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8.4. In a convex quadrilateral $A B C D$, angle $C B D$ is equal to angle $C A B$, and angle $A C D$ is equal to angle $B D A$. Prove that then angle $A B C$ is equal to angle $A D C$. | Solution: Let the intersection of the diagonals of the quadrilateral be point $O$. In triangles $BCO$ and $ABC$, $\angle CBO = \angle BAC$ by the given condition, and $\angle BCO = \angle BCA$ as a common angle. Therefore, $\angle BOC = \angle ABC$, since the sum of the angles in a triangle is 180 degrees. Similarly, b... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,481 |
8.5. Each digit of the natural number $N$ is strictly greater than the one to its left. What is the sum of the digits of the number $9 N$? | Answer: 9.
Solution: Note that $9 N=10 N-N$. Let's perform this subtraction in a column. In the units place, there will be the difference between 10 and the last digit of the number $N$, in the tens place - the last and the second-to-last digit, decreased by 1. In all subsequent places, there will be the difference be... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,482 |
9.1. In the morning, a dandelion blooms, it flowers yellow for three days, on the fourth morning it turns white, and by the evening of the fifth day, it withers. On Monday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and on Wednesday - 15 yellow and 11 white. How many white dandelions will the... | Answer. Six dandelions.
Solution. A blooming dandelion is white on the fourth and fifth day. This means that on Saturday, the dandelions that bloomed on Tuesday or Wednesday will be white. Let's determine how many there are. The dandelions that were white on Monday had already flown away by Wednesday, and 20 yellow on... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,483 |
9.3. Inside a semicircle of radius 12, there are a circle of radius 6 and a small semicircle, each touching the others pairwise, as shown in the figure. Find the radius of the small semicircle. | Answer: 4.
Solution. Let the radius of the small semicircle be $x$, the center of the large semicircle be $A$, the center of the circle be $B$, and the center of the small semicircle be $C$. The centers of the tangent circles and semicircle, and the corresponding points of tangency, lie on the same straight line, so $... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,484 |
9.5. a) Divide all natural numbers from 1 to 12 inclusive into six pairs, the sums of the numbers in which are six different prime numbers.
b) Can all natural numbers from 1 to 22 inclusive be divided into eleven pairs, the sums of the numbers in which are eleven different prime numbers? | Answer. a) for example, 1 and 4, 2 and 5, 3 and 8, 6 and 7, 9 and 10, 11 and 12, b) it is impossible.
Solution. The sum of all numbers from 1 to 22 inclusive is 253. There are 13 prime numbers that can be represented as the sum of two numbers from 1 to 22 - these are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,485 |
10.1. Find all three-digit natural numbers $A$, the square of which ends in $A$. | Answer: $A=376,625$.
Solution. By the condition, $A^{2}-A=A(A-1)$ is divisible by $1000=2^{3} \cdot 5^{3}$. Due to the mutual simplicity of $A$ and $A-1$, one of them is divisible by $2^{3}=8$, and the other by $5^{3}=125$.
If $A$ is divisible by 125, then $A \in\{125,250,375,500,625,750,875\}$ and $A-1 \in\{124,249,... | 376,625 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,486 |
10.3. Find all solutions in four-digit natural numbers for the equation $1+2013 x+2015 y=x y$. | Answer: (3020,6043$), (4027,4029), (6041,3022)$.
Solution. Transform the equation to the form $(x-2013)(y-2015)=2013 \cdot 2015+1$, the right side can be factored into primes as $2014^{2}=2^{2} \cdot 19^{2} \cdot 53^{2}$.
The right side can be decomposed into the product of two natural factors in 27 different ways, b... | (3020,6043),(4027,4029),(6041,3022) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,487 |
11.4. In a semicircle with a radius of 18 cm, a semicircle with a radius of 9 cm is constructed on one of the halves of the diameter, and a circle is inscribed, touching the larger semicircle from the inside, the smaller semicircle from the outside, and the second half of the diameter. Find the radius of this circle.
... | # Answer: 8 cm.
Solution. Let O, O_1, O_2 be the centers of the large semicircle, the small semicircle, and the inscribed circle, respectively, and let P, Q, R be the points of tangency of the inscribed circle with the diameter of the large semicircle, the small semicircle, and the large semicircle, respectively. Then... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,490 |
11.5. In how many ways can a table of size $n \times n$ cells be filled with zeros and ones so that each row and each column contains an even number of ones? Each cell of the table must contain either a zero or a one. | Answer: $2^{(n-1)^{2}}$.
Solution. We will prove that any arrangement of zeros and ones in the cells of the lower left corner square of size $(n-1) \times(n-1)$ can be uniquely extended to fill the entire table with zeros and ones, satisfying the conditions of the problem. Consider any of the lower $n-1$ rows of the t... | 2^{(n-1)^{2}} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,491 |
7.2. The road from point A to point B first goes uphill and then downhill. A cat takes 2 hours and 12 minutes to travel from A to B, and the return trip takes 6 minutes longer. The cat's speed going uphill is 4 km/h, and downhill is 5 km/h. How many kilometers is the distance from A to B? (Provide a complete solution, ... | Solution: When a cat goes uphill, it takes 15 minutes for 1 km, and when it goes downhill, it takes 12 minutes. That is, when the direction changes, the time spent on 1 km changes by 3 minutes. Since the cat spent 6 minutes more on the return trip, the uphill section on the return trip is 2 km longer.
Let the length o... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,493 |
7.3. Which of the numbers is greater:
$$
2017^{2022} \cdot 2018^{2021} \cdot \ldots \cdot 2022^{2017} \text { or } 2017^{2017} \cdot 2018^{2018} \cdot \ldots \cdot 2022^{2022} \text { ? }
$$
(Justify your answer.) | Solution 1: Let's write both products without powers, arranging the factors in each of them in non-decreasing order. Place these records one under the other.
Then it is obvious that each fac... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,494 |
7.4. On a certain island, there live 100 people, each of whom is either a knight, who always tells the truth, or a liar, who always lies. One day, all the inhabitants of this island lined up, and the first one said: "The number of knights on this island is a divisor of the number 1." Then the second said: "The number o... | Solution: If there are no knights, then all the speakers are lying, since 0 is not a divisor of any natural number.
If there are knights, let there be $a$ of them. Then only the people with numbers $a k$ for $k=1,2, \ldots$ are telling the truth. On the other hand, since exactly $a$ people are telling the truth, $k$ c... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,495 |
7.5. A $2 \times 2$ square was divided by lines parallel to its sides into several rectangles (not necessarily equal). Then these rectangles were painted yellow and blue in a checkerboard pattern. It turned out that the total area of the blue rectangles coincided with the total area of the yellow ones. Prove that the b... | Solution: Rearrange the columns so that all blue rectangles in the first row are consecutive, starting from the left edge. It is easy to understand that in this case, blue rectangles will also be consecutive in all other rows: in odd rows - from the left edge, in even rows - from the right edge. Yellow rectangles will ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,496 |
8.2. The product of two natural numbers $a$ and $b$ is a three-digit number, which is the cube of some natural number $k$. The quotient of the numbers $a$ and $b$ is the square of this same number $k$. Find $a, b$, and $k$. | Solution: Let's write the conditions as follows: $a \cdot b=k^{3}, a / b=k^{2}$. Dividing the first by the second, we get $b^{2}=k$. Since $k^{3}$ is a three-digit number, then $b^{6}$ is also a three-digit number. Clearly, $b$ cannot be 1 or 2, since $1^{6}=1, 2^{6}=64$. Next, $b$ can be 3, since $3^{6}=729$. Finally,... | =243,b=3,k=9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,498 |
8.3. Krosch, Losyash, and Sovunya participated in races. Krosch started first, but during the race, he was overtaken, or he overtook others exactly 12 times. Sovunya started last, but during the race, she was overtaken, or she overtook others exactly 10 times. In what order did the participants finish, given that Losya... | Solution: Let's number the positions where the Smeshariki are located. Initially, Kros has position 1, Losyash - 2, and Sovunya - 3. When overtaking, the parity of the position number of those involved in the overtaking changes. Therefore, after an even number of overtakes, the parity of the positions of Sovunya and Kr... | Sovunya,Losyash,Krosch | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,499 |
8.4. On a cubic planet, there live cubic mice, and they live only on the faces of the cube, not on the edges or vertices. It is known that different numbers of mice live on different faces, and the number on any two adjacent faces differs by at least 2. What is the minimum number of cubic mice that can live on this pla... | Solution: We will prove that no three consecutive numbers can be the number of mice on the faces. Indeed, if there were $x, x+1$, and $x+2$ mice on some three faces, then $x$ and $x+1$ would have to be on opposite faces. But then $x+2$ mice could not be anywhere.
Consider the first 8 natural numbers. Among the first t... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,500 |
8.5. In an isosceles triangle, the bisector drawn to the base turned out to be twice as short as the bisector drawn to the lateral side. Find the angles of this triangle.
 | Solution: Let in the isosceles triangle $ABC$ the sides $AB = BC$ and the condition for the bisectors from the problem statement $AE = 2BD$ is satisfied. Extend $ABC$ to the rhombus $ABCF$. Then, since $BD$ is the bisector in the isosceles triangle, it is also a median. Therefore, $BD$ is half of the diagonal of our rh... | 36,36,108 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,501 |
11.1. What three digits should be appended to the right of the number 579 so that the resulting six-digit number is divisible by 5, 7, and 9? Find all possible solutions. | Answer. $(6,0,0),(2,8,5)$ or $(9,1,5)$. The obtained number will be, respectively, 579600, 579285, or 579915.
Solution. Let the digits to be appended be denoted as $x, y, z$ in order from left to right. The resulting 6-digit number will be $\overline{579 x y z}$. By the divisibility rule for 5, we have $z=0$ or $z=5$.... | (6,0,0),(2,8,5),(9,1,5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,502 |
11.2. Find all triples of real numbers such that each of these numbers is equal to the square of the difference of the other two. | Answer. $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$.
Solution. Write the condition as a system of equations in the unknowns $x, y, z$ as follows: $x=(y-z)^{2}, y=(x-z)^{2}, z=(x-y)^{2}$. Subtract the second equation from the first, the third from the first, and the third from the second, to obtain three equations - consequences... | (0,0,0),(1,1,0),(1,0,1),(0,1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,503 |
11.3. The perimeter of triangle $ABC$ is 24 cm, and the segment connecting the point of intersection of its medians with the point of intersection of its angle bisectors is parallel to side $AC$. Find the length of $AC$. | Answer: 8 cm.
Solution. Let AK be the median from vertex A, M - the point of intersection of the medians ABC, and I - the point of intersection of its angle bisectors AA1, BB1, CC1. Draw a line through K parallel to AC, intersecting the angle bisector BB1 at point P - its midpoint. By Thales' theorem, $PI: IB1 = KM: M... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,504 |
11.4. On a sphere, a point $M$ is marked. Consider all triples of points $A, B, C$ on the sphere, different from $M$, such that the segments $M A, M B, M C$ are pairwise perpendicular, and for each such triple, consider the plane passing through $A, B, C$. Prove that all these planes pass through some common point. | Solution. Consider an arbitrary triple of points $A, B, C$ satisfying the condition and a point $N$ lying in the plane $MAB$, which, together with points $M, A, B$, forms the fourth vertex of the rectangle $MABN$. The point $N$ lies on the circle with diameter $AB$, that is, on the circumcircle of triangle $MAB$. This ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,505 |
11.5. Prove that among five arbitrary vertices of a regular (all sides and all angles of which are equal) 15-gon, there will always be three that are the vertices of an isosceles triangle. | Solution. The lengths of the sides and diagonals of a regular 15-gon can take 7 possible values. Five chosen vertices are connected by ten segments, so there will be coincidences among their lengths.
1) If the lengths of three of these segments are equal, then two of them have a common vertex, and their ends form an i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,506 |
9.1. Let the numbers $x, y, u, v$ be distinct and satisfy the relation $\frac{x+u}{x+v}=\frac{y+v}{y+u}$. Find all possible values of the sum $x+y+u+v$. | Answer. $x+y+u+v=0$.
Solution. Let's bring the difference between the left and right parts of the expression from the condition, equal to zero, to a common denominator and factor the numerator: $\frac{x+u}{x+v}-\frac{y+v}{y+u}=\frac{x u+y u+u^{2}-y v-x v-v^{2}}{(x+v)(y+u)}=\frac{(x+y)(u-v)+(u+v)(u-v)}{(x+v)(y+u)}=\fra... | x+y+u+v=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,507 |
9.2. Represent the number 100 as the sum of the maximum possible number of pairwise coprime natural numbers. Explanation: the condition means that the greatest common divisor of any two numbers used in the sum is 1. | Answer: 100 is the sum of all nine of the first prime numbers from 2 to 23 inclusive.
Solution. Indeed, $100=2+3+5+7+11+13+17+19+23$, the number of addends is 9, which is the maximum possible. If 100 were the sum of not less than ten different pairwise coprime natural numbers, then due to their pairwise coprimality, t... | 100=2+3+5+7+11+13+17+19+23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,508 |
9.5. What is the smallest number of colors needed to color all the cells of a 6 by 6 square so that in each row, column, and diagonal of the square, all cells have different colors? Explanation: a diagonal of the square is understood to mean all rows of at least two cells running diagonally from one edge of the square ... | Answer: In 7 colors.
Solution. Let's provide an example of coloring in 7 colors that satisfies the condition of the problem. Consider a 7 by 7 square, and color it in the required way using 7 colors with a known technique: the coloring of each subsequent row is obtained from the coloring of the previous row by a cycli... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,509 |
7.1. In the bakery, there are pies with two fillings (apple and cherry) and of two types (fried and baked). Prove that it is possible to buy two pies that will differ both in filling and in the method of preparation. | Solution: We have only 4 types of pies: fried with cherry, fried with apple, baked with cherry, baked with apple, which are divided into two pairs, differing from each other in both properties. If we have at least three types for sale, then two are definitely opposite, and we buy them. If there are 2 "non-opposite" typ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,510 |
7.3. A scalene triangle is divided into two parts by some line. Prove that these parts cannot be congruent figures. | Solution: It is clear that it makes sense to consider only the case when the line divides the triangle into two other triangles. Let triangle $ABC$ be divided by segment $AD$ into two equal triangles $ABD$ and $ADC$.
If in these triangles angle $B$ equals angle $C$, as corresponding equal elements, then triangle $ABC$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,512 |
7.5. In the city, there are 9 bus stops and several buses. Any two buses have no more than one common stop. Each bus has exactly three stops. What is the maximum number of buses that can be in the city | Answer: 12.
Solution: if some stop is common for 5 routes, then no two of them have any more common stops, which means there are at least $1+5*2=11$ stops, which contradicts the condition. Therefore, each stop is the intersection of no more than 4 routes, then the total number of stops made by buses is no more than 94... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,514 |
8.1. Two mountain goats, Gennady and Nikolai, arranged a competition. Gennady makes two jumps of 6 meters each in the same time it takes Nikolai to make 3 jumps of 4 meters each. The goats agreed to jump along a straight line, not to turn around until they have traveled at least 2 kilometers, and then return. Who will ... | Answer: Nikolai.
Solution: Both goats cover 12 meters in the same amount of time, but one makes jumps of 6 meters, while the other makes jumps of 4 meters. They need to jump at least 2000 meters in one direction. Nikolai can jump exactly 2000 meters, but Gennady cannot (2000 is not divisible by 6), so he will have to ... | Nikolai | Other | math-word-problem | Yes | Yes | olympiads | false | 9,515 |
8.2. In a 3 by 3 square, place nine consecutive integers such that numbers in cells adjacent by side and diagonal do not have common divisors other than 1. | Solution: for example, as follows:
| 8 | 9 | 10 |
| :--- | :--- | :--- |
| 5 | 7 | 11 |
| 6 | 13 | 12 |
Criteria: any correctly filled table without explanations - 7 points. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,516 |
8.3. Anya drew a square $A B C D$. Then she constructed an equilateral triangle $A B M$ such that the vertex $M$ ended up inside the square. The diagonal $A C$ intersects the triangle at point $K$. Prove that $C K = C M$. | Solution: Let's find the angles $\angle C K M$ and $\angle C M K$, if they turn out to be equal, then the triangle is isosceles.
$$
\begin{aligned}
\angle C K M & =180^{\circ}-\angle B K C=180^{\circ}-\left(180^{\circ}-\angle B C K-\angle C B K\right)= \\
& =\angle B C K+\angle C B K=45^{\circ}+\left(90^{\circ}-60^{\c... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,517 |
8.4. Let $p$ be an odd prime number. Prove that for some pair of distinct natural numbers $m$ and $n$, the equation $2 / p = 1 / n + 1 / m$ holds, and that such a pair of numbers is unique (up to the permutation of $n$ and $m$). | Solution: Multiply the equation by $2 n m p$, we get $4 n m=2 m p+2 n p$. Move everything to the left side, add $p^{2}$ to both sides, and group the terms on the left side, we get $(2 n-p)(2 m-p)=p^{2}$. Since $n$ and $m$ are different, $(2 n-p)$ and $(2 m-p)$ are different, and without loss of generality, $2 n-p=1, 2 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,518 |
8.5. In the city, there are 9 bus stops and several buses. Any two buses have no more than one common stop. Each bus has exactly three stops. What is the maximum number of buses that can be in the city? | Answer: 12.
Solution: if some stop is common for 5 routes, then no two of them have any more common stops, which means there are at least $1+5*2=11$ stops, which contradicts the condition. Therefore, each stop is the intersection of no more than 4 routes, then the total number of stops made by buses is no more than 94... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,519 |
9.1. The electronic clock on the building of the station shows the current hours and minutes in the format HH:MM from 00:00 to 23:59. How much time in one day will the clock display four different digits? | Answer: 10 hours 44 minutes.
Solution. Each possible combination of four digits burns on the clock for one minute. Consider separately the time of day from 00:00 to 19:59 and from 20:00 to 23:59. In the first case, the number of valid combinations according to the problem's conditions will be: 2 (the tens digit of the... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,520 |
9.2. In an arbitrary triangle $ABC$, find a point $M$ such that if circles are constructed on the segments $MA$, $MB$, and $MC$ as diameters, the lengths of their pairwise common chords will be equal. Answer. The desired point is the intersection of the angle bisectors, which is also the center of the inscribed circle ... | Solution. Let the second intersection points of the circles constructed on segments $M A$ and $M B$ be denoted by $P$, on segments $M B$ and $M C$ by $Q$, and on segments $M C$ and $M A$ by $R$. Since the angles $A P M$ and $B P M$, subtending the diameters $M A$ and $M B$ of the respective circles, are right angles, t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,521 |
9.5. Find all natural $n$ such that from sticks of lengths $1,2, \ldots, n$ (all numbers from 1 to $n$ exactly once) an equilateral triangle can be formed. All sticks must be used. | Answer. All natural $n \geq 5$, the remainders of which when divided by 6 are $0, 2, 3$ or 5.
Solution. The sum of the lengths of all sticks, equal to $\frac{n(n+1)}{2}$, must be divisible by 3, so the product $n(n+1)$ must be divisible by 6, taking into account the mutual simplicity of the factors, we get that one of... | Allnaturaln\geq5,theremaindersofwhichwhendivided60,2,3or5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,523 |
10.1. The area of the quadrilateral formed by the midpoints of the bases and diagonals of a trapezoid is four times smaller than the area of the trapezoid itself. Find the ratio of the lengths of the bases of the trapezoid. | Answer: $3: 1$.
Solution. Let the vertices of the trapezoid be $A, B, C, D$, the midpoints of the bases $A D$ and $B C$ be $K$ and $M$, and the midpoints of the lateral sides $A B$ and $C D$ be $P$ and $Q$ respectively. Let the lengths of $A D$ and $B C$ be $a$ and $b$, with $a > b$. By Thales' theorem, the midline $P... | 3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,524 |
5. Let's consider $A B C$ as the largest angle of the triangle. Denote by $M$ the point of intersection of the circles that touch the extensions of the sides $C B$ and $A C$ and pass through the vertices $A$ and $B$. The second circle forms angles equal to $C$ with the side $B C$ at points $B$ and $C$, so its arc betwe... | Answer. $(a, b)=(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)$
Solution. Without loss of generality, assume $b \geq a$. Then $b^{2} \geq b \geq a$, excluding the case $b^{2}=a, b=a=1$, we have $\frac{a^{2}+b}{b^{2}-a}>0$, hence $\frac{a^{2}+b}{b^{2}-a} \geq 1$. From the last inequality, $b-a \leq 1$, so in the case we are consi... | (,b)=(1,2),(2,1),(2,2),(2,3),(3,2),(3,3) | Number Theory | proof | Yes | Yes | olympiads | false | 9,526 |
11.1. Find all solutions in natural numbers for the equation: $x!+9=y^{3}$. | Answer. $x=6, y=9$
Solution. If $x \geq 9$ then $x$ ! is divisible by 27, $x$ ! + 9 is divisible by 9 and not divisible by 27. Then $y^{3}$ is divisible by 3, which means $y$ is divisible by 3 and $y^{3}$ is divisible by 27 - a contradiction. Therefore, $x \leq 8$. Checking the values $x=1,2,3, \ldots, 8$ gives the on... | 6,9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,527 |
11.2. Find the number of different ways to arrange all natural numbers from 1 to 9 inclusive in the cells of a 3 by 3 table, one number per cell, such that the sums of the numbers in each row and each column are equal. The table cannot be rotated or reflected. | Answer: 72 ways
Solution. Among the numbers from 1 to 9, there are 5 odd numbers. Since the sums in all rows and columns are $\frac{1}{3}(1+2+\ldots+9)=15$ - which are odd, there must be an odd number of odd numbers in each row and each column. This is only possible if one row contains three odd numbers, and the other... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,528 |
11.4. The lengths of the sides of the pentagon $A B C D E$ are equal to 1. Let points $P, Q, R, S$ be the midpoints of sides $A B, B C, C D, D E$ respectively, and points $K$ and $L$ be the midpoints of segments $P R$ and $Q S$ respectively. Find the length of segment $K L$. | Answer: $\frac{1}{4}$
Solution 1. Under $X Y$ in this solution, we will understand the vector with the origin at point $X$ and the end at point $Y$. Then $A K=\frac{1}{2}(A P+A R)=\frac{1}{2}\left(\frac{1}{2} A B+A B+B C+\frac{1}{2} C D\right)=\frac{3}{4} A B+\frac{1}{2} B C+\frac{1}{4} C D$. Similarly, $A L=\frac{1}{... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,529 |
8.1. A corner cell has been cut out of a 5 by 5 square. Cut it into six polygons of equal area along the grid lines in such a way that there is exactly one pair of identical ones. | Solution: An example of the cutting is shown in the figure. Criteria: Any correct example - 7 points.
No correct example, but it is proven that the T-shaped figure repeats -1 point.
Note: It is not difficult to notice that there are only five different polygons of area 4, which means that in a correct
| 2 | 2 | 1 | ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,530 |
8.2. In a fairy-tale country, every piglet either always lies or always tells the truth, and each piglet reliably knows whether every other piglet is a liar. One day, Nif-Nif, Naf-Naf, and Nuf-Nuf met for a cup of tea, and two of them made statements, but it is unknown who exactly said what. One of the three piglets sa... | # Answer: Two
Solution: If at least one of the statements is true, then the piglets mentioned in it are liars, which means there are at least two liars. At the same time, the one making this true statement must be telling the truth. Therefore, there are no more than two liars. In total, if at least one of the spoken p... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,531 |
8.3. In the city, there are four jewelers to whom the tsar sent 13 bags of gold. The first bag contained one gold ingot, the second - two, the third - three, ..., the
thirteenth - 13 gold ingots. One of the bags was immediately lost somewhere, and
the remaining jewelers distributed them so that each received an equal ... | Answer: $2,9,10$
Solution: In total, there were $1+2+3+\ldots+13=91$ ingots. When one bag was lost, the remaining number of ingots became a multiple of 4. The number 91 gives a remainder of 3 when divided by 4, so the lost bag must have contained either 3, 7, or 11 ingots. From the condition, we know that the bags wit... | 2,9,10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,532 |
8.4. On the segment $A B$, a point $M$ is chosen, and isosceles triangles $A M C$ and $M B N$ are constructed with bases $A M$ and $M B$ respectively. It turns out that points $B, N$, and $C$ lie on the same line, and $A B = B C$. The perpendicular from $B$ to segment $A C$ intersects segment $C M$ at point $H$. Prove ... | Solution: Triangle $ABC$ is isosceles. Let the base angles be $x$, and the remaining angle be $y$. Thus, $2x + y = 180^{\circ}$.
Triangle $AMC$ is isosceles, so $\angle AMC = \angle CAM = x$.
Triangle $MNB$ is isosceles, so $\angle MBN = \angle BMN = y$.
Since points $A, M, B$ lie on a straight line, $\angle AMC + \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,533 |
8.5. An archipelago consists of $n$ islands, between any two of which there is a ferry. The fare on each ferry is the same in both directions, but the cost is different on any two different ferries. A traveler wants to fly to one of the islands by helicopter and then sail on $n-1$ ferries in such a way that each time h... | Solution 1: Remove all ferries, and then start reintroducing them one by one in ascending order of price (the first one being the cheapest, the second one being the cheapest of the remaining, and so on). At each moment, we will write down the maximum number of ferries that can be sequentially taken from each island, st... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,534 |
7.1. Provide an example of a natural number that is a multiple of 2020, such that the sum of its digits is also a multiple of 2020. | Solution: For example, the number 20202020...2020, where the fragment 2020 repeats 505 times, fits. Such a number is obviously divisible by 2020, and the sum of its digits is $505 \times 4=2020$.
Criteria: Any correct answer with or without verification - 7 points.
Correctly conceived example, but with a calculation ... | 2020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,535 |
7.2. Lada and Lera each drew a triangle. Lada chose two angles of her triangle, and Lera noticed that among the angles of her triangle, there was one equal to the sum of these two angles. After this, Lada chose another pair of angles in her triangle, and Lera again found an angle in her triangle equal to the sum of the... | Solution: Notice that in the two pairs of angles chosen by Lada, there is exactly one common angle (otherwise, there would be at least 4 angles in the triangle). Let's call the angle that participates in both pairs $\alpha$, and the other two angles of Lada's triangle $\beta$ and $\gamma$. In Lera's triangle, there is ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,536 |
7.3. In a certain city, there is a street in the shape of a regular hexagon, with a person living at each of its vertices. Each resident is either a knight, who always tells the truth, or a liar, who always lies. Each of them knows their neighbors and the names of the others (but not where they live), however, no one k... | Solution: Let the other people be called A, B, C, D, E, and Lёshka knows that his neighbors are called A and E. Then he will send two letters to people B and C with questions about people A and E.
If a person answers "Yes" to both letters, he is lying, because only Lёshka himself lives simultaneously next to A and D. ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,537 |
7.4. Two swimmers started simultaneously on adjacent lanes in a 50-meter pool for a 1 km distance. Each swam at their own constant speed. The first swimmer, upon completing the swim, got out of the pool. The second continued swimming the distance. In total, the swimmers met 16 times (when the first catches up with the ... | Answer: In $3 / 2.5 / 3$ or 5 times.
Solution: When the first swimmer swam the pool in one direction (which happened 20 times), he met the second swimmer exactly once (possibly at the edge). The meeting at the start is not counted, so there were 19 minus the number of meetings at the edge before the first swimmer fini... | 3/2.5/3or5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,538 |
7.5. An archipelago consists of $n$ islands, between any two of which there is a ferry. The fare on each ferry is the same in both directions, but the cost is different on any two different ferries. A traveler wants to fly to one of the islands by helicopter and then sail on $n-1$ ferries in such a way that each time h... | Solution 1: Remove all ferries, and then start reintroducing them one by one in ascending order of price (the first one being the cheapest, the second one being the cheapest of the remaining, and so on). At each moment, we will write down the maximum number of ferries that can be sequentially taken from each island, st... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,539 |
7.1. A hundred elves of different ages stood in a circle. Each of them who turned out to be older than both of their neighbors closed their left eye. Each one who turned out to be younger than both neighbors closed their right eye. In the end, it turned out that all the elves were standing with one eye closed. Provide ... | Solution. Let's number the seats in the circle from 1 to 100, and let the very short elves stand at the even positions, while the very tall elves stand at the odd positions. Then it is clear that each will either be shorter than both neighbors or taller, and the condition of the problem will be satisfied.
Criteria. An... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,540 |
7.2. On a certain island, only knights, who always tell the truth, and liars, who always lie, live. One day, 99 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: 9 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 99th in the sequence. Then, the 10 people with numbers from 1 to 10 are li... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,541 |
7.3. Professor Fortran wrote a program that takes a natural number, multiplies all its digits, and then either subtracts this product from the initial number or adds it to the initial number at random. Thus, from the number 239, the program can get either $239-2 \cdot 3 \cdot 9=239-54=185$ or $239+54=293$. The result i... | Solution. The product of the digits of the number 141 is 4, so the next number to appear on the screen will be either 145 or 137.
In the first case, all subsequent numbers will be divisible by 5, and the number 141 will not appear. Indeed, a number is divisible by 5 if and only if its last digit is 0 or 5. This means ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,542 |
7.4. Vera Aleksandrovna urgently needed to cut out three 20-sided polygons (not necessarily identical) from one rectangular sheet of paper. She can take this sheet and cut it along a straight line into two parts. Then take one of the resulting parts and cut it along a straight line. Then take one of the available piece... | Answer: 50 cuts.
Solution: With each cut, the total number of paper pieces increases by 1 (one piece turns into two new pieces), so after $n$ cuts, there will be $(n+1)$ pieces of paper. Let's calculate how many vertices all the pieces together can have after $n$ cuts. With each cut, the total number of vertices incre... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,543 |
7.5. At Anton Pavlovich's house, there are 198 cats. All of them have different weights and different speeds. It is known that in any group of 100 cats, the heaviest of them is also the fastest of them. Prove that it is possible to choose a group of 100 cats such that the lightest of them will also be the slowest of th... | Solution. Consider the fattest cat. No matter which 99 cats are added to it, it will be the fastest among them. This means that it is both fatter and faster than each of the other cats. Assign this cat the number 1 and lock it on the balcony. From the remaining cats, choose the fattest one again. No matter which 99 cat... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,544 |
10.1. Find all triples of real numbers $(x, y, z)$ that are solutions to the system of equations $\left\{\begin{array}{l}x^{2}+y+z=1, \\ x+y^{2}+z=1, \\ x+y+z^{2}=1 .\end{array}\right.$ | Answer. $(1,0,0),(0,1,0),(0,0,1),(-1-\sqrt{2},-1-\sqrt{2},-1-\sqrt{2}),(-1+\sqrt{2},-1+\sqrt{2},-1+\sqrt{2})$.
Solution. By factoring the difference of the first and second equations, we get $(x-y)(x+y-1)=0$, which means that for the pair of variables $x, y$ from the triple of solutions of the system $(x, y, z)$, one ... | (1,0,0),(0,1,0),(0,0,1),(-1-\sqrt{2},-1-\sqrt{2},-1-\sqrt{2}),(-1+\sqrt{2},-1+\sqrt{2},-1+\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,545 |
10.3. Let $\mathrm{M}$ be the smallest set of numbers such that a) $\mathrm{M}$ contains 1, b) if $\mathrm{M}$ contains a number $x$, then $\mathrm{M}$ must also contain the numbers $\frac{1}{1+x}, \frac{x}{1+x}$. Exactly which numbers does $\mathrm{M}$ consist of? | Answer. M is precisely the set of all positive rational numbers not exceeding 1.
Solution. On the one hand, it is clear that if $x$ is a positive rational number, then both numbers $\frac{1}{1+x}, \frac{x}{1+x}$ are also rational and lie in the interval (0,1), so the minimal set $\mathrm{M}$ consisting only of numbers... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,546 |
9.1. Which natural numbers can be represented as the fraction $\frac{x^{2}}{y^{3}}$, where $x$ and $y$ are some natural numbers? | Answer. Any natural number.
Solution. Let $n$ be any natural number. Set $x=n^{2}, y=n$, then $\frac{x^{2}}{y^{3}}=\frac{n^{4}}{n^{3}}=n$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,547 |
9.2. Solve the system of equations in real numbers: $\left\{\begin{array}{l}x^{2}=4 y^{2}+19, \\ x y+2 y^{2}=18 .\end{array}\right.$ | Answer. $(x, y)= \pm\left(\frac{55}{\sqrt{91}}, \frac{18}{\sqrt{91}}\right)$.
Solution. Factorize both equations: $\left\{\begin{array}{l}(x-2 y)(x+2 y)=19, \\ y(x+2 y)=18 .\end{array}\right.$
Since $x+2 y \neq 0$, dividing one by the other, we get: $18 x=55 y, y=\frac{18}{55} x$, substituting into the second equatio... | (x,y)=\(\frac{55}{\sqrt{91}},\frac{18}{\sqrt{91}}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,548 |
9.3. In triangle $A B C$, the measure of angle $A$ is 30 degrees, and the length of the median drawn from vertex $B$ is equal to the length of the altitude drawn from vertex $C$. Find the measures of angles $B$ and $C$. | Answer. $\angle B=90^{\circ}, \angle C=60^{\circ}$.
Solution. Given that $\angle A=30^{\circ}$, the height drawn from vertex $C$ is half of $A C$, therefore, the median drawn from vertex $B$ is also half of $A C$. It follows immediately that vertex $B$ lies on the circle with diameter $A C$, hence, $\angle B=90^{\circ... | \angleB=90,\angleC=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,549 |
9.4. Prove that from any 18 consecutive three-digit numbers, it is always possible to select a number that is divisible by the sum of its digits. Provide an example of 17 consecutive three-digit numbers, none of which is divisible by the sum of its digits. | Solution. Among 18 consecutive three-digit numbers, there are always exactly two numbers divisible by 9, their difference is 9, so one of them must be even, meaning it is divisible by 18. On the other hand, the sum of the digits of a three-digit number, except for 999, does not exceed 26. Therefore, the sum of the digi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,550 |
9.5. One hundred numbers are written in a circle with a total sum of 100. The sum of any 6 consecutive numbers does not exceed 6. One of the written numbers is 6.
Find the other numbers. | Answer. Considering the first number mentioned in the condition is 6, all numbers with odd indices are equal to 6, and all numbers with even indices are equal to -4. Solution. Let's denote the numbers in order as $a_{1}, a_{2}, \ldots, a_{100}$. Adding all 100 inequalities $a_{i}+a_{i+1}+\ldots+a_{i+5} \leq 6, i=1,2, \... | 6,-4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,551 |
10.2. Find the number of different arrangements of 8 rooks on different white squares of an 8 by 8 chessboard such that no rook attacks another. A chess rook attacks all squares in the row and column it occupies. | Answer. $576=(4!)^{2}$.
Solution. From the fact that no rook attacks another, it follows that there is exactly one rook in each row and each column of the board. Note that the white cells of the columns with odd numbers (first type) are located in rows with even numbers, while the white cells of the columns with even ... | 576=(4!)^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,552 |
10.5. Can four different natural numbers be found such that each of them is divisible by the difference of any two of the remaining three numbers? | Answer. It is impossible.
Solution. Suppose the required quadruples of numbers exist, and choose among them the quadruple with the smallest sum. First, assume that one of our four numbers is odd. Among the remaining three, choose two of the same parity. According to the problem's condition, the selected odd number mus... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,553 |
8.1. Find all three-digit numbers divisible by 4, in which the ratio of the first digit to the second is equal to the ratio of the second digit to the third. | Answer: $124,248,444,888,964$.
Solution. Let the desired number be $n=\overline{a b c}$. From the condition, it follows that the square of the middle digit is equal to the product of the extreme digits, and the last digit is even. From this, it immediately follows that the middle digit is also even. Let's consider the... | 124,248,444,888,964 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,554 |
8.2. From two cities, the distance between which is 105 km, two pedestrians set out simultaneously towards each other at constant speeds and met after 7.5 hours. Determine the speed of each of them, knowing that if the first walked 1.5 times faster and the second 2 times slower, they would have met after $8 \frac{1}{13... | Answer: 6 and 8 km per hour.
Solution: Let their speeds be $x$ and $y$ km per hour, respectively. From the condition, we get: $\frac{15}{2}(x+y)=105, \frac{105}{13}\left(\frac{3}{2} x+\frac{1}{2} y\right)=105$, from which $\quad x=6, y=8$. | 6,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,555 |
8.3. What can be the sum of the digits of a number divisible by 7? | Answer. Any natural number, greater than or equal to 2.
Solution. Note that the numbers 21 and 1001 are divisible by 7, and the sums of their digits are 3 and 2, respectively. Therefore, to get a sum of digits equal to an even number $n$, we need to take a number whose decimal representation consists of $\frac{n}{2}$ ... | Anynatural,greaterthanorequalto2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,556 |
8.5. On a 10 by 10 cell board, some 10 cells are marked. For what largest $n$ is it always possible, regardless of which cells are marked, to find a rectangle consisting of several cells, the perimeter of which will be at least $n$? The length or width of the rectangle can be equal to one cell. | Answer: $n=20$.
Solution. First, we prove that when $n=20$, it is always possible to find such a rectangle. Suppose 10 cells are colored. If there is a column or row without colored cells, then a rectangle of $1 \times 9$ with a perimeter of 20 (or even $1 \times 10$ with a perimeter of 22) can be cut from it. Now, le... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,557 |
9.1. Let's call a four-digit number $\overline{a b c d}$ curious if the sum of the two-digit numbers $\overline{a b}$ and $\overline{c d}$ equals the two-digit number $\overline{b c}$. For example, the number 1978 is curious because 19+78=97. Find the number of curious numbers. Answer. 36. | Solution. Let's form the equation $\overline{a b}+\overline{c d}=10(a+c)+b+d=\overline{b c}=10 b+c$, from which we get $10 a+9 c+d=9 b$. The difference $9(b-c)$ is divisible by 9, so the sum $10 a+d=9(b-c)$ is also divisible by 9, which is equivalent to the divisibility by 9 of the sum $a+d=9(b-c-a)$. The sum of two di... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,558 |
9.3. We consider all possible tilings of an 8 by 8 chessboard with dominoes, each consisting of two adjacent squares. Determine the maximum natural number \( n \) such that for any tiling of the 8 by 8 board with dominoes, one can find some rectangle composed of \( n \) squares of the board that does not contain any do... | Answer. $n=4$.
Solution. 1) We will prove that $n \leq 4$. Consider the following tiling of an 8 by 8 chessboard with dominoes. Divide the board into 2 by 2 squares, color each of them in red and blue in a checkerboard (relative to the 4 by 4 board) pattern, and divide the red squares into pairs of horizontal dominoes... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,559 |
9.5. On a circle, $n>1$ points, called positions, are marked, dividing it into equal arcs. The positions are numbered clockwise from 0 to $n-1$. Vasya places a chip in one of them. Then the following actions, called moves, are repeated an unlimited number of times: Petya names some natural number, and Vasya moves the c... | Answer. For $n=2^{k}$, for all natural $k$.
Solution. 1) Let first $n=2^{k}$, for some natural $k$. A winning strategy for Petya is as follows: if after Vasya's move the chip is on a position with number $m=2^{a}(2 b+1)$, then Petya names the number $m=2^{a}$. After this, Vasya will move the chip either to the positio... | 2^{k} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,560 |
9.1. What is the minimum sum of digits in the decimal representation of the number $f(n)=17 n^{2}-11 n+1$, where $n$ runs through all natural numbers?
# Answer. 2. | Solution. When $n=8$, the number $f(n)$ is 1001, so the sum of its digits is 2. If $f(n)$ for some $n$ had a sum of digits equal to 1, it would have the form $100, \ldots 00$ and would either be equal to 1 or divisible by 10. The function of a real variable $f(x)$ reaches its minimum at $x=\frac{11}{34}1$, and $f(n)$ c... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,561 |
9.2. What digit can the number $f(x)=[x]+[3 x]+[6 x]$ end with, where $x$ is an arbitrary positive real number? Here $[x]$ denotes the integer part of the number $x$, that is, the greatest integer not exceeding $x$.
| Answer: $0,1,3,4,6,7$.
Solution. Let $x=[x]+a$, where $0 \leq a<1$ is the fractional part of $x$. Then it is easy to understand that $f(x)=[x]+[3 x]+[6 x]=10[x]+[a]+[3 a]+[6 a]$, so the last digit does not depend on the integer part of $x$. Consider the possible values of its fractional part, dividing the interval $[0... | 0,1,3,4,6,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,562 |
9.3. Inside an isosceles triangle $\mathrm{ABC}$ with equal sides $\mathrm{AB}=\mathrm{BC}$ and an angle of 80 degrees at vertex $\mathrm{B}$, a point $\mathrm{M}$ is taken such that the angle
$\mathrm{MAC}$ is 10 degrees, and the angle $\mathrm{MCA}$ is 30 degrees. Find the measure of angle $\mathrm{AMB}$. | Answer: 70 degrees.
Solution. Draw a perpendicular from vertex B to side AC, and denote the points of its intersection with lines AC and CM as P and T, respectively. Since angle MAC is less than angle MCA, side CM of triangle MAC is shorter than side AM, so point M is closer to C than to A, and therefore T lies on the... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,563 |
8.2. It is known that $70 \%$ of mathematicians who have moved to IT regret their change of activity. At the same time, only $7 \%$ of all people who have moved to IT regret the change. What percentage of those who have moved to IT are mathematicians, if only they regret the change of activity? | Solution. Let a total of $x$ people went into IT, and $y$ of them are mathematicians. According to the condition of the change in activity, on the one hand, $0.07 x$ people regret, and on the other - $0.7 y$. From this, we get that $0.07 x=0.7 y$, from which $y / x=0.1$, that is, $10 \%$.
Criteria. Only the answer - 1... | 10 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,566 |
8.3. Given two numbers $a$ and $b$. It is known that out of the four numbers $a b, a+b, a-b, a / b$ exactly three are equal to each other ( $b$ is not equal to 0 ). Find $a$ and $b$. | Solution. Obviously, $a+b \neq a-b$, so exactly one of these two numbers is different from all the others. Then $a b=a / b$. If $a=0$, then the numbers from the condition are equal to 0, $b, -b, 0$ respectively, and the condition cannot be satisfied. Then $a \neq 0$, and it can be canceled, from which $b=1 / b$, which ... | =\1/2,b=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,567 |
8.4. Once in medieval England, 2021 people gathered around a round table. Each of them was either a knight, who always told the truth, or a liar, who always lied, and among those present, there was at least one knight and at least one liar. Moreover, it is known that on that day there was a dense fog, and each person c... | Solution. Suppose everyone answered "no". Then let's find two neighboring knight and liar (for example, we can take a knight who exists by the condition and move clockwise until we find a liar). Consider 26 people $\mathrm{A}_{1}, \ldots \mathrm{A}_{26}$, with the central ones being these very knight $\mathrm{A}_{13}$ ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 9,568 |
8.5. On the table, there are $n>3$ identical coins arranged in a circle, which can lie either heads up or tails up. If next to some coin there are two heads or two tails, then this coin can be flipped. Such an operation is allowed to be performed an unlimited number of times. For which $n$ is it possible, regardless of... | Solution. Let $n$ be odd. Divide the entire circle into groups of consecutive identical coins. There will be a group of an odd number of coins (otherwise, the total number of coins in the circle would be even as the sum of even numbers). Consider this group. Flip the second, fourth, and so on - all the even-numbered co... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,569 |
7.1. Form three irreducible (not necessarily proper) fractions, the product of which is 1, using six numbers from the set $\{1,2,3,4,5,6,7,8,9\}$ as numerators and denominators. (Each number can be used once or not at all). | Answer: for example, $3 / 8, 4 / 9, 6 / 1$ or $3 / 4, 6 / 1, 2 / 9$.
Path to solution: Obviously, 5 and 7 cannot be used because there would be nothing to cancel them with, the remaining numbers consist of twos and threes in their factorization, by distributing them, we construct the example.
Criterion: Any correct s... | \frac{3}{8},\frac{4}{9},\frac{6}{1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,570 |
7.2. Olya, Oleg, Polya, and Pasha participated in a competition and took the first 4 places. After the competition, Polya left immediately, and the others made 2 statements, with only one child telling the truth both times, while the others lied both times. Each said that they took first place. In addition, Olya said t... | Answer: Oleg - I, Olya - II, Polya - III, Pasha - IV
Solution: Olya and Pasha lied, as each of them made two contradictory statements (placing themselves and someone else simultaneously in the first place). Therefore, Oleg told the truth, he took the first place, and Olya took the second place. Olya lied when she said... | Oleg-I,Olya-II,Polya-III,Pasha-IV | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,571 |
7.3. It is known that all krakozyabrs have horns or wings (possibly both). According to the results of the world census of krakozyabrs, it turned out that $20 \%$ of the krakozyabrs with horns also have wings, and $25 \%$ of the krakozyabrs with wings also have horns. How many krakozyabrs are left in the world, if it i... | Answer: 32.
Solution: Let $n$ be the number of krakozyabrs with both wings and horns. Then the number of horned krakozyabrs is $-5 n$, and the number of winged krakozyabrs is $-4 n$. Using the principle of inclusion-exclusion, the total number of krakozyabrs is $5 n + 4 n - n = 8 n$. There is only one integer between ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,572 |
7.5. Vasya wrote down all natural numbers from 1 to 2014 on the board, after which Petya erased 1006 of them. Prove that among the remaining numbers, there will be two such that one is a divisor of the other. | Solution: Consider the greatest odd divisors of the remaining numbers. Among the numbers from 1 to 2014, there are exactly 1007 different greatest odd divisors (the numbers $1, 3, \ldots, 2013$). And we have 1008 numbers left. Therefore, by the Pigeonhole Principle, two of the remaining numbers have the same greatest o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,574 |
8.1. Olya, Oleg, and Pasha took the first three places in a competition. Each of them said they took first place. Olya, in addition to this, said that all odd places were taken by boys, and Oleg said that Olya was wrong. It is known that the children either always lie or always tell the truth. Who took which place? | Answer: I - Oleg, II - Pasha, III - Olya.
Solution: Olya could not have told the truth because according to her statement, she took the I place, which could only be occupied by a boy. Therefore, she is not first. Moreover, if she took the II place, her second statement would be true, which is not allowed. Therefore, O... | I-Oleg,II-Pasha,III-Olya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,575 |
8.2. Write down 20 numbers in a row such that the sum of any three consecutive numbers is positive, while the sum of all 20 numbers is negative. | Answer: for example, $-3 ;-3 ; 6.5 ;-3 ;-3 ; 6.5 ;-3 ;-3 ; 6.5 ;-3 ;-3 ; 6.5 ;-3 ;-3 ; 6.5 ;-3 ;-3 ; 6.5 ;-3 ;-3$.
Path to solution: one can assume that the sequence will be periodic
$$
a, a, b, a, a, b, a, a, b, a, a, b, a, a, b, a, a, b, a, a
$$
From this, it follows that $2 a+b>0$ and $14 a+6 b<0$. It is not diff... | -3;-3;6.5;-3;-3;6.5;-3;-3;6.5;-3;-3;6.5;-3;-3;6.5;-3;-3;6.5;-3;-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,576 |
8.3. What is the maximum number of rooks that can be placed on an 8x8 chessboard so that each rook attacks no more than one other? A rook attacks all squares on the same row and column it occupies. | Answer: 10.
Solution: It is clear that in each column and row there are no more than two rooks. Let $k$ rooks be placed while satisfying the condition. On each square where a rook is placed, write the number 0. In each of the 8 columns, perform the following operation: if there are two numbers in the column, add 1 to ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,577 |
8.4. On the sides $AB, BC$, and $AC$ of triangle $ABC$, points $C_{1}, A_{1}, B_{1}$ are chosen arbitrarily. Let $K_{1}, K_{2}, K_{3}$ be the midpoints of $A A_{1}, B B_{1}, C C_{1}$. Prove that these points cannot lie on the same line. | Solution: Assume the opposite. First, we will prove that points $K_{1}, K_{2}, K_{3}$ lie on the midlines of the triangle (first part of the proof). If this is true, then $K_{1}, K_{2}, K_{3}$ lie, firstly, on different sides of the triangle, and secondly, on one straight line. Therefore, there exists a line intersecti... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,578 |
8.5. In a certain Italian city, 20 mafia clans are conducting their dark affairs, and it is known that each clan is at odds with at least 14 others. Will there always be 4 clans, each pair of which are at odds with each other? | Answer: Yes, always.
Solution: Gather all clans in one room. Choose one of them, denote it as 1, and leave only this clan and the clans that are enemies with it in the room. This means no more than 5 clans have left the room. Choose a new clan, denote it as 2, and leave only it and its enemies in the room. Again, no m... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,579 |
9.1. At the school for slackers, a competition on cheating and giving hints was organized. It is known that $75 \%$ of the students did not show up for the competition at all, and all the rest participated in at least one of the competitions. When the results were announced, it turned out that $10 \%$ of all those who ... | Answer: 200.
Solution. Let the number of students in our school be $n$ people. $\frac{n}{4}$ people attended the competitions. Taking into account $\frac{n}{40}$ people who participated in both competitions, the number of participants in the competition with hints was $\frac{3}{5}\left(\frac{n}{4}+\frac{n}{40}\right)=... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,580 |
9.5. Find all natural $n$ such that for any partition of the set of all natural numbers from 1 to $n$ inclusive into two subsets, there will be two distinct numbers in one of the subsets whose sum is a square of a natural number. | Answer. $n \geq 15$.
Solution. Let's give the process a scientific flavor. We will construct a graph where the vertices are the numbers from 1 to $n$, and vertices $a$ and $b$ are connected by an edge if and only if the sum of these numbers is a perfect square of a natural number. For any partition of the numbers from... | n\geq15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,581 |
10.1. A motorist and a cyclist set off from points A and B, not at the same time, towards each other. Meeting at point C, they immediately turned around and continued with the same speeds. Upon reaching their respective points A and B, they turned around again and met for the second time at point D. Here they turned ar... | Answer: At point C.
Solution. Let the speeds of the motorist and the cyclist be $x$ and $y$ respectively. Suppose the motorist left earlier than the cyclist by $t$ hours and, by the time the cyclist set off, was at point P, a distance of $xt$ from A. After meeting at C and turning around, when the cyclist returned to ... | C | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,582 |
10.2. Find all values of parameters $a, b, c$, for which the system of equations $\left\{\begin{array}{l}a x+b y=c \\ b x+c y=a \\ c x+a y=b,\end{array}\right\}$
has at least one negative solution (when $x, y<0$). | Answer. When $a+b+c=0$.
Solution. Assume the contrary, that $a+b+c \neq 0$ and the system has a negative solution. Add all the equations: $(a+b+c)(x+y)=a+b+c$ and divide by $a+b+c \neq 0$, we get
$x+y=1$, which is impossible for $x, y<0$. Therefore, the condition $a+b+c=0$ is necessary. If this condition is satisfied,... | +b+=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,583 |
11.1. A pedestrian left point A for point B, which are 40 km apart, at 4:00 AM, and a cyclist set off from A at 7:20 AM, catching up with the pedestrian exactly halfway between A and B, after which both continued their journey. A second cyclist, traveling at the same speed as the first, left B for A at 8:30 AM and met ... | Answer: 5 km/h and 30 km/h.
Solution: Let the speeds of the pedestrian and the cyclists be $x$ and $y$ km/h, respectively. By 7:20 AM, the pedestrian has walked $\frac{10}{3} \cdot x$ km. After that, the cyclist started chasing him and caught up after $\frac{10 x}{3(y-x)}$ hours, during which the pedestrian continued ... | x=5,y=30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,584 |
11.2. On the sides $A B$ and $A C$ of an equilateral triangle $A B C$ with side length 10, points $P$ and $Q$ are taken such that the segment $P Q$ is tangent to the inscribed circle of the triangle and its length is 4. Find the area of triangle $A P Q$. | Answer: $\frac{5}{\sqrt{3}}$.
Solution: Let the lengths of segments $A P$ and $A Q$ be $x$ and $y$, respectively, and the point of tangency of segment $P Q$ with the inscribed circle of the triangle be $S$. The points of tangency of $A B$ and $A C$ with the inscribed circle are denoted as $R$ and $T$. By the cosine ru... | \frac{5}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,585 |
11.3. Find the set formed by the solutions of the system of equations $\left\{\begin{array}{l}a x+y=2 a+3, \\ x-a y=a+4,\end{array}\right\}$ for all possible values of the parameter $a$. (For each value of $a$, a solution $(x, y)$ of the given system is found, and all these solutions together form the desired set of po... | Answer. A circle with center $(3,1)$ and radius $\sqrt{5}$, with the point $(2,-1)$ removed.
Solution. Consider an arbitrary solution $(x, y)$ of the given system. If $x \neq 2$, express $a$ from the first equation in terms of $x, y: a=\frac{y-3}{2-x}$. Similarly, if $y \neq-1$, express $a$ from the second equation in... | (x-3)^{2}+(y-1)^{2}=5, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,586 |
11.5. Find all solutions in natural numbers of the equation: $2^{x}+3^{y}=z^{2}$. | Answer: $x=4, y=2, z=5$.
Solution. 1) The power of two is not divisible by 3, so the left side of the equation is not divisible by 3, hence $z^{2}$ and $z$ are not divisible by 3. Therefore, the remainder of the division of the right side of the equation by 3 is 1.
2) Consider the left side modulo 3, it is equal to $... | 4,2,5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,587 |
7.2. Ellie and Toto painted daisies in the field. On the first day, Ellie painted one fourteenth of the entire field. On the second day, she painted twice as much as on the first day, and on the third day, she painted twice as much as on the second day. Toto, in total, painted 7000 daisies. How many daisies were there ... | Solution. Let there be $n$ daisies in the field. Then Ellie painted $n / 14 + 2 n / 14 + 4 n / 14 = 7 n / 14 = n / 2$ daisies in total. This means Toto also painted half of the field, from which it follows that half of the field is 7000 daisies, and the entire field is 14000.
Criteria. Only the answer - 1 point.
The ... | 14000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,589 |
7.3. Three eleventh-graders played tic-tac-toe against two ninth-graders (in each match, an eleventh-grader played against a ninth-grader). It is known that Veronika won against Rita, then Yulia won against Svetlana, and Veronika - against Maria, and finally, Maria won against Yulia. What were the names of the eleventh... | Solution. Veronica won against Rita and Maria. This means Rita and Maria are from the same class. Maria won against Yulia, so Yulia is not in the same class as Maria, which means she is with Veronica. Similarly, Yulia won against Svetlana, so Svetlana is also with Maria. Therefore, we get that Veronica and Yulia are in... | Rita,Maria,Svetlana | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,590 |
7.4. Anton from the village was given several zucchini, and he decided to give them to his friends. He gave half of the received zucchini to Arina, and a third (also from the received amount) to Vera. It turned out that after this, the number of zucchini Arina had became a square of some natural number, and the number ... | Solution. Let Anton receive $n$ zucchinis. Since both half and a third of $n$ are integers, $n$ is divisible by 6, that is, $n=6k$ for some natural number $k$. Then it is known that $3k$ is the square of a natural number, and $2k$ is a cube. Let $k=2^p 3^q m$. In other words, let $p$ and $q$ be the highest powers of tw... | 648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,591 |
7.5. At a party, 20 people arrived. It is known that each of them has exactly 14 friends among those who came (friendship is mutual). In addition, during the party, 10 people went out to the balcony, and it turned out that all of them are friends with each other. Prove that all the people who came to the party can be d... | Solution. Let each pair of friends give each other a slap on the back. Fix ten of them, any two of whom are friends (those who went out on the balcony). Call them blue, and the other ten - green. Each of the blue ones is acquainted with nine other blue ones and, therefore, with five green ones. Thus, in total, the blue... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,592 |
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