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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 |
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 |
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 |
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 |
9.5. 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 |
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 |
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 |
9.1. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
9.1. 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 |
9.3. 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 |
9.1. 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 |
9.3. 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 |
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 |
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 |
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 |
9.1. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
9.1. 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 |
9.4. 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 |
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 |
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 |
9.1. 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 |
9.2. 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 |
9.5. 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 |
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 |
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 |
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 |
9.2 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 |
9.5. 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
9.4. 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 |
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 |
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 |
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 |
7.1. Lesha wrote down a number, and then replaced the same digits with the same letters, and different digits with different letters. He ended up with the word NOVOSIBIRSK. Could the original number be divisible by 3? | Answer: Yes, it could.
Solution: For example, 10203454638.
Criteria: Any correct example without verification - 7 points. | 10203454638 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. The steamship "Raritet" after leaving the city moves at a constant speed for three hours, then drifts for an hour, moving with the current, then moves for three hours at the same speed, and so on. If the steamship starts its journey from city A and heads to city B, it takes 10 hours. If it starts from city B and h... | Answer: 60 hours.
Solution: Let the speed of the steamboat be $U$, and the speed of the river be $V$. When the steamboat travels from B to A, it arrives at the destination just before the fourth engine stop, meaning it travels 12 hours at a speed of $U-V$ towards A, and then 3 hours back at a speed of $V$ towards B. T... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. A square box 3 by 3 is divided into 9 cells. It is allowed to place balls in some cells (possibly a different number in different cells). What is the minimum number of balls that need to be placed in the box so that each row and each column of the box contains a different number of balls? | Answer: 8.
Solution: Add up the number of balls in all rows and columns. Since these are 6 different non-negative numbers, this sum is at least $0+1+\ldots+5=15$.
Now notice that the sum of the numbers in the rows is equal to the sum of the numbers in the columns, since these sums are equal to the total number of bal... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find all natural numbers $n$ such that $\frac{1}{n}=\frac{1}{p}+\frac{1}{q}+\frac{1}{p q}$ for some primes $p$ and $q$ | Answer. $n=1$.
Solution. Bringing the expression in the condition to a common denominator, we get: $n(p+q+1)=pq$. From the simplicity of $p$ and $q$, it follows that the divisors of the right-hand side can only be the numbers $1, p, q$, and $pq$, one of which must equal $p+q+1$. Since $1, p$, and $q$ are less than $p+... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. A kilogram of meat with bones costs 165 rubles, a kilogram of meat without bones costs 240 rubles, and a kilogram of bones costs 40 rubles. How many grams of bones are there in a kilogram of meat with bones? | Answer: 375 grams.
Solution: Let $x$ kilos of bones be in a kilogram of meat with bones. Then $40 x + 240(1-x) = 165$, from which $x=0.375$. | 375 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3. From the highway, four roads sequentially lead to four villages A, B, C, D. It is known that the route by road/highway/road from A to B is 9 km, from A to C - 13 km, from B to C - 8 km, from B to D - 14 km. Find the length of the route by road/highway/road from A to D. Explain your answer. | Answer: 19 km.
Solution. Let's add the distances from $A$ to $C$ and from $B$ to $D$. Then the highway segment from the turn to $B$ to the turn to $C$ will be counted twice, while the highway segments from the turn to $A$ to the turn to $B$ and from the turn to $C$ to the turn to $D$, as well as the four roads from th... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. What is the maximum number of different rectangles that an 8 by 8 chessboard can be cut into? All cuts must follow the grid lines. Rectangles are considered different if they are not equal as geometric figures. | Answer: 12.
Solution. Let's list the possible sizes of different integer rectangles of minimal areas that can fit along the grid lines on an 8 by 8 board in ascending order of these areas: 1 by 1, 1 by 2, 1 by 3, 1 by 4, 2 by 2, 1 by 5, 1 by 6, 2 by 3, 1 by 7, 1 by 8, 2 by 4, 3 by 3, 2 by 5. There are already 13 recta... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. It is known that the values of the quadratic trinomial $a x^{2}+b x+c$ on the interval $[-1,1]$ do not exceed 1 in absolute value. Find the maximum possible value of the sum $|a|+|b|+|c|$. Answer. 3. | Solution. Substituting the values $x=0,1,-1$ from the interval $[-1,1]$ into the polynomial $a x^{2}+b x+c$, we obtain three inequalities: $-1 \leq c \leq 1$, $-1 \leq a+b+c \leq 1$, and $-1 \leq a-b+c \leq 1$. Adding the second and third inequalities, we also get $-1 \leq a+c \leq 1$. Subtracting the second from the t... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. What is the maximum number of integers that can be written in a row so that the sum of any five consecutive ones is greater than zero, and the sum of any seven consecutive ones is less than zero? | Answer. Ten.
Solution. The sum of any seven written numbers is negative, while the sum of the five outermost numbers from this seven is positive, which means the sum of the two leftmost and the sum of the two rightmost numbers from this seven are negative. Therefore, the sum of any two adjacent numbers, with at least ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. There are 100 boxes numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all ... | Answer: 99.
Solution: To be able to definitively determine which of the 100 boxes contains the prize, it is necessary to have the possibility of receiving at least 100 different answers to one set of questions. Since the host's answers for different prize positions can only differ by the number of "yes" responses, it ... | 99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.2. Find the number of all five-digit numbers $\overline{a b c d e}$, all digits of which are distinct and $ad>e$. | Answer: 1134.
Solution. For the correct notation of a number satisfying the condition of the problem, one needs to arbitrarily select a quintet of different digits from 10 possible ones, and then arrange two of them to the left of the maximum in ascending order and the two remaining to the right of the maximum in desc... | 1134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Prove that for any $0 \leq x, y \leq 1$ the inequality $\frac{x}{1+y}+\frac{y}{1+x} \leq 1$ holds.
Proof 1. Replace the ones in the denominators of the fractions on the left side of the inequality with $0 \leq x \leq 1$ and $0 \leq y \leq 1$ respectively. In this case, the denominators of the fractions will not ... | Answer. $A P M=90^{\circ}$.
Solution. Mark a point T on the extension of CM beyond M such that MT=MP. In this case, segments $\mathrm{AB}$ and TP are bisected by their intersection point M, so quadrilateral ATBP is a parallelogram. In particular, segments AP and BT are equal and parallel. Consider triangles $\mathrm{B... | 90 | Inequalities | proof | Yes | Yes | olympiads | false |
9.1. Petya wrote 10 integers on the board (not necessarily distinct).
Then he calculated the pairwise products (that is, he multiplied each of the written numbers by each other). Among them, there were exactly 15 negative products. How many zeros were written on the board? | Answer: 2.
Solution. Let there be $A$ positive numbers and $B$ negative numbers on the board. Then $A+B \leq 10$ and $A \cdot B=15$. Since a negative product is obtained when we multiply a negative and a positive number. From this, it is easy to understand that the numbers $A$ and $B$ are 3 and 5 (1). Therefore, $A+B=... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.2. In a trapezoid, one lateral side is twice as large as the other, and the sum of the angles at the larger base is 120 degrees. Find the angles of the trapezoid. | Answer: 90 and 30 degrees.
Solution: Let the vertices of the trapezoid be A, B, C, D, with the larger base AD, and assume CD is twice as long as AB. Choose a point E on AD such that BE is parallel (and equal) to CD, and let M be the midpoint of segment BE. Then triangle ABM is isosceles with a 60-degree angle at verte... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.4. $N$ different natural numbers, not exceeding 1000, are written in a circle such that the sum of any two of them, standing one apart, is divisible by 3. Find the maximum possible value of $N$.
# | # Answer: 664.
Solution. Consider the remainders of the numbers when divided by 3. Divisibility by 3 means that in each pair of numbers standing one apart, either both numbers are divisible by 3, or one has a remainder of 1 and the other has a remainder of 2 when divided by 3. Among the numbers from 1 to 1000, 333 are... | 664 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. For what smallest $n$ is the following condition satisfied: if $n$ crosses are placed in some cells of a $6 \times 6$ table in any order (no more than one per cell), then there will definitely be three cells forming a strip of length 3, either vertical or horizontal, each containing a cross? | Answer. $n=25$.
Solution. If there are no fewer than 25 crosses, then one of the rows of the table contains no fewer than 5 crosses, and no more than one empty cell. Then either the three left cells of this row, or the three right cells of it, all contain crosses and form the desired strip.
If there are fewer than 25... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.4. Find all natural numbers $x$ such that the product of all digits in the decimal representation of $x$ equals $x^{2}-10 x-22$ | Answer: $x=12$.
Solution: First, the product of all digits of a natural number is non-negative, so $x^{2}-10 x-22$, from which $x \geq \frac{10+\sqrt{188}}{2}$, that is, $x \geq 12$. Second, if in the product of all digits of a natural number, all digits except the first are replaced by tens, the product will not decr... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. Lesha wrote down a number, and then replaced the same digits with the same letters, and different digits with different letters. He ended up with the word NOVOSIBIRSK. Could the original number be divisible by 9? | Answer: Yes, it could.
Solution: For example, 10203454638.
Criteria: Any correct example without verification - 7 points. | 10203454638 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.4. The steamship "Raritet" after leaving the city moves at a constant speed for three hours, then drifts for an hour, moving with the current, then moves for three hours at the same speed, and so on. If the steamship starts its journey from city A and heads to city B, it takes 10 hours. If it starts from city B and h... | Answer: 60 hours.
Solution: Let the speed of the steamboat be $U$, and the speed of the river be $V$. When the steamboat travels from B to A, it arrives at the destination just before the fourth engine stop, meaning it travels 12 hours at a speed of $U-V$ towards A, and then 3 hours back at a speed of $V$ towards B. T... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find all four-digit numbers $\overline{x y z t}$, where all digits $x, y, z, t$ are distinct and not equal to 0, such that the sum of all four-digit numbers obtained from $\overline{x y z t}$ by all possible permutations of the digits is 10 times the number $\overline{x x x x}$. | Answer. The number 9123 and all numbers obtained from it by permuting the last three digits, a total of 6 answers.
Solution. The number of four-digit numbers obtained from $\overline{x y z t}$ by all possible permutations of the digits is 24, in each of which each of the digits $x, y, z, t$ appears exactly 6 times in ... | 9123 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. Around a circle, 32 numbers $a_{1}, a_{2}, \ldots, a_{32}$ are written, each of which is either -1 or 1. In one operation, each number $a_{n}, n=1,2, \ldots, 32$ is replaced by the product $a_{n} a_{n+1}$ of it and the next number in the cycle, with indices considered cyclically, $a_{33}=a_{1}, a_{34}=a_{2}$, and... | Answer: 32.
Solution: We will prove by induction on $n$ that if $2^n$ numbers are written in a circle, the answer to the problem is $2^n$. The base case $n=1$ is straightforward: either $\{1,-1\} \rightarrow\{-1,-1\} \rightarrow\{1,1\}$, or $\{-1,-1\} \rightarrow\{1,1\}$, in any case, two operations are always suffici... | 32 | Combinatorics | proof | Yes | Yes | olympiads | false |
11.4. Find all natural numbers $n$ that can be represented as $n=\frac{x+\frac{1}{x}}{y+\frac{1}{y}}$, for some natural numbers $x$ and $y$. | Answer. $n=1$
Solution. Transform the equality in the condition to the form $n=\frac{\left(x^{2}+1\right) y}{\left(y^{2}+1\right) x}$. Note that the numbers $x^{2}+1$ and $x$, as well as the numbers $y^{2}+1$ and $y$ are coprime, so $x$ in the denominator can only cancel out with $y$ in the numerator, meaning $y$ is d... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2 Sergey arranged several (more than two) pairwise distinct real numbers in a circle so that each number turned out to be equal to the product of its neighbors. How many numbers could Sergey have arranged? | Answer: 6.
Solution: Let's denote the two adjacent numbers as $a$ and $b$. Then, next to them stands $b / a$, followed by $1 / a, 1 / b, a / b$, and again $a$. Thus, it is impossible to arrange more than 6 numbers.
If 3 numbers can be arranged, then $a=1 / a$, which means $a$ is 1 or -1. In the first case, $b$ and $b... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. A number is called good if any two adjacent digits in its notation differ by at least 4. Vera wrote some good number, and then replaced identical digits with identical letters, and different ones with different letters. Could she have ended up with the word NOVOSIBIRSK? | Answer: For example, the number 82729161593 could work ( $\mathrm{H}=8, \mathrm{O}=2, \mathrm{~B}=7, \mathrm{C}=9$, I = 1, B = 6, $\mathrm{P}=5, \mathrm{~K}=3$).
Criterion: any valid example without verification - 7 points. | 82729161593 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.3. Find the angle $D A C$, given that $A B=B C$ and $A C=C D$, and the lines on which points $A, B, C, D$ lie are parallel, with the distances between adjacent lines being equal. Point $A$ is to the left of $B$, $C$ is to the left of $B$, and $D$ is to the right of $C$ (see figure).
 + 8 + 9, k \geq 5, n \geq 21 \), in the second case - \( n = 4k + 3 = 4(k-3) + 6 + 9, k \geq 4, n \geq 19 \). On the other hand, the three smallest compos... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. In the cells of an 8 by 8 board, tokens are placed such that for each token, the row or column of the board in which it lies contains no more than three tokens. What is the maximum possible number of tokens on the board? | Answer: 30.
Solution: By swapping the verticals and horizontals, we assume that only the left $x$ verticals and the bottom $y$ horizontals contain more than 3 chips. From the condition, it follows that there are no chips at all in the lower left rectangle at the intersection of these verticals and horizontals. Each ve... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. Given a triangle $A B C$, side $A B$ is divided into 4 equal segments $A B_{1}=B_{1} B_{2}=B_{2} B_{3}=B_{3} B$, and side $A C$ into 5 equal segments $A C_{1}=C_{1} C_{2}=C_{2} C_{3}=C_{3} C_{4}=C_{4} C$. How many times larger is the area of triangle $A B C$ compared to the sum of the areas of triangles $C_{1} B_{... | Answer: 2 times.
Solution: Let the area of $A B_{1} C_{1}$ be $S$. Then the area of $B_{1} C_{1} C_{2}$ is also $S$, since $B_{1} C_{1}$ is the median in $A B_{1} C_{2}$. Similarly, the area of $B_{1} B_{2} C_{2}$ is $2S$, as $C_{2} B_{1}$ is the median in $A B_{2} C_{2}$. The area of $B_{2} C_{2} C_{3}$ is half the a... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.4. Masha and Misha set out to meet each other simultaneously, each from their own house, and met one kilometer from Masha's house. Another time, they again set out to meet each other simultaneously, each from their own house, but Masha walked twice as fast, and Misha walked twice as slow as the previous time. This ti... | # Answer: 3 km.
Solution: We will prove that they spent the same amount of time on the first and second occasions. Suppose this is not the case, and they spent less time on the second occasion. Then, on the first occasion, Masha walked 1 km, and on the second occasion, less than 2 km (her speed was twice as high, but ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Given the number 1836549, you can take two adjacent non-zero digits and swap their places, after which you subtract 1 from each of them. What is the smallest number that can result from these operations? | Answer: 1010101
Solution: The digits in the number alternate in parity: odd, even, etc. Note that with the described operation, even and odd numbers swap places, and then 1 is subtracted from them, thereby not disrupting the order of parity. Thus, it is impossible to obtain a number less than 1010101 (in each place, t... | 1010101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Find all positive integer solutions of the equation $(n+2)!-(n+1)!-(n)!=n^{2}+n^{4}$. Answer. $n=3$. | Solution. Rewrite the equation as $n!=\left(n^{*}\left(n^{2}+1\right)\right) /(n+2)$. Transforming the right side, we get $n!=n^{2}-2 n+5-10:(n+2)$. The last fraction will be an integer for $n=3$ and $n=8$, but the latter number is not a solution (substitute and check!)
Grading criteria. Acquiring extraneous solutions... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2 To buy an apartment, you need to take either exactly 9 small, 6 medium, and one large loan, or exactly 3 small, 2 medium, and 3 large loans. How many only large loans would be required to buy an apartment? | Answer: 4 large loans.
Solution: Let's denote small, medium, and large loans by the letters m, c, and b, respectively. Rewrite the condition using these notations:
$$
9 \mathrm{M}+6 \mathrm{c}+\sigma=3 \mathrm{~m}+2 \mathrm{c}+3 b
$$
Simplifying, we get $6 \mathrm{~m}+4 \mathrm{c}=2$ b or $3 \mathrm{~m}+2 \mathrm{c}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. It is known that the sum of the digits of number A is 59, and the sum of the digits of number B is 77. What is the minimum sum of the digits that the number A+B can have? | Answer. 1.
Solution. It is sufficient to consider A=9999995, B=999999990000005, then $\mathrm{A}+\mathrm{B}=1000000000000000$, the sum of the digits is 1 - the minimum possible.
Grading Criteria. Correct answer and example: 7 points. Presence of arithmetic errors: minus 1-2 points. Any other answer: 0 points. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. On an island, 20 people live, some of whom are knights who always tell the truth, and the rest are liars who always lie. Each islander knows for sure who among the others is a knight and who is a liar. When asked by a visitor how many knights live on the island, the first islander answered: "None," the second: "No... | Answer: 10.
Solution: If the first islander were a knight, he would have lied in his answer, which cannot be the case. Therefore, the first is a liar, and there are no more than 19 knights on the island. This means the twentieth islander told the truth, so he is a knight, and there is at least one knight on the island... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find the value of the expression $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$, given that $\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}=5$ and $x+y+z=2$. | Answer: 7.
Solution. Transform: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x+y+z}{y+z}+\frac{y+x+z}{x+z}+\frac{z+x+y}{x+y}-3=$ $=(x+y+z)\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3=2 \cdot 5-3=7$.
Grading Criteria. Presence of arithmetic errors: minus 1-2 points. Correct answer calculated on some ex... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. There is a steamship route between the cities of Dzerzhinsk and Lviv. Every midnight, a steamship departs from Dzerzhinsk, arriving exactly eight days later in Lviv. How many steamships will the steamship "Raritet" meet on its way to Dzerzhinsk if it departs from Lviv exactly at midnight and spends the same eight ... | Answer: 17.
Solution: As "Raritet" departs from Lviv, a steamer arrives there, which left Dzerzhinsk 8 days ago. By the time "Raritet" arrives at its final destination, 8 days have passed since the initial moment, and at this moment, the last steamer departs from Dzerzhinsk, which "Raritet" meets on its way. Thus, "Ra... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser gave up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play? | Answer: 14.
Solution: On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The bathtub fills up in 23 minutes from the hot water tap, and in 17 minutes from the cold water tap. Pete first opened the hot water tap. After how many minutes should he open the cold water tap so that by the time the bathtub is full, one and a half times more hot water has been added than cold water? | # 9.2. Answer. In 7 minutes.
In one minute, the hot water tap fills $\frac{1}{23}$ of the bathtub, and the cold water tap fills $\frac{1}{17}$ of the bathtub. After filling the bathtub, the hot water should make up $\frac{3}{5}$ of the bathtub, and the cold water should make up $\frac{2}{5}$ of the bathtub. Therefore,... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Among all natural numbers from 1 to 20 inclusive, some 10 numbers were painted blue, and the other 10 - red, then all possible sums of pairs of numbers, one of which is blue and the other is red, were counted. What is the maximum number of different sums that can be among the hundred obtained numbers? | 9.6. Answer. A maximum of 35 different numbers.
The sum of a blue and a red number can be a natural number from $1+2=3$ to $19+20=39$ inclusive, so there cannot be more than 37 different numbers. Moreover, note that one of the numbers $3,4, . ., 13$ must not be the sum of a blue and a red number. Otherwise, if the num... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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