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1. A necklace consists of 30 blue and a certain number of red beads. It is known that on both sides of each blue bead there are beads of different colors, and one bead away from each red bead there are also beads of different colors. How many red beads can be in this necklace? (The beads in the necklace are arranged cy... | Answer: 60.
Solution. It is obvious that blue beads appear in the necklace in pairs, separated by at least one red bead. Let there be $n$ red beads between two nearest pairs of blue beads. We will prove that $n=4$. Clearly, $n \leqslant 4$, since the middle one of five consecutive red beads does not satisfy the condit... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 175 beads of red, blue, and green colors. It is known that each red bead has neighbors of different colors, and on any segment of the necklace between two green beads, there is at least one blue bead. What is the minimum number of blue beads that can be in this necklace? (The beads in the neck... | Answer: 30.
Solution 1. We will show that any block of six consecutive beads contains a blue bead. We can assume that there is no more than one green bead in it, otherwise there is nothing to prove. If the block contains 5 red beads, then at least 3 of them are consecutive, and the middle one does not satisfy the prob... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given a right triangle $ABC$ with a right angle at $C$. On its leg $BC$ of length 26, a circle is constructed with $BC$ as its diameter. A tangent $AP$ is drawn from point $A$ to this circle, different from $AC$. The perpendicular $PH$, dropped from point $P$ to segment $BC$, intersects segment $AB$ at point $Q$. Fi... | Answer: 24.
Solution. Let $O$ be the center of $\omega$. Note that
$$
B H=\frac{4}{13} B C=8, \quad C H=18, \quad O H=\frac{1}{2} B C-B H=5, \quad P H=\sqrt{O P^{2}-O H^{2}}=12
$$
Right tr... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 100 beads of red, blue, and green colors. It is known that among any five consecutive beads, there is at least one blue one, and among any seven consecutive beads, there is at least one red one. What is the maximum number of green beads that can be in this necklace? (The beads in the necklace ... | Answer: 65.
Solution. Let there be a set of beads $A$ such that in every set of $n$ consecutive beads, there is at least one from $A$. We will show that $A$ contains no fewer than $\frac{100}{n}$ elements. Indeed, between any two adjacent beads from $A$, there are no more than $n-1$ beads. If the set $A$ contains $m$ ... | 65 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. For a natural number ending not in zero, one of its digits (not the most significant) was erased. As a result, the number decreased by 9 times. How many numbers exist for which this is possible? | Answer: 28.
Solution. Let's represent the original number in the form $m+10^{k} a+10^{k+1} n$, where $a$ is a decimal digit, and $k, m, n$ are non-negative integers, with $m>0$. By erasing the digit $a$, we get the number $m+10^{k} n$. According to the condition,
$$
m+10^{k} a+10^{k+1} n=9\left(m+10^{k} n\right) \Lon... | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A square $4 \times 4$ is divided into 16 squares $1 \times 1$. We will call a path a movement along the sides of the unit squares, in which no side is traversed more than once. What is the maximum length that a path connecting two opposite vertices of the large square can have? | Answer: 32.
Solution. Let's call the sides of the $1 \times 1$ squares edges, the vertices of these squares nodes, and the number of edges adjacent to a node the multiplicity of the node. Notice that the $1 \times 1$ squares generate 40 distinct edges. If a path passes through a node of multiplicity 3, it enters the n... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A string is threaded with 150 beads of red, blue, and green. It is known that among any six consecutive beads, there is at least one green, and among any eleven consecutive beads, there is at least one blue. What is the maximum number of red beads that can be on the string? | Answer: 112.
Solution. We can choose $\left[\frac{150}{11}\right]=13$ consecutive blocks of 11 beads each. Since each block contains at least one blue bead, there are at least 13 blue beads on the string. In addition, we can group all the beads into 25 consecutive blocks of 6 beads each. Each block contains at least o... | 112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. A rectangle $3 \times 5$ is divided into 15 squares $1 \times 1$. We will call a path a movement along the sides of the unit squares, such that no side is traversed more than once. What is the maximum length that a path connecting two opposite vertices of the rectangle can have? | Answer: 30.
Solution. Let the rectangle be denoted as $A B C D$, and let the path connect its vertices $A$ and $C$. We will call the sides of the $1 \times 1$ squares edges, the vertices of these squares - nodes, and the number of edges adjacent to a node - the multiplicity of the node. Note that the $1 \times 1$ squa... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 50 blue and a certain number of red beads. It is known that in any segment of the necklace containing 8 blue beads, there are at least 4 red ones. What is the minimum number of red beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one is a... | Answer: 29.
Solution. Note that any segment of the necklace consisting of 11 beads contains no more than 7 blue and no fewer than 4 red beads (otherwise, it would contain 8 blue beads and no more than 3 red ones). Fix a red bead in the necklace. The 7 consecutive segments of 11 beads adjacent to it do not cover the en... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 100 red and a certain number of blue beads. It is known that in any segment of the necklace containing 10 red beads, there are at least 7 blue ones. What is the minimum number of blue beads that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning the last one i... | Answer: 78.
Solution. Note that any segment of the necklace containing 16 beads has no more than 9 red and no fewer than 7 blue beads (otherwise, it would contain 10 red beads and no more than 6 blue ones). Fix a blue bead in the necklace. The 11 consecutive segments of 16 beads adjacent to it do not cover the entire ... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 50 blue, 100 red, and 100 green beads. We will call a sequence of four consecutive beads good if it contains exactly 2 blue beads and one each of red and green. What is the maximum number of good quartets that can be in this necklace? (The beads in the necklace are arranged cyclically, meaning... | Answer: 99.
Solution. Blue beads make up one fifth of the total. Therefore, there will be two consecutive blue beads (let's call them $a$ and $b$), separated by at least three beads. Note that $a$ and $b$ are part of no more than three good quartets, while the other blue beads are part of no more than four. If we add ... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A thread is strung with 75 blue, 75 red, and 75 green beads. We will call a sequence of five consecutive beads good if it contains exactly 3 green beads and one each of red and blue. What is the maximum number of good quintets that can be on this thread? | Answer: 123.
Solution. Note that the first and last green beads are included in no more than three good fives, the second and second-to-last - in no more than four fives, and the rest - in no more than five fives. If we add these inequalities, we get $2 \cdot 3 + 2 \cdot 4 + 71 \cdot 5 = 369$ on the right side, and th... | 123 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the table, there are three cones standing on their bases, touching each other. The radii of their bases are $2 r, 3 r$, and $10 r$. A frustum of a cone is placed on the table with its smaller base down, and it shares a common generatrix with each of the other cones. Find $r$ if the radius of the smaller base of t... | Answer: 29.
Solution. Let $C$ be the center of the smaller base of the truncated cone, and $O_{1}, O_{2}, O_{3}$ be the centers of the bases of the other cones, with $R=15$. Denote by $\math... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A necklace consists of 80 beads of red, blue, and green colors. It is known that on any segment of the necklace between two blue beads, there is at least one red bead, and on any segment of the necklace between two red beads, there is at least one green bead. What is the minimum number of green beads that can be in ... | Answer: 27.
Solution. If the blue beads are arranged in a circle, the number of pairs of adjacent beads is equal to the number of beads. Since there is a red bead between any two blue beads, there are no fewer red beads in the necklace than blue ones. Similarly, it can be proven that there are no fewer green beads tha... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the cells of an $80 \times 80$ table, pairwise distinct natural numbers are placed. Each of them is either a prime number or a product of two prime numbers (possibly the same). It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. ... | Answer: 4266.
Solution. We will say that a composite number $a$ serves a prime number $p$ if $a$ and $p$ are not coprime (i.e., $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than two distinct prime divisors, it serves n... | 4266 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Around a round table, 50 schoolchildren are sitting: blondes, brunettes, and redheads. It is known that in any group of schoolchildren sitting in a row, between any two blondes there is at least one brunette, and between any two brunettes - at least one redhead. What is the minimum number of redheads that can sit at... | Answer: 17.
Solution. If only blondes were sitting at the table, the number of pairs of neighbors would be equal to the number of blondes. Since there is a brunette between any two blondes, there are no fewer brunettes than blondes sitting at the table. Similarly, it can be proven that there are no fewer redheads than... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the cells of a $75 \times 75$ table, pairwise distinct natural numbers are placed. Each of them has no more than three distinct prime divisors. It is known that for any number $a$ in the table, there is a number $b$ in the same row or column such that $a$ and $b$ are not coprime. What is the maximum number of pri... | Answer: 4218.
Solution. We will say that a composite number $a$ serves a prime number $p$ if the numbers $a$ and $p$ are not coprime (that is, $a$ is divisible by $p$). For each prime number in the table, there is a composite number that serves it. Since each composite number has no more than three distinct prime divi... | 4218 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) We will say that a number has the form $\overline{a b a}$ if its first and third digits are the same; the second must be different. For example, 101 and 292 have this form, while 222 and 123 do not. Similarly, we define the form of the number $\overline{a b c a b d}$. How many odd numbers of the form $\o... | Answer: 448.
Solution. Odd numbers divisible by 5 are numbers ending in 5, so for $d$ we have only one option. For $a$ we have 8 options, as the number cannot start with zero, and $a$ cannot be equal to $d$. The digit $b$ cannot be equal to $a$ or $d$, and there are no other restrictions on it - we get 8 possible valu... | 448 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) The numbers $s_{1}, s_{2}, \ldots, s_{1008}$ are such that their sum is $2016^{2}$. It is known that
$$
\frac{s_{1}}{s_{1}+1}=\frac{s_{2}}{s_{2}+3}=\frac{s_{3}}{s_{3}+5}=\ldots=\frac{s_{1008}}{s_{1008}+2015}
$$
Find $s_{17}$. | Answer: 132.
Solution. Note that none of the $s_{i}$ is equal to zero (otherwise, all the fractions $\frac{s_{i}}{s_{i}+2 i-1}$ would be equal to zero, and, consequently, all $s_{i}$ would have to be equal to zero, which contradicts the fact that their sum is $2016^{2}$). Therefore, the original condition is equivalen... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell. | Answer: 416.
Solution. A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will be no fewer than 416 such cells.
We will sh... | 416 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In some cells of a $1 \times 2021$ strip, one chip is placed in each. For each empty cell, the number equal to the absolute difference between the number of chips to the left and to the right of this cell is written. It is known that all the written numbers are distinct and non-zero. What is the minimum number of ch... | Answer: 1347.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 1347 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In some cells of a $1 \times 2100$ strip, one chip is placed. In each of the empty cells, a number is written that is equal to the absolute difference between the number of chips to the left and to the right of this cell. It is known that all the written numbers are distinct and non-zero. What is the minimum number ... | Answer: 1400.
Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}... | 1400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given an acute scalene triangle $A B C$. In it, the altitudes $B B_{1}$ and $C C_{1}$ are drawn, intersecting at point $H$. Circles $\omega_{1}$ and $\omega_{2}$ with centers $H$ and $C$ respectively touch the line $A B$. From point $A$ to $\omega_{1}$ and $\omega_{2}$, tangents other than $A B$ are drawn. Denote th... | Answer: $180^{\circ}$.
Solution. Since right triangles $A C C_{1}$ and $A C E$ are equal by leg and hypotenuse, line $C A$ is the bisector of isosceles triangle $C_{1} C E$. Therefore, it i... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A circle is circumscribed around an acute-angled triangle $A B C$. Point $K$ is the midpoint of the smaller arc $A C$ of this circle, and point $L$ is the midpoint of the smaller arc $A K$ of this circle. Segments $B K$ and $A C$ intersect at point $P$. Find the angle between the lines $B C$ and $L P$, given that $B... | Answer: $90^{\circ}$.
Solution. Let $\angle A B L=\varphi$. Then $\angle K B L=\varphi$ and $\angle K B C=2 \varphi$. Since $B K=B C$, we get $\angle B K C=\angle B C K=90^{\circ}-\varphi$.... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Circles $\omega_{1}$ and $\omega_{2}$ intersect at points $A$ and $B$, and a circle with center at point $O$ encompasses circles $\omega_{1}$ and $\omega_{2}$, touching them at points $C$ and $D$ respectively. It turns out that points $A, C$, and $D$ lie on the same line. Find the angle $A B O$. | Answer: $90^{\circ}$.
Solution. Let $O_{1}$ and $O_{2}$ be the centers of circles $\omega_{1}$ and $\omega_{2}$. Triangles $C O D$, $C O_{1} A$, and $A O_{2} D$ are isosceles. Since points ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $A$. Segment $O_{2} A$ intersects circle $\omega_{1}$ again at point $K$, and segment $O_{1} A$ intersects circle $\omega_{2}$ again at point $L$. The line passing through point $A$ parallel to $K L$ intersects cir... | Answer: $90^{\circ}$.
Solution. Triangles $O_{1} A K$ and $O_{2} A L$ are isosceles and have a common base angle. Therefore, their vertex angles are equal. Let the common value of these ang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Circles $\omega_{1}$ and $\omega_{2}$ with centers $O_{1}$ and $O_{2}$ respectively intersect at point $B$. The extension of segment $O_{2} B$ beyond point $B$ intersects circle $\omega_{1}$ at point $K$, and the extension of segment $O_{1} B$ beyond point $B$ intersects circle $\omega_{2}$ at point $L$. The line pa... | Answer: $90^{\circ}$.
Solution. Since $\angle O_{1} B K = \angle O_{2} B L$, the isosceles triangles $O_{1} B K$ and $O_{2} B L$ are similar. Therefore, their vertex angles are equal. Let th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on a $10 \times 11$ board such that any placement of a three-cell corner on the board touches at least one marked cell? | Answer: 50.
Solution. It is not hard to notice that in any $2 \times 2$ square, there are at least two marked cells. Since 25 such squares can be cut out from a $10 \times 11$ board, there must be no fewer than 50 marked cells in it. An example with 50 marked cells is obtained if the cells with an even first coordinat... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A warehouse stores 400 tons of cargo, with the weight of each being a multiple of a centner and not exceeding 10 tons. It is known that any two cargos have different weights. What is the minimum number of trips that need to be made with a 10-ton truck to guarantee the transportation of these cargos from the warehous... | Answer: 51.
Solution. We will show that it is always possible to transport the goods in 51 trips, even if the warehouse contains all weights from 1 to 100 tons. Indeed, we can divide all the goods, except for the 50-ton and 100-ton ones, into 49 pairs as follows:
$$
(1,99), \quad(2,98), \quad(3,97), \quad \ldots, \qu... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on a $9 \times 9$ board such that any placement of a three-cell corner piece will touch at least two marked cells? | Answer: 56.
Solution. It is not hard to notice that in any $2 \times 2$ square, at least three cells must be marked, and in each $1 \times 2$ rectangle, at least two cells must be marked. Since from a $9 \times 9$ board, 16 $2 \times 2$ squares and 8 $1 \times 2$ rectangles can be cut out, at least $16 \cdot 3 + 8 = 5... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A cinema was visited by 50 viewers, the total age of whom is 1555 years, and among them, there are no viewers of the same age. For what largest $k$ can we guarantee to select 16 viewers whose total age is not less than $k$ years? | Answer: 776.
Solution. We will show that $k \geqslant 776$, that is, the total age of the 16 oldest viewers is always not less than 776. Arrange the viewers in order of increasing age, and let $a_{i}$ be the age of the $i$-th viewer. Since there are no viewers of the same age, we get
$$
a_{1} \leqslant a_{2}-1 \leqsl... | 776 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on a $12 \times 12$ board such that any placement of a four-cell figure $\square \square$ on the board touches at least one marked cell? (The figure can be rotated and flipped.) | Answer: 48.
Solution. It is not hard to notice that in any $2 \times 3$ rectangle, there are at least two marked cells. Since the $12 \times 12$ board is divided into 24 such rectangles, there must be no fewer than 48 marked cells. An example with 48 marked cells can be obtained by marking the cells where the sum of t... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In the stands of the hockey arena, there are several rows with 168 seats in each row. For the final match, 2016 students from several sports schools were invited as spectators, with no more than 40 from each school. Students from any school must be seated in one row. What is the minimum number of rows that must be i... | Answer: 15.
Solution. Suppose that 57 schools sent 35 students each to the final match, and one school sent 21 students. Since only four groups of 35 students can fit in one row, the number of rows required to accommodate the students from 57 schools should be no less than $\frac{57}{4}=14 \frac{1}{4}$, which means at... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given a natural number $x=9^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 13. Find $x$. | Answer: 728.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 13. The remainders of the powers of 9 when divided by 13 are $9, 3, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 13 means ... | 728 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. For what smallest $k$ can $k$ cells be marked on an $8 \times 9$ board such that for any placement of a four-cell figure on the board, it can be rotated and flipped. | Answer: 16.
Solution. It is not hard to notice that in any $2 \times 4$ rectangle, there are at least two marked cells. Since from an $8 \times 9$ board, 8 non-overlapping $2 \times 4$ rectangles can be cut out, there must be at least 16 marked cells in it. An example with 16 marked cells is shown in the figure.
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. A warehouse stores 1500 tons of various goods in containers. The weight of any container is a multiple of a ton and does not exceed $k$ tons. A train consisting of 25 platforms, each with a load capacity of 80 tons, has been dispatched to the warehouse. For what maximum $k$ can this train guarantee the transportatio... | Answer: 26.
Solution. If $k=27$, then it is not always possible to transport all the goods. Indeed, suppose there are 55 containers, each weighing 27 tons, and one weighing 15 tons. Only two containers of 27 tons each can fit on one platform. Therefore, to transport 55 such containers, at least 28 platforms are requir... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Given a natural number $x=8^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 31. Find $x$. | Answer: 32767.
Solution. One of the simple divisors of $x$ is obviously 7. First, let's prove that a number of the form $y=8^{k}-1$ is a power of 7 only in the case $k=1$. Indeed, for $k>1$, write $y=a b$, where $a=2^{k}-1, b=2^{2 k}+2^{k}+1$. Since
$$
b=2^{2 k}-1+2^{k}-1+3=a\left(2^{k}+2\right)+3
$$
the numbers $a$... | 32767 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the theater, there are $k$ rows of seats. 770 spectators came to the theater and sat down (possibly not occupying all the seats). After the intermission, all the spectators forgot which seats they had and sat down differently. For what largest $k$ will there definitely be 4 spectators who sat in the same row both... | Answer: 16
Solution. If the audience is seated on 16 rows, then on some row there are no fewer than 49 spectators (otherwise, there would be no more than 48 spectators on each row, and the total would not exceed $16 \cdot 48=7683$, making it impossible to seat the spectators of this row after the intermission such tha... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given a natural number $x=9^{n}-1$, where $n$ is a natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 59048.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 11. First, let's prove that the number $3^{s}+1$ cannot be a power of two when $s>1$, and the number $3^{s}-1$ cannot be a power of two whe... | 59048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a natural number $x=9^{n}-1$, where $n-$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 61. Find $x$. | Answer: 59048.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 61. The remainders of the powers of 9 when divided by 61 are $9, 20, 58, 34, 1$ and then repeat cyclically. Thus, the divisibility of $x$ b... | 59048 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $ABC$, the median $AM$ is drawn. Circle $\alpha$ passes through point $A$, touches line $BC$ at point $M$, and intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively. On the arc $AD$ that does not contain point $E$, a point $F$ is chosen such that $\angle BFE = 72^{\circ}$. It turns out that $... | Answer: $36^{\circ}$.
Solution. Inscribed angles $D A F$ and $D E F$ subtend the arc $D F$ and are therefore equal. Given that $\angle F A B = \angle A B C$, it follows that $B C \parallel ... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given a natural number $x=7^{n}+1$, where $n$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 16808.
Solution. Since the number $x$ is even, one of its prime divisors is 2. Therefore, we need to find all $x$ that have exactly two odd prime divisors, one of which is 11. The divisibility of $x$ by 11 is equivalent to $n$ being an odd multiple of 5, that is, $x=7^{5m}+1$ for some odd $m$. From here, using... | 16808 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A circle $\omega$ is circumscribed around triangle $ABC$. A line tangent to $\omega$ at point $C$ intersects the ray $BA$ at point $P$. On the ray $PC$ beyond point $C$, a point $Q$ is marked such that $PC = QC$. The segment $BQ$ intersects the circle $\omega$ again at point $K$. On the smaller arc $BK$ of the circl... | Answer: $30^{\circ}$.
Solution. Inscribed angles $L A K$ and $L B K$ subtend the arc $L K$ and are therefore equal. Given that $\angle L B Q = \angle B Q C$, it follows that $B L \| P Q$. Th... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. What is the maximum number of vertices of a regular 2016-gon that can be marked so that no four marked vertices are the vertices of any rectangle? | Answer: 1009.
Solution. Note that an inscribed quadrilateral is a rectangle if and only if its diagonals are diameters of the circumscribed circle. The 2016-gon has exactly 1008 pairs of diametrically opposite vertices. If no rectangle can be formed from the marked vertices, then only in one pair can both vertices be ... | 1009 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given a natural number $x=6^{n}+1$, where $n-$ is an odd natural number. It is known that $x$ has exactly three distinct prime divisors, one of which is 11. Find $x$. | Answer: 7777.
Solution. The divisibility of $x$ by 11 is equivalent to $n$ being an odd multiple of 5, that is, $x=6^{5m}+1$ for some odd $m$. From here, using the formula for the sum of a geometric progression,
$$
x=p \cdot q, \quad \text{where } p=6^{m}+1, q=1-6^{m}+6^{2m}-6^{3m}+6^{4m}
$$
Note that $p$ and $q$ ar... | 7777 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) The rules of the game are as follows: from 64 different items, on each turn, the player must form a set of items that has not been mentioned in the game before, with the number of items equal to the player's age in years. Players take turns; either player can start the game; the player who cannot make a ... | Answer: 34 years
Solution: The game move is the selection of 64 elements from a certain subset, which contains as many elements as the player's age. Thus, the problem requires comparing the number of ways to make such a selection: the player with a greater number of ways to choose their "own" subset, i.e., $C_{64}^{x}... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) In math class, each dwarf needs to find a three-digit number without zero digits, divisible by 3, such that when 297 is added to it, the result is a number with the same digits but in reverse order. What is the minimum number of dwarfs that should be in the class so that among the numbers they find, ther... | Answer: 19.
Solution: Let the number of numbers that satisfy the condition of the problem be $N$. Then, by the Pigeonhole Principle, the minimum number of gnomes must be $N+1$.
Let's write a three-digit number as $\overline{x y z}$, where $x$ is the number of hundreds, $y$ is the number of tens, and $z$ is the number... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) Find the number of pairs of natural numbers $m$ and $n$, satisfying the equation $\frac{1}{m}+\frac{1}{n}=\frac{1}{2020}$. | Answer: 45.
Solution: Transform the given equation
\[
\begin{gathered}
\frac{1}{m}+\frac{1}{n}=\frac{1}{2020} \Leftrightarrow \frac{n+m}{m \cdot n}=\frac{1}{2020} \Leftrightarrow m n=2020(n+m) \Leftrightarrow \\
\Leftrightarrow m n-2020 n-2020 m+2020^{2}-2020^{2}=0 \Leftrightarrow(n-2020)(m-2020)=2020^{2}
\end{gather... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) The ballroom in the palace of the thirtieth kingdom is a region on the plane, the coordinates of which satisfy the conditions $|x| \leqslant 4,|y| \leqslant 6$. How many identical parquet tiles, having the shape of a rectangle with sides 1.5 and 2, are needed to tile the floor of the room? Tiling is cons... | Answer: 32.
Solution: From the condition, we find that the magic room is a rectangle with sides parallel to the coordinate axes, and the lengths of the sides parallel to the x-axis are 8, while the lengths of the sides parallel to the y-axis are 12. It is not hard to see that a tiling exists (along the side of the roo... | 32 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there are two tribes: the tribe of knights, who always tell the truth, and the tribe of liars, who always lie. On the main holiday, 2017 islanders sat around a large round table. Each islander said the phrase: "my neighbors are from the same tribe." It turned out that two liars made a mistake and accid... | Answer: 1344 liars
Solution. Note that no two knights can sit next to each other, meaning the neighbors of each knight are liars. Indeed, consider a chain of knights sitting in a row, surrounded by liars. Suppose there are at least two knights in this chain. The neighbors of the extreme knight are a knight and a liar,... | 1344 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Given various natural numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ and $a_{7}$. Prove that $\left(a_{1}-a_{2}\right)^{4}+\left(a_{2}-a_{3}\right)^{4}+\left(a_{3}-a_{4}\right)^{4}+\left(a_{4}-a_{5}\right)^{4}+\left(a_{5}-a_{6}\right)^{4}+\left(a_{6}-\right.$ $\left.a_{7}\right)^{4}+\left(a_{7}-a_{1}\right)^{4} ... | Solution. Note that all the terms are no less than 1. If one of the brackets is at least 3, then the entire sum is certainly greater than $3^{4}+1=82$. Therefore, we will assume from now on that the absolute value of any bracket does not exceed 2. Mark the numbers $a_{1}, a_{2}, \ldots, a_{7}$ on a line. Let $a_{j}$ be... | 82 | Inequalities | proof | Yes | Yes | olympiads | false |
5. What is the maximum number of rooks that can be placed on the cells of a $300 \times 300$ board so that each rook attacks no more than one other rook? (A rook attacks all cells it can reach according to chess rules, without passing through other pieces.)
# | # Answer: 400
First solution. We will prove that no more than 400 rooks can be placed on the board. In each row or column, there are no more than two rooks; otherwise, the rook that is not at the edge will attack at least two other rooks. Suppose there are $k$ columns with two rooks each. Consider one such pair. They ... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. What is the maximum number of chips that can be placed in the cells of a chessboard (no more than one chip per cell) so that no more than three chips are placed on each diagonal? | Answer: 38
Solution. Consider the eight-cell diagonal consisting of black cells, and the parallel diagonals consisting of black cells. These diagonals have $2, 4, 6, 8, 6, 4,$ and 2 cells. On each of the outermost diagonals, no more than two chips can be placed, and on each of the middle diagonals, no more than three.... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On an island, there live knights who always tell the truth, and liars who always lie. On the main holiday, 100 islanders sat around a large round table. Half of those present said: "both my neighbors are liars," and the rest said: "one of my neighbors is a liar." What is the maximum number of knights that can sit at... | # Answer: 67
Solution. Note that three knights cannot sit in a row, since then the middle knight would not be able to utter any of the required phrases. Let $k$ be the number of pairs of neighboring knights. Then each of the knights said the phrase "among my neighbors there is exactly one liar," so $2 k \leqslant 50$.... | 67 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. (20 points) Young marketer Masha was supposed to survey 50 customers in an electronics store throughout the day. However, there were fewer customers in the store that day. What is the maximum number of customers Masha could have surveyed, given that according to her data, 7 of the respondents made an impulse purchas... | Answer: 47.
Solution: Let the number of customers surveyed be $x$. Then, the number of customers who made a purchase under the influence of advertising is $(x-7) \cdot 3 / 4$, and the number of customers who made a purchase on the advice of a sales consultant is $(x-7)/4$. Since the number of customers can only be an ... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) The reception hall in the palace of the tritriandiat kingdom consists of points on the plane whose coordinates satisfy the condition $4|x|+5|y| \leqslant 20$. How many identical two-sided parquet tiles, shaped as right triangles with legs of 1 and 5/4, are needed to tile the floor of the hall? Tiling is ... | Answer: 64.
Solution: It is not hard to see that the reception hall is a rhombus with vertices at points $(-5,0),(0,4),(5,0)$ and $(0,-4)$, and each quarter of the hall (bounded by the coordinate axes and one of the sides, i.e., forming a right-angled triangle) is similar to one parquet tile with a similarity ratio of... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (40 points) At the "Horizon" base, 175 university students arrived. Some of them are acquainted with each other, while others are not. It is known that any six students can be accommodated in two three-person rooms such that all those in the same room are acquainted with each other. What is the minimum number of pai... | Answer: 15050.
Solution: It is clear that each student has no more than three unfamiliar students. If each student has no more than two unfamiliar students, then each is familiar with at least 172 students, and the total number of pairs of familiar students is no less than $175 \times 172 / 2 = 15050$. Suppose there i... | 15050 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) In the Martian calendar, a year consists of 5882 days, and each month has either 100 or 77 days. How many months are there in the Martian calendar? | Answer: 74.
Solution: Let $x$ be the number of months with 100 days, and $y$ be the number of months with 77 days. According to the problem, $100 x + 77 y = 5882$. It is obvious that $y \leqslant 66$, otherwise $x < 0$. Notice that
$$
x \bmod 11 = 100 x \quad \bmod 11 = 5882 \quad \bmod 11 = 8
$$
Thus, $x = 11 k + 8... | 74 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Professor K., wishing to be known as a wit, plans to tell no fewer than two but no more than three different jokes at each of his lectures. At the same time, the sets of jokes told at different lectures should not coincide. How many lectures in total will Professor K. be able to give if he knows 8 jokes? | Answer: 84.
Solution: The professor can use all possible triplets of anecdotes (the number of which is $C_{8}^{3}=56$), as well as all pairs of anecdotes (which is $C_{8}^{2}=28$). Therefore, the maximum number of sets of anecdotes the professor can use in lectures is $56+28=84$. | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) We will call a word any finite sequence of letters of the Russian alphabet. How many different five-letter words can be formed from the letters of the word САМСА? And from the letters of the word ПАСТА? In your answer, indicate the sum of the found numbers. | Answer: 90.
Solution: In the word SAMSA, the letter A appears twice and the letter S appears twice. Therefore, the number of different words will be $\frac{5!}{2!\cdot 2!}=30$. In the word PASTA, only the letter A appears twice. Therefore, the number of different words in this case will be $\frac{5!}{2!}=60$. In total... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. (40 points) The numbers from 1 to 600 are divided into several groups. It is known that if a group contains more than one number, then the sum of any two numbers in this group is divisible by 6. What is the minimum number of groups that can be formed? | Answer: 202.
Solution: For $k=0,1, \ldots, 5$, let $G_{k}$ be the set of numbers from 1 to 600 that give a remainder of $k$ when divided by 6. Suppose a number $a$ from $G_{k}$ is included in some group. If another number $b$ is also in this group, then it belongs to $G_{6-k}$, otherwise $a+b$ is not divisible by 6. S... | 202 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. (40 points) Five boys played a word game: each of them wrote down 7 different words. It turned out that each boy had exactly 2 words that were not found in any of the other boys' lists. What is the maximum number of different words that the boys could have written in total? | Answer: 22.
Solution: In total, $5 \times 7=35$ words were written. Since each boy wrote exactly 2 words that did not appear in any of the other boys' writings, there were a total of $5 \times 2=10$ such unique words. From the remaining 25 words (repeated), each was written at least twice. Therefore, there are no more... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. (40 points) In a store where all items cost a whole number of rubles, two special offers are in effect:
1) A customer who buys at least three items can choose one item as a gift (free of charge), the cost of which does not exceed the minimum cost of the paid items;
2) A customer who buys exactly one item for no les... | Answer: 504.
Solution: Let $S$ denote the total cost of the four items that interested the customer, and let $X$ be the minimum possible cost of the items. Being interested in exactly four items means that the customer can take advantage of either offer 1 or offer 2, but not both.
Let the items cost $s_{1}, s_{2}, s_... | 504 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Triangle $A B C$ with angle $\angle A B C=135^{\circ}$ is inscribed in circle $\omega$. The lines tangent to $\omega$ at points $A$ and $C$ intersect at point $D$. Find $\angle A B D$, given that $A B$ bisects segment $C D$. Answer: $90^{\circ}$
=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secan... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) In math class, each dwarf needs to find a three-digit number such that when 198 is added to it, the result is a number with the same digits but in reverse order. For what maximum number of dwarfs could all the numbers they find be different? | Answer: 70.
Solution: By the Pigeonhole Principle, the maximum number of gnomes is equal to the number of numbers that satisfy the condition of the problem.
Let the three-digit number be denoted as $\overline{x y z}$, where $x$ is the hundreds digit, $y$ is the tens digit, and $z$ is the units digit. Since the number... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (40 points) On an island, there live only 50 knights, who always tell the truth, and 15 commoners, who can either tell the truth or lie. A scatterbrained professor, who came to the island to give a lecture, forgot what color hat he was wearing. How many of the local residents should the professor ask about the color... | Answer: 31.
Solution: Since the professor will only ask about the color of the hat, to accurately determine the color of the hat, it is necessary to survey more knights than commoners. In the worst case, among those surveyed, there could be all the commoners on the island, i.e., 15 people; therefore, it is necessary t... | 31 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (30 points) In the royal dining hall, there are three tables, and three identical pies are served. For lunch, the king invited six princes to his table. At the second table, one can seat from 12 to 18 courtiers, and at the third table, from 10 to 20 knights. Each pie is cut into equal pieces according to the number ... | Answer: Both guests and courtiers will be 14 people on that day.
First solution: Let the number of courtiers at the table be $a$, and the number of knights be $-b$, then the dining rule can be written as $\frac{1}{a} + \frac{1}{b} = \frac{1}{7} \Leftrightarrow \frac{1}{b} = \frac{1}{7} - \frac{1}{a}$. Maximizing the v... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) Before the math lesson, the teacher wrote nine consecutive numbers on the board, but the duty students accidentally erased one of them. When the lesson began, it turned out that the sum of the remaining eight numbers is 1703. Which number did the duty students erase? | Answer: 214.
Solution: Let the average of the original numbers be $a$. Then these numbers can be written in a symmetric form:
$$
a-4, a-3, a-2, a-1, a, a+1, a+2, a+3, a+4
$$
The erased number has the form $a+b$, where $-4 \leqslant b \leqslant 4$, and the sum of the remaining numbers is $9a - (a+b) = 8a - b$. On the... | 214 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Alexei came up with the following game. First, he chooses a number $x$ such that $2017 \leqslant x \leqslant 2117$. Then he checks if $x$ is divisible by 3, 5, 7, 9, and 11 without a remainder. If $x$ is divisible by 3, Alexei awards the number 3 points, if by 5 - then 5 points, ..., if by 11 - then 11 p... | Answer: $2079=11 \cdot 9 \cdot 7 \cdot 3$.
Solution: Note that the divisibility of $x$ by 9 immediately gives 12 points, as it also implies divisibility by 3. If the number $x$ is not divisible by 11, it will score no more than $9+7+5+3=24$ points, and if it is not divisible by 9, it will score no more than $11+7+5+3=... | 2079 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (30 points) In triangle $K I A$, angles $K$ and $I$ are equal to $30^{\circ}$. On the line passing through point $K$ perpendicular to side $K I$, a point $N$ is marked such that $A N$ is equal to $K I$. Find the measure of angle $KAN$. | Answer: $90^{\circ}$ or $30^{\circ}$.
Solution: Let $K I=2 a$, and point $H$ be the foot of the perpendicular from vertex $A$ to the
line $K N$. According to the condition, $\triangle K I A... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the country of Alphia, there are 150 cities, some of which are connected by express trains that do not stop at intermediate stations. It is known that any four cities can be divided into two pairs such that there is an express train running between the cities of each pair. What is the minimum number of pairs of c... | Answer: 11025.
Solution. Suppose that some city (let's call it Alfsk) is connected by express trains to no more than 146 cities. Then a quartet of cities, consisting of Alfsk and any three with which it is not connected, does not satisfy the problem's condition, since Alfsk cannot be paired with any of the three remai... | 11025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the country of Betia, there are 125 cities, some of which are connected by express trains that do not stop at intermediate stations. It is known that any four cities can be toured in a circle in some order. What is the minimum number of pairs of cities connected by express trains? | Answer: 7688.
Solution. Suppose that some city (let's call it Betsk) is connected by express trains to no more than 122 cities. Take a quartet of cities consisting of Betsk and any three other cities, two of which Betsk is not connected to. This quartet cannot be traveled in a circle, otherwise Betsk must be connected... | 7688 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the village of Sosnovka, there are 240 residents, some of whom are acquainted with each other, while others are not. It is known that any five residents can be seated at a round table such that each of them is acquainted with both of their neighbors. What is the minimum number of pairs of acquaintances that can e... | Answer: 28440.
Solution. Suppose that some resident (let's call him Petya) is not acquainted with at least three other residents. Choose the following five people: Petya, the three residents he is not acquainted with, and one more arbitrary person. They cannot be seated around a round table in the required manner, sin... | 28440 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a tennis tournament, 254 schoolchildren are participating. A win earns 1 point, a loss earns 0 points. If the winner initially had fewer points than the opponent, the winner additionally receives one point from the loser. In each round, participants with the same number of points compete, but in one of the pairs,... | Answer: 56.
Solution. Let $f(m, k)$ denote the number of students who have $k$ points after $m$ rounds. If $f(m, k)$ is even for any $k \in \{0, \ldots, m\}$, then participants with different numbers of points will not meet in the $(m+1)$-th round (otherwise, there would be at least two such meetings). Also, note that... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In the village of Beryozovka, there are 200 residents, some of whom are acquainted with each other, while others are not. It is known that any six residents can be seated at a round table such that each of them is acquainted with both of their neighbors. What is the minimum number of pairs of acquainted residents in... | Answer: 19600.
Solution. Suppose that some resident (let's call him Vasya) is not acquainted with at least four other residents. We will select the following six people: Vasya, four residents he is not acquainted with, and one more arbitrary person. They cannot be seated around a round table in the required manner, si... | 19600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a tennis tournament, 1152 schoolchildren are participating. 1 point is awarded for a win, and 0 points for a loss. Before each round, pairs are formed by drawing lots among participants with the same number of points (those who do not find a pair are awarded a point without playing). A player is eliminated after ... | Answer: 14.
Solution. Note that $1152=1024+128$. Let's prove two statements first.
1) In a tournament with $2^{n}$ participants, for any $m \in\{1, \ldots, n\}$, after the $m$-th round, there will be $2^{n-m}$ participants without any losses and $m \cdot 2^{n-m}$ participants with one loss. We will use induction on $... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Inside triangle $ABC$, a point $P$ is chosen such that $AP=BP$ and $CP=AC$. Find $\angle CBP$, given that $\angle BAC = 2 \angle ABC$.
---
Here is the translation of the provided text, maintaining the original formatting and structure. | Answer: $30^{\circ}$.
Solution 1. Draw the perpendicular bisector of side $A B$. Obviously, it will pass through point $P$. Let $C^{\prime}$ be the point symmetric to $C$ with respect to th... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. A health camp was visited by 175 schoolchildren. Some of the children know each other, while others do not. It is known that any six schoolchildren can be accommodated in two three-person rooms so that all schoolchildren in the same room know each other. What is the minimum number of pairs of schoolchildren who coul... | Answer: 15050.
Solution. Suppose that some student (let's call him Vasya) is not acquainted with at least four students. We will select the following six children: Vasya, four students he is not acquainted with, and one more arbitrary student. It is impossible to seat them in two rooms, since out of this six, Vasya is... | 15050 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a triangle $ABC$ with the largest side $BC$. The bisector of its angle $C$ intersects the altitudes $AA_{1}$ and $BB_{1}$ at points $P$ and $Q$ respectively, and the circumcircle of $ABC$ at point $L$. Find $\angle ACB$, if it is known that $AP=LQ$.
| Answer: $60^{\circ}$.
Solution. Let $\alpha=\angle B C L, \beta=\angle A L C, \gamma=\angle B L C$ (see the figure). We will prove the equality of triangles $A L P$ and $B L Q$. Note that $A... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In a health camp, 225 schoolchildren are resting. Some of the children know each other, while others do not. It is known that among any six schoolchildren, there are three non-intersecting pairs who know each other. What is the minimum number of pairs of schoolchildren who could be resting in the camp? | Answer: 24750.
Solution. Suppose that some student (let's call him Vasya) is not acquainted with at least five other students. Then Vasya and the five students he is not acquainted with form a group of six that does not satisfy the condition of the problem. Therefore, each student can have no more than four unknowns. ... | 24750 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a tennis tournament, 512 schoolchildren are participating. 1 point is awarded for a win, and 0 points for a loss. Before each round, pairs are formed by lottery among participants with the same number of points (those who do not find a pair are awarded a point without playing). The tournament ends as soon as a so... | Answer: 84.
Solution. We will show that in a tournament with $2^{n}$ participants, no one will score points without playing, and for any $k \in\{0, \ldots, n\}$, exactly $C_{n}^{k}$ participants will end up with $k$ points. We will use induction on $n$. For $n=1$, this is obvious. Suppose that for some $n$ both statem... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $P$ and $Q$ are chosen such that $A P: P B = B Q: Q C = 2: 1$. Find $\angle A K B$, where $K$ is the intersection point of segments $A Q$ and $C P$. | Answer: $90^{\circ}$.
Solution 1. Let $B H$ be the height and median of triangle $A B C$. Draw a line through vertex $B$ parallel to $A C$ and denote the point of its intersection with line... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The government has decided to privatize civil aviation. For each pair of 202 cities in the country, the airline connecting them is sold to one of the private airlines. The mandatory condition for the sale is as follows: each airline must ensure the possibility of flying from any city to any other (possibly with seve... | Answer: 101.
Solution. First, we prove that for $k \in \{1, \ldots, 100\}$, the remainders of the numbers
$$
0, k, 2k, 3k, \ldots, 99k, 100k
$$
when divided by the prime number 101 are distinct. Indeed, if for some $0 \leqslant a < b \leqslant 100$ the numbers $ak$ and $bk$ give the same remainder when divided by 10... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the hypotenuse $A B$ of an isosceles right triangle $A B C$, points $K$ and $L$ are marked such that $A K: K L: L B=1: 2: \sqrt{3}$. Find $\angle K C L$. | Answer: $45^{\circ}$.
Solution 1. Let $A K=1$. Then $K L=2$ and $L B=\sqrt{3}$, so $A B=3+\sqrt{3}$ and $A C=B C=\frac{3+\sqrt{3}}{\sqrt{2}}$. Therefore,
$$
A C \cdot B C=6+3 \sqrt{3}=3(2+\... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The government has decided to privatize civil aviation. For each pair of the country's 127 cities, the connecting airline is sold to one of the private airlines. Each airline must make all the purchased air routes one-way, but in such a way as to ensure the possibility of flying from any city to any other (possibly ... | # Answer: 63.
Solution. For integers $k$ and $n$, let $N_{k, n}$ denote the remainder when $k \cdot n$ is divided by the prime number 127. First, we prove that if $k$ is not divisible by 127, then the numbers $N_{k, 0}, \ldots, N_{k, 126}$ are distinct. Indeed, if for some $0 \leqslant a < b \leqslant 126$ the numbers... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Sasha chooses a natural number \( n \) and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers \( a \) and \( b \) are not connected by a segment, then the sum \( a + b \) must be coprime with... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among seven numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $x$ in a base $r$ system ( $r \leqslant 36$ ) has the form $\overline{p p q q}$, and $2 q=5 p$. It turns out that the $r$-ary representation of the number $x^{2}$ is a seven-digit palindrome with a zero middle digit. (A palindrome is a number that reads the same from left to right and from right to ... | Answer: 36.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. Let $p=2 s, q=5 s$. Then $x=\overline{p p q q}_{r}=\left(p r^{2}+q\right)(r+1)=s\left(2 r^{2}+5\right)(r+1)$. From the condition on $x^{2}$, we get the equality
$$
s^{2}\left(2 r^{2}+5\right)^{2}\left(r^{2}+2 r+1\right)=a\left(1+r^{... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Sasha chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a+b$ must be coprime with $n$; if ... | Answer: $n=35$.
Solution. We will make two observations.
1) $n$ is odd. Indeed, let $n$ be even. Among eight numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $q$ is a prime divisor of $n$, then ... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $x$ in a base $r(r \leqslant 400)$ numeral system has the form $\overline{p p q q}$, and $7 q=17 p$. It turns out that the $r$-ary representation of the number $x^{2}$ is a seven-digit palindrome with a zero middle digit. (A palindrome is a number that reads the same from left to right and from righ... | Answer: 400.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. Let $p=7 s, q=17 s$. Then $x=\overline{p p q q}_{r}=\left(p r^{2}+q\right)(r+1)=s\left(7 r^{2}+17\right)(r+1)$. From the condition on $x^{2}$, we get the equality
$$
s^{2}\left(7 r^{2}+17\right)^{2}\left(r^{2}+2 r+1\right)=a\left(1... | 400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the picture, several circles are drawn, connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if ... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, t... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a square table of size $2021 \times 2021$. Each of its cells is colored in one of $n$ colors. It is known that for any four cells of the same color located in the same column, there are no cells of the same color to the right of the top one and to the left of the bottom one. What is the smallest $n$ for which ... | Answer: 506.
Solution. Consider some specific color (say, blue). In each column, mark the three lower blue cells with a cross, and the rest with a zero (if there are fewer than four blue cells in the column, mark all cells with a cross). Note that in any row, there is no more than one blue cell with a zero (otherwise,... | 506 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The picture shows several circles connected by segments. Tanya chooses a natural number $n$ and places different natural numbers in the circles so that for all these numbers the following property holds:
if numbers $a$ and $b$ are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with $n$; if $... | Answer: $n=65$.
Solution. First, let's make three observations.
1) $n$ is odd. Indeed, suppose $n$ is even. Among six numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, th... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The board has the number 5555 written in an even base $r$ ($r \geqslant 18$). Petya found out that the $r$-ary representation of $x^2$ is an eight-digit palindrome, where the difference between the fourth and third digits is 2. (A palindrome is a number that reads the same from left to right and from right to left).... | Answer: $r=24$.
Solution. Let's agree to write $u \equiv v(\bmod w)$ if $(u-v) \vdots w$. According to the condition, there exist such $r$-ary digits $a, b, c, d$ that $d-c=2$ and
$$
25(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$
... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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