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9.2. Several consecutive natural numbers are written on the board. It is known that $48 \%$ of them are even, and $36 \%$ of them are less than 30. Find the smallest of the written numbers. | Answer: 21.
Solution. $\frac{48}{100}=\frac{12}{25}, \frac{36}{100}=\frac{9}{25}$ - these are irreducible fractions, so the total number of numbers is divisible by 25. If there were 50 or more, then, by the condition, there would be at least 2 fewer even numbers than odd numbers, which is impossible for consecutive na... | 21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,921 |
9.4. Let's call a natural number suitable if it is the smallest among all natural numbers with the same digit sum as it has. Find all suitable numbers that are perfect squares of natural numbers. | Answer: $1,4,9,49$.
Solution. In any suitable number, all digits except the first one are 9. Otherwise, by carrying the one from the highest place to the place where it is not 9, we get a smaller number with the same sum of digits (possibly with fewer digits). All single-digit numbers are suitable, so $1,4,9$ satisfy ... | 1,4,9,49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,922 |
9.5. On a 10 by 10 grid, several cells are marked such that each 3 by 3 square contains exactly one marked cell. How many cells can be marked? | Answer. Any number from 9 to 16.
Solution. Divide the board with horizontal lines at the 3rd, 6th, and 9th positions and vertical lines at the 3rd, 6th, and 9th positions into nine 3x3 squares, three vertical strips 1x3 in the rightmost column, three horizontal strips in the top row, and the top right corner cell. In ... | 9to16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,923 |
8.2. In a family of 4 people. If Masha's scholarship is doubled, the total income of the entire family will increase by $5 \%$, if instead the mother's salary is doubled - by $15 \%$, if the father's salary is doubled - by $25 \%$. By what percentage will the family's total income increase if the grandfather's pension ... | Answer: by $55 \%$.
Solution: When Masha's scholarship is doubled, the family's total income increases by the amount of this scholarship, so it constitutes $5 \%$ of the income. Similarly, the salaries of Masha's mother and father constitute $15 \%$ and $25 \%$. Therefore, the grandfather's pension constitutes $100-5-... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,925 |
8.3. Find the perimeter of a parallelogram if the bisector of one of its angles divides a side of the parallelogram into segments of 7 and 14. | Answer: 56 or 70.
Solution: Let the bisector of the angle at vertex $A$ of parallelogram $A B C D$ intersect side $B C$ at point $M$, such that $B M=7$ and $C M=14$. Then $\angle B M A=\angle M A D=\angle A M B$, so triangle $A B M$ is isosceles. Therefore, $A B=B M=7, C D=A B=7, A D=B C=7+14$. The perimeter is 56.
S... | 56or70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,926 |
8.4. A square was divided into 100 rectangles by nine vertical and nine horizontal lines (parallel to its sides). Among these rectangles, there were exactly 9 squares. Prove that there are at least two identical squares. | Solution: Assume that all squares are of different sizes. Then no two squares are in the same row or column, because the side of a square is equal to the width of the column and the height of the row in which it is located. The total width of the nine columns containing squares is equal to the sum of the side lengths o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,927 |
8.5. There are 100 boxes numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all ... | Answer: 99.
Solution: To be able to definitively determine which of the 100 boxes contains the prize, it is necessary to have the possibility of receiving at least 100 different answers to one set of questions. Since the host's answers for different prize positions can only differ by the number of "yes" responses, it ... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,928 |
8.1. Five buses stand in a row behind each other in a traffic jam, and in any two of them, there is a different non-zero number of passengers. Let's call two different people sympathizers if they are either in the same bus or in adjacent ones. It turned out that each passenger has either exactly 20 or exactly 30 sympat... | Solution. For example, let there be 12, 9, 10, 2, and 19 people in the buses respectively. This example is not unique. It is easy to construct if you realize that on average, there should be 10 people in a bus. Indeed, there should be 21 or 31 passengers in the first two buses and in the first three buses. Therefore, t... | 12,9,10,2,19 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,929 |
8.2. On a certain island, only knights, who always tell the truth, and liars, who always lie, live. One day, 1001 inhabitants of this island stood in a circle, and each of them said: "All ten people following me in a clockwise direction are liars." How many knights could there be among those standing in the circle? | Answer: 91 knights.
Solution. Note that all people cannot be liars, because then it would mean that each of them is telling the truth. Therefore, among these people, there is at least one knight. Let's number all the people so that the knight is the 1001st in the sequence. Then the 10 people with numbers from 1 to 10 ... | 91 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,930 |
8.3. In square $ABCD$, point $H$ is the midpoint of side $CD$, and $K$ is a point on side $BC$ such that $KC = 2KB$. Prove that $KA$ is the angle bisector of $\angle BKH$.
 | Solution. Let the side of the square be 6. Then $D H=H C=3, B K=2, K C=4$. Extend $B C$ beyond point $B$ to point $E$ such that $B E=H C=3$. Note that right triangles $A B E$ and $A D H$ are equal by two legs, so $A E=A H$. Moreover, $K E=K B+B E=5$, and by the Pythagorean theorem for triangle $H K C$, we have $H K^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,931 |
8.4. We will call a number remarkable if it can be decomposed into the sum of 2023 addends (not necessarily distinct), each of which is a natural composite number. Find the largest integer that is not remarkable. | Answer. $4 \times 2023+3=8095$.
Solution. Replace 2023 with $n$ and solve the problem in the general case for a sum of $n \geqslant 2$ composite addends. We will prove that the answer is $4 n+3$, from which we will obtain the answer to the original problem.
Claim 1. The number $4 n+3$ is not remarkable.
Proof of Cla... | 8095 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,932 |
8.5. After a successful train robbery, 102 bandits divided the obtained rubies, sapphires, and emeralds in such a way that each of them received exactly 100 precious stones in total. Prove that at least one of the following two statements is true:
- There are two bandits who have the same number of rubies, sapphires, ... | Solution. Assume the opposite, then any two have the same number of stones of exactly one type (exactly two quantities cannot coincide, because then the third would also coincide due to the fact that their sums are equal to one number 100$).
Let's single out one bandit $A$, suppose he got $r$ rubies, $s$ sapphires, an... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,933 |
11.2. Let $n$ be a natural number not ending in 0, and $R(n)$ be the four-digit number obtained from $n$ by reversing the order of its digits, for example $R(3257)=7523$: Find all natural four-digit numbers $n$ such that $R(n)=4n+3$. | Answer: 1997.
Solution. Consider the decimal representation of the original four-digit number $n=\overline{a b c d}$, then $R(n)=\overline{d c b a}=4 n+3$ is also a four-digit number. Therefore, $4 a \leq 9$, so $a=1$ or $a=2$. Moreover, the number $R(n)=\overline{d c b a}=4 n+3$ is odd and ends in $a$, hence $a=1$. I... | 1997 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,934 |
11.3. Two circles intersect at points A and B. A tangent to the first circle is drawn through point A, intersecting the second circle at point C. A tangent to the second circle is drawn through point B, intersecting the first circle at point D. Find the angle between the lines $\mathrm{AD}$ and $\mathrm{BC}$. | Answer. The lines are parallel, the angle is $0^{\circ}$.
Solution. Mark point $\mathrm{P}$ on the extension of $\mathrm{CA}$ beyond point A and point M on the extension of DB beyond point B. The inscribed angle ABD in the first circle, which subtends the chord AD, is equal to the angle PAD between the chord AD and th... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,935 |
11.4. The sequence of natural numbers $a_{n}, n=1,2, \ldots$ is such that $a_{1}=1$ and $a_{n}=\frac{n+1}{n-1}\left(a_{1}+a_{2}+\ldots+a_{n-1}\right)$ for all $n=2,3, \ldots$. Find the formula for the "general term" of the sequence, that is, a formula that explicitly expresses $a_{n}$ in terms of $n$ for any $n$. | Answer. $a_{n}=(n+1) 2^{n-2}$
Solution. Let's find the first few terms of the sequence: $a_{1}=1, a_{2}=\frac{3}{1} \cdot 1=3 \cdot 1$, $a_{3}=\frac{4}{2} \cdot(1+3)=8=4 \cdot 2, a_{4}=\frac{5}{3} \cdot(1+3+8)=20=5 \cdot 4$ and make the assumption that $a_{n}=(n+1) 2^{n-2}$. We also calculate the sums of the first ter... | a_{n}=(n+1)2^{n-2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,936 |
11.5. Let $M$ be some set of pairs of natural numbers $(i, j), 1 \leq i<j \leq n$ for a fixed $n \geq 2$. If a pair $(i, j)$ belongs to $M$, then no pair $(j, k)$ belongs to it. What is the largest set of pairs that can be in the set $M$? | Answer: $\frac{n^{2}}{4}$ for even $n$, $\frac{n^{2}-1}{4}$ for odd $n$.
Solution. Let's call a natural number $k$ from the interval from 1 to $n$ initial if it is the smaller number in some pair $(k, j)$ from $M$, and a natural number $l$ from the interval from 1 to $n$ final if it is the larger number in some pair $... | \frac{n^{2}}{4}forevenn,\frac{n^{2}-1}{4}foroddn | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,937 |
11.2. In the described quadrilateral $A B C D$, the measures of angles CDA, DAB, and ABC are $90^{\circ}, 120^{\circ}$, and $120^{\circ}$ respectively, and the length of side BC is 1 cm. Find the length of side AB. | Answer. $A B=2-\sqrt{3}$
Solution. Let the length of side AB be $x$. Extend sides DA and CB until they intersect at point E. Triangle EAB is isosceles with angles of 60 degrees at the base AB, making it equilateral. Triangle ECD is a right triangle with a 30-degree angle at vertex C, so $E D=\frac{1}{2} E C=\frac{x+1}... | 2-\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,938 |
11.3. Each cell of a square board of size $n$ by $n$ is colored blue or red. A row or column is called bluish if it contains more blue cells than red ones. Conversely, a row or column is called reddish if it contains more red cells than blue ones. What is the maximum value that the sum of the number of reddish rows and... | Answer. The maximum value of the sum of the number of reddish rows and the number of bluish columns is: a) if $n$ is odd, then $2 n-2$, b) if $n$ is even and $n \geq 8$, then $2 n-4$, c) for $n=2,4,8$ respectively $2,5,9$.
Solution. Consider an arbitrary coloring of the board in two colors, denote by $x$ the number of... | )ifnisodd,then2n-2,b)ifnisevenn\geq8,then2n-4,)forn=2,4,8respectively2,5,9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,939 |
9.1. There are two ingots of different copper and tin alloys weighing 6 and 12 kg respectively. From each of them, a piece of the same weight was cut off and the first piece was alloyed with the remainder of the second ingot, and the second piece - with the remainder of the first ingot, after which the ratio of copper ... | Answer: 4 kilograms.
Solution: Let the weight of each of the cut pieces be $x$ kg, and the proportions of tin in the first and second ingots be $a \neq b$ respectively. Then, the proportion of tin after remelting in the first ingot will be $\frac{b x + a(6 - x)}{6}$, and in the second ingot $\frac{a x + b(12 - x)}{12}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,940 |
9.3. Find five different numbers if all possible sums of triples of these numbers are equal to $3,4,6$, $7,9,10,11,14,15$ and 17. The numbers do not have to be integers. | Answer: $-3,2,4,5,8$.
Solution. Let the required numbers be $a<b<c<d<e$.
Notice that each number appears in exactly six triplets of numbers, so by adding all ten sums of triplets and dividing by 6, we get the sum of all five numbers, which is 16. Clearly, the smallest sum is $a+b+c=3$, from which $d+e=16-a-b-c=13$, a... | -3,2,4,5,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,941 |
9.4. In the tournament, each of the six teams played against each other exactly once. In the end, the teams scored 12, 10, 9, 8, 7, and 6 points respectively. a) How many points were awarded for a win in a match, if 1 point was awarded for a draw, and 0 points for a loss? The answer, of course, should be a natural numb... | Answer. a) 4 points.
b) The first team had three wins, the second team had two wins and two draws, the third team had two wins and one draw, the fourth team had two wins, the fifth team had one win and three draws, the sixth team had one win and two draws. The rest of the matches were lost by the teams.
c) One exampl... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,942 |
9.5. Find all integer solutions of the equation: $x^{3}+y^{3}=2^{30}$. Answer. $\left(0,2^{10}\right),\left(2^{10}, 0\right)$. | Solution. We will show that if $x^{3}+y^{3}$ is a power of two not less than the second, then both variables must be even numbers. Indeed, $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=2^{n}, n \geq 2$, if one of the variables is odd, then the other is also odd, since the sum of their cubes is even. In this case, the ... | (0,2^{10}),(2^{10},0) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,943 |
11.1. Mikhail leaves from Berdsk to Cherapanovo at $8:00$ AM; On the same day, at the same time, and on the same road, Khariton and Nikolay leave from Cherapanovo to Berdsk. At 9:30 AM, Khariton was exactly halfway between Mikhail and Nikolay; at 10:00 AM, Mikhail was exactly halfway between Khariton and Nikolay. Deter... | Answer. Mikhail and Khariton met at 9:48 AM, Mikhail and Nikolay - at 10:15 AM.
Solution. Let $S$ be the distance between Berdsk and Cherepanovo, and $x, y, z$ be the speeds of Mikhail, Khariton, and Nikolay, respectively. From the fact that at 9:30 AM, Khariton was exactly halfway between Mikhail and Nikolay, it foll... | MikhailKharitonmetat9:48AM,MikhailNikolay-at10:15AM | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,944 |
11.2. Find all quadruples of real numbers $(a, b, c, d)$ such that $a(b+c)=b(c+d)=c(d+a)=d(a+b)$. | Answer. 8 infinite series of solutions, obtained by multiplying each of the numbers $(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1),(1,1,1,1),(1,-1,1,-1),(1,-1 \pm \sqrt{2},-1,1 \mp \sqrt{2})$ by any real number.
Solution. Subtracting the first expression from the third, we get $a b=c d$, subtracting the fourth expression fr... | 8\inftyiniteseriesofsolutions,obtainedmultiplyingeachofthe(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1),(1,1,1,1),(1,-1,1,-1),(1,-1\\sqrt{2},-1,1\\sqrt{2})anyreal | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,945 |
11.3. Prove that for any three positive real numbers $x, y, z$, the inequality $\left(x^{2}+y^{2}\right)^{2} \geq(x+y+z)(x-y+z)(x+y-z)(y+z-x)$ holds. Identify all triples $(x, y, z)$ for which equality is achieved. | Answer. All triples of positive numbers $x, y, z$, which are the lengths of the sides of an isosceles right triangle with legs $x, x$, i.e., triples $x, x, x \sqrt{2}, x>0$.
Proof. 1. Note that if $x, y, z$ are the lengths of the sides of some triangle, then when each of the three parentheses in the product on the rig... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 9,946 |
11.5. In one of the vertices of a cube, there are $N$ butterflies, and the other seven vertices are empty. Every minute, from one of the vertices of the cube, one butterfly flies to each of the three adjacent vertices along the edges of the cube, one to the opposite vertex (relative to the center) of the cube, and one ... | Answer. Any $N$, divisible by 45.
Solution. 1. We will show that for any $N=45k, k \in N$ the situation where after some time the same number of butterflies can be in each vertex is possible. Let's number the vertices of the cube, first numbering the vertices of the lower base clockwise from 1 to 4, then the vertices ... | 45k,k\in\mathbb{N} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,947 |
8.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya? | Solution: If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adj... | 18 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,949 |
8.3. Find the largest four-digit number, all digits of which are different, and which is divisible by each of its digits. Of course, zero cannot be used. | Solution: Since the number of digits is fixed, the number will be larger the larger the digits in its higher places are. We will look for the number in the form $\overline{98 a}$. It must be divisible by 9. Therefore, the sum $a+b$ must give a remainder of 1 when divided by 9. At the same time, this sum does not exceed... | 9864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,950 |
8.4. Point $A$ is located midway between points $B$ and $C$. The square $A B D E$ and the equilateral triangle $C F A$ are in the same half-plane relative to the line $B C$. Find the angle between the lines $C E$ and $B F$.
, \pm(1,1), \pm(\sqrt{3},-\sqrt{3})$ - a total of 5 solutions.
Solution. Consider the cases.
1) $x=y$, then $x^{3}=x=y$, from which $(x, y)=(0,0), \pm(1,1)$ - the first three solutions.
2) $x=-y$, then $x^{3}=3 x=-y$, from which $(x, y)=(0,0), \pm(\sqrt{3},-\sqrt{3})$ - the first and the last two soluti... | (0,0),\(1,1),\(\sqrt{3},-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,953 |
10.2. Three people are playing table tennis, with the player who loses a game giving way to the player who did not participate in it. In the end, it turned out that the first player played 21 games, and the second - 10. How many games did the third player play? | Answer: 11.
Solution: According to the problem, the first player played 21 games, so there were at least 21 games in total. Out of any two consecutive games, the second player must participate in at least one, which means there were no more than \(2 \cdot 10 + 1 = 21\) games. Therefore, a total of 21 games were played... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,954 |
10.4. Let's call a snake in a convex $n$-gon an open, non-self-intersecting broken line with $n-1$ segments, the set of vertices of which coincides with the set of all vertices of the $n$-gon. Find the number of different snakes in an $n$-gon. (Snakes are equal if they coincide as geometric loci of points. For example,... | Answer. $n \cdot 2^{n-3}$.
Solution. 1) We will prove by induction that the number of snakes, one of whose ends is a fixed vertex $\mathrm{A}$, is $2^{n-2}$. The base case for $n=3$ is obvious. For an arbitrary $n$, we have two possibilities for the segment with end $\mathrm{A}$ - these are the sides $\mathrm{AB}$ and... | n\cdot2^{n-3} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,955 |
10.5. On the board, 10 natural numbers are written, some of which may be equal, and the square of each of them divides the sum of all the others. What is the maximum number of different numbers that can be among the written ones? | Answer. Four
Solution. An example for four different numbers: 1,1,1,2,2,3,5,5,5,5.
Let among the listed numbers there are exactly $n \geq 2$ different ones, the maximum of which we denote by $x$, and the sum of all numbers by $\mathrm{S}$. Then the sum of all numbers except the maximum does not exceed $(9-n) x + x - ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,956 |
9.1. The bathtub fills up from the hot water tap in 17 minutes, and from the cold water tap in 11 minutes. After how many minutes from opening the hot water tap should the cold water tap be opened so that by the time the bathtub is full, there is one third more hot water than cold water? | Answer. In 5 minutes.
Solution. Let the volume of the bathtub be $V$, then by the time it is filled, there should be $\frac{3}{7} V$ of cold water and $\frac{4}{7} V$ of hot water in it. The filling rates of cold and hot water are $\frac{V}{11}$ and $\frac{V}{17}$, respectively. Therefore, the desired time is the diff... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,957 |
9.2. Can the number 99... 99 (with a total of 9 nines) be represented as the sum of two natural numbers, the sums of whose digits are the same? | Answer: No.
Solution: Suppose we can represent the number A=99... 99 (with a total of 9 nines) as the sum of two numbers B and C, the sums of whose digits are equal. If the addition of the digits in the last place of B and C results in a carry to the previous place, then the last digit of the sum does not exceed 8. Ho... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,958 |
9.3. Find the smallest natural number $n$ such that in any set of $n$ distinct natural numbers, not exceeding 1000, it is always possible to select two numbers, the larger of which does not divide the smaller one. | Answer. $n=11$.
Solution. Among the first 10 powers of two $1=2^{0}, 2=2^{1}, 4=2^{2}, \ldots, 512=2^{9}$, in each pair of numbers, the larger number is divisible by the smaller one, hence $n \geq 11$.
On the other hand, let there be some set of $n \geq 11$ numbers where the larger number of each pair is divisible by... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,959 |
9.5. In each cell of a 10 by 10 table, a minus sign is written. In one operation, it is allowed to simultaneously change the signs to their opposites in all cells of some column and some row (plus to minus and vice versa). What is the minimum number of operations required to make all the signs in the table pluses? | Answer. In 100 operations.
Solution. There are 19 cells in total in the row and column passing through a given cell, so if we perform operations on all pairs of rows and columns of the table (a total of $10 \times 10=100$ operations), each sign in the table will change 19 times, turning from minus to plus, so 100 oper... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,960 |
7.1. Egor, Nikita, and Innokentiy received three grades each for three tests, and all the grades turned out to be threes, fours, and fives. Egor said: "I have higher grades for two tests than Nikita." Nikita replied: “But I have better grades for two tests than Innokentiy." Innokentiy countered: “And I wrote two tests ... | Answer: yes, they could
Solution: Let Egor receive grades 5,4,3; Nikita - 4,3,5; Innokentiy - 3,5,4. Then the condition of the problem is satisfied.
Criteria: only answer - 0 points.
any correct example without verification - 7 points. | yes,theycould | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,961 |
7.2. Anton Pavlovich baked bread for his birthday and invited Vladimir Alekseevich and Fyodor Alekseevich to visit. It turned out that the bread and Fyodor together weigh as much as Anton and Vladimir. After the bread was eaten, Vladimir's weight became equal to the combined weight of Anton and Fyodor. Prove that the p... | Solution: Let Anton weigh A before his birthday, Vladimir - B, Fyodor - F, and they ate pieces of bread weighing ХA, ХB, ХF respectively. Then from the first condition, we get the equation $\mathrm{X}+\mathrm{XB}+\mathrm{X} \Phi+\Phi=\mathrm{A}+\mathrm{B}$, and from the second $\mathrm{B}+\mathrm{XB}=\mathrm{A}+\mathrm... | XB | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 9,962 |
7.3. Anya wrote down 100 numbers in her notebook. Then Sonya wrote down in her notebook all the pairwise products of the numbers written by Anya. Artem noticed that there were exactly 2000 negative numbers in Sonya's notebook. How many zeros did Anya initially write down in her notebook? | Answer: 10 zeros.
Solution: Let Anya write down $n$ positive numbers, $m$ negative numbers, and $100-n-m$ zeros in her notebook. Then, by the condition, $n m=2000$, since a negative number can only be obtained by multiplying numbers of different signs.
Let's list all the divisors of the number $2000=2^{4} * 5^{3}$:
... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,963 |
7.4. A point inside a convex pentagon was connected to its vertices, as a result of which the pentagon was divided into five equal non-isosceles triangles. Prove that these triangles are right-angled. | Solution: Let there be a pentagon $ABCDE$ and a point $O$ inside it. Consider the segments connecting $O$ with the vertices of the pentagon. By the condition, no two consecutive segments are equal (otherwise, there would be an isosceles triangle). We will prove that there are three consecutive different segments. Indee... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,964 |
7.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone h... | Answer: 120
Solution: Consider any pair of friends. Their "friendship" was counted exactly once, as it was included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Si... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,965 |
7.2. Bored Yura added two numbers and got a third one. Then, he changed each digit in this example by 1 in either direction (for example, from the number 239 he could get the number 148, but he could not get 140). Could the new example be correct? | Answer: No.
Solution: We will track the parity of the numbers. Notice that all numbers have changed their parity. Thus, the parity of the sum has changed, on the one hand, twice (along with the parity of each of the addends), and on the other hand, once (the sum itself). Therefore, the new example cannot be correct.
... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,967 |
7.3. In a school, there are 1000 students and 35 classes. On the forehead of each student, the number of students in their class is written. What can the sum of the reciprocals of these numbers be? List all the options and prove that there are no others. Recall that the reciprocal of a number $a$ is the number $1 / a$. | Answer: 35.
Solution: Let there be a people in the class, then the sum of the fractions corresponding to the numbers from this class is 1 (a fractions, each equal to 1/a). There are 35 classes in total. Therefore, the total sum is 35.
Criteria:
Answer - 0 points.
Answer with examples - 0 points.
The idea of partit... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,968 |
7.4. Arseny sat down at the computer between 4 and 5 PM, when the hour and minute hands were pointing in opposite directions, and got up from it on the same day between 10 and 11 PM, when the hands coincided. How long did Arseny sit at the computer? | Solution: Let's see where the hands will be 6 hours after Arseny sat down at the computer. The minute hand will go around the clock 6 times and return to its place. The hour hand will move exactly half a circle. Therefore, the angle between the hands will change by 180 degrees, i.e., the hands will coincide. It is obvi... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,969 |
7.5. Several seventh-graders were solving problems. The teacher does not remember how many children there were and how many problems each of them solved. However, he remembers that, on the one hand, each solved more than a fifth of what the others solved, and on the other hand, each solved less than a third of what the... | Answer: 5 seventh-graders.
Solution: Let one seventh-grader solve $a$ problems, and the rest solve $S - a$. Then
$$
\begin{gathered}
a < (S-a) / 3 \\
3a < S - a \\
4a < S \\
a < S / 4
\end{gathered}
$$
Similarly,
$$
\begin{gathered}
(S-a) / 5 < a \\
S - a < 5a \\
S < 6a \\
S / 6 < a
\end{gathered}
$$
Thus, if ther... | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,970 |
8.1. Arrange the digits from 1 to 9 in a circle such that any two adjacent digits, when read clockwise, form a composite two-digit number. One example is sufficient.
Place the digits from 1 to 9 in a circle so that any two adjacent digits, when read clockwise, form a composite two-digit number. One example is sufficie... | Solution: For example, the arrangement 12745639 8 is suitable. Other solutions are possible. Criteria:
Any correct example without explanation - 7 points. | 127456398 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,971 |
8.2. In a competition, there are 2018 Dota teams, all of different strengths. In a match between two teams, the stronger one always wins. All teams paired up and played one game. Then they paired up differently and played another game. It turned out that exactly one team won both games. How could this be | Solution: Let's number the teams in ascending order of strength from 1 to 2018. In the first round, we will have the matches 1 - 2, $3-4, \ldots, 2017$ - 2018, and in the second round - $2018-1, 2-3, 4$ - 5, ..., 2016 - 2017. It is obvious that only the team with the number 2018 will win in both rounds.
## Criteria:
... | 2018 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,972 |
8.3. Yura chose three distinct integers $a, b, c$. Then he added the numbers $a$ and $b$ and got the number $c$. After that, he multiplied the numbers $b$ and $c$ and got $a$. Find all such triples of numbers and prove that there are no others. | Answer: $a=-4, b=2, c=-2$.
Solution: So, $a+b=c, bc=a$. Therefore, $bc+b=c$. Transform this expression into $b(c+1)=c$. The left side is divisible by $c+1$, so the right side is also divisible. That is, $c$ is a multiple of $c+1$. In this case, if the number $c+1$ is divisible by $m$, then $c$ gives a remainder of $m-... | =-4,b=2,=-2 | Algebra | proof | Yes | Yes | olympiads | false | 9,973 |
8.4. In a right triangle $A B C$ with a right angle $B$ and angle $A$ equal to 30, the height $B D$ was drawn. Then, in triangle $B D C$, the median $D E$ was drawn, and in triangle $D E C$ - the bisector $E F$. Find the ratio of $F C$ to $A C$. | Solution: Since angle $BDC$ is a right angle, segment $DE$ is the median in the right triangle $BDC$, drawn from the vertex of the right angle. Therefore, $BE = ED = EC$. In particular, triangle $EFC$ is isosceles, so $EF$ is a median, and $2FC = DC$. Since angle $BAD$ is 30 degrees, angle $ABD$ is 60 degrees, and cons... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,974 |
8.5. Eleven of the best football teams each played one match against each other. It turned out that each team scored 1 goal in their first match, 2 goals in their second match, ..., 10 goals in their tenth match. What is the maximum number of matches that could have ended in a draw? | Solution: According to the condition, each team scored 1 goal in the first game. In the case of a draw, it also conceded 1 goal. Then for the other team, this match was also the first. Since the number of teams is odd, they cannot be paired. Therefore, at least one of the teams played a non-draw in their first match. T... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,975 |
11.1. In triangle $\mathrm{ABC}$, the bisectors of angles $\mathrm{BAC}$ and $\mathrm{BCA}$ intersect sides ВС and АВ at points К and Р, respectively. It is known that the length of side АС is equal to the sum of the lengths of segments АР and СК. Find the measure of angle $\mathrm{ABC}$. | Answer: $60^{\circ}$.
Solution 1. Let the angles of triangle ABC be denoted by the corresponding letters $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$, and the intersection point of the angle bisectors by I. Reflect points P and K with respect to the angle bisectors AK and CP, respectively; their images will be points $\... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,976 |
11.2. A triplet of real numbers $A, B, C$ is such that $\operatorname{Sin} A+\operatorname{Sin} B+\operatorname{Sin} C=0$ and $\cos A+\operatorname{Cos} B+\operatorname{Cos} C=0$. Find the value of the expression
$\operatorname{Cos}(A-B)+\operatorname{Cos}(B-C)+\operatorname{Cos}(C-A)$. | Answer: $-\frac{3}{2}$.
Solution. Squaring the first two expressions, we get $(\operatorname{Sin} A+\operatorname{Sin} B+\operatorname{Sin} C)^{2}=\operatorname{Sin}^{2} A+\operatorname{Sin}^{2} B+\operatorname{Sin}^{2} C+2(\operatorname{Sin} A \cdot \operatorname{Sin} B+\operatorname{Sin} B \cdot \operatorname{Sin} C... | -\frac{3}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,977 |
11.3. Find all solutions of the system of equations in real numbers:
$$
\left\{\begin{array}{l}
x^{5}=y^{3}+2 z \\
y^{5}=z^{3}+2 x \\
z^{5}=x^{3}+2 y
\end{array}\right.
$$ | Answer. Three solutions: $(0,0,0), \pm(\sqrt{2}, \sqrt{2}, \sqrt{2})$.
Solution. 1. Let's find the easily guessable solutions of the system, when all variable values are equal to each other, that is, when $x=y=z$. Then $x^{5}-x^{3}-2 x=0$, from which $x=0$ or $x= \pm \sqrt{2}$. This leads to the three solutions indica... | (0,0,0),\(\sqrt{2},\sqrt{2},\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,978 |
11.5. On one side of each of 100 cards, one of the natural numbers from 1 to 100 inclusive was written (each number is written on exactly one card), after which they were turned face down and laid out in a random order on the table. In one question, Vasya can point to any two cards, after which he receives an answer fr... | Answer. In 98 questions.
Solution. Let Vasya choose some card A and ask 98 questions, each of which will be about A and one of the 99 cards different from A. The total number of numbers not adjacent to the number written on A does not exceed 98 if 1 or 100 is written on A, and 97 if numbers from 2 to 99 are written on... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,979 |
8.2. Cat Behemoth starts eating ice cream, and a minute later, a girl named Rita starts eating the same kind of ice cream. Two minutes later, Rita has eaten as much as what is left for Cat Behemoth to eat. Could they together finish such an ice cream in three minutes? | 8.2. Answer: Yes.
Solution 1: The cat ate the ice cream for three minutes, after which some part was left to eat. Rita ate the ice cream for two minutes and finished eating the remaining part. That is, in three minutes, starting at different times, they finished one ice cream. Therefore, starting at the same time, the... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,980 |
8.3. Can the numbers $1,2, \ldots, 7$ be placed on the picture below so that the sums of the numbers on each side and each median are the same?
 | # 8.3. Answer: No.
Suppose it is possible, and the sum on each side is $x$. Consider the sum of the numbers on two sides and one median, having a common vertex. This sum is equal to the sum of all the numbers in the triangle, increased by twice the value at the vertex, and thus depends on the choice of the vertex, on ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,981 |
8.4. $P$ and $Q$ are the midpoints of the bases $AD$ and $BC$ of trapezoid $ABCD$ respectively. It turns out that $AB=BC$, and point $P$ lies on the bisector of angle $B$. Prove that $BD=2PQ$. | # 8.4.
Solution 1: Notice that $B P A$ and $B P C$ are equal ($B P$ is common, $A B=B C$, $\angle A B P=\angle C B P$). Therefore, $\angle A=\angle B C P$, from which $180^{\circ}=\angle A+\angle B=\angle B+\angle B C P$, hence $C P \| A B$. This means $A B C P$ is a rhombus ($A B=B C$) and $B C=A P=P D$. Then $B C D ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,982 |
8.5. Find all integer solutions \(x\) and \(y\) for which \(x^{2} + xy - y = 2\). Justify your answer. | 8.5. Answer: $(2,-2),(0,-2)$
Let's transform the original equation as follows:
$$
\begin{gathered}
x^{2}+x y-y=2 \\
x^{2}-1+y(x-1)=1 \\
(x-1)(x+1)+y(x-1)=1 \\
(x-1)(x+1+y)=1
\end{gathered}
$$
The product of two integers equals 1 if both numbers are 1 or if both numbers are -1.
In the first case, we have the solutio... | (2,-2),(0,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,983 |
11.2. Find all natural $n$ that can be represented as the sum $n=a^{2}+b^{2}$, where $a$ is the smallest divisor of $n$ different from 1, and $b$ is some divisor of $n$. | Answer: 8 and 20.
Solution. If $n$ is odd, then all its divisors are odd. Therefore, the right side of the equation $n=a^{2}+b^{2}$ is even - a contradiction. Hence, $n$ is even and its smallest non-unit divisor $a$ is 2, so $n=4+b^{2}$. By the condition, $b$ divides $n=4+b^{2}$, which means it also divides the differ... | 820 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,984 |
11.3. Find all real numbers $a$ for which there exist three distinct real numbers $x, y, z$ such that $a=x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}$. | Answer. $a= \pm 1$.
Solution. Suppose that $a$ satisfies the condition of the problem. From the first equality, express $x=a-\frac{1}{y}=\frac{a y-1}{y}$ and substitute it into the third: $z=a-\frac{1}{x}=a-\frac{y}{a y-1}=\frac{a^{2} y-y-a}{a y-1}$. Substitute the last expression into the second equality: $a=y+\frac{... | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,985 |
11.5. On some cells of a rectangular board of size 101 by 99, there is one turtle each. Every minute, each of them simultaneously crawls to one of the cells adjacent to the one they are in, by side. In doing so, each subsequent move is made in a direction perpendicular to the previous one: if the previous move was hori... | Answer: 9800.
Solution. Examples of unlimited movement on the board of 9800 turtles.
Example 1. Place the reptiles in the cells of a rectangle consisting of cells at the intersection of the 98 lower horizontals and 100 leftmost verticals. They will all move in the same way: first all to the right, then all up, then a... | 9800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,986 |
9.1. Let $a<b<c<d$ be real numbers. Arrange in ascending order the sums $a b+c d, a c+b d, a d+b c$. | Answer. $a d+b c, a c+b d, a b+c d$.
Solution. The difference between the first and second numbers from the condition is $a b+c d-(a c+b d)=d(c-b)+a(b-c)=(d-a)(c-b)>0$. Similarly, the difference between the second and third numbers from the condition is $a c+b d-(a d+b c)=b(d-c)+a(c-d)=(b-a)(d-c)>0$. Therefore, in the... | +,+,+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,987 |
9.2. Non-zero real numbers $x, y$ satisfy the equation $2 x^{2}+2 y^{2}=5 x y$. Find all possible values of the expression $\frac{x+y}{x-y}$. | Answer: 3 and -3.
Solution. Add $4xy$ to both sides of the equality in the condition, we get $2x^2 + 4xy + 2y^2 = 2(x + y)^2 = 9xy$, from which $x + y = \pm \frac{3}{\sqrt{2}} \sqrt{xy}$. Similarly, subtract $4xy$ from both sides of the equality in the condition, we get $2x^2 - 4xy + 2y^2 = 2(x - y)^2 = xy$, from whic... | 3-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,988 |
9.4. Find all triples of prime numbers $p, q, r$ such that the numbers $|q-p|,|r-q|,|r-p|$ are also prime. | Answer: $2,5,7$.
Solution. We can assume that $p2$, then all numbers $p, q, r$ are odd, and their differences $|q-p|,|r-q|,|r-p|$ are even. Then $|r-p|=r-p=r-q+q-r \geq 2+2=4$ cannot be a prime number, which contradicts the problem's condition. Therefore, $p=2$.
Then $q$ and $r$ are odd prime numbers, and their diffe... | 2,5,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,989 |
9.5. In each cell of a $5 \times 5$ table, one of the numbers $1,2,3,4,5$ is written in such a way that each row, each column, and each of the two diagonals of the table contains each of the numbers from 1 to 5. What is the maximum value that the sum of the five numbers written in the cells marked with dots on the diag... | Answer: 22.
Solution. If all 4 numbers marked with a dot and not in the top right corner are different, then the sum of all numbers marked with a dot does not exceed $5+5+4+3+2=19$. Let's f... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,990 |
8.1. A bus with programmers left Novosibirsk for Pavlodar. When it had traveled 70 km, Pavel Viktorovich set off from Novosibirsk in a car along the same route, and caught up with the programmers in Karasuk. After that, Pavel drove another 40 km, while the bus traveled only 20 km in the same time. Find the distance fro... | Solution. Since by the time the car has traveled 40 km, the bus has traveled half that distance, its speed is exactly half the speed of the car. However, when the bus has traveled 70 km after the car's departure, the car will have traveled 140 km and will just catch up with the bus. According to the problem, this happe... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,991 |
8.3. Anton, Borya, Vova, Grisha, and Dima competed in eating buuz, khinkali, and pelmeni. In each of the three competitions, the first place was taken by the boy in gray pants, the second place by the boy in brown pants, and the third place by the boy in raspberry pants (each wears exactly one pair of pants). In additi... | Solution. Note that there are five boys, while there are only three colors being considered. This means that one of the colors must appear no more than once. Let's assume this color is gray (other cases are considered similarly). Suppose Anton has gray pants. But then the boy in gray pants could not have won the buuz-e... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,993 |
8.4. In triangle $A B C$, a point $D$ is marked on side $A C$ such that $B C = C D$. Find $A D$, given that $B D = 13$, and angle $C A B$ is three times smaller than angle $C B A$.
 | Solution. Let $\angle C A B=x$. Then $\angle C B A=3 x$ and $\angle A C B=180^{\circ}-4 x$. According to the problem, triangle $B C D$ is isosceles, so $\angle C D B=\angle C B D=\left(180^{\circ}-\angle B C D\right) / 2=2 x$. Therefore, $\angle D B A=\angle A B C-\angle D B C=3 x-2 x=x=\angle D A B$. Hence, triangle $... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,994 |
8.5. Vasya wrote a set of different natural numbers on the board, each of which does not exceed 2023. It turned out that for any two written numbers $a$ and $b$, the number $a+b$ is not divisible by $a-b$. What is the maximum number of numbers Vasya could have written? | Solution. We will prove that the answer is $\left\lceil\frac{2023}{3}\right\rceil=675$.
Estimate. We will prove that Vasya could not write more than 675 numbers. For this, consider any three consecutive numbers $a, b=a+1$, and $c=a+2$. Note that adjacent numbers cannot be written (otherwise their difference would be o... | 675 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,995 |
1. There are 5 pieces of transparent glass of the same square shape and the same size. Each piece of glass is conditionally divided into 4 equal parts (right triangles) by its diagonals, and one of these triangles is painted with an opaque paint of its individual color, different from the colors of the painted parts of... | Solution. First, consider some fixed vertical order of laying the glasses (from bottom to top). It is clear that by rotating the entire layout by some angle, we will not change the layout (we will not get a new layout). Therefore, we can assume that the bottom glass in the layout is always fixed (does not rotate). Then... | 7200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,996 |
2. Solve the equation $\sin ^{4}(2025 x)+\cos ^{2019}(2016 x) \cdot \cos ^{2018}(2025 x)=1$.
# | # Solution:
$\sin ^{4} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1 \Leftrightarrow\left(1-\cos ^{2} 2025 x\right)^{2}+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1$
$\Leftrightarrow \cos ^{4} 2025 x-2 \cos ^{2} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=0 \Leftrightarrow$
$\cos ^{2} 2025 x\lef... | \frac{\pi}{4050}+\frac{\pin}{2025},n\inZ,\frac{\pik}{9},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,997 |
3. All members of an infinite geometric progression are natural numbers. The sum of the third, fifth, and seventh terms of this progression is equal to $819 \cdot 6^{2016} \cdot$ Find the common ratio of the progression.
(16 points) | Solution: We have a geometric progression $b_{1}, b_{1} q, b_{1} q^{2}, \cdots, b_{1} q^{n-1}, \cdots$, and $b_{1} q^{n-1} \in N$ for any number $n \in N$. Thus, $b_{1}$ and $q$ are natural numbers. According to the condition $b_{3}+b_{5}+b_{7}=819 \cdot 6^{2016}$, or $b_{1} q^{2}+b_{1} q^{4}+b_{1} q^{6}=2^{2016} \cdot... | q=1,q=2,q=3,q=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,998 |
4. On the sides $AB$ and $AC$ of triangle $ABC$, points $D$ and $E$ are chosen such that the area of triangle $ADE$ is 0.5. The inscribed circle of quadrilateral $BDEC$ touches side $AB$ at point $K$, and $AK=3$. Find the tangent of angle $BAC$, if a circumcircle can be circumscribed around quadrilateral $BDEC$, and $B... | Solution: Let $O_{1}$ be the center of the inscribed circle in triangle $ABC$, $R$ its radius, $O_{2}$ the center of the inscribed circle in triangle $ADE$, $r$ its radius. Let $p_{1}$ be the semiperimeter of triangle $ABC$, and $p_{2}$ the semiperimeter of triangle $ADE$. Then $A K=p_{2}$, $A K=p_{1}-BC$. Indeed, by t... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,999 |
5. Specify all values of $a$ for which the system of equations
$\left\{\begin{array}{c}\log _{\mid x+3}(a x+4 a)=2 \log _{\mid x+3}(x+y), \\ x+1+\sqrt{x^{2}+2 x+y-4}=0\end{array}\right.$ has two distinct solutions, and find these solutions
for each $a$.
(20 points) | Solution: Consider the second equation of the system $x+1+\sqrt{x^{2}+2 x+y-4}=0 \Leftrightarrow\left\{\begin{array}{l}x \leq-1, \\ y=5 .\end{array}\right.$
Therefore, $\left\{\begin{array}{c}\log _{|x+3|}(a x+4 a)=2 \log _{|x+3|}(x+y), \\ x+1+\sqrt{x^{2}+2 x+y-4}=0\end{array} \Leftrightarrow\left\{\begin{array}{c}x^{... | \in(4;4,5)\cup(4,5;16/3],solutionsx_{1/2}=\frac{-10\\sqrt{^{2}-4}}{2},y_{1/2}=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,000 |
6. Find the volumes of the parts into which the plane, parallel to the diagonal $A C_{1}$ of the lateral face $A A_{1} C_{1} C$, passing through vertex $C$ and the center of symmetry of the lateral face $A A_{1} B_{1} B$, divides the regular triangular prism $A B C A_{1} B_{1} C_{1}$, if the area of the cross-section o... | # Solution:
1) Construction of the section.
In the lateral face $A A_{1} C_{1} C$, draw a line $A_{2} C$ through point $C$ parallel to $A C_{1}$, where point $A_{2}$ is the intersection of $A A_{1}$ and $A_{2} C$, and $A A_{1}=A A_{2}$. Let $D$ be the center of symmetry of the lateral face $A A_{1} B_{1} B$. Draw a l... | \frac{112}{3},\frac{154}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,001 |
1. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 19, green - 23, blue - 53, blue - 55, purple - 83, black - 128.... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$, where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)$ - the $k$-fold composition of the function $f(n)$, which is equal to $3 n-2$ for $n \leq 17$, and equal to $|129-2 n|$ for $n \geq 18$. Let's compute and write down the first... | blue | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,002 |
3. Find all pairs of natural numbers $a$ and $b$, for which out of the four statements:
1) $a^{2}+4 a+3$ is divisible by $b$;
2) $a^{2}+a b-6 b^{2}-2 a-16 b-8=0$;
3) $a+2 b+1$ is divisible by 4;
4) $a+6 b+1$ is a prime number
three are true, and one is false. | Solution. Statements 3) and 4) cannot both be true. If 3) is true, then $a+6b+1=(a+2b+1)+4b$ is divisible by 4 and is not a prime number. Therefore, one of the statements 3) or 4) is false.
Let's consider statement 2$): a^{2}+a(b-2)-6b^{2}-16b-8=0$. Consider this equation as a quadratic in $a$. Calculate the discrimin... | =6,b=1;=18,b=7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,004 |
4. In triangle $ABC$ with angle $A$ equal to $60^{\circ}$, the angle bisector $AD$ is drawn. The radius of the circumcircle of triangle $ADC$ with the center at point $D$ is $2 \sqrt{3} / 3$. Find the length of segment $BM$, where $M$ is the intersection point of segments $AD$ and $BO$, if $AB=1$. | Solution: 1) $D C=2 \sqrt{3} / 3$
$\triangle O D C$
is equilateral,
since
$2 \angle D A C=\angle D O C=60^{\circ}$.
2) Let $A D=l, B D=x, A C=z$. By the properties of the angle bisector, we have $\frac{A B}{A C}=\frac{B D}{D C} \Rightarrow \frac{1}{z}=\frac{x \sqrt{3}}{2}$. Since $S_{\triangle A B D}+S_{\triangle ... | \frac{\sqrt{21}}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,005 |
5. For each value of the parameter $a$, solve the equation $\log _{2} \frac{3 \sqrt{3}+\cos a(\sin x+4)}{3 \sin a \cos x}=|3 \sin a \cos x|-|\cos a(\sin x+4)+3 \sqrt{3}|$. | Solution: Domain of definition (ODZ): $\frac{3 \sqrt{3}+\cos a(\sin x+4)}{3 \sin a \cos x}>0$. On the ODZ, we have
$\log _{2} \frac{3 \sqrt{3}+\cos a(\sin x+4)}{3 \sin a \cos x}=\log _{2}|\cos a(\sin x+4)+3 \sqrt{3}|-\log _{2}|3 \sin a \cos x|$. Then the original equation on the ODZ is equivalent to the following $\log... | \frac{\pi}{6}+2\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,006 |
6. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to $\sqrt{3}$, and the height of the pyramid coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane passing through the center of the sphere circumscribed around the pyramid, forming an angle of $... | Solution: The center of the sphere $O$ lies on the perpendicular to the base plane, drawn through the center of the base $E$; $O E=A T / 2$. The distance from point $A$ to the line of intersection $M P(P \in B C)$ of the cutting plane with the base plane $A S=A R / \sin \angle A S N$, where $N \in A T, A R$ is the heig... | \frac{11\sqrt{3}}{30} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,007 |
1. Find a natural number that has six natural divisors (including one and the number itself), two of which are prime, and the sum of all its natural divisors is 78. | Solution: The desired natural number $n$ can be represented as $n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}, 12,1+p_{2}+p_{2}^{2}>7$, then no factorization of the number 78 into a product of two natural factors fits $78=1 \cdot 78,78=2 \cdot 39,78=3 \cdot 26,78=6 \cdot 13$ up to the order).
2) $\alpha_{1}=2, \alpha_{... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,008 |
2. Find all natural values of $n$ for which $\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9}$, and $\log _{3}^{2} n+14<\log _{3} 9 n^{7}$. | Solution: $\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9} \Leftrightarrow$ $2 \sin \frac{\pi}{9} \cos \frac{2 \pi}{9}+2 \sin \frac{\pi}{9} \cos \frac{4 \pi}{9}+\cdots+2 \sin \frac{\pi}{9} \cos \frac{2 \pi n}{9}=2 \sin \frac{\pi}{9} \cos \frac{\pi}{9} \Leftrightarrow$ $\sin \f... | n=2+9,=3,4,5,6,7,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,009 |
3. Find the set of values of the function $y=f^{[2019]}(x)$, where $f(x)=\log _{0.5}\left(\frac{\sin x}{\sin x+15}\right)$,
 | Solution: First, let's find the range of the function \( y_{1}=f(x) \). Let \( t=\sin x, t \in[-1 ; 1] \Rightarrow y_{1}=\log _{0.5}(z), \quad z=\frac{t}{t+15}=1-\frac{15}{t+15} \quad \) (the function \( z(t) \) is increasing). For \( t \in[-1 ; 1] \) we have \( z \in\left[-\frac{1}{14} ; \frac{1}{16}\right] \), but by... | [4;+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,010 |
4. The lateral sides $AB$ and $CD$ of trapezoid $ABCD$ are equal to 2 and 3, respectively, and angles $A$ and $D$ are acute. The bisectors of angles $A$ and $B$ of the trapezoid intersect at point $M$, and the bisectors of angles $C$ and $D$ intersect at point $N$. The length of segment $MN$ is 4. Find the radius of th... | Solution: Draw the angle bisectors of angles $A M$ and $B M$ of angles $A$ and $B$ respectively. Since $\angle A + \angle B = 180^{\circ}$, triangle $A B M$ is a right triangle, $\angle A M B = 90^{\circ}$. Let $L$ be the intersection of $B M$ with the base $A D$. Then triangle $A B L$ is isosceles, as $A M$ is both a ... | \frac{16\sqrt{2}}{15+\sqrt{129}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,011 |
5. Find all values of the parameter $a$ for which the equation $\left(2 \sin x+a^{2}+a\right)^{3}-(\cos 2 x+3 a \sin x+11)^{3}=12-2 \sin ^{2} x+(3 a-2) \sin x-a^{2}-a$ has two distinct solutions on the interval $\left[-\frac{\pi}{6} ; \frac{3 \pi}{2}\right]$. Specify these solutions for each found $a$. | Solution: $\quad\left(2 \sin x+a^{2}+a\right)^{3}-(\cos 2 x+3 a \sin x+11)^{3}=-2 \sin ^{2} x+(3 a-2) \sin x-a^{2}-a+12 \Leftrightarrow$ $\left(2 \sin x+a^{2}+a\right)^{3}+2 \sin x+a^{2}+a=(\cos 2 x+3 a \sin x+11)^{3}+1-2 \sin ^{2} x+3 a \sin x+11 \Leftrightarrow$ $\left(2 \sin x+a^{2}+a\right)^{3}+2 \sin x+a^{2}+a=(\c... | \in[2.5;4),x_{1}=\arcsin(-3),x_{2}=\pi-\arcsin(-3);\in[-5;-2),x_{1}=\arcsin(0.5+2),x_{2}=\pi-\arcsin(0.5+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,012 |
6. The base of the pyramid $TABC$ is the triangle $ABC$, all sides of which are equal to 3, and the height of the pyramid, equal to $\sqrt{3}$, coincides with the lateral edge $TA$. Find the area of the section of the pyramid by a plane that passes through the center of the sphere circumscribed around the pyramid, is p... | Solution: The center of the sphere $O$ lies on the perpendicular to the base plane, drawn through the center of the base $E$; $O E=A T / 2$. The distance from point $E$ to the line of intersection of the cutting plane with the base plane $E L=O E \cdot \operatorname{ctg} \angle O L E$. In all cases, the conditions are ... | \frac{11\sqrt{3}}{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,013 |
1. The automatic line for processing body parts initially consisted of several identical machines. The line processed 38880 parts daily. After the production was modernized, all the machines on the line were replaced with more productive but also identical ones, and their number increased by 3. The automatic line began... | Solution: Let $x$ be the number of machines before modernization, $y$ be the productivity of each machine, i.e., the number of parts processed per day, and $z$ be the productivity of the new machines. Then we have $x y = 38880 = 2^{5} \cdot 3^{5} \cdot 5, (x+3) z = 44800 = 2^{8} \cdot 5^{2} \cdot 7, x > 1, y \frac{3888... | 1215 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,014 |
4. In an isosceles triangle $ABC$ with base $AC$, the lateral side $AB$ is equal to 2, and segment $AD$ is the angle bisector. Through point $D$, a tangent $DH$ is drawn to the circumcircle of triangle $ADB$, and point $H$ lies on side $AC$. Find the area of triangle $ABC$ if $CD = \sqrt{2} CH$. | Solution: Let $\angle C A D=\angle B A D=\alpha$, then by the properties of the tangent $\angle C D P=\alpha$. Triangles $C D H$ and $A C D$ are similar by two angles. We have $\frac{C H}{C D}=\frac{C D}{A C}=\frac{1}{\sqrt{2}}, C H=\frac{C D}{\sqrt{2}}, A C=C D \sqrt{2}, C H=\frac{A C}{2}$.
Let $C H=A H=x, D H=y$. Th... | \sqrt{4\sqrt{2}-5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,017 |
5. Find all values of the parameter $a$ for which the equation $(\cos 2 x+14 \cos x-14 a)^{7}-\left(6 a \cos x-4 a^{2}-1\right)^{7}=(6 a-14) \cos x+2 \sin ^{2} x-4 a^{2}+14 a-2 \quad$ has $\quad$ two distinct solutions on the interval $\left[-\frac{2 \pi}{3} ; \pi\right]$. Specify these solutions for each found $a$.
S... | Answer: $a \in[3.25 ; 4), x_{1}=\arccos (2 a-7), x_{2}=-\arccos (2 a-7)$;
$$
a \in[-0.5 ; 1), x_{1}=\arccos (a), x_{2}=-\arccos (a)
$$
6 Find the area of the section of a regular triangular pyramid TABC by a plane passing through the center of the sphere circumscribed around the pyramid, and through the midpoint of t... | 2\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,018 |
1. The numbers $u, v, w$ are roots of the equation $x^{3}-3 x-1=0$. Find $u^{9}+v^{9}+w^{9}$. (12 points) | Solution. $\quad$ According to Vieta's theorem, $u+v+w=0, uv+vw+uw=-3, uvw=1$. Consider the sequence $S_{n}=u^{n}+v^{n}+w^{n}$. We have $S_{0}=3, S_{1}=0$. Let's find $S_{2}$: $S_{2}=u^{2}+v^{2}+w^{2}=(u+v+w)^{2}-2(uv+vw+uw)=6. \quad$ Since $\quad u^{3}=3u+1, v^{3}=$ $3v+1, w^{3}=3w+1, \quad$ then $S_{3}=u^{3}+v^{3}+w^... | 246 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,019 |
2. In the laboratory, there are flasks of two sizes (volume $V$ and volume $V / 3$) in a total of 100 pieces, with at least 2 flasks of each size. The lab assistant randomly selects two flasks in sequence, and fills the first one with a 70% salt solution, and the second one with a 40% salt solution. Then, he pours the ... | Solution. Let $N$ be the number of large flasks in the laboratory, $N=2,3, \ldots, 98$, $n=100-N$ be the number of small flasks in the laboratory, $n=2,3, \ldots, 98$, and $\mathrm{P}(A)$ be the probability of the event $A=\{$ the salt content in the dish is between $50 \%$ and $60 \%$ inclusive\}. It is necessary to f... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,020 |
3. In a convex quadrilateral $A B C D$, the lengths of sides $A B$ and $B C$ are equal, $D B$ is the bisector of angle $A D C$, and $A D: D C=4: 3$. Find the cosine of angle $A K B$, where $K$ is the intersection point of diagonals $A C$ and $B D$, and $B K: K D=1: 3$. (16 points) | # Solution.
1) $A D: D C=4: 3$, let $A D=4 x, D C=3 x$, $B K: K D=1: 3$, let $B K=y, K D=3 y$.
2) $D K$ is the bisector of triangle $A D C$, $A K: K C=A D: D C=4: 3, A K=4 z, K C=3 z$.
3) Point $B$ is the intersection of the perpendicular bisector of diagonal $A C$ and the bisector of angle $D$ in the convex quadrilat... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,021 |
5. In tetrahedron $ABCD$, the sums of three plane angles at each vertex are $180^{\circ}$. Find the volume of the tetrahedron if $BC=4, \cos \angle BAC=3/4, \sin \angle CBD=5\sqrt{7}/16$. (20 points) | Solution. Let's make a development of the tetrahedron on the plane $ABC$ relative to the sides of the triangle $ABC$. Since the sums of the three planar angles at each vertex are $180^{\circ}$, we will get a triangle $D_{1} D_{2} D_{3}$, for which the segments $AB, AC$, and $BC$ are the midlines. From this, we obtain t... | \frac{15\sqrt{6}}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,023 |
6. As part of the renovation program, it was decided to demolish an old house and build a new one in its place. During the demolition of the old house, two problems arose.
1) The accumulation of construction debris blocked access to certain points on the construction site, between which it was necessary to measure the ... | # Solution.
1) Using the similarity of triangles ACB and RCV:
$$
\frac{AC}{CR}=\frac{AB}{VR} \quad \Rightarrow \quad AB=\frac{AC}{CR} \cdot VR=\frac{4}{1} \cdot 3=12
$$
The area of quadrilateral CADB will be found as the sum of the areas of triangles CAB and ABD: the area of the first triangle will be found as half ... | 12,\frac{57\sqrt{15}+3\sqrt{39}}{2}\approx119.74,\quad2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,024 |
1. The fraction $\frac{1}{5}$ is written as an infinite binary fraction. How many ones are there among the first 2022 digits after the decimal point in such a representation? (12 points) | Solution. The smallest number of the form $2^{n}-1$ divisible by 5 is 15. Then
$$
\frac{1}{5}=\frac{3}{15}=\frac{3}{16-1}=\frac{3}{2^{4}-1}=\frac{3}{16} \cdot \frac{1}{1-2^{-4}}=\left(\frac{1}{16}+\frac{1}{8}\right)\left(1+2^{-4}+2^{-8}+2^{-12}+\cdots\right)=
$$
$\left(2^{-3}+2^{-4}\right)\left(1+2^{-4}+2^{-8}+2^{-12... | 1010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 10,025 |
2. There is a cube fixed on legs, and six different paints. In how many ways can all the faces of the cube be painted (each in one color, not all paints have to be used) so that adjacent faces (having a common edge) are of different colors? (16 points) | Solution. Let's consider 4 cases of coloring a cube.
1) The top and bottom faces are the same color, and the left and right faces are the same color. We choose the color for the top and bottom faces in 6 ways, then the color for the left and right faces in 5 ways, then the color for the front face in 4 ways, and the c... | 4080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 10,026 |
3. Point $M$ lies on the leg $A C$ of the right triangle $A B C$ with a right angle at $C$, such that $A M=2, M C=16$. Segment $M H$ is the altitude of triangle $A M B$. Point $D$ is located on the line $M H$ such that the angle $A D B$ is $90^{\circ}$, and points $C$ and $D$ lie on the same side of the line $A B$. Fin... | Solution. 1. A circle can be circumscribed around quadrilateral $A B C D$ with diameter $A B$ (angles $A D B$ and $A C B$ are right angles). Then $\angle A B D=\angle A C D$,
$$
\angle H A D=90^{\circ}-\angle A B D, \angle A D H=\angle A B D=\angle A C D
$$
Triangles $A C D$ and $A D M$ are similar, and $\frac{A D}{A... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 10,027 |
4. Find all values of the parameter $a$ for which the system
$$
\left\{\begin{array}{c}
\left(a\left|y^{3}\right|+a\left|x^{3}\right|-8\right)\left(x^{6}+y^{6}-3 a^{2}\right)=0 \\
\sqrt{x^{6} y^{6}}=a
\end{array}\right.
$$
has eight distinct solutions. (16 points) | Solution. Let's make a change of variables: $x^{3}=u, y^{3}=v$. Variables $u$ and $v$ can take any values, and $x$ and $y$ are uniquely determined. Then we get
$$
\left\{\begin{array} { c }
{ ( a | v | + a | u | - 8 ) ( u ^ { 2 } + v ^ { 2 } - 3 a ^ { 2 } ) = 0 } \\
{ | u v | = a , a > 0 }
\end{array} \Leftrightarrow... | \in(0;2/3)\cup{2}\cup(2\sqrt[3]{2};+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 10,029 |
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