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4. Variant 1.
The numbers $96, 28, 6, 20$ were written on the board. One of them was multiplied, another was divided, a third was increased, and a fourth was decreased by the same number. As a result, all the numbers became equal to one number. Which one? | Answer: 24.
Solution: Addition and multiplication increase numbers, while subtraction and division decrease them, so the two larger numbers have decreased, and the smaller ones have increased. Now let's consider the two middle numbers 20 and 28. It is clear that a number was subtracted from 28, and the same number was... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
Petya wrote the numbers from 1 to 10 on cards and laid them out along the edge of a $3 \times 4$ rectangle. At first, one of the cards - with the number 7 - was opened (see the figure). When the other cards were opened, it turned out that the sum of the numbers in the top and bottom horizontal rows is th... | Answer: 23.
Solution. Let $B$ be the last number in the second horizontal row. Since the sum of all numbers from 1 to 10 is 55, we have $2A + B + 7 = 55$ or $B = 48 - 2A$. It follows that $B$ is an even number. To make $A$ the largest, $B$ needs to be the smallest. The smallest even number among the given numbers is 2... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 7. Variant 1.
100 natural numbers are written in a circle. It is known that among any three consecutive numbers, there is an even number. What is the smallest number of even numbers that can be among the written numbers? | Answer: 34.
Solution. Consider any 3 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 99 into 33 groups of 3 consecutive numbers. In each such group, there is at least one even number. Thus, the total number of even numbers is at least $1+33=34$. Such a situation is possible. ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.3. Grandma has two balls of wool: a large one and a small one. From the large one, she can knit either a sweater and three socks, or five identical caps. $A$ from the small one - either half a sweater, or two caps. (In both cases, all the wool will be used.) What is the maximum number of socks that Grandma can knit u... | # Solution:
Method 1. Half a sweater requires as much wool as 2 hats, so a sweater requires as much wool as 4 hats. Then 4 hats and 3 socks require as much wool as 5 hats. Therefore, one hat is equivalent to three socks. In total, 5 + 2 = 7 hats or 21 socks can be knitted.
Method 2. Let the amount of wool required fo... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. Ten different natural numbers are such that the product of any five of them is even, and the sum of all ten is odd. What is the smallest possible sum of all these numbers? Justify your answer. | Solution: There cannot be more than four odd numbers (five odd numbers in a product give an odd number), and, moreover, their quantity must be odd. Therefore, there are either 3 or 1 odd numbers, and 7 or 9 even numbers, respectively. The sets with the smallest sum are 1, 3, 5, 2, 4, 6, 8, 10, 12, 14 and 1, 2, 4, 6, 8,... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On the faces $BCD, ACD, ABD$, and $ABC$ of the tetrahedron $ABCD$, points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are marked, respectively. It is known that the lines $AA_{1}, BB_{1}, CC_{1}$, and $DD_{1}$ intersect at point $P$, and $\frac{AP}{A_{1}P}=\frac{BP}{B_{1}P}=\frac{CP}{C_{1}P}=\frac{DP}{D_{1}P}=r$. Find all po... | Solution. Let V be the volume* of tetrahedron ABCD. We introduce the consideration of the partition of the original tetrahedron ABCD into tetrahedra PBCD, PACD, PABD, and PABC. Then for the volumes of the specified tetrahedra, the following is true:
$$
V=V_{\mathrm{PBCD}}+V_{\mathrm{PACD}}+V_{\mathrm{PABD}}+V_{\mathrm... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 7.1. (7 points)
Find the value of the expression
$$
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)
$$ | Answer: 1.
Solution: Notice that
$$
\begin{array}{r}
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=\frac{3}{2} \cdot \frac{2}{3}=1,\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{5}{4} \cdot \frac{4}{5}=1, \ldots \\
\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)=\frac{2 m+1}{2 m} \cdot \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7.2. (7 points)
Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon and, without stopping, arrived: one - in $B$ at 4 PM, and the other - in $A$ at 9 PM. At what hour was dawn that day? | Answer: at 6 o'clock.
Solution: Let $x$ be the number of hours from dawn to noon. The first pedestrian walked $x$ hours before noon and 4 after, the second - $x$ before noon and 9 after. Note that the ratio of times is equal to the ratio of the lengths of the paths before and after the meeting point, so $\frac{x}{4}=\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7.5. (7 points)
In a family, there are six children. Five of them are older than the youngest by 2, 6, 8, 12, and 14 years, respectively. How old is the youngest if the ages of all the children are prime numbers? | Answer: 5 years.
Solution. First, let's check the prime numbers less than six. It is obvious that the number we are looking for is odd. The number 3 does not satisfy the condition because $3+6=9-$ is not a prime number. The number 5 satisfies the condition because $5+2=7, 5+6=11, 5+8=13, 5+12=17, 5+14=19$, which means... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The second term of an infinite decreasing geometric progression is 3. Find the smallest possible value of the sum $A$ of this progression, given that $A>0$. | Answer: 12.
Solution. Let the first term of the progression be $a$, and the common ratio be $q$. The sum of the progression $A$ is $\frac{a}{1-q}$. From the condition, we have $3=a q$, from which $a=3 / q$. Therefore, we need to find the minimum value of $A=\frac{3}{q(1-q)}$. Note that from the condition it follows: $... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A positive integer $N$ and $N^{2}$ end with the same sequence of digits $\overline{a b c d}$, where $a-$ is a non-zero digit. Find $\overline{a b c}$. | Answer: 937.
Solution. Let's represent the number $N$ as $10000 M+k$, where $k$ and $M$ are natural numbers, and $1000 \leq k \leq 9999$. Since $N$ and $N^{2}$ end with the same sequence of digits, the difference
$$
N^{2}-N=\left(10^{4} M+k\right)^{2}-\left(10^{4} M+k\right)=10^{4}\left(10^{4} M^{2}+2 M k-M\right)+k^... | 937 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. In the city of Perpendicularinsk, it was decided to build new multi-story houses (some of them may be single-story), but in such a way that the total number of floors would be 30. The city architect, Parallelnikov, proposed a project according to which, if after construction one climbs to the roof of each new hou... | Solution: 1) We will show that the project does not involve building houses with more than two floors. Assume the opposite, that such houses are planned. Take the lowest of them and reduce it by one floor, building an additional one-story house as a result. The sum of the numbers in question will decrease by the number... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.6. The sheriff believes that if the number of bandits he catches on a certain day is a prime number, then he is lucky. On Monday and Tuesday, the sheriff was lucky, and starting from Wednesday, the number of bandits he caught was equal to the sum of the number from the day before yesterday and twice the number from ... | Solution: Let's say the sheriff caught 7 bandits on Monday and 3 on Tuesday. Then on Wednesday, Thursday, and Friday, he caught 13, 29, and 71 bandits, respectively. All these numbers are prime, and the sheriff has been lucky for five days in a row.
We will show that the sheriff cannot be lucky for six days in a row. ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.9. In the company, there are 100 children, some of whom are friends (friendship is always mutual). It is known that by selecting any child, the remaining 99 children can be divided into 33 groups of three such that in each group, all three are pairwise friends. Find the minimum possible number of pairs of friends.
... | Answer: 198.
Solution: Let's translate the problem into the language of graphs, associating each child with a vertex and each friendship with an edge. Then we know that in this graph with 100 vertices, after removing any vertex, the remaining vertices can be divided into 33 triples such that the vertices in each tripl... | 198 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. 9 knights and liars stood in a row. Each said that there is exactly one liar next to him. How many liars are there among them, if knights always tell the truth, and liars always lie? | Answer: 3 liars.
Sketch of the solution. Consider the partition of all into groups of people of the same type standing in a row, with people of different types in adjacent groups. In such a group, there can only be one liar. If there are at least two, the extreme liars tell the truth. There are no more than two knight... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Will Katya be able to write a ten-digit number on the board where all digits are different and all differences between two adjacent digits are different (when finding the difference, the larger is subtracted from the smaller)? | Answer: will be able to.
Example: 9081726354.
Criteria. 7 points for any correct example. | 9081726354 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. There is a paper rectangle $3 \times 100$, divided into 300 cells $1 \times 1$. What is the maximum number of pairs consisting of one corner and one $2 \times 2$ square that can be cut out along the grid lines? (A corner is obtained from a $2 \times 2$ square by removing one of its corner cells). | Answer: 33.
Sketch of the solution. The square in the middle row occupies two cells, and the corner - at least one, so the pair occupies at least three cells in the middle row. If there are no fewer than 34 pairs, then they occupy at least $34 \times 3=102$ cells in the middle row, while there are only 100 there.
Exa... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.
(A. Yaku... | Answer: $60^{\circ}$.
Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.
^{2} + 1$ we verify the correctness of the answer.
Comment. The correct answer without a rigorous justification up to 2 points. The inequality is solved, but the wrong answer is chosen 2 points. | -4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.2. Two cars simultaneously departed from the same point and are driving in the same direction. One car was traveling at a speed of 50 km/h, the other at 40 km/h. Half an hour later, a third car departed from the same point and in the same direction, which overtook the first car one and a half hours later than the sec... | Answer: 60 km/h.
In half an hour, the first car will travel 25 km, and the second car will travel 20 km. Let $x$ be the speed of the third car. The time it takes for the third car to catch up with the first car is $\frac{25}{x-50}$, and the time to catch up with the second car is $\frac{20}{x-40}$. We get the equation... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. At a drama club rehearsal, 8 people gathered. Some of them (honest people) always tell the truth, while the others always lie. One of those present said: "There is not a single honest person here." The second said: "There is no more than one honest person here." The third said: "There is no more than two honest pe... | 3. The standard evaluation methodology for solutions is provided below.
| Points | Correctness (Incorrectness) of the Solution |
| :---: | :--- |
| 7 | Fully correct solution. |
| $6-7$ | Correct solution, but with minor flaws that do not significantly affect the solution. |
| $5-6$ | The solution is generally correct... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Two-headed and seven-headed dragons came to a meeting. At the very beginning of the meeting, one of the heads of one of the seven-headed dragons counted all the other heads. There were 25 of them. How many dragons in total came to the meeting? | Answer: 8 dragons. Solution. Subtract the 6 heads belonging to the seven-headed dragon from the 25 heads counted by the seven-headed dragon. 19 heads remain. The remaining dragons cannot all be two-headed (19 is an odd number). There can only be one more seven-headed dragon (if there were two, an odd number of heads wo... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Vovodya is running on a circular track at a constant speed. There are two photographers standing at two points on the track. After the start, Vovodya was closer to the first photographer for 2 minutes, then closer to the second photographer for 3 minutes, and then closer to the first photographer again. How long did... | Solution. Let's divide both arcs of the circle between the photographers in half. The halves of the arcs adjacent to the second photographer make up half the distance. Vasya ran this half in 3 minutes, so he will run the entire circle in 6 minutes.
Answer: in 6 minutes.
Grading criteria: Full solution - 7 points. Onl... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. What is the minimum number of factors that need to be crossed out from the number $99!=1 \cdot 2 \cdot \ldots \cdot 99$ so that the product of the remaining factors ends in $2?$ | Answer: 20 factors.
Solution. From the number 99! it is necessary to remove all factors that are multiples of 5, otherwise the product will end in 0. There are a total of 19 such factors (ending in 0 or 5).
The product of the remaining factors ends in 6. Indeed, since the product $1 \times 2 \cdot 3 \cdot 4 \cdot 6 \... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. Does there exist a ten-digit number, divisible by 11, in which all digits from 0 to 9 appear? | Answer. Yes, for example, 9576843210.
Solution. Let's consider a possible way to find the required number. Note that a number is divisible by 11 if the difference between the sum of the digits in the odd positions and the sum of the digits in the even positions is divisible by 11. If we write all ten digits in descend... | 9576843210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are pan scales without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings? | Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the count... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. From the set of numbers $1,2,3,4, \ldots, 2021$, one number was removed, after which it turned out that the sum of the remaining numbers is divisible by 2022. Which number was removed? | Answer: 1011.
Solution. Let's write out the sum and perform grouping:
$1+2+3+\cdots+2019+2020+2021=(1+2021)+(2+2020)+$ $\cdots+(1012+1010)+1011$.
All the terms enclosed in parentheses are divisible by 2022. If we remove the last ungrouped term from the sum, the sum will be a multiple of 2022.
Comment. Answer only -... | 1011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Petya came home from school today at $16:45$, looked at the clock and wondered: after what time will the hands of the clock be in the same position for the seventh time since he came home from school? | Answer: 435 minutes.
Solution. The speed of the minute hand is 12 divisions/hour (one division here refers to the distance between adjacent numbers on the clock face), and the hour hand is 1 division/hour. Before the seventh meeting of the minute and hour hands, the minute hand must first "lap" the hour hand 6 times, ... | 435 | Other | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) There are apples in five boxes, with an equal number of apples in each. When 60 apples were taken out of each box, after that, the total number of apples left was the same as the number of apples that were originally in two boxes. How many apples were in each box?
Answer: 100. | Solution. In total, $60 \cdot 5=300$ apples were taken out, and this is equal to the number of apples that were in three boxes. Therefore, there were 100 apples in each box.
Criteria. Any correct solution: 7 points.
If it is not justified that there were 100 apples in each box, but it is verified that the condition i... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) A gnome in shoes weighs 2 kg more than a gnome without shoes. If you put five identical gnomes in shoes and five such gnomes without shoes on the scales, the scales will show 330 kg. How much does a gnome in shoes weigh? | Answer: 34 kg.
Solution. Let's put boots on five gnomes, then the weight will increase by 10 kg. It turns out that ten gnomes in boots weigh 340 kg. Therefore, one gnome in boots weighs 34 kg.
Criteria. Correctly found the weight of a gnome in boots by any method: 7 points.
Correctly found the weight of a gnome with... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Two runners, starting simultaneously at constant speeds, run on a circular track in opposite directions. One of them runs the loop in 5 minutes, while the other takes 8 minutes. Find the number of different meeting points of the runners on the track, if they ran for at least an hour. | Answer: 13 points.
Solution.
Let the length of the track be $\mathrm{S}$ meters. Then the speeds of the runners are $\mathrm{S} / 5$ and S/8 meters per minute, respectively. | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out from the plane?
Answer: 150 triangles
# | # Solution.
Consider 100 nodes - the intersection points of lines in the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second
... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.1. What is the sum of the digits of the number $A=10^{50}-10^{40}+10^{30}-10^{20}+10^{10}-1$? | # Answer: 270.
Solution. The number is the sum of three numbers: a number composed of 10 nines followed by 40 zeros, a number composed of 10 nines followed by 20 zeros, and finally, a number composed of 10 nines. All the nines fall on the zeros in the other addends, so there is no carry-over, and the answer is $90+90+... | 270 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Find the largest natural number with all distinct digits such that the sum of any two of its digits is a prime number. | Answer: 520.
Solution: If the desired number is at least a four-digit number, then it either has three digits of the same parity or two pairs of digits of the same parity. In each of these cases, we get that two of the sums of the digits are even. Therefore, they must equal 2. The number 2 can be represented as the su... | 520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. A round table was sat at by 10 people - liars and knights. Liars always lie, and knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, each one said: “I have more coins than my right neighbor.” What is the maximum number of knig... | Answer: 6.
Solution: After the coins are passed, each person sitting at the table can have 0, 1, or 2 coins. Note that 3 knights cannot sit in a row. Indeed, let knights $A, B, C$ sit next to each other, with $B$ sitting to the right of $A$, $C$ to the right of $B$, and $D$ to the right of $C$. If $A$ has $x$ coins, $... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. In the class, there are more than 20 but fewer than 30 students. In the class, those who attend the chess club are half the number of those who do not attend. And those who attend the checkers club are one-third the number of those who do not attend. How many students are in the class? Provide all possible answers... | Solution: Let $n$ be the number of students in the class who attend the chess club, then $2n$ students do not attend, and the total number of students in the class is $3n$, meaning the total number of students in the class is divisible by 3. Similarly, from the fact that the number of people attending the checkers club... | 24 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.3. On the island of knights and liars, each resident was asked about each of the others: is he a knight or a liar. In total, 42 answers of "knight" and 48 answers of "liar" were received. What is the maximum number of knights that could have been on the island? Justify your answer. (It is known that knights always te... | Solution: Each of the $n$ residents gave $n-1$ answers; in total, there were $n(n-1)$ answers, which, according to the problem, equals $42+48=90$. Hence, $n=10$, meaning there are 10 residents on the island. Let the number of knights be $x$, then the number of liars is $10-x$. Answers of "liar" arise in two cases: when... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1 How many five-digit natural numbers are there that are divisible by 9, and for which the last digit is 2 more than the second last digit? | Answer: 800. Hint The second last digit (the tens digit) can be any from 0 to 7 (so that after adding two, the last digit makes sense). The third and second digits can be any (from 0 to 9). After choosing the specified three digits (the second, third, and fourth), the last digit is uniquely determined by the second las... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. A tourist goes on a hike from $A$ to $B$ and back, and completes the entire journey in 3 hours and 41 minutes. The route from $A$ to $B$ first goes uphill, then on flat ground, and finally downhill. Over what distance does the road run on flat ground, if the tourist's speed is 4 km/h when climbing uphill, 5 km/h on ... | Solution. Let $x$ km of the path be on flat ground, then $9-x$ km of the path (uphill and downhill) the tourist travels twice, once (each of the ascent or descent) at a speed of 4 km/h, the other at a speed of 6 km/h, and spends $(9-x) / 4 + (9-x) / 6$ hours on this part. Since the tourist walks $2x / 5$ hours on flat ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1-0. The number 111123445678 is written on the board, and several digits (not all) need to be erased to get a number that is a multiple of 5. In how many ways can this be done? | Answer: 60
Solution. The digits 6, 7, and 8 must be crossed out, and 5 must be kept (otherwise, the number will not be divisible by 5). Each digit before the five can be crossed out or not. There are two options for each of the digits 2 and 3 (cross out or not), three options for the digit 4 (do not cross out, cross o... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2-0. The number $n$ is such that $8n$ is a 100-digit number, and $81n$ is a 102-digit number. What can the second digit from the beginning of $n$ be? | Answer: 2
Solution. Since $8 n10^{101}$ (equality here is obviously impossible), it means $n>123 \cdot 10^{97}$. Therefore, the second digit from the beginning of the number $n$ is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4-0. Kisa and Busya came to the cafeteria during the break, where only muffins and doughnuts were sold, costing a whole number of rubles. Kisa bought 8 muffins and 3 doughnuts, spending less than 200 rubles, while Busya bought 4 muffins and 5 doughnuts, spending more than 150 rubles. Name the highest possible price of ... | Answer: 19
Solution. Let the price of a cake be $k$, and the price of a bun be $p$ rubles. Then $8 k+3 p<150$. Multiplying the first inequality by 5, and the second by 3, we get $40 k+15 p<1000, -12 k-15 p<-450$. Adding these inequalities: $28 k<550$, from which, taking into account the integer nature, $k \leqslant 19... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-0. A certain quadratic trinomial $x^{2}-p x+q$ has integer roots $x_{1}$ and $x_{2}$. It turns out that the numbers $x_{1}$, $x_{2}$, and $q$ form a decreasing arithmetic progression. Find the sum of all possible values of $x_{2}$. | Answer: -5
Solution. By Vieta's theorem $q=x_{1} x_{2}$. Then $2 x_{2}=x_{1}+x_{1} x_{2}$, from which $x_{1}=\frac{2 x_{2}}{1+x_{2}}$. Since the roots are integers, $2 x_{2}$ is divisible by $1+x_{2}$. But $2+2 x_{2}$ is divisible by $1+x_{2}$, which means 2 is also divisible by $1+x_{2}$. Therefore, $x_{2}=-3,-2,0$ o... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7-0. The number $n$ has exactly six divisors (including 1 and itself). They were arranged in ascending order. It turned out that the third divisor is seven times greater than the second, and the fourth is 10 more than the third. What is $n$? | Answer: 2891
Solution. If $n$ has six divisors, then either $n=p^{5}$ or $n=p \cdot q^{2}$ (where $p$ and $q$ are prime numbers). In the first case, $p=7$ (since the third divisor is seven times the second), but then the second condition is not satisfied.
Therefore, $n$ has two prime divisors (one of which is squared... | 2891 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8-0. Once, King Shahryar said to Scheherazade: "Here is a paper circle with 1001 points on its boundary. Each night, you must cut the figure you have along a straight line containing any two of the marked points, keeping only one fragment and discarding the other. And make sure that the figure you keep is not a polygon... | # Answer: 1999
Solution. First, let's describe the strategy that Scheherazade will use to meet Shahryar's conditions for 1998 nights. Initially, she will cut along the lines connecting adjacent vertices and discard the smaller part (let's call such a part a segment). She will do this for 1000 nights. After this, a 100... | 1999 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A right triangle ABC is inscribed in a circle with hypotenuse AB. On the larger leg BC, a point D is taken such that AC = BD, and point E is the midpoint of the arc AB containing point C. Find the angle DEC. | 3. Point $\mathrm{E}$ is the midpoint of arc $\mathrm{AB}$, so $\mathrm{AE}=\mathrm{BE}$. Moreover, the inscribed angles $\mathrm{CAE}$ and $\mathrm{EBC}$, which subtend the same arc, are equal. Also, by the given condition, $\mathrm{AC}=\mathrm{BD}$. Therefore, triangles $\mathrm{ACE}$ and $\mathrm{BDE}$ are congruent... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. For numbers $a, b, c, d$, it is known that $a^{2}+b^{2}=1, c^{2}+d^{2}=1, a c+b d=0$. Calculate $a b+c d$.
| 4. Consider the equality: $(a c+b d)(a d+b c)=0$, since $a c+b d=0$.
We get $a^{2} c d+b^{2} c d+c^{2} a b+d^{2} a b=0$ or $\left(a^{2}+b^{2}\right) c d+\left(c^{2}+d^{2}\right) a b=0$.
Since $a^{2}+b^{2}=1$ and $c^{2}+d^{2}=1$ we obtain $a b+c d=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. There are 40 visually identical coins, among which 3 are counterfeit - they weigh the same and are lighter than the genuine ones (the genuine coins also weigh the same). How can you use three weighings on a balance scale without weights to select 16 genuine coins? | 5. First solution: Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, the... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Find the value of the expression $a^{3}+b^{3}+12 a b$, given that $a+b=4$. | 1. $a^{3}+b^{3}+12 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+12 a b=4\left(a^{2}-\right.$ $\left.a b+b^{2}\right)+12 a b=4 a^{2}+4 b^{2}+8 a b=4(a+b)^{2}=4 \cdot 16=64$ | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The boy went to the shooting range with his father. The father bought him 10 bullets. Later, the father took away one bullet for each miss and gave one additional bullet for each hit. The son fired 55 times, after which he ran out of bullets. How many times did he hit the target? | 2. Each time the boy hit the target, the number of bullets he had remained the same (he used one and received one from his father). Each time the boy missed, the number of bullets he had decreased by 2 (he used one and his father took one). This means that the son missed $10: 2=5$ times out of 55 shots, so he hit 55 - ... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.5. How many solutions in integers \(x, y\) does the equation \(6 x^{2}+2 x y+y+x=2019\) have? | Answer. 4 solutions. Solution. Express $y$ from the given equation: $y=\frac{2019-6 x^{2}-x}{2 x+1}$. Dividing $6 x^{2}+x$ by $2 x+1$ with a remainder, we get $6 x^{2}+x=(3 x-1)(2 x+1)+1$. Thus, the expression for $y$ will take the form: $y=\frac{2019-(3 x-1)(2 x+1)-1}{2 x+1}=\frac{2018}{2 x+1}-3 x+1$. Therefore, for $... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.2. From a square with a side of 10, a green square with a side of 2, a blue square, and a yellow rectangle were cut out (see figure). What is the perimeter of the remaining figure?
The perimeter of a figure is the sum of the lengths of all its sides.
 & :(x+5)=4 \\
5 x+13 & =4(x+5) \\
5 x+13 & =4 x+20 \\
x & =7
\end{aligned}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. On some trees in the magical forest, coins grow. The number of trees that do not grow any coins at all is twice as many as the trees that grow three coins. On three trees, two coins grow, on four trees - four coins, and no tree grows more than four coins. By how much is the total number of coins in the mag... | Answer: 15.
Solution. Let $x$ be the number of trees in the forest on which three coins grow, and on which one coin grows. Then in the forest, $2 x$ trees do not grow any coins at all.
Thus, the total number of coins is
$$
2 x \cdot 0+y \cdot 1+3 \cdot 2+x \cdot 3+4 \cdot 4=3 x+y+22
$$
and the total number of trees... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.6. In the zoo, there are red, yellow, and green parrots (there is at least one parrot of each of the listed colors; there are no parrots of other colors in the zoo). It is known that among any 10 parrots, there is definitely a red one, and among any 12 parrots, there is definitely a yellow one. What is the ma... | Answer: 19.
Solution. Let there be $x$ red, $y$ yellow, and $z$ green parrots in the zoo.
Since among any 10 parrots there is a red one, the number of non-red parrots does not exceed 9, that is, $y+z \leqslant 9$. By similar reasoning, we get that the number of non-yellow parrots does not exceed 11, that is, $x+z \le... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. A boastful fisherman says the same phrase every day: "Today I caught more perch than I did the day before yesterday (2 days ago), but less than I did 9 days ago." What is the maximum number of days in a row that he can tell the truth? | Answer: 8.
Solution. First, let's provide an example where he tells the truth for 8 days in a row. The numbers of perch he caught on consecutive days are indicated:
$$
2020202020202061 \underbrace{728394105}_{\text {truth }} \text {. }
$$
Now let's prove that he could not have told the truth for 9 days in a row. Sup... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. Borya found the smallest prime number \( p \) such that \( 5 p^{2} + p^{3} \) is a square of some natural number. What number did Borya find? | Answer: 11.
Solution: Since $5 p^{2}+p^{3}=p^{2}(5+p)$, the original number is a perfect square if and only if the number $5+p$ is a perfect square. Since $p$ is a prime number, $p+5 \geq 7$. It is sufficient to verify that if $p+5=9$, then $p=4$ is not a prime; if $p+5=16$, then $p=11$ satisfies the condition, and th... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a convex quadrilateral $\mathrm{ABCD}$, the bisector of angle $\mathrm{B}$ passes through the midpoint of side $\mathrm{AD}$, and $\angle \mathrm{C}=\angle \mathrm{A}+\angle \mathrm{D}$. Find the angle $\mathrm{ACD}$.
$, from which $\angle \mathrm{AEB}=180-\angle \mathrm{A}-\angle \mathrm{B} / 2=\angle \mathrm{D}$. Therefor... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. On the board, 100 pairwise distinct natural numbers $a_{1}, a_{2}, \ldots, a_{100}$ were written. Then, under each number $a_{i}$, a number $b_{i}$ was written, obtained by adding to $a_{i}$ the greatest common divisor of the remaining 99 original numbers. What is the smallest number of pairwise distinct numbers t... | # Answer. 99.
First solution. If we set $a_{100}=1$ and $a_{i}=2 i$ for $i=1,2, \ldots, 99$, then $b_{1}=b_{100}=3$, so there will be no more than 99 different numbers among the numbers $b_{i}$. It remains to prove that there will always be 99 different numbers among the numbers $b_{i}$.
Without loss of generality, w... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 9.4. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars alw... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
=1$, that is, $A+B=90^{\circ}$.
Second method. Adding the original equalities, we get $\sin \left(A+45^{\circ}\right)+\sin \left(B+45^{\circ}\right)=2$, from which $\sin \left(A+45^{\circ}\right)=\si... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two parks with a total area of 110 hectares are divided into the same number of plots, and in each park, the plots have the same area, but differ from those in the other park. If the first park were divided into plots of the same area as the second, it would result in 75 plots. If the second park were divided into p... | 2. Answer. 50 and 60 hectares. Solution. Let x be the area of the first park. According to the condition, we will construct a table (for the new division):
| Park | In the new division | | |
| :---: | :---: | :---: | :---: |
| | Area of the park | Number of plots | Area of the plot |
| First | x | 75 | x/75 |
| Sec... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given the functions $f(x)=x^{2}+4 x+3$ and $g(x)=x^{2}+2 x-1$. Find all integer solutions to the equation $f(g(f(x)))=g(f(g(x)))$. | 4. Answer. $x=-2$. Solution. Let's represent the functions as $f(x)=x^{2}+4 x+3=(x+2)^{2}-1$ and $g(x)=x^{2}+2 x-1=(x+1)^{2}-2 . \quad$ Then $\quad f(g(x))=\left((x+1)^{2}-2+2\right)^{2}-1=(x+1)^{4}-1$, $g(f(x))=\left((x+2)^{2}-1+1\right)^{2}-2=(x+2)^{4}-2$. Performing similar operations, we get $g(f(g(x)))=(x+1)^{8}-2... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.3. Given a triangle $A B C$. It is known that $\angle B=60^{\circ}, \angle C=75^{\circ}$. On side $B C$ as the hypotenuse, an isosceles right triangle $B D C$ is constructed inside triangle $A B C$. What is the measure of angle $D A C$? | Answer: $30^{0}$.
Solution 1: From the problem statement, it follows that $\angle A=45^{0}$. Draw a circle with center $\mathrm{M}$ and radius $\mathrm{MB}=\mathrm{MC}$. Since $\angle \mathrm{BDC}=90^{\circ}$, the major arc BC is seen at an angle of $45^{0}$. Therefore, vertex A lies on this circle. This means that tr... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Sasha chose a natural number $N>1$ and wrote down in ascending order all its natural divisors: $d_{1}<\ldots<d_{s}$ (so that $d_{1}=1$ and $d_{s}=N$). Then, for each pair of adjacent numbers, he calculated their greatest common divisor; the sum of the resulting $s-1$ numbers turned out to be $N-2$. What values cou... | Answer: $N=3$.
Solution. Note immediately that $d_{s+1-i}=N / d_{i}$ for all $i=$ $=1,2, \ldots, s$.
The number $d_{i+1}-d_{i}$ is divisible by the GCD $\left(d_{i}, d_{i+1}\right)$, so the GCD $\left(d_{i}, d_{i+1}\right) \leqslant d_{i+1}-d_{i}$. For $i=1, \ldots, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)-$ GCD $... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Two-digit numbers are written on the board. Each number is composite, but any two numbers are coprime. What is the maximum number of numbers that can be written? | Answer: four numbers.
Solution. Evaluation. Since any two of the written numbers are coprime, each of the prime numbers 2, 3, 5, and 7 can appear in the factorization of no more than one of them. If there are five or more numbers on the board, then all prime factors in the factorization of some of them must be at leas... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. The altitudes of an acute-angled, non-isosceles triangle \(ABC\) intersect at point \(H\). \(O\) is the center of the circumcircle of triangle \(BHC\). The center \(I\) of the inscribed circle of triangle \(ABC\) lies on the segment \(OA\). Find the angle \(BAC\). | Answer: $60^{\circ}$.
Solution. From the problem statement, it follows that point O lies at the intersection of the angle bisector of angle $A$ and the perpendicular bisector of side $BC$. Since these lines intersect on the circumcircle of triangle $ABC$, point O lies on this circle and is the midpoint of arc $BC$ (se... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. Given two five-digit numbers without the digits 0 and 1 in their notation. The absolute value of their difference is a four-digit number \( S \). It is known that if each digit of one of the original numbers is decreased by 1, then the absolute value of the difference becomes 10002. What values can the number \( ... | Answer: 1109.
Solution: Let $A$ and $B$ be the two given numbers, and $C$ be the number obtained from $B$ by decreasing each of its digits by 1, that is, $C = B - 11111$. If $A10000 > B - A$ (this number is four-digit), then $C > B$. This is a contradiction. Therefore, $A > C$. Also, the case $A > B$ is impossible (th... | 1109 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In volleyball competitions, where there are no ties, 5 teams participate. All teams played against each other. The team that took 1st place won all their matches, and the teams that took 2nd and 3rd place each won exactly two matches. In the case of equal points, the position is determined by the result of the match... | Solution. Let's denote a team by a point. If team A won against team B, we draw an arrow from A to B.
The number of outgoing arrows equals the number of wins. In total, 10 arrows can be draw... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Little One gave a big box of candies to Karlson. Karlson ate all the candies in three days. On the first day, he ate 0.2 of the entire box and 16 more candies. On the second day, he ate -0.3 of the remainder and 20 more candies. On the third day, he ate -0.75 of the remainder and the last 30 candies. How many candie... | Solution. Let $x$ be the number of candies in the box. On the first day, $(0.2x + 16)$ candies were eaten; on the second and third days, $(0.8x - 16)$ candies were eaten. On the second day, $(0.3(0.8x - 16) + 20) = (0.24x + 15.2)$ candies were eaten; on the third day, $(0.56x - 31.2)$ candies remained. Since on the thi... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The figure "archer" on a grid board attacks along a ray - along cells upwards, downwards, to the right, or to the left (exactly one of the four directions; the directions for different archers are independent). What is the maximum number of non-attacking archers that can be placed on an $8 \times 8$ chessboard? | Solution. Answer: 28. We will prove that it is impossible to place more than 28 archers. Consider any arrangement with the maximum possible number of archers. Perform the following two operations sequentially:
1) turn those archers who are standing at the edge of the board so that they shoot "outward" (this will not "... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. In the parliament of the island state of Promenade-and-Tornado, only the indigenous inhabitants of the island can be elected, who are divided into knights and liars: knights always tell the truth, liars always lie. In the last convocation, 2020 deputies were elected to the parliament. At the first plenary session ... | Solution. Note that the statement was made by more than half of the deputies. If all those who made the statement are knights, then their statements turn out to be false; if all those who made the statement are liars, then their statements turn out to be true - both are impossible. Therefore, among those who made the s... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. A trading organization wholesale purchased exotic fruits, the moisture content of which was $99\%$ of their mass. After delivering the fruits to the market, the moisture content dropped to $98\%$. By what percentage should the trading organization increase the retail price of the fruits (the price at which it will... | Solution. In 100 kg of fresh fruits, there was 99 kg of water and 1 kg of solid mass. After drying, 1 kg of solid mass constituted $2 \%$ of the mass of 50 kg, i.e., every 100 kg of fresh fruits dried down to 50 kg, i.e., by half, and the retail price, compared to the wholesale price, should be increased by two times.
... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Find the smallest natural number ending in the digit 6 (on the right) which, when this digit is moved to the beginning (to the left), increases exactly four times. | Solution. Multiplication "in column" is performed from the end, so we can start the process of multiplication by sequentially finding all the digits: *****6 ( $6 \times 4=24$ - "4 we write, 2 in mind" - the 4 written under the line is the 4th digit that stood before 6; by writing 4 in the top line before 6, we will con... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. The following numbers are written on the wall: $1,3,4,6,8,9,11,12,16$. Four of these numbers were written by Vova, four numbers were written by Dima, and one number was simply the house number of the local police officer. The police officer found out that the sum of the numbers written by Vova is three times the s... | Solution. The sum of all numbers on the wall is $1+3+4+6+8+9+11+12+16=70$. Let $n$ be the sum of the numbers written by Dima, then $3n$ is the sum of the numbers written by Vova, $4n$ is the sum of the numbers written by both, $70-4n-$ is the house number of the district police officer, which when divided by 4 leaves a... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Masha left the house for school. A few minutes later, Vanya ran out of the same house to school. He overtook Masha at one-third of the way, and when he arrived at school, Masha still had half of the way left to go. How many times faster does Vanya run compared to how Masha walks? | Solution. At one third of the way, Masha and Vanya were at the same time. After that, Vanya ran $2 / 3$ of the way, while Masha walked $1 / 2 - 1 / 3 = 1 / 6$ of the way in the same time. This means that he runs $2 / 3$ : $1 / 6 = 4$ times more in the same amount of time than Masha. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. What is the smallest identical number of pencils that need to be placed in each of 6 boxes so that any 4 boxes contain pencils of any of the 26 pre-specified colors (there are enough pencils available)? Prove that fewer is impossible. | Solution. Let's assume that we have fewer than 3 pencils of some color. Then, if we take 4 boxes in which such pencils are not present (and such boxes can be found, since there are no more than $2 y x$ pencils of that color), the condition of the problem will not be met. This means that there are at least 3 pencils of ... | 78 | Combinatorics | proof | Yes | Yes | olympiads | false |
1. Real numbers $a, b, c$ are such that $a+1 / b=9, b+1 / c=10$, $c+1 / a=11$. Find the value of the expression $a b c+1 /(a b c)$. | Answer: 960.
Sketch of the solution. By multiplying the equations, expanding the brackets, and grouping, we get: $a b c + 1/(a b c) + a + 1/b + b + 1/c + c + 1/a = 990$. From this, $a b c + 1/(a b c) = 990 - 9 - 10 - 11 = 960$. | 960 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a football tournament, 17 teams participate, and each team plays against each other exactly once. A team earns 3 points for a win. For a draw, 1 point is awarded. The losing team gets no points. What is the maximum number of teams that can accumulate exactly 10 points? | Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 10 points each. The total points include all points scored by these teams in matches against each other (at least 2) and possibly in matches against other teams: $10 n \geq 2(n-1) n / 2$. Hence, $n \leq 11$.
Example: All eleven teams pl... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given three non-zero real numbers $a, b, c$ such that the equations: $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ each have two roots. How many of the roots of these equations can be negative? | Answer: 2.
Sketch of the solution.
If the numbers $a, b, c$ are replaced by their opposites, then the "new" equations will have the same set of roots as the original ones. There are two possible cases: the numbers $a, b, c$ are of the same sign; among them, there are both positive and negative numbers.
First case. D... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Several cells on a $14 \times 14$ board are marked. It is known that no two marked cells are in the same row and the same column, and also, that a knight can, starting from some marked cell, visit all marked cells in several jumps, visiting each exactly once. What is the maximum possible number of marked cells? | Answer: 13.
Since there is no more than one marked cell (field) in each row, there are no more than 14 marked cells. Suppose there are 14 marked cells. Number the rows and columns from 1 to 14 from bottom to top and from left to right, and color the cells in a checkerboard pattern, where the sum of the row and column ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 8.1. Can a cube be cut into 71 smaller cubes (of any non-zero size)? | Answer: Yes, it is possible.
Solution: Let's divide each edge of the cube in half and cut the cube into 8 smaller cubes. Now, take one of these smaller cubes and divide it into 8 even smaller cubes. One cube disappears, but 8 new ones appear, increasing the total number of parts by \(8-1=7\) to 15. Repeating this oper... | 71 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.2. Two cyclists, Andrey and Boris, are riding at a constant and identical speed along a straight highway in the same direction, so that the distance between them remains constant. There is a turnoff to a village ahead. At some point in time, the distance from Andrey to the turnoff was equal to the square of the dista... | Answer: 2 or 0 km.
Solution: Let the first mentioned distances be $a$ and $b$, then $a=b^{2}$. When each of them had traveled another kilometer, the remaining distance to the turn was $a-1$ and $b-1$ km, respectively, so $a-1=3(b-1)$, which means $b^{2}-1=3(b-1),(b-1)(b+1)=3(b-1)$, from which $b=1$ or $b=2$. When $b=1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$ with angle $C$ equal to $30^{\circ}$, median $A D$ is drawn. Angle $A D B$ is equal to $45^{\circ}$. Find angle $B A D$.
# | # Answer: $30^{\circ}$.
Solution. Draw the height $B H$ (see the figure). In the right triangle $B H C$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B H=\frac{B C}{2}=B D$. The angle $H B C$ is $180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$. Triangle $B H D$ is isosceles with an angle of $60^{\circ}$, so i... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. It is known that for real numbers $a$ and $b$ the following equalities hold:
$$
a^{3}-3 a b^{2}=11, \quad b^{3}-3 a^{2} b=2
$$
What values can the expression $a^{2}+b^{2}$ take? | Answer: 5.
Solution. We have
$$
\left(a^{2}+b^{2}\right)^{3}=a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}=\left(a^{3}-3 a b^{2}\right)^{2}+\left(b^{3}-3 a^{2} b\right)^{2}=11^{2}+2^{2}=125
$$
From this, $a^{2}+b^{2}=5$.
Comment. A correct and justified solution - 7 points. The correct answer obtained based on a simple e... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. What is the maximum number of members that can be in a sequence of non-zero integers, for which the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative? | Answer: 16.
Solution: Estimation. Assume there are no fewer than 17 numbers. Construct a table with 7 columns and 11 rows, in which the first 17 numbers are arranged.
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{2}$ | $a_{3}$ |... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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