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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 | 9,809 |
3. Find all solutions of the system of equations: $x+y=4,\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)=280$. | 9.3. Answer. (3,1) and $(1,3)$.
From the condition $x+y=4$ - not equal to 0, so in the second equation, we can cancel $x+y=4$, obtaining $\left(x^{2}+y^{2}\right)\left(x^{2}-x y+y^{2}\right)=70$. Further, squaring the first equation, we get $x^{2}+y^{2}=16-2 x y$. Substituting into the second, we get $3(x y)^{2}-40 x ... | (3,1)(1,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,810 |
4. Let $A B C D$ be a parallelogram, the incircle of triangle $A B D$ touches sides $A B$ and $A D$ at points $M$ and $N$, respectively, and the incircle of triangle $A C D$ touches sides $A D$ and $D C$ at points $P$ and $Q$, respectively. Prove that lines $M N$ and $P Q$ are perpendicular. | 9.4. By the property of tangents drawn from one point, triangles $A M N$ and $D P Q$ are isosceles, so lines $M N$ and $P Q$ are perpendicular to the bisectors of angles $B A D$ and $A D C$ respectively. Let the point of intersection of these bisectors be $S$. The sum of the measures of adjacent angles $B A D$ and $A D... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,811 |
5. For which natural numbers $n$ can a board of size $n$ by $n$ cells be divided along the grid lines into dominoes of size 1 by 2 cells so that there are an equal number of vertical and horizontal dominoes | 9.5. Answer. If and only if $n$ is divisible by 4.
If $n$ is divisible by 4, the required partition is almost obvious. For example, divide the board into two equal halves vertically, the left half into horizontal dominoes, and the right half into vertical dominoes. The dimensions of both halves are divisible by 2, so ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,812 |
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 | 9,813 |
10.1. Find all pairs of natural numbers $x \geq 2, y \geq 2$ such that the remainder of the division of the number $3 \boldsymbol{x}$ by $\boldsymbol{y}$ is 1, the remainder of the division of the number $3 \boldsymbol{y}$ by $\boldsymbol{x}$ is 1, and the remainder of the division of the number $\boldsymbol{x} \boldsy... | Answer. Four solutions: $x=2, y=5, x=5, y=2$.
Solution. The numbers $x$ and $y$ cannot be equal, otherwise the remainder of the division of the number $3x$ by $y$ would be 0. Therefore, due to symmetry, we can assume $x<y$, so $3x-1<3y$ and $3x-1$ is divisible by $y$, from which $3x-1=y$ or $3x-1=2y$. In the first cas... | 2,5,5,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,814 |
10.2. Vasya and Petya are playing "Take or Split." In this game, there is initially one large pile of stones. On each turn, the player whose turn it is either splits one of the existing piles into two smaller piles in any way, or takes one of the existing piles. The player who leaves no stones on the field after their ... | Answer. Vasya will always win.
Solution. Let by the time the game ends, both players have made $\boldsymbol{x}$ moves of the type taking a pile and $\boldsymbol{y}$ moves of the type dividing a pile into two. Each move of the first type reduces the number of piles on the field by one, and each move of the second type ... | Vasyawillalwayswin | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,815 |
10.5. What is the maximum number of 4-element subsets that can be selected from a set of 8 elements such that the intersection of any three of the selected subsets contains no more than one element? | Answer. Eight.
Solution. We will provide two different examples of choosing eight 4-element subsets in a set $\mathrm{X}$ of eight elements, satisfying the condition of the problem. Both examples are constructed as geometric objects.
Example 1. We will consider the elements of $\mathrm{X}$ as the vertices of a unit c... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,816 |
8.1. At the Olympiad, students from gymnasiums, lyceums, and regular schools met. Some of them stood in a circle. Gymnasium students always lie to regular school students, lyceum students lie to gymnasium students, and regular school students lie to lyceum students. In all other cases, the students tell the truth. Each... | Answer: There were no ordinary school students in the circle.
Solution: Let's assume there was an ordinary school student in the circle. Consider his left neighbor. This neighbor could not be another ordinary school student or a lyceum student, because they would tell the truth. But he also cannot be a gymnasium stude... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,817 |
8.2. In the bus, there are single and double seats. In the morning, 13 people were sitting in the bus, and there were 9 completely free seats. In the evening, 10 people were sitting in the bus, and 6 seats were completely free. How many seats are there in the bus? | Answer: 16.
Solution: If in the morning passengers sat on 6 double seats (i.e., as densely as possible), then they occupied 7 seats, with 9 seats still free. In total: 16 seats. If they did not sit as densely, then they would have occupied more seats. That is, there are no fewer than 16 seats in the bus, on the one ha... | 16 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,818 |
8.3. The natural numbers from 1 to 15 are written on the board. Lera chooses two numbers and finds their product, while Lada gets the remaining thirteen numbers and finds their sum. Can the results of the girls be the same? | Answer: They cannot.
Solution: Let Lera take the numbers $a$ and $b(a<b)$. Then, if the numbers of the girls are equal, we have:
$$
1+2+3+4+\ldots+15-a-b=a b
$$
Transforming the equation:
$$
\begin{gathered}
16 * 15: 2=a b+a+b \\
8 * 15+1=a b+a+b+1 \\
121=(a+1)(b+1)
\end{gathered}
$$
Since the left side is divisib... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,819 |
8.4. In the country, there are 15 cities, some of which are connected by roads. Each city is assigned a number equal to the number of roads leading out of it. It turned out that there are no roads between cities with the same number. What is the maximum number of roads that can be in the country? | Answer: 85.
Solution: Let's order the city numbers in non-increasing order:
$$
a_{1} \geq a_{2} \geq \cdots \geq a_{15}
$$
Notice that the number of cities with number $15-i$ is no more than $i$. Indeed, if there are at least $i+1$ such cities, then they cannot be connected to each other, and thus can be connected t... | 85 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,820 |
8.5. Given a convex quadrilateral $A B C D$ with side $A D$ equal to 3. Diagonals $A C$ and $B D$ intersect at point $E$, and it is known that the areas of triangles $A B E$ and $D C E$ are both 1. Find the side $B C$, given that the area of $A B C D$ does not exceed 4. | # Answer: 3.
Solution: Triangles $A B D$ and $A C D$ have the same area, as they are composed of the common $A E D$ and equal-area $A B E$ and $D C E$. Since $A B D$ and $A C D$ have the same base $A D$, their heights to this base are equal. Therefore, $B C$ is parallel to $A D$, meaning our quadrilateral is a trapezo... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,821 |
9.3. Find all triples of distinct natural numbers, the least common multiple of which is equal to their sum. The least common multiple of several numbers is the smallest natural number that is divisible by each of these numbers. | Answer. All triples of the form $\{n, 2 n, 3 n\}$ for any natural number $n$.
Solution. Let the required numbers be $a<b<c$. By the condition, their least common multiple, which is equal to $a+b+c$, is divisible by $c$, so $c$ divides the sum $a+b<2 c$. Therefore, $a+b=c$. Next, $a+b+c=2 a+2 b$ is divisible by $b$, so... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,822 |
9.1. For non-negative numbers $a, b, c, d$, the following equalities are satisfied:
$\sqrt{a+b}+\sqrt{c+d}=\sqrt{a+c}+\sqrt{b+d}=\sqrt{a+d}+\sqrt{b+c}$. What is the maximum number of distinct values that can be among the numbers $a, b, c, d$? | Answer. Two.
Solution. Square the first equality, combine like terms, cancel by 2, square again, combine like terms once more, ultimately obtaining the equality $a c+b d=a b+c d$, equivalent to $(a-d)(c-b)=0$, from which either $a=d$ or $b=c$. Performing similar manipulations with the second equality, we get $a=b$ or ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,823 |
9.2. If Petya gives two of his notebooks to Vasya, then Vasya will have $n$ times more notebooks than Petya, and if Vasya gives $n$ of his notebooks to Petya, then Petya will have twice as many notebooks as Vasya. Find all natural values of $n$ for which this is possible. | Answer. $n=1,2,3,8$.
Solution. Let the number of notebooks that Petya and Vasya have be $x$ and $y$ respectively. After Petya gives Vasya two notebooks, we get the equation $n(x-2)=y+2$, and after Vasya gives Petya $n$ notebooks, we get the equation $x+n=2(y-n)$. From the first equation, we express $y=n x-2 n-2$ and s... | 1,2,3,8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,824 |
9.3. Can all natural numbers from 1 to 100 inclusive be divided into ten sets, each containing a different number of elements, such that the greater the number of elements in a set, the smaller the sum of its elements? | Answer. No.
Solution. Suppose the partition specified in the condition is possible. The sum of all numbers from 1 to 100 is 5050, so the sum of the numbers in the set with the maximum sum is not less than $5050: 10=505$, therefore, it contains no fewer than 6 numbers. According to the condition, each of the nine remai... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,825 |
11.2. A sequence of real numbers $a_{n}, n=1,2,3, \ldots$ is such that $a_{n+1}=a_{n}+\sqrt{a_{n}+a_{n+1}}, n=1,2,3, \ldots$ and $a_{1}=1$. Find an explicit formula expressing the number $a_{n}$ in terms of $n$. | Answer. $a_{n}=1+2+3+\ldots+n=\frac{n(n+1)}{2}$.
Solution. Let's calculate the first terms of the sequence $a_{n}, n=2,3$. For $a_{2}$, we have by the condition $a_{2}=1+\sqrt{1+a_{2}}$, which means $a_{2}^{2}-3 a_{2}=0$ and $a_{2}>1$, hence $a_{2}=3=1+2$. For $a_{3}$, we then have $a_{3}=3+\sqrt{3+a_{3}}$, which mean... | a_{n}=\frac{n(n+1)}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,826 |
11.3. What is the maximum number of colors needed to color all cells of a 4 by 4 square so that for every pair of different colors, there are two cells of these colors that are either in the same row or in the same column of the square? | Answer: In 8 colors.
Solution. If the cells were painted in 9 or more colors, there would be a color in which only one cell is painted. There are only 6 cells located in the same row or column with it, so there are no more than 6 other colors forming a pair with it, as required by the condition - a contradiction. Ther... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,827 |
11.5. Find all natural numbers $a$ such that the product $n(n+a)$ is not a square of a natural number for any natural $n$. | Answer. $a=1,2,4$.
Solution. The fact that the product $n(n+a)$ is not a square of a natural number for $a=1,2,4$ follows from the inequalities $n^{2}<n(n+1)<n(n+2)<(n+1)^{2}$ and $n^{2}<n(n+4)<(n+2)^{2}, n(n+4) \neq(n+1)^{2}$. We will prove that for all other natural $a$, the product $n(n+a)$ is always a square of a ... | 1,2,4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,828 |
7.1. Numbers from 1 to 9 are arranged in some order in a circle. Can it happen that the sum of any two consecutive numbers is divisible by at least one of the numbers 2 and 9? | Answer: It can.
Solution: Yes, let's arrange the numbers as follows: -7-1-9-3-5-4-6-8-2-.
Criteria: Any correct arrangement - 7 points. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,829 |
7.2. A square with a side of $100 \mathrm{~cm}$ was drawn on the board. Alexei crossed it with two lines parallel to one pair of sides of the square. Then Danil crossed the square with two lines parallel to the other pair of sides of the square. As a result, the square was divided into 9 rectangles, and it turned out t... | Answer: 2400.
Solution 1: Without loss of generality, we will assume that the central rectangle has a width of 60 cm and a height of 40 cm. Let $x$ and $y$ be the width and height, respectively, of the lower left rectangle. Then the upper left rectangle has sides $x$ and $(60-y)$, the upper right rectangle has sides $... | 2400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,830 |
7.3. In the entrance, Lera rides the elevator, while Lada walks down (the speeds of the elevator and Lada are constant). Once, when they were descending from the 4th floor to the 1st, Lada was faster and waited for Lera at the bottom for some time. Another time, the girls were descending from the 8th floor, and Lada do... | Answer: In $11 / 4$ times.
Solution: Let $U, V$ be the speeds of Lada and the elevator, respectively, and $S$ be the distance between two floors. Then, the first time the girls covered a distance of 3S, so Lada waited at the bottom for $(3 S / V) - (3 S / U)$. Similarly, the second time it was $(7S/V) - (7S/2U)$. Sinc... | \frac{11}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,831 |
7.4. On a square board with a side of 2017, a chess bishop stands in the lower left corner. Alexei and Danil take turns moving this bishop, with Alexei moving first, and players are not allowed to move to cells where the bishop has already been (the bishop moves diagonally to any distance). The player who cannot make a... | Answer: Danil.
Solution: Let's describe Danil's strategy and prove that he is the one who wins.
1) If Alexei's move lands him on the main diagonal, we move to any cell on the same diagonal. This is always possible because initially, there is an even number of unvisited cells on it, and after Alexei's move, there is a... | Danil | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,832 |
7.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser would give 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 | 9,833 |
8.2. A duckling and a gosling were competing in a triathlon. The distance consisted of equally long running, swimming, and flying segments. The duckling ran, swam, and flew at the same speed. The gosling ran twice as slow as the duckling, but swam twice as fast. Who and by how many times faster did they fly, if they st... | Answer: The gosling flew twice as fast.
Solution: Let's take the time spent by the duckling on each segment of the distance as a two. Then the time spent by the gosling on running and swimming is 4 and 1, respectively. Therefore, he spent $6-4-1=1$ on flying, i.e., he flew twice as fast as the duckling.
Criteria: Onl... | 2 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,835 |
8.3. 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 f... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,836 |
8.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed into 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 sam... | # 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 be no 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 | 9,838 |
10.1. Which natural numbers can be represented as the fraction $\frac{x^{3}}{y^{4}}$, where $x$ and $y$ are some natural numbers? | Answer. Any natural number.
Solution. Let $n$ be any natural number. Set $x=n^{3}$, $y=n^{2}$, then $\frac{x^{3}}{y^{4}}=\frac{n^{9}}{n^{8}}=n$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,839 |
10.2. In triangle $A B C$, the measure of angle $A$ is 30 degrees, and the length of the median drawn from vertex $B$ is equal to the length of the altitude drawn from vertex $C$. Find the measures of angles $B$ and $C$. | Answer. $\angle B=90^{\circ}, \angle C=60^{\circ}$.
Solution. Given that $\angle A=30^{\circ}$, the height drawn from vertex $C$ is half of $A C$, therefore, the median drawn from vertex $B$ is also half of $A C$. It follows that vertex $B$ lies on the circle with diameter $A C$, hence $\angle B=90^{\circ}, \angle C=9... | \angleB=90,\angleC=60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,840 |
10.3. Find the smallest natural number divisible by 99, all digits of which are even. | Answer: 228888.
Solution. Let the sum of the digits of the desired number $X$, located in even-numbered positions (tens, thousands, etc.), be denoted by $A$, and the sum of the digits in odd-numbered positions (units, hundreds, etc.) be denoted by $B$. According to the divisibility rules for 9 and 11, $A+B$ is divisib... | 228888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,841 |
8.3. Given a triangle $A B C$, a point $D$ is chosen on side $B C$ and a point $H$ on side $A C$. Additionally, $D K$ is the angle bisector of triangle $B D A$. It turns out that angles $C H D$ and $H D K$ are right angles. Find $H C$, if $A C=2$. | Answer: $H C=1$.
Solution: Since angles $C H D$ and $H D K$ are right angles, line $K D$ is parallel to line $A C$. Therefore, angles $H A D$ and $A D K$ are equal as alternate interior angles, and angles $D C A$ and $B D K$ are equal as corresponding angles. Since $D K$ is the angle bisector, angles $D A C$ and $D C ... | HC=1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,846 |
8.4. Vsevolod has 20 identical squares: one white and the rest black. Vsevolod placed the white square on the table and is going to cover it with 19 black squares such that the sides of all black squares are parallel to the sides of the white square (black squares can overlap). Yaroslav claims that no matter how Vsevol... | Answer: No.
Solution: Let's assume that each square has dimensions $10 x 10$, and mentally divide them all into small 1 x 1 squares. In the white square, highlight a diagonal of 10 squares running from the bottom left corner to the top right corner. Let's call this diagonal $A$. The diagonal immediately above it consi... | proof | Geometry | proof | Yes | Yes | olympiads | false | 9,847 |
8.5. A blanket-pulling tournament was held at the school, consisting of several rounds. In each round, two teams participated, each consisting of a non-zero number of students, of course, a person could not be in both teams at the same time. After the tournament, it turned out that every possible team that could be for... | Solution 1: Let's call two teams complementary if one of them consists of all students who did not enter the other. Each student in the school is in only one of the two complementary teams, so he is in exactly half of all teams. But half the number of all teams is equal to the number of all rounds. Thus, the number of ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,848 |
7.2. Find the value of the expression (1+1/2)(1+1/3)...(1+1/2012). | 7.2. Answer: $\frac{2013}{2}=1006.5$
Perform the operation in each parenthesis:
$\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{2012}\right)=\frac{3}{2} * \frac{4}{3} * \frac{5}{4} * \ldots * \frac{2012}{2011} * \frac{2013}{2012}$.
Notice that the numerator of each fraction coincides wit... | 1006.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,850 |
7.3. In the math test, each of the 25 students could receive one of four grades: $2, 3, 4$ or 5. It turned out that the number of students who received fours was 4 more than those who received threes. How many people received twos, if it is known that the sum of all grades for this test is 121? | # 7.3. Answer: 0.
Let's assume that the 4 students who received fours skipped school, then 21 students should have scored 105 points on the test, which is only possible if all of them received fives. Therefore, in the real situation, no one received a two.
## Comments on Evaluation.
Answer only: 1 point. Answer with... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,851 |
7.4. Find the smallest natural number ending in the digit 4 that quadruples when its last digit is moved to the beginning of the number. | # 7.4. Answer: 102564
If a number ending in 4 is multiplied by 4, the result is a number ending in 6, so the last digits of the desired number are 64. If such a number is multiplied by 4, the result is a number ending in 56, so the desired number ends in 564. Continuing to restore the number in this way, it turns out ... | 102564 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,852 |
7.5. Ten girls each had ten envelopes. Each girl signed and sent some envelopes to other girls (an envelope cannot be signed twice). As a result, each girl ended up with a different number of envelopes. Prove that at least one girl ended up with more envelopes than she sent out. | # 7.5.
Solution 1: We will prove that there is a girl who ended up with no fewer than 11 envelopes. Suppose this is not the case. Then each girl ended up with no more than 10 envelopes. Therefore, the total number of envelopes in the end does not exceed \(10+9+\ldots+1=55\). This leads to a contradiction. Note that th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,853 |
11.3. Find the maximum natural number $A$ such that for any arrangement of all natural numbers from 1 to 100 inclusive in a row in some order, there will always be ten consecutively placed numbers whose sum is not less than $A$. | Answer: 505.
Solution: The sum of all numbers from 1 to 100 is 5050. Let's divide the 100 numbers in a row into 10 segments, each containing 10 numbers. Clearly, the sum of the numbers in one of these segments is not less than 505, so A is not less than 505.
We will show that among the numbers arranged in the followi... | 505 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,854 |
9.1. A merchant bought several bags of salt in Tver and sold them in Moscow with a profit of 100 rubles. With all the money earned, he again bought salt in Tver (at the Tver price) and sold it in Moscow (at the Moscow price). This time the profit was 120 rubles. How much money did he spend on the first purchase? | Answer: 500 rubles.
Solution: The additional 100 rubles spent the second time brought the merchant an additional 20 rubles in profit. Therefore, the first time, to earn $5 \cdot 20=100$ rubles in profit, the merchant must have paid $5 \cdot 100=500$ rubles.
Second solution: Let the amount of the first purchase be $x$... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,855 |
9.2. Ten numbers are written around a circle, the sum of which is 100. It is known that the sum of any three consecutive numbers is not less than 29. Indicate the smallest number $A$ such that in any set of numbers satisfying the condition, each number does not exceed $A$. | Answer. $A=13$
Solution. Let $X$ be the largest of the listed numbers. The remaining numbers can be divided into 3 "neighbor" triplets. The sum of the numbers in each such triplet is no less than 29, therefore, $X \leq 100 - 3 \cdot 29 = 13$. An example of a set with the maximum number 13: $13,9,10,10,9,10,10,9,10,10$... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 9,856 |
9.4. Find the smallest natural number in which each digit occurs exactly once and which is divisible by 990. | Answer: 1234758690.
Solution. The number 990 is the product of coprime numbers 2, 5, 9, and 11. Any ten-digit number composed of different digits, each used once, is divisible by 9, since their sum, which is 45, is divisible by 9. According to the divisibility rule for 10, the desired number must end in 0. It remains ... | 1234758690 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,857 |
9.5. Let $M$ - be a finite set of numbers (distinct). It is known that among any three of its elements, there will be two whose sum belongs to $M$. What is the maximum number of elements that can be in $M$? | Answer: 7.
Solution: An example of a set with 7 elements: $\{-3,-2,-1,0,1,2,3\}$.
We will prove that a set $M=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ with $n>7$ numbers does not have the required property.
We can assume that $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$ and $a_{4}>0$ (changing the signs of all elements does ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,858 |
8.1. In the wagon, several kilograms of apple jam were loaded, of which $20 \%$ was good and $80 \%$ was bad. Every day, half of the existing bad jam rotted, and it was thrown away. After several days, it turned out that $20 \%$ of the jam in the wagon was bad and $80 \%$ was good. How many days have passed since the l... | Answer: 4 days.
Solution 1: Let the initial total amount be $x$, and the final amount be $y$ kilograms of jam. Then, since the amount of good jam did not change, $0.2 x = 0.8 y$, which means $x = 4 y$. Therefore, initially, the amount of bad jam was $0.8 x = 3.2 y$, and it became $0.2 y$, meaning the mass of bad jam d... | 4 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,859 |
8.2. Once Alexei and Daniil were playing such a game. If a number \( x \) is written on the board, it can be erased and replaced with \( 2x \) or \( x - 1000 \). The player who gets a number not greater than 1000 or not less than 4000 loses. Both players aim to win. At some point, the boys stopped playing. Who lost if ... | Answer: no one lost.
Solution: note that if a number is less than 2000 but greater than 1000, then by multiplying by 2, you can get a number that is less than 4000. If a number is less than 4000 but greater than 2000, then by subtracting 1000 (possibly twice), you can get a number between 1000 and 2000. Thus, the only... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,860 |
8.3. The cells of a 4028 by 4028 board 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 are necessarily painted in a checkerboard pattern? | Answer: No, it is not necessary
Solution: Let's color the entire board in a checkerboard pattern, and the central 4x4 square as shown in the picture. We will check that in any corner of the specified size, there are an equal number of black and white cells. If it does not touch the central 4x4 square, this is obvious.... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,861 |
8.5. In the list $1,2, \ldots, 2016$, two numbers $a<b$ were marked, dividing the sequence into 3 parts (some of these parts might not contain any numbers at all). After that, the list was shuffled in such a way that $a$ and $b$ remained in their places, and no other of the 2014 numbers remained in the same part where ... | Answer: $1+2+\ldots+1008=1009 * 504=508536$ ways.
## Solution:
Hypothesis. We will prove that the question is equivalent to counting the number of ways to split 2014 into three ordered non-negative addends $2014=x+y+z$, for which the non-strict triangle inequality holds, i.e., $x+y \geq z, x+z \geq y, y+z \geq x$.
N... | 508536 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,862 |
9.2. Around a round table, 15 boys and 20 girls sat down. It turned out that the number of pairs of boys sitting next to each other is one and a half times less than the number of pairs of girls sitting next to each other. Find the number of boy - girl pairs sitting next to each other. | Answer: 10.
Solution. Let's call a group several children of the same gender sitting in a row, with children of the opposite gender sitting to the left and right of the outermost ones. Let $X$ be the number of groups of boys, which is equal to the number of groups of girls sitting in a row. It is easy to see that the ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,863 |
9.3. Can the number 2017 be represented as the sum of two natural numbers, the sum of the digits of one of which is twice the sum of the digits of the other? | Answer: No.
Proof. Suppose the opposite, that 2017 can be represented as the sum of natural numbers A and B, and the sum of the digits of A is twice the sum of the digits of B. When adding two digits of the same place value, the sum remains in that place if it is less than 10, or the sum minus 10 if it is greater than... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,864 |
9.4. What is the maximum number of triangles with vertices at the vertices of a regular 18-gon that can be marked so that no two different sides of these triangles are parallel? The triangles can intersect and have common vertices, coinciding segments are considered parallel. | Answer: 5.
Solution. Estimating the number of triangles. Let's number the vertices of the 18-gon from 1 to 18 clockwise. The sides of the triangles are the sides and diagonals of the regular 18-gon. We will call a diagonal even if an even number of sides lies between its ends, and odd otherwise. The parity of a diagon... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,865 |
9.5. Provide any way to arrange all natural numbers from 1 to 100 inclusive in a row in some order so that the sum of any $n$ consecutive numbers does not divide $p$ for all $2 \leq n \leq 100$. | Solution. Let's write down all numbers from 1 to 100 from left to right in order: $1,2,3,4, \ldots, 99,100$, and divide them into 50 pairs of adjacent numbers: 1 and 2, 3 and $4, \ldots, 99$ and 100, and in each pair, swap the numbers: $2,1,4,3, \ldots, 100,99$. We will prove that the resulting permutation satisfies th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,866 |
11.1. The decimal representation of a natural number $N$ contains each digit from 0 to 9 exactly once. Let $A$ be the sum of five two-digit numbers formed from the first and second, third and fourth, ..., ninth and tenth digits of $\mathrm{N}$, and let $B$ be the sum of four two-digit numbers formed from the second and... | Answer. No.
Solution. Let $N=\overline{a_{1} a_{2} \ldots a_{8} a_{9} a_{10}}$, where $a_{1}, a_{2}, \ldots, a_{8}, a_{9}, a_{10}$ are some permutation of the numbers $0,1,2, \ldots, 8,9$. Then
$A=\overline{a_{1} a_{2}}+\overline{a_{3} a_{4}}+\ldots+\overline{a_{9} a_{10}}=10\left(a_{1}+a_{3}+\ldots+a_{9}\right)+\lef... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,867 |
11.2. Find all solutions in real numbers of the system of equations
$$
\left\{\begin{array}{l}
x(1+y z)=9 \\
y(1+x z)=12 \\
z(1+x y)=10
\end{array}\right.
$$ | Answer: $x=1, y=4, z=2$.
Solution. Subtract the first equation from the second and third, we get: $y-x=3, z-x=1$. Substitute the expressions $y=x+3, z=x+1$ into the first equation, we get $x(x+3)(x+1)+x=9$, which after expanding the brackets leads to the cubic equation $x^{3}+4 x^{2}+4 x-9=0$. One of its roots is $x=1... | 1,4,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,868 |
11.3. Let Q and P be the bases of the perpendiculars dropped from vertex B of triangle ABC to the bisectors of its angles A and C, respectively. Prove that line PQ is parallel to side AC.
Proof. Let I be the point of intersection of the bisectors AA1, BB1, CC1 of triangle ABC, and let the measures of its angles be A, ... | Answer. All two-digit numbers ending in $9: 19,29, \ldots, 99$.
Solution. Let $N=\overline{a_{n} a_{n-1} \ldots a_{0}}$ be the decimal representation of an $n+1$-digit number $N$, obviously $N$ is at least two digits, so $n \geq 1$. By the condition, $N=\overline{a_{n} a_{n-1} \ldots a_{0}}=10^{n} a_{n}+10^{n-1} a_{n-... | 19,29,\ldots,99 | Geometry | proof | Yes | Yes | olympiads | false | 9,870 |
11.1. Find all solutions of the system of equations in real numbers:
$$
\left\{\begin{array}{c}
x y+z+t=1 \\
y z+t+x=3 \\
z t+x+y=-1 \\
t x+y+z=1
\end{array}\right.
$$ | Answer. $(1,0,-1,2)$.
Solution. Subtract the second equation from the first, the third from the second, the fourth from the third, and the first from the fourth, factorize each difference, and we get: $\quad(x-z)(y-1)=-2,(y-t)(z-1)=4,(z-x)(t-1)=-2,(y-t)(x-1)=0 . \quad$ From the second equality, $y-t \neq 0$, so from t... | (1,0,-1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,871 |
11.2. Let $a, b, c$ be natural numbers. Can the greatest common divisors of the pairs of numbers $a$ and $b, b$ and $c, c$ and $a$ be equal to $30!+111, 40!+234$ and $50!+666$ respectively | Answer. No, they cannot.
Solution. Suppose the situation described in the condition is possible. Note that the numbers $30!, 40!$, and $50!$ are clearly divisible by 9. The numbers 234 and 666 are divisible by 9 by the rule, as the sums of their digits are divisible by 9, while 111 is divisible by 3 but not by 9. Ther... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,872 |
11.3. Find the maximum length of a horizontal segment with endpoints on the graph of the function $y=x^{3}-x$ | Answer: 2.
Solution 1. A horizontal segment of length $a>0$ with endpoints on the graph of the function $y=x^{3}-x$ exists if and only if the equation $(x+a)^{3}-(x+a)=x^{3}-x$ has at least one solution for the given value of the parameter $a$. Expanding the brackets, combining like terms, and dividing by $a>0$, we ob... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 9,873 |
11.4. Let points $\mathrm{O}$ and $\mathrm{I}$ be the centers of the circumcircle and incircle of triangle $\mathrm{ABC}$, respectively. It is known that angle $\mathrm{AIO}$ is a right angle, and the measure of angle $\mathrm{CIO}$ is $45^{\circ}$. Find the ratio of the sides $\mathrm{AB:BC:CA}$. | Answer: 3:4:5
Solution. 1. The measure of angle AIC is $180^{\circ}-\frac{A+C}{2}=90^{\circ}+\frac{B}{2}>90^{\circ}$. If ray IO were outside angle AIC, the measure of angle AIO would equal the sum of the measures of angles AIC and CIO and would be greater than 90 degrees, which contradicts the condition. Therefore, ra... | 3:4:5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,874 |
10.1. Find all solutions in non-negative real numbers of the system of equations $a(a+b)=b(b+c)=c(c+a)$. | Answer. All triples of equal numbers ( $a, a, a$ ) for any non-negative $a$.
Solution. 1) Triples of numbers $(a, a, a)$ for any $a$ are, obviously, solutions to the system from the condition.
2) If exactly one of the variables is zero, for example $a=0, b \neq 0, c \neq 0$, then $c+a=0$, hence $c=0$ - a contradictio... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,876 |
10.2. Let $A$ be a set of ten distinct positive numbers (not necessarily integers). Determine the maximum possible number of arithmetic progressions consisting of three distinct numbers from the set $A$. | Answer: 20.
Solution: Let the elements of set $A$ be denoted as $a_{1}<a_{2}<\ldots<a_{10}$. Three numbers $a_{k}<a_{l}<a_{m}$ form a three-term arithmetic progression if and only if $a_{l}-a_{k}=a_{m}-a_{l}$. Let's see how many times each element of $A$ can be the middle term $a_{l}$ of such a progression. It is easy... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,877 |
10.4. Let's consider all 7! seven-digit numbers obtained from the number 1234567 by all possible permutations of its digits. How many of them give a remainder of 5 when divided by 7? Answer: 6!. | Solution. Let $A_{n}, n=0,1, \ldots, 6$ be the sets of the considered numbers that give remainders $0,1,2, \ldots 6$ respectively when divided by 7. We will prove that each of these sets contains the same number of elements, equal to $7!/ 7=6!$. Note that the number 1111111 gives a remainder of 1 when divided by 7. Wit... | 6! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,878 |
10.5. How many cells need to be marked on an 8 by 8 grid so that each cell on the board, including the marked ones, is adjacent by side to some marked cell? Find all possible answers. Note that a cell is not considered adjacent to itself. | Answer: 20.
Solution: First, let's gather our strength and mark twenty cells on an 8 by 8 board as required by the problem. For example, as shown in the figure. In this case, the board naturally divides into 10 parts, as indicated by the bold lines in the figure. Each part consists of cells adjacent to the given pair ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,879 |
8.1. Pasha and Sasha made three identical toy cars. Sasha did one-fifth of the total work. After selling the cars, they divided the proceeds in proportion to the work done. Pasha noticed that if he gave Sasha 400 rubles, and Sasha made another such car and sold it, they would have the same amount of money. How much doe... | Answer: 1000 rubles.
Solution: Sasha did one-fifth of the entire work, which means he made 0.6 of one car, while Pasha did the remaining 2.4. That is, the difference is 1.8 cars. If Sasha makes another car, the difference will be 0.8 of one car. Pasha gave 400 of his rubles, thereby reducing the amount of money he had... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,880 |
8.2. Mark has 2020 stones. He plans to divide them into 5 piles so that no two piles have the same number of stones. Additionally, Mark wants to be able to remove any pile and distribute all the stones from it among the remaining four piles so that all of them end up with an equal number of stones. Will Mark be able to... | Answer: It will work.
Solution: For example, piles of 402, 403, 404, 405, 406.
Let's prove that these piles will work. For convenience, consider each pile as 396 main stones and an additional $6, 7, 8, 9, 10$ stones, respectively. The part with 396 stones can be divided into 4 piles equally (99 stones each). Thus, in... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,881 |
8.3. The number 4 is written on the board. In one move, Anya is allowed to choose any proper divisor of the number written on the board, add it to this number, and write the sum instead of the old number. This operation is allowed to be performed an unlimited number of times. Prove that Anya can obtain any composite nu... | # Solution:
Any even number Anya can obtain from 4 by adding 2 the necessary number of times.
An odd composite number can be represented as $p k$, where $p$ is the smallest prime divisor of this odd number. Let Anya first obtain the even number $2 k$, and then by adding $k$ the necessary number of times, obtain $p k$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 9,882 |
8.4. From identical isosceles triangles, where the angle opposite the base is $45^{\circ}$ and the lateral side is 1, a figure was formed as shown in the diagram. Find the distance between points $A$ and $B$. | Answer: 2.
Solution: Let's denote the points $K, L, M$, as shown in the figure. We will construct an isosceles triangle $A K C$ equal to the original one. Connect vertex $C$ to other points as shown in the figure.
In the original triangles, the angle at the vertex is $45^{\circ}$. Therefore, the other two angles are ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,883 |
8.5. Yura and Roma found 2019 consecutive natural numbers and squared each of them. After that, Yura took 1010 of the 2019 resulting squares, while Roma took the remaining 1009. It turned out that the sum of the numbers Yura took is equal to the sum of the numbers Roma took. Provide an example of the numbers Yura and R... | Solution: For example, the numbers from 2037171 to 2039189.
To verify, let's rewrite these numbers in another form. Let $x$ be the number $2020 * 1009$, then the numbers in the example will be represented as $x-1009, \ldots, x, \ldots, x+1009$. Suppose Yura takes the squares of the first 1010 numbers, and Roma takes t... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 9,884 |
7.1. Pasha and Sasha made three identical toy cars. Sasha made a third of the first car, a fifth of the second car, and a fifteenth of the third car. What part of the entire work did Sasha do? | Answer: one fifth
Solution 1: In total, Sasha made $1 / 3+1 / 5+1 / 15=3 / 5$ of a car. This constitutes $3 / 5: 3=1 / 5$ of three cars.
Solution 2: Divide each car into 15 parts. In total, there are 45 parts. Sasha made five parts of the first car, 3 parts of the second car, and 1 part of the third car. That is, a t... | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,885 |
7.2. Yura and Roma found five consecutive natural numbers and squared each of them. After that, Yura took three of the five resulting squares, while Roma took the two remaining ones. It turned out that the sum of the numbers Yura took is equal to the sum of the numbers Roma took. Provide an example of the numbers Yura ... | For example, 10, 11, 12, 13, 14. Yura took the squares of the first three numbers $100+121+144=365$. Roma got the remaining two squares $169+196=365$.
Criteria: Any suitable set of five numbers without verification - 3 points. | 100+121+144=365169+196=365 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,886 |
7.3. On a plane, there are points $A, B, C, D, X$. Some segment lengths are known: $A C=2$, $A X=5$, $A D=11$, $C D=9$, $C B=10$, $D B=1$, $X B=7$. Find the length of the segment $C X$. | Answer: 3.
Solution: Note that $A D=11=2+9=A C+C D$. Therefore, points $A, C, D$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Note that $C B=10=9+1=C D+D B$. Therefore, points $C, D, B$ lie on the same line $C D$ (since the triangle inequality becomes an equality). Therefore, all fou... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,887 |
7.4. Petya and the Wolf are playing a game. Initially, the number 0 is written on the board, and on each turn, the written number must be increased by 2, 3, 8, or 9. Moves are made alternately, with Petya going first. The player who, after their move, makes the number a multiple of 2020 wins. Who can ensure their victo... | Answer: Petya.
Solution: Let Petya first write 3, and then complement the Wolf's move so that the sum of their moves adds up to 11 (if the Wolf adds 3, then Petya adds 8, and vice versa; if the Wolf adds 2, then Petya adds 9, and vice versa). Thus, after Petya's move, the number will be of the form $3 + 11k$. In this ... | Petya | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,888 |
7.5. In a club, 50 schoolchildren attended, who sometimes came to classes. It turned out that any two schoolchildren met at some class exactly once. In addition, it is known that not all schoolchildren came to any class at the same time. Prove that there is a schoolchild who attended at least 8 classes. | Solution: Suppose this is not the case. Take an arbitrary student, let's call him Yura. Yura attended no more than 7 classes and met all other students. That is, there will be a class, let's say the first one, where, besides Yura, at least 7 other students attended (if this is not the case, then the total number of stu... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 9,889 |
11.1. Which natural numbers can be represented as the fraction $\frac{x^{3}}{y^{4}}$, where $x$ and $y$ are some natural numbers? | Answer. Any natural number.
Solution. Let $n$ be any natural number. Set $x=n^{3}$, $y=n^{2}$, then $\frac{x^{3}}{y^{4}}=\frac{n^{9}}{n^{8}}=n$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,890 |
11.2. In an isosceles triangle \(ABC\) with base \(AC\), the angle between the bisector and the altitude drawn from vertex \(A\) is 60 degrees. Find the angles of triangle \(ABC\). | Answer: 20, 20, and 140 degrees.
Solution. Let $A P$ and $A M$ be the altitude and angle bisector, respectively, drawn from vertex $A$. Denote the measure of angle $BAC$ as $x$, then
$\angle PAC=90^{\circ}-x$ and $\angle MAC=\frac{x}{2}$.
1) Suppose first that point $M$ lies on segment $PC$, which occurs when $\angl... | 20,20,140 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,891 |
11.4. It is known that the lengths of the sides of a triangle are consecutive natural numbers, and the radius of its inscribed circle is 4. Find the radius of the circumscribed circle of this triangle. | Answer: $\frac{65}{8}$.
Solution: Let the lengths of the sides be $n-1, n, n+1$ for some natural number $n$. By Heron's formula, the area of the triangle $S=\sqrt{\frac{3 n^{2}}{4}\left(\frac{n^{2}}{4}-1\right)}=\frac{1}{2} P \times r=\frac{3}{2} n \times 4=6 n$, from which $n=14, S=84$. Substituting this into the for... | \frac{65}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,893 |
11.5. Solve the equation in integers: $\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3$. | Answer. $(\mathrm{x}, \mathrm{y}, z)=(2,2,3),(1,3,8),(1,4,5),(-4,1,1)$ and all permutations of these triples. In total, there are 18 solutions.
Solution. Without loss of generality, we can assume that $a \leq b \leq c$.
Notice that if $x<0$, then $1+\frac{1}{x}<1$, and if $x \geq 1$, then $1+\frac{1}{x} \leq 2$. Ther... | (\mathrm{x},\mathrm{y},z)=(2,2,3),(1,3,8),(1,4,5),(-4,1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,894 |
9.1. The company receives apple and grape juices in standard barrels and produces a cocktail (mixture) of these juices in standard cans. Last year, one barrel of apple juice was enough for 6 cans of cocktail, and one barrel of grape juice was enough for 10. This year, the proportion of juices in the cocktail (mixture) ... | Answer: 15 cans.
Solution. Last year, one barrel of apple juice was enough for 6 cans of cocktail, which means each can contained $1 / 6$ of a barrel of apple juice. Similarly, one barrel of grape juice was enough for 10 cans, which means each can contained $1 / 10$ of a barrel of grape juice. Therefore, the capacity ... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,895 |
9.2. It is known that $\frac{a^{2} b^{2}}{a^{4}-2 b^{4}}=1$. Find all possible values of the expression $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$. | Answer: $\frac{1}{3}$.
Solution. Multiply by the denominator and transform the equality in the condition to $b^{2}\left(a^{2}+b^{2}\right)=a^{4}-b^{4}=\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)$. Here, $a, b$ cannot both be zero at the same time, otherwise the denominator of the fraction in the condition would t... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,896 |
9.4. Is it possible to place one chip in some cells of an 8 by 8 chessboard so that the number of chips in any two adjacent rows differs by a factor of 3, and in any two adjacent columns by a factor of 4? There must be at least one chip on the board. | Answer: No.
Solution. A row or column of an 8x8 board cannot contain more than 8 chips, so the minimum number of chips in a row is 1 or 2; otherwise, one of the adjacent rows would contain at least 9 chips. From the condition, it easily follows that in the first case, the rows contain $1,3,1,3,1,3,1,3$ or $3,1,3,1,3,1... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,897 |
9.5. For what minimum $\boldsymbol{n}$ in any set of $\boldsymbol{n}$ distinct natural numbers, not exceeding 100, will there be two numbers whose sum is a prime number? | Answer. $\boldsymbol{n}=51$.
Solution. The sum of two even natural numbers is always even and greater than two, hence it cannot be a prime number. Therefore, the example of a set of all fifty even numbers not exceeding 100 shows that the minimum $\boldsymbol{n}$ is not less than 51.
On the other hand, let's divide al... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,898 |
10.4. Find all sets of four real numbers such that each number, when added to the product of the other three, equals 2. | Answer. $\{1,1,1,1\},\{-1,-1,-1,3\},\{-1,-1,3,-1\},\{-1,3,-1,-1\}$ and $\{3,-1,-1,-1\}$.
Solution. Let the numbers of the sought quadruple be $a, b, c, d$. By the condition, $a+bcd=b+acd=2$, from the first equation we have $(a-b)(1-cd)=0$, and if $a \neq b, cd=1$ and from the second equation $a+b=2$. Therefore, for an... | {1,1,1,1},{-1,-1,-1,3},{-1,-1,3,-1},{-1,3,-1,-1},{3,-1,-1,-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,900 |
10.5. In each cell of a 5 by 5 table, a letter is written such that in any row and in any column there are no more than three different letters. What is the maximum number of different letters that can be in such a table | Answer: 11.
Solution: If each row contains no more than two different letters, then the total number of letters does not exceed $10=5 * 2$. Further, we can assume that the first row contains exactly three different letters. If each of the remaining rows has at least one letter in common with the first, then the total ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,901 |
11.1. Let $a^{2}+b^{2}=c^{2}+d^{2}=1$ and $a c+b d=0$ for some real numbers $a, b, c, d$. Find all possible values of the expression $a b+c d$. | Answer: 0.
Solution: Let's first assume $b \neq 0$. From the second equation, express $d=\frac{-a c}{b}$ and substitute it into the equation $c^{2}+d^{2}=1$. Eliminating the denominator, we get $c^{2}\left(a^{2}+b^{2}\right)=b^{2}$, from which, given $a^{2}+b^{2}=1$, we obtain $b= \pm c$. Substituting this into the eq... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,902 |
11.2. Solve the equation: $\sin ^{2} x+\sin ^{2} 2 x+\sin ^{2} 3 x=2$. | Answer. $x_{1}=\frac{\Pi}{4}+\frac{\Pi n}{2}, x_{2}=\frac{\Pi}{2}+\Pi n, x_{1}= \pm \frac{\Pi}{6}+\Pi n$.
Solution. We will use the formula $\sin ^{2} t=\frac{1-\cos 2 t}{2}$, and transform the equation to $\cos 6 x+\cos 4 x+\cos 2 x=-1$. We will add the first and third cosines in it: $2 \cos 2 x \operatorname{Cos} 4 ... | x_{1}=\frac{\Pi}{4}+\frac{\Pin}{2},x_{2}=\frac{\Pi}{2}+\Pin,x_{1}=\\frac{\Pi}{6}+\Pin | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,903 |
11.3. Can the sum of the volume, the lengths of all edges, and the areas of all faces of a certain rectangular parallelepiped, the lengths of the edges of which are integers, equal 866? | Answer. No.
Solution. Let the lengths of the edges of the original parallelepiped be $x, y, z$. Then the sum of the volume, the lengths of all edges, and the areas of all its faces is $x y z + 2(x y + x z + y z) + 4(x + y + z) = 866$. If we add 8 to both sides, this equation can be written as $(x+2)(y+2)(z+2)=874$. Th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,904 |
11.4. In a set $X$ of 17 elements, a family of $N$ distinct non-empty subsets is selected such that each element of the set $X$ is contained in exactly two subsets from this family. What is the maximum value of $N$? Find the number of all possible different types of such families for the maximum $N$. Two families of su... | Answer. The maximum $N$ is 25, and there exist two different types of families of 25 subsets that satisfy the condition of the problem.
Solution. Consider an arbitrary family of $N$ distinct non-empty subsets such that each element of the set $X$ is contained in exactly two subsets of this family. If there are $x$ one... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,905 |
8.1. Large sandglasses measure an hour, and small ones measure 11 minutes. How can you use these sandglasses to measure a minute? | Solution: We will run the large hourglass twice in a row and the small one eleven times in a row. A minute will be measured between the second time the large hourglass finishes (120 minutes) and the 11th time the small one finishes (121 minutes).
Criteria: Any correct example - 7 points. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,906 |
8.2. Students of the 8th grade exchanged stickers on December 31 and January 1. It turned out that on December 31, each received as many stickers as all the others combined on January 1. Prove that all students received an equal number of stickers. | Solution: Let all students receive a total of $N$ stickers on the second day. Consider any student; let him receive $m$ stickers on the second day. Then all the others received a total of $N-m$ stickers on the second day. Therefore, this participant received $N-m$ stickers on the first day, which means he received a to... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 9,907 |
8.3. 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 tips if the weight on it is greater than three times the weight on the left pan. The left pan tips if three times the weight on it is greater than t... | 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 | 9,908 |
8.4. Given a triangle $\mathrm{ABC}$ with angle $\mathrm{BAC}$ equal to $30^{\circ}$. In this triangle, the median $\mathrm{BD}$ was drawn, and it turned out that angle $\mathrm{BDC}$ is $45^{\circ}$. Find angle $\mathrm{ABC}$. | Answer: $45^{\circ}$
Solution: Draw the height $C H$. Then $H D=A D=C D$ as the median to the hypotenuse. Moreover, $\angle H C D=\angle C H A-\angle H A C=60^{\circ}$, so triangle $C H D$ is equilateral, which means $\angle H D C=60^{\circ}$ (from which it follows, in particular, that $H$ lies between $A$ and $B$). T... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 9,909 |
8.5. In the city of Omsk, a metro has been built, represented by a straight line. On this same line is the house where Nikita and Egor live. Every morning they leave the house for classes at the same time, after which Egor runs to the nearest metro station at a speed of 12 km/h, while Nikita walks along the metro line ... | Answer: 23 km/h
Solution: Obviously, this is only possible if the subway train first arrives at the nearest station A, where Egor runs to, and then goes to station B, where Nikita is heading.
Let $v$ be the speed of the subway, $S$ be the distance between two adjacent stations, and $R$ be the distance between this su... | 23 | Other | math-word-problem | Yes | Yes | olympiads | false | 9,910 |
10.1. Vikentiy walked from the city to the village, and at the same time Afanasiy walked from the village to the city. Find the distance between the village and the city, given that the distance between the pedestrians was 2 km twice: first, when Vikentiy had walked half the way to the village, and then, when Afanasiy ... | Answer: 6 km.
Solution. Let the distance between the village and the city be denoted as $S$ km, the speeds of Vikentiy and Afanasy as $x$ and $y$, and calculate the time spent by the travelers in the first and second cases. In the first case, we get: $\frac{S / 2}{x}=\frac{S / 2-2}{y}$, in the second case $\frac{2 S /... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,912 |
10.2. Can the number 199... 99 (one one and 10 nines) be represented as the sum of two natural numbers, the sums of whose digits are the same? | Answer: No.
Solution: Suppose we can represent the number $A=99...99$ (with a total of 9 nines) as the sum of two numbers B and C, the sums of whose digits are equal. If, when adding the digits of the last place of B and C, a carry-over to the previous place occurs, then the last digit of the sum does not exceed 8. Ho... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,913 |
10.4. Is it possible to place different integers at the vertices of a cube so that the number at each vertex is equal to the sum of the three numbers at the ends of the edges emanating from that vertex? Answer. Yes, it is possible, for example: in the vertices of the lower face clockwise $6,1,-3,2$, in the vertices of ... | Solution. Let's denote the vertices of the cube as $\mathrm{ABCDA}_{1} \mathrm{~B}_{1} \mathrm{C}_{1} \mathrm{D}_{1}$, and call vertices $\mathrm{A}, \mathrm{C}, \mathrm{B}_{1}$, and $\mathrm{D}_{1}$ black, and vertices $\mathrm{B}, \mathrm{D}, \mathrm{A}_{1}, \mathrm{C}_{1}$ white. One end of each edge is white, and t... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 9,914 |
10.5. Natural numbers $a, b, c, d$ are such that $a+c=1000, b+d=500$. Find the maximum value of the sum $\frac{a}{b}+\frac{c}{d}$. | Answer. $\frac{1}{499}+\frac{999}{1}$.
Solution. Due to symmetry, we can assume that $b \geq d$. Then, by replacing the pair $a, c$ with the pair $a-1, c+1$, we get $\left(\frac{a-1}{b}+\frac{c+1}{d}\right)-\left(\frac{a}{b}+\frac{c}{d}\right)=\frac{1}{d}-\frac{1}{b} \geq 0$ - an increase in the desired expression, so... | \frac{1}{499}+\frac{999}{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,915 |
9.1. From points A and B towards each other with constant speeds, a motorcyclist and a cyclist started simultaneously from A and B, respectively. After 20 minutes from the start, the motorcyclist was 2 km closer to B than the midpoint of AB, and after 30 minutes, the cyclist was 3 km closer to B than the midpoint of AB... | Answer: In 24 minutes.
Solution: In 10 minutes, the motorcyclist travels $1 / 4$ of the distance from A to B plus 1 km, while the cyclist travels $1 / 6$ of the distance from A to B minus 1 km. Therefore, in 10 minutes, both of them, moving towards each other, cover $1 / 4 + 1 / 6 = 5 / 12$ of the distance from A to B... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,916 |
9.2. Can a number ending in 222 be a non-trivial power of some natural number, that is, be represented in the form $x^{y}$, where $x, y>1$ are natural numbers? | Answer: No, it cannot.
Solution. A number ending in 2 is even, so the base $x$ of the power in question must be even. In this case, its $y \geq 2$-th power must be divisible by $2^{y}$ and, consequently, divisible by 4. However, according to the divisibility rule by 4, a number ending in 222 is not divisible by 4, sin... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,917 |
9.3. Vasya must write one digit on each face of several dice so that any ordered combination of three digits from 000 to 999 inclusive can be obtained by selecting some three different dice and placing them with the appropriate sides up in the correct order. At the same time, the digits 6 and 9 do not transform into ea... | Answer: 5.
Solution. Since among the possible combinations there must be $000, 111, 222, \ldots, 999$, each digit must appear on at least three different faces (of different cubes), so there must be at least 30 digits on the faces in total, hence there must be at least $30: 6=5$ cubes.
On the other hand, we will show... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 9,918 |
9.5. On the board, there are 10 numbers: $1,2,3,4,4,5,5,11,12,13$. You can perform two types of operations with them: either subtract 1 from any nine of them and add 9 to the remaining one, or vice versa, subtract 9 from one and add 1 to each of the others. Negative numbers cannot be obtained. Can all ten numbers be ma... | Answer. No.
Solution. Note that with any of the described operations, the difference between any two written numbers either does not change or changes by 10. We can divide all numbers into 5 pairs: 1 and 11, 2 and 12, 3 and 13, 4 and 4, 5 and 5, where the differences between the numbers in each pair are 0 or 10. If af... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 9,919 |
9.1. From two cities, the distance between which is 105 km, two pedestrians set out simultaneously towards each other at constant speeds and met after 7.5 hours. Determine the speed of each of them, knowing that if the first walked 1.5 times faster, and the second 2 times slower, they would have met after $8 \frac{1}{1... | Answer: 6 and 8 km per hour.
Solution: Let their speeds be $x$ and $y$ km per hour, respectively. From the condition, we get: $\frac{15}{2}(x+y)=105, \frac{105}{13}\left(\frac{3}{2} x+\frac{1}{2} y\right)=105$, from which $x=6, y=8$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 9,920 |
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