problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
10.1. What two digits need to be appended to the right of the number 2013 so that the resulting six-digit number is divisible by 101? Find all possible solutions. | Answer: 94, the obtained number will be equal to 201394.
Solution: The remainder of dividing the number $\overline{2013 x y}$ by 101 is $\overline{x y}+7$ and this must be divisible by 101. This is greater than 0 but less than 202, so $\overline{x y}+7=101, \overline{x y}=94, x=9, y=4$.
Grading: Just the answer with ... | 94 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,593 |
10.2. Solve the equation: $\sqrt[3]{20-x}-\sqrt[3]{13-x}=\sqrt[3]{7}$. | Answer: 20 and 13.
Solution. If everything is simply raised to the cube and like terms are combined, we get $\sqrt[3]{20-x} \sqrt[3]{13-x}(\sqrt[3]{20-x}-\sqrt[3]{13-x})=0$. The bracket is not equal to 0, so either $x=20$ or $x=13$.
Evaluation. Just the answer with verification: 1 point. Loss of a solution - minus 3 ... | 2013 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,594 |
10.3. Circles with centers $O_{1}$ and $O_{2}$ intersect at two points $A$ and $B$. Let $P$ and $Q$ be the points of intersection of the circumcircle of triangle $O_{1} A_{2}$ with the first and second circles, respectively. Prove that the segments $O_{1} Q$ and $O_{2} P$ intersect at point $B$. | Solution. It is sufficient to prove that point $B$ belongs to segments $O_{1} Q$ and $O_{2} P$.
Triangles $O_{2} O_{1} B$ and $O_{2} O_{1} A$ are equal by three sides, so $\angle O_{2} O_{1} B=\angle O_{2} O_{1} A$. On the other hand, segments $O_{2} A$ and $O_{2} Q$ are equal, as radii, so $\angle O_{2} O_{1} A=\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,595 |
10.4. On the cells of an 8 by 8 board, chips are placed such that for each chip, the row or column of the board in which it lies contains only one chip. What is the maximum possible number of chips on the board? | Answer: 14.
Solution: Let's match each chip to the row or column of the board in which it is the only one. If it is the only one in both, we match it to the row. From the condition, it follows that different chips are matched to different rows and columns. If not all rows and columns are matched, then their total numb... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,596 |
10.5. In an acute-angled triangle $A B C$, points $A_{1}, B_{1}, C_{1}$ are the feet of the altitudes dropped from vertices $A, B, C$ respectively, and $H$ is the orthocenter. Point $M$ is the midpoint of $A H$, $Q$ is the intersection point of segments $B H$ and $A_{1} C_{1}$, and $P$ is the intersection point of line... | Solution. It is sufficient to show that $P Q$ is parallel to $A A_{1}$. The latter is equivalent to $A P: P C_{1}=A_{1} Q: Q C_{1}$. It is a well-known fact that the triangles $A B C, A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$ are similar to each other, i.e., the original triangle and the three triangles, each of whi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,597 |
7.1. Is it possible that 1000 participants in the olympiad will solve this problem correctly, and among them, there will be 43 more boys than girls? | Answer: No.
Solution: Let $x$ be the number of girls who solved this problem. Then $x+43$ boys solved it. Therefore, in total, $2x+43=1000$ people solved it. But then $2x=957$ - an odd number, which is a contradiction.
Criteria: 6 points - no explanation of where the equation comes from. 6 points - an arithmetic erro... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,598 |
7.2. Lada and Lera each thought of a natural number. If Lada's number is decreased by $5 \%$, and Lera's number is increased by $5 \%$, the results will be equal. It turned out that the numbers they thought of are the smallest with this property. What numbers did the girls think of? | Answer: 21 and 19.
Solution: Let Lada have guessed the number $a$, and Lera the number $b$. Then the new numbers $0.95 a$ and $1.05 b$ are equal, that is, $0.95 a = 1.05 b$. Simplifying this equation, we get $19 a = 21 b$. The left side is divisible by 19, so the right side must also be divisible by 19. Since 21 and 1... | 2119 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,599 |
7.3. Anna has a 2015 by 2015 grid, in which she wrote real numbers. It turned out that in any three cells forming a corner (see the figure, the corner can be rotated), the sum of the numbers is 3. Prove that Anna put 1 in all cells.
$ and drop a perpendicular from $C$ to the line $AO$. It will fall on point $H$, and $O$ will lie between $H$ and $A$, since $\angle AOC$ is obtuse. Let $L_1$ be an arbitrary point between $O$ and $H$. Const... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,602 |
11.3. Is it possible to place ten consecutive natural numbers at the vertices and on the edges of a regular tetrahedron so that for each edge, the sum of the three numbers on the edge and at its ends is constant? | Answer: No.
Solution. Suppose the required arrangement is possible. Let the sum of the three numbers on each edge and at its ends be $S$. Adding these sums over all edges, we get 6S - an even number. On the other hand, each number on an edge is counted once in the sum $6 \mathrm{~S}$, and each number at a vertex is co... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,604 |
11.5. At a diplomatic reception, there are 99 persons, each of whom has heard of no fewer than $n$ other attendees. If A has heard of B, it does not automatically mean that B has heard of A. For what minimum $n$ is it guaranteed that there will be two attendees who have heard of each other? | Answer. When $n=50$.
Solution. Let's call a situation where one of the guests has heard about another a half-acquaintance. If each guest has heard about at least 50 other participants at the reception, then there are at least $99 \cdot 50$ half-acquaintances, which is more than the total number of pairs of guests at t... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,605 |
10.3. Find all natural $n$, for which on a square grid of size $n$ by $n$ cells, it is possible to mark $n$ cells, each in different rows and different columns, that can be sequentially visited by a knight's move in chess, starting from some cell, without landing on the same cell twice, and returning to the starting ce... | Answer. $n=4$.
Solution. An example for $n=4$ is not difficult:
We will prove that for $n \neq 4$, the required set of cells does not exist.
Suppose that for a given $n$, it is possible t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,607 |
10.5. On the segment AB, a point M is marked, different from A and B. A point C is chosen on one side of the line AB, and points D and E are chosen on the other side such that triangles ABC, AMD, and MBE are equilateral. Let P, Q, R be the points of intersection of the medians of triangles ABC, AMD, and MBE, respective... | Solution. a) Let $T$ be the intersection point of the lines AQ and BR. The quadrilateral APBT is a rhombus with angles of $60^{\circ}$ and $120^{\circ}$ at vertices A and P, and triangles APT and BPT are equilateral. Note that $A P=\frac{A B}{\sqrt{3}}, A Q=\frac{A M}{\sqrt{3}}, B R=\frac{B M}{\sqrt{3}}$ and $A M+B M=A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,608 |
9.2. In each cell of a 3 by 3 table, an integer is written such that all eight sums of triples of numbers, written in the cells of each row, each column, and each of the two diagonals, are equal to one number $S$ (i.e., the table is a 3 by 3 magic square). Prove that $\mathrm{S}$ is divisible by 3. | 9.2. In each cell of a 3 by 3 table, an integer is written such that all eight sums of triples of numbers, written in the cells of each row, each column, and each of the two diagonals, are equal to one number \( S \) (i.e., the table is a 3 by 3 magic square). Prove that \( S \) is divisible by 3.
Proof. Let the sum o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,609 |
9.3. Let $\mathrm{P}$ be the foot of the altitude dropped from vertex $\mathrm{A}$ of the right triangle $\mathrm{ABC}$ to its hypotenuse $\mathrm{BC}$, and let $\mathrm{M}$ be the midpoint of segment $\mathrm{CP}$. Denote by $\mathrm{E}$ the point on the extension of side $\mathrm{AB}$ beyond point $\mathrm{B}$ such t... | 9.3. Let $\mathrm{P}$ be the foot of the altitude dropped from vertex $\mathrm{A}$ of the right triangle $\mathrm{ABC}$ to its hypotenuse $\mathrm{BC}$, and let $\mathrm{M}$ be the midpoint of segment $\mathrm{CP}$. Denote by $\mathrm{E}$ the point on the extension of side $\mathrm{AB}$ beyond point $\mathrm{B}$ such t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,610 |
9.5. On an 8x8 chessboard, 8 rooks are arranged in such a way that none of them can attack each other. Prove that each of them can be simultaneously moved to one of the diagonally adjacent cells in such a way that none of them will be able to attack each other after the move. Recall that a chess rook attacks all cells ... | 9.5. On an 8x8 chessboard, 8 rooks are arranged in such a way that none of them can attack each other. Prove that each of them can be simultaneously moved to one of the diagonally adjacent cells in such a way that none of them will be able to attack each other after the move. Recall that a chess rook attacks all cells ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,612 |
10.1. Two athletes with constant speeds run on an oval track of a sports ground, the first of them runs the track completely 5 seconds faster than the second. If they run on the track from the same starting point in the same direction, they will meet again for the first time after 30 seconds. How many seconds will it t... | Answer. In 6 seconds.
Solution. Let the length of the track be $S$ meters, and the speeds of the first and second runners be $x$ and $y$ meters per second, respectively. From the first condition: $\frac{S}{x} + 5 = \frac{S}{y}$, and from the second condition $\frac{S}{x-y} = 30$, since in this case the first runner ca... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,613 |
10.2. Find all pairs of real values $a$ and $b$, for which both equations $x^{2} + a x + b^{2} = 0$ and $x^{2} + b x + a^{2} = 0$ have at least one common root. | Answer. $a=b=0$.
Solution. For the equations to have a common root, it is necessary that each of them has roots, that is, their discriminants must be non-negative. Therefore, $a^{2} \geq 4 b^{2}, b^{2} \geq 4 a^{2}$, from which $a^{2} \geq 16 a^{2}, b^{2} \geq 16 b^{2}$, which is possible only when $a=b=0$. In this ca... | =b=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,614 |
10.3. In a row from left to right, all natural numbers from 1 to 37 are written in such an order that each number, starting from the second to the 37th, divides the sum of all numbers to its left: the second divides the first, the third divides the sum of the first and second, and so on, the last divides the sum of the... | Answer: 2.
Solution. If the first number is the prime number 37, then the second must be 1, and the third must be a divisor of the number $37+1=38$, that is, 2 or 19. However, 19 must be in the last position, since the number 37 minus an even number must divide the sum of all the other numbers and itself, that is, div... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,615 |
10.5. What is the maximum number of 2 by 2 squares that can be placed on a 7 by 7 grid of squares so that any two placed squares share no more than one common cell? The 2 by 2 squares are placed along the grid lines such that each covers exactly 4 cells. The squares do not extend beyond the boundaries of the board. | Answer: 18 squares.
Solution: First, let's provide an example of laying out 18 squares of 2 by 2: 9 of them cover the left bottom square of size 6 by 6 of the board, and the other 9 cover the right top square of size 6 by 6 of the board.
We will prove that it is impossible to lay out 19 squares correctly. Note that i... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,616 |
7.1. In a correct numerical equality, identical digits were replaced with identical letters, and different ones with different letters. It is known that the result is
$$
\text { Я + ДЕД = ТЫ + HЕТ. }
$$
Provide a version of the original equality. (It is sufficient to provide one example.) | Solution. For example, the following works
$$
3+202=96+109
$$
Criteria. Any correct example -7 points. | 3+202=96+109 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,617 |
7.2. 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 | 9,618 |
7.4. On a certain island, there are 2022 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 stood in a circle, and each was asked in turn, "Is the person to your left a liar?" The total number of "Yes" answers was 2, and the numbe... | Solution. We will denote knights and liars as K and L, respectively. Note that in pairs of adjacent KK or LL, the person on the right cannot say that the person on the left is a liar. Therefore, the positive answer to the first question could only come from pairs KL and LK. This means that since there were two "Yes" an... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,620 |
7.5. Given an empty $3 \times 3$ grid. In one move, it is allowed to choose any three cells forming an L-shape (oriented in any direction) and place one checker in each of them. Can it happen after several moves that each cell contains the same non-zero number of checkers? (Justify your answer.) | Solution. Suppose it is possible, and after several moves, there will be $n$ checkers in each cell. Then there are a total of $9 n$ checkers on the board, and since three checkers are added with each move, $3 n$ moves were made. However, note that in each of the four corner cells, we can only place checkers in separate... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,621 |
7.1. Come up with at least one three-digit PAU number (all digits are different) such that $(П+\mathrm{A}+\mathrm{У}) \times \Pi \times \mathrm{A} \times \mathrm{Y}=300$ (it is sufficient to provide one example) | Solution: For example, PAU = 235 is suitable. There are other examples as well. Criteria: Any correct example - 7 points. | 235 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,622 |
7.2. Students in the seventh grade send each other New Year's stickers on Telegram. It is known that exactly 26 people received at least one sticker, exactly 25 - at least two stickers, ..., exactly 1 - at least 26 stickers. How many stickers did the students in this class receive in total, if it is known that no one r... | # Answer: 351
Solution: Note that exactly 1 sticker was received by one person, as it is precisely in this case that the difference between those who received at least 1 and those who received at least 2. Similarly, one person received exactly $2, 3, \ldots, 26$ stickers, so the total number of stickers is $1+\ldots+2... | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,623 |
7.3. There are broken balance scales. The pans of the scales are in equilibrium if the weight on the right pan equals three times the weight on the left pan. The right pan outweighs if the weight on it is greater than three times the weight on the left pan. The left pan outweighs if three times the weight on it is grea... | Solution: Number the coins from 1 to 7 and for the first weighing, place coin 1 on the left pan, and coins -2, 3, and 4 on the right pan. Consider the following scenarios:
1) The right pan is heavier. Then the counterfeit coin is number 1.
2) The left pan is heavier. Then the counterfeit coin is among 2, 3, and 4. Now... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,624 |
7.4. Vsevolod somehow placed numbers equal to one or minus one at the vertices of a cube. After that, Yaroslav calculated the product of the numbers at the vertices of each face. Can it be that the sum of the eight numbers of Vsevolod and the six numbers of Yaroslav is zero?
# | # Answer: No
Solution: The product of all 14 numbers is equal to the fourth power of the product of the numbers at the vertices (since each number is counted four times) and is therefore positive. Consequently, among the 14 numbers, there is an even number of minus ones, and their sum equals 0 only when there are exac... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,625 |
7.5. Each cell of a $5 \times 5$ table is painted in one of several colors. Lada shuffled the rows of this table so that no row remained in its original position. Then Lera shuffled the columns so that no column remained in its original position. To their surprise, the girls noticed that the resulting table was the sam... | # Answer: 7.
Solution: Let's renumber the colors and reason about numbers instead. Both columns and rows could have been cyclically permuted or divided into a pair and a triplet. If a cyclic permutation of columns was used, then all columns consist of the same set of numbers, i.e., no more than five different numbers.... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,626 |
11.1. In each of the four volleyball teams, there are six players, including a captain and a setter, and these are different people. In how many ways can a team of six players be formed from these four teams, such that there is at least one player from each team and there must be a pair of captain and setter from at le... | Answer: 9720.
Solution. Case 1. Three players, including the captain and the setter, are chosen from one of the teams, and one player is chosen from each of the remaining three teams. The team is chosen in four ways, the third player from it in another four ways, and the three players from the remaining teams in $6 \c... | 9720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,627 |
11.3. In a football tournament, 17 teams participated, each of which played against each of the others once. Could it be that for each team, the number of victories it achieved equaled the number of matches it played to a draw? | Answer. No, it could not.
Solution. Suppose the opposite, that the number of victories each team achieved equals the number of matches it played to a draw. Let's find the sum $\mathrm{S}$ of the total number of all victories, draws, and defeats of all teams. In this sum, the total number of all victories will equal th... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,628 |
11.5. For what minimum natural $n$ can $n$ distinct natural numbers $s_{1}, s_{2}, \ldots, s_{n}$ be found such that $\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{7}{66} ?$ | Answer. $n=9$.
Solution. We can assume that $1<s_{1}<s_{2}<\ldots<s_{n}$, then for any $k=1, \ldots, n$ the inequality $s_{k} \geq k+1$ holds. Therefore, $\frac{7}{66}=\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right) \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\rig... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,629 |
9.1. A finite set of distinct real numbers $X$ is called good if each number in $X$ can be represented as the sum of two other distinct numbers in $X$. What is the minimum number of elements that a good set $X$ can contain?
# | # Answer: 6.
Solution. From the condition, it follows that $X$ contains no less than three numbers, which means there are non-zero numbers in it. By multiplying all numbers by minus one if necessary, we can assume that $X$ contains positive numbers. Let's choose the largest number $M$ from them. According to the condi... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,630 |
9.4. What is the maximum number of colors in which all cells of a 4 by 4 square can be painted so that any 2 by 2 square of cells necessarily contains at least two cells of the same color? | Answer: In 11 colors.
Solution. We will prove that the maximum number of colors under the conditions of the problem does not exceed 11. Consider in a 4x4 square five 2x2 squares: four corner ones and the central one. The corner 2x2 squares do not intersect, and the central one shares one common cell with each of the c... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,631 |
9.5. In a company of $n>1$ people, some of its members are acquainted with each other, while others are not. If X is acquainted with Y, then Y is also acquainted with X, and a person is neither considered their own acquaintance nor non-acquaintance. Find all $n$ for which in the company there will always be two people ... | Answer. All $\boldsymbol{n}$, not equal to 2 and 4.
Proof. 1. Examples. When $\boldsymbol{n}=2$, in any case, both participants have the same number of acquaintances, 0 or 1, but there are no candidates for a common acquaintance or stranger. When $\boldsymbol{n}=4$, the only example where it is impossible to find V or... | All\boldsymbol{n},notequalto24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,632 |
11.2. Find the number of five-digit numbers containing two digits, one of which is divisible by the other. | Answer. $89760=9 \cdot 10^{4}-2 \cdot 5!$.
Solution. We will count the number of five-digit numbers in which no digit is divisible by any other digit, and then subtract this number from the total number of five-digit numbers, which is 90000, to get the answer to the problem.
Note that a number, none of whose digits a... | 89760 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,633 |
11.5. Find all pairs of natural numbers $a$ and $b$ such that both numbers $\frac{a^{2}+b}{b^{2}-a}$ and $\frac{b^{2}+a}{a^{2}-b}$ are integers. | Answer. $(a, b)=(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)$ - six pairs.
Solution. When $a$ is replaced by $b$ and $b$ by $a$, the fractions in the condition swap places, so we can immediately assume that $a \leq b$, adding the symmetric solutions obtained later. In this case, the case $a=b=1$ is excluded due to the inequali... | (2,2),(3,3),(1,2),(2,1),(2,3),(3,2) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,634 |
7.2. One morning at 9:00, pedestrian Fedia left the village of Fedino for the village of Novoverandovo. At the same time, cyclist Vera set off towards him from Novoverandovo. It is known that by the time they met, Fedia had walked one-third of the distance between the villages. However, if Fedia had left an hour earlie... | Answer: $10: 20$
Solution 1: Understand that if Fedya has walked some distance before Vera sets off, then from the moment Vera starts until they meet, he will walk a third of the remaining distance. This means that the midpoint of the entire distance is a third of the remaining distance, so Fedya covers one quarter of... | 10:20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,636 |
7.3. Arseny is going to draw several (more than two) non-overlapping squares of possibly different sizes on a plane, which may share parts of their boundaries. Ivan claims that in any case, he will be able to color all these squares using 3 colors in such a way that no two squares of the same color share a common bound... | Answer: Arseny.
Solution: We can consider the configuration shown in the figure. Suppose it can be properly colored with three colors. Then let the central square have color number 1. All the others then have colors 2 and 3, and in the chain of five squares around the central one, these colors must alternate. However,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,637 |
7.4. There are 100 coins, 99 of which are genuine and weigh the same, and 1 is counterfeit and lighter than the others. Dmitry has a pair of balance scales without weights, which always show incorrect results (for example, if the left pan is heavier, they will show either balance or the right pan being heavier, but it ... | # Solution:
Let's number the coins from 1 to 100. Weigh the first coin against the second. If the scales show equality, then one of them is fake, and all the other coins are genuine, and we have achieved the desired result. If the scales do not balance, assume the coin numbered 1 is heavier. Then the second coin is de... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,638 |
7.5. Given a 4x2017 board. Two players take turns placing rooks on it. The first player must place a rook so that it attacks an even number of already placed rooks, while the second player must place a rook so that it attacks an odd number of already placed rooks. The player who cannot make a move loses. Who can ensure... | Answer: second
Solution: Let's mentally divide the $4 \times 2017$ board into two parts, each $2 \times 2017$. Then, let the second player mirror the first player's moves in the other part. For example, if the first player moves to the upper left cell of the first part, the second player moves to the other part, but a... | second | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,639 |
9.1. How much of a $5 \%$ and a $20 \%$ salt solution in water should be taken to obtain 90 kg of a $7 \%$ solution | Answer: 78 kg of 5% solution and 12 kg of 20% solution.
Solution. Let the mass of the 5% solution be $x$ kg, the mass of the 20% solution will be $90-x$ kg, and the total mass of salt in the 5% and 20% solutions is equal to the mass of salt in 90 kg of 7% solution: $\frac{5}{100} x+\frac{20}{100}(90-x)=\frac{7}{100} 9... | 78 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,640 |
9.2. In a kindergarten, each child was given three cards, each of which had either "MA" or "NYA" written on it. It turned out that 20 children could form the word "MAMA" from their cards, 30 children could form the word "NYANYA," and 40 children could form the word "MANYA." How many children had all three cards the sam... | Answer: 10 children.
Solution. Let's denote the number of children who received three "MA" cards as $x$, two "MA" cards and one "NA" card as $y$, two "NA" cards and one "MA" card as $z$, and three "NA" cards as $t$. Then, the word "MAMA" can be formed by all children from the first and second groups and only them, the... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,641 |
9.4. On a plane, there is a segment $\mathrm{AB}$ of length 1 and an arbitrary point M on it. On the segments AM and MB as sides, squares AMCD and MBEF are constructed, lying on the same side of AB. Let $\mathrm{P}$ and $\mathrm{Q}$ be the points of intersection of the diagonals of these squares, respectively. Find the... | Answer. The midline ST of the isosceles right triangle AKB, constructed on $\mathrm{AB}$ as the hypotenuse on the same side of $\mathrm{AB}$ as the squares. This is a segment of length $1 / 2$ at a height of $1 / 4$ from $A B$, parallel to $A B$, whose projection on $A B$ coincides with the interval $\left[\frac{1}{4} ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,642 |
9.5. For seven natural numbers $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ it is known that all of them are different prime numbers. Find all values that the smallest of these seven numbers can take. | Answer: 3.
Solution. From the condition, it follows that $a, b, c$ are also primes. If the smallest of the seven numbers were equal to two, the last four numbers would be different even numbers, which means they could not all be prime. If all seven numbers are greater than three, due to their primality, they are not d... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,643 |
7.2. Kay has an ice plate in the shape of a "corner" (see figure). The Snow Queen demanded that Kay cut it into four equal parts. How can he do this?
 | Answer: for example, the corner can be cut as follows:
Criteria: any correct cutting is scored 7 points. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,644 |
11.1. Paramon set out from point A to point B. At $12^{00}$, when he had walked half the way to B, Agafon ran out from A to B, and at the same time, Solomon set out from B to A. At $13^{20}$, Agafon met Solomon, and at $14^{00}$, he caught up with Paramon. At what time did Paramon and Solomon meet? | Answer: At 13 o'clock.
Solution: Let the distance between A and B be $\mathrm{S}$ km, and the speeds of Paramon, Solomon, and Agafon be $x, y, z$ km per hour, respectively. Then from the condition, we get: $\frac{S / 2}{z-x}=2, \frac{S}{y+z}=\frac{4}{3}$, from which $x+y=\frac{1}{2} S$. Therefore, Paramon and Solomon ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,645 |
11.2. The median $A M$ of triangle $A B C$ divides the segment $P R$, parallel to side $A C$, with endpoints on sides $\mathrm{AB}$ and $\mathrm{BC}$, into segments of lengths 5 cm and 3 cm, starting from side $\mathrm{AB}$. What is the length of side AC? | Answer: 13 cm.
Solution. Let the ends of the segment be denoted as $\mathrm{P}$ and $\mathrm{R}$, and the point of intersection with the median $\mathrm{AM}$ as $\mathrm{Q}$, with $\mathrm{P}$ lying on side $\mathrm{AB}$ and $\mathrm{R}$ on side $\mathrm{BC}$. Draw the midline $\mathrm{MN}$ of the triangle, its length... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,646 |
11.3. Can three fractions be chosen from $\frac{1}{100}, \frac{2}{99}, \frac{3}{98}, \ldots, \frac{100}{1}$ (all fractions with natural numerators and denominators, the sum of which is 101) such that their product equals 1? | Answer. No, it cannot.
Solution. Suppose the opposite, and there exist three such fractions $\frac{a}{101-a}, \frac{b}{101-b}, \frac{c}{101-c}, 1 \leq a<b<c \leq 100 \quad, \quad$ whose product equals 1. Then $2 a b c=101^{3}-101^{2}(a+b+c)+101(a b+b c+a c)$, the right side of the equation is divisible by 101, therefo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,647 |
10.1. Ivan was walking from Pakhomovo to Vorobyevo. At noon, when he had covered $4 / 9$ of the entire distance, Foma set off after him from Pakhomovo on a bicycle, and Erema set off towards him from Vorobyevo. Foma overtook Ivan at 1 PM, and met Erema at 1:30 PM. When will Ivan and Erema meet? | Answer: At 14:30
Solution. Let the distance from Pakhomo to Vorobyovo be denoted as P, and the speeds of the performance participants as I, F, and E km per hour, respectively, using the first letters of their names. Then, according to the problem, Foma and Ivan are moving in the same direction, while Foma and Yerema a... | 14:30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,648 |
10.4. A proper divisor of a natural number is any of its divisors, different from 1 and the number itself. Find all natural numbers for which the maximum proper divisor is 2 more than the square of the minimum proper divisor. | Answer: $n=12$ and $n=33$.
Solution. Let the required number be $n$, and its smallest proper divisor, which is a prime number, be $p$. The largest proper divisor of $n$ is $\frac{n}{p}$, which, according to the condition, equals $p^{2}+2$, hence $n$ has the form $n=p\left(p^{2}+2\right)$, where $p$ is its smallest pro... | n=12n=33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,649 |
7.1. The steamship "Rarity" is rapidly sinking. If Captain Alexei gives exactly $2017^{2017}$ instructions to his 26 sailors, the steamship can be saved. Each subsequent sailor can receive 2 fewer or 2 more instructions than the previous one. Can Alexei save the steamship? | Answer: No.
Solution: It is obvious that the parity of the number of orders received by all sailors is the same. However, in this case, the total number of orders received is the sum of 26 numbers of the same parity, i.e., an even number. But $2017^{2017}$ is odd, so this is impossible.
## Criteria:
Answer only, ans... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,650 |
7.2. Daniil has 6 cards with letters, from which he managed to form the word WNMWNM shown in the picture. Note that this word has a remarkable property: if you rotate it 180 degrees, you get the same word. How many words with such a property can Daniil form using all 6 cards at once?
 According to the problem, Danil has 2 cards with the letter $\mathrm{N}$, which remains the same when flipped, and 4 cards with the letter M, which turns into the letter W when flipped. Clearly, to get a word with the desired properties, we need to arrange 2 letters M and one $\math... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,651 |
7.3. Cells of an infinite grid plane are painted black and white in such a way that in any corner of 2018 cells (even rotated and/or flipped), there are an equal number of white and black cells. Is it true that all cells of the plane are painted in a checkerboard pattern?
2017
 notice that if the corner lies as shown in the figure, then it can be unfolded with the protruding cell facing up. In this case, all other cells will remain in their places. This means that in any column, cells alternate in color. The same reasoning applies to rows.
(2) Suppose the colo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,652 |
7.4. In a triangle with sides of length $a, b$ and with side $c$ opposite an angle of 120 degrees. Prove that a triangle can be formed from segments of length $a, c$ and $a+b$.
# | # Solution:
Let there be a triangle $ABC$, where $AB=c, AC=$ $b, BC=a$. Extend the segment $AC$ beyond point $C$ to point $D$ at a distance $a$. We obtain triangle $CBD$. In it, $BC=CD$, so it is isosceles, and angle $BCD=$ 60 degrees as the adjacent angle to $BCA$. But then triangle $BCD$ is equilateral, and $BD=a$.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,653 |
10.1. Find all solutions of the equation: $\sqrt{1-x}+\sqrt{2-y}=\sqrt{3 x+2 y-7}$. | Answer. $x=1, y=2$.
Solution. From considering the domain, it immediately follows that ${ }^{x \leq 1, y \leq 2}$ and $y \geq \frac{7-3 x}{2}$. Note that, for $x \leq 1$, we have $y \geq \frac{7-3 x}{2} \geq \frac{7-3}{2}=2$, and the inequalities become strict when $x<1$, so the only point in the domain will be $x=1, ... | 1,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,655 |
10.2. A proper divisor of a natural number is any of its divisors, different from one and the number itself. Find all natural numbers that have at least two distinct proper divisors and are divisible by the difference of any two of them. | Answer: $6,8,12$.
Solution. Let the required number be ${ }^{n}$ and show that it must be even. Otherwise, all its divisors must be odd, and their difference must be even. But an even number cannot divide an odd number - a contradiction.
Due to the evenness of ${ }^{n}$, it must have proper divisors 2 and $\frac{n}{2... | 6,8,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,656 |
10.3. Different lines ${ }^{a}$ and ${ }^{b}$ intersect at point O. Consider all possible segments $\mathrm{AB}$ of length $l$, with endpoints A and B lying on ${ }^{a}$ and $^{b}$ respectively, and denote by $\mathrm{P}$ the point of intersection of the perpendiculars to the lines ${ }^{a}$ and ${ }^{b}$, dropped from... | Answer. The circle with center O and radius $\frac{l}{\sin x}$, where $x$ is the angle between the lines $a$ and $b$. Solution. Angles RAO and RBO are right angles, so points A, B, O, and R lie on a circle with diameter OR. Therefore, R belongs to the circumcircle of triangle AOB, the radius of which, by the sine theor... | \frac{}{\sinx} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,657 |
10.4. Real numbers ${ }^{a}$ and ${ }^{b}$ are such that $a^{3}+b^{3}=1-3 a b$. Find all values that the sum ${ }^{a+b}$ can take. | Answer: -2 and 1.
Solution. Let $a+b=x$ and $a b=y$, then the equation from the condition can be written as $x^{3}-3 x y=1-3 y$, from which $x^{3}-1=3 x y-3 y=3 y(x-1)$, we easily obtain the first possible value $a+b=x=1$. Dividing by the factor $x-1$, we have $x^{2}+x+1-3 y=0$. Substituting $x=a+b$ and $y=a b$ here, ... | -21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,658 |
10.5. Find the number of all possible arrangements of chips in some cells of an 8 by 8 chessboard such that the number of chips in each row is different and the number of chips in each column is different. | Answer: $2 \cdot(8!)^{2}$.
Solution. A row or column can contain from 0 to 8 chips, so if the quantities in the rows and columns are different, then the set of values of the number of chips in the rows and the set of values of the number of chips in the columns both represent a sequence of numbers $0,1,2$, $\ldots, 7,... | 2\cdot(8!)^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,659 |
9.2 The banker leaves home, and at that exact moment, a car from the bank arrives to take him to the bank. The banker and the car always leave at the same time, and the car always travels at a constant speed. One day, the banker left home 55 minutes earlier than usual and, for fun, started walking in the direction oppo... | Answer. The speed of the car is 12 times greater than the speed of the banker.
Solution. Indeed, the car delivered the banker to the bank 10 minutes later than usual, which means it caught up with him 5 minutes later than usual, i.e., from the moment he usually leaves his house. Therefore, the car traveled from the ba... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,660 |
9.5. In a row from left to right, there are $n$ coins. It is known that two of them are counterfeit, they lie next to each other, the left one weighs 9 grams, the right one 11 grams, and all the remaining ones are genuine and each weighs 10 grams. The coins are weighed on balance scales, which either show which of the ... | Answer. $n=28$.
Solution. Let $n=28$. Divide all 28 coins into three piles: the first pile contains coins numbered $11,13,15,17,19,21,23,25,27$, the second pile contains coins numbered $12,14,16,18,20,22,24,26,28$, and the third pile contains coins numbered from 1 to 10. The first weighing compares the first and secon... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,661 |
10.1. Find all solutions of the equation: $\sqrt{x+\sqrt{2 x-1}}+\sqrt{x-\sqrt{2 x-1}}=\sqrt{2}$. | Answer: $\frac{1}{2} \leq x \leq 1$.
Solution. First, as required, we find the domain of the variable $x$. From the non-negativity of the expressions under the square roots, we have: $2 x-1 \geq 0, x \geq \frac{1}{2}$ and $x-\sqrt{2 x-1} \geq 0, x \geq 0, x \geq \frac{1}{2}$, the second is equivalent to $(x-1)^{2} \ge... | \frac{1}{2}\leqx\leq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,662 |
10.2. In a rectangular trapezoid $A B C D$, the sum of the lengths of the bases $A D$ and $\mathrm{BC}$ is equal to its height $\mathrm{AB}$. In what ratio does the bisector of angle $\mathrm{ABC}$ divide the lateral side $\mathrm{CD}$? | Answer: 1:1.
Solution. Let the intersection of the bisector of angle $\mathrm{ABC}$ with line $\mathrm{AD}$ be $\mathrm{Q}$. Since angle $\mathrm{ABC}$ is a right angle, triangle $\mathrm{ABQ}$ is a right isosceles triangle, and the length of segment $\mathrm{AQ}$ is equal to the length of the height $\mathrm{AB}$, wh... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,663 |
10.3. At first, the balls were distributed among several white and black boxes so that each white box contained 31 balls, and each black box contained 26 balls. Then, three more boxes were brought in, and the balls were rearranged so that each white box contained 21 balls, and each black box contained 16 balls. Is it p... | Answer: No.
Solution. If the required condition were possible, the total number of balls would be divisible by 5. Then, since the remainders of the numbers $31, 26, 21$, and 16 when divided by 5 are all 1, the total number of boxes in both cases would be divisible by 5. This cannot be, because the number of boxes in t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,664 |
10.5. Can the number 2016 be represented as the sum of several pairwise distinct natural numbers such that among all possible pairwise sums of these numbers there are exactly 7 different ones? | Answer: No.
Solution: Suppose that 2016 can be represented as the sum of pairwise distinct natural numbers $a_{1}<a_{2}<\ldots<a_{n}$ such that among all possible pairwise sums of these numbers, there are exactly 7 different sums. The total number of pairs from $n$ numbers is $\frac{n(n-1)}{2}$ and must be at least 7,... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,665 |
11.1. Find the value of the expression $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{2}{1+x y}$, given that $x \neq y$ and the sum of the first two terms of the expression is equal to the third. | Answer: 2.
Solution. Let's write the condition of the equality of the sum of the first two terms to the third one as: $\frac{1}{1+x^{2}}-\frac{1}{1+x y}=\frac{1}{1+x y}-\frac{1}{1+y^{2}} \quad$ and bring it to a common denominator:
$\frac{x(y-x)}{(1+x^{2})(1+x y)}=\frac{y(y-x)}{(1+y^{2})(1+x y)}$. Given $x \neq y$, we... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,666 |
11.2. On a coordinate plane, a grasshopper jumps starting from the origin. The first jump, one cm long, is directed along the OX axis, each subsequent jump is 1 cm longer than the previous one, and is directed perpendicular to the previous one in either of two directions at its choice. Can the grasshopper end up at the... | Answer: No.
Solution: The grasshopper will make 50 vertical and 50 horizontal jumps. The lengths of the vertical jumps are 2, 4, ..., 100 cm. Among these numbers, 25 are divisible by 4 and 25 give a remainder of 2 when divided by 4. Therefore, regardless of the arrangement of plus and minus signs between them, the sum... | No | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,667 |
11.3. Find all natural numbers that can be represented simultaneously as the sum of several (more than one) natural numbers and as the product of the same natural numbers. | Answer. All, except for one and prime numbers.
Solution. We will prove that prime numbers and one do not fit. Obviously, one cannot be represented as the sum of more than one natural number. If a prime number $p$ is equal to the product of several naturals, then one of the factors is equal to $p$ itself, and the other... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,668 |
7.1. Anton baked such a cake that Arina was able to divide it into 4 parts with one straight cut. How could this be? (it is enough to give one example) | Solution: For example, a cake in the shape of the letter Ш, where all the vertical bars are cut off. There are many other possible examples.
Criteria: Any correct example - 7 points. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,669 |
7.3. At a large round table, 100 people are sitting. Each of them is either a knight, a liar, or a fool. Knights always tell the truth, liars always lie. A fool tells the truth if a liar is sitting to their left; lies if a knight is sitting to their left; and can say anything if a fool is sitting to their left. Each pe... | Answer: 0 or 50 liars.
Solution: If there is a liar at the table, then to the right of the liar is either a fool or a knight. In this situation, both of them tell the truth, so the next one after them is again a liar. We get that liars and non-liars alternate, i.e., there are exactly half liars.
If there are no liars... | 0or50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,671 |
7.4. On the table, 28 coins of the same size but possibly different masses are arranged in a triangular shape (see figure). It is known that the total mass of any triplet of coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle. | Answer: 60 g.
Solution 1: Take a rhombus made of 4 coins. As can be seen from the diagram, the masses of two non-touching coins in it are equal. Considering such rhombi, we get that if we color the coins in 3 colors, as shown in the diagram, then the coins of the same color will have the same mass.
Now it is easy to ... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,672 |
7.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed among several bags (each bag contains more than one barrel). We will call a bag good if the number of one of the barrels in it equals the product of the numbers of the other barrels in the sa... | Answer: 8.
Solution: In each good bag, there are no fewer than three barrels. The smallest number in each good bag must be unique; otherwise, the largest number in this bag would not be less than $10 \times 11=110$, which is impossible. For the same reason, if a good bag contains a barrel with the number 1, it must al... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,673 |
10.1. Let $a, b, c$ be not necessarily distinct natural numbers such that the fractions $\frac{a+b}{c}, \frac{a+c}{b}, \frac{b+c}{a}$ are also natural numbers. Find all possible values of the sum $\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}$. | Answer: 6, 7 or 8.
Solution. Due to symmetry, we can assume that $a \leq b \leq c$. Then $a+b \leq 2c$, so the integrality of the fraction $\frac{a+b}{c}$ implies $a+b=2c$ or $a+b=c$. In the first case, $a=b=c$ and the sum of the three fractions in the condition is 6. Let's consider further the case $a+b=c$. From the ... | 6,7,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,674 |
10.3. Find the minimum and maximum values of the expression $3 x^{2} y-2 x y^{2}$, where $x, y$ take arbitrary values from the interval $[0,1]$ | Answer: $-\frac{1}{3}$ and $\frac{9}{8}$
Solution 1. Let $f(x, y)=3 x^{2} y-2 x y^{2}$, then $f(x, y)=x y(3 x-2 y)$. The obtained expression equals 0 when $x=0$, or $y=0$, or $y=\frac{3}{2} x$. Therefore, in the region $0 \leq x, y \leq 1$, the function $f(x, y)$ is positive below (or to the right of) the line $y=\fra... | -\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,675 |
10.5. Represent the number 1000 as the sum of the maximum possible number of natural numbers, the sums of the digits of which are pairwise distinct. | Answer: 19.
Solution: Note that the smallest natural number with the sum of digits A is a99..99, where the first digit is the remainder, and the number of nines in the record is the incomplete quotient of the division of A by 9. From this, it follows that if A is less than B, then the smallest number with the sum of d... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,676 |
10.1. On an 8 by 8 chessboard, two arbitrary cells are marked. Is it true that the board can always be cut along the grid lines into two identical parts, each containing one of the marked cells? | Answer. Yes.
Solution. First, divide the board with two middle grid lines into 4 squares of 4 by 4 cells. If the marked cells lie in different squares, then by cutting the board along one of the middle lines, we obtain the required partition. Now suppose both marked cells lie in one 4 by 4 square, which we can conside... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 9,677 |
10.2. Find the necessary and sufficient conditions that natural numbers $x, y, z$ must satisfy such that it is possible to write $x \geq 1$ times the letter $A$, $y \geq 1$ times the letter $B$, and $z \geq 1$ times the letter $C$ in some order around a circle so that no two identical letters are written next to each o... | Answer. Each of the numbers $x, y, z$ is not greater than the sum of the other two.
Solution. 1) Necessity of the condition. According to the condition, between any two nearest letters $A$ on the circle, there should be at least one of the letters $B$ or $C$. Therefore, the total number of $y+z$ letters $B$ and $C$ is... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,678 |
10.5. Which natural numbers $n$ can be represented as $n=[a, b]+[a, c]+[b, c]$ for some natural numbers $a, b, c$? Here $[x, y]$ denotes the least common multiple of natural numbers $x$ and $y$. | Answer. For all $n$ that are not powers of two.
Solution. 1) Let $a=2^{k} \cdot a_{1}, a=2^{l} \cdot b_{1}, a=2^{m} \cdot c_{1}$, where the numbers $a_{1}, b_{1}, c_{1}$ are odd. Due to symmetry, we can assume that $k \leq l \leq m$. Then $[a, b]=\left[2^{k} \cdot a_{1}, 2^{l} \cdot b_{1}\right]=2^{l}\left[a_{1}, b_{1... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,679 |
8.1. Experienced sawyer Garik knows how to make cuts. In one day of continuous work, he can saw 600 nine-meter logs into identical three-meter logs (they differ from the original only in length). How much time will the experienced sawyer Garik need to saw 400 twelve-meter logs (they differ from the nine-meter logs only... | Answer: one day
Solution: to turn a 9-meter log into 3 three-meter pieces, 2 cuts are needed. Therefore, Garik makes $2 * 600=1200$ cuts per day. To turn a 12-meter log into three-meter pieces, 3 cuts are needed, and for 400 logs, $400 * 3=1200$ cuts are needed, which means it will take the same amount of time.
Crite... | 1 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,681 |
8.2. Once a month, Uncle Chernomor gathers all his 33 bogatyrs and proposes a new salary list, which he himself has compiled. Chernomor himself does not vote. The bogatyrs whose pay is proposed to be increased vote in favor, while the others vote against. The proposal is adopted by a majority vote. Can Chernomor achiev... | Answer: Yes, he can
Solution: Let the first 33 months Chernomor propose to one of the bogatyrs to make the salary equal to 0, and increase the others by one-thousandth of their initial salary. All these proposals will be accepted. After this, the salary of all bogatyrs will be no more than 33/1000 of the initial, i.e.... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,682 |
8.4. Find all solutions to the equation $4[x]=25\{x\}-4.5$. Here $[x]$ denotes the integer part of the number $x$, that is, the greatest integer not exceeding the number $x$, and $\{x\}$ is the fractional part of the number $x$, which by definition is equal to $x-[x]$. | Answer: $x=k+(8 k+9) / 50$ for $k=-1,0, \ldots, 5-$ a total of 7 fractions.
Solution: multiply the equation by 2 and transform it. We get the equation $8[x]+9=50\{x\}$. Notice that the right side is a non-negative number, and it is no more than 50, that is, $0 \leq 8[x]+9 \leq 50$, from which $-9 / 8 \leq[x] \leq 41 /... | k+\frac{8k+9}{50}fork=-1,0,\ldots,5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,684 |
8.5. 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 is 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. Sim... | 120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,685 |
7.2. 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? (Provide a complete solution, not just the answer.) | 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 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,687 |
7.3. Katya wrote a four-digit number on the board that was divisible by each of its digits without a remainder (there were no zeros in the number). Then she erased the first and last digits, and the number 89 remained on the board. What could have been written on the board initially? (Find all options and show that the... | Solution: The original number was divisible by 8. Therefore, the number formed by its last three digits is also divisible by 8. From the condition, this number has the form $\overline{89 x}$. Clearly, $x=6$ fits, and other numbers divisible by 8 do not fit into this decade. The entire number is divisible by 9, which me... | 4896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,688 |
7.4. Vsevolod assembled an octahedron from eight identical triangles (depicted in the image), after which he painted each of the twelve segments composing it in red, blue, or green. It turned out that at each of the six vertices of the octahedron, segments of each color arrive. How many segments could have been painted... | Solution: Consider one of the colors. According to the condition, an edge of this color leaves each vertex. The total number of such exits is at least 6, and each edge leaves exactly two vertices. Therefore, the total number of edges of each color is at least 3. From this, it also follows that the number of edges of ea... | 3to6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,689 |
7.5. TikTok-House represents a $3 \times 3$ square of nine rooms, each inhabited by a blogger. On Monday, the bloggers randomly swapped rooms, after which every two people who ended up in adjacent rooms recorded a collaborative TikTok. On Tuesday, the bloggers again swapped rooms and recorded TikToks according to the s... | Solution: There are 9 people in total. From them, we can form $9 \cdot 8 / 2 = 36$ pairs. In one day, 12 TikToks are shot. Therefore, if everyone managed to shoot a collaborative video, each person shot a TikTok with every other person exactly once.
Each person shoots either 2, 3, or 4 TikToks per day, and everyone ne... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,690 |
9.1. Over 100 days, each of the six friends visited the pool exactly 75 times, no more than once a day. Let $n$ be the number of days when the pool was visited by at least five of them. Determine the maximum and minimum possible values of the number $n$. | Answer: 90 and 25, respectively.
Solution. The total number of visits to the swimming pool by six friends over 100 days is $75 \cdot 6 = 450$, so the number of days when the pool was visited by at least five of them is no more than $\frac{450}{5} = 90$. On the other hand, if the number of such days is $n$, then the to... | 9025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,691 |
9.2. Vasya swapped the digits of a three-digit number $A$ so that no digit of the new three-digit number $B$ coincided with the digit of number $\mathrm{A}$ in the same place. It turned out that the difference $A-B$ is a two-digit number, which is a perfect square. What can the number $A$ be? Find all possible options. | Answer. There are 20 solutions in total: $A=218,329, A=213, . ., 879, A=706,817,928, A=201,312, . ., 978$.
Solution. Let $A=\overline{a b c}$, where $a, b, c$ are the digits of its hundreds, tens, and units respectively. From the pigeonhole principle and the condition, it follows that all of them are distinct, as it i... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,692 |
9.4. Vikentiy has two jars, a red one and a blue one, and a pile of 20 pebbles. Initially, both jars are empty. A move in Vikentiy's game consists of transferring a pebble from the pile to one of the jars or returning a pebble from one of the jars to the pile. The number of pebbles in the jars determines the game posit... | Answer: 110.
Solution: The position in Vikenty's game is uniquely defined by a pair of non-negative integers $(x, y)$, and $x+y \leq 20$, where $0 \leq y \leq x \leq 20$ are the numbers of pebbles in the blue and red jars, respectively. In total, there are $21+19+17 \ldots+1=121$ positions in Vikenty's game. We will c... | 110 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,693 |
9.5. Can it be that in some company everyone has exactly 5 friends, and any two people have exactly 2 common friends? | Answer: No, it cannot.
Solution: Let A be one of the members of this company, who has 5 friends. Each member of the company, except A, has two common friends, who are two of the five friends of A. Suppose the pairs of common friends of A and B, as well as A and C, coincide, denoted by D and E. Then D and E have at lea... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,694 |
11.1. The ore contains $21 \%$ copper, enriched - $45 \%$ copper. It is known that during the enrichment process, $60 \%$ of the mined ore goes to waste. Determine the percentage content of ore in the waste. | Answer: $5 \%$.
Solution. In the extracted 100 kg of ore, there are 21 kg of copper. From these 100 kg of enriched ore, 40 kg will be obtained, containing 18 kg of copper. Therefore, the 60 kg of waste that went to the dump contain 3 kg of copper, that is, $5 \%$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,695 |
11.4. Inside a right-angled triangle with sides 3, 4, and 5 cm, there are two circles, the ratio of whose radii is 9 to 4. The circles touch each other externally, both touch the hypotenuse, one touches one leg, and the other touches the other leg. Find the radii of the circles. | Answer: $\frac{20}{47}$ cm and $\frac{45}{47}$ cm.
Solution. Let the radii of the circles be $4x$ and $9x$ respectively, and they touch the legs of the triangle as shown in Figure 2. Express through $x$ the distance $y$ between the points of tangency of the circles with the hypotenuse of the triangle (see Figure 1).
... | \frac{20}{47} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,697 |
11.5. Find the number of different arrangements in a row of all natural numbers from 1 to 10 such that the sum of any three consecutive numbers is divisible by 3. | Answer: $4! \cdot 2 \cdot 3! \cdot 3! = 1728$
Solution. From the condition, it follows that the remainders of the numbers standing two apart when divided by 3 are equal. Therefore, the numbers standing at positions 1, 4, 7, and 10, as well as those at positions 2, 5, and 8, and at positions 3, 6, and 9, have equal rem... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,698 |
10.1. Salt was added to the water twice, with the percentage increase in the mass of the solution being the same each time. What was the initial mass of the water and what was the percentage increase in the mass of the solution each time, if the final mass of the solution was 850 g and the final concentration was $36 \... | Answer. The initial mass of water is 544 g, the mass increase each time is $25 \%$.
Solution. Let the initial mass of water be $x$ g, and the mass increase of the solution each time be $p \%$. After the second time, the mass will be $\left(1+\frac{p}{100}\right)^{2} x$, and the mass of salt in it $\left(\left(1+\frac{... | 544,p=25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,699 |
10.2. Prove that for any $x \in\left(0, \frac{\pi}{2}\right) \quad$ the inequality holds: $1+\operatorname{tg} x<\frac{1}{1-\sin x}$. | Solution. All expressions are positive, we get rid of the denominators, combine like terms, and divide by $\sin (x)$, leading to the equivalent expression $1<\sin (x)+\cos (x)$. The right-hand side is positive, squaring it, we obtain the equivalent inequality: $1<(\sin (x)+\cos (x))^{2}=1+2 \sin (x) \cos (x)$, which is... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 9,700 |
10.3. Find all points on the plane, with coordinates $(x, y)$, that satisfy the equation: $\max \{|x+y|,|x-y|\}=1$ | Answer. Rhombus with vertices $((-1,0),(1,0),(0,-1),(0,1))$.
Solution. Let's expand the absolute values of the expressions. In the diagram, the upper one indicates $|x+y|$, the lower one - the modulus $|x-y|$, and the maximum of the expressions is framed:
,(1,0),(0,-1),(0,1)) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,701 |
10.4. On the line $l$ there are two distinct points A and B. We consider all possible pairs of circles that are tangent to each other and to the line $l$ at points A and B. For each pair, let M denote the midpoint of the segment of the external tangent to these circles, not lying on $l$. Find the geometric locus of poi... | Answer. The circle with the center at the midpoint of $AB$ and a radius equal to $AB$, excluding points lying on $l$.
Solution. The external tangents $AB$ and $CD$ are equal due to axial sym... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,702 |
10.5. All natural numbers from 1 to 100 are written in some order in a circle. For each pair of adjacent numbers, the sum is calculated. Out of the hundred resulting numbers, what is the maximum number that can be divisible by 7? | Answer: 96.
Solution. For the sum of a pair of adjacent numbers to be divisible by 7, their remainders when divided by 7 must sum to 7: 0+0, 1+6, 2+5, and 3+4. Let's call such pairs of remainders suitable pairs. Among the numbers from 1 to 100, there are 14, 15, 15, 14, 14, 14, and 14 numbers with remainders 0, 1, 2, ... | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,703 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.