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7. There is a checkered board of size $2 n \times 2 n$. Petya placed $(n+1)^{2}$ chips on it. The cat can, with one swipe of its paw, knock any one chip or two chips standing in adjacent side-by-side or corner-by-corner cells onto the floor. What is the minimum number of swipes the cat can definitely use to knock all t... | Answer. $n^{2}+n$. Solution. Estimation. Divide the board into $n^{2}$ squares $2 \times 2$. The cat can sweep away any two chips that are in the same square. As long as there are more chips on the board than $n^{2}$, the cat has the opportunity to sweep away two chips, meaning he can do this at least $n+1$ times (afte... | n^{2}+n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,775 |
8. On the board, there are 100 natural numbers, among which exactly 33 are odd. Every minute, the sum of all pairwise products of all the numbers already on the board is added (for example, if the numbers 1, 2, 3, 3 were on the board, the next move would add the number $1 \cdot 2 + 1 \cdot 3 + 1 \cdot 3 + 2 \cdot 3 + 2... | Answer: Yes. Solution: Since there are 33 odd numbers written on the board, in the first minute, the sum of pairwise products, among which there are 33*32/2 odd ones, will be appended. Therefore, the appended number will be even, and there will be exactly 33 odd numbers left on the board. Repeating this reasoning, we g... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,776 |
1. At 9:00, a pedestrian set off on a journey. An hour later, a cyclist set off after him from the same starting point. At 10:30, the cyclist caught up with the pedestrian and continued on, but after some time, the bicycle broke down. After $3a$ minutes of repair, the cyclist resumed his journey, following the pedestri... | Answer: 100 minutes. Solution: The cyclist caught up with the pedestrian half an hour after his start and one and a half hours after the pedestrian's start. This means he is moving three times faster than the pedestrian. By the time of the second meeting with the cyclist, the pedestrian had been walking for 4 hours = 2... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,777 |
2. Can several cells be marked in a 9x9 table so that in any two adjacent rows of the table, at least 6 cells are marked, and in any two adjacent columns - no more than 5 cells are marked? (S. Berlov) | Answer. Yes. Solution. One of the examples is shown in the figure on the right.
Comment. Let's show how one can come up with such an example. In the pairs of rows 1-2, 3-4, 5-6, 7-8, there are no fewer than \(6 \times 4 = 24\) marked cells. On the other hand, in the pairs of columns 2-3, 4-5, 6-7, 8-9, there are no mo... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,778 |
3. $A B C D$ — a convex quadrilateral, where $A B=7, B C=4, A D=D C, \angle A B D=\angle D B C$. Point E is on segment $A B$ such that $\angle D E B=90^{\circ}$. Find the length of segment $A E$. (Spain, local phase, 2020-2021) | Answer: 1.5. Solution: From point $B$, on ray $B A$, we lay off segment $B F = B C$. Triangles $D B F$ and $D B C$ are congruent by two sides and the included angle. Therefore, $D F = D C = D A$, which means $D E$ is the height drawn from the vertex of isosceles triangle $A D F$. Since it is also the median, we have $A... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,779 |
4. Prove that if the lines $y=k x+m, y=m x+n$ and $y=n x+k$ on the coordinate plane have a common point, then they coincide. (S. Tokarev) | Solution. Let $k \geq m$ and $k \geq n$ (cases where the largest coefficient is $m$ or $n$ are analogous). Let $a$ be the abscissa of some common point of the lines. Then $k a + m = m a + n = n a + k$, from which $(k-m) a = n - m$, $(m-n) a = k - n$, and $(n-k) a = m - k$.
Suppose the coefficients $k, m, n$ are pairwi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 6,780 |
5. Given natural numbers $a$ and $b$ ($a > 1$), and $b$ is divisible by $a^2$. Moreover, any divisor of the number $b$, smaller than $\sqrt{a}$, is also a divisor of the number $a$. Prove that the number $a$ has no more than three distinct prime divisors. (A. Gолованов, I. Богданов) | Solution. Suppose the number $a$ has four distinct prime divisors $p, q, r$, and $s$. Then $a=p^{k} q^{l} r^{m} s^{n} c$, where $k, l, m, n \geq 1$ and $c$ is some natural number not divisible by $p, q, r$, and $s$. Up to the choice of notation, we can assume that $p^{k}$ is the smallest of the first four factors in th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,781 |
6. Can the number 240 be represented as the sum of nine two-digit numbers (some of which may be the same), each of which has a nine in its decimal representation? (I. Bogdanov) | Answer: No. Solution: Suppose it is possible. At least one of the addends must start with a nine, otherwise all nines would be in the units place, and the sum would end in 1. This addend is at least 90, and each of the others is at least 19. Therefore, their sum is at least $90+8 \cdot 19=242>240$. Contradiction. Remar... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,782 |
7. Inside parallelogram $A B C D$, point $E$ is marked on the bisector of angle $A$, and point $F$ is marked on the bisector of angle $C$. It is known that the midpoint of segment $B F$ lies on segment $A E$. Prove that the midpoint of segment $D E$ lies on line $C F$. (A. Kuznetsov) | Solution. Let $\angle B A D=\angle B C D=2 \alpha$, and the bisector $A E$ intersects the line $B C$ at point $K$. Then $\angle B A K=\angle K A D=\alpha=\angle F C B$. Therefore, the bisectors of angles $A$ and $C$ are parallel. Let $O$ be the midpoint of segment $B F$. Since by the condition it lies on $A E$, and $A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,783 |
8. We will call two numbers almost equal to each other if they are equal to each other or differ from each other by no more than one. A rectangular grid with sides of natural numbers a and b is such that it is impossible to cut out a rectangle along the grid lines, the area of which is almost equal to half the area of ... | Answer: 4. Solution: If one of the sides of the rectangle is even, a rectangle of half the area can be cut from it along the midline. Therefore, it is sufficient to consider the case where both sides are odd. In this case, $|b-a|$ is even. If $|b-a|=0$, then $a=b=2n+1$, half the area of the rectangle is $2n^2+2n+0.5$, ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,784 |
9. We will say that we have shortened a number if we erased its last digit. A natural number greater than a million is such that if we shorten it, we get the square of a natural number, if we shorten this square, we get the cube of a natural number, if we shorten this cube, we get the fourth power of a natural number, ... | Solution. Let the number after the first shortening be $n^{2}$, and after the third $-m^{4}$. Then $100 m^{4}$ differs from $n^{2}$ only in the last two digits, that is, $100 \geq n^{2}-\left(10 m^{2}\right)^{2}=\left(n-10 m^{2}\right)\left(n+10 m^{2}\right) \geq 0$. Since by the condition $m^{4} \geq 1000,10 m^{2}>100... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,785 |
10. There are two piles of stones on the table, containing 100 and 101 stones respectively. Two players play a game, taking turns. On a turn, a player is allowed to take a pile, remove some number of stones from it (at least one), and split the remaining stones in this pile into two non-empty piles. The player who cann... | Answer. The one who makes the first move. Solution. Let the first player be called Petya, and the second Vasya. Petya's first move will be to remove one stone from the pile of 101, and then divide the remaining into piles of 1 (which can be forgotten) and 99 stones. Now let's prove a more general fact: if there are pil... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,786 |
1. All natural numbers from 1 to 2018 are written in a row on the board: 1, 2, 3, ..., 2018. Find any two of them, the removal of which will result in the sum of all numbers between the erased ones being half the sum of all the remaining non-erased numbers? (Problem author - A. Gолованов) | Answer. For example, 673 and 1346. Solution. Note that the sums of numbers equally distant from the ends of the sequence are equal: $1+2018=2+2017=\ldots=1009+1010$. If we erase one such pair of numbers, the number of pairs left will be $1008=336 \cdot 3$. Therefore, if there are 336 pairs between the erased numbers, t... | 6731346 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,787 |
2. In triangle $ABC$, the bisector $BD$ is drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \parallel AC$. Find the angle $DEF$. (I. Rubanov) | Answer: 45 degrees. Solution: Let segments $B D$ and $E F$ intersect at point $G$. From the condition, we have $\angle E D G = \angle E D A = \angle D E G$, hence $G E = G D$. Similarly, $G F = G D$. Therefore, $G E = G F$, which means $B G$ is the bisector and median, and thus the altitude in triangle $B E F$. Therefo... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,788 |
3. For each pair of different natural numbers a and b, not greater than 20, Petya drew on the board the line $y=a x+b$ (that is, he drew the lines $y=x+2, \ldots, y=x+20, y=2 x+1, y=2 x+3, \ldots$, $y=2 x+20, \ldots, y=3 x+1, y=3 x+2, y=3 x+4, \ldots, y=3 x+20, \ldots, y=20 x+1, \ldots, y=20 x+19)$. Vasya drew on the s... | Answer: 190. First solution. The graph $y=a x+b$ intersects the coordinate axes at points $A(-b / a, 0)$ and $B(0, b)$. If $ba$. Then points $K(-1,0)$ and $N(0,1)$ lie on the legs of triangle $O A B$, where $O$ is the origin, and point $M(-1,1)$ lies inside this triangle, since $a \cdot(-1)+b=b-a \geq 1$. Thus, the ang... | 190 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,789 |
4. A square with a side of 100 was cut into squares (not necessarily the same) with sides parallel to the sides of the original square and less than 10. Prove that the sum of the perimeters of the resulting squares is not less than 4400. (I. Rubanov) | Solution. Draw 11 parallel segments, two of which are sides of the $100 \times 100$ square, and the other nine divide this square into rectangles $10 \times 100$. Then each square of our dissection intersects exactly one of these segments with a segment equal to its side. Therefore, the sum of the sides of the squares ... | 4400 | Geometry | proof | Yes | Yes | olympiads | false | 6,790 |
5. On each of five cards, a number is written. The cards lie on the table numbers down. We can, by paying a ruble, point to any three cards, and we will be told the sum of the numbers written on them. For what minimum price can we surely find out the sum of all five numbers? (I. Rubanov) | Answer: For 4 rubles. Solution: Let the numbers written be $a, b, c, d, e$. We ask about the sums $a+b+c, a+b+d$, $a+b+e, c+d+e$. Then, by adding the first three sums and subtracting the fourth from the result, we get $3(a+b)$, then $a+b$ and, by adding $c+d+e$ to the result, the sum of all five numbers.
Suppose we ma... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,791 |
1. Kolya, Vasya, and Petya went shopping. They have 2200 rubles in total, and none of them has coins smaller than a ruble. Kolya has 18 times less money than Vasya. Prove that Petya can buy an ice cream for 15 rubles. | Solution. Let Kolya have $n$ rubles. Then Vasya has $18n$ rubles, and Kolya and Vasya together have $19n$ rubles. Since no one has coins smaller than a ruble, $n$ is an integer. Dividing 2200 by 19 with a remainder, we get $2200 = 115 \cdot 19 + 15$. Thus, Kolya and Vasya together have no more than $115 \cdot 19 = 2185... | 15 | Algebra | proof | Yes | Yes | olympiads | false | 6,792 |
2. On the side AC of triangle ABC with an angle of 120 degrees at vertex B, points D and E are marked such that $AD = AB$ and $CE = CB$. A perpendicular $DF$ is dropped from point D to the line BE. Find the ratio $BD / DF$. | Answer. 2. Solution. Let $\angle C A B=\alpha, \angle A C B=\beta$. Since $A D=A B$ and $C E=C B$, we have
$\angle D B E=\angle D B A+\angle E B C-\angle A B C=\left(180^{\circ}-\alpha\right) / 2+\left(180^{\circ}-\beta\right) / 2-120^{\circ}=60^{\circ}-(\alpha+\beta) / 2=30^{\circ}$.
Thus, in the right triangle $B F... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,793 |
3. 30 people are lined up in six rows of five people each. Each of them is either a knight, who always tells the truth, or a liar, who always lies, and they all know who among them is a knight and who is a liar. A journalist asked each of them: “Is it true that there will be at least 4 rows in which liars are in the ma... | Answer: 21. Solution: Let's call a row blue if it contains more than half (that is, no less than three) liars and red if there are no more than two liars in it.
Suppose the knights said "yes." Then we have no more than two red and no less than four blue rows. In the red rows, there are no more than 10 knights, and in ... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,794 |
4. Misha and Masha were traveling by train to Kirov. Misha was lying on the bunk, while Masha was looking out the window. "How far is it to Kirov?" Misha asked Masha at 12:00. "73 kilometers," Masha replied. To the same question asked at 12:15 and 12:45, Masha answered: "62 kilometers" and "37 kilometers." It is known ... | Answer: 48 km/h. Solution: Since Masha rounded the distance to the nearest whole number each time, at 12:00 the distance to Kirov was no less than 72.5 and no more than 73.5 km, at 12:15 - no less than 61.5 and no more than 62.5 km, and at 12:45 - no less than 36.5 and no more than 37.5 km. Therefore, in 15 minutes fro... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,795 |
5. Two players are playing a game. Initially, they have a rectangular sheet of paper of size $m \times n$, where $m$ and $n$ are natural numbers greater than 1. The players take turns. On each turn, a player cuts the existing rectangle into two, one of which has an area of 1, and discards the rectangle of unit area. Th... | Answer. If the area of the original rectangle is odd, the first player wins; if it is even, the second player wins. Solution. Lemma. If the area $S$ of a rectangle is not less than 3, and both of its sides are longer than 1, then by cutting off a unit rectangle parallel to the shorter side, we will obtain a rectangle w... | If\the\area\of\the\original\rectangle\is\odd,\the\first\player\wins;\if\it\is\even,\the\\player\wins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,796 |
1. During the physical education lesson, all students of grade 8 lined up. It turned out that boys and girls alternated in the line. It is known that exactly $52\%$ of the students in grade 8 are boys. Find the number of girls in grade 8. Don't forget to justify your answer. | Answer: 12. Solution. If there were an even number of students in the row, the number of boys and girls would be equal. But according to the condition, there are more boys. Therefore, the number of students in the row is odd, and boys stand in odd positions. Let there be $n$ girls in the class. Then there are $n+1$ boy... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,797 |
2. On an island, there are 1000 villages, each with 99 residents. Each resident of the island is either a knight, who always tells the truth, or a liar, who always lies. It is known that there are exactly 54054 knights on the island. One fine day, every resident of the island was asked the question: "Who is more in you... | Answer: 638. Solution. There cannot be an equal number of knights and liars in any village, because then all its inhabitants would have lied. If there are more knights in a village, then obviously 66 people told the truth and 33 people lied, and if there are more liars - then the opposite. Let there be $n$ villages whe... | 638 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,798 |
3. In triangle $ABC$, the bisector $BL$ is drawn, and a point $K$ is chosen on its extension beyond point $L$ such that $LK = AB$. It turns out that $AK \parallel BC$. Prove that $AB > BC$. | Solution. Let angle $ABC$ be $2x$. Then each of the angles $ABL, CBL$, and $AKB$ is $x$ (the latter as an alternate interior angle with $CBL$ when parallel lines $AK$ and $BC$ are intersected by $BK$). Therefore, triangle $BAK$ is isosceles, from which $AK = AB$. Since $LK = AB$ by the problem's condition, triangle $AK... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,799 |
4. A square $15 \times 15$ is divided into $1 \times 1$ squares. Some of these squares were chosen, and one or two diagonals were drawn in each of the chosen squares. It turned out that no two drawn diagonals share a common endpoint. What is the maximum number of diagonals that can be drawn? (In your solution, provide ... | Answer: 128. Solution. Let's number the rows and columns of the square in order with numbers from 1 to 15 and draw two diagonals in the cells located at the intersection of odd-numbered rows with odd-numbered columns. There are 64 such cells, meaning 128 diagonals will be drawn. On the other hand, the diagonals should ... | 128 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,800 |
5. In a row, 2011 consecutive five-digit numbers are written. It turned out that the sum of the digits of the 21st number is 37, and the sum of the digits of the 54th number is 7. Find the sum of the digits of the 2011th number. (Provide all possible answers and prove that there are no other answers). | Answer: 29. Solution. When transitioning from the 21st number to the 54th, the sum of the digits decreases by 30. It is easy to verify that this is possible only if there is a transition through a number divisible by 10000 along the way. This means that the thousands and hundreds places of the 21st number are nines, an... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,801 |
1. A mushroom is called bad if it contains no fewer than 10 worms. In the basket, there are 90 bad and 10 good mushrooms. Can all the mushrooms become good after some worms crawl from the bad mushrooms to the good ones? | Answer. They can. Solution. Let there be exactly 10 worms in each bad mushroom, and no worms in good ones. Then, let one worm from each bad mushroom crawl into the good ones, 9 into each. As a result, there will be 9 worms in each mushroom, and all mushrooms will be good. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,802 |
2. Point $K$ is the midpoint of the hypotenuse $AB$ of a right isosceles triangle $ABC$. Points $L$ and $M$ are chosen on the legs $BC$ and $AC$ respectively such that $BL = CM$. Prove that triangle $LMK$ is also a right isosceles triangle. | Solution. The median SK of triangle ABC is also the altitude and bisector, as the triangle is isosceles. Therefore, $\angle K B C=\angle K C B=\angle K C A=45$ degrees. Hence, $K C=$ $\mathrm{KB}$, and thus, triangles $\mathrm{KBL}$ and $K C M$ are equal by two sides ($K C=K B, B L=C M$) and the angle between them. The... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,803 |
3. The numbers 1, 2, 3, ..., 10 are written in a circle in some order. Petya calculated 10 sums of all triples of adjacent numbers and wrote the smallest of the calculated numbers on the board. What is the largest number that could have been written on the board? | Answer: 15. Solution: First, we prove that the written number is not greater than 15. We single out the number 10, and divide the remaining 9 numbers into three groups of three consecutive numbers. The sum of the numbers in these three groups is $1+2+3+\ldots+9=45$, so in at least one of the considered groups, the sum ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,804 |
4. We have a balance scale and 100 coins, among which several (more than 0, but less than 99) are counterfeit coins. All counterfeit coins weigh the same, and all genuine coins also weigh the same, with a counterfeit coin being lighter than a genuine one. We can perform weighings on the scale, paying one coin (either c... | Solution. Let's set aside one coin. Since the condition states that there are at least two genuine coins, among the remaining 99 coins there is at least one genuine one. We will number these coins and weigh the first against the second, paying for the weighing with the set-aside coin. If one of the coins outweighs the ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,805 |
5. On the table, there are 7 cards with digits from 0 to 6. Two players take turns picking one card each. The player who first forms a natural number divisible by 17 from their cards wins. Who will win with correct play - the starting player or their opponent? (I. Rubanov) | Answer. Beginner. Solution. Let the players be A (the beginner) and B (his opponent). We will present a strategy that allows A to win guaranteed. Suppose he takes the digit 3 on his first move; then B is forced to take 4 (otherwise, A will take it on his second move and win, forming the number 34). Note that in this ca... | Beginner | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,806 |
6. In three cells of a grid sheet, numbers are written, while all other cells are empty. It is allowed to choose two numbers from different non-empty cells and write their sum in an empty cell; also, one can choose numbers $a$, $b$, and $c$ from three different non-empty cells and write the number $ab + c^2$ in an empt... | Solution. Let the numbers $a, b, c$ be written. Sequentially form $a+b, b+c, a+c, (a+b)c+a^2$, $(b+c)a+b^2, (c+a)b+c^2, (a+b)c+a^2+b+c)a+b^2$ and $\left((a+b)c+a^2+b+c\right)a+b^2)+(c+a)b+c^2=(a+b+c)^2$. The problem can also be solved in many other ways. | (+b+)^2 | Algebra | proof | Yes | Yes | olympiads | false | 6,807 |
7. In a convex quadrilateral $A B C D$, some point on the diagonal $A C$ belongs to the perpendicular bisectors of the sides $A B$ and $C D$, and some point on the diagonal $B D$ belongs to the perpendicular bisectors of the sides $A D$ and $B C$. Prove that $A B C D$ is a rectangle. | Solution. Let $E$ be a point on the diagonal $AC$, belonging to the perpendicular bisectors of the sides $AB$ and $CD$, and $F$ be some point on the diagonal $BD$, belonging to the perpendicular bisectors of the sides $AD$ and $BC$. By the triangle inequality, we have $EB + ED \geq BD$, $FA + FC \geq AC$. From the cond... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,808 |
8. In a football tournament, 8 teams participated, and each played against each other exactly once. It is known that any two teams that drew with each other ended up with a different number of points. Find the maximum possible total number of draws in this tournament. (A team earns 3 points for a win, 1 point for a dra... | Answer: 22. Solution: We will prove that no more than two teams can have exactly 6 draws. Indeed, any such team has either 6 or $6+$ 3 points (depending on whether they won or lost their decisive match). If there are three such teams, then two of them have the same number of points, meaning they did not play to a draw ... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,809 |
5. In a convex pentagon $A B C D E$, the diagonals $A D$ and $C E$ intersect at point $X$. It turns out that $A B C X$ is a parallelogram and $B D = C X ; B E = A X$. Prove that $A E = C D$. (C. Berlov) | First solution. Note that $A B=C X=B D$; $B C=A X=B E$ and $\angle A B D=180^{\circ}-2 \angle B A D=180^{\circ}-2 \angle B C E=\angle C B E$, from which $\angle A B E=\angle C B D$. Therefore, triangles $A B E$ and $D B C$ are equal by two sides and the angle between them, so $A E=C D$.
 | Solution. For $k=2022 \cdot 2021 / 2$, take the numbers $2, 2^{2}, \ldots, 2^{2022}$. All their pairwise sums are distinct: if the equality $2^{a}+2^{b}=2^{c}+2^{d}$ were to hold, then dividing it by the smallest of the powers of two involved, we would get that an even number equals an odd number. Now let $0 \leq s_{1}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,811 |
7. Positive numbers $a, b$, $c$ and $d$ do not exceed one. Prove the inequality $\frac{1}{a^{2}+b^{2}+c^{2}+d^{2}} \geq \frac{1}{4}+(1-a)(1-b)(1-c)(1-d)$. (A. Khryabrov) | Solution. Let $s=(a+b+c+d) / 4$. Since $0 \leq a, b, c, d \leq 1$, the inequality $a^{2}+b^{2}+c^{2}+d^{2} \leq a+b+c+d$ holds. Furthermore, the inequality $(1-a)(1-b) \leq(1-(a+b) / 2)^{2}$ is true, which simplifies to the obvious inequality $0 \leq(a-b)^{2} / 4$. Therefore, $(1-a)(1-b)(1-c)(1-d) \leq(1-(a+b) / 2)^{2}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,812 |
8. In the club, there are 42 people, any two of whom have at least ten common friends among the club members. Prove that there will be two people who have at least twelve common friends among the club members. (M. Antipov) | Solution. Let's calculate how many pairs of common acquaintances each pair of club members has, i.e., how many cycles of length 4 exist in the acquaintance graph with these two opposite vertices. Each cycle of length 4 will be counted twice, so the sum of all the results of the count will be even. Suppose the statement... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,813 |
1. A car is driving at a constant speed in one direction along a straight road, near which there are two houses. At noon, when the car had not yet reached the houses, the sum of the distances from it to these houses was 10 km. After 10 minutes, when the car had already passed both houses, it turned out that the sum of ... | Answer: 60 km/h. Solution: Let the houses be at points $A$ and $B$, the car was at point $C$ at noon, and at point $D$ 10 minutes later (see figure). According to the problem, $C A + C B = D A + D B = 10$. Note that $C A + C B = 2 C A + A B$, and $D A + D B = 2 D B + A B$, from which we get $C A = D B$.
 so that there are $k$ rows, in each of which the sum of the numbers is at least 3, and $k$ columns, in each of which the sum of the numbers is at most 2? (O. Nechaeva,... | Answer. For $k=4$. Solution. For $k=5$, it is impossible to fill the table in the required manner, since when calculating the sum of the numbers in the table by adding the sums of the rows, it turns out that it is not less than $3 \cdot 5=15$, while when calculating the same sum by the columns, it turns out that it is ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,815 |
3. Inside triangle ABC, there is a point $P$. On side BC, a point $H$ is chosen, which does not coincide with the midpoint of the side. It turns out that the bisector of angle AHP is perpendicular to side BC, angle ABC equals angle HCP, and BP = AC. Prove that $B H = A H .(\mathrm{O}$. Nechaeva based on folk motifs) | Solution. Let's extend segment $\overline{A H}$ beyond point $H$ and mark segment $\overline{H Q}=H P$. Since the bisectors of adjacent angles are perpendicular to each other, line $B C$ is the bisector of angle $P H Q$, which is adjacent to angle $A H P$, and thus the perpendicular bisector of $P Q$. Therefore, $\angl... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,816 |
4. Find all natural numbers $n$ for which the number $n^{7}+n^{6}+n^{5}+1$ has exactly three natural divisors. (O. Nechaeva, I. Rubanov) | Answer. 1. Solution. For $n=1, n^{7}+n^{6}+n^{5}+1=4$. The number 4 has exactly three divisors: 1, 2, 4. Notice further that $n^{7}+n^{6}+n^{5}+1=\left(n^{7}+n^{5}\right)+\left(n^{6}+1\right)=n^{5}\left(n^{2}+1\right)+\left(n^{2}+1\right)\left(n^{4}-n^{2}+1\right)=\left(n^{2}+1\right)\left(n^{5}+n^{4}-n^{2}+1\right)=\l... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,817 |
1. A smaller square was cut out from a larger square, one of its sides lying on the side of the original square. The perimeter of the resulting octagon is $40 \%$ greater than the perimeter of the original square. By what percentage is its area less than the area of the original square? | Answer: By $64 \%$. Solution. Let $ABCD$ be the original square with side 1 (and area 1), $KLMN$ be the cut-out square with side $x$, where $KL$ lies on $AB$ (point $K$ is closer to $A$ than $L$). Then the perimeter of the octagon $AKNMLBCD$ exceeds the perimeter of the square $ABCD$ by $KN + LM = 2x = 0.4 \cdot 4AB = ... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,819 |
2. There are three consecutive even numbers. For the first one, we found the largest even proper divisor, for the second one - the largest odd proper divisor, and for the third one - again the largest even proper divisor. Can the sum of the three obtained divisors be equal to 2013? (A divisor of a natural number is cal... | Answer: Yes, it can. Solution. Here is an example: 1340, 1342, and 1344. For the first number, the largest even divisor is 670, for the third number, it is 672, and for the second number, the largest odd divisor is 671. $670+671+672=2013$. Remark. There are two natural ways to come up with this example. One can try to ... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,820 |
3. In parallelogram $A B C D$ with side $A B=1$, point $M$ is the midpoint of side $B C$, and angle $A M D$ is 90 degrees. Find the side $B C$. | Answer: 2. First solution. Draw the median $M N$ in triangle $A M D$. $A B M N$ is a parallelogram since $B M$ and $A N$ are equal and parallel, thus $M N = B C = 1$. But the median $M N$ is half the hypotenuse $A D$ in the right triangle $A M D$, from which it follows that $B C = A D = 2$. Second solution. Extend the ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,821 |
4. Sharik and Matroskin are skiing on a circular track, half of which is an uphill climb and the other half is a downhill descent. Their speeds on the climb are the same and four times slower than their speeds on the descent. The minimum distance by which Sharik lags behind Matroskin is 4 km, and the maximum distance i... | Answer: 24 km. Solution. The minimum lag of Sharik behind Matroskin occurs when Sharik is at the lowest point of the track, and Matroskin is climbing the hill (if Matroskin were descending at this time, it would mean that half of the track is less than 4 km, which is obviously impossible). And it remains until Matroski... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,822 |
5. In a country of liars and knights (knights always tell the truth, liars always lie), ten people were given different numbers from 1 to 10. Then each person was asked: "Is your number divisible by 2?". Three people gave an affirmative answer. To the question "Is your number divisible by 4?" six people gave an affirma... | Answer: The liars received the numbers $2,5,6,10$. Solution. Consider the people with numbers $4,5,8$, 10. Only these people could have given different answers to the second and third questions. Four more people answered affirmatively to the second question than to the third. This means that all the people with the spe... | 2,5,6,10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,823 |
1. The teacher drew a rectangle $A B C D$ on the board. Student Petya divided this rectangle into two rectangles with a line parallel to side AB. It turned out that the areas of these parts are in the ratio 1:2, and the perimeters are in the ratio 3:5 (in the same order). Student Vasya divided this rectangle into two p... | Answer: 20:19. Solution. Let $A B=a, B C=b$, and the first line divides side $B C$ into segments $x$, $b-x$. Then, according to the condition, $b-x=2 x$ and $5(x+a)=3(b-x+a)$, from which $b=6 a$. Now let the second line divide side $A B$ into segments $y, a-y$. According to the condition, $y=2(a-y)$, that is, $y=2 a / ... | 20:19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,824 |
3. Solve the rebus UHA = LCM(UX, UA, HA). Here U, X, A are three different digits. Two-digit and three-digit numbers cannot start with zero. Recall that the LCM of several natural numbers is the smallest natural number that is divisible by each of them. | Answer. $150=\operatorname{HOK}(15,10,50)$. Solution. Since UHA is divisible by UX, A $=0$, that is, the number has the form UX0. Since UX0 is divisible by U0 and X0, UX is divisible by U and X, from which it follows that X is divisible by U, and 10 U is divisible by X. Let $\mathrm{X}=\mathrm{M} \cdot \mathrm{U}$. The... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,826 |
4. Twenty-eight lyamziks weighing 2, 3, 4, and 5 kg (7 lyamziks of each weight) crossed a river in a rowing boat that can hold 10 kg. It is known that each lyamzik rowed no more than twice. Prove that at least 12 lyamziks had to row. The boat has one rower, and without a rower, the boat cannot move. | Solution. Let the rowing not exceed 11 lyamziks. Then there were no more than 22 trips. Therefore, the number of trips "there" would not exceed eleven. To the other bank, it is necessary to transport $(2+3+4+5) \cdot 7=98$ kg, so the number of trips "there" cannot be less than 10. Therefore, there are two possible opti... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,827 |
5. Petya marks four points on a plane so that all of them cannot be crossed out by two parallel lines. From the lines passing through pairs of points, Vasya chooses two, measures the angle between them, and pays Petya an amount equal to the degree measure of the angle. What is the largest sum Petya can guarantee for hi... | Answer: 30. Solution: Estimation. There are 6 pairs of points, that is, 6 possible lines. According to the condition, there are no parallel pairs among them. Draw six lines parallel to them through one point. They will divide the plane into 12 angles, so there is an angle no more than 30 degrees. Example. Take an arbit... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,828 |
1. We will call a non-negative integer a zebra if even and odd digits strictly alternate in its representation. Can the difference of two 100-digit zebras be a 100-digit zebra? | Answer: Yes. Solution. Here is one of many possible examples: $5050 \ldots 50-2525 \ldots 25=2525 \ldots 25$
Grading Guidelines. An obviously correct example, or a correct example with justification of its correctness - 7 points. A correct example, for which additional calculations are required for justification, not ... | 5050\ldots50-2525\ldots25=2525\ldots25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,829 |
2. A regular triangle with a side of 2 is divided into triangles with a side of 1. Six identical-looking coins are placed at the vertices of these triangles. It is known that two of them are counterfeit, lighter than the genuine ones, and lie at the ends of a unit segment. How can you find both counterfeit coins in 2 w... | Solution. Let coins $a, b, c$ lie at the vertices of the triangle, and $a_{1}, b_{1}, c_{1}$ lie at the midpoints of the opposite sides, respectively. We weigh the pairs $\left(a, a_{1}\right)$ and $\left(b, b_{1}\right)$. If their weights are equal, then each of these pairs contains one counterfeit coin (it is impossi... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,830 |
3. Here are four properties of quadrilaterals:
(1) opposite sides are equal in pairs;
(2) two opposite sides are parallel;
(3) some two adjacent sides are equal;
(4) the diagonals are perpendicular and are divided by the point of intersection in the same ratio.
One of the two given quadrilaterals has any two of th... | Solution. It is not difficult to verify that quadrilaterals with properties (1)+(3), or (3)+(4), or (1)+(4) are rhombuses (in the participants' work, this, of course, should be proven). It remains to note that in any partition of the given four properties into two pairs, one of the three listed above will be present.
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,831 |
4. On the car market, you can exchange three Ladas for one Volga and one Mercedes, and three Volgas for two Ladas and one Mercedes. Will collector Vasya, having 700 Ladas, be able to get 400 Mercedes? | Answer: No. Solution. Let's estimate the Zhiguli at 4 rubles, the Volga at 5 rubles, and the Mercedes at 7 rubles. Then, the total value of Vasya's cars will be preserved with any exchanges. The total value of 400 Mercedes is exactly equal to the total value of 700 Zhigulis, but it is impossible to exchange Zhigulis fo... | No | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,832 |
5. Petya found the sum of all odd divisors of some even number, and Vasya - the sum of all even divisors of this number. Could the product of these two sums be a square of a natural number? | Answer: No. Solution. Let's write the original even number in the form $N=2^{n}(2 m+1)$. It is clear that the sum of all odd divisors of this number is equal to the sum of all divisors of the number $2 m+1$. Let's denote this sum by $S$. Then the sum of all even divisors of the number $N$ equals $2 S+4 S+\ldots+2^{n} S... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,833 |
1. The robbers filled a chest to the brim with gold and silver sand, with twice as much gold sand by volume as silver sand. Ali-Baba calculated that if half of the silver sand were removed and the chest topped up with gold sand, the value of the chest would increase by 20 percent. By what percentage and how would the v... | Answer. It will decrease by $40 \%$. Solution. Let the initial volume of silver sand in the chest be $A$, the cost of a unit volume of gold sand be $x$, and the cost of a unit volume of silver sand be $y$. Then the initial cost of the chest is $2 A x+A y$. The cost of the chest according to Ali-Baba's calculations is $... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,834 |
2. On the sides $AB, BC, CD$, and $DA$ of quadrilateral $ABCD$, points $K, L, M, N$ are chosen respectively such that $AK = AN, BK = BL, CL = CM, DM = DN$, and $KLMN$ is a rectangle. Prove that $ABCD$ is a rhombus. | Solution. Triangles $A N K, B K L, C L M$ and $D M N$ are isosceles by condition. In isosceles triangles, the angles at the base are equal. Let $\angle A K N=\angle A N K=\alpha, \angle B K L=\angle B L K=\beta$. Since angle $N K L$ is right, $\alpha+\beta=90^{\circ}$. But $\angle K L B+\angle M L C=90^{\circ}$, so $\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,835 |
3. Princess Sonya and 7 bogatyrs (Russian epic heroes) approached the ferry. The bogatyrs lined up in such a way that every two adjacent bogatyrs are friends, while bogatyrs who are not standing next to each other do not get along. The princess is friends with everyone except the middle bogatyr. There is one boat that ... | Answer: They can. Solution: Let C denote the princess, and number the heroes from 1 to 7 in order (Sonya does not get along with the 4th). Letters and numbers indicate who is in the boat, the arrow $\rightarrow$ denotes the crossing to the other bank, and the arrow $\leftarrow-$ the return trip. The following algorithm... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,836 |
4. Pete and Vasya are playing on a $20 \times 20$ grid. Each turn, a player selects a cell whose all 4 sides are not colored and paints all sides in red and blue in any order (for example, they can paint all in one color). At the same time, there should not be segments of the same color longer than one side of a cell. ... | Answer. Vasya wins. Solution. Vasya divides all the cells of the board into pairs of centrally symmetric cells, and in response, paints the sides of the cell from the same pair so that the centrally symmetric sides are painted in different colors. As long as Vasya's strategy works, after his move, in each pair of centr... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,837 |
5. What is the maximum number of two-digit numbers that can be written in a row so that any two adjacent numbers are not coprime, while any two non-adjacent numbers are coprime? | Answer: 10. Solution: Ten numbers can be written: 11, 77, 91, 65, 85, 51, 57, 38, 46, 23. Suppose we managed to write 11 such numbers. Let's list the common divisor, different from 1, for each pair of adjacent numbers. This will result in 10 numbers, any two of which are coprime. All these numbers are either single-dig... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,838 |
1. In the encrypted equation $А Б+A Б+A Б+A Б+A Б+A Б+A Б+A Б+A Б=A A Б$, the digits have been replaced by letters: the same digits are replaced by the same letter, and different digits by different letters. Find all possible decryptions. (I. Rubanov) | Answer. $25+\ldots+25=225$. Solution. Add AB to both sides of the given equality. We get the equality $10 \cdot A B=$ AAB+AB. Therefore, B+B ends in 0, that is, $B=0$ or $B=5$. If $B=0$, then $100 A=120 A$, from which $A=0$, which is impossible. If, however, B $=5$, then from the equation $100 A+50=120 A+10$ we find th... | 25+\ldots+25=225 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,839 |
2. Given a triangle $ABC$, in which $AB = BC$. On the side $BC$, there is a point $D$ such that $CD = AC$. Point $E$ on the ray DA is such that $DE = AC$. Which segment is longer - $EC$ or $AC$? (I. Rubanov) | Answer. Segment $A C$ is not shorter than segment $E C$. Solution. If triangle $A B C$ is equilateral, segments $A C$ and $E C$ are equal. Consider the case when triangle $A B C$ is not equilateral. Since $B C>C D=A C, A C-$ is the smallest side of the isosceles triangle $A B C$, and the angle opposite to it, $\angle A... | AC\geqCE | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,840 |
3. Natural numbers $n(n>1)$ and $k$ are such that for any natural divisor $d$ of the number $n$, at least one of the numbers $d+k$ or $d-k$ is also a natural divisor of the number $n$. Prove that the number $n$ is prime. (A. Gолованов) | Solution. Since the numbers 1 and $n$ are divisors of the number $n$, the numbers $n-k$ and $1+k$ must also be its divisors. Since all divisors of the number $n$, except for $n$ itself, do not exceed $n / 2$, we have $n-k \leq n / 2$, from which $k \geq n / 2$. But then $1+k>n / 2$, from which $1+k=n$, that is, $k=n-1$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,841 |
4. Do there exist such positive numbers $a, b, c$ that $a+b+c=ab+ac+bc=abc$? (I. Rubanov) | Answer. Such numbers do not exist. First solution. Dividing the equality $a b+a c+b c=a b c$ by $a b c$, we get $1 / a+1 / b+1 / c=1$, from which $a>1, b>1, c>1$. But then $a b>a, a c>c, b c>b$ and $a b+a c+b c>a+b+c$. Second solution. If the equality from the condition is true, then the equality $(a b+a c+b c)^{2}=a b... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,842 |
5. The sides of 100 identical equilateral triangles are painted in 150 colors such that exactly two sides are painted in each color. If two triangles are placed with their same-colored sides together, the resulting rhombus will be called good. Petya wants to form as many good rhombi as possible from these triangles, wi... | Answer: 25. Solution: Why is it always possible to get 25 good rhombuses? Let's construct the maximum possible number of good rhombuses. Suppose there are fewer than 25. Then there are at least 52 triangles that have not entered any good rhombus. Let's call such triangles single. For each triangle, choose a color, exac... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,843 |
6. The sum of four integers is 0. The numbers are arranged in a circle, and each is multiplied by the sum of its two neighbors. Prove that the sum of these four products, multiplied by -1, is equal to twice the square of an integer. (N. Agakhanov) | Solution. Let the numbers $a, b, c, d$ stand in a circle (in the given order). Then the sum of the four products, multiplied by -1, is $-a(b+d)-b(a+c)-c(b+d)-d(a+c)=-2(a+c)(b+d)=2(a+c)^{2}$, which is what we needed to prove. The last equality here follows from the condition that $a+c=-(b+d)$. | 2(+)^2 | Algebra | proof | Yes | Yes | olympiads | false | 6,844 |
7. We will call two cells of a grid neighbors if they share a side. Is it possible to color 32 cells black in a $10 \times 10$ white table so that each black cell has an equal number of black and white neighbors, and each white cell does not have an equal number? (O. Yuzhakov) | Answer. It is possible. Solution. Divide the table into four $5 \times 5$ squares and in each, paint black 8 cells adjacent by side or corner to its central cell. Each black cell will be adjacent to two white and two black cells, and it is not difficult to verify that each white cell has an unequal number of adjacent b... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,845 |
8. Point $N$ is the midpoint of side $B C$ of triangle $A B C$, where $\angle A C B=60^{\circ}$. Point $M$ on side $A C$ is such that $A M=B N$. Point $K$ is the midpoint of segment $B M$. Prove that $A K=K C$. (E. Bakayev, A. Kuznetsov) | First solution. Complete the triangle $M C N$ to a parallelogram NCML. In triangle $A M L$, $L M=N C=B N=A M, \angle A M L=\angle B C M=60^{\circ}$. Therefore, triangle $A M L$ is equilateral. Hence, $A L=N C$ and $\angle A L K=\angle A L M+\angle N L M=60^{\circ}+60^{\circ}=120^{\circ}=\angle K N C$. Moreover, segment... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,846 |
9. There are 70 switches and 15 lamps. Each lamp is connected to 35 switches. No two switches are connected to the same set of lamps. Pressing a switch changes the state of all the lamps it is connected to (turning on the off ones and vice versa). Initially, all lamps are off. Prove that it is possible to press some 19... | Solution. Consider all possible sets of 19 switches. Also, consider any light bulb. Divide all the switches into 35 pairs in such a way that in each pair, exactly one switch is connected to the chosen light bulb. Note that then all sets of 19 switches are also divided into pairs obtained by replacing all switches with ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,847 |
10. Petya chooses non-negative numbers $x_{1}, x_{2}, \ldots, x_{11}$ such that their sum is 1. Vasya arranges them in a row as he sees fit, calculates the products of adjacent numbers, and writes the largest of the resulting ten products on the board. Petya wants the number on the board to be as large as possible, whi... | Answer: $\frac{1}{40}$. Solution: If Petya chooses the numbers $\frac{1}{2}, \frac{1}{20}, \frac{1}{20}, \ldots, \frac{1}{20}$, then no matter how Vasya arranges these numbers, the adjacent numbers will be $\frac{1}{2}$ and $\frac{1}{20}$. Therefore, one of the products will be $\frac{1}{40}$, and the others will not e... | \frac{1}{40} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,848 |
1. Foma and Yerema were walking in the same direction along a road with kilometer markers. In one hour, Foma passed by five markers, while Yerema passed by six. Could Foma's speed have been greater than Yerema's? | Answer. Could. Solution. Let Erema be 50 m from the first of the poles he passed at the beginning of the hour, and 50 m beyond the sixth pole at the end of the hour. Then he walked 5100 m in an hour. Let Foma be 600 m from the first of the poles he passed at the beginning of the hour, and 600 m beyond the fifth pole he... | 5200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,849 |
2. Three people gathered in a room. Each of them is either a knight, who always tells the truth, or a liar, who always lies, or a trickster, who can either tell the truth or lie as they please. One of them said: "Among us there is a liar." Another said: "Among any two of us there is a liar." The third said: "We are all... | Solution. Suppose there is no trickster among those gathered. Then each of them is either a knight or a liar. The third one could not have told the truth: otherwise, it would mean that a liar told the truth. Therefore, he is a liar. Then the first one told the truth, and he is a knight. The second one is left. Suppose ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 6,850 |
3. Can the median and bisector drawn from vertex $A$ of an acute-angled triangle $ABC$ divide the altitude $BH$ of this triangle into three equal parts? | Answer. They cannot. Solution. Suppose they can. Mark points $F$ and $G$ on the height $B H$ such that $B F = F G = G H = B H / 3$. Let the angle bisector of $\angle A$ intersect $B H$ at point $K$. From triangle $A B H$, by the property of the angle bisector, we have $B K / K H = A B / A H > 1$, hence $K = G$. This me... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,851 |
4. Vasya thought of six natural numbers: $a, b, c, d, e, f$. For one ruble, you can indicate any two of them and find out their product. Petya knows that any two of the thought-of numbers are coprime (that is, they do not have common divisors greater than 1). What is the smallest amount he can spend to find out all the... | Answer. For 4 rubles. Solution. We will show how to manage with four rubles. First, we find the products $a b$ and $b c$. Since the numbers $a$ and $c$ have no common prime divisors, the greatest common divisor of these products is $b$. Thus, we find the number $b$, and with it the numbers $a=a b / b$ and $c=b c / b$. ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,852 |
5. Igor wants to cut out $1 \times 6$ grid rectangles from a grid square of size $11 \times 1117$. Is it possible to mark one cell in the square so that it will definitely remain uncut, no matter how hard Igor tries? | Answer. Yes. Solution. The central cell of the square will remain uncut. Indeed, suppose it is cut. For convenience, let's position the square so that the $1 \times 6$ rectangle containing this cell is horizontal. Then, in the six columns where it is located, a vertical rectangle cannot be placed. Two vertical rectangl... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,853 |
1. Two runners are running at equal constant speeds along the diagonals AC and BD, respectively, of a square $ABCD$. Upon reaching the end of the diagonal, the runner immediately turns back. They started simultaneously from two randomly chosen points on their respective diagonals. Prove that there will be a moment when... | Solution. Consider the moment when the runner on diagonal $A C$ is at point $O$, the intersection of the diagonals. If the second runner is not at point $B$ or $D$ at this moment, the distance between the runners will be less than half the diagonal, and the problem is solved. Otherwise, for definiteness, let's assume t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,854 |
2. Let $p_{1}, p_{2}, \ldots, p_{100}$ be one hundred distinct prime numbers. Natural numbers $a_{1}, \ldots, a_{k}$, greater than 1, are such that each of the numbers $p_{1} p_{2}^{3}, p_{2} p_{3}^{3}, \ldots, p_{99} p_{100}^{3}, p_{100} p_{1}^{3}$ is equal to the product of some two of the numbers $a_{1}, \ldots, a_{... | Solution. We will call a natural number white if it is divisible by the square of some prime number, and black otherwise. Let's say that the number $a_{i}$ serves a number from the list $p_{1} p_{2}^{3}, p_{2} p_{3}^{3}, \ldots, p_{99} p_{100}^{3}, p_{100} p_{1}^{3}(*)$, if it gives it in the product with some $a_{j}$.... | 150 | Number Theory | proof | Yes | Yes | olympiads | false | 6,855 |
3. In the cells of a $10 \times 10$ table, the natural numbers 1, 2, ..., 99, 100 are arranged. Let's call a corner the figure that results from removing one cell from a 2x2 square. We will call a corner good if the number in its cell, which borders by sides with two other cells, is greater than the numbers in these tw... | Answer: 162. Solution: Let's call the center of a corner the cell that borders by sides with two other cells of the corner. In any $2 \times 2$ square, no more than two of the four corners contained in it can be good: those with the two largest numbers in the center of this square. There are 81 $2 \times 2$ squares in ... | 162 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,856 |
4. On the side $A C$ of triangle $A B C$, a point $E$ is chosen. The bisector $A L$ intersects the segment $B E$ at point $X$. It turns out that $A X=X E$ and $A L=B X$. What is the ratio of the angles $A$ and $B$ of the triangle? (S. Berlov) | Answer. 2. Solution. Let the line passing through point $E$ parallel to $A L$ intersect lines $B C$ and $B A$ at points $P$ and $Q$ respectively. From the similarity of triangles $A B L$ and $Q B P$, we have $P Q / A L = B E / B X = B E / A L$, from which it follows that $P Q = B E$. Due to the parallelism of lines $A ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,857 |
5. Ninety-nine positive numbers are arranged in a circle. It turns out that for any four consecutive numbers, the sum of the first two of them in a clockwise direction is equal to the product of the last two of them in a clockwise direction. What can the sum of all 99 arranged numbers be? (S. Berlov) | Answer: 198. Solution: Let the consecutive numbers be $a, b, c, d, e$. Notice that if $a > c$, then $a + b > b + c$, which implies $c < a$. Similarly, if $c d > d e$, then $c > e$. Continuing this reasoning, we get that each number is greater than the number that is two positions clockwise from it. However, writing a c... | 198 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,858 |
5. Three athletes ran a distance of 3 kilometers. The first kilometer they ran at constant speeds $v_{1}, v_{2},$ and $v_{3}$ respectively, such that $v_{1}>v_{2}>v_{3}$. After the 1-kilometer mark, each of them changed their speed: the first from $v_{1}$ to $v_{2}$, the second from $v_{2}$ to $v_{3}$, and the third fr... | Answer. Second. Solution. The first athlete ran the distance faster than the second, as his speed was higher than the second's both on the first kilometer of the distance and on the last two. The third athlete spent as much time on the first kilometer as the second did on the second kilometer, and ran the second and th... | Second | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,859 |
6. Find all natural numbers that can be represented in the form $\frac{x y+y z+z x}{x+y+z}$, where $x, y$ and $z$ are three distinct natural numbers. (D. Khramtsov) | Answer. All natural numbers except 1. Solution. Any natural number $n>1$ can be obtained by setting $x=1, y=n$ and $z=n^{2}: \frac{x y+y z+z x}{x+y+z}=\frac{n\left(1+n^{2}+n\right)}{1+n+n^{2}}=n$. We will show that the number 1 cannot be obtained. For this, it is sufficient to show that if the natural numbers $x, y$ an... | Allnaturalexcept1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,860 |
7. A row of 100 coins is laid out. All the coins look the same, but somewhere among them are 50 counterfeit ones (and the rest are genuine). All genuine coins weigh the same, and the counterfeit ones may weigh differently, but each counterfeit is lighter than a genuine one. Can at least 34 genuine coins be found with a... | Answer. It is possible. Solution. Number the coins from left to right with numbers from 1 to 100. Compare coins 17 and 84. At least one of them is genuine. Therefore, if the scales are in balance, both coins are genuine; in this case, 34 coins with numbers 1-17 and 84-100 will be genuine, since 50 counterfeit coins do ... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,861 |
8. Points $M$ and $N$ are the midpoints of the angle bisectors $AK$ and $CL$ of triangle $ABC$, respectively. Prove that angle $ABC$ is a right angle if and only if $\cdot M B N=45^{\circ}$. (A. Kuznetsov, method commission) | Solution. Let angle $ABC$ be a right angle (see the figure on the right). Then $BM$ and $BN$ are medians in the right triangles $ABK$ and $CBL$, from which it follows that $MBA = MAB = BAC / 2$ and $NBC = NCB = BCA / 2$. Therefore,
- $MNB = 90^\circ - (BAC + BCA) / 2 = 45^\circ$.
![](https://cdn.mathpix.com/cropped/2... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,862 |
2. Given two non-zero numbers. If you add one to each of them, as well as subtract one from each of them, then the sum of the reciprocals of the four resulting numbers will be equal to 0. What number can result if you subtract the sum of their reciprocals from the sum of the original numbers? Find all possibilities. (S... | Answer: 0. Solution: Let the given numbers be a and b. We transform the sum of the four numbers given in the condition: $0=\frac{1}{a+1}+\frac{1}{a-1}+\frac{1}{b+1}+\frac{1}{b-1}=\frac{2 a}{a^{2}-1}+\frac{2 b}{b^{2}-1}$. Let $x=\frac{a}{a^{2}-1}=-\frac{b}{b^{2}-1}$. Then
$$
a-\frac{1}{a}+b-\frac{1}{b}=\frac{a^{2}-1}{a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,864 |
3. A hundred people are sitting in a circle. Some of them are knights, who always tell the truth, and the rest are liars, who always lie. For some natural number $k<100$, each of those sitting said: "The next $k$ people sitting to my right are liars." What could the number $k$ be? (S. Berlov) | Answer. Any divisor of the number 100, except 1, decreased by $1: 1,3,4,9,19,24,49$, 99. Solution. If all the people sitting were liars, they would all tell the truth. Therefore, there must be a knight among them. After him, there are $k$ liars. The one sitting next to the knight, the next $k-1$ people are liars, and s... | 1,3,4,9,19,24,49,99 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,865 |
4. Inside parallelogram $A B C D$, a point $E$ is chosen such that $A E=D E$ and $\angle A B E=90^{\circ}$. Point $M$ is the midpoint of segment $BC$. Find the angle $DME$. (A. Kuznetsov) | Answer: $90^{\circ}$. Solution: Let $N$ be the midpoint of segment $A D$. Since triangle $A E D$ is isosceles, $E N \perp A D$. Since $A B \| M N$ and $\angle A B E=90^{\circ}$, then $B E \perp M N$. Therefore, $E$ is the orthocenter of triangle $B M N$. This means $M E \perp B N$. Since $B M D N$ is a parallelogram, $... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,866 |
5. In the Thirtieth Kingdom, from each city, 30 roads lead out, and each road connects two cities without passing through other cities. The Thirtieth Tsar wanted to place a Road Maintenance Office (RMO) in some cities, which would service all the roads leading out of that city, so that each road is serviced by at least... | Answer: It cannot. Solution: If the king's wish is impossible, then everything is already proven. Let's assume it is possible. Take any arrangement of DUEs (Department of Engineering and Utilities) that meets the king's requirements. Let $S$ be the total number of roads in the kingdom. Since each road connects two citi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,867 |
1. Before the sale, a spoon and a fork cost the same. During the sale, the price of the spoon was reduced by 1 ruble, and the price of the fork was reduced to one-tenth of its original price. Could it happen that the spoon was sold cheaper than the fork during the sale? | Answer: It could. Solution. Let the initial price be 1 p. 10 kop. Then on sale, the spoon cost 10 kop., and the fork - 11 kop. Remark. Let $x$ be the initial price in kopecks. Then $x$ must be greater than 100 (since the sale price of the spoon must be positive), divisible by 10 (since the sale price of the fork must b... | Itcould | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,868 |
2. Find all such triples of numbers $m, n, k$, such that each of the equations $m x^{2}+n=0, n x^{2}+k=0$ and $k x^{2}+m=0$ has at least one solution. | Answer: $m=n=k=0$. Solution. If $m=0$, then from $m x^{2}+n=0$ we get $n=0$, and from $n x^{2}+k=0$ we get $k=0$. Similarly for $n=0$ and $k=0$. Thus, if one of the numbers $m, n, k$ is 0, then the other two are also 0. Suppose none of the numbers $m, n, k$ are 0. Then from the first equation it follows that the number... | =n=k=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,869 |
3. In 1001, a flying carpet at the Baghdad market cost 1 dinar. Then, over the course of 99 years, it increased in price by 1 dinar every year, except for one year when it tripled in price. Could a similar flying carpet have cost 152 dinars in 1100? | Answer: Could not. First solution. If the flying carpet increased in price by 1 dinar every year except one, and in one year it did not increase at all, then in the year 1100 it would have cost 1 + 98 = 99 dinars. Therefore, as a result of the tripling of the carpet's price, 152 - 99 = 53 dinars were added to its cost.... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,870 |
4. Point $D$ lies inside triangle ABC. Can it happen that the shortest side of triangle BCD is 1, the shortest side of triangle ACD is 2, and the shortest side of triangle ABD is 3? | Answer: It cannot. Solution. Since by condition $A D \geq 3$ and $C D \geq 2$, in triangle $B C D$ the unit can only equal $B C$, and in triangle $A C D$ the two can equal $C D$ or $A C$. But in both cases, the triangle inequality is not satisfied: if $A C=2$, then $A B \geq 3=B C+A C$ in triangle $A B C$, and if $C D=... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,871 |
5. A six-digit number $N$ coincides with each of the five six-digit numbers $A, B, C, D, E$ in three digits. Prove that among the numbers $A, B, C, D, E$ there will be two that coincide in at least two digits. | Solution. Write down the number $N$ and put a cross under the digits in the positions where the number $N$ matches the number $A$. Then put a cross under the digits in the positions where the number $N$ matches the number $B$, and so on. In the end, we will put 15 crosses. This means there will be a position with no le... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,872 |
2. The school table tennis championship was held according to the Olympic system. The winner won 6 matches. How many participants in the championship won more matches than they lost? (In the first round of the championship held according to the Olympic system, participants are paired. Those who lost the first game are ... | Answer: 16. Solution. Since in each round every player found a partner and in each pair one of the players was eliminated, the total number of players decreased by half after each round. The winner participated in every round and won, so there were a total of six rounds. Since the winner was determined unequivocally af... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,874 |
3. In triangle $ABC$, the median $BM$ is twice as short as side $AB$ and forms an angle of 40 degrees with it. Find angle $ABC$.
| Answer: $110^{\circ}$. Solution. Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a parallel... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,875 |
4. Represent the numerical expression $2 \cdot 2009^{2}+2 \cdot 2010^{2}$ as the sum of squares of two natural numbers. | Answer: $4019^{2}+1^{2}$ or $2911^{2}+2771^{2}$ Solution. It is sufficient to note that $2 \cdot 2009^{2}+2 \cdot 2010^{2}=(2010+2009)^{2}+(2010-2009)^{2}$.
Note. It can be shown that there are no other answers.
Grading Guidelines. For full credit, it is sufficient to provide and justify one of the two possible answe... | 4019^{2}+1^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,876 |
5. Tom Sawyer took on the task of painting a very long fence, adhering to the condition: any two boards, between which there are exactly two, exactly three, or exactly five boards, must be painted in different colors. What is the smallest number of different colors he can manage with. | Answer: Three. Solution. Note that between the first and fourth, and the fourth and seventh boards, there are two boards each, and between the first and seventh boards, there are five boards. Therefore, the first, fourth, and seventh boards of the fence must be painted in different colors, so Tom will need at least thr... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,877 |
1. The length of a rectangle was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$? | Answer: By 18%. Solution. Let the length be $a$, and the width be $b$. According to the condition, $2(0.1 a + 0.2 b) = 0.12(2 a + 2 b)$, from which we get $a = 4 b$. If the length is reduced by $20\%$, and the width by $10\%$, then the perimeter will decrease by $2(0.2 a + 0.1 b) = 1.8 b$, and it was $10 b$. Therefore,... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 6,878 |
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