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26. In the tetrahedron $A B C D$, the median $A E$ of face $A B C$ is perpendicular to edge $B D$, and the median $A F$ of face $A B D$ is perpendicular to edge $B C$. Prove that edge $A B$ is perpendicular to edge $C D$.
(A. Golev) | 26. Let $\vec{b}=\overrightarrow{A B}, \vec{c}=\overrightarrow{A C}, \vec{d}=\overrightarrow{A D}$. Then $\overrightarrow{B D}=\vec{d}-\vec{b}, \overrightarrow{B C}=\vec{c}-\vec{b}, \overrightarrow{C D}=\vec{d}-\vec{c}$. The vectors along the medians are $\overrightarrow{A E}=\frac{1}{2}(\vec{b}+\vec{c})$ and $\overrig... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,873 |
28. When preparing a district olympiad, each jury member participated in no more than 10 discussions. Discussions can be large or small. In a small discussion, 7 jury members participate, each sending exactly one email to each of the other 6. In a large discussion, 15 jury members participate, each sending exactly one ... | 28. Answer: The secretary participated in 6 small discussions and 2 large ones.
In a small discussion, 7 jury members participate, each sending exactly one email to each of the 6 others, so in the end, 42 emails are sent in a small discussion. Similarly, participants in a large discussion send 210 emails.
Let a total... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,875 |
2. In an acute-angled triangle $A B C$ with an angle of $45^{\circ}$ at vertex $A$, altitudes $A D, B E$, and $C F$ are drawn. Ray $E F$ intersects line $B C$ at point $X$. It turns out that $A X \| D E$. Find the angles of triangle $A B C$. | Answer: 45, 60, 75. From the condition, it follows that $A F=F C$. From the inscribed angles $\angle C D E=45$, and now from the parallelism $\angle A X D=45$. In triangle $A X C$, point $F$ turned out to be the center of the circumscribed circle (equidistant from vertices $A$ and $C$ and the central angle is twice the... | 45,60,75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,876 |
3. Ivan and Kasya are playing the following game. Initially, the polynomial $x-1$ is written on the board. In one move, the polynomial $f(x)$ written on the board can be replaced with the polynomial $a x^{n+1}-f(-x)-2$, where $n$ is the degree of the polynomial $f(x)$, and $a$ is one of its real roots. The players take... | Answer: Ivan will not lose; Koshchei can play forever. Ivan gets a polynomial of odd degree, so he cannot lose. However, Koshchei can also avoid losing: for this, it is enough for him to choose a positive root each time. Note that the constant term is always equal to -1. It is easy to verify that with such a strategy, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,877 |
5. The altitudes $A A_{1}, B B_{1}, C C_{1}$ of an acute triangle $A B C$ intersect at point $H$. A perpendicular $H Q$ is dropped from point $H$ to the tangent line drawn from point $C$ to the circumcircle of triangle $A B_{1} C_{1}$ (point $Q$ lies inside triangle $A B C$). Prove that the circle passing through point... | Solution. Let the circle $(A B_{1} C_{1})$ be denoted by $s_{1}$, and the circle passing through point $B_{1}$ and tangent to the line $A B$ at point $A$ - by $s_{2}$. The chords $B_{1} C_{1}$ and $B_{1} A$ of these circles cut off arcs of the same angular magnitude. Indeed, the halves of these arcs in both cases are e... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,878 |
6. Find all pairs of non-zero (not necessarily positive) rational numbers $x, y$ that have the following property: any positive rational number can be represented as $\{r x\} / \{r y\}$ with a positive rational $r$.
---
Note: The translation preserves the original formatting and structure of the text. | Answer: all pairs where $x y<0$ are suitable. Divide the plane into unit squares by lines of the integer lattice. Draw a line $\ell$ through the origin $O$ and the point $(x, y)$. Since $x, y \in \mathbb{Q}$, it has a rational slope and passes through some lattice node. A point of the form $(r x, r y)$ is any rational ... | allpairswherexy<0suitable | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,879 |
7. There are $2 n$ cards, each with a number from 1 to $n$ (each number appears on exactly two cards). The cards are lying on the table face down. A set of $n$ cards is called good if each number appears exactly once. Baron Munchausen claims that he can point out 80 sets of $n$ cards, at least one of which is guarantee... | Answer: $n=7$. We will present an algorithm for how to indicate $2^{n-1}$ sets on $2n$ cards, one of which is suitable. (In our case, this is $2^{6}=64<80$ sets.) Imagine that identical cards are connected by (invisible to us for now) red edges. We will arbitrarily pair the cards with blue edges. The red-blue graph is ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,880 |
29. On September 1st, children brought roses and carnations to school. 101 boys and 3 girls stood in a circle, each holding all their flowers. It turned out that each boy had exactly 50 flowers. At the principal's signal, each child passed all their carnations to the child on their left. After this, it turned out that ... | 29. If a boy received $k$ nails from the neighbor on the right, then he himself passed $k+1$ nails to the neighbor on the left. Since each boy has no more than 50 nails, no more than 50 boys can stand in a row. Therefore, there are at least three groups of boys standing in a row. Since there are only 3 girls in total, ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 18,881 |
30. There are 111 children. They all weigh the same number of grams and always tell the truth, except for one who weighs less and always lies. The nearsighted caregiver places 55 children on each pan of the scales, after which the child who did not participate in the weighing tells the caregiver which pan was heavier (... | 30. Let's conduct all possible weighings. The fake child is the one who, in all weighings (in which they participated), ended up on the lighter side. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,882 |
31. Let's call a number complex if it has at least two different prime divisors. Find the largest natural number that cannot be represented as the sum of two complex numbers. | 31. The numbers $6, 12, 15$, and 21 are composite and give all possible remainders when divided by 4. Therefore, from any number $n>23$, one of these numbers can be subtracted so that the result is a number of the form $4k+2=2(2k+1)$. Clearly, this difference will be a composite number. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,883 |
32. A ten-digit number is written on the board. You can take any digit of this number that is less than 8, add 1 or 2 to it, and write the resulting new number on the board instead of the old one. This operation was performed 55 times. Prove that at least one of the 56 numbers written on the board in this experiment wa... | 32. To ensure the number never divides by 3 at any point, one must alternately add ones and twos, i.e., either $82=1+2+\cdots+1$, or $83=2+1+\cdots+2$. But the last digit could increase by no more than 2 (from 1 to 3, or from 7 to 9), the first digit by no more than 8, and the rest by no more than 9. Thus, no more than... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,884 |
33. Dima and Gosha are playing "noughts-noughts" on a $14 \times 441$ board. In one move, a player can place one nought in any empty cell. They take turns, with Gosha going first. The player who, after their move, forms 7 consecutive noughts either vertically or horizontally wins. Who among the players can win, regardl... | 33. Dima will win. In each column, we will divide the cells into pairs of the form $(a, a+7)$. Dima should complete Gosha's move to a pair, except in the case when he sees 7 zeros in a row. It is clear that Gosha will not be able to win in a row. Suppose that Gosha wins by placing seven zeros in a row vertically. Note ... | Dimawillwin | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,885 |
34. In the gym, 200 schoolchildren gathered. Every pair of acquaintances shook hands. It turned out that any two strangers made at least 200 handshakes in total. Prove that there were at least 10000 handshakes in total. | 34. Let's write the solution in the language of graphs. If the degree of a vertex is not less than 100, we call it rich; if it is less than 100, we call it poor. From the condition, it follows that all poor vertices are pairwise adjacent to each other.
We will prove that each poor vertex can be paired with a non-adjac... | 10000 | Combinatorics | proof | Yes | Yes | olympiads | false | 18,886 |
12. Solution 1. Mark a point \( E' \) on the segment \( AD \) such that \( DE' = DE \). Notice that then triangles \( BDE' \) and \( CDE \) are equal by two sides and the angle, which means \( BE' = CE \). We need to check that the perimeter of triangle \( ADC \) is greater than the perimeter of quadrilateral \( ABDE \... | Solution 2. Construct triangle $D A^{\prime} C$ symmetric to triangle $D A B$ with respect to the perpendicular bisector of segment $B C$. Then $A D=A^{\prime} D=A^{\prime} E+E D$, $A B=A^{\prime} C$. Remove the equal segments $B D=D C$ from the perimeters of triangle $A D C$ and quadrilateral $A B D E$. Then we need t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,888 |
3. Given a rectangular table $14 \times n$. In its top-left cell, there is a chip. Sasha and Olya take turns (Sasha starts) moving the chip one cell to the right or down. We will say that a turn has occurred if the direction of the player's move differs from the direction of the move made by the opponent before that. I... | Answer: 13, 14, 15. The presence of an even number of turns is equivalent to the directions of the first and last moves coinciding. If $n>15$, then Sasha can make the first move down, and then always move right. Similarly, in the case of $n<13$: Sasha can make the first move right, and then move down. It is easy to see... | 13,14,15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,891 |
4. A sequence of 300 natural numbers is written in a row. Each number, starting from the third, is equal to the product of the two preceding numbers. How many perfect squares can there be among these numbers? (Provide all answers and prove that there are no others.) | Answer: 0, 100 or 300. If after number a, number $b$ is written, then the numbers аb and $а b^{2}$ follow them. The number $a b^{2}$ is a perfect square if and only if a is a perfect square. Among the first three numbers, squares can be all three, one, or none (examples of all three situations are trivial). Hence, the ... | 0,100,300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,892 |
6. Find the minimum value of the expression $\left[\frac{7(a+b)}{c}\right]+\left[\frac{7(a+c)}{b}\right]+\left[\frac{7(b+c)}{a}\right]$, where $a, b$ and $c$ are arbitrary natural numbers. | Answer: 40. Evaluation: the sum of the fractions under the integer parts is no less than $14 \cdot 3=42$ (we add three inequalities of the form $7(a / b+b / a) \geqslant 14$, and each integer part is greater than the fraction reduced by 1. Therefore, the desired value is greater than $42-3=39$, i.e., not less than 40. ... | 40 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 18,893 |
1. Does there exist a 100-digit number without zeros in its representation, which is divisible by all possible sums of its digits (in particular, by all its digits) | Answer: No. One of the digits appears no less than 10 times, so the number is divisible by 10, i.e., it contains 0 in its representation. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,894 |
3. Vladik thought of a natural number $N$. He divided it by one of its prime divisors and wrote the result on the board. Then he divided this result by one of its prime divisors and wrote the new result on the board. He continued this process until he wrote 1 on the board. This 1 turned out to be the 22nd number writte... | Answer: $2 \cdot 3^{21}$. Let's write the sum starting from the beginning: $1+p_{1}+p_{1} p_{2}+p_{1} p_{2} p_{3}+\cdots=N / 2$. If $p_{1}$ were an odd divisor of $N$, the right side would be divisible by $p_{1}$, which cannot be the case since the left side is not divisible by $p_{1}$ due to the first term. Therefore,... | 2\cdot3^{21} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,895 |
4. Is it possible to form some rectangle by taking squares $1 \times 1, 3 \times 3, 5 \times 5, \ldots, 85 \times 85$ and $2021 \times 2021$ (each one time) and adding several $2 \times 2$ squares to them? | Answer: No. The total area is even, so one of the sides of the formed rectangle (let's say the horizontal one) must be even. We will paint the cells of the rectangle in two colors using a vertical checkerboard pattern, and there will be an equal number of each color. However, the large square gives a bias of one color ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,896 |
5. In 1685, having found himself on an uninhabited island, the native Friday met Robinson Crusoe on the very first day. Good spirits taught Friday the European calendar and hinted that only one day a year, February 13, Robinson tells the truth, while on other days he lies. Every day, Friday asks Robinson Crusoe one que... | Answer: Yes. Let's number the days of the year from the zeroth (February 13) to the 364th. Mark the 183rd day on the calendar, and call it "day $X$". Between the zeroth day and day $X$, 182 days pass, and between day $X$ and the zeroth day, 181 days pass.
Suppose Friday asks for the first 181 times whether today is da... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,897 |
6. At the pier, 9 trucks arrived. Each of them brought no more than 10 tons of cargo, and it is known that the weight of each individual cargo does not exceed 1 ton. At the pier, there are 10 barges with a carrying capacity of $k$ tons each. For what minimum $k$ can all the delivered cargo be guaranteed to be transport... | Answer: when $k=180 / 19=10-10 / 19$ tons. Evaluation. If all items weigh $10/19$ tons each, and there are a total of $9 \cdot 19=171$ items, then on some barge there will be no fewer than 18 items, and they will weigh no less than $180 / 19 \mathrm{~m}$.
Example. To load the barges, items weighing between $10/19$ and... | 10-\frac{10}{19} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,898 |
24. A polynomial of degree 10 has three distinct roots. What is the maximum number of zero coefficients it can have?
(A. Khryabrov) | 24. Answer: Answer: 9 zero coefficients. For example, the polynomial $x^{10}-x^{8}$ has roots $0,1,-1$. If a polynomial has only one non-zero coefficient, it is of the form $a x^{10}$, and therefore has exactly one root. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,899 |
25. All fishermen are divided into ordinary and honest ones. An honest fisherman exaggerates the weight of the fish he caught exactly by 2 times, while an ordinary fisherman exaggerates it by an integer greater than six times (these coefficients can be different for different ordinary fishermen). 10 fishermen caught a ... | 25. Each honest fisherman caught exactly 30 kg, while each ordinary one caught less than $60 / 6=10$ kg of fish. Therefore, there could not have been zero honest fishermen (10 ordinary fishermen would have caught less than 100 kg). Similarly, there could not have been exactly one honest fisherman (9 ordinary fishermen ... | 2or3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,900 |
26. In a kindergarten group, there are 26 children. When the children went out for a walk, each had two mittens of the same color, and the mittens of different children were of different colors. During the walk, the children formed pairs three times (not necessarily in the same way). During the first formation, the chi... | 26. If child A is wearing mittens of colors x and y (x on the left hand, y on the right), we will denote this fact by the notation $\mathrm{A}(\mathrm{x}, \mathrm{y})$.
If any child exchanges mittens twice with the same child, they will end up with two identical mittens. Therefore, we will assume that this did not hap... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,901 |
27. Points $X$ and $Y$ are the midpoints of arcs $A B$ and $B C$ of the circumcircle of triangle $A B C$. $B L$ is the angle bisector of this triangle. It turns out that $\angle A B C=2 \angle A C B$ and $\angle X L Y=90^{\circ}$. Find the angles of triangle $A B C$. | 27. Answer: $\angle B=90^{\circ}, \angle A=\angle C=45^{\circ}$.
By the condition $\angle C=\angle B / 2=\angle L B C$, so triangle $B L C$ is isosceles: $B L=C L$. In addition, $B Y=Y C$, so triangles $B L Y$ and $C L Y$ are equal by three sides, and $L Y-$ is the bisector of angle $B L C$. By the condition, line $L ... | \angleB=90,\angleA=\angleC=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,902 |
28. Given natural numbers $m$ and $n (m < n)$ and a large chocolate bar, the sides of which are divisible by $n^5$. (All chocolate bars in this problem are grid rectangles, with the side of each cell equal to 1.) Lesha ate several cells five times such that each time a smaller chocolate bar was left, the area of which ... | 28. Let the original chocolate bar have dimensions $a n^{5} \times b n^{5}$. Then its area was $a b n^{10}$. Suppose after five bites, the resulting chocolate bar is $x \times y$ with an area of $x y = a b m^{5} n^{5}$. Since $x \leqslant a n^{5}, y \leqslant b n^{5}$, then $x \geqslant b m^{5}, y \geqslant a m^{5}$. T... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,903 |
4. There are two piles of stones: 444 stones in one pile and 999 in the other. Sasha and Fedia are playing a game, taking turns, with Sasha starting. Let the piles contain $a$ and $b$ stones before a player's move, with $a \geqslant b$. Then, on their turn, the player is allowed to take any number of stones from the pi... | Answer: first. Suppose the position $a>b$ is a losing position. Then all positions $(a-k, b)$ for $k=1, \ldots, b$ are winning. But since $a-1 \geqslant b-$ is winning, from it one can reach some losing position, and it can only be the position $(a-b-1, b)$. By such procedures, we can obtain losing positions: $(999,444... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,905 |
5. Let $n>2$ be a natural number, $1=a_{1}<\ldots<a_{k}=n-1$ be all numbers from 1 to $n$ that are coprime with $n$. Denote by $f(n)$ the greatest common divisor of the numbers $a_{1}^{3}-1, \ldots, a_{k}^{3}-1$. What values can the function $f(n)$ take? | Answer: 1, 2, 7, 26, 124 (achieved for $n=5,10,3,4,6$ respectively). Cases $n \leqslant 7$ are considered directly, and from now on we assume that $n>7$. We call numbers of the form $a_{i}^{3}-1$ special.
Let $n>7$ be odd. Then among the special numbers there are both $2^{3}-1=7$ and either $7^{3}-1$ (if $n$ is not di... | 1,2,7,26,124 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,906 |
4. On the bank of the river stand 10 sheikhs, each with a harem of 100 wives. Also at the bank stands an $n$-person yacht. By law, a woman should not be on the same bank, on the yacht, or even at a transfer point with a man if her husband is not present. What is the smallest $n$ for which all the sheikhs and their wive... | Answer: 10. Example: first, all 1000 wives move quietly, then one returns and 10 sheikhs leave. Finally, another returns and picks up the first. Evaluation: let the number of places not exceed 9. Consider the moment when the first sheikh appears on the other shore - or several at once, but not all, for they could not h... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,907 |
6. We will call a set of natural numbers not exceeding $10^{100}$ conservative if none of its elements divides the product of all the other elements, and progressive if together with any of its elements it contains all multiples of that element not exceeding $10^{100}$. Which sets are more numerous: progressive or cons... | Answer: there are more progressive sets. Consider the conservative set $A$ and the set $P(A)$ of all numbers not exceeding $10^{100}$ that are multiples of at least one of the numbers in $A$. Clearly, $P(A)$ is progressive. It is not difficult to establish that the mapping $P$ is injective, and therefore, there are no ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,909 |
1. Given a prime number $p$. All natural numbers from 1 to $p$ are written in a row in ascending order. Find all $p$ for which this row can be divided into several blocks of consecutive numbers so that the sums of the numbers in all blocks are equal. | Answer: $p=3$. Let $k-$ be the number of blocks, $S-$ be the sum in each block. Since $p(p+1) / 2=k S$, either $k$ or $S$ is divisible by $p$. Clearly, $k<p$, so $S$ is a multiple of $p$. Let the leftmost group consist of numbers from 1 to $m$. Then $m(m+1) / 2$ is a multiple of $p$, from which $m \geqslant p-1$, i.e.,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,910 |
2. The cells of a $100 \times 100$ table are painted white. In one move, it is allowed to select any 99 cells from one row or one column and repaint each of them to the opposite color - from white to black, and from black to white. What is the minimum number of moves required to obtain a table with a checkerboard patte... | Answer: in 100 moves. Evaluation: to repaint each of the cells on the black diagonal, a move is required. Example: repaint all rows and columns with odd numbers. In this case, in the $k$-th row and $u$-th column, we will not repaint their common cell. It is easy to see that as a result, we will get a chessboard colorin... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,911 |
7. A square is cut into red and blue rectangles. The sum of the areas of the red rectangles is equal to the sum of the areas of the blue ones. For each blue rectangle, we write down the ratio of the length of its vertical side to the length of its horizontal side, and for each red rectangle - the ratio of the length of... | Answer: $1 / 2$. Example: two rectangles $1 / 2 \times 1$.
We can assume that the side of the original square is 1. The ratio of the sides of any rectangle is greater than or equal to its area, so each of the two sums ("red" and "blue") is not less than $1 / 2$. It remains to prove that one of these two sums is not le... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,912 |
42. Hooligan Vasya is dissatisfied with his average math grade, which has dropped below 3. As a measure to sharply improve his grade, he managed to get to the school journal and changed all his twos to threes. Prove that after this, his average grade will still not exceed 4.
(K. Tyschuk $)$ | 42. Let Vasya receive $a$ units, $b$ twos, $c$ threes, $d$ fours, and $e$ fives. Then, by the condition, $a+2b+3c+4d+5e+2e$, from which $a+b>e$. It is required to prove that $2b+3(a+c)+4d+5e \leqslant 4(a+b+c+d+e)$, which is equivalent to the inequality $a+2b+c \geqslant e$. This easily follows from the derived inequal... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,913 |
43. Triangles $A B C$ and $A_{1} B_{1} C_{1}$ are such that $\angle A=\angle A_{1}$ and $\angle B+\angle B_{1}=180^{\circ}$. Prove that if $A_{1} B_{1}=$ $A C+B C$, then $A B=A_{1} C_{1}-B_{1} C_{1}$. | 43. Let's mark a point \( B' \) on the ray \( AC \) (beyond point \( C \)) such that \( CB' = CB \). Then, by the condition, \( AB' = A_1B_1 \). Let's mark a point \( C' \) on the ray \( AB \) such that \( \triangle AB'C' = \triangle A_1B_1C_1 \). The point \( C' \) will fall on the extension of side \( AB \) beyond po... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,914 |
44. Andrey, Borya, Vitya, and Gena are playing on a $100 \times 2019$ board (100 rows, 2019 columns). They take turns - first Andrey, then Borya, followed by Vitya, and finally Gena, then again Andrey, and so on. Each turn, a player must color two uncolored cells forming a rectangle of two cells, with Andrey and Borya ... | 44. Let's show how players A, B, and C can beat D. We will divide the board into vertical dominoes of size $2 \times 1$. Let players A and B cover any dominoes of the partition with their moves, and C complements D's move to a pair of dominoes of the partition. Then, before D's move, several dominoes of the partition w... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,915 |
45. A natural number $n$ has exactly 1000 natural divisors (including 1 and $n$ itself). These 1000 divisors are listed in ascending order. It turns out that any two consecutive divisors have different parity. Prove that the number $n$ has more than 150 digits.
( $\Phi$. Petrov $)$ | 45. Let the divisors be $d_{1}2 d_{998}=2^{2} d_{997}>\ldots>2^{499} d_{2}=2^{500} d_{1}=2^{500}$, i.e., $n>2^{500}=\left(2^{10}\right)^{50}>\left(10^{3}\right)^{50}=10^{150}$, which is what we needed to prove. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,916 |
35. The exchange rate of the cryptocurrency Chuhoin on March 1 was one dollar, and then it increased by one dollar every day. The exchange rate of the cryptocurrency Antonium on March 1 was also one dollar, and then each subsequent day it was equal to the sum of the previous day's rates of Chuhoin and Antonium, divided... | 35. Answer: 92/91 dollars. It is not difficult to prove by induction that on each day with an odd number, Antonium costs 1 dollar, and on an even day with number $2 n$, he costs $2 n /(2 n-1)$ dollars. | \frac{92}{91} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,919 |
36. In a convex quadrilateral $A B C D$, the angles at vertices $A, B$ and $C$ are equal. A point $E$ is marked on side $A B$. It is known that $A D=C D=B E$. Prove that $C E$ is the bisector of angle $B C D . \quad(A$. Kuznetsov) | 36. Since $A D=C D$, then $\angle D A C=\angle D C A$, this together with $\angle A=\angle C$ means that $\angle B A C=\angle B C A$, so $A B=B C$. Then triangles $B D A$ and $B D C$ are equal by three sides. From this it follows that $B D$ is the bisector of angle $A B C$, and moreover, triangles $B C D$ and $C B E$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,920 |
37. Several people are standing in a row, some of them are knights who always tell the truth, while the others are liars who always lie. Each of them said one of two phrases: There are more knights to the right of me than to the left or There are more knights to the left of me than to the right, and the number of peopl... | 37. We will call the phrase "To my right, there are more knights than to my left" the first phrase, and the phrase "To my left, there are more knights than to my right" the second phrase. The number of knights cannot be odd, because in that case, the central knight (who has an equal number of knights on both sides) cou... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 18,921 |
38. Do there exist pairwise distinct natural numbers $a, b$, and $c$ such that
$$
2 a+\operatorname{LCM}(b, c)=2 b+\operatorname{LCM}(a, c)=2 c+\operatorname{LCM}(a, b) ?
$$
(S. Berlov, A. Kuznetsov) | 38. Answer: such numbers do not exist. Let such $a, b, c$ exist, without loss of generality, $a>b>c$. From the equality $2 b+\operatorname{LCM}(a, c)=2 c+\operatorname{LCM}(a, b)$, we conclude that $2(b-c)$ is divisible by $a$. But $2(b-c)<2(a-c)$, so either $2(a-c)=3 b$, or $2(a-c)=2 b$, or $2(a-c)=b$. In the first ca... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,922 |
39. In front of Andrey, there are 2019 plates, on which there are a total of 2019 pastries. Andrey can perform two operations.
I. If there are an equal number of pastries on any two plates, he can eat all the pastries on one of these plates.
II. He can transfer one pastry from each non-empty plate to an empty plate.
... | 39. Let it be that sad moment when Andryusha can no longer eat another pastry. Having nothing else to do, let Andryusha arrange the non-empty plates in ascending order of the number of pastries. Since there are no plates with the same number of pastries, the last non-empty plate (let it have the number $k$ and be calle... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,923 |
40. Sasha chose four natural numbers $x, y, z, t$ and wrote down 12 fractions:
$$
\begin{array}{cccccccccccc}
\frac{x}{y}, & \frac{x}{z}, \quad \frac{x}{t}, \quad \frac{y}{x}, \quad \frac{y}{z}, \quad \frac{y}{t}, & \frac{z}{x}, \quad \frac{z}{y}, \quad \frac{z}{t}, \quad \frac{t}{x}, & \frac{t}{y}, & \frac{t}{z}
\end... | 40. Note that if Sasha chose two identical numbers (for example, $x$ and $y$), then he wrote two fractions equal to 1 (in our example, $x / y$ and $y / x$), and the statement of the problem is obvious. Therefore, we can assume that all of Sasha's numbers are distinct.
Let's list Sasha's fractions in ascending order: $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,924 |
41. On the side $AC$ of triangle $ABC$, a point $K$ is marked. It turns out that $\angle ABK=7^{\circ}$ and $\angle ABC=77^{\circ}$. Prove that $2AK + AC > BC$. | 41. Let point $K^{\prime}$ be symmetric to $K$ with respect to line $A B$, and point $A^{\prime}$ be symmetric to $A$ with respect to $B K^{\prime}$. Then $A K=A K^{\prime}=A^{\prime} K^{\prime}$ and $\angle A^{\prime} B C=7^{\circ}+7^{\circ}+77^{\circ}=91^{\circ}>90^{\circ}$. Therefore, $B C<A^{\prime} C \leqslant A^{... | 2AK+AC>BC | Geometry | proof | Yes | Yes | olympiads | false | 18,925 |
1. A $10 \times 10$ table is filled with numbers from 1 to 100: in the first row, the numbers from 1 to 10 are written in ascending order from left to right; in the second row, the numbers from 11 to 20 are written in the same way, and so on; in the last row, the numbers from 91 to 100 are written from left to right. C... | 1. Answer: Yes. If $x$ is in the center of the fragment, then the sum of the numbers in it is $7x$ and then with 646566 $x=65$ the sum will be exactly 455 (see figure). | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,926 |
2. In childhood, Kostya was taught to add natural numbers incorrectly: he believes that after the usual addition, the digits of the sum should be rearranged in descending order. Let's denote Kostya's addition by the symbol $\oplus$ (for example, $99 \oplus 2=110$.) Do there exist such natural numbers $a$ and $b$ for wh... | 2. Answer: No. With regular addition, the number will increase, and with the rearrangement of digits, it will increase even more (or at least not decrease). Therefore, $a \oplus b$ is always strictly greater than $a$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,927 |
3. At a large round table, 100 people are sitting. Each of them is either a knight, a liar, or a fool. A knight always tells the truth, a liar always lies. A fool tells the truth if a liar is sitting to their left; lies if a knight is sitting to their left; and can say anything if a fool is sitting to their left. Each ... | 3. Answer: 50 or 0 liars. The first scenario is realized when knights and liars sit alternately around the table. The second scenario is when all those sitting at the table are eccentrics who tell lies.
Let's prove that there cannot be a different number of liars at the table. Indeed, if there is a liar at the table, ... | 50or0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,928 |
4. The teacher considers some students in class $6^{a}$ to be excellent students, while the rest are poor students. During the quarter, there were 6 math tests in the class (grades from 2 to 5 were given). All students were present at each test, and they sat in pairs at desks (possibly different pairs for different tes... | 4. In fact, there is even an excellent student who received a two.
Suppose that such an excellent student does not exist. Then all the twos were received by poor students. The total number of twos for all six tests is even, because they were received by poor students sitting at the same desk. Let the number of twos be... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 18,929 |
4. We will say that a point on the plane $(u, v)$ lies between the parabolas $y=f(x)$ and $y=g(x)$ if $f(u) \leqslant$ $v \leqslant g(u)$. Find the smallest real $p$, for which the following statement is true: any segment, the ends and the midpoint of which lie between the parabolas $y=x^{2}$ and $y=x^{2}+1$, lies enti... | Answer: 9/8. Consider a segment $s$, the ends and midpoint of which lie between the original parabolas, and the segment itself intersects the parabola. Let it lie on the line $y=k x+\ell$. The two original parabolas cut three segments $A B, B C$ and $C D$ on it. From Vieta's theorem, it immediately follows that $A B=C ... | \frac{9}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,931 |
5. There are two piles of stones: 1703 stones in one pile and 2022 in the other. Sasha and Olya are playing a game, taking turns, with Sasha starting. Let the piles contain $a$ and $b$ stones before the player's move, with $a \geqslant b$. Then, on their turn, the player is allowed to take any number of stones from the... | Answer: Sasha will win. Suppose the position $a>b$ is a losing position. Then all positions $(a-k, b)$ for $k=1, \ldots, b$ are winning. But since $a-1 \geqslant b-$ is winning, from it one can obtain some losing position, and it can only be the position $(a-b-1, b)$. By such procedures, one can obtain losing positions... | Sashawillwin | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,932 |
7. Given $n$ distinct natural numbers, any two of which can be obtained from each other by permuting the digits (zero cannot be placed in the first position). For what largest $n$ can all these numbers be divisible by the smallest of them? | Answer: 9. It is clear that there cannot be more than nine numbers.
We will use a known property of the period of a purely periodic rational fraction $\alpha=$ $a / b<1$ with coprime $(a, b)$: the length of the period is the smallest natural $t$ for which $\left(10^{t}-1\right) \vdots b$, and the period $T$ itself is ... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,933 |
52. Solution 1 (complete residue system). Let $p-$ be the smallest of the prime numbers. Then $p<q<2 p$. Consider the integers from $-\frac{p-1}{2}$ to $\frac{p-1}{2}$. These are $p$ consecutive numbers, so they all give different remainders when divided by $p$. Now multiply these $p$ numbers by $q$. We will show that ... | Solution 2 (Chinese Remainder Theorem). Let $p-$ be the smallest of the prime numbers. By the Chinese Remainder Theorem, there exists a number $a$ that is divisible by $q$ and gives a remainder of 1 when divided by $p$. Note that any number of the form $a-k p q$ for integer $k$ also satisfies this condition. Choose $k$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,936 |
52. Solution 1 (complete residue system). Let $p-$ be the smallest of the prime numbers. Then $p<q<2 p$. Consider the integers from $-\frac{p-1}{2}$ to $\frac{p-1}{2}$. These are $p$ consecutive numbers, so they all give different remainders when divided by $p$. Now multiply these $p$ numbers by $q$. We will show that ... | Solution 2 (Chinese Remainder Theorem). Let $p-$ be the smallest of the prime numbers. By the Chinese Remainder Theorem, there exists a number $a$ that is divisible by $q$ and gives a remainder of 1 when divided by $p$. Note that any number of the form $a-k p q$ for integer $k$ also satisfies this condition. Choose $k$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,939 |
29. The cells of a $4 \times 4$ table are filled with integers. In each row and each column, the product of all numbers is calculated. Could the resulting numbers be $1,5,7,2019,-1,-5,-7,-2019$ in some order? | 29. Answer: yes, they could. One of the possible examples is shown on the right. | yes,theycould | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,940 |
30. We will call all odd prime numbers and the number 1 "simple". 2019 knights and liars are standing in a row. Each of them said: "The number of knights to my right and to my left differ by a simple number." How many knights can there be in this row?
( $A$. Kuznetsov) | 30. Answer: $0,2,4,6$ or 8 knights.
Suppose there are $n>8$ knights in a row. Then, for the leftmost one, the difference between the number of knights to the right and to the left of him is $(n-1)$, for the second from the left it is $(n-3)$, and for the third from the left it is $-(n-5)$. All these numbers have diffe... | 0,2,4,6,8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,941 |
31. For each number from 1 to 1000, all its natural divisors were listed (as a result, some numbers were listed multiple times). Determine which is greater: the sum of all listed numbers or one million? | 31. Answer: a million is greater than the sum of the written numbers.
Let's add separately all the written ones, all the written twos, all the written threes, etc.
Consider any number $n$, where $1 \leqslant n \leqslant 1000$. It is a divisor of the numbers $n, 2 n, 3 n, \ldots, k n$, where $k n$ is the largest numbe... | 1000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,942 |
32. In the Martian calendar, the days of the week are named the same as ours, but each week can have from one to seven Fridays, which is why a week can last from 7 to 13 days. Different weeks can have a different number of Fridays. The minister has been brought a calendar for the next 2019 weeks. The minister decides w... | 32. The considered period of 2019 weeks will be called a year.
Let the minister try to make all Mondays, Tuesdays, Wednesdays, and Thursdays holidays. Then, in each week, there will be 4 holidays and a maximum of 13 days, meaning the proportion of holidays is not less than $\frac{4}{13} > \frac{2}{7}$. If in this case... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,943 |
33. A sports hall in the shape of a square with a side length of 100 meters is paved with square tiles 1 m $\times 1$ m. Two players take turns covering the floor of the sports hall with mats. Each mat covers two adjacent tiles on the floor. The player who starts the game places one mat on their turn, while the second ... | 33. Let's outline a strategy for the second player. First, he mentally divides the hall into 2x2 squares (we will call them "blocks"), and then each time he moves so that after his move, in each block all cells are covered by mats the same number of times.
We will prove that the second player can always make a move ac... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,944 |
34. A company gathered for a meeting. Let's call a person sociable if in this company they have at least 20 acquaintances, and at least two of them are acquainted with each other. Let's call a person shy if in this company they have at least 20 strangers, and at least two of them are strangers to each other. It turned ... | 34. Answer: 40.
Evaluation: Suppose that a certain person (let's call him Kostya) is acquainted with at least 20 people. If some of Kostya's acquaintances know each other, then Kostya is sociable. Let's consider the case where all of Kostya's acquaintances do not know each other. In this case, if Kostya is acquainted ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,945 |
50. In the city, 2019 metro stations have been built. Some pairs of stations are connected by tunnels, and from any station, you can reach any other station via the tunnels. The mayor ordered the organization of several metro lines: each line must include several different stations, sequentially connected by tunnels (t... | 50. Answer: $k=1008$. Let's provide an example of a connected graph with 2019 vertices that cannot be covered by 1008 simple paths: one vertex is connected to 2018 vertices of degree 1. Any simple path contains no more than two pendant vertices, so at least 1009 paths are required for coverage.
We will prove that 1009... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,947 |
51. Prove that the distance between the midpoint of side $B C$ of triangle $A B C$ and the midpoint of arc $A B C$ of its circumscribed circle is not less than $A B / 2$. | 51. Consider the point $B^{\prime}$, symmetric to $B$ with respect to the perpendicular bisector of the segment $A C$. Then $A B B^{\prime} C$ is an isosceles trapezoid inscribed in the same circle. Let $P$ and $Q$ be the midpoints of the arc $A B C$ and the side $B C$, and $R$ be the midpoint of the segment $B B^{\pri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,948 |
52. Olya wrote fractions of the form $1 / n$ on cards, where $n-$ are all possible divisors of the number $6^{100}$ (including one and the number itself). She arranged these cards in some order. After that, she wrote the number on the first card on the board, then the sum of the numbers on the first and second cards, t... | 52. Answer: two denominators. Let's represent all fractions in the form $a_{n} / 6^{100}$, then $a_{1}, a_{2}, \ldots$ are again all divisors of the number $6^{100}$, each appearing once. Let the partial sums be denoted by $S_{n} / 6^{100}$. Then the denominator of the irreducible representation of the partial sum depe... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,949 |
53. Let's call an improvement of a positive number its replacement by a power of two (i.e., one of the numbers 1, $2, 4, 8, \ldots$), such that it increases but not more than three times. Given $2^{100}$ positive numbers with a sum of $2^{100}$. Prove that it is possible to erase some of them, and improve each of the r... | 53. First, let's prove the following lemma: from any set of powers of two not exceeding \(2^{100}\) and summing to at least \(2^{100}\), we can select several such that their sum is exactly \(2^{100}\). To prove this, note that two equal powers of two in this set can be replaced by one that is twice as large. Repeating... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,950 |
54. The bisectors $B B_{1}$ and $C C_{1}$ of an acute-angled triangle $A B C$ intersect at point $I$. Points $B^{\prime}$ and $C^{\prime}$ are marked on the extensions of segments $B B_{1}$ and $C C_{1}$, respectively, such that the quadrilateral $A B^{\prime} I C^{\prime}$ is a parallelogram. Prove that if $\angle B A... | 54. Note that $\angle B^{\prime} I C^{\prime}=\angle B I C=120^{\circ}$, so the sum of angles $I B^{\prime} C^{\prime}$ and $I C^{\prime} B^{\prime}$ is $60^{\circ}$. Mark a point $P$ on the segment $B^{\prime} C^{\prime}$ such that $\angle P A B_{1}=\angle I B^{\prime} C^{\prime}$ and $\angle P A C_{1}=\angle I C^{\pr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,951 |
55. There are 2019 plates arranged in a circle, each with one pastry on it. Petya and Vasya are playing a game. In one move, Petya points to a pastry and names a number from 1 to 16, and Vasya moves the indicated pastry to the specified number of plates clockwise or counterclockwise (Vasya chooses the direction each ti... | 55. Answer: 32 pastries.
Let's show how Petya can make it so that one of the plates ends up with no fewer than 32 pastries. Number the plates in a circle from 0 to 2018. Paint 32 plates in red with numbers $0, 32, 2 \cdot 32, \ldots, 31 \cdot 32$. Petya will only point to plates with numbers from 0 to $31 \cdot 32$, a... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,952 |
9. Amateur Dima and professional Fedya both chopped wood and boasted to each other about how much they had chopped. In doing so, Dima exaggerated the result of his work by 2 times, Fedya by 7 times, and in total, it turned out to be three times more wood than in reality. Who chopped more wood and by how many times?
(D... | 9. Answer: Dima chopped 4 times more wood than Fedya. If Dima chopped $D$ wood, and Fedya chopped $F$ wood, then from the condition, it is easy to form the equation: $2 D+7 F=3(D+F)$. From this, we immediately get $D=4 F$. | 4F | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,953 |
13. A grasshopper starts moving in the top-left cell of a $10 \times 10$ square. It can jump one cell down or to the right. Additionally, the grasshopper can fly from the bottom cell of any column to the top cell of the same column, and from the rightmost cell of any row to the leftmost cell of the same row. Prove that... | 13. Consider the diagonal running from the bottom-left corner to the top-right corner. We will paint all 10 cells on this diagonal red. Note that from any red cell, without making any jumps, you can only move to cells that are to the right of it, below it, or to the right and below it. Therefore, it is impossible to mo... | 9 | Combinatorics | proof | Yes | Yes | olympiads | false | 18,956 |
5. A $70 \times 70$ table is filled with numbers from 1 to 4900: in the first row, from left to right, the numbers from 1 to 70 are written in ascending order; in the second row, the numbers from 71 to 140 are written in the same way, and so on; in the last row, from left to right, the numbers from 4831 to 4900 are wri... | 5. Answer: such a fragment does not exist. If $x$ is in the center of the fragment, the sum of the numbers in it is $5 x$, i.e., divisible by 5, and therefore cannot be equal to 2018. | such\\fragment\does\not\exist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,957 |
7. Points $M$ and $N$ are the midpoints of the equal sides $A B$ and $B C$ of triangle $A B C$, respectively. On the extension of segment $M N$ beyond point $N$, point $X$ is marked, and on segment $N X$, point $Y$ is marked such that $M N = X Y$. Prove that $B Y = C X$. | 7. Since the sides of the triangle are equal, their halves are also equal: $B M=C N$. In addition, $M Y=M N+N Y=X Y+N Y=N X$. Finally, $\angle B M Y=\angle B N M=\angle C N X$. Thus, triangles $B M Y$ and $C N X$ are equal by the first criterion. Therefore, their sides $B Y$ and $C X$ are equal. | BY=CX | Geometry | proof | Yes | Yes | olympiads | false | 18,958 |
8. On a parking lot, there are cars. Among them, there are cars of the brands "Toyota", "Honda", "Skoda", as well as cars of other brands. It is known that the number of non-"Honda" cars is one and a half times the number of non-red cars; the number of non-"Skoda" cars is one and a half times the number of non-yellow c... | 8. Let's call all other brands Fords, and all other colors - green. By the condition, t+sh+f=1.5(zh+z), t + x + φ=1.5(k+3), x + m + φ=0.5(k+ zh). Adding the first two equations and subtracting the tripled third: 2 t -2 sh-2 x-φ=3. From this, it is clear that 2 t -2 sh -2 x is a non-negative number, as required. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 18,959 |
63. Given a polynomial $f(x)$ of degree 2000. The polynomial $f(x-1)$ has exactly 3400 roots, and the polynomial $f\left(1-x^{2}\right)$ has exactly 2700 roots. Prove that the distance between some two roots of $f(x)$ is less than 0.002. | 63. Let $f(x)$ have roots $c_{1}, \ldots, c_{k}$, where $k \leqslant 2000$. Since $f\left(x^{2}-1\right)$ has exactly 3400 roots, and the equation $x^{2}-1=c_{i}$ has no more than two solutions, then no more than $k-1700$ roots of $f$ lie outside $[-1, \infty)$ - the set of values of $x^{2}-1$. Similarly, since $f\left... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,960 |
64. On the board, 100 different natural numbers are written. To each of these numbers, the GCD of all the others is added. Could it be that among the 100 numbers obtained as a result of these actions, there are three identical ones? | 64. Answer: they could not. Suppose that the numbers $a<b<c$, initially written on the board, turned into three identical numbers. Note that the GCD, added to the number $a$, is a divisor of the numbers $b$ and $c$, and therefore, of their difference $c-b$. Hence, it does not exceed $c-b$, and thus is certainly less th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,961 |
65. At the beginning of the game, Little Boy and Karlson have one piece of chocolate in the form of a $2019 \times 2019$ square grid. Each move, Little Boy divides some piece along the grid lines into three rectangular pieces, and Karlson eats one of these three pieces at his choice. The game ends when no more moves ca... | 65. Answer: Karlson will win. Let's call a piece of chocolate big if it can be cut, and small if it cannot. Initially, there is only one big piece, and at the end of the game, there are 0. Karlson can play in such a way that the parity of the number of big pieces after his move will definitely change: one big piece is ... | Karlsonwins | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,962 |
66. An isosceles triangle \(ABC\) with a perimeter of 12 is inscribed in a circle \(\omega\). Points \(P\) and \(Q\) are the midpoints of the arcs \(ABC\) and \(ACB\) respectively. The tangent to the circle \(\omega\) at point \(A\) intersects the ray \(PQ\) at point \(R\). It turns out that the midpoint of segment \(A... | 66. Answer: 4. Let $I_{A}, I_{B}, I_{C}$ be the centers of the excircles of triangle $ABC$, touching sides $BC, CA$, and $AB$ respectively. Then the lines $A I_{A}, B I_{B}, C I_{C}$ will be the angle bisectors of triangle $ABC$, and the lines $I_{B} I_{C}, I_{C} I_{A}, I_{A} I_{B}$ will be its external angle bisectors... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,963 |
67. Baron Münchhausen has a set of 1000 weights of different integer values, with $2^{1000}$ weights of each value. The baron claims that if you take one weight of each value, the total weight of these 1000 weights will be less than $2^{1010}$, and this weight cannot be achieved by any other combination of weights from... | 67. Answer: Yes, they do exist. Let $k=1000$. We will prove that the set of weights $a_{k-1}=2^{k}-2^{k-1}$, $a_{k-2}=2^{k}-2^{k-2}, \ldots, a_{0}=2^{k}-2^{0}$ works. The sum of their weights is $S=k \cdot 2^{k}-2^{k-1}-\cdots-2^{1}-2^{0}=(k-1) \cdot 2^{k}+1$.
For $k=1000$, it is clear that $s<2^{1010}$. Moreover, $s ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,964 |
68. Let there be roads $A B$ and $C D$ between cities $A, B, C$, and $D$, but no roads $B C$ and $A D$. We will call a reconstruction the replacement of the pair of roads $A B$ and $C D$ with the pair of roads $B C$ and $A D$. Initially, in the country, there were several cities, some pairs of which were connected by r... | 68. Consider the set $M$ consisting of all possible 100-regular graphs on a given set of vertices $V$ (our two road schemes are among them). We will prove that any two graphs in $M$ can be transformed into each other through a series of reconstructions. For two graphs $G, G' \in M$, let $F(G, G')$ be the set of non-com... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,965 |
69. Triangle $ABC$ is inscribed in circle $\omega$ with center $O$. Line $AO$ intersects circle $\omega$ again at point $A'$. $M_B$ and $M_C$ are the midpoints of sides $AC$ and $AB$ respectively. Lines $A'M_B$ and $A'M_C$ intersect circle $\omega$ again at points $B'$ and $C'$, and intersect side $BC$ at points $D_B$ ... | 69. Let's perform a rotation centered at point $O$, which maps $B^{\prime}$ to $A$, and denote the image of point $C$ under this rotation as $B^{\prime \prime}$. Let $X$ be the intersection point of the lines $A A^{\prime}$ and $B B^{\prime \prime}$. From the equality of arcs $A B^{\prime \prime}$ and $B^{\prime} C$, i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,966 |
49. What is the maximum number of solutions that the equation $\max \left\{a_{1} x+b_{1}, \ldots, a_{10} x+b_{10}\right\}=0$ can have, if $a_{1}, \ldots, a_{10}, b_{1}, \ldots, b_{10}$ are real numbers, and all $a_{i}$ are not equal to 0? | 49. Answer: 2 solutions. For example, 5 functions $-x-1$ and 5 functions $x-1$.
Suppose this equation has three roots: $u<v<w$. At point $v$, one of the linear functions $a_{i} x+b_{i}$ is zero. On the other hand, its values at points $u$ and $w$ do not exceed zero. However, such a linear function can only be a consta... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,967 |
50. We will call a triplet of natural numbers $a, b, c$ interesting if $\left(a^{2}+1\right)\left(b^{2}+1\right)$ is divisible by $c^{2}+1$, but neither of the two factors is divisible by $c^{2}+1$.
Given an interesting triplet $a, b, c$. Prove that there exist natural numbers $u, v$, such that the triplet $u, v, c$ i... | 50. If the product $\left(a^{2}+1\right)\left(b^{1}+1\right)$ is divisible by $c^{2}+1$, then $c^{2}+1$ can be factored into a product of two factors $X Y$ such that $a^{2}+1$ is divisible by $X$, and $b^{2}+1$ is divisible by $Y$. (For this, it is sufficient, for example, to set $X=\left(a^{2}+1, c^{2}+1\right)$.) We ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,968 |
51. On the side $AD$ of the convex quadrilateral $ABCD$ with an acute angle $B$, a point $E$ is marked. It is known that $\angle CAD = \angle ADC = \angle ABE = \angle DBE$. Prove that $BE + CE < AD$. | 51. Let $\alpha$ be the angle $C A D$ and the angles equal to it. Let $C^{\prime}$ be the point symmetric to point $C$ with respect to the line $A D$. Then $A C D C^{\prime}$ is a rhombus. We write the sum of the angles of triangle $A D C^{\prime}: 180^{\circ}=\angle A C^{\prime} D+2 \alpha=$ $\angle A C^{\prime} D+\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,969 |
52. In the table, there are 25 columns and 300 rows, and Kostya painted all its cells in three colors. Then Lesha, looking at the table, names one of the three colors for each row and marks all the cells of this color in that row. (If there are no cells of the specified color in the row, he marks nothing in it.) After ... | 52. Answer: Two columns. To leave at least two columns, Lesha must for each row name a color that does not appear in the first two cells of that row. With this strategy, the first two columns will not be crossed out.
Note now that $300=C_{25^{2}}$. Using this observation, we can associate each row with its own pair of... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,970 |
53. The point $I_{a}$ is the center of the excircle of triangle $ABC$, touching side $BC$ at point $X$, and point $A^{\prime}$ is diametrically opposite to point $A$ on the circumcircle of this triangle. On the segments $I_{A} X, B A^{\prime}, C A^{\prime}$, points $Y, Z, T$ are chosen respectively such that $I_{A} Y=B... | 53. We will prove that the perpendicular bisectors of segments $YZ$, $YT$, and $YX$ intersect at one point, which is the center of the desired circle.
From the condition, it immediately follows that $\angle ABZ = 90^\circ$. Additionally, if $I$ is the incenter of the inscribed circle, then $\angle IBI_a = 90^\circ$. F... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,971 |
54. On the coordinate plane, points are marked: $(1,1),(2,3),(4,5),(999,111)$. If a point $(a, b)$ is marked, then points ( $b, a$ ) and ( $a-b, a+b$ ) can also be marked; if points $(a, b)$ and $(c, d)$ are marked, then the point $(a d+b c, 4 a c-4 b d)$ can be marked. Will it be possible to mark a point on the line $... | 54. Answer: it will not succeed. Let's associate each point $(a, b)$ with the number $f(a, b)=a^{2}+b^{2}$. It is easy to verify that $f(b, a)=f(a, b)$ and $f(a-b, a+b)=2 f(a, b)$, and $f(a d+b c, 4 a c-4 b d) \equiv f(a, b) f(c, d)(\bmod 5)$. For each of the initial points, the value of $f$ is not divisible by 5. Mult... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,972 |
55. In a graph with 400 vertices, for any edge $A B$, let's call the set of all edges emanating from vertices $A$ and $B$ (including the edge $A B$ itself) a "catshark". Each edge in the graph is labeled with either 1 or -1. It is known that the sum of the numbers on the edges of any catshark is greater than or equal t... | 55. Let's call the weight of a vertex $A$ the sum of the numbers on all edges outgoing from it; denote this number by $v(A)$. Note that the sum of the weights of all vertices in the graph is exactly twice the sum of the numbers on all its edges. It is sufficient to prove that the sum of all weights is not less than -20... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,973 |
49. Prove that for each natural number $N$ there exists an integer $k \geqslant 0$ such that $N$ can be written as a sum of the numbers $2^{0}, 2^{1}, 2^{2}, \ldots, 2^{k}$, each of which appears in this sum 1 or 2 times. (For example, $\left.12=2^{0}+2^{0}+2^{1}+2^{2}+2^{2}.\right)$
(M. Antipov) | 49. Let's write the number $N+1$ in binary. Suppose the largest power of two in this expansion is $2^{k+1}, k \geqslant 0$. By replacing this term with the sum
$$
2^{0}+2^{1}+\ldots+2^{k}=2^{k+1}-1
$$
we obtain the expansion of the number $N$, in which each of the terms $2^{0}, 2^{1}, \ldots, 2^{k}$ will appear 1 or ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,974 |
50. Given an odd natural number $n>1$. On the board, the numbers $n, n+1, n+2, \ldots, 2 n-1$ are written. Prove that one of them can be erased so that the sum of the remaining numbers is not divisible by any of the remaining numbers. | 50. Let $S$ be the sum of all numbers in the given set.
For any two numbers $a$ and $b$ from this set, we draw an arrow from $a$ to $b$ if $(S-a)$ is divisible by $b$. If the statement of the problem is false, then from each number $a$ there is at least one arrow to a number $b \neq a$. Moreover, it is easy to see tha... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,975 |
51. Given an acute triangle $ABC$. On the segment $AC$ and on the extension of side $BC$ beyond point $C$, variable points $X$ and $Y$ are chosen such that $\angle ABX + \angle CXY = 90^\circ$. Point $T$ is the projection of point $B$ onto the line $XY$. Prove that all such points $T$ lie on one line.
(S. Berlov)
(S.... | 51. Let point $H$ be the foot of the altitude dropped from point $B$ to side $AC$. Extend the right triangle $AHB$ to form a rectangle $AHBK$. We will show that all points $T$ lie on the line $KH$.
To do this, we take a point $T'$ on it and draw a line through $T'$ perpendicular to $BT'$. Let this line intersect side ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,976 |
52. On a circular necklace, there are $n>3$ beads, each painted in red or blue. If ${ }^{B}$ the neighboring beads of a certain bead are painted the same, it can be repainted (from red to blue or from blue to red). For which $n$ can the necklace be made such that all beads are painted the same color from any initial co... | 52. Answer: for all odd $n$.
We will prove that with an odd length of the necklace, all beads can be made the same color. Any necklace can be divided into blocks of consecutive beads of the same color. Among them, there must be a block of odd length (let's say red). First, we repaint all the beads with even numbers in... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,977 |
53. Can a triangle $ABC$ be drawn on a plane and two points $X$ and $Y$ be marked on the same plane such that
$$
\begin{aligned}
& A X=B Y=A B \\
& B X=C Y=B C \\
& C X=A Y=C A ?
\end{aligned}
$$
(M. Ivanov) | 53. Answer: Yes, it is an isosceles triangle $30^{\circ}, 75^{\circ}, 75^{\circ}$, and points $X$ and $Y$ are obtained by reflecting the vertices of the base of this triangle across the opposite sides.
That the example fits is obvious. | 30,75,75 | Geometry | proof | Yes | Yes | olympiads | false | 18,978 |
54. Given two odd natural numbers $a$ and $b$. Prove that there exists a natural number $k$ such that at least one of the numbers $b^{k}-a^{2}$ and $a^{k}-b^{2}$ is divisible by $2^{2018}$. | 54. We will solve the generalized problem. Given a natural number $n$ and two odd natural numbers $a$ and $b$. Prove that there exists a natural number $k$ such that at least one of the numbers $b^{2 k}-a^{2}$ and $a^{2 k}-b^{2}$ is divisible by $2^{n}$.
We will use the following known statement: if the number $c-1$ g... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,979 |
55. In a $10 \times 10$ grid (the sides of the cells have a unit length), $n$ cells were chosen, and in each of them, one of the diagonals was drawn and an arrow was placed on this diagonal in one of two directions. It turned out that for any two arrows, either the end of one coincides with the beginning of the other, ... | 55. Answer: when $n=48$.
For each arrow, consider the three-cell corner obtained by removing from a $2 \times 2$ square, centered at the end of the arrow, a $1 \times 1$ square, the diagonal of which is this arrow. Note that such corners do not intersect and are contained within a $12 \times 12$ square. Therefore, the... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,980 |
42. There are 300 people sitting in a circle: some of them are knights, and the rest are liars. Anton asked each of them: "How many liars are among your neighbors?" and added up the numbers he received. Then Anya did the same. When answering the question, knights always tell the truth, while liars always lie, but they ... | 42. Answer: no, this difference cannot be greater than 400.
Clearly, each knight gave two identical answers, so the difference between the sums given to Anton and Anya could only arise because some liars gave different answers to Anton and Anya. The difference in any liar's answers can only be 1 or 2.
Let there be $R... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,981 |
43. Inside angle $ABC$, which is equal to $105^{\circ}$, a point $X$ is marked such that $\angle CBX=70^{\circ}$ and $BX=BC$. On segment $BX$, a point $Y$ is chosen such that $BY=BA$. Prove that $AX + AY \geqslant CY$. | 43. Let's mark point $A^{\prime}$, symmetric to point $A$ with respect to line $B X$. Then $\angle A B X=\angle X B A^{\prime}=\angle A^{\prime} B C=35^{\circ}$ and $A^{\prime} Y=A Y$. Note that triangles $A B X$ and $A^{\prime} B C$ are equal by two sides and an angle, from which $A X=A^{\prime} C$. Therefore, $A X+A ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,982 |
44. Vasya placed numbers from 1 to $101^{2}$ once each in all the cells of a $101 \times 101$ board. Petya chooses a cell on the board, places a token on it, and wants to make as many moves as possible so that the number under the token keeps increasing. In one move, Petya can move the token to any cell within a $5 \ti... | 44. Answer: 8 moves (you can visit 9 cells).
In any $3 \times 3$ square, regardless of the arrangement of numbers, all cells can be visited, which gives us 8 moves. We will color the board i... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,983 |
45. A natural number is written on the board. Every minute, the sum of the first 100 digits of the number on the board is added to it. Prove that after some time, three times in a row, a number that is not divisible by 3 will be obtained. | 45. Let's choose a natural number $n>103$, which is also greater than the number of digits in the original number. Clearly, each minute no more than 900 is added to the number on the board. Therefore, at some point, a number from the interval from $\underbrace{10 \ldots 0000}_{n}$ to $\underbrace{10 \ldots 0900}_{n}$ w... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,984 |
46. In a convex quadrilateral $A B C D$, point $M$ is the midpoint of side $A D$, $C M \| A B$, $A D=B D$ and $3 \angle B A C=\angle A C D$. Find the angle $A C B$.
(S. Berlov) | 46. Answer: $\angle A C B=90^{\circ}$.
Notice that $\angle B A C=\angle A C M$, hence $\angle D C M=\angle A C D-\angle A C M=2 \angle A C M$. Let $N$ be the midpoint of segment $A B$. Then $D N$ is the median, bisector, and altitude of the isosceles triangle $A B D$. Next, point $C$ lies on the midline of this triang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,985 |
48. In the country, there are 600 cities, some pairs of which are connected by roads, and from any city, you can reach any other city via these roads. For any two cities $A$ and $B$ connected by a road, there will be two more cities $C$ and $D$ such that any two of these four cities are connected by a road. Prove that ... | 48. Let's build a graph: cities are vertices, roads are edges. This graph is connected. We need to construct a spanning tree with fewer than 100 vertices of degree 2. We will build the tree by adding new vertices in such a way that, if possible, no new vertices of degree 2 appear. We start with an arbitrary vertex (its... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,987 |
29. In a notebook, all natural divisors of a natural number $a$ were written down, and then all natural divisors of a natural number $b$. As a result, an even number of numbers were recorded in the notebook. Katya divided all these numbers into pairs and calculated the product of the numbers in each pair. (For example,... | 29. Suppose that the numbers $a$ and $b$ are different: let, for example, $a > b$. Among the numbers written on the board, there are two ones: one is a divisor of $a$, the other is a divisor of $b$. Since the products in all of Katya's pairs are equal, the two smallest numbers (the ones) must be paired with the two lar... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,988 |
30. What is the minimum number of colors needed to color the cells of a $5 \times 5$ square so that among any three consecutive cells in a row, column, or diagonal, there are no cells of the same color?
(M. Antipov) | 30. Answer: in five colors.
Consider a cross of five cells, the central cell of which coincides with the central cell of the square. By the condition, three cells in its column have different colors. Similarly, the colors of the three cells in its row are also different. Finally, any two cells at the "ends" of the cro... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,989 |
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