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4. Among 49 visually identical coins - 25 are genuine and 24 are counterfeit. To identify the counterfeit coins, there is a tester. You can place any number of coins in it, and if more than half of these coins are counterfeit, the tester will signal. How can you find two counterfeit coins in five tests? (K. Knop) | Solution. Let's call "working" those coins among which we continue to search for a pair of fake ones. We will denote the situation "N coins are working, among which there are at least M fake ones" as $N: M$. If the current test gives a signal (let's denote this as "+"), then after this, we consider working only those w... | 2:2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,669 |
5. A certain natural number \( a \) was divided with a remainder by the numbers \( 1, 2, 3, \ldots, 1000 \). Could it happen that among the remainders, the numbers \( 0, 1, 2, 3, \ldots, 99 \) each appear exactly 10 times? (S. Berlov) | Answer: It could not. Solution: From the condition, it follows that all remainders of the division of the number $a$ by numbers from 1 to 1000 are less than 100. Let the remainder of the division of the number $a$ by 100 be $r$. Then, when dividing $a$ by any number that is a multiple of 100, the remainder should be of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,670 |
6. In a convex quadrilateral $ABCD$, angles $A$ and $C$ are both 100°. Points $X$ and $Y$ are chosen on sides $AB$ and $BC$ respectively such that $AX = CY$. It turns out that line $YD$ is parallel to the bisector of angle $ABC$. Find angle $AXY$. (A. Kuznetsov, S. Berlov) | Solution. Draw a line through point $Y$ parallel to $AB$. Let it intersect $AD$ at point $K$. Then $\angle DYC = \angle DYK$ and $\angle C = 100^{\circ} = \angle BAD = \angle YKD$, so triangles $DYC$ and $DYK$ are congruent by two angles and a side. Therefore, $YK = YC = AX$ and $AXYK$ is a parallelogram. Hence, $\angl... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,671 |
7. Given a circle of length 90. Is it possible to mark 10 points on it so that among the arcs with endpoints at these points there are arcs of all integer lengths from 1 to 89? (K. Knop) | Answer: No. Solution: 10 points divide the circle into 90 arcs. We need to get 45 arcs of different odd lengths. Since there are 44 even arcs, the number of odd arcs is no more than $90-44=46$, and an arc of length 45 cannot be the only one because it is a semicircle. Therefore, we need exactly 46 odd arcs.
Take any o... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,672 |
8. Given an odd natural number \( a \), greater than 100. On the board, all natural numbers of the form \( \frac{a-n^{2}}{4} \), where \( n \) is a natural number, were written. It turned out that for \( n \leq \sqrt{a / 5} \) all of them are prime. Prove that each of the other natural numbers written on the board is a... | Solution. If $a$ when divided by 4 gives 3, there are no integers on the board because squares when divided by 4 can only give remainders of 0 or 1. In this case, the statement of the problem is obviously true. From now on, we assume that $a$ gives a remainder of 1 when divided by 4. Then $a=4p+1$, where $p=\frac{a-1}{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,673 |
1. Can five identical rectangles with a perimeter of 10 be arranged to form one rectangle with a perimeter of 22? | Answer. Yes. Solution. You can arrange 5 rectangles of size $1.5 \times 3.5$ vertically in a row. Another example: place 3 rectangles of size $3 \times 2$ vertically, and below them, arrange two such rectangles horizontally. There are probably other examples as well. | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,674 |
2. In triangle $ABC$, $AC=1$, $AB=2$, $O$ is the point of intersection of the angle bisectors. A segment passing through point $O$ parallel to side $BC$ intersects sides $AC$ and $AB$ at points $K$ and $M$ respectively. Find the perimeter of triangle $AKM$. | Answer: 3. Solution. Note that $\angle K C O=\angle B C O=\angle K O C$ (alternate interior angles). Therefore, $O K=K C$. Similarly, $B M=O M$. Therefore, $A K+A M+K M=A K+K C+A M+B M=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,675 |
3. The company "Horns and Hooves" split into the company "Horns" and the company "Hooves" with different numbers of employees. The director of the company "Horns" receives the same salary as the director of the company "Hooves," and the average salary of all other employees of the company "Horns" matches the average sa... | Answer: The salary of the company director is equal to the average salary of all other employees.
Solution. Let the company "Horns" have $m$ employees, and the company "Hooves" have $n$ employees, with the directors' salaries being $x$ rubles, and the average salaries of the other employees being $y$ rubles. Then, the... | y | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,676 |
4. A divisor of a natural number is called proper if it is greater than 1 and less than the number itself. A natural number is called elegant if it has at least two proper divisors and is divisible by the difference of any two of them. Find all elegant numbers. | Answer: 6, 8, 12. Solution. It is easy to verify that the numbers 6, 8, and 12 are elegant. Let $\mathrm{N}$ be an arbitrary elegant number. An odd number has only odd divisors, the differences between divisors are even, and an odd number cannot be divided by an even number. Therefore, $N=2 n$. But then $2 n$ must be d... | 6,8,12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,677 |
5. On a grid paper, there is a 1x2011 strip, with the number 1 written in the first cell and the number 2 in the last cell. Petya and Vasya take turns writing the numbers 1 and 2 in any of the free cells. Petya goes first and writes only ones, while Vasya writes only twos. When the free cells run out, Petya counts the ... | Answer: Vasya. Solution: Let's divide the cells of the strip from the second to the 2011th into pairs of adjacent cells («dominoes»). Additionally, mark the cell 2010. Vasya's strategy: 1) as long as the marked cell is empty, after each of Petya's moves into a cell of some domino, move into the other cell of the same d... | Vasya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,678 |
1. On the shores of a lake, there are 5 piers arranged in a circle, with one person at each pier, and one of them has a one-person boat. People on adjacent piers are in a dispute and do not want to meet each other. How can each of them move to the next pier clockwise, if they can only travel across the lake? | Solution. It is sufficient to travel twice counterclockwise along the star-shaped path formed by the diagonals. The first 5 trips: $1 \rightarrow 4,4 \rightarrow 2,2 \rightarrow 5,5 \rightarrow 3,3 \rightarrow 1$. Each person has moved two positions counterclockwise, so on piers 1 through 5 are people numbered $3,4,5,1... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,679 |
2. In the Martian parliament, deputies from three parties are in session: "Vowels," "Consonants," and "Sibilants" - 50 deputies from each party. A bill "On the Reconstruction of Martian Canals" was put to a vote. After the vote, 30 deputies from each party said they voted "for," 10 said they voted against, and the rest... | Answer: No. Solution: Those from the "vowels" or "sibilants" who voted "for" were lying, meaning they answered that they voted "against" or abstained. In both parties, 20 people answered this way. Even if all those who answered this way voted "for," it would be no more than 40 people. Those from the "consonants" who vo... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,680 |
3. In triangle $ABC$, angle $C$ is twice as large as angle $B$, and $CD$ is the angle bisector. From the midpoint $M$ of side $BC$, a perpendicular $MH$ is dropped to segment $CD$. On side $AB$, a point $K$ is found such that $KMN$ is an equilateral triangle. Prove that points $M$, $H$, and $A$ lie on the same line. | First solution. Since $\angle D C B=\angle C / 2=\angle B, D M$ is the median, bisector, and altitude in the isosceles triangle $B D C$. Drop a perpendicular $M E$ from point $M$ to line $A B$. Then $M E=M H=M K$, from which it follows that $K=E$. Next, in the right triangles $C H M$ and $B K M$, the sum of the angles ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,681 |
4. Ten natural numbers were written on the board. If any three of the written numbers are marked, then the sum of all three will be divisible by two numbers from this triplet. Prove that among the written numbers there are equal ones. | First solution. If $a \leq b \leq c \leq d \leq e$ are five of the given numbers, then among the sums $a+d+e, b+d+e, c+d+e$, two must be divisible by one of the numbers $d$ and $e$. But the difference between these two sums is equal to the difference between some two of the numbers $a, b, c$, and is therefore less than... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,682 |
5. Do there exist two numbers such that the first is 2016 times greater than the second, and the sum of its digits is less than the sum of the digits of the second by 2016 times? | Answer. Yes, they do exist. Solution. Consider, for example, a number $A$ of the form $11...15$, where the number of ones is greater than 10. Multiplying $A$ by 2016, we get:
$$
11 \ldots 15 \times 2016=10080+20160+201600+2016000+\ldots+201600 \ldots 0=2240 \ldots .007840
$$
(here, many nines form due to $2+0+1+6=9$,... | 27\times2016-5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,683 |
1. Four numbers are written on the board, none of which are equal to 0. If each of them is multiplied by the sum of the other three, the results are four identical values. Prove that the squares of the numbers written on the board are equal. (I. Rubanov). | Solution. Let the numbers $a, b, c$ and $d$ be written. By the condition $a(b+c+d)=b(a+c+d)$, from which $(a-b)(c+d)=0$. Similarly, from the equality $c(a+b+d)=d(a+b+c)$ we get $(c-d)(a+b)=0$. Since either $c+d$ or $c-d$ is not equal to 0, then either $a=b$ or $a=-b$. In both cases, the squares of the numbers $a$ and $... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,684 |
2. It is allowed to cut out any 18 cells from a $20 \times 20$ chessboard, and then place several rooks on the remaining cells so that they do not attack each other. What is the maximum number of rooks that can be placed in this way? Rooks attack each other if they stand on the same row or column of the board and there... | Answer: 38 rooks. Solution: Let's call the cut-out cells holes. In addition to them, add a hole at the bottom of each vertical line of the board, and a hole to the right of each horizontal line; a total of $2 \cdot 20=40$ holes are added. Suppose several rooks are placed on the board, not attacking each other. We will ... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,685 |
3. A divisor of a natural number is called proper if it is less than the number but greater than 1. For a natural number \( n \), all proper divisors (there were at least three of them) were found and all possible pairwise sums of these divisors were recorded (repeated identical sums were not recorded). Prove that the ... | Solution. Suppose the obtained set of numbers turned out to be the set of all proper divisors of some number $m$. Since the number $n$ has at least three proper divisors, among them there will be two divisors of the same parity. Then their sum is even. The number $m$ is divisible by this sum, so it is also even. Theref... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,686 |
4. The perpendicular bisectors of sides $AB$ and $BC$ of a convex quadrilateral $ABCD$ intersect sides $CD$ and $DA$ at points $P$ and $Q$ respectively. It turns out that $\cdot A P B=\cdot B Q C$. Inside the quadrilateral, a point $X$ is chosen such that $Q X \| A B$ and $P X \| B C$. Prove that the line $B X$ bisects... | Solution. It is sufficient to prove that the distances from points $A$ and $C$ to the line $B X$ are equal. This is equivalent to $S_{A B X}=S_{B C X}$, since triangles $A B X$ and $B C X$ share the same base $B X$. Since $Q X \| A B$, we have $\quad S_{A B X}=S_{A B Q}$. Similarly, $\quad S_{C B X}=S_{C B P}$. Note th... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,687 |
5. A stick 10 cm long lies on the table. Petya breaks it into two pieces and puts both resulting sticks on the table. Vasya performs the same operation with one of the sticks lying on the table, then Petya does the same, and so on, taking turns. Petya wants all the resulting sticks to be shorter than 1 cm after 18 brea... | Answer: Vasya. Solution: Mark 9 points dividing the stick into segments of 1 cm. Vasya needs to play in such a way that all these points are broken. Since he has 9 moves, he can do this. In the end, after 18 breaks, 10 sticks of 1 cm in length will be obtained, some of which are broken into smaller pieces. Since the nu... | Vasya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,688 |
6. Viewers rate a movie with an integer number of points from 0 to 10. At any given time, the movie's rating is calculated as the sum of all the given ratings divided by their number. At some point in time $T$, the rating was an integer, and then with each new voting viewer, it decreased by one. What is the maximum num... | Answer: 5. Solution: Consider a moment when the rating has decreased by 1. Suppose that before this, $n$ people had voted, and the rating was an integer $x$. This means the sum of the scores was $n x$. Let the next viewer give a score of $y$. Then the sum of the scores becomes $n x + y = (n + 1)(x - 1)$, from which we ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,689 |
7. In trapezoid $A B C D$, where $A D \| B C$, angle $B$ is equal to the sum of angles $A$ and $D$. On the extension of segment $C D$ beyond vertex $D$, segment $D K=B C$ is laid out. Prove that $A K=B K$. (B. Obukhov) | Solution. On the ray $DA$, we mark off the segment $DE = BC$. Then the quadrilateral $DCBE$ is a parallelogram, so $\cdot CBE = \cdot CDE$. Using the given condition, we get $\cdot ABE = \cdot ABC - \cdot CBE = \cdot ABC - \cdot CDE = \cdot BAE$; hence, triangle $ABE$ is isosceles, $AE = BE$. Furthermore, since $ED = B... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,690 |
8. On a $20 \times 20$ chessboard, 220 knights are placed, each attacking all the free cells. Prove that it is possible to remove 20 knights such that the remaining knights still attack all the free cells. Recall that a knight attacks in an "L" shape (see figure). (S. Berlov) | Solution. We will remove knights from the board one by one as follows. If at
any step, we can remove a knight such that the remaining knights still cover all free squares, we will do so. If ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,691 |
1. Let's enumerate all prime numbers in ascending order: $p_{1}=2, p_{2}=3, \ldots$. Can the arithmetic mean $\left(p_{1}+\ldots+p_{n}\right) / n$ for some $n \geq 2$ be a prime number? (S. Volchonkov) | Answer: No. Solution. Let $\left(p_{1}+\ldots+p_{n}\right) / n=q \Leftrightarrow p_{1}+\ldots+p_{n}=n q(*)$, where $q$ is a prime number. Obviously, if $n>1$, then $q>2$. Therefore, for even $n$, the left side of the equation $\left(^{*}\right)$ is odd, while the right side is even, and for odd $n$ - the opposite. Cons... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,692 |
2. In Shvambania, some cities are connected by two-way non-stop flights. The flights are divided among three airlines, and if one airline serves a route between cities A and B, then planes of other airlines do not fly between these cities. It is known that from each city, flights of all three airlines depart. Prove tha... | Solution. Fly out of any city A with the first company, arrive in city B, from there fly with the second company, then the third, and again the first, second, third, and so on. Consider the very first moment when we encounter a city C that we have already visited. Then the segment of our path from the first visit to ci... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,693 |
3. In quadrilateral $A B C D$, side $A B$ is equal to diagonal $A C$ and is perpendicular to side $A D$, and diagonal $A C$ is perpendicular to side $C D$. A point $K$ is taken on side $A D$ such that $A C=A K$. The bisector of angle $A D C$ intersects $B K$ at point $M$. Find the angle $A C M$. (R. Zhenodarov) | Answer: $\angle A C M=45^{\circ}$. Solution. Since triangle $B A K$ is a right isosceles triangle, $\angle A K B=45^{\circ}$. Let the bisector of angle $C A D$ intersect segment $B K$ at point $N$. Triangles $A N K$ and $A N C$ are equal: $A N$ is common, $A C=A K, \angle C A N=\angle K A N$. Therefore, $\angle N C A=\... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,694 |
4. At the vertices of a cube, the numbers $1^{2}, 2^{2}, \ldots, 8^{2}$ (one number per vertex) are placed. For each edge, the product of the numbers at its ends is calculated. Find the maximum possible sum of all these products. | Answer: 9420. Solution. Let's color the vertices of the cube in two colors so that the ends of each edge are of different colors. Let the numbers $a_{1}, a_{2}, a_{3}, a_{4}$ be placed in the vertices of one color, and the numbers $b_{1}, b_{2}, b_{3}, b_{4}$ in the vertices of the other color, with numbers having the ... | 9420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,695 |
5. Is it possible to place 12 natural numbers on the edges of a cube such that the sums of the numbers on any two opposite faces differ by exactly one? (D. Khramtsov) | Answer: No. Solution: Suppose it is possible to arrange the numbers in such a way. Then the sum of the numbers on any two opposite faces would be odd. Adding up all three such sums, we would get an odd result. However, in this result, each edge is counted exactly twice, since it lies on two faces of the cube, so it sho... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,696 |
6. Do there exist such distinct natural numbers $a, b$, and $c$ that the number $a + 1/a$ is equal to the half-sum of the numbers $b + 1/b$ and $c + 1/c$? (A. Golev) | Answer. No. Solution. Suppose such numbers were found. Note that if $m$ and $n$ are natural numbers and $m<n$, then $m+1 / m \leq m+1 \leq n<n+1 / n$. Therefore, we can (if necessary, by swapping the numbers $b$ and $c$) assume that $b<a<c$. Rewrite the condition as $(a-b)+(1 / a-1 / b)=(c-a)+(1 / c-1 / a)$. Since each... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,697 |
7. The angles of triangle $ABC$ satisfy the condition $2 \angle A + \angle B = \angle C$. Inside this triangle, on the bisector of angle $A$, a point $K$ is chosen such that $BK = BC$. Prove that $\angle KBC = 2 \angle KBA$. (S. Berlov) | First solution. Since $\angle C > \angle A + \angle B$, then $\angle C > 90^{\circ}$. Choose a point $T$ on the ray $A C$ such that $B C = B T$. Note that $\angle A T B = \angle B C T = \angle A + \angle B$. Since $2 \angle A + \angle B = \angle C$, we have $3 \angle A + 2 \angle B = \angle A + \angle B + \angle C = 18... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,698 |
8. Let $n$ be a natural number greater than 1. Kostya has a device that, if given $2 n+1$ coins of different weights, will indicate which coin is the median among those placed. Baron Munchausen gave Kostya $4 n+1$ coins of different weights and said that one of them is the median by weight. How can Kostya, using the de... | Solution. Let $M$ be the coin that the baron considers to be of medium weight. Kostya can act according to the following algorithm: | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,699 |
5. (P. Zhenedarov) Can the asterisks in the expression LCM $(*, *, *) - \text{LCM}(*, *, *) = 2009$ be replaced with six consecutive natural numbers in some order so that the equation becomes true
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Solution. The least common multiple (LCM) of several numbers is divisible by each of them and, consequently, by each of their divisors. Therefore, if there is an even number among the numbers for which the LCM is being found, the LCM will also be even. The difference between two even numbers is an even number, while 20... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,700 |
6. (S. Berlov) In a convex quadrilateral $ABCD$, the following relations are satisfied: $AB=BD; \angle ABD=\angle DBC$. On the diagonal $BD$, there is a point $K$ such that $BK=BC$. Prove that $\angle KAD=\angle KCD$.
| Solution. On side $A B$, we mark off a segment $B E = B C$. The isosceles triangles $E B K$ and $K B C$ are equal by two sides and the angle between them. Therefore, $E K = K C$, and
$$
\angle A E K = 180^{\circ} - \angle B E K = 180^{\circ} - \angle B K C = \angle C K D
$$
Moreover, $K D = B D - B K = B A - B E = E ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,701 |
7. (I. Rubanov, A. Shapovalov) On the table, there are 10 piles with 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 nuts. Two players take one nut at a time, taking turns. The game ends when there are 3 nuts left on the table. If these are three piles of one nut each, the player who moved second wins; otherwise, his opponent wins. ... | Solution 1. Let's call piles with one nut units, and those with two nuts doubles. The first player should adhere to the following rules: 1) if there are units on the board, remove one of them; 2) do not take from doubles. Otherwise, the first player's moves can be any. Note that the number of nuts at the beginning of t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,702 |
8. (I. Bogdanov) On an infinite tape, numbers are written in a row. The first is one, and each subsequent number is obtained from the previous one by adding to it the smallest non-zero digit of its decimal representation. How many digits are in the decimal representation of the number standing in this sequence at the $... | Answer: 3001. Solution. Since each number in the sequence, starting from the second, is at least one greater than the previous one, the $9 \cdot 1000^{1000}$-th number in the sequence is greater than $9 \cdot 1000^{1000}$, meaning it has at least 3001 digits. Let's denote the $n$-th number in the sequence by $a_{n}$, a... | 3001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,703 |
1. Given two numbers (not necessarily integers), not equal to 0. If each of them is increased by one, their product will double. By what factor will their product increase if each of the original numbers is squared and then decreased by one? (D. Shiryayev, I. Rubanov) | Answer. 4 times. Solution. Let the numbers be denoted by $a$ and $b$. According to the condition, $(a+1)(b+1)=a b+a+b+1=2 a b$. By combining like terms in the last equality, we get $a b-a-b-1=0$, from which $(a-1)(b-1)=a b-a-b+1=2$ and $\left(a^{2}-1\right)\left(b^{2}-1\right)=(a-1)(b-1)(a+1)(b+1)=2 \cdot 2 a b=4 a b$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,704 |
2. The device KK42 works as follows: if you put four balls into it, the second heaviest ball (i.e., the ball of weight $b$, if $a>b>c>d$) will fall into the first tray, and the rest will fall into the second tray. The device does not work with a different number of balls. There are 100 balls that look identical but hav... | Solution. First, each time we put 4 balls that have not been set aside into the device and set aside the one that falls into the first tray. After 97 trials, the heaviest and the two lightest balls remain, as none of them can fall into the first tray. Let their numbers be $x, y, z$. We choose any three balls $a, b, c$ ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,705 |
3. Given a 1000-digit number with no zeros in its representation. Prove that from this number, several (possibly none) of the last digits can be erased so that the resulting number is not a natural power of a number less than 500. (S. Berlov) | Solution. We will erase at the end zero, one, two, three, ..., 499 digits. If all the time the results are powers of numbers less than 500, then the bases of some two of them must coincide. Let these be $a^{x}$ and $a^{y}(x<y)$. Multiply the number $a^{x}$ by a power of ten so that its notation has as many digits as th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,706 |
4. Given a convex quadrilateral $A B S C$. A point $P$ is chosen on the diagonal $B C$ such that $A P=C P>B$. Point $Q$ is symmetric to point $P$ with respect to the midpoint of diagonal $B C$, and point $R$ is symmetric to point $Q$ with respect to the line $A C$. It turns out that $\angle S A B=\angle Q A C$ and $\an... | Solution. Mark a point $L$ on the segment $A C$ such that $Q L \| A P$. Then triangles $A P C$ and $L Q C$ are similar, and $L Q=Q C=B P$. Moreover, $B Q=P C=A P$ and $\angle A P B=\angle L Q B$, so triangles $A B P$ and $B L Q$ are equal by two sides and the angle between them. Therefore, $B A=B L$. Further,
$$
\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,707 |
1. Squares with sides of 11, 9, 7, and 5 are arranged approximately as shown in the figure. It turned out that the area of the gray parts is twice the area of the black parts. Find the area of the white parts. | Answer: 42. Solution: Let the area of the white parts be $x$, and the area of the black parts be $y$. The total area of the white and black parts is $9^{2}+5^{2}=106=x+y$, and the total area of the white and gray parts is $11^{2}+7^{2}=170=x+2 y$. By subtracting the first equation from the second, we find that $y=64$, ... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,708 |
2. In the underwater kingdom, there live octopuses that can have 6, 7, or 8 legs. Those with 7 legs always lie, while the others always tell the truth. Upon meeting four octopuses, Blue said: "We have a total of 25 legs," Green countered: "No, there are 26 legs in total," Red said that the total number of legs is 27, a... | Answer: Red has 6, the others have 7 each. Solution. If all four were lying, they would have 28 legs in total, and Yellow would be right - a contradiction. Therefore, there is a truthful one among the four. There is only one, as any two statements of the octopuses contradict each other. Therefore, three octopuses have ... | Red\has\6,\the\others\have\7\each | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,709 |
3. In triangle $A B C$, point $M$ is the midpoint of $A C$, moreover, $B C=2 A C / 3$ and $\angle B M C=2 \angle A B M$. Find the ratio $A M / A B$. | Answer. $\frac{A M}{A B}=\frac{3}{2 \sqrt{5}}$. Solution. Let $\angle A B M=\alpha$. Then $\angle B M C=2 \alpha, \angle B M A=180^{\circ}-2 \alpha$, $\angle B A M=180^{\circ}-\angle A B M-\angle A M B=\alpha=\angle A B M$, from which $B M=A M=M C$. It follows that the median $B M$ of triangle $A B C$ is equal to half ... | \frac{3}{2\sqrt{5}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,710 |
4. From a square grid with a side length of 2015, several squares with a side length of 10 were cut out. Prove that from the remaining part of the large square, you can cut out:
a) a rectangle with sides 1 and 10;
b) five rectangles with sides 1 and 10. | Solution. We will prove a stronger statement b). Consider a $1 \times 10$ rectangle adjacent to the left side of the $2015 \times 2015$ square by its short side. If any of its cells falls into the cut-out $10 \times 10$ square, then the rightmost cell of this rectangle also falls into it. Therefore, if the rightmost ce... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,711 |
5. It is known that both the sum and the product of two natural numbers $a$ and $b$ are squares of natural numbers. Prove that the number $|16a - 9b|$ is not prime. | Solution. If $16 a-9 b=0$, the statement of the problem is obvious. From now on, we assume that $16 a-9 b \neq 0$.
Let $d=\operatorname{GCD}(a, b)$, and let $a=d m, b=d n$. Then $a b=d^{2} m n=c^{2}$. Since the numbers $m$ and $n$ are coprime, in their factorization into prime factors, all prime numbers enter with eve... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,712 |
1. Find all five-digit numbers where the second digit is five times the first, and the product of all five digits is 1000. | Answer: 15855, 15585, 15558. Solution: The first digit can only be 1, otherwise the second digit will be greater than 9. Then the second digit is 5, and it follows that the product of the last three digits must equal 200. Since $200=5 \times 5 \times 8$, two of these three digits must be 5. But then the third digit mus... | 15855,15585,15558 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,713 |
2. The numbers $a, b$, and $c$ are such that $a>b$ and $(a-b)(b-c)(c-a)>0$. Which is greater: $a$ or $c$? | Answer. $a>c$. Solution. Obviously, among the three given numbers, there are no equal ones. Since $a>b, a-b>0$, from which $(b-c)(c-a)>0$. Suppose $c>a$. Then $c-a>0$ and $b-c>0$, that is, $b>c>a$ - contradiction. | > | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,714 |
3. Two players take turns drawing lines on a plane, with no line being drawn twice. The player who, after their move, causes the number of pieces the plane is divided into by the drawn lines to be divisible by 5 for the first time, wins. Who will win with correct play: the player who moves first, or their opponent, and... | Answer. Second. First solution. The second player, on their first move, draws a line parallel to the one drawn by the first. If the first player, on their second move, draws a line parallel to the two already drawn, after their move, the plane will be divided into 4 parts. Then the second player draws a line parallel t... | Second | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,715 |
4. On the side $B C$ of triangle $A B C$, a point $E$ is marked, and on the bisector $B D$ - a point $F$ in such a way that $E F \| A C$ and $A F=A D$. Prove that $A B=B E$. | Solution. From the condition of the problem, it follows that $\angle E F D=\angle A D F=\angle A F D$ (the first equality is true because $E F \| A C$, the second - since $A F=A D$). Therefore, the angles $A F B$ and $E F B$, adjacent to the angles $A F D$ and $E F D$, are also equal. In addition, by the condition $\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,716 |
5. In a football tournament where each team played against each other once, teams $A$, B, V, G, D, and E participated. Teams received 3 points for a win, 1 point for a draw, and 0 points for a loss. In the end, it turned out that teams $A$, B, V, G, and D each scored 7 points. What is the maximum number of points that ... | Answer: 7 points. Solution: In a match where one of the teams won, the teams together score 3 points, in a match that ended in a draw - 2 points. Since 7 is not divisible by 3, the team that scored 7 points must have at least one draw. Since there are five such teams, there were at least three draws in the tournament. ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,717 |
1. 10 runners start simultaneously: five in blue T-shirts from one end of the running track, and five in red T-shirts from the other. Their speeds are constant and different, with each runner's speed being greater than 9 km/h but less than 12 km/h. Upon reaching the end of the track, each runner immediately turns aroun... | Answer: 50. Solution: We will show that by the time the fastest runner finishes, any two runners of different colors have met exactly twice, from which the answer 2$\cdot$5$\cdot$5 = 50 will follow.
Let $s$ (km) be the length of the track. Set $T=2 s / 12$ (hours). Since the speed of the fastest runner is less than 12... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,718 |
2. Provide an example of six different natural numbers such that the product of any two of them does not divide the sum of all the numbers, while the product of any three of them does divide. (S. Volchenkov) | Solution. Let $p-$ be a sufficiently large odd prime number. Represent the number $p^{2}$ as the sum $a_{1}+\ldots+a_{6}$ of distinct natural numbers, none of which are divisible by $p$. The numbers $p a_{1}, \ldots, p a_{6}$ will be the desired ones: the product of any two of them does not divide their sum, equal to $... | 5,10,15,20,30,45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,719 |
3. On the sides $A B$ and $A C$ of the isosceles triangle $A B C$, points $P$ and $Q$ are chosen respectively such that $P Q \| B C$. On the bisectors of triangles $A B C$ and $A P Q$, starting from vertices $B$ and $Q$, points $X$ and $Y$ are chosen respectively such that $X Y \| B C$. Prove that $P X = C Y$. (A. Kuzn... | Solution. Let $M$ and $N$ be the points of intersection of the line $X Y$ with the sides
$A B$ and $A C$ respectively. Note that $\cdot M X B=\cdot X B C=\cdot M B X=\cdot N Q Y=\cdot Y Q P=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,720 |
4. In a country, direct non-stop flights are organized between cities in such a way that from each city to any other city, one can travel (possibly with transfers). Moreover, for each city A, there exists a city B such that any other city is directly connected to A or B. Prove that from any city, one can reach any othe... | Solution. Let $X$ and $Y$ be any two cities, and $X, Z_{1}, \ldots, Z_{k}, Y$ be the route between them with the fewest number of transfers. Suppose $k \geq 3$. Then $Z_{2}$ is not connected by a flight to either $X$ or $Y$, otherwise our route could be shortened by using this flight. By the condition, there exists a c... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,721 |
5. On a shelf, ten volumes of an encyclopedia, numbered from 1 to 10, are arranged in a random order. It is allowed to swap any two volumes if there are at least four other volumes between them. Is it always possible to arrange all the volumes in ascending order of their numbers? (D. Khramtsov) | Answer: Yes. Solution. Take any volume. It is removed by no less than 4 volumes, either from the one standing first or from the one standing last. Therefore, we can place it either in the first or in the last position, and then, if we want, move it from the last to the first or vice versa. Since the position where it s... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,722 |
6. In a convex quadrilateral $A B C D$, angles $B$ and $D$ are equal, $C D=4 B C$, and the bisector of angle $A$ passes through the midpoint of side $C D$. What can the ratio $A D / A B$ be? (S. Berlov) | Answer: 2:3. Solution. Let $M$ be the midpoint of side $CD$. Consider point $K$ on ray $AB$, symmetric to point $D$ with respect to line $AM$. Since $\angle ABC = \angle ADM = \angle AKM$, then $BC \| KM$ and point $K$ lies on segment $AB$. Since $CM = DM = KM$, then $\angle DKC = 90^{\circ}$ and $KC \| AM$. Therefore,... | 2:3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,723 |
7. Prove that for arbitrary $a, b, c$ the equality $\frac{a(b-c)}{b+c}+\frac{b(c-a)}{c+a}+\frac{c(a-b)}{a+b}=0$ holds if and only if the equality $\frac{a^{2}(b-c)}{b+c}+\frac{b^{2}(c-a)}{c+a}+\frac{c^{2}(a-b)}{a+b}=0$ holds. | Solution. By expanding the brackets on the left-hand side and combining like terms, it is easy to show that
$$
\left(\frac{a(b-c)}{b+c}+\frac{b(c-a)}{c+a}+\frac{c(a-b)}{a+b}\right)(a+b+c)=\frac{a^{2}(b-c)}{b+c}+\frac{b^{2}(c-a)}{c+a}+\frac{c^{2}(a-b)}{a+b}\left(^{*}\right)
$$
Therefore, if $\frac{a(b-c)}{b+c}+\frac{b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,724 |
8. Among 100 coins, there are 4 counterfeit ones. All genuine coins weigh the same, and the counterfeit ones also weigh the same, but a counterfeit coin is lighter than a genuine one. How can you find at least one genuine coin in two weighings using a balance scale without weights? ( | Solution. Let's present one of the possible methods. Divide 100 coins into two groups (number 1 and number 2) of 33 coins each and one group (number 3) of 34 coins. For the first weighing, place groups 1 and 2 on the scales. If one of the pans is heavier than the other, then it contains no more than one counterfeit coi... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,725 |
5. Can a rectangle $1000 \times 2016$ be cut into rectangles $1 \times 2015$ and three-cell "corners" so that both types of figures are present? (E. Bakayev) | Answer: No. Solution: Suppose it is possible. Clearly, each of the rectangles $1 \times 2015$ adjoins one of the short sides of the rectangle $1000 \times 2016$, and is one cell away from the other side. Let's call a rectangle $1 \times 2015$ black (respectively, white) if this cell is covered by a corner, the other tw... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,726 |
6. In a school, there are 30 clubs, each attended by 40 children. For each $i=1,2, \ldots, 30$, let $n_{i}$ denote the number of children attending exactly $i$ clubs. Prove that in the same school, it is possible to organize 40 clubs with 30 children in each such that the numbers $n_{i}$ for these new clubs would be th... | Solution. Let's write down the members of the first circle in a row, followed by those of the second, and so on. In this process, if a child is in both the $i$-th and the ( $i+1$ )-th circle, we record them in the same position in the list of the ( $i+1$ )-th circle as in the list of the $i$-th. By cutting the resultin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,727 |
7. The sum of non-negative numbers $a, b, c$, and $d$ is 4. Prove that $(a b+c d)(a c+b d)(a d+b c) \leq 8$. (A. Khryabrov) | Solution. $\sqrt{(a b+c d)(a c+b d)} \leq \frac{a b+c d+a c+b d}{2}=\frac{(a+d)(b+c)}{2} \leq \frac{1}{2}\left(\frac{a+b+c+d}{2}\right)^{2}=2$. Similarly, $\sqrt{(a b+c d)(a d+b c)} \leq 2$ and $\sqrt{(a c+b d)(a d+b c)} \leq 2$. Multiplying the three obtained inequalities, we get the desired result. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,728 |
8. Given a parallelogram $A B C D$. On sides $A B$ and $B C$ and the extension of side $C D$ beyond point $D$, points $K$, $L$, and $M$ are chosen respectively such that triangles $K L M$ and $B C A$ are equal (with this correspondence of vertices). Segment $K M$ intersects segment $A D$ at point $N$. Prove that $L N \... | First solution. Draw a line through $L$ parallel to $KM$; let it intersect lines $AB$ and $CD$ at points $P$ and $Q$ respectively. Note that the heights of triangles $BCA$ and $KLM$, dropped from the corresponding vertices $C$ and $L$, are equal; in turn, they are equal to the distances between the pairs of parallel li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,729 |
2. Twenty-two people are standing in a circle, each of them is either a knight (who always tells the truth) or a liar (who always lies). Each of them said: "The next 10 people clockwise after me are liars." How many of these 22 people are liars? | Answer: 20 liars
Solution: If more than 10 liars stand in a row, then one of them is telling the truth, which is impossible. There are 22 people in total, so there must be a knight among them. Consider the knight, who tells the truth, meaning that the 10 people following him are liars. Since 11 liars cannot stand in a... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,731 |
3. Given an isosceles triangle \(ABC (AC = BC)\). Points \(A_1\), \(B_1\), and \(C_1\) are marked on the sides \(BC\), \(AC\), and \(AB\) respectively. It turns out that \(C_1 B_1\) is perpendicular to \(AC\), \(B_1 A_1\) is perpendicular to \(BC\), and \(B_1 A_1 = B_1 C_1\). Prove that \(A_1 C_1\) is perpendicular to ... | Solution. Let $\angle A=\angle B=\alpha$. Then $\angle C=\pi-2 \alpha$. Triangle $C A_{1} B_{1}$ is a right triangle, so $\angle C B_{1} A_{1}=\pi / 2-(\pi-2 \alpha)=2 \alpha-\pi$. Then $\angle A_{1} B_{1} C_{1}=\pi / 2-(2 \alpha-\pi / 2)=\pi-2 \alpha$. Since $B_{1} A_{1}=B_{1} C_{1}$, $\angle B_{1} C_{1} A_{1}=\angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,732 |
5. Is it possible to divide the numbers from 1 to 100 into three groups such that the sum of the numbers in the first group is divisible by 102, in the second group by 203, and in the third group by 304? | Answer: No.
Solution: Suppose it was possible to divide the numbers from 1 to 100 into groups with sums of $102A$, $203B$, and $304C$. Then the equation $102A + 203B + 304C = 5050$ or $A + B + C + 101(A + 2B + 3C) = 101 \cdot 50$ must hold. Therefore, the expression $A + B + C$ must be divisible by 101, which implies ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,734 |
2. In triangle $ABC$, the bisector $BD$ is drawn, in triangle $BDC$ - the bisector $DE$, and in triangle $DEC$ - the bisector $EF$. It turned out that lines $BD$ and $EF$ are parallel. Prove that angle $ABC$ is twice the angle $BAC$. | Solution. From the condition, it follows that $\angle A B D=\angle E B D=\angle C E F=\angle D E F=\angle B D E$. Thus, the alternate interior angles $A B D$ and $B D E$ formed by the intersection of line $B D$ with lines $A B$ and $D E$ are equal. Therefore, $A B \| D E$, from which $\angle B A C=\angle E D F=\angle E... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,736 |
3. Vasya wrote one number on each of 99 cards (among these numbers there could be equal ones) and placed the cards in a circle numbers down. For each pair of adjacent cards, he told Petya what numbers were written on these cards, but did not tell which number was on which card. Could Vasya choose the numbers in such a ... | Answer. Could not. Solution. Note that it is enough for Pete to determine the number on one card: by considering pairs of it and adjacent cards, we can learn the numbers on the adjacent cards, and by proceeding in the same way further, we can learn the numbers on all the cards. We will show that Pete can do this.
Cons... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,737 |
4. Before starting to solve the problem, Kolya looked at the clock. It was two o'clock. Spending exactly an hour on solving it, Kolya looked at the clock again and noticed that the angle between the hour and minute hands remained the same. When did Kolya start solving the problem? | Answer. At 1 hour $8^{2} /{ }_{11}$ min or at 1 hour $40^{10} / 11$ min. Solution. Let it be $x$ minutes past two when Petya looked at the clock. Since the minute hand moves $6^{\circ}$ per minute, and the hour hand $-0,5^{\circ}$, the hour hand at this moment formed an angle of $30^{\circ}+0,5 x^{\circ}$ with the 12 o... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,738 |
5. What is the smallest number of different numbers that can be chosen so that each chosen number is equal to the sum of some three other different chosen numbers?
---
Note: The translation maintains the original text's formatting and line breaks. | Answer. Seven. Solution. Example. $-3,-2,-1,0,1,2,3$. Evaluation. Let $n-$ be the largest of the chosen numbers. If $n \leq 0$, all other chosen numbers are negative, and the sum of any two of them is less than each of the addends, and therefore cannot equal $n$. Therefore, $n>0$. The two of the chosen numbers whose su... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,739 |
1. A student received 13 grades (from the set 2, 3, 4, 5) over one week, the arithmetic mean of which is an integer. Prove that the student received some grade no more than twice. (N. Agakhanov). | Solution. Assume the opposite. Then each of the grades $2,3,4,5$ was received by the student no less than three times. Let's take three grades of each kind. The sum of the 12 taken grades is 42, while the sum of all 13 grades is 44, 45, 46, or 47, depending on the remaining grade. But none of the numbers 44, 45, 46, 47... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,740 |
2. The expert was presented with 12 coins that look identical, among which there might be counterfeit ones. All genuine coins weigh the same, and all counterfeit coins also weigh the same, but a counterfeit coin is lighter than a genuine one. The expert has a balance scale and standard coins: 5 genuine and 5 counterfei... | Answer: Yes. Solution: It is obvious that it is enough to show that the number of fake coins among the six given can be determined in two weighings. Let's call these six coins unknown. We take three genuine coins and three fake ones and weigh them against the unknowns. If the scales are in balance, then among the unkno... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,741 |
3. We took four natural numbers. For each pair of these numbers, we wrote down their greatest common divisor. Six numbers were obtained: 1, 2, 3, 4, 5, N, where $N>5$. What is the smallest value that the number $N$ can take? (O. Dmitriev) | Answer: 14. Solution: The number $N$ can equal 14, as shown, for example, by the quartet of numbers 4, $15, 70, 84$. It remains to show that $N \geq 14$.
Lemma. Among the pairwise GCDs of four numbers, there cannot be exactly two numbers divisible by some natural number $k$. Proof. If among the original four numbers t... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,742 |
4. On the side $AC$ of triangle $ABC$, a point $D$ is chosen such that $BD = AC$. The median $AM$ of this triangle intersects the segment $BD$ at point $K$. It turns out that $DK = DC$. Prove that $AM + KM = AB$. (S. Berlov) | Solution. Let $L$ be the point symmetric to $K$ with respect to $M$. Then
$$
A D=A C-C D=B D-D K=B K=C L .
$$
Since angles $B D A$ and $A C L$ are equal as corresponding angles, and $B D=A C$ by the condition, triangles $B D A$ and $A C L$ are equal by two sides and the included angle. Hence, $A B=A L=A M+M L=A M+K M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,743 |
5. On a circle, 2013 points are marked and each is connected to its two neighbors. Also, the center of the circle is marked and connected to all other marked points. Is it possible to paint 1007 of the marked points red and the other 1007 blue so that each red point is connected to an odd number of blue points, and eac... | Answer: No. Solution: Suppose the center of the circle is red. Then on the circumference, there are 1006 red and 1007 blue points, so there will be two adjacent blue points. But then the next point in the clockwise direction must also be blue - otherwise, there would be a blue point connected to one blue point. Continu... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,744 |
6. Given a convex pentagon $A B C D E$, with line $B E$ parallel to line $C D$ and segment $BE$ shorter than segment $C D$. Inside the pentagon, points $F$ and $G$ are chosen such that $ABCF$ and $AGDE$ are parallelograms. Prove that $CD = BE + FG$. (K. Knop, S. Berlov) | Solution. Mark a point $H$ on the segment $C D$ such that $C H=B E$. Then $B E H C$ is a parallelogram. This means that segment $E H$ is parallel and equal to segment $B C$, and thus to segment $A F$. Therefore, $AFHE$ is a parallelogram. Now we obtain that segment $F H$ is parallel and equal to segment $A E$, and thus... | CD=BE+FG | Geometry | proof | Yes | Yes | olympiads | false | 6,745 |
7. On a grid board of size $2014 \times 2014$, several (no less than one) cells are colored such that in each $3 \times 3$ square, an even number of cells are colored. What is the smallest possible number of colored cells? (M. Antipov) | Answer: 1342. Solution: Example. We will color the second, third, fifth, sixth, ..., 2012th, and 2013th cells in the first vertical column of the board. Then, in all 3x3 squares adjacent to the left edge of the board, exactly two cells are colored, and in all other colored cells, there are none. In this case, a total o... | 1342 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,746 |
8. Given 2014 pairwise distinct natural numbers such that the product of any two of them is divisible by their sum. Prove that none of these numbers can be equal to the product of six pairwise distinct prime numbers. (S. Berlov, V. Senderov, I. Rubanov) | Solution. Assume the opposite: among the given numbers, there is a number $a$ equal to the product of six pairwise distinct primes. From the equality $a b /(a+b)=b-b^{2} /(a+b)$, it follows that $a b$ is divisible by $a+b$ if and only if $b^{2}$ is divisible by $a+b$. Therefore, if the number $a$ is on the board, then ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,747 |
3. In triangle $ABC$, the angle bisectors $BK$ and $CL$ are drawn. A point $N$ is marked on segment $BK$ such that $LN \parallel AC$. It turns out that $NK = LN$. Find the measure of angle $ABC$. (A. Kuznetsov) | Answer: $120^{\circ}$. Solution. In the isosceles triangle $L N K$, $\angle K L N = \angle L K N$. Moreover, the angles $\angle K L N$ and $\angle L K A$ are equal as alternate interior angles when the lines $L N$ and $A C$ are parallel. Thus, $\angle K L N = \angle L K A$, which means that the ray $K L$ is the bisecto... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,750 |
4. The numbers 1, 2, ..., 1000 are divided into two sets of 500 numbers each: red $k_{1}, k_{2}, \ldots, k_{500}$ and blue $s_{1}, s_{2}, \ldots, s_{500}$. Prove that the number of pairs $(m, n)$ for which the difference $k_{m} - s_{n}$ gives a remainder of 7 when divided by 100 is equal to the number of pairs $(m, n)$... | Solution. Write down all numbers from 1 to 1000 on the board, and draw an arrow from number $a$ to number $b$ if the difference $a-b$ gives a remainder of 7 when divided by 100. Then, 10 arrows will enter each number and 10 arrows will exit each number. Therefore, the number of arrows with a blue start is the same as t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,751 |
5. For what largest $p$ does there exist a convex $n$-gon in which the lengths of the diagonals take on no more than two distinct values? (I. Rubanov) | Answer. For $n=7$. Solution. Example. A regular heptagon. It has diagonals of exactly two types: connecting vertices one apart and two apart. Evaluation. Let $A B$ be a side of a convex polygon $M$, which has diagonals of only two possible lengths $x$ and $y$. Then for any vertex $C$, not adjacent to $A$ and $B$, the s... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,752 |
1. Find the quotient if it is known that it is 6 times larger than the dividend and 15 times larger than the divisor. | Answer: 2.5. Solution. Let the quotient be $a$. Then the dividend is $a / 6$, the divisor is $a / 15$, and the quotient is $(a / 6):(a / 15)=15 / 6=2.5$. | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,753 |
2. Three brothers returned from fishing. Mother asked each of them how many fish they caught together. Vasya said: "More than ten", Petya: "More than eight", Kolya: "More than fifteen". How many fish could have been caught (list all possibilities), if it is known that two brothers told the truth, while the third one li... | Answer: 16, 17 or 18. Solution. If the brothers caught more than 18 fish, then all of them told the truth. If the brothers caught no more than 15 fish, then Petya and Kolya lied. In both cases, we get a contradiction with the problem's condition. If, however, the brothers caught more than 15 but no more than 18 fish, V... | 16,17,18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,754 |
3. Is it possible to number the faces of a cube with the numbers 1, 2, 3, 4, 5, and 6 such that the number on each face is a divisor of the sum of the numbers on the adjacent faces? If yes, how? If no, why? | Answer: No. Solution. Suppose we managed to number the faces of the cube in accordance with the condition of the problem. Consider the face numbered 6. The sum of the numbers of the four adjacent faces must be divisible by 6. This sum is no less than $1+2+3+4=10$ and no more than $2+3+4+5=14$, so it must be equal to 12... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,755 |
4. For each pair of numbers $x$, let us denote by $s(x, y)$ the smallest of the numbers $x, 1-y, y-x$. What is the largest value that the number $s(x, y)$ can take? | Answer: 1/3. Solution. The sum of the three numbers mentioned in the condition is 1. Therefore, the smallest of them is not greater than $1 / 3$: otherwise, the sum of these numbers would be greater than $1 / 3+1 / 3+1 / 3=1$. On the other hand, when $x=1 / 3$, $y=2 / 3$, the smallest of the three given numbers is $1 /... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,756 |
5. At the vertices of a hexagon, numbers are written, and on each side, the sum of the numbers at its ends. Let's call rounding the replacement of a non-integer number with one of the two nearest integers (the nearest larger or the nearest smaller), and let an integer remain unchanged when rounded. Prove that all 12 nu... | Solution. Let's number the vertices of the hexagon in a circle from 1 to 6. Non-integer numbers in vertices with even numbers will be rounded to the nearest not smaller integer, while numbers in vertices with odd numbers will be rounded to the nearest not larger integer. Now, take the number standing on the side. It is... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,757 |
1. Vintik and Shpuntik built a machine called "Tug-Push," which moves forward on syrup with a fuel consumption of Zl/km, and backward on orange juice with a fuel consumption of 5l/km. Leaving home, they drove the machine in turns. Vintik drove 12 km in both directions. Shpuntik drove forward half as much as Vintik, and... | Answer: 9 km. Solution: Let Vintik travel $2 x$ km forward and $y$ km back, then $2 x+y=12$ and $9 x+15 y=75$ (since together they traveled $3 x$ km forward and $3 y$ km back). Solving the system, we get $x=5, y=2$. It remains to calculate $3 x-3 y=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,758 |
2. A natural number ends with zero, and the greatest of its divisors, not equal to itself, is a power of a prime number. Find the second-to-last digit of this number. | Answer: 1 or 5. Solution: A natural number is divisible by 2 and 5. Then its greatest proper divisor is half of the number, and the number itself has the form $2 \cdot 5^{k}$. For $k=1$, the second-to-last digit will be 1, and for $k>1$ it will be 5, since $5^{k}$ in this case ends in 25. | 1or5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,759 |
3. In a convex pentagon $A B C D E$, $A B$ is parallel to $D E$, $C D = D E$, $C E$ is perpendicular to $B C$ and $A D$. Prove that the line passing through $A$ parallel to $C D$, the line passing through $B$ parallel to $C E$, and the line passing through $E$ parallel to $B C$, intersect at one point. | Solution. Triangle $CDE$ is isosceles, and $AD$ is the height to its base. Therefore, $AD$ is the bisector of triangle $CDE$, and angles $ADE$ and $ADC$ are equal. Angles $ADE$ and $BAD$ are equal as alternate interior angles when parallel lines $AB$ and $DE$ are intersected by the transversal $AD$. Thus, angles $ADC$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,760 |
4. In the city of liars and knights, there are 366 residents, all born on different days of a leap year. All residents of the city answered two questions. To the question “Were you born in February?” 100 people answered affirmatively, and to the question “Were you born on the 30th?” 60 people answered affirmatively. Ho... | Answer: 29. Solution: To the first question, knights born in February and liars born in other months answered affirmatively. Let $x$ be the number of knights born in February, where $x$ does not exceed 29. Then, $29-x$ liars were born in February, and $100-x$ liars were born in other months. The total number of liars i... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,761 |
5. Triangles are inscribed in some cells of an $8 \times 8$ board, with one side coinciding with the side of the cell, and the third vertex lying on the opposite side of the cell. The triangles have no common points. What is the least possible number of empty cells? | Answer: 24. Solution: Estimation. On each side of a triangle, there are no fewer than two vertices of cells, with a total of $9 * 9=81$ vertices. Then, the total number of triangles is no more than 40, and the number of free cells is no less than 24. Example. Filled and unfilled concentric rings alternate (see figure).... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,762 |
1. Given the fraction 2/3. It is allowed to perform the following operations multiple times: add 2013 to the numerator or add 2014 to the denominator. Is it possible to obtain a fraction equal to $3 / 5$ using only these operations? | Answer: No. Solution. Suppose there exist such non-negative integers $a$ and $b$ that $(2+2013a)/(3+2014b)=3/5$. Then after simplification, the numerator must become three. But this is impossible, because 2+2013a is not divisible by 3, since 2013 is divisible by 3, but 2 is not. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,763 |
2. A rectangular grid with sides 629 and 630 is cut into several squares (all cuts follow the grid lines). What is the smallest number of squares with an odd side that can be in such a partition? Don't forget to explain why there cannot be a smaller number of squares with an odd side in the partition. | Answer: Two. Solution. An example when there are exactly two squares: two squares with a side of 315 adjoin the side of the rectangle with a length of 630, and the remaining rectangle $630 \times 314$ is cut into squares $2 \times 2$. A smaller number of squares with an odd side cannot be: to each of the sides of lengt... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,764 |
3. A mad constructor created a clock with 150 hands. The first hand rotates at a speed of one revolution per hour, the second hand makes 2 revolutions per hour, ..., the 150th hand makes 150 revolutions per hour. The clock was started from a position where all hands were pointing straight up. When two or more hands mee... | Answer: In 20 minutes. Solution. The first meeting of the hands will occur when the fastest 150th hand catches up with the slowest first hand. After this, they will fall off, and we can forget about them. The second meeting will happen when the fastest of the remaining, the 149th, catches up with the slowest of the rem... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,765 |
4. In the election in Sunny City, one could vote for Vintik, Shpuntik, or Knopochka. After the results were announced, it turned out that all candidates together received $146\%$ of the votes. The vote counter, Neznaika, explained that by mistake he calculated the percentage of votes for Vintik not from the total numbe... | Solution. Let Shpuntik receive $a$ votes, Vintik - $k a$ votes, and Knopochka - $b$ votes. According to the condition, $k a /(a+k a)+(a+b) /(a+k a+b)=1.46 \Rightarrow k a /(a+k a)>0.46 \Rightarrow k>0.46(1+k) \Rightarrow k>46 / 54>0.85$. Since Shpuntik received more than 1000 votes, Vintik received $k a>1000 k>1000 \cd... | 850 | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,766 |
5. Diagonals $A D$ and $B E$ of a convex pentagon $A B C D E$ intersect at point P. It is known that $A C=C E=A E, \angle A P B=\angle A C E$ and $A B+B C=C D+D E$. Prove that $A D=B E$. | Solution. According to the condition, triangle $A C E$ is equilateral, hence $\angle A P B=\angle A C E=60^{\circ}$ and $\angle A P E=120^{\circ}$. Let $\phi=\angle B E A$. Then $\angle D A E=180^{\circ}-\angle A P E-\angle B E A=60^{\circ}-\phi$, from which $\angle C A D=\phi$. Therefore, when rotating by $120^{\circ}... | AD=BE | Geometry | proof | Yes | Yes | olympiads | false | 6,767 |
1. What is the smallest number of digits that can be erased from the number 20162016 so that the result is divisible by 2016 (it is not allowed to erase nothing)? Please note that you need to not only provide an example but also explain why it is impossible to do with fewer digits. | Answer. Three. Solution. Since 2016 is divisible by 9, the sum of the digits of the resulting number after erasing some digits must also be divisible by 9. The sum of the digits of the number 20162016 is 18. Erasing one or two zeros does not yield the desired result: the numbers 2162016, 2016216, and 216216 are not div... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,768 |
3. In a football tournament where each team played against each other once, 16 teams participated. Three points were awarded for a win, one point for a draw, and zero points for a loss. After the tournament, it was found that each team won at least a third of their matches and lost at least a third of their matches. Pr... | First solution. Each team played 15 matches in the championship. According to the condition, each team won at least five of them and lost at least five, so they scored no less than 15 and no more than 30 points. At the same time, no team could have scored 29 points. Indeed, let's assume such a team exists. Then it must... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,770 |
4. The diagonals of parallelogram $A B C D$ intersect at point $O$. Point $P$ is such that DOCP is also a parallelogram (with $C D$ as its diagonal). Let $Q$ be the point of intersection of $B P$ and $A C$, and $R$ be the point of intersection of $D Q$ and $C P$. Prove that $P C=C R$. | Solution. Note that segments $D P$ and $B C$ are parallel and equal. Therefore, $B O P C$ is a parallelogram, from which $Q C=O C / 2=P D / 2$. Thus, segment $Q C$ with endpoints on sides $R D$ and $R P$ of triangle $D R P$ is parallel to side $D P$ of this triangle and equal to its half. Therefore, it is the midline o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,771 |
5. Do there exist natural numbers $m, n, k$ such that all three numbers $m^{2}+n+k, n^{2}+k+m, k^{2}+m+n$ are squares of natural numbers? | Answer. No. Solution. Suppose the statement of the problem is true. Then $m^{2}+n+k \geq(m+1)^{2}$, from which $n+k \geq 2 m+1$. Similarly, $m+k \geq 2 n+1, n+m \geq 2 k+1$. Adding the three obtained inequalities, we get $2(n+m+k) \geq 2(n+m+k)+3$. Contradiction. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,772 |
5. On an island, there live only knights, who always tell the truth, and liars, who always lie. One day, they all sat in a circle, and each one said: "Among my two neighbors, there is a liar!" Then they sat in a circle in a different order, and each one said: "Among my two neighbors, there is no knight!" Could there ha... | Answer. It could not. Solution. In the first round, both neighbors of each liar were knights. By matching each liar with their right neighbor in this round, we can be convinced that there are no fewer knights than liars on the island. In the second round, both neighbors of each knight were liars. By matching each knigh... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,773 |
6. In a convex quadrilateral $A B C D$, the bisector of angle $B$ passes through the midpoint of side $A D$, and $\angle C=\angle A+\angle D$. Find the angle $A C D$. (S. Berlov) | Answer. $\angle A C D=90^{\circ}$. Solution. Let $E$ be the midpoint of side $A D$, and $F$ be the
intersection point of $B E$ and $A C$. From the given condition, we have: $\angle B=360^{\ci... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,774 |
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