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10.7. In a math club, 24 schoolchildren are participating. Each team of 6 schoolchildren is considered either played or not played by the leader. For a math battle tournament, the leader plans to divide the children into 4 teams of 6. Can it happen that in any division of the schoolchildren into 4 teams, either exactly... | Answer: Yes, it can.
Solution. Let's provide one of the possible examples. We will highlight three students. We will call teams played if they contain 1 or 3 highlighted students, and the rest - unplayed.
The highlighted students can either end up in three different teams, in which case we will get three played teams... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,323 |
10.8. Given a non-constant polynomial $P(x)$ with integer coefficients and a natural number $n$. Let $a_{0}=n, a_{k}=P\left(a_{k-1}\right)$ for all natural $k$. It turns out that for any natural number $b$, the sequence $a_{0}, a_{1}, a_{2}, \ldots$ contains a number that is a $b$-th power of a natural number greater t... | Solution. Note immediately that for each natural number $b$, the sequence $a_{0}, a_{1}, a_{2}, \ldots$ will contain infinitely many $b$-th powers of natural numbers greater than one. Indeed, if their number is finite, and the largest one is $N=x^{b}$, then the sequence will not contain any $N b$-th power, which is imp... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,324 |
3. The beetle is 3 times heavier than the cat, the mouse is 10 times lighter than the cat, the turnip is 60 times heavier than the mouse. How many times heavier is the turnip than the beetle? Justify your answer. | # Answer. 2 times.
Solution. Cat $=10$ mice, turnip $=60$ mice. Therefore, the turnip is 6 times heavier than the cat. That is, the turnip $=6$ cats. According to the condition, Zhuchka $=3$ cats. Therefore, the turnip is 2 times heavier than Zhuchka.
## Grading Criteria.
- Correct solution - 7 points.
- Correct ans... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,327 |
5. Each of two ants, Fat and Thin, needs to transport 150 g of cargo from point $A$ (where they are currently located) to point $B$, the distance between which is 15 meters. The Fat ant walks at a speed of 3 m/min but can carry 5 g of cargo, while the Thin ant walks at a speed of 5 m/min but can carry only 3 g of cargo... | Answer: The Fat one will finish 2 minutes earlier.
Solution: To deliver the cargo, the Fat one needs to make 30 trips from point $A$ to point $B$ and 29 return trips from point $B$ to point $A$. One trip takes him 5 minutes, so the total time will be $5 \cdot (30 + 29) = 295$ minutes. The Thin ant needs to make 50 tri... | The\Fat\one\will\finish\2\\earlier | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,329 |
11.5. Let $P(x)$ be a polynomial of degree $n \geqslant 2$ with non-negative coefficients, and let $a, b$, and $c$ be the lengths of the sides of some triangle. Prove that the numbers $\sqrt[n]{P(a)}, \sqrt[n]{P(b)}$, and $\sqrt[n]{P(c)}$ are also the lengths of the sides of some triangle. (N. Agakhanov) | Solution. Let, without loss of generality, $a \geqslant b \geqslant c$; these three positive numbers are the lengths of the sides of a triangle if and only if $a < b + c$; this means we need to check that $\sqrt[n]{P(a)} < \sqrt[n]{P(b)} + \sqrt[n]{P(c)}$.
Let $P(x) = p_{n} x^{n} + p_{n-1} x^{n-1} + \ldots + p_{0}$. D... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,330 |
11.6. In some cells of a $200 \times 200$ square, there is one chip - either red or blue; the other cells are empty. A chip can see another if they are in the same row or column. It is known that each chip sees exactly five chips of the other color (and possibly some chips of its own color). Find the maximum possible n... | Answer: 3800 chips.
Solution: An example containing 3800 chips can be constructed as follows. Highlight the "border" of width 5 in a $200 \times 200$ square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place the chips in these four rectangles: red chips in the left and ... | 3800 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,331 |
11.7. Initially, a natural number $N$ is written on the board. At any moment, Misha can choose a number $a>1$ on the board, erase it, and write down all natural divisors of $a$, except for $a$ itself (the same numbers can appear on the board). After some time, it turned out that there are $N^{2}$ numbers on the board. ... | Answer. Only for $N=1$.
Solution. Lemma. For any natural $n>1$, the inequality holds
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<1
$$
Proof. Consider the function $f(t) = \frac{1}{t^{2}}$ for $t > 1$. The function $f(t)$ is decreasing for $t > 1$. Let $1 = d_{1} < d_{2} < \ldots < d_{k} < d_{k+1} = N$ ... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,332 |
1. Find all natural numbers such that if you add their smallest divisor greater than one to them, the result is 30. | Answer: $25 ; 27 ; 28$.
Solution. The smallest divisor greater than one is a prime number. If each of the addends is divisible by it, then the sum is also divisible. The number 30 has three prime factors: $2,3,5$. Possible options for the desired numbers: $30-2=28,30-3=27,30-5=25$. Checking shows that all these number... | 25,27,28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,334 |
3. Eight aborigines - representatives of three tribes - stood in a circle. They tell the truth to their tribesmen and lie to representatives of other tribes. Could it happen that each of them said to the neighbor on the right: "My neighbor on the left is from a different tribe"? | Answer: could.
Solution. Let's denote the representatives of the tribes by numbers: $1,2,3$. We will arrange them as follows: $\rightarrow 1 \rightarrow 1 \rightarrow 2 \rightarrow 2 \rightarrow 1 \rightarrow 1 \rightarrow 3 \rightarrow 3 \rightarrow$ (the arrow indicates who is speaking to whom). It is clear that eac... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,336 |
4. A square $10 \times 10$ is divided into unit squares. How many triangles are formed after drawing one diagonal? | Answer: 110.
Solution. The figure shows one of the obtainable triangles. All such triangles are right-angled, with the right angle vertex being any lattice node except those lying on the diagonal. There are a total of $11 \times 11$ nodes, and 11 of them are on the diagonal, so the number of triangles is $11 \times 11... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,337 |
5. Masha bought three pies of each of two types: with apples and with cherries. While she was carrying them home, they got mixed up. Masha has a device at home into which several pies can be placed, and if there are pies of different types among them, a green light will turn on. How can all the pies be divided into two... | Solution. Choose one of the pies. Compare this pie with four of the remaining ones. According to the results of the comparison in the device, send the pie to the group of the same type as the chosen one or to the other group. Add the sixth pie so that there are three pies of each type.
Criteria. Any correct solution: ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,338 |
6.1. Find the largest six-digit number, all digits of which are different, and each of the digits, except for the extreme ones, is either the sum or the difference of the adjacent digits. | Answer: 972538.
Solution. Let $A$ be the desired number. Let's try to find the number $A$ with the first digit being 9. We will try different options for the second digit. If the second digit is 8, then we get: $A=98176-$ it does not form a six-digit number. If the second digit is 7, then we get: $A=972538-$ a six-dig... | 972538 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,339 |
6.2. Two cyclists decided to travel from point $A$ to point $B$. The speed of the first one is 35 km/h, and the second one's speed is 25 km/h. It is known that each of them only rode when the other was resting (standing still), and in total, they covered the same distance in 2 hours. Could they have reached point $B$, ... | Answer: Could not.
Solution: We will assume that the cyclists did not rest simultaneously. Since they covered the same distance, and the ratio of their speeds is $7:5$, the time of movement for the first cyclist is 5 parts, and for the second cyclist - 7 parts of time. In total, they rode for 2 hours $=$ 120 minutes. ... | Couldnot | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,340 |
6.3. Find all solutions to the puzzle: ТУК + ТУК + ТУК + ТУК + ТУК = СТУК. The same letters correspond to the same digits, different letters correspond to different digits. | Answer. Two solutions $(250+\ldots+250=1250$ and $750+\ldots+750=3750)$.
Solution. Subtract TUK from both sides of the equation. We get: TUK + TUK + TUK + $T U K=C 000$. That is, $4 \cdot T U K=C \cdot 1000$. Dividing by 4, we get: $T U K=C \cdot 250$. The number $T U K-$ is a three-digit number, so $C<4$. Checking sh... | 250+\ldots+250=1250750+\ldots+750=3750 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,341 |
6.5. In the castle, there are 16 identical square rooms forming a $4 \times 4$ square. Sixteen people, who are either liars or knights (liars always lie, knights always tell the truth), moved into these rooms, one person per room. Each of these 16 people said: "At least one of the rooms adjacent to mine is occupied by ... | Answer: 8 liars.
Solution: Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's divide the rooms into 8 pairs of adjacent rooms. Then, in each pair, there can be no more than one liar. Therefore, there can be no more than 8 liars in total. Consider a chessboard coloring of ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,343 |
9.1. Given quadratic trinomials $f_{1}(x)=x^{2}+2 a_{1} x+b_{1}, f_{2}(x)=$ $=x^{2}+2 a_{2} x+b_{2}, f_{3}(x)=x^{2}+2 a_{3} x+b_{3}$. It is known that $a_{1} a_{2} a_{3}=$ $=b_{1} b_{2} b_{3}>1$. Prove that at least one of these trinomials has two roots.
(N. Agakhanov) | Solution. Suppose the opposite; then the discriminants of all trinomials are non-positive, that is, $a_{k}^{2} \leqslant b_{k}(k=1,2,3)$. The left (and therefore the right) parts of these inequalities are non-negative, so they can be multiplied, yielding $\left(a_{1} a_{2} a_{3}\right)^{2} \leqslant$ $\leqslant b_{1} b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,344 |
9.2. Seven skiers with numbers $1,2, \ldots, 7$ left the start one after another and completed the distance - each at their own constant speed. It turned out that each skier was involved in overtakes exactly twice. (In each overtake, exactly two skiers participate - the one who overtakes and the one who is overtaken.) ... | Answer. Note that the parity of each skier's position changed with every overtake; therefore, their finishing position is of the same parity as their starting position.
Solution. Since the speeds are constant, any two skiers met no more than once. We will denote the skiers by their starting numbers.
The winner could ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,345 |
9.3. Is it possible for some natural $k$ to divide all natural numbers from 1 to $k$ into two groups and write down the numbers in each group in a row in some order so that two identical numbers are obtained?
(N. Agakhanov) | Answer: No
Solution. Assume the opposite. Clearly, $k \geqslant 10$, since the set of digits from 1 to 9 contains no repetitions. Consider the highest power of ten $10^{n}$ that does not exceed $k$. The sequence of digits of the number $10^{n}$ will entirely fit into one of the formed numbers. But then the same sequen... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,346 |
9.4. In triangle $A B C$, angle $A$ is equal to $60^{\circ}$. Let $B B_{1}$ and $C C_{1}$ be the angle bisectors of this triangle. Prove that the point symmetric to vertex $A$ with respect to the line $B_{1} C_{1}$ lies on side $B C$.
(D. Prokopenko) | Solution. Let $I$ be the point of intersection of the angle bisectors of triangle $ABC$. Then $\angle B_{1}IC_{1} = \angle BIC = 180^{\circ} - \angle IBC - \angle ICB = 180^{\circ} - (\angle ABC + \angle ACB) / 2 = 180^{\circ} - 60^{\circ} = 120^{\circ} = 180^{\circ} - \angle B_{1}AC_{1}$. Therefore, the quadrilateral ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,347 |
9.1. Initially, there were 10 piles of candies on the table, containing $1, 2, \ldots, 10$ candies respectively. The Child decided to redistribute the candies. On each odd minute, he chooses one pile and divides it into two piles, each containing at least one candy. On each even minute, he chooses two piles and merges ... | Answer: Yes.
Solution. We will provide an example of how Little One can achieve such a distribution. On the first minute, he divides the pile of 10 candies into two piles of 5 candies each. Then, on the 2nd, 4th, 6th, and 8th minutes, he combines the piles of 1+9, 2+8, 3+7, 4+6 respectively, and on the 3rd, 5th, 7th, ... | 11 | Combinatorics | proof | Yes | Yes | olympiads | false | 13,348 |
9.2. On the board, there are $n$ different integers, any two of which differ by at least 10. The sum of the squares of the three largest of them is less than three million. The sum of the squares of the three smallest of them is also less than three million. For what largest $n$ is this possible? | Answer. For $n=202$.
Solution. Notice immediately that $990^{2}+1000^{2}+1010^{2}=(1000-10)^{2}+1000^{2}+(1000+10)^{2}=3 \cdot 1000^{2}+2 \cdot 10^{2}$, which is greater than three million. On the other hand, $989^{2}+999^{2}+1009^{2}=(1000-11)^{2}+(1000-1)^{2}+(1000+9)^{2}=3 \cdot 1000^{2}-6 \cdot 1000+(9^{2}+1^{2}+1... | 202 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,349 |
9.3. Kolya and Dima are playing a game on an $8 \times 8$ board, taking turns, with Dima starting. Kolya draws crosses in the cells, while Dima covers pairs of adjacent cells (by sides) with $1 \times 2$ rectangles (dominoes). On his turn, Kolya must place one cross in any empty cell (i.e., a cell that does not yet hav... | Answer. Kolya.
Solution. We will present a winning strategy for Kolya. Mentally color the board in a chessboard pattern and place crosses only in the black cells. Dima covers exactly one of the black cells with each of his moves; therefore, Kolya will be able to make 16 moves. We will show that Dima will not be able t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,350 |
9.4. Let $p$ be a prime number greater than 3. Prove that there exists a natural number $y$, less than $p / 2$, such that the number $p y+1$ cannot be represented as the product of two integers, each greater than $y$.
(M. Antipov) | Solution. Let $p=2k+1$. Assume the opposite: for each of the numbers $y=1,2, \ldots, k$ there exists a factorization $p y+1=a_{y} b_{y}$, where $a_{y}>y, b_{y}>y$. Note that each of the numbers $a_{y}$ and $b_{y}$ is strictly greater than 1, and that $a_{y} > p y + 1$. Therefore, each of the $p-1$ numbers in the set $a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,351 |
9.5. Quadrilateral $ABCD$ is circumscribed around a circle $\omega$. Prove that the diameter of the circle $\omega$ does not exceed the length of the segment connecting the midpoints of sides $BC$ and $AD$.
(O. Yuzhakov) | Solution. We have $S_{ABCD} = pr$, where $p$ is the semiperimeter of the quadrilateral, and $r$ is the radius of $\omega$. From the tangential property, it follows that $AB + CD = BC + DA$, hence
$$
S_{ABCD} = (BC + AD) \cdot r
$$
On the other hand, if $M$ and $N$ are the midpoints of sides $BC$ and $AD$ respectively... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,352 |
1. Does there exist a 2020-digit number (not containing the digits 0 and 1), divisible by the sum of its digits? | Answer. Yes.
Solution. For example, such a number is 222... $2223333444 . . .4444$ (454$\cdot$4 twos, 4 threes, and $50 \cdot 4$
on 1111 (it is composed of blocks of 4 digits) and by 4, and t... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,353 |
2. Scheherazade tells the Shah mathematical tales. A mathematical tale is an equation of the form $a x+b y=c$, where $a, b, c$ are natural numbers ranging from 1 to 17 (inclusive). A tale is sad if the corresponding equation has no solutions in integers (for example, the tale $2 x+4 y=7$ is sad, while the tale $3 x+5 y... | Answer: will be able to.
Solution. The following fairy tales will be sad:
a) a and b are even, c is odd. There are 8$\cdot$8$\cdot$9=576 such fairy tales.
b) a and b are divisible by 3, c is not divisible by 3. There are 5$\cdot$5$\cdot$12 such fairy tales. However, we need to account for the overlap with a): a and ... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,354 |
3. Kolya and Olya place bishops on a $5 \times 5$ board so that they do not attack each other. The player who cannot make a move loses. Kolya moves first, but he is not allowed to place a bishop in the center of the board on his first move. Who will win with the best play from both sides? | Answer: Kolya will win.
Solution. We will use a chessboard coloring (center - black). Kolya can make the first move, for example, to the cell c1. Thereafter, when Olya moves to a white cell, Kolya responds symmetrically (such a move will be valid; the position of the bishops on the white cells will remain symmetrical,... | Kolyawillwin | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,355 |
5. In triangle $ABC$ (angle B is obtuse), the altitude $BH$ and the bisector $AK$ are drawn. Find the angle $AKB$, if the angle $KHC$ is $45^{\circ}$. | Answer: $45^{\circ}$.
Solution. Point K is equidistant from lines AB and AC (since AK is the bisector), and from lines HB and HC (since HK is the bisector of the right angle). Therefore, point K is equidistant from lines AB and BH. This means that BK is the bisector of the angle external to angle ABH. Therefore, angle... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,357 |
11.1 A quirky person didn't mind the effort and recorded 2023 numbers in a circle such that each number is equal to the product of its two neighbors. What is the maximum number of different numbers that could have been used? | Solution: If there is a zero among the numbers, then its neighbors are also zeros, so all the recorded numbers will be equal to zero. We will assume that there are no zeros among the numbers. Let $a$ and $b$ be two adjacent numbers. Then on the other side of $a$ stands $a / b$, and on the other side of $b$ stands $b / ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,358 |
11.2 About numbers \(a\) and \(b\) it is known that the system of equations
\[
\left\{\begin{array}{l}
y=x^{2}+a x+b \\
x=y^{2}+a y+b
\end{array}\right.
\]
has a unique solution. Prove that \(a^{2}=2(a+2 b)-1\). | Solution 1: Note that if $(x, y)$ is a solution to the system, then $(y, x)$ is also a solution. Since solutions with unequal $x$ and $y$ are paired, the system has a solution with $x=y$, and there is exactly one such solution. Then the quadratic equation $x=x^{2}+a x+b$ has one solution. Therefore, its discriminant $(... | ^{2}=2(+2b)-1 | Algebra | proof | Yes | Yes | olympiads | false | 13,359 |
11.3 A weirdo chose 677 different natural numbers from the list $1,2,3, \ldots, 2022$. He claims that the sum of no two of the chosen numbers is divisible by 6. Did he go too far with his claim? | Solution: Suppose the eccentric is right. From the list specified in the condition, there are exactly 377 remainders of division by 6 of each type from 0 to 5. Numbers with remainders 0 and 3 can be taken no more than one each. It is impossible to take numbers with remainders 1 and 5 simultaneously, so in the eccentric... | 676 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,360 |
11.4 The eccentric has $N$ unit squares, from which he managed to form a rectangle with sides differing by 9. The eccentric did not stop there and formed another rectangle from the same $N$ squares, but this time with sides differing by 6. Find $N$. | Solution 1: If we denote the smaller sides of the constructed rectangles by $x$ and $y$, we get the equation $N=x(x+9)=y(y+6)$. Multiply both sides by 4.
$$
\begin{gathered}
4 x^{2}+36 x=4 y^{2}+24 y \\
(2 x+9)^{2}-81=(2 y+6)^{2}-36 \\
(2 x+2 y+15)(2 x-2 y+3)=45
\end{gathered}
$$
Both brackets on the left side are in... | 112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,361 |
11.5 The tangents to the circumcircle of triangle $A B C$ at points $A$ and $C$ intersect at point $T$. The rays $A B$ and $T C$ intersect at point $S$. It is known that the areas of triangles $\triangle A C T, \triangle A B C$ and $\triangle B C S$ are equal. Prove that triangle $\triangle A B C$ is a right triangle. | Solution: First, note that points $B$ and $T$ must lie on opposite sides of line $A C$, otherwise $S(\triangle A B C) < S(\triangle A C T)$. Triangles $\triangle A B C$ and $\triangle B C S$ have a common height, so the equality $S(\triangle A B C) = S(\triangle B C S)$ means that $A B = S B$ (1 point). Triangles $\tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,362 |
9.1. In the country, some pairs of cities are connected by one-way direct flights (there is no more than one flight between any two cities). We say that city $A$ is accessible from city $B$ if it is possible to fly from $B$ to $A$, possibly with layovers. It is known that for any two cities $P$ and $Q$, there exists a ... | First solution. Let us renumber all cities in the country as $A_{1}, A_{2}, \ldots, A_{n}$. By the condition, there exists a city $B_{2}$, from which cities $A_{1}$ and $A_{2}$ are accessible. Next, cities $A_{3}$ and $B_{2}$ are accessible from some city $B_{3}$. Since $A_{1}$ and $A_{2}$ are accessible from $B_{2}$, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 13,363 |
9.2. Given an isosceles trapezoid \(ABCD\) with bases \(BC\) and \(AD\). A circle \(\omega\) passes through vertices \(B\) and \(C\) and intersects side \(AB\) and diagonal \(BD\) again at points \(X\) and \(Y\) respectively. The tangent to circle \(\omega\) at point \(C\) intersects ray \(AD\) at point \(Z\). Prove th... | Solution. Since $B C \| A D$, and the line $Z C$ is tangent to the circle $\omega$, we have $\angle A D B=\angle Y B C=\angle Y C Z$. Consequently, $\angle Y D Z+\angle Y C Z=180^{\circ}$, which means that the quadrilateral $C Y D Z$ is cyclic (see Fig. 1).
Thus, $\angle C Y Z=\angle C D Z=$ $=\angle X B C=180^{\circ}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,364 |
9.3. One hundred gnomes, whose weights are $1, 2, 3, \ldots, 100$ pounds, have gathered on the left bank of a river. They cannot swim, but on the same bank, there is a rowboat with a capacity of 100 pounds. Due to the current, it is difficult to row back, so each gnome has enough strength to row from the right bank to ... | Answer: No.
First solution. Suppose the gnomes managed to cross the river. Let's call the boat trips from the left bank to the right direct, and from the right bank to the left - reverse. Let there be $k$ reverse trips; then there were $k+1$ direct trips.
In $k$ reverse trips, $k$ different gnomes rowed; thus, the to... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,365 |
9.4. Does there exist an infinite increasing sequence $a_{1}, a_{2}, a_{3}, \ldots$ of natural numbers such that the sum of any two
distinct terms of the sequence is coprime with the sum of any three distinct terms of the sequence?
(S. Berlov) | Answer. Yes, it exists.
Solution. We will construct an example of such a sequence. Let $a_{1}=1, a_{2}=7, a_{n+1}=\left(3 a_{n}\right)!+1$. To show that it meets the requirements, we need to strengthen these requirements somewhat. We will say that a pair (triple) of numbers is good if all its elements, different from ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,366 |
1. The natural numbers from 1 to $n$, a multiple of 50, are written on the board. Vasya claims that if all numbers divisible by 50 are erased from the board, the sum of the remaining numbers is a square of some natural number. Is Vasya right? | Answer: Vasya is right.
Solution. Let $n=50 m$. Then the sum of the remaining numbers is
$$
(1+2+\cdots+50 m)-50(1+2+\cdots+m)=25 m(50 m+1)-25 m(m+1)=25 \cdot 49 m^{2}=(35 m)^{2}
$$
Comment. The correct answer is obtained based on the consideration of specific cases - 2 points. If the correct approach is used but co... | (35)^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,367 |
2. Find all integer solutions of the equation $3 \cdot 2^{x}+1=y^{2}$. | Answer. (0,2), $(3,5),(4,7)$.
Solution. Note that $x \geq 0$ (otherwise $y$ is not an integer). If $x=0$, then $y=2$. Let $x$ be a natural number, then $y$ is odd, denote $y=2k+1$. We get $3 \cdot 2^{x}+1=(2k+1)^{2}$, or $3 \cdot 2^{x}=4k^{2}+4k$, from which $3 \cdot 2^{x-2}=k(k+1)$. Since $k$ and $k+1$ are coprime, t... | (0,2),(3,5),(4,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,368 |
3. In an acute-angled triangle $ABC$, two altitudes $AD$ and $CE$ are drawn. Perpendiculars $AM$ and $CN$ are dropped from points $A$ and $C$ to the line $DE$. Prove that $ME = DN$. | Solution. Since $\angle A D C=\angle A E C$, quadrilateral $A E D C$ is cyclic. By the property of a cyclic quadrilateral, $\angle N D C=\angle B A C=\alpha$, $\angle M E A=\angle B C A=\gamma$. Then, using right triangles $A M E$ and
 \rightarrow (2; 4), (5; 7) \rightarrow (6; 8), \ldots, (4k-3; 4k-1) \rightarrow (4k-2; 4k)$.
N... | Foralln\neq4k+2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,371 |
1. It is known that the quadratic function $f(x)=a x^{2}+b x+c$ takes integer values for each integer $x$. Does it follow that the numbers $a, b$, and $c$ are all integers? | Answer: No.
Solution: The function $f(x)=\frac{1}{2} x^{2}+\frac{1}{2} x$ takes an integer value for every integer $x$, since $f(x)=\frac{x(x+1)}{2}$ and one of the numbers $x$ and $x+1$ is even.
Remark. This example is not unique.
Comment. A correct answer without an example - 0 points.
A correct example is given,... | No | Algebra | proof | Yes | Yes | olympiads | false | 13,372 |
3. Given a $101 \times 101$ grid, all cells of which are painted white. It is allowed to choose several rows and repaint all cells in these rows to black. Then, choose exactly the same number of columns and repaint all cells in these columns to the opposite color (i.e., white to black, and black to white). What is the ... | Answer: 5100.
Solution. Let $k$ rows be repainted first, then $k$ columns. After the first stage of repainting, each column will contain $k$ black and $101-k$ white cells. Since $101-k$ columns will remain untouched, the total number of black cells in these columns will be $k(101-k)$. In each of the repainted columns,... | 5100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,373 |
4. In an equilateral triangle $A B C$, points $N, T$, and $F$ are chosen on sides $A C, A B$, and $B C$ respectively, such that $A N=T B$ and $C F=F B$. Prove that the area of quadrilateral $T A N F$ is half the area of triangle $A B C$. | The first solution. The area of triangle $ABC$ is equal to the sum of the areas of triangles $CNF$ and $FTB$ and quadrilateral $TANF$. We will prove that $S_{\triangle CNF} + S_{\triangle FTB} = \frac{1}{2} S_{\triangle ABC}$, from which it will immediately follow that the area of $TANF$ is the remaining half.
Let the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,374 |
5. Non-zero real numbers $x, y, z$ and $t$ are such that $x+y+z+t=0$ and $\frac{1}{x}+\frac{1}{y}+$ $\frac{1}{z}+\frac{1}{t}=0$. Prove that among the numbers $x, y, z$ and $t$, two numbers can be chosen with a zero sum. | Solution. Suppose that the sum of any two of the given numbers is not zero and arrive at a contradiction. From the first equality, it follows that $z+t=-(x+y)$. Since $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=0$, then $\frac{x+y}{x y}+\frac{t+z}{z t}=0$, which means $(x+y)\left(\frac{1}{x y}-\frac{1}{z t}\right)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,375 |
1. (7 points) In an $8 \times 8$ frame with a width of 2 cells (see figure), there are a total of 48 cells. How many cells are in a $254 \times 254$ frame with a width of 2 cells?
 | Answer: 2016.
Solution. First method. Cut the frame into four identical rectangles as shown in the figure. The width of the rectangles is equal to the width of the frame, i.e., 2 cells. The length of each rectangle is 2 less than the side of the frame: $254-2=252$ cells. Then the area of one rectangle is $2 \cdot 252=... | 2016 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,376 |
2. (7 points) Anya multiplied 20 twos, and Vanya multiplied 17 fives. Now they are going to multiply their huge numbers. What will be the sum of the digits of the product? | Answer: 8.
Solution. In total, 20 twos and 17 fives are multiplied. Let's rearrange the factors, alternating twos and fives. This results in 17 pairs of $2 \cdot 5$ and three additional twos, which multiply to 8. Thus, the number 8 needs to be multiplied by 10, 17 times. This results in a number consisting of the digi... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,377 |
4. (7 points) In the room, there are 10 lamps. Petya said: «In this room, there are 5 lamps on». Vasya replied: «You are wrong». And added: «In this room, there are three lamps off». Kolya then said: «An even number of lamps are on». It turned out that only one of the four statements is true. How many lamps are on? | # Answer. 9.
Solution. The first and third statements cannot both be false at the same time, otherwise there would be fewer than five lights on and fewer than three lights off, i.e., fewer than eight lights in total, which contradicts the condition. The first and second statements also cannot both be false at the same... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,379 |
5. (7 points) Nезнayka measured the lengths of the sides and diagonals of his quadrilateral plot of land, wrote down the results of six measurements in his notebook, and immediately forgot which numbers referred to the diagonals and which to the sides. Then he noticed that among the written numbers, there were four ide... | Answer. No, not necessarily.
Solution. Construct an equilateral triangle $ABC$ and on the bisector of its angle $B$, mark off a segment $BD$ equal to $AB$. In the quadrilateral $ABCD$, we have $AB = BC = CA = BD$ (by construction) and $AD = DC$ (for example, from the congruence of triangles $BAD$ and $BCD$ by two side... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,380 |
6. (7 points) Four fleas are playing leapfrog on a large sheet of graph paper.
Every second, one of the fleas jumps over another and, flying along the same line, covers a distance twice as long as the initial distance between the fleas. Currently, the fleas are sitting at the four vertices of one cell. Can all four fl... | Solution. Suppose this did happen, and consider the moment when all four fleas first ended up on the same line. We will ask the flea that made the last jump to jump back. In doing so, it must again jump over one of the other fleas along the segment connecting them, i.e., it must remain on the same line. This means that... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,381 |
10.1. On the coordinate plane, the graphs of two reduced quadratic trinomials and two non-parallel lines $\ell_{1}$ and $\ell_{2}$ are drawn. It is known that the segments cut off by the graphs on $\ell_{1}$ are equal, and the segments cut off by the graphs on $\ell_{2}$ are also equal. Prove that the graphs of the tri... | The first solution. Let $f_{1}(x)$ and $f_{2}(x)$ be the given reduced quadratic trinomials, and let $\Gamma_{1}$ and $\Gamma_{2}$ be their graphs. Then there exists a unique vector $\vec{a}$ such that when the parabola $\Gamma_{1}$ is translated by this vector, it transforms into $\Gamma_{2}$ (the vector $\vec{a}$ con... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,382 |
10.2. An acute isosceles triangle \(ABC (AB = AC)\) is inscribed in a circle with center at point \(O\). The rays \(BO\) and \(CO\) intersect the sides \(AC\) and \(AB\) at points \(B'\) and \(C'\) respectively. A line \(\ell\) is drawn through point \(C'\) parallel to the line \(AC\). Prove that the line \(\ell\) is t... | Solution. Let the line $A O$ intersect $\ell$ at point $T$ (see Fig. 2). By symmetry with respect to $A O$, we have $\angle B^{\prime} T O = \angle C^{\prime} T O$. Since $\ell \| A C$, we get $\angle C^{\prime} T O = \angle O A C = \angle O C A$. Therefore, $\angle B^{\prime} T O = \angle B^{\prime} C O$, which means ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,383 |
10.3. Initially, there are three piles of 100, 101, and 102 stones on the table, respectively. Ilya and Kostya are playing the following game. On each turn, each of them can take one stone from any pile except the one from which they took a stone on their previous turn (on their first turn, each player can take a stone... | Answer. Ilya.
Solution. We will show how Ilya can move to guarantee a win. Let $A, B$, and $C$ denote the piles that initially contain 100, 101, and 102 stones, respectively. Ilya's first move will be to take a stone from pile $B$. There are two possible cases.
Case 1. Kostya takes a stone from a pile other than $B$ ... | Ilya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,384 |
10.4. On a board, there are $n$ positive numbers $a_{1}, a_{2}, \ldots, a_{n}$ written in a row. Vasya wants to write a number $b_{i} \geqslant a_{i}$ under each number $a_{i}$ such that for any two of the numbers $b_{1}, b_{2}, \ldots, b_{n}$, the ratio of one to the other is an integer. Prove that Vasya can write the... | Solution. We will prove that there exist even numbers $b_{1}, b_{2}$, $\ldots, b_{n}$, satisfying the following (stronger) conditions:
(1) $b_{i} \geqslant a_{i}$ for all $i \leqslant n$
(2) $b_{1} b_{2} \ldots b_{n} \leqslant 2^{(n-1) / 2} a_{1} a_{2} \ldots a_{n}$
(3) the ratio of any two of the numbers $b_{i}$ is... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 13,385 |
2. (7 points) The bisector of the exterior angle of triangle $ABC$ at vertex $C$ intersects line $AB$ at point $D$. Prove that $AD: BD = AC: BC$.
# | # Solution.
Draw a line $B K \| C D, K \in A C$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 13,387 |
3. (7 points) The graph of a reduced quadratic trinomial is shown in the figure (the y-axis is erased, the distance between adjacent marked points is 1). What is the discriminant of this trinomial?
. From the condition, it follows that $x_{2}-x_{1}=2$. Since $x_{2}=\frac{-b+\sqrt{D}}{2}, x_{1}=\frac{-b-\sqrt{D}}{2}$, we get that $x_{2}-x_{1}=\sqrt{D}$, hence $D=4$.
## Answer. 4. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,388 |
4. (7 points) A ribbon is wound around a cylindrical column that is 20 meters high and 3 meters in diameter, rising from the base to the top in seven full turns. What is the length of the ribbon? | Solution. By cutting the cylinder along the generatrix of its lateral surface passing through the beginning of the tape and unfolding this surface, we obtain a rectangle $A B C D$ with dimensions $20 \times 3 \pi$ (see figure).
 Prove that the expression $5 x^{2}+5 y^{2}+5 z^{2}+6 x y-8 x z-8 y z$ is positive, provided that $x^{2}+y^{2}+z^{2}$ does not equal zero.
# | # Solution.
Factor out the multiplier 5: $5\left(x^{2}+y^{2}+z^{2}+\frac{6}{5} x y-\frac{8}{5} x z-\frac{8}{5} y z\right)$.
Estimate the sign of the expression in parentheses. For this, we complete the square for the variable $x$, and then for the variable $y$:
$x^{2}+y^{2}+z^{2}+\frac{6}{5} x y-\frac{8}{5} x z-\fra... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,390 |
1. (7 points) Replace each letter with a digit so that the operations performed horizontally and vertically are valid.
| $a b$ | + | $c d$ | $=$ | $e f$ |
| :---: | :---: | :---: | :---: | :---: |
| $\times$ | | $:$ | | + |
| $k$ | $:$ | $m$ | $=$ | $m$ |
| $m n d$ | $:$ | $e$ | $=$ | $e m$ |
Identical letters corr... | # Solution.
| 42 | + | 18 | $=$ | 60 |
| :---: | :---: | :---: | :---: | :---: |
| $\times$ | | $:$ | | + |
| 9 | $:$ | 3 | $=$ | 3 |
| 378 | $:$ | 6 | $=$ | 63 | | 63 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,391 |
3. (7 points) From the airplane window, I can see part of an island, part of a cloud, and a bit of the sea. The cloud occupies half of the landscape visible from the window and thus hides a quarter of the island, which therefore takes up only a quarter of the observed landscape. What part of the landscape does the part... | # Solution.
Three quarters of the island occupy a quarter of the landscape visible through the porthole. Therefore, the entire island constitutes $\frac{1}{3}$, and the sea constitutes $-\fra... | \frac{5}{12} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,393 |
5. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the x-coordinate of the point of intersection.
# | # Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,395 |
1. Buratino left Papa Carlo's house and arrived at the Field of Wonders exactly at $22:00$. If his walking speed had been $25\%$ faster, he would have arrived at $21:30$. At what time did he leave the house? | Answer: at 19:30.
Solution: If Buratino spent $t$ (hours) on his journey, then with the increased speed, he would have spent 1.25 times less, i.e., $\frac{4}{5} t$. Therefore, he would have saved $\frac{1}{5} t$, which amounted to 30 minutes. Thus, he spent 2.5 hours on the way to the Field of Wonders, and he left hom... | 19:30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,396 |
2. A three-digit number, all digits of which are different, will be called balanced if it is equal to the sum of all possible two-digit numbers formed from the different digits of this number. Provide an example of any balanced number. Justify your answer. | Solution: For example, the number $132=13+12+32+21+31+23$ is balanced (there are other options).
Grading criteria: Any suitable number with verification - 7 points, incorrect solution or only answer - 0 points. | 132 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,397 |
3. In the tournament, each participant was supposed to play exactly one game with each of the remaining participants, but two participants dropped out during the tournament, having played only 4 games each. In the end, the total number of games played turned out to be 62. How many participants were there in total? | Answer: 13.
Solution: Let the total number of participants be $n$. Then, excluding the two who dropped out, the remaining participants played $\frac{(n-2)(n-3)}{2}$ matches. If these two managed to play against each other, then 7 matches were played with their participation, and if they did not, then 8 matches. Thus, ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,398 |
4. Do there exist positive numbers $a, b, c$ such that the numbers $d$ and $\sqrt{d}$ are respectively roots of the equations $a x^{2}+b x-c=0$ and $\sqrt{a} x^{2}+\sqrt{b} x-\sqrt{c}=0 ?$ | Answer: No.
Solution: Suppose such numbers exist, substitute the values of $d$ and $\sqrt{d}$ for $x$ in the equations. Then we have $c=a d^{2}+b d$ and $\sqrt{c}=\sqrt{a} d+\sqrt{b} \sqrt{d}$. In the last equation, both sides are positive, square both sides: $c=(\sqrt{a} d+\sqrt{b} \sqrt{d})^{2}=a d^{2}+2 d \sqrt{a b... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,399 |
5. In triangle $\mathrm{ABC}$ with sides $\mathrm{AB}=5, \mathrm{BC}=\sqrt{17}$, and $\mathrm{AC}=4$, a point $\mathrm{M}$ is taken on side $\mathrm{AC}$ such that $\mathrm{CM}=1$. Find the distance between the centers of the circumcircles of triangles $\mathrm{ABM}$ and $\mathrm{BCM}$. | Answer: 2.
Solution: Draw the height BH to side AC. Let $\mathrm{CH}=x$, then $\mathrm{BH}=$ 4 - $x$. By the Pythagorean theorem from two triangles, we have $B H^{2}=B C^{2}-C H^{2}=17-x^{2}$ and $B H^{2}=A B^{2}-A H^{2}=25-(4-x)^{2}$. Equating the right sides of both equations, we get $x=1$, which means points M and ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,400 |
6. In a school quiz, 100 students participated. After the results were tallied, it turned out that any 66 of them together earned no less than $50 \%$ of the total prize points. What is the highest percentage of points that one participant could have earned? | Answer: $25 \%$.
Solution: Suppose participant $\mathrm{X}$ scored the highest percentage of points $-x \%$. Divide the remaining participants into three groups A, B, and C, each with 33 people. Let the total percentages of points scored by these groups be $a, b$, and $c$ respectively. Then,
$$
2(100-x)=2(a+b+c)=(a+b... | 25 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 13,401 |
1. On a sheet of paper, a parabola - the graph of the function $y=a x^{2}+b x+c$ with $a>0, b>0$, and $c<0,$ was drawn, but the coordinate axes were erased. How could they have been positioned? (Illustrate any example that corresponds to the given signs of the coefficients without changing the position of the parabola ... | Answer: see Fig. 10.1.
Since $a>0$, the branches of the parabola are "opened" in the positive direction of the y-axis. Since $c<0$, the point of intersection of the graph with the y-axis has a negative ordinate. Since $-\frac{b}{2 a}<0$, the vertex of the parabola is located in the half-plane $x<0$.
Evaluation criter... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,402 |
2. The sum of two integers is $S$. Masha multiplied the left number by an integer $a$, the right number by an integer $b$, added these products, and found that the resulting sum is divisible by $S$. Alyosha, on the contrary, multiplied the left number by $b$, and the right number by $a$. Prove that his analogous sum wi... | Solution. Let $x$ be the left number and $y$ be the right number; according to the condition: $x+y=S$.. Then Masha got the number $a x+b y$, and Alyosha got the number $b x+a y$. The sum of these numbers is $a x+b y+b x+a y=(a+b)(x+y)=(a+b) S$, which means it is divisible by $S$. Since one of the two addends (Masha's n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,403 |
3. In the zoo, there are 10 elephants and large balance scales. It is known that if any four elephants stand on the left pan and any three on the right, the left pan will outweigh. Three elephants stood on the left pan and two on the right. Will the left pan definitely outweigh? | Answer: necessarily
Solution. First method. Suppose three elephants stand on the left pan of the scales, and two - on the right, and in this case, the left pan does not outweigh the right. Then, let's ask the lightest of the five elephants not standing on the scales to stand on the left pan, and the heaviest - on the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,404 |
4. From the vertex of the obtuse angle $A$ of triangle $A B C$, the altitude $A D$ is dropped. A circle with center $D$ and radius $D A$ is drawn, intersecting sides $A B$ and $A C$ again at points $M$ and $N$ respectively. Find $A C$, if $A B=c, A M=m$ and $A N=n$. | Answer: $\frac{m c}{n}$.
Solution. We will prove that $A M \cdot A B = A N \cdot A C$. This can be done in different ways.
First method. In the right triangles $A D B$ and $A D C$, draw the altitudes $D P$ and $D Q$ respectively (see Fig. 10.4a). Then $A P \cdot A B = A D^{2} = A Q \cdot A C$. Since triangles $A D M$... | \frac{}{n} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,405 |
5. Vasya disassembled the frame of a triangular pyramid in the math classroom and wants to form two triangles from its six edges so that each edge is a side of exactly one triangle. Will Vasya always be able to do this? | Answer: always.
Solution. Note that if Vasya manages to form a triangle from the edges emanating from one vertex of the tetrahedron, then the second triangle is already formed, and the problem is solved.
Let $AB$ be the longest edge of the tetrahedron $DABC$ (see Fig. 10.5). Suppose that neither from the triplet of e... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,406 |
6. 100 lit and 100 unlit flashlights are randomly distributed between two boxes. Each flashlight has a button that, when pressed, turns off a lit flashlight and turns on an unlit one. Your eyes are blindfolded, and you cannot see whether a flashlight is lit. However, you can move the flashlights between the boxes and p... | Solution. First, move all the flashlights to the right box without touching the switches. Then, move any hundred flashlights from the right box to the left box, switching each one, and the goal will be achieved. Let's prove this.
When moving (and switching) one flashlight, the difference between the number of lit flas... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,407 |
# 3. CONDITION
There is an unlimited number of chips in six colors. What is the smallest number of chips that need to be arranged in a row so that for any two different colors, there are two adjacent chips of these colors in the row? | Solution. From the condition, it follows that for each fixed color A, a chip of this color must be paired with a chip of each of the other 5 colors. In a row, a chip has no more than two neighbors, so a chip of color A must appear at least 3 times. Similarly for each other color. Thus, there should be no less than $3 \... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,408 |
# 4. CONDITION
In a plane, there are n identical gears such that the first is engaged with the second, the second with the third, and so on, finally, the n-th gear is engaged with the first. Can the gears in such a system rotate? | Solution. Any two adjacent wheels rotate in opposite directions, meaning that the first and last wheels also have opposite directions of rotation, which is possible if and only if the number of wheels is even. Thus, the wheels of such a system can rotate if the number of wheels is even, and cannot in the opposite case. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,409 |
# 6. CONDITION
There are 2001 coins on the table. Two players play the following game. They take turns. On a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. Who wins with correct play? ... | Solution. Let's describe the first player's strategy. The first move should be to take 81 coins from the table. For each subsequent move, if the second player takes $x$ coins, the first player should take $101-x$ coins. He can always do this because if $x$ is an even number from 2 to 100, then ($101-x$) is an odd numbe... | Thefirstplayerwins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,410 |
Problem 8.1. Gleb placed the numbers $1,2,7,8,9,13,14$ at the vertices and the center of a regular hexagon such that in any of the 6 equilateral triangles, the sum of the numbers at the vertices is divisible by 3. What number could Gleb have written in the center? It is sufficient to provide one suitable example.
, and the centr... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,411 |
Problem 8.2. Misha suggested that Yulia move a chip from cell $A$ to cell $B$. In one step, you can move the chip to an adjacent cell by side or by corner. To make it more interesting, Misha put 30 candies in the prize fund, but said that he would take 2 candies for each horizontal or vertical move and 3 candies for ea... | Answer: 14.
Solution. From $A$ to $B$, one can get through the top or the bottom. If going through the top, the first 2 moves are diagonal (a diagonal move is more advantageous than 2 horizontal moves), and the next 5 moves are horizontal. Misha will take $2 \cdot 3 + 5 \cdot 2 = 16$ candies, and Yulia will win 14. If... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,412 |
Problem 8.4. The diagonals of quadrilateral $A B C D$ intersect at point $K$. It turns out that $A B=B K=K D$. On segment $K C$, a point $L$ is marked such that $A K=L C$. Find $\angle B L A$, given that $\angle A B D=52^{\circ}$ and $\angle C D B=74^{\circ}$.
. We have
$$
\begin{aligned}
\angle BLA & =180^{\circ}-\angle BAL-\angle ABL=180^{\circ}-\frac{180^{\circ}-\angle ABD}{2}-\angle CDB= \\
& =90^{\circ}+\frac{1}{2} \angle ABD-\angle CDB=42^{\circ... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,414 |
Problem 8.5. A field was partially planted with corn, oats, and millet. If the remaining part is completely planted with millet, then millet will occupy half of the entire field. If the remaining part is equally divided between oats and corn, then oats will occupy half of the entire field. By what factor will the amoun... | Answer: 3.
Solution. Let the area of the entire field be 1, and the empty part be $x$. Then, from the first condition, millet occupies $\frac{1}{2}-x$, and from the second condition, oats occupy $\frac{1}{2}-\frac{1}{2} x$.
Corn is left with $1-x-\left(\frac{1}{2}-x\right)-\left(\frac{1}{2}-\frac{1}{2} x\right)=\frac... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,415 |
Problem 8.6. Six princesses have a magic chest. Every minute, a dress of one of 10 colors and one of 9 styles can be taken out of it. However, within one hour, it is impossible to take out two dresses from the chest that match both in color and style. What is the minimum number of dresses the princesses will have to ta... | Answer: 46.
Solution. Note that 45 dresses would not be enough, as the chest can issue exactly 5 dresses of each of the 9 styles.
We will prove that if the princesses take out 46 dresses, there will definitely be 6 dresses of the same style, and thus of different colors (identical dresses could not have occurred over... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,416 |
Problem 8.7. Along an alley, maples and larches were planted in one row, a total of 75 trees. It is known that there are no two maples between which there are exactly 5 trees. What is the maximum number of maples that could have been planted along the alley? | Answer: 39.
Solution. Let's divide all the trees into groups of 12 standing in a row. There will be 6 complete groups and 3 more trees at the end. In each group, we will divide the trees into 6 pairs: the first with the seventh, the second with the eighth, ..., the sixth with the twelfth. Note that there are exactly 5... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,417 |
Problem 8.8. In trapezoid $A B C D(A D \| B C) \angle A B C=108^{\circ}$ and $\angle A D C=54^{\circ}$. On ray $B A$ beyond point $A$, point $K$ is marked such that $A K=B C$. Find the angle $D K C$, given that $\angle B K C=27^{\circ}$.
Fig. 1: to the solu... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,418 |
1. Semyon has 20 numbers: $1,2,3, \ldots, 19,20$. He formed 10 fractions by writing ten of these numbers in some order as numerators, and the remaining ten in some order as denominators. What is the maximum number of integers Semyon could obtain after simplifying all the written fractions? | Answer: 8 numbers.
## Solution:
For the fractions to have an integer value, the prime numbers $11, 13, 17, 19$ can only be numerators with a denominator of 1. Therefore, to form fractions equal to an integer, no more than 17 numbers can be used, meaning no more than 8 fractions can be formed. Example: 20/10, 19/1, $1... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,419 |
2. Before the lesson, Nestor Petrovich wrote several words on the board. When the bell rang for the lesson, he noticed a mistake in the first word. If he corrects the mistake in it, the words with mistakes will make up $24 \%$, and if he erases the first word from the board altogether, the words with mistakes will make... | Answer: $28 \%$.
## Solution:
Let there be $n$ words written on the board before the lesson, of which $x$ have errors. If the error in the first word is corrected, then there will be $x-1$ words with errors out of $n$, and by the condition $x-1=0.24 n$. If a word with an error is erased, then there will be $x-1$ word... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 13,420 |
3. From point $O$, rays $O A, O B, O C$ and $O D$ were laid out, and $\angle A O B=\angle B O C=\angle C O D=3 \angle A O D$. What can $\angle A O D$ be equal to? | Answer: $36^{\circ}, 45^{\circ}$.
## Solution:
Notice that the angles $A O B, B O C$ and $C O D$ follow each other in one direction (otherwise, some rays would coincide).
Let $\angle A O D=X$. Then $\angle A O B=\angle B O C=\angle \underline{\mathrm{COD}}=3 X$.
If the sum of the angles $A O B, B O C$ and $C O D$ i... | 36,45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,421 |
4. In 1919, Pinocchio buried three gold coins on the Field of Wonders. Each year, the number of buried coins increased by three, except for one very lucky year when the number of coins tripled. In 2018 (that is, after 99 years), Pinocchio dug up all the gold coins. Could there have been exactly 456 of them? | Answer: it could not.
## Solution:
## First method.
Let the number of coins increase by three over $n$ years (and then it became $3+3n$), after 1 year it immediately increased by 3 times (and became $9+9n$), and then for another $k$ years it increased by 3 (and became $9+9n+3k$). 99 years have passed, so $n+1+k=99$,... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,422 |
10.1. Can we find twelve different integers, among which exactly six are prime numbers, exactly nine are odd numbers, exactly ten are non-negative numbers, and exactly seven are greater than ten? | Solution. Yes, for example: $-8,-4,2,5,9,11,13,15,21,23,37,81$.
Comments. Any correct example is given - $\underline{7 \text{ points. }}$ | -8,-4,2,5,9,11,13,15,21,23,37,81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,424 |
10.3. The integer 23713 has the following two properties:
(1) any two adjacent digits form a prime two-digit number,
(2) all these prime two-digit numbers are pairwise distinct.
Find the largest of all integers with properties (1) and (2). | Answer: 617371311979.
Solution. Since even digits and the digit 5 can only be in the highest place (and only one of these digits can form a single prime number), and the remaining prime two-digit numbers are $11, 13, 17, 19, 31, 37, 71, 73, 79$, 97, the maximum number of digits in the desired number is 12. In the desi... | 617371311979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 13,425 |
10.4 Find the smallest positive integer $k$, for which in any coloring of the numbers of the set $M=\{1,2,3, \ldots, k\}$ in two colors, there will be ten not necessarily distinct numbers of the same color from the set $M$ | Answer: 109.
Solution. Let $k \geq 100$. Suppose that for any 10 numbers of one color, their sum is either a number of the other color or does not belong to M. Let the number 1 be of the first color, and the number 2 be of the second color. Then the numbers 10 and 20 are of the second and first colors, respectively. N... | 109 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 13,426 |
9.5. The numbers $b>0$ and $a$ are such that the quadratic trinomial $x^{2}+a x+b$ has two distinct roots, exactly one of which lies in the interval $[-1 ; 1]$. Prove that exactly one of these roots lies in the interval $(-b ; b)$.
(A. Khryabrov) | First solution. Let $f(x)=x^{2}+a x+b$. If exactly one root lies on the interval $[-1,1]$, then the quadratic polynomial changes sign on this interval, that is,
$$
(1+a+b)(1-a+b)=f(1) f(-1) \leqslant 0
$$
Then
$0 \geqslant b^{2}(1+a+b)(1-a+b)=\left(b^{2}+a b+b\right)\left(b^{2}-a b+b\right)=f(b) f(-b)$. Therefore, t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 13,427 |
9.6. Inside an acute scalene triangle $A B C$, where $\angle A B C=60^{\circ}$, a point $T$ is marked such that $\angle A T B = \angle B T C = \angle A T C = 120^{\circ}$. The medians of the triangle intersect at point $M$. The line $T M$ intersects the circumcircle of triangle $A T C$ again at point $K$. Find $T M / M... | Answer: $1 / 2$.
First solution. Let $O$ be the center of the circumcircle $\Omega$ of triangle $ABC$. Since $\angle AOC = 2 \angle ABC = 120^{\circ}$, point $O$ lies on the circumcircle $\gamma$ of triangle $ATC$. Let the line $BT$ intersect the circle $\gamma$ again at point $X$, and the circle $\Omega$ at point $P$... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,428 |
9.7. Natural numbers $n>20$ and $k>1$ are such that $n$ is divisible by $k^{2}$. Prove that there exist natural numbers $a, b$, and $c$ such that $n=a b+b c+c a$.
(A. Khryabrov) | Solution. Note that from the equality $n+a^{2}=(a+b)(a+c)$ follows the equality $n=a b+b c+c a$. Therefore, to solve the problem, it is sufficient to find a natural number $a$ such that the number $n+a^{2}$ can be factored into the product of two natural numbers $x$ and $y$, both greater than $a$ (then we can set $b=x-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 13,429 |
9.8. A hundred sages were offered the following trial. They are brought into a hall one by one (in a predetermined order). In the hall, the warden offers the sage a choice of two different numbers from the set $1,2,3$. The sage chooses exactly one of them, reports the chosen number to the warden, and leaves the hall. B... | Solution. Let's present one of the possible agreements. Each sage will use one of two strategies: either choose the odd number (strategy H) from the two proposed numbers, or choose the larger of the two numbers (strategy B). They will choose them as follows:
(1) The first sage acts according to strategy B. The second ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 13,430 |
Problem 5.3. In five of the nine circles in the picture, the numbers 1, 2, 3, 4, 5 are written. Replace the digits $6, 7, 8, 9$ in the remaining circles $A, B, C, D$ so that the sums of the four numbers along each of the three sides of the triangle are the same.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 13,435 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
![](https://cdn.mathpix.com/cropped/2024_05_06_ddc... | Answer: On face $A$ there are 3 points, on face $B-5$, on face $C-6$, on face $D-5$. Solution. Let's consider the arrangement of the faces on one die. We will denote the faces by numbers corresponding to the number of dots on them.
From the picture, it is clear that face 1 borders with faces $2,3,4$ and 5. Therefore, ... | 3,5,6,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,436 |
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 | 13,437 |
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. What is the total number of emeralds in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 13,438 |
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