd number. Since the product $a \cdot b$ is even, at least one of the numbers must be even. However, if both numbers were even, t... | 4057 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,900 |
4. Brothers Petya and Vasya decided to shoot a funny video and post it on the internet. First, they filmed each of them walking from home to school - Vasya walked for 8 minutes, and Petya walked for 6 minutes. Then they came home and sat down at the computer to edit the video: they started Vasya's video from the beginn... | # Problem 4.
Answer: 6 minutes.
Let $t$ minutes be the time from the start of the video until the brothers meet at the same point ($t$ does not have to be an integer). Then, from Petya's video, there were $6-t$ minutes left until the end of the viewing (i.e., until the start of the clip, since the video is playing in... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,902 |
6. On the side $AB$ of an equilateral triangle $ABC$, a point $K$ is chosen, and on the side $BC$, points $L$ and $M$ are chosen such that $KL = KM$, with point $L$ being closer to $B$ than $M$.
a) Find the angle $MKA$ if it is known that $\angle BKL = 10^{\circ}$.
b) Find $MC$ if $BL = 2$ and $KA = 3$.
Justify your... | # Problem 6.
Answer: a) $130^{\circ} ;$ b) 5 .
First solution. Mark a point $N$ on the segment $MC$ such that $NM = BL$. Since triangle $LKM$ is isosceles, the angles $KLM$ and $KML$ at its base are equal. Therefore, the adjacent angles $KLB$ and $KMN$ are also equal. This implies the equality of triangles $BKL$ and ... | 130 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,904 |
Task 1. The figure shows two code locks. The lock will open if different digits are entered in the circles so that the number inside each of the triangles matches either the sum or the product of the digits at its vertices. What combination of a) four b) five different digits will open the lock?
a)
 The figure shows two possible sets of digits that can unlock the lock. In each example, the top number can be swapped with the bottom number, and the right number with the left number.
 and \(BC\) of square \(ABCD\) with side length 10, points \(K\) and \(L\) are marked such that \(AK = CL = 3\). On segment \(KL\), a point \(P\) is chosen, and on the extension of segment \(AB\) beyond point \(B\), a point \(Q\) is chosen such that \(AP = PQ = QL\) (see figure).
a) Prove... | Solution. Drop a perpendicular $PH$ from point $P$ to side $AB$. Since triangle $KBL$ is isosceles and right-angled, $\angle PKH = 45^\circ$. Therefore, triangle $KPH$ is also isosceles and $KH = HP$.
, while at the same time pointing out a property characteristic of all such examples.
All roads can be divided into two types: some roads connect the top-left corner of a county with the bott... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,911 |
Problem 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:
a) in 5 or fewer;
b) in 4 or fewer;
.
Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)
Comment. It is impossible to paint all cells ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,912 |
Problem 4. Dima saw strange clocks in a museum (see figure). They differ from ordinary clocks in that their dial has no numbers and it is generally unclear where the top of the clock is; moreover, the second, minute, and hour hands are of the same length. What time did the clocks show?
![](https://cdn.mathpix.com/crop... | Answer: Ten to five.
Solution: Let's think about which of the hands is the hour hand. If the hour hand were pointing exactly at an hour mark, the minute and second hands would be pointing exactly at the "12" mark - but there are no matching hands in the picture. Therefore, the hour hand must be hand B.
The remaining ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,913 |
Problem 1. Tanya photographed four kittens eating sausages (Fig. 1). Soon she took another shot (Fig. 2). Each kitten eats its sausages continuously and at a constant speed, and does not touch others' sausages. Who will finish first and who will finish last? Explain your answer.
Fig. 1
, and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$),... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,916 |
Problem 4. Misha built a $3 \times 3 \times 3$ cube from smaller cubes. Then he glued some adjacent cubes together. The resulting solid structure consisted of 16 cubes, and Misha removed the rest. After dipping the structure in ink, he pressed it against a piece of paper with three different faces. The result was the w... | (M.A. Evdokimov, O.A. Zaslavsky, A.V. Shapovalov) Answer. See the figure.
Solution. Since the letter T can be printed, two corner cubes are removed. The remaining six corner cubes must remai... | M | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,917 |
Problem 5. In the forest, there live 40 animals - foxes, wolves, hares, and badgers. Every year they organize a masquerade ball: each one wears a mask of another animal type, and they do not wear the same mask for two consecutive years. Two years ago, at the ball, there were 12 "foxes" and 28 "wolves", last year there ... | Answer. The most numerous are the badgers.
Solution. Let's record the data from the problem in a table.
| | "Wolves" | "Foxes" | "Rabbits" | "Badgers" |
| :--- | :---: | :---: | :---: | :---: |
| Two years ago | 28 | 12 | | |
| Last year | | 10 | 15 | 15 |
| This year | | 25 | 15 | |
Let's look at the "rabbits... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,918 |
Problem 6. Vanya is coming up with a number consisting of non-repeating digits without zeros - a password for his phone. The password works as follows: if, without lifting his finger from the screen, he sequentially connects the points corresponding to the digits of the password with line segments, the phone will unloc... | Answer. For example, 12769. See the figure.
This password meets Vanya's requirements. Let's see how we can connect its digits without any intersections. The digit 7 must be connected to some digit, which can be either 2 or 6. Suppose, for example, we draw the segment $7-6$. Now 9 can only be connected to 6. Next, it i... | 12769 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,919 |
Problem 5. Can the digits $1,2, \ldots, 8$ be placed in the cells of a) the letter Ш; b) the strips (see figure), so that for any cutting of the figure into two parts, the sum of all the digits in one part is divisible by the sum of all the digits in the other? (Cuts can only be made along cell boundaries. Each cell mu... | Solution. Let the sum of the numbers in one of the parts be $x$, in the other $y$, and $y$ is divisible by $x$. Then $x+y$ is also divisible by $x$, and this is the sum of all the numbers, which is $1+2+3+4+5+6+7+8=$ $=36$. Therefore, the smaller of the sums of the parts is a divisor of the number 36. The converse is a... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,922 |
Problem 1. In a multicolored family, there were an equal number of white, blue, and striped octopus children. When some blue octopus children became striped, the father decided to count the children. There were 10 blue and white children in total, while white and striped children together amounted to 18. How many child... | Answer: 21.
First solution. Note that the white octopuses were one third of the total number, and they did not change color. If we add 10 and 18, we get the total number of all children, plus the number of white ones, which is $4 / 3$ of the total number of all children. Thus, $4 / 3$ of the number of children in the ... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,923 |
Task 2. Using each of the digits from 0 to 9 exactly once, write down 5 non-zero numbers such that each divides the previous one. $\quad[6$ points] (A.V. Shapovalov) | Answer. For example, $1,2,4,8,975360$.
Comment. It is easier to check divisibility when most numbers are written with 1-2 digits, and for this, most of the quotients should be quite small ( $2,3, \ldots$ ). Start with the smallest sequence: $1,2,4,8$. Whether the remaining number is divisible by 8 depends only on its ... | 1,2,4,8,975360 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,924 |
Problem 3. All cells of the top row of a $14 \times 14$ square are filled with water, and one cell contains a bag of sand (see figure). In one move, Vasya can place bags of sand in any 3 cells not occupied by water, after which the water fills each of the cells that border the water (by side), if there is no bag of san... | Solution. We will prove that no matter how Vasya acts, the water will fill at least 37 cells.
No matter how Vasya acts on the first move, after it, there will be no more than 3 bags in the second row, which means that the water will fill no fewer than 11 cells in the second row. No matter how Vasya acts on the second ... | 37 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,925 |
Problem 4. Two squares and an isosceles triangle are arranged as shown in the figure (vertex $K$ of the larger square lies on the side of the triangle). Prove that points $A, B$, and $C$ lie on the same line.
. Therefore, $AC$ forms the same angle with the base as the diagonal of the square $KD$, which is $45^{\circ}$. But $AB$ also forms an angle of $45^{\circ}$ with the b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,926 |
Problem 5. Figures made of four cells are called tetraminoes. There are five types of them (see figure). Is there such a figure that, for any choice of tetramino type, this figure can be formed using only tetraminoes of the chosen type? (Tetraminoes can be flipped.)
[10 points] (Y.S. Markelov, 8th grade student)
.
Comments. 1. It is difficult to keep track of whether the figure is being cut into 5 different types of pieces. However, one can notice that from two square... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,927 |
Problem 6. Robin Hood captured seven rich men and demanded a ransom. A servant of each rich man brought a purse of gold, and they all lined up in a queue in front of the tent to pay the ransom. Each servant entering the tent places the purse they brought on the table in the center of the tent, and if no one has previou... | Answer. a) Seventh; b) sixth or seventh.
Solution. If the servant brought a new purse, and since he was last in the tent, no one has been released, then he is definitely in for a stroke of luck (his purse is the heaviest and this time his master will be released). That is, if there were $N$ rich men in captivity and o... | Seventh;sixthorseventh | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,928 |
Problem 4. Two squares are arranged as shown in the figure, the marked segments are equal. Prove that triangle $B D G$ is isosceles.
$[6$ points]
 | Solution 1. Consider triangles $D E G$ and $D E B$. They share a common side $D E$, and have equal sides $E G$ and $E B$ (as two sides of the square). It remains to prove that angles $D E G$ and $D E B$ are equal, - then the specified triangles will be equal (by two sides and the angle between them), and therefore, the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,930 |
Problem 5. The figure "violinist" attacks the cell to the left on the side (with the elbow) and the cell to the upper right diagonally (with the bow), if he is right-handed, and, conversely, the right cell on the side and the upper left cell diagonally, if he is left-handed (all violinists are facing us). Place as many... | Solution. Placing 32 violinists is not difficult: for example, you can fill four columns every other one (it doesn't matter whether they are right-handed or left-handed). However, it is possible to place more. An example with 34 violinists is shown in the figure.
| $\mathrm{p}$ | $\Lambda$ | | | $\mathrm{p}$ | $\Lam... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,932 |
Problem 2. Along the path between the houses of Nезнайка (Nезнayka) and Синеглазка (Sineglazka), there were 15 peonies and 15 tulips growing in a row, mixed together.
Setting out from home to visit Nезнайка, Синеглазка watered all the flowers in a row. After the 10th tulip, the water ran out, and 10 flowers remained u... | Answer: 19 flowers.
Solution: 10 flowers were left unwatered, which means $30-10=20$ flowers were watered. Consider the last flower watered by Blue-Eyes, which is a tulip. Since there are 15 tulips in total, there are $15-10=5$ tulips after this tulip.
Therefore, Nезнайка will pick these 5 tulips and finish picking f... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,934 |
Problem 4. A rectangular sheet of paper was folded, aligning a vertex with the midpoint of the opposite shorter side (Fig. 12). It turned out that triangles I and II are equal. Find the longer side of the rectangle if the shorter side is 8.
[6 points] (A. V. Khachatryan) | Answer: 12.
Fig. 12
Solution. Let's mark the equal segments (Fig. 13 - here we used the fact that in congruent triangles, sides opposite equal angles are equal). We see that the length of t... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,936 |
Task 5. In the handbook "Magic for Beginners" it is written:
Replace identical letters in the word EARTHQUAKE with identical digits, and different letters with different digits. If the resulting number turns out to be prime, a real earthquake will occur.
Is it possible to cause an earthquake this way? (A natural numb... | Answer. No.
Solution. Let's count the letters in the word "EARTHQUAKE". The letter $\mathrm{E}$ appears 4 times, and the other 9 letters appear once each. This means that in the number, all 10 digits will be present once each, and one digit (corresponding to the letter E) - 3 more times. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,937 |
Task 2. Pete liked the puzzle, he decided to glue it together and hang it on the wall. In one minute, he glued together two pieces (initial or previously glued). As a result, the entire puzzle was joined into one complete picture in 2 hours. How long would it have taken to assemble the picture if Pete had glued togethe... | Answer: In one hour.
Solution 1. Each gluing reduces the number of pieces on the table by 1. Since after 120 gluings one piece (the complete puzzle) was obtained, there were 121 pieces at the beginning. Now, if three pieces are glued together per minute (i.e., the number of pieces is reduced by 2), one piece will rema... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,940 |
Problem 3. The inhabitants of the Island of Misfortune, like us, divide the day into several hours, an hour into several minutes, and a minute into several seconds. But they have 77 minutes in a day and 91 seconds in a minute. How many seconds are there in a day on the Island of Misfortune?
[5 points] (I. V. Raskina) | Answer: 1001.
Solution: If you divide 77 by the number of minutes in an hour, you get the number of hours in a day. If you divide 91 by the number of minutes in an hour, you get the number of seconds in a minute. Therefore, both 77 and 91 are divisible by the number of minutes in an hour. Since there are obviously mor... | 1001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,941 |
Problem 4. The cake is packed in a box with a square base. The height of the box is half the side of this square. A ribbon of length 156 cm can be used to tie the box and make a bow on top (as shown in the left image). To tie it with the same bow on the side (as shown in the right image), a ribbon of length 178 cm is n... | Answer: 22 cm $\times$ 22 cm $\times$ 11 cm.
Solution: In the first method of tying, the ribbon encircles the box twice along the length, twice along the width, and four times along the height, meaning its length is equal to six sides of the base plus the bow. In the second method of tying, the ribbon encircles the bo... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,942 |
Task 5. Replace in the equation
$$
\text { PIE = SLICE + SLICE + SLICE + ... + SLICE }
$$
identical letters with identical digits, and different letters with different digits, so that the equation is true, and the number of "slices of pie" is the largest possible. | Answer. The maximum number of "pieces" is seven, for example: ПИРОГ $=95207$, КУСОК $=13601$.
Solution. An example for seven "pieces" is given above. We will show that there cannot be more than seven "pieces". For this, it is convenient to rewrite the condition as a multiplication example: ПИРОГ $=$ КУСОК $\cdot n$, w... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,943 |
Problem 6. It is known that the Jackal always lies, the Lion tells the truth, the Parrot simply repeats the last heard answer (and if asked first, will answer randomly), and the Giraffe gives an honest answer but to the previous question asked to him (and to the first question, answers randomly). The wise Hedgehog, in ... | Answer: Parrot, Lion, Giraffe, Jackal.
Solution. To the first question "Are you a Jackal?" the Lion and the Jackal would definitely say "no." Therefore, the Hedgehog could only identify the Giraffe and not the Parrot in one case: if the Giraffe answered "Yes" and the Parrot answered "No." The same can be said about th... | Parrot,Lion,Giraffe,Jackal | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,944 |
3. Solve the inequality $\log _{\frac{3 x-6}{3 x+5}}(3 x-9)^{10} \geq-10 \log _{\frac{3 x+5}{3 x-6}}(3 x+6)$.
Answer. $x \in\left(-2 ;-\frac{5}{3}\right) \cup(2 ; 3) \cup(3 ;+\infty)$. | The domain of definition is given by the inequalities $\frac{3 x-6}{3 x+5}>0, \quad \frac{3 x-6}{3 x+5} \neq 1, \quad 3 x+6>0, \quad 3 x-9 \neq 0,$ from which $\mathrm{x} \in\left(-2 ;-\frac{5}{3}\right) \cup(2 ; 3) \cup(3 ;+\infty)$. On the domain of definition, the inequality is equivalent to the following:
$$
\begi... | x\in(-2;-\frac{5}{3})\cup(2;3)\cup(3;+\infty) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,945 |
4. One of the lateral sides of the trapezoid is perpendicular to the bases and equals $2 R$. A circle is constructed on this side as a diameter, which divides the other lateral side into three segments. The ratio of the lengths of these segments is 7:21:27 (counting from the upper base). Find the area of the trapezoid.... | Let the given trapezoid be $ABCD$ ($AD=2R$, $AB$ - the upper base, points $P$ and $Q$ - the points of intersection of the circle with the lateral side $BC$, $BP: PQ: QC=7: 21: 27$) and let $BP=7x$. Then $PQ=21x$, $QC=27x$, and by the tangent-secant theorem, we find that $BA=\sqrt{BP \cdot BQ}=14x$, $CD=\sqrt{CQ \cdot C... | \frac{100R^{2}}{11\sqrt{21}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,946 |
5. Find the maximum and minimum values of the function $g(x)=\sin ^{8} x+8 \cos ^{8} x$.
Answer. $g_{\min }=\frac{8}{27}, g_{\max }=8$. | Solution: We find the derivative: $g^{\prime}(x)=8 \sin ^{7} x \cos x-64 \cos ^{7} x \sin x=8 \sin x \cos x\left(\sin ^{6} x-8 \cos ^{6} x\right)$. The critical points of this function are the roots of the equations $\sin x=0, \cos x=0$ and $\sin ^{6} x=8 \cos ^{6} x$, from which $x=k \pi, x=\frac{\pi}{2}+k \pi, x= \pm... | g_{\}=\frac{8}{27},g_{\max}=8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,947 |
6. First-year students admitted to the university were distributed into study groups so that each group had the same number of students. Due to a reduction in the number of specialties, the number of groups decreased by 9, and all first-year students were redistributed into groups; as a result, the groups were again eq... | Let the new number of groups be $n$, then initially there were ( $n+9$ ) groups. In each group, there was an equal number of students, so $2376: n$ and $2376:(n+9)$. We factorize 2376 into prime factors ( $2376=2^{3} \cdot 3^{3} \cdot 11$ ) and list all divisors: $1,2,4,8,3,6,12,24,9,18,36,72,27,54$, $108,216,11,22,44,... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,948 |
7. Three parallel lines touch a sphere with center at point O at points $K$, $L$, and $M$. It is known that the radius of the sphere is 5, the area of triangle $O K L$ is 12, and the area of triangle $K L M$ is greater than 30. Find the angle $K M L$.
Answer. $\angle K M L=\arccos \frac{3}{5}$. | Solution If a line is tangent to a sphere, then the plane passing through the center of the sphere perpendicular to this line passes through the point of tangency. Consider the plane $\alpha$, passing through point 0 perpendicular to the three given lines. The section of the sphere by the plane $\alpha$ - a circle of r... | \arccos\frac{3}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,949 |
8. It is known that the equation $2 x^{3}-7 x^{2}+7 x+p=0$ has three distinct roots, and these roots form a geometric progression. Find p and solve this equation. | Answer. $p=-2$; roots of the equation: $x=1, x=\frac{1}{2}, x=2$.
Solution Let $x, k x, k^{2} x$ be the roots of the given equation, with $x \neq 0, k \neq 0, k \neq \pm 1$ (since the roots are distinct). From Vieta's theorem, we have
$$
\left\{\begin{array}{l}
x+k x+k^{2} x=\frac{7}{2} \\
k x^{2}+k^{2} x^{2}+k^{3} x... | \frac{3\pi}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,950 |
4. One of the lateral sides of the trapezoid is perpendicular to the bases and equals $2 R$. A circle is constructed on this side as a diameter, which divides the other lateral side into three segments. The ratio of the lengths of these segments is 12:15:5 (counting from the lower base). Find the area of the trapezoid.... | Let the given trapezoid be $ABCD$ ($AD=2R$, $AB$ - the upper base, points $P$ and $Q$ - the points of intersection of the circle with the lateral side $BC$, $BP: PQ: QC=5: 15: 12$) and let $BP=5x$. Then $PQ=15x$, $QC=12x$, and by the tangent-secant theorem, we find that $BA=\sqrt{BP \cdot BQ}=10x$, $CD=\sqrt{CQ \cdot C... | \frac{7R^{2}}{\sqrt{15}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,952 |
7. Three parallel lines touch a sphere with center at point O at points K, L, and M. It is known that the radius of the sphere is $5 \sqrt{2}$, the area of triangle OKL is 7, and the area of triangle KLM is greater than 50. Find the angle KML.
Answer. $\angle \mathrm{KML}=\arccos \frac{1}{5 \sqrt{2}}$. | Solution If a line is tangent to a sphere, then the plane passing through the center of the sphere perpendicular to this line passes through the point of tangency. Consider the plane $\alpha$, passing through point 0 perpendicular to the three given lines. The section of the sphere by the plane $\alpha$ is a circle of ... | \arccos\frac{1}{5\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,954 |
8. It is known that the equation $x^{3}+7 x^{2}+14 x-p=0$ has three distinct roots, and these roots form a geometric progression. Find p and solve this equation. | Answer. $p=-8$; roots of the equation: $x=-1, x=-2, x=-4$.
Solution Let $x, k x, k^{2} x$ be the roots of the given equation, with $x \neq 0, k \neq 0, k \neq \pm 1$ (since the roots are distinct). From Vieta's theorem, it follows that
$$
\left\{\begin{array}{l}
x+k x+k^{2} x=-7 \\
k x^{2}+k^{2} x^{2}+k^{3} x^{2}=14 ... | -8;rootsoftheequation:-1,-2,-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,955 |
1. Find the sum of the first thirty-three terms of an arithmetic progression, given that the sum of the seventh, eleventh, twentieth, and thirtieth terms of this progression is 28. | Answer. $S_{33}=231$.
Solution Let $a_{k}$ be the $k$-th term of the arithmetic progression, and $d$ be its common difference. Then, according to the condition, $a_{7}+a_{11}+a_{20}+a_{30}=28$, from which we have $a_{7}+\left(a_{7}+4 d\right)+\left(a_{7}+13 d\right)+\left(a_{7}+23 d\right)=28 \Leftrightarrow a_{7}+10 ... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,956 |
2. The radius of the circle inscribed in the triangle is 2. One of the sides of the triangle is divided by the point of tangency into segments equal to 1 and 8. Find the other sides of the triangle.
Answer. 10 and 17. | Let K, L, M be the points of tangency of the inscribed circle with the sides AB, BC, AC of triangle $ABC$ respectively; thus, $AK=8, BK=1$. Denote $CL=x$. By the equality of the segments of tangents drawn from a point to a circle, we get $\mathrm{BL}=1, \mathrm{AM}=8$, $\mathrm{CM}=\mathrm{x}$. We express the area of t... | 1017 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,957 |
3. Solve the inequality $\sqrt{2 x+\frac{7}{x^{2}}}+\sqrt{2 x-\frac{7}{x^{2}}}<\frac{6}{x}$.
Answer. $x \in\left[\sqrt[3]{\frac{7}{2}} ; \sqrt[3]{\frac{373}{72}}\right)$. | The domain of definition (ODZ) is determined by the inequalities $2 x+\frac{7}{x^{2}} \geq 0, 2 x-\frac{7}{x^{2}} \geq 0$, solving which, we get $x \geq \sqrt[3]{\frac{7}{2}}$.
On the ODZ, both sides of the given inequality are positive, so we square them:
$$
\begin{aligned}
& 2 x+\frac{7}{x^{2}}+2 \sqrt{\left(2 x+\f... | x\in[\sqrt[3]{\frac{7}{2}};\sqrt[3]{\frac{373}{72}}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,958 |
4. It is known that $3 \operatorname{tg}^{2} \gamma+3 \operatorname{ctg}^{2} \gamma+\frac{2}{\sin ^{2} \gamma}+\frac{2}{\cos ^{2} \gamma}=19$. Find the value of the expression $\sin ^{4} \gamma-\sin ^{2} \gamma$.
Answer. $-\frac{1}{5}$. | Transform the given equality in the condition. On the domain of definition, it is equivalent to the following:
$$
\begin{aligned}
& \frac{3 \sin ^{2} \gamma}{\cos ^{2} \gamma}+\frac{3 \cos ^{2} \gamma}{\sin ^{2} \gamma}+\frac{2}{\sin ^{2} \gamma}+\frac{2}{\cos ^{2} \gamma}=19 \Leftrightarrow 3 \sin ^{4} \gamma+3 \cos ... | -\frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,959 |
5. A boat departed downstream from pier A. At the same time, a motorboat departed upstream from pier B towards the boat. After some time, they met. At the moment of their meeting, a second boat departed downstream from A and after some time met the motorboat. The distance between the points of the first and second meet... | Let $S$ be the distance between the piers, $u$ be the speed of the current, $v$ be the own speed of the boat, and $5v$ be the own speed of the motorboat. Then the speeds of the boat and the motorboat on the river are $V_{n} = v + u$ and $V_{k} = 5v - u$, respectively. The first meeting occurred after time $t_{1} = \fra... | \frac{56}{81} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,960 |
6. In trapezoid $\mathrm{PQRS}$, it is known that $\angle \mathrm{PQR}=90^{\circ}, \angle Q R S>90^{\circ}$, diagonal $S Q$ is 24 and is the bisector of angle $S$, and the distance from vertex $R$ to line $Q S$ is 5. Find the area of trapezoid PQRS. | Answer: $\frac{27420}{169}$.
Solution $\angle \mathrm{RQS}=\angle \mathrm{PSQ}$ as alternate interior angles, so triangle $\mathrm{RQS}$ is isosceles. Its height $\mathrm{RH}$ is also the median and equals 5. Then $S R=\sqrt{R H^{2}+S H^{2}}=13$. Triangles SRH and SQP are similar by two angles, the similarity coeffici... | \frac{27420}{169} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,961 |
7. Represent the number 2015 as the sum of some number of natural numbers so that their product is the largest.
Answer. $2015=3+3+\ldots+3+2$ (671 threes and one two). | Solution If the sought representation contains at least one addend greater than 4, then the product is not maximal. Indeed, if $n \geq 5$ is replaced by the sum $2+(n-2)$, then the product $2 \cdot(\mathrm{n}-2)$ is greater than $\mathrm{n}$. Also, note that if the representation contains an addend 4, it can be replace... | 2015=671\cdot3+2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,962 |
8. Solve the system of equations $\left\{\begin{array}{l}x^{2}-y^{2}+z=\frac{27}{x y} \\ y^{2}-z^{2}+x=\frac{27}{y z} \\ z^{2}-x^{2}+y=\frac{27}{x z}\end{array}\right.$.
Answer. $(3 ; 3 ; 3),(-3 ;-3 ; 3),(-3 ; 3 ;-3),(3 ;-3 ;-3)$. | By adding all three equations of the system, we get $x+y+z=\frac{27}{x y}+\frac{27}{y z}+\frac{27}{x z} \Leftrightarrow x+y+z=\frac{27(x+y+z)}{x y z}$. There are two possible cases: $x+y+z=0 \quad$ or $x y z=27$.
a) $x+y+z=0$, then $z=-x-y$. Substituting into the first two equations of the original system:
$$
\left\{... | (3;3;3),(-3;-3;3),(-3;3;-3),(3;-3;-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,963 |
2. The radius of the circle inscribed in the triangle is 2. One of the sides of the triangle is divided by the point of tangency into segments equal to 6 and 14. Find the other sides of the triangle.
Answer. 7 and 15. | Let K, L, M be the points of tangency of the inscribed circle with the sides AB, BC, AC of triangle $ABC$ respectively; thus, $AK=6, BK=14$. By the property of the equality of tangent segments drawn from a point to a circle, we get $\mathrm{BL}=14, \mathrm{AM}=6$, CM $=x$. We express the area of the triangle in two way... | 715 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,964 |
3. Solve the inequality $\sqrt{3 x-\frac{5}{x^{2}}}+\sqrt{3 x+\frac{5}{x^{2}}}<\frac{8}{x}$.
 | The domain of definition (ODZ) is determined by the inequalities $3 x+\frac{5}{x^{2}} \geq 0, 3 x-\frac{5}{x^{2}} \geq 0$, solving which, we get $x \geq 3 \sqrt{\frac{5}{3}}$.
On the ODZ, both sides of the given inequality are positive, so we square them:
$$
\begin{aligned}
3 x+ & \frac{5}{x^{2}}+2 \sqrt{\left(3 x+\f... | x\in[\sqrt[3]{\frac{5}{3}};\sqrt[3]{\frac{1049}{192}}) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,965 |
4. It is known that $4 \operatorname{tg}^{2} Y+4 \operatorname{ctg}^{2} Y-\frac{1}{\sin ^{2} \gamma}-\frac{1}{\cos ^{2} \gamma}=17$. Find the value of the expression $\cos ^{2} Y-\cos ^{4} \gamma$.
Answer. $\frac{3}{25}$. | Transform the given equality in the condition. On the domain of definition, it is equivalent to the following:
$$
\begin{aligned}
\frac{4 \sin ^{2} \gamma}{\cos ^{2} \gamma}+\frac{4 \cos ^{2} \gamma}{\sin ^{2} \gamma}-\frac{1}{\sin ^{2} \gamma}-\frac{1}{\cos ^{2} \gamma}= & 17 \Leftrightarrow 4 \sin ^{4} \gamma+4 \cos... | \frac{3}{25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,966 |
5. A motorboat departed downstream from pier A. At the same time, a boat departed upstream from pier B towards the motorboat. After some time, they met. At the moment of their meeting, a second boat departed from B and after some time met the motorboat. The distance between the points of the first and second meetings i... | Let $S$ be the distance between the piers, $u$ be the speed of the current, $2v$ be the own speed of the motor boat, and $3v$ be the own speed of the motorboat. Then the speeds of the boat and the motorboat on the river are $V_{n}=2v+u$ and $V_{\kappa}=3v-u$, respectively. The first meeting occurred after time $t_{1}=\... | \frac{161}{225} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,967 |
6. In trapezoid $P Q R S$, it is known that $\angle P Q R=90^{\circ}, \angle Q R S<90^{\circ}$, diagonal $S Q$ is 24 and is the bisector of angle $S$, and the distance from vertex $R$ to line $Q S$ is 16. Find the area of trapezoid PQRS.
Answer. $\frac{8256}{25}$. | Given $\angle R Q S=\angle P S Q$ as alternate interior angles, therefore triangle $R Q S$ is isosceles. Its height $R H$ is also a median and equals 16. Then $S R=\sqrt{R H^{2}+S H^{2}}=20$. Triangles $S R H$ and $S Q P$ are similar by two angles, the similarity coefficient is $\frac{S R}{S Q}=\frac{5}{6}$. Therefore,... | \frac{8256}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,968 |
7. Represent the number 2017 as the sum of some number of natural numbers so that their product is the largest.
Answer. $2017=3+3+\ldots+3+2+2$ (671 threes and two twos). | Solution If the sought representation contains at least one addend greater than 4, then the product is not maximal. Indeed, if $n \geq 5$ is replaced by the sum $2+(n-2)$, then the product $2 \cdot(\mathrm{n}-2)$ turns out to be greater than $\mathrm{n}$. Also, note that if the representation contains an addend 4, it c... | 2017=3+3+\ldots+3+2+2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,969 |
8. Solve the system of equations $\left\{\begin{array}{l}x^{2}-y^{2}+z=\frac{64}{x y} \\ y^{2}-z^{2}+x=\frac{64}{y z} \\ z^{2}-x^{2}+y=\frac{64}{x z}\end{array}\right.$. | Answer: $(4 ; 4 ; 4),(-4 ;-4 ; 4),(-4 ; 4 ;-4),(4 ;-4 ;-4)$.
Solution: Adding all three equations of the system, we get $x+y+z=\frac{64}{x y}+\frac{64}{y z}+\frac{64}{x z} \Leftrightarrow x+y+z=\frac{64(x+y+z)}{x y z}$. There are two possible cases: $x+y+z=0$ or $x y z=64$.
a) $x+y+z=0$, then $z=-x-y$. Substituting i... | (4;4;4),(-4;-4;4),(-4;4;-4),(4;-4;-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,970 |
3. It is known that for positive numbers $a, b$, and $c$, each of the three equations $a x^{2} +$ param $1 b x + c = 0$, $b x^{2} +$ param $1 c x + a = 0$, $c x^{2} +$ param $1 a x + b = 0$ has at least one real root.
What is the smallest possible value for the product of the roots of the second equation if the produc... | # Solution
Let $10 / 2=p, 6=q$. By Vieta's formulas, $c / a=q$. The discriminant of the third equation is non-negative, so $a^{2} p^{2} \geq b c$. Replacing $c$ in this inequality, we get: $a^{2} p^{2} \geq a b q$. Therefore, $a / b \geq q / p^{2}$. Hence, $\min \{a / b\}=q / p^{2}=0.24$, which is the minimum value of... | 0.24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,971 |
4. An infinite geometric progression consists of natural numbers. It turned out that the product of the first four terms equals param1. Find the number of such progressions.
The infinite geometric progression consists of positive integers. It turned out that the product of the first four terms equals param1. Find the ... | Solution
If $b_{1}$ is the first term of the progression and $q$ is its common ratio, then the product of the first four terms of the progression is $b_{1}^{4} q^{6}$. Therefore, $b_{1}^{2} q^{3}=2^{100} \cdot 3^{150}$. Hence, $b_{1}=2^{a} 3^{b}, q=2^{c} 3^{d}$, and we obtain the system: $2 a+3 c=100, 2 b+3 d=150$. Th... | 442 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,972 |
5. Given a function $f: \sqcup \rightarrow \sqcup$ such that $f(1)=1$, and for any $x \in \sqcup, y \in \sqcup$ the equality $f(x)+f(y)+x y+1=f(x+y)$ holds. Find all integers $n$ for which the equality $f(n)=$ param 1 holds. In the answer, write down the sum of cubes of all such values of $n$.
Function $f: \sqcup \rig... | # Solution
Substituting $y=1$ into the given functional equation, we get: $f(x)+f(1)+x+1=f(x+1)$, which means $f(k)-f(k-1)=k+1$. Therefore, $f(n)-f(1)=(f(n)-f(n-1))+(f(n-1)-f(n-2))+\ldots$ $+f(2)-f(1)=n+1+n+\ldots+2$. Hence, $f(n)=\frac{n^{2}+3 n-2}{2}$. Reasoning similarly,
we get $f(n)=\frac{n^{2}+3 n-2}{2}$ for $n ... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,973 |
7. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. What is the smallest possible area of the pentagon $A B C D E$, given that the quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and... | # Solution
Let the areas of parallelograms ABPM and DCPN, triangle BCP, and quadrilateral MPNE be $S_{1}, S_{2}, S_{3}$, and $S_{4}$, respectively. According to the problem, $A M \sqcup B \mathcal{B}$ and $P N \sqcup C \mathcal{D}$. Therefore, $\mathrm{AE} \sqcup \mathcal{B} N \cup C D$. Similarly, $D E \sqcup C M \sq... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,974 |
8. Let param1. What is the largest possible value of param2?
It is given that param1. Find the largest possible value of param2.
| param 1 | param 2 | Answer |
| :---: | :---: | :---: |
| $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=3$ | $6 \sin x$ | 4 |
| $\frac{25 \sin ^{2} x-37+40 \cos x}{... | # Solution
Notice that $\frac{9 \cos ^{2} x-7+12 \sin x}{16-9 \sin ^{2} x+6 \sqrt{5} \cos x}=\frac{6-(3 \sin x-2)^{2}}{2+(3 \cos x+\sqrt{5})^{2}}$. The obtained expression can equal 3 only if $3 \sin x-2=0$ and $3 \cos x+\sqrt{5}=0$. Therefore, the expression $6 \sin x$ can only take the value 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,975 |
The clock hand points to 12. Jack writes a sequence consisting of param 1 symbols, each symbol being plus or minus. After that he gives this sequence to a robot. The robot reads it from right to left. If he sees a plus he turns the clock hand $120^{\circ}$ clockwise and if he sees a minus he turns it $120^{\circ}$ coun... | # Solution
Let $a_{n}$ be the number of sequences of length $n$ that result in the arrow pointing at 12 o'clock, and $b_{n}$ be the number of sequences of length $n$ that result in the arrow pointing at 4 or 8 o'clock. It is not difficult to understand that $a_{n+1}=2 b_{n}, b_{n+1}=a_{n}+b_{n}$. From this, we get tha... | 682 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,976 |
11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits... | # Solution
Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, whi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,977 |
14. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and param2, respectivel... | # Solution
Let the areas of parallelograms ABP M and DCPN, triangle BCP, and quadrilateral
$A E \sqcup B N N C D$. Similarly, $D E \sqcup C M \sqcup\{B$. Then M PNE is also a parallelogram,... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,978 |
17. The difference of the squares of two different real numbers is param 1 times greater than the difference of these numbers, and the difference of the cubes of these numbers is param 2 times greater than the difference of these numbers. By how many times is the difference of the fourth powers of these numbers greater... | # Solution
From the equalities $a^{2}-b^{2}=k(a-b)$ and $a^{3}-b^{3}=m(a-b)$ (in the problem $k=37, m=1069$), it follows that $a+b=k$ and $a^{2}+a b+b^{2}=m$. Squaring the first equality and subtracting the second from it, we get: $a b=k^{2}-m$. Then the ratio $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{2}\right)=a^{2}+b... | 769 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,979 |
1. Simplify the fraction $\frac{\sqrt{-x}-\sqrt{-3 y}}{x+3 y+2 \sqrt{3 x y}}$.
Answer. $\frac{1}{\sqrt{-3 y}-\sqrt{-x}}$. | Let $-x=u,-y=v$, where $u \geq 0, v \geq 0$. Then the given fraction is equal to $\frac{\sqrt{u}-\sqrt{3 v}}{-u-3 v+2 \sqrt{3 u v}}=\frac{\sqrt{u}-\sqrt{3 v}}{-(\sqrt{u}-\sqrt{3 v})^{2}}=\frac{1}{\sqrt{3 v}-\sqrt{u}}=\frac{1}{\sqrt{-3 y}-\sqrt{-x}}$. | \frac{1}{\sqrt{-3y}-\sqrt{-x}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,980 |
2. The segment connecting the centers of two intersecting circles is divided by their common chord into segments equal to 5 and 2. Find the common chord, given that the radii of the circles are in the ratio $4: 3$.
Answer. $2 \sqrt{23}$. | Let 0 be the center of the circle with radius $3 R$, $Q$ be the center of the circle with radius $4 R$; $AB$ be the common chord of the circles; $A B \cap O Q=N$.
Since $A N \perp 0 Q$, we have $A N^{2}=9 R^{2}-4=16 R^{2}-25$, from which we find that $R^{2}=3$. Therefore, $A N^{2}=23$, $A B=2 \cdot A N=2 \sqrt{23}$. | 2\sqrt{23} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,981 |
5. Out of two hundred ninth-grade students, 80% received excellent grades on the first exam, 70% on the second exam, and 59% on the third exam. What is the smallest number of participants who could have received excellent grades on all three exams?
Answer: 18. | Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the problem,
$$
\left\{\begin... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,982 |
6. Piglet ran down a moving escalator and counted 66 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 198 steps. How many steps would he have counted if he had gone down a stationary escalator?
Answer. 99. | Let \( u \) be the speed of Piglet, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Piglet spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v} \cdot u \) steps.... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,983 |
7. The diagonals of a trapezoid are perpendicular to each other, and one of them is 13. Find the area of the trapezoid if its height is 12.
Answer. $\frac{1014}{5}$. | Let $ABCD$ be the given trapezoid, $AB \| CD, BD=13$, its height $BH$ is 12. Mark a point $M$ on the line $CD$ such that $BM \| AC$. Then $ABMC$ is a parallelogram and $CM=AB$. Triangles $ABD$ and $BCM$ are equal in area (since they have equal heights and bases). Therefore, the area of trapezoid $ABCD$ is equal to the ... | \frac{1014}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,984 |
8. Solve the equation $\frac{x^{4}}{2 x+1}+x^{2}=6(2 x+1)$.
Answer. $x=-3 \pm \sqrt{6}, x=2 \pm \sqrt{6}$. | By dividing both sides of the equation by $(2 x+1)$, we get $\left(\frac{x^{2}}{2 x+1}\right)^{2}+\frac{x^{2}}{2 x+1}-6=0$, from which $\frac{x^{2}}{2 x+1}=2$ or $\frac{x^{2}}{2 x+1}=-3$. In the first case, we get $x^{2}-4 x-2=0, x=2 \pm \sqrt{6}$, and in the second case $x^{2}+6 x+3=0, x=-3 \pm \sqrt{6}$.
1 Simplify ... | -3\\sqrt{6},2\\sqrt{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,985 |
2. The segment connecting the centers of two intersecting circles is divided by their common chord into segments equal to 4 and 1. Find the common chord, given that the radii of the circles are in the ratio $3: 2$.
Answer. $2 \sqrt{11}$. | Let 0 be the center of the circle with radius $2R$, $Q$ be the center of the circle with radius $3R$; $AB$ be the common chord of the circles; $A B \cap 0 Q=N$.
Since $A N \perp O Q$, we have $A N^{2}=4 R^{2}-1=9 R^{2}-16$, from which we find that $R^{2}=3$. Therefore, $A N^{2}=11$, $A B=2 \cdot A N=2 \sqrt{11}$. | 2\sqrt{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,986 |
3. The remainder of dividing a certain natural number $n$ by 22 is 7, and the remainder of dividing $n$ by 33 is 18. Find the remainder of dividing $n$ by 66. | Answer: 51.
Solution: According to the condition $n=22l+7, \quad l \in Z$ and $n=33m+18, m \in Z$. By equating these two expressions, we get $22l+7=33m+18, 2l=3m+1$. Since the left side of the equation is even, the right side must also be divisible by 2, so $m$ is an odd number, i.e., $m=2q+1, q \in Z$. Then $n=33(2q+... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,987 |
4. Solve the inequality $\frac{|x+3|+|1-x|}{x+2016}<1$.
Answer. $x \in(-\infty ;-2016) \cup(-1009 ; 1007)$. | Solution If $x+20160$, then we multiply both sides by $x+2016$ and get $|x+3|+|x-1|-1009$. Considering the constraint, the interval that fits is $x \in(-1009 ;-3]$.
3) $-3<x<1$. Then $-x-3+x-1<2016 \Leftrightarrow x \in \mathbf{R}$. Considering the constraint, the interval that fits is $x \in(-3 ; 1)$.
Combining all t... | x\in(-\infty;-2016)\cup(-1009;1007) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 22,988 |
5. Out of three hundred eleventh-grade students, excellent and good grades were received by $77 \%$ on the first exam, $71 \%$ on the second exam, and $61 \%$ on the third exam. What is the smallest number of participants who could have received excellent and good grades on all three exams?
Answer: 27. | Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the conditions,
\[
\left\{\be... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,989 |
6. Winnie the Pooh ran down a moving escalator and counted 55 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 1155 steps. How many steps would he have counted if he had gone down a stationary escalator?
Answer: 105. | Let \( u \) be the speed of Winnie the Pooh, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Winnie the Pooh spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v}... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,990 |
7. The diagonals of a trapezoid are perpendicular to each other, and one of them is 17. Find the area of the trapezoid if its height is 15.
Answer. $\frac{4335}{16}$. | Let $ABCD$ be the given trapezoid, $AB \| CD, BD=17$, its height $BH$ is 15. Mark a point $M$ on the line $CD$ such that $BM \| AC$. Then $ABMC$ is a parallelogram and $CM=AB$. Triangles $ABD$ and $BCM$ are equal in area (since they have equal heights and bases). Therefore, the area of trapezoid $ABCD$ is equal to the ... | \frac{4335}{16} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,991 |
1. A disk rotates around a vertical axis with frequency p. The axis passes through the center of the disk. The surface of the disk is horizontal and smooth. A small washer is tied to the axis with a light thread, lies on the disk, and rotates with it. The radius of the disk is $R$, the length of the thread is $R / 4$. ... | ```
\begin{aligned}
& V_{0}=2 \pi n \cdot \frac{R}{4} \\
& l=\sqrt{R^{2}-(R / 4)^{2}}=\frac{\sqrt{15}}{4} R \\
& T=\frac{l}{V_{0}}=\frac{\sqrt{15} R}{4} \cdot \frac{4}{2 \pi n R}=\frac{\sqrt{15}}{2 \pi n}
\end{aligned}
``` | \frac{\sqrt{15}}{2\pin} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,992 |
3. A thermally insulated vessel is divided by a heat-conducting partition into two parts of different volumes. In the first part, there is helium at a temperature of $127{ }^{\circ} \mathrm{C}$ in an amount of $v_{1}=0.2$ moles. In the second part, there is helium at a temperature of $7{ }^{\circ} \mathrm{C}$ in an amo... | 3. $v_{1} C_{V}\left(t_{1}-t\right)=v_{2} C_{V}\left(t-t_{2}\right) ; \quad v_{1} t_{1}-v_{1} t=v_{2} t-v_{2} t_{2}$
$t=\frac{v_{1} t_{1}+v_{2} t_{2}}{v_{1}+v_{2}}=\frac{0.2 \cdot 127+0.8 \cdot 7}{0.2+0.8}=31^{\circ} \mathrm{C} \quad(304 \mathrm{~K})$
$\frac{P_{1}^{\prime}}{P_{1}}=\frac{T}{T_{1}}=\frac{v_{1} T_{1}+v_... | 31\mathrm{C},0.76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,994 |
4. In the circuit shown in the diagram, the EMF of the source $\varepsilon=69 \mathrm{~V}, R_{1}=R=10 \mathrm{~\Omega}, R_{2}=3 R$, $R_{3}=3 R, R_{4}=4 R$. The internal resistance of the source and the resistance of the ammeter can be neglected.
1) Find the current through the resistor $R_{1}$.
2) Find the current thro... | 4. $R_{13}=\frac{R_{1} R_{3}}{R_{1}+R_{3}}=\frac{3}{4} R ; \quad R_{24}=\frac{R_{2} R_{4}}{R_{2}+R_{4}}=\frac{12}{7} R ;$
$U_{13}=\frac{\varepsilon R_{13}}{R_{13}+R_{24}}=\frac{7}{23} \varepsilon ; \quad I_{1}=\frac{U_{13}}{R}=\frac{7}{23} \frac{\varepsilon}{R}=2.1 \mathrm{~A}$
$U_{24}=\varepsilon-U_{13}=\frac{16}{23... | 2.1\mathrm{~A} | Other | math-word-problem | Yes | Yes | olympiads | false | 22,995 |
5. A capacitor with capacitance $C$, charged to a voltage $U_{0}$, is connected through a resistor with resistance $R$ to an uncharged capacitor with capacitance $4 C$.
1) Find the current in the circuit immediately after connection.
2) Find the steady-state (after a long time) voltage on the capacitor with capacitance... | 5.
1) $I_{0}=\frac{U_{0}}{R}$
2) Law of charge conservation:
$$
C U_{0}=C U+4 C U \Rightarrow U=\frac{U_{0}}{5}
$$
3) $Q=\frac{C U_{0}^{2}}{2}-\frac{(C+4 C) U^{2}}{2}=\frac{C U_{0}^{2}}{2}-... | I_{0}=\frac{U_{0}}{R},\quadU=\frac{U_{0}}{5},\quadQ=\frac{2}{5}CU_{0}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,996 |
11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits... | # Solution
Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, whi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,997 |
1. A boy standing on a horizontal ground surface at a distance of $L=5$ m from a vertical wall of a house kicked a ball lying in front of him on the ground. The ball flew at an angle of $\alpha=45^{\circ}$ to the horizon and, after an elastic collision with the wall, fell back to the same place where it initially lay.
... | 1. $V_{0} \sin \alpha-g \tau=0 \Rightarrow V_{0}=\frac{g \tau}{\sin \alpha}$
$$
L=V_{0} \cos \alpha \cdot \tau=\frac{g \tau}{\sin \alpha} \cdot \cos \alpha \cdot \tau=\frac{g \tau^{2}}{\operatorname{tg} \alpha} \Rightarrow \tau=\sqrt{\frac{\operatorname{Ltg} \alpha}{g}}
$$
![](https://cdn.mathpix.com/cropped/2024_05_... | \sqrt{2}\approx1.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,998 |
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