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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 cyclically, that is, the last one is adjacent to the first.)
|
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 condition of the problem. For $n<4$, there are three possible situations:
CCKKKCC, CCKKCC, CCKCC.
In the first case, the middle red bead does not satisfy the condition, and in the other cases, all red beads do not satisfy the condition.
Thus, pairs of blue beads must be separated by four red beads. It is clear that such a necklace satisfies the condition of the problem. In it, the number of red beads is twice the number of blue beads, that is, there are 60.
|
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 necklace are arranged cyclically, meaning the last one is adjacent to the first.)
|
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 problem's condition. Therefore, the block contains no more than four red beads, and thus, there is a blue one.
Fix a blue bead in the necklace, and divide the rest into 29 blocks of 6 beads each. By the proven fact, each block contains at least one blue bead. Therefore, there are at least 30 blue beads in the necklace.
It remains to provide an example of a necklace with exactly 30 blue beads:
$$
\text {RRGRBS; RRGRBS; ..; RRGRBS (29 times); B. }
$$
Solution 2. Fix any green or blue bead, and divide the rest into 58 consecutive triplets. Each triplet contains no more than two red beads, otherwise the middle red bead would have same-colored neighbors. Therefore, there are no more than $2 \cdot 58 = 116$ red beads in the necklace, and thus, there are at least $175 - 116 = 59$ blue and green beads together. If we mentally remove all the red beads, we notice that the blue beads now make up at least half, as there is a blue bead between any two green beads. Therefore, there are at least $\frac{59}{2} = 29 \frac{1}{2}$ blue beads, meaning there are at least 30.
It remains to provide an example of a necklace with exactly 30 blue beads:
$$
\text {RRGRBS; RRGRBS; ...; RRGRBS (29 times); B. }
$$
|
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 $Q$. Find the area of triangle $BPQ$, given that $BH: CH = 4: 9$.
|
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 triangles $B H P$ and $O C A$ are similar because
$$
\angle C B P=\frac{1}{2} \angle C O P=\frac{1}{2}\left(180^{\circ}-\angle C A P\right)=90^{\circ}-\angle C A O=\angle C O A
$$
Then
$$
\frac{A C}{B C}=\frac{1}{2} \cdot \frac{A C}{O C}=\frac{1}{2} \cdot \frac{P H}{B H}=\frac{3}{4}
$$
From the similarity of triangles $B H Q$ and $B C A$, we get $Q H=\frac{3}{4} B H=6$. Therefore,
$$
S_{B P Q}=\frac{1}{2} \cdot P Q \cdot B H=\frac{1}{2} \cdot(P H-Q H) \cdot B H=\frac{1}{2} \cdot 6 \cdot 8=24
$$
|
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 are arranged cyclically, that is, the last one is adjacent to the first.)
|
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 $m$ elements, then
$$
100 \leqslant m+m(n-1)=m n, \quad \text { hence } \quad m \geqslant \frac{100}{n}
$$
By the proven result, the number of blue beads is no less than $\frac{100}{5}=20$, and the number of red beads is no less than $\frac{100}{7}=14 \frac{2}{7}$, which means at least 15. Therefore, the number of green beads is no more than $100-20-15=65$.
Let's provide an example of a necklace containing 65 green beads. Place the blue beads at positions that are multiples of 5, and the red beads at positions with the following numbers:
$$
1,8,14 ; 21,28,34 ; 41,48,54 ; 61,68,74 ; 81,88,94
$$
Fill the remaining positions with green beads.
|
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,
$$
m+10^{k} a+10^{k+1} n=9\left(m+10^{k} n\right) \Longleftrightarrow 8 m=10^{k}(a+n)
$$
Note that $k>0$, otherwise $m=0$ and $a=n=0$. Then the number $8 m$ is divisible by 10 and thus ends in 0. Given the condition, the number $m$ does not end in 0. Therefore, the last digit of $m$ is 5 and the number $m$ is odd. Therefore, $8 m$ is not divisible by 16, which implies $k \leqslant 3$. Let's consider three cases.
1) Suppose $k=3$. Then $m=125(a+n)$. Since the number $m$ is odd and less than $1000, a+n$ can take the values $1,3,5,7$. Note that the pair $(a, n)$ uniquely determines the original number, and each value of $a+n$ gives $a+n$ different pairs. Thus, we get $1+3+5+7=16$ options.
2) Suppose $k=2$. Then $m=25 \cdot \frac{a+n}{2}$. Since the number $m$ is odd and less than $100, a+n$ equals 2 or 6. These values give us $2+6=8$ options.
3) Suppose $k=1$. Then $m=5 \cdot \frac{a+n}{4}$. Since the number $m$ is odd and less than 10, we get $a+n=4$, which gives us 4 options.
Note that in 1) we get four-digit numbers, in 2) — three-digit numbers, in 3) — two-digit numbers. Therefore, each number satisfying the condition of the problem falls into exactly one of the sets 1) - 3). Thus, the total number of options is $16+8+4=28$.
|
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 node along one edge and exits along another, while the third edge adjacent to the node cannot belong to the path. Consider a triplet of nodes of multiplicity 3 lying on one side of the $4 \times 4$ square. Each node has an edge adjacent to it that is free from the path (regardless of whether the path passes through the node or not), and for non-adjacent nodes, these edges are different. Therefore, there are at least $2 \cdot 4 = 8$ edges free from the path, and the length of the path does not exceed $40 - 8 = 32$. An example of a path of length 32 is shown in the figure.
|
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 one green bead, so there are at least 25 of them on the string. Therefore, the number of red beads is no more than $150-25-13=112$.
Let's provide an example where the string contains exactly 112 red beads. Place the green beads at positions that are multiples of 6, and the blue beads at positions
$$
11,22,33,44,55 ; 65,76,87,98,109 ; 119,130,141
$$
Fill the remaining positions with red beads.
|
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 $1 \times 1$ squares generate 38 different edges. If the path passes through a node of multiplicity 3, then it enters the node along one edge and exits along another, while the third edge adjacent to the node cannot belong to the path. Let $X$ be the set of nodes lying on the broken line $B A D$, excluding points $B$ and $D$. It consists of point $A$ and six nodes of multiplicity 3. Note that point $A$ has multiplicity 2, and the path does not enter it. Therefore, for each node in $X$, there is an edge free from the path (regardless of whether the path passes through the given node or not). For non-adjacent nodes, these edges are certainly different, and there cannot be more than 3 pairs of adjacent nodes. Thus, the set $X$ provides at least 4 edges free from the path. The same reasoning applies to the broken line $B C D$. Therefore, there are at least $2 \cdot 4 = 8$ edges free from the path, and the length of the path does not exceed $38 - 8 = 30$. An example of a path of length 30 is shown in the figure.
|
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 adjacent to the first.)
|
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 entire necklace, as these segments contain no more than 49 blue beads out of 50. Therefore, the total number of red beads is no less than $7 \cdot 4 + 1 = 29$.
Let's provide an example of a necklace containing exactly 29 red beads. Consider the block
$$
\text{B = RC RC RC RC; CCC,}
$$
consisting of 4 red and 7 blue beads. Then the desired necklace has the form
$$
\text{B, B, .., B (7 times); RC.}
$$
|
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 is adjacent to the first.)
|
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 necklace, as these segments contain no more than 99 red beads out of 100. Therefore, the total number of blue beads is no less than $11 \cdot 7+1=78$.
Let's provide an example of a necklace containing exactly 78 blue beads. Consider the block
$$
\text { B = CB CB CB CB CB CB CB; RR, }
$$
consisting of 7 blue and 9 red beads. Then the desired necklace has the form
$$
\text { B, B, .., B (11 times); CR. }
$$
|
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 the last bead is adjacent to the first.)
|
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 these inequalities, the right side will be $48 \cdot 4 + 2 \cdot 3 = 198$, and the left side will be twice the number of good quartets, since each will be counted twice. Therefore, there can be no more than 99 good quartets.
Let's provide an example of a necklace that gives exactly 99 good quartets:
$$
\text{RBYB; RBYB; ..; RBYB (25 times); RZ ...}
$$
(the ellipsis at the end means any combination of green and red beads). The good quartets will start at positions $1, 2, \ldots, 99$, and only these.
|
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 three times the number of good fives on the left side, since each will be counted three times. Therefore, there can be no more than 123 good fives.
Let's provide an example of bead placement that gives exactly 123 good fives:
$$
\text { KSGGG; KSGGG; ...; KSGGG (25 times); KS ... }
$$
(the ellipsis at the end means an arbitrary combination of blue and red beads). The good fives will be those starting at positions $1, 2, \ldots, 123$, and only those.
|
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 the frustum is 15.
|
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 $\mathcal{K}_{0}$ the cone that completes the truncated cone to a regular cone, and by $\mathcal{K}_{1}$ the cone with the center of its base at $O_{1}$. On the left diagram, the section of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ by the plane $\Pi$ passing through points $O_{1}$ and $C$ and perpendicular to the table is shown. By the problem's condition, $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ have a common generatrix, which lies in $\Pi$ since it passes through the vertices of the cones. Let $B$ be the point of intersection of this generatrix with the table. Then $B$ lies on the boundaries of the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$, as well as on the segment $C O_{1}$ connecting the centers of the bases. Hence, the bases of $\mathcal{K}_{0}$ and $\mathcal{K}_{1}$ touch each other at point $B$, i.e., $B C=R$. Similarly, it can be verified that the distance from $C$ to the bases of the other two cones is also $R$. Note that
$$
O_{1} O_{2}=5 r, \quad O_{1} O_{3}=12 r, \quad O_{2} O_{3}=13 r
$$
which means that triangle $O_{1} O_{2} O_{3}$ is a right triangle. Direct the coordinate axes along the rays $O_{1} O_{3}$ and $O_{1} O_{2}$ (see the right diagram). Let the coordinates of point $C$ be $(x, y)$. Since
$$
C O_{1}=B O_{1}+B C=2 r+R, \quad C O_{2}=3 r+R, \quad C O_{3}=10 r+R
$$
the following equalities hold:
$$
(2 r+R)^{2}-x^{2}=y^{2}=(10 r+R)^{2}-(12 r-x)^{2} \Longleftrightarrow 4 r^{2}+4 r R=20 r R+24 r x-44 r^{2} \Longleftrightarrow 3 x=6 r-2 R
$$
and also
$$
(2 r+R)^{2}-y^{2}=x^{2}=(3 r+R)^{2}-(5 r-y)^{2} \Longleftrightarrow 4 r^{2}+4 r R=6 r R+10 r y-16 r^{2} \Longleftrightarrow 5 y=10 r-R
$$
Since $x^{2}+y^{2}=C O_{1}^{2}=(2 r+R)^{2}$, we get
$$
225(2 r+R)^{2}=(30 r-10 R)^{2}+(30 r-3 R)^{2} \Longleftrightarrow 225 r^{2}-420 R r-29 R^{2}=0
$$
This equation in terms of $r$ has a unique positive solution $r=\frac{29}{15} R=29$.
## Variant 9
|
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 this necklace? (The beads in the necklace are arranged cyclically, that is, the last one is adjacent to the first.)
|
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 than red ones. Therefore, there are no fewer than $\frac{80}{3}=26 \frac{2}{3}$ green beads, which means there are at least 27. An example of a necklace containing exactly 27 green beads is as follows:
GKR 3 GKR ... ; GKR (26 times); GK.
|
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. What is the maximum number of prime numbers that can be in the table?
|
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 no more than two prime numbers. Thus, if the table contains $n$ composite numbers, then the number of primes does not exceed $2n$. Therefore, the total number of numbers in the table does not exceed $3n$. Then
$$
3n \geqslant 80^{2} \Longrightarrow n \geqslant \frac{80^{2}}{3}=2133 \frac{1}{3} \Longrightarrow n \geqslant 2134 \Longrightarrow 80^{2}-n \leqslant 80^{2}-2134=4266
$$
Thus, the number of prime numbers in the table does not exceed 4266.
Now, let's show how 4266 prime numbers can be placed in the table. We will use the following algorithm for filling rows and columns.
1) The first 52 positions are filled with different prime numbers $p_{1}, p_{2}, \ldots, p_{52}$. These numbers must be new, i.e., not used before in the table.
2) In the next 26 cells, we place the numbers $p_{1} p_{2}, p_{3} p_{4}, \ldots, p_{51} p_{52}$.
3) The last two positions are left unfilled.
We apply this algorithm sequentially to the rows $1,2, \ldots, 80$, and then to the last two columns. Thus, we will place $80 \cdot 52 + 2 \cdot 52 = 4264$ prime numbers. It remains to fill the $2 \times 2$ square in the bottom right corner. In it, on one diagonal, we will place a pair of new prime numbers, and on the other diagonal - their squares. In the end, we will place 4266 different prime numbers.
|
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 this table?
|
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 brunettes. Therefore, there are no fewer than $\frac{50}{3}=16 \frac{2}{3}$ redheads sitting at the table, which means there are at least 17. Let's provide an example of seating with exactly 17 redheads:
$$
\text { RCB; RCB; ... RCB; (16 times); RC }
$$
(The letters R, C, and B denote redheads, brunettes, and blondes, respectively).
|
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 prime numbers that can be in the table?
|
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 divisors, it serves no more than three prime numbers. Thus, if the table contains $n$ composite numbers, then the number of primes does not exceed $3n$. Therefore, the total number of numbers in the table does not exceed $4n$. Then
$$
4n \geqslant 75^{2} \Longrightarrow n \geqslant \frac{75^{2}}{4}=1406 \frac{1}{4} \Longrightarrow n \geqslant 1407 \Longrightarrow 75^{2}-n \leqslant 75^{2}-1407=4218
$$
Thus, the number of prime numbers in the table does not exceed 4218.
Now let's show how 4218 prime numbers can be placed in the table. We will use the following algorithm for filling rows and columns.
1) The first 54 positions are filled with different prime numbers $p_{1}, p_{2}, \ldots, p_{54}$. These numbers must be new, that is, not used before in the table.
2) In the next 18 cells, we place the numbers $p_{1} p_{2} p_{3}, p_{4} p_{5} p_{6}, \ldots, p_{52} p_{53} p_{54}$.
3) The last three positions are left unfilled.
We apply this algorithm sequentially to the rows $1,2, \ldots, 75$, and then to the last three columns. Thus, we will place $75 \cdot 54 + 3 \cdot 54 = 4212$ prime numbers. It remains to fill the cells of the $3 \times 3$ square in the lower right corner. We choose new prime numbers $q_{1}, \ldots, q_{6}$ and place them as follows:
$$
\left(\begin{array}{ccc}
q_{1} q_{2} & q_{3} q_{4} & q_{5} q_{6} \\
q_{1} & q_{3} & q_{5} \\
q_{2} & q_{4} & q_{6}
\end{array}\right)
$$
In the end, we will place 4218 different prime numbers.
|
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 $\overline{\text { adabcd }}$ are divisible by 5?
|
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 values. Similarly, for the digit $c$ - 7 options. Therefore, the total number of such numbers is $1 \cdot 8 \cdot 8 \cdot 7=448$.
|
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 equivalent to the condition
$$
\frac{s_{1}+1}{s_{1}}=\frac{s_{2}+3}{s_{2}}=\frac{s_{3}+5}{s_{3}}=\ldots=\frac{s_{1008}+2015}{s_{1008}} \Leftrightarrow \frac{1}{s_{1}}=\frac{3}{s_{2}}=\frac{5}{s_{3}}=\ldots=\frac{2015}{s_{1008}}
$$
From the first equality, express $s_{2}$ in terms of $s_{1}: s_{2}=3 s_{1}$; then, similarly, $s_{3}$ in terms of $s_{1}$: $s_{3}=5 s_{1}; \ldots s_{1008}=2015 s_{1}$. Now we have
$$
\sum_{i=1}^{1008} s_{i}=\sum_{i=1}^{1008}(2 i-1) s_{1}=s_{1} \cdot \frac{(1+2015) \cdot 1008}{2}=1008^{2} s_{1}=2016^{2}
$$
From this, $s_{1}=4$ and, consequently, $s_{17}=(2 \cdot 17-1) \cdot 4=132$.
|
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 show how to mark 416 cells in the required manner. Mark all parallel diagonals with lengths of $5, 11, 17, 23, 29, 35, 41$, and $47$. In total, there will be $2 \cdot (5+11+17+23+29+35+41+47)=416$ marked cells.
|
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 chips that can be placed in the cells?
|
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 $\left[\frac{n+1}{2}\right]$, otherwise the numbers placed in them would repeat. Then
$$
2021-n \leqslant\left[\frac{n+1}{2}\right] \leqslant \frac{n+1}{2}, \quad \text { hence } \quad n \geqslant \frac{4041}{3}=1347
$$
We will show that the value $n=1347$ is achievable. A suitable arrangement is
$$
\underbrace{0101 \ldots 01}_{674 \text { pairs }} \underbrace{111 \ldots 1}_{673 \text { numbers }}
$$
where ones represent chips, and zeros represent empty cells. In this case, the numbers in the positions of the zeros will be:
$$
1347,1345,1343, \ldots, 3,1
$$
Remark. The implementation provided in the solution is not unique. For example, the following also works:
$$
1011011 \ldots 0110
$$
The corresponding numbers are:
$$
1345,1341, \ldots, 5,1,3,7, \ldots, 1343,1347
$$
|
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 of chips that can be placed in the cells?
|
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 $\left[\frac{n+1}{2}\right]$, otherwise the numbers placed in them would repeat. Then
$$
2100-n \leqslant\left[\frac{n+1}{2}\right] \leqslant \frac{n+1}{2}, \quad \text { hence } n \geqslant \frac{4199}{3}=1399 \frac{2}{3} \text { and } n \geqslant 1400
$$
We will show that the value $n=1400$ is achievable. The arrangement
$$
\underbrace{0101 \ldots 01}_{700 \text { pairs }} \underbrace{111 \ldots 1}_{700 \text { numbers }}
$$
where 1s represent chips and 0s represent empty cells, will work. In this case, the numbers in the positions of the zeros will be:
$$
1400,1398,1396, \ldots, 4,2
$$
Remark. The implementation provided in the solution is not unique. For example, the following arrangement also works:
$$
\underbrace{101101 \ldots 101}_{350 \text { triples }} \underbrace{110110 \ldots 110}_{350 \text { triples }}
$$
The corresponding numbers are:
$$
1398,1394, \ldots, 6,2,4,8, \ldots, 1396,1400
$$
|
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 the points of tangency with these circles by $D$ and $E$ respectively. Find the angle $B_{1} D E$.
|
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 is the perpendicular bisector of segment $C_{1} E$, meaning points $C_{1}$ and $E$ are symmetric with respect to line $A C$. From the equality of triangles $A H D$ and $A H C_{1}$, it follows that $\angle A H D = \angle A H C_{1}$. The quadrilateral $A B_{1} H C_{1}$ has two opposite right angles. Therefore, it is cyclic, from which $\angle A B_{1} C_{1} = \angle A H C_{1}$. By the symmetry of points $C_{1}$ and $E$,
$$
\angle A B_{1} E = \angle A B_{1} C_{1} = \angle A H C_{1}.
$$
Moreover, $\angle A D H = 90^{\circ} = \angle A B_{1} H$. Therefore, quadrilateral $A D B_{1} H$ is also cyclic, and
$$
\angle A B_{1} D = \angle A H D = \angle A H C_{1}
$$
Thus, $\angle A B_{1} D = \angle A B_{1} E$, which means $\angle B_{1} D E = 180^{\circ}$.
|
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 $B K = B C$.
|
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 $\angle B K C=\angle B C K=90^{\circ}-\varphi$. Moreover,
$$
\angle L C K+\angle B K C=\angle L B K+\angle B K C=\varphi+\left(90^{\circ}-\varphi\right)=90^{\circ},
$$
which means that lines $B K$ and $C L$ are perpendicular. On the other hand,
$$
\angle B L C+\angle A C L=\angle B K C+\angle A B L=\left(90^{\circ}-\varphi\right)+\varphi=90^{\circ},
$$
so lines $B L$ and $A C$ are also perpendicular. Therefore, point $P$ is the orthocenter of triangle $B C L$, from which it follows that $L P \perp B C$.
|
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 $C, A, D$ lie on the same line, we get
$$
\angle O_{1} A C = \angle O_{1} C A = \angle O C D = \angle O D C
$$
Therefore, $O_{1} A \| O O_{2}$ and, similarly, $O_{2} A \| O O_{1}$. Thus, $O O_{1} A O_{2}$ is a parallelogram, from which we have
$$
O_{1} A = O_{2} O \quad \text{and} \quad \angle A O_{1} O_{2} = \angle O_{1} O_{2} O
$$
Since triangles $A O_{1} B$ and $A O_{2} B$ are isosceles, segment $O_{1} O_{2}$ is the perpendicular bisector of $A B$. Then
$$
O_{1} B = O_{1} A = O_{2} O \quad \text{and} \quad \angle B O_{1} O_{2} = \angle A O_{1} O_{2} = \angle O_{1} O_{2} O
$$
Therefore, $O_{1} O_{2} O B$ is an isosceles trapezoid, from which $B O \| O_{1} O_{2}$ and, hence, $B O \perp A B$.
|
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 circles $\omega_{1}$ and $\omega_{2}$ again at points $C$ and $D$ respectively. Segments $C K$ and $D L$ intersect at point $N$. Find the angle between the lines $O_{1} A$ and $O_{2} N$.
|
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 angles be $2 \alpha$. Since an inscribed angle is half the corresponding central angle, we have $\angle A D L = \angle A C K = \alpha$. Since $K L \| C D$, we also get $\angle K L N = \angle L K N = \alpha$. Then
$$
\angle L N K = 180^{\circ} - 2 \angle K L N = 180^{\circ} - 2 \alpha = 180^{\circ} - \angle A O_{2} L = 180^{\circ} - \angle K O_{2} L
$$
Therefore, the quadrilateral $L N K O_{2}$ is cyclic, from which we get
$$
\angle K O_{2} N = \angle K L N = \alpha = \frac{1}{2} \angle A O_{2} L
$$
This means that the line $O_{2} N$ is the angle bisector of the vertex angle in the isosceles triangle $A O_{2} L$, and therefore also the altitude. Thus, the lines $O_{2} N$ and $O_{1} A$ are perpendicular.
|
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 passing through point $B$ parallel to $K L$ intersects circles $\omega_{1}$ and $\omega_{2}$ again at points $A$ and $C$ respectively. The rays $A K$ and $C L$ intersect at point $N$. Find the angle between the lines $O_{1} N$ and $O_{2} B$.
|
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 the common value of these angles be $2 \alpha$. Since an inscribed angle is half the corresponding central angle, we have $\angle B C L = \angle B A K = \alpha$. Since $K L \| A C$, we also get $\angle K L N = \angle L K N = \alpha$. Then,
$$
\angle L N K = 180^{\circ} - 2 \angle L K N = 180^{\circ} - 2 \alpha = 180^{\circ} - \angle B O_{1} K = 180^{\circ} - \angle L O_{1} K
$$
Therefore, the quadrilateral $L N K O_{1}$ is cyclic, from which we get
$$
\angle L O_{1} N = \angle L K N = \alpha = \frac{1}{2} \angle B O_{1} K
$$
Thus, the line $O_{1} N$ is the angle bisector of the vertex angle in the isosceles triangle $B O_{1} K$, and therefore also the altitude. Hence, the lines $O_{1} N$ and $O_{2} B$ are perpendicular.
|
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 coordinate are marked.
|
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 warehouse?
|
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:
$$
(1,99), \quad(2,98), \quad(3,97), \quad \ldots, \quad(49,51)
$$
For their transportation, 49 trips are sufficient, since each pair fits into the truck. Two more trips are needed to transport the 50-ton and 100-ton goods.
Now we will show that it is not always possible to transport the goods in 50 trips. Suppose the warehouse stores goods weighing 31 and $47,48, \ldots, 100$ tons. Their total weight is $31+\frac{54 \cdot(47+100)}{2}=4000$ tons. No two
goods weighing from 50 to 100 tons can be transported together in one truck. But there are 51 such goods, so at least 51 trips are required.
|
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 $16 \cdot 3 + 8 = 56$ cells must be marked in total. An example with 56 marked cells is shown in the figure.
|
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
$$
a_{1} \leqslant a_{2}-1 \leqslant a_{3}-2 \leqslant \ldots \leqslant a_{50}-49
$$
that is, the numbers $b_{i}=a_{i}-(i-1)$ are increasing. Moreover,
$$
b_{1}+b_{2}+\ldots+b_{50}=a_{1}+a_{2}+\ldots+a_{50}-(0+1+2+\ldots+49)=1555-\frac{49 \cdot 50}{2}=330
$$
from which the arithmetic mean of the numbers $b_{k}$ is $\frac{330}{50}=6.6$. Due to the increasing nature of the numbers $b_{i}$, there exists an index $m$ such that $b_{i} \leqslant 6$ for $i \leqslant m$ and $b_{i} \geqslant 7$ for $i>m$. If $m \leqslant 34$, then
$$
b_{35}+b_{36}+\ldots+b_{50} \geqslant 16 \cdot 7=112
$$
and if $m>34$
$$
b_{1}+b_{2}+\ldots+b_{34} \leqslant 34 \cdot 6=204 \quad \text { and } \quad b_{35}+b_{36}+\ldots+b_{50} \geqslant 330-204=126>112
$$
In both cases
$$
a_{35}+a_{36}+\ldots+a_{50}=b_{35}+b_{36}+\ldots+b_{50}+(34+35+\ldots+49) \geqslant 112+\frac{16 \cdot(34+49)}{2}=776
$$
Now we will show that $k \leqslant 776$. We need to provide an example where the total age of the 16 oldest viewers is 776. Suppose people aged $6,7, \ldots, 25$ and $27,28, \ldots, 56$ years came to the cinema. Their total age is $\frac{51 \cdot(6+56)}{2}-26=1555$, and the total age of the 16 oldest of them is $41+42+\ldots+56=\frac{16 \cdot(41+56)}{2}=776$.
|
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 the coordinates is divisible by 4.
|
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 in the arena to ensure this can always be done?
|
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 least 15 rows are needed.
We will show that 15 rows are sufficient. Assume that there is one large row on the stand, divided by aisles into sectors of 168 seats each. We will seat the students on this large row, ignoring the aisles, starting with the students from the first school, then the second, and so on. As a result, 12 sectors will be fully occupied. With this seating arrangement, students from no more than 11 schools can be seated in two sectors. Since any four schools can fit in one sector, it is sufficient to have three sectors for seating the remaining students.
|
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 that $n$ is a multiple of 3, that is, $x=9^{3m}-1$. From this,
$$
x=p \cdot q, \quad \text{where } p=9^{m}-1, \quad q=9^{2m}+9^{m}+1
$$
Note that the numbers $p$ and $q$ are coprime. Indeed, if a number $r$ divides both $p$ and $q$, then it also divides 3, since
$$
3=q-(9^{2m}-1)-(9^{m}-1)=q-p(9^{m}+2)
$$
But $p$ is not divisible by 3, hence $r=1$.
We will prove that the number $p$ is a power of two only when $m=1$. Indeed, let $m>1$. Write $p=(3^{m}-1)(3^{m}+1)$. In the right-hand side, we have the product of two consecutive even numbers, both greater than 4. Therefore, at least one of them is not divisible by 4 and, thus, is not a power of 2.
If $m=1$, we get $x=9^{3}-1=728=2^{3} \cdot 7 \cdot 13$, which fits our criteria. We will show that there are no solutions when $m>1$. One of the numbers $p$ and $q$ is divisible by 13. Consider two cases.
1) $p$ is divisible by 13. Then $m \vdots 3$, that is, $m=3k$. If $k=1$, then $p$ is divisible by 7 and 13. For $k>1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will be factored into two coprime factors, neither of which is a power of 2. Therefore, $x$ has at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 13. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 13, that is, $q=13^{s}$ for some natural number $s$. The remainder of $q$ when divided by 8 is 5 for odd $s$ and 1 for even $s$. On the other hand, this remainder must be the same as that of $9^{2m}+9^{m}+1$, which is 3, which is impossible.
|
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.
|
16
|
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 45 from each school. Students from any school need to be seated in one row. What is the minimum number of rows that must be in the arena to ensure this can always be done?
|
Answer: 16.
Solution. Suppose that 46 schools sent 43 students each to the final match, and one school sent 34 students. Since only three groups of 43 students can fit in one row, the number of rows required to accommodate the students from 46 schools should be no less than $\frac{46}{3}=15 \frac{1}{3}$, which means at least 16 rows are needed.
We will show that 16 rows are sufficient. Assume that there is one large row on the stand, divided by aisles into sectors of 168 seats each. We will seat the students on this large row, ignoring the aisles, starting with the students from the first school, then the second, and so on. As a result, 12 sectors will be fully occupied. With this seating arrangement, students from no more than 11 schools can be on two sectors. Since any three schools can fit in one sector, four sectors are sufficient for seating the remaining students.
|
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 7. 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 7. The remainders of the powers of 9 when divided by 7 are $2, 4, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 7 means that $n$ is a multiple of 3, that is, $x=9^{3m}-1$. From this,
$$
x=p \cdot q, \quad \text{where } p=9^{m}-1, \quad q=9^{2m}+9^{m}+1
$$
Note that the numbers $p$ and $q$ are coprime. Indeed, if a number $r$ divides both $p$ and $q$, then it also divides 3, since
$$
3=q-(9^{2m}-1)-(9^{m}-1)=q-p(9^{m}+2)
$$
But $p$ is not divisible by 3, hence $r=1$.
We will prove that the number $p$ is a power of two only when $m=1$. Indeed, let $m>1$. Write $p=(3^{m}-1)(3^{m}+1)$. In the right-hand side, we have the product of two consecutive even numbers, both greater than 4. Therefore, at least one of them is not divisible by 4 and, thus, is not a power of 2.
If $m=1$, we get $x=9^{3}-1=728=2^{3} \cdot 7 \cdot 13$, which fits our criteria. We will show that there are no solutions when $m>1$. One of the numbers $p$ and $q$ is divisible by 7. Consider two cases.
1) $p$ is divisible by 7. Then $m \vdots 3$, that is, $m=3k$. If $k=1$, then $p$ is divisible by 7 and 13. For $k>1$, we can apply the same reasoning to $p$ as we did to $x$. In both cases, $p$ will be factored into two coprime factors, neither of which is a power of 2. Therefore, $x$ has at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 7. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 7, that is, $q=7^{s}$ for some natural number $s$. Thus, the remainder of $q$ when divided by 8 is 7 for odd $s$ and 1 for even $s$. On the other hand, this remainder must be the same as that of $9^{2m}+9^{m}+1$, which is 3, which is impossible.
|
728
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In the cells of a $10 \times 10$ table, the numbers $1,2,3, \ldots, 100$ are arranged such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$.
|
Answer: 202.
Solution. Divide the $10 \times 10$ table into 25 squares of $2 \times 2$. Since the sum of the numbers in the entire table is
$$
1+2+\cdots+100=\frac{100 \cdot 101}{2}=5050
$$
the arithmetic mean of the sums of the numbers in these 25 squares is 202. Therefore, in at least one square, the sum of the numbers is not less than 202, that is, $S \geqslant 202$. An example of an arrangement where the value $S=202$ is achieved is shown in the figure.
| 100 | 99 | 98 | 97 | 96 | 95 | 94 | 93 | 92 | 91 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 90 | 89 | 88 | 87 | 86 | 85 | 84 | 83 | 82 | 81 |
| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 80 | 79 | 78 | 77 | 76 | 75 | 74 | 73 | 72 | 71 |
| 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 |
| 70 | 69 | 68 | 67 | 66 | 65 | 64 | 63 | 62 | 61 |
| 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40 |
| 60 | 59 | 58 | 57 | 56 | 55 | 54 | 53 | 52 | 51 |
| 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50 |
|
202
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a school, there are 920 students, and in each class, there are no more than $k$ students. All students need to go on a bus tour. For this, 16 buses, each with 71 seats, have been booked. The students need to be seated on the buses so that students from each class end up in the same bus. For what largest $k$ is this guaranteed to be possible?
|
Answer: 17.
Solution. If $k=18$, then it is not always possible to seat the students in the buses. Indeed, suppose the school has 50 classes of 18 students each and two classes of 10 students each. Only three classes of 18 students can fit in one bus. Therefore, it would require at least 17 buses to transport students from such classes.
We will show that if there are no more than 17 students in each class, it is always possible to seat the students in the buses. Number the seats in the buses consecutively: from 1 to 71 in the first bus, from 72 to 142 in the second, and so on. Seat the students in the order of increasing seat numbers, starting with all students from the first class, then from the second, and so on. As a result, the students will fit into 13 buses, since $13 \cdot 71=923>920$. With this seating arrangement, students from no more than 12 classes can end up in two buses. It is sufficient to use three buses to accommodate these classes, since any four classes can fit in one bus.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Given a natural number $x=5^{n}-1$, 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: 3124.
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 remainders of the powers of 5 when divided by 11 are $5, 3, 4, 9, 1$ and then repeat cyclically. Thus, the divisibility of $x$ by 11 means that $n$ is a multiple of 5, that is, $x = 5^{5m} - 1$. From here, using the formula for the sum of a geometric progression,
$$
x = p \cdot q, \quad \text{where } p = 5^m - 1, \quad q = 1 + 5^m + 5^{2m} + 5^{3m} + 5^{4m}
$$
Note that the numbers $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then
$$
5 = q - (5^m - 1) - (5^{2m} - 1) - (5^{3m} - 1) - (5^{4m} - 1)
$$
Differences of the form $5^{km} - 1$ are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But $p$ is not divisible by 5, hence $r = 1$.
We will prove that the number $p$ is a power of two only when $m = 1$. Indeed, let $m > 1$. If $m = 2k$, then write $p = (5^k - 1)(5^k + 1)$. On the right side, we have the product of two consecutive even numbers. At least one of them is greater than 4 and does not divide by 4, so it is not a power of 2. Now let $m = 2k + 1$. Then
$$
p = 5^m - 5 + 4 = 5 \cdot (5^k - 1)(5^k + 1) + 4.
$$
The right side is not divisible by 8, since $(5^k - 1)(5^k + 1)$ is divisible by 8 as the product of two consecutive even numbers. Since $p > 4$, the number $p$ cannot be a power of two.
When $m = 1$, we get $x = 3124 = 4 \cdot 11 \cdot 71$, which fits. We will show that there are no solutions when $m > 1$. One of the numbers $p$ and $q$ is divisible by 11. Consider two cases.
1) $p$ is divisible by 11. Then $m \vdots 5$, that is, $m = 5k$. If $k = 1$, then $p$ is divisible by 11 and 71. For $k > 1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will decompose into two coprime factors that are not powers of two. Therefore, $x$ has at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 11. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 11. The remainder of $5^m$ when divided by 8 is 5 for odd $m$ and 1 for even $m$. If $m$ is odd, the remainder of $q$ when divided by 8 is the same as that of $1 + 5 + 1 + 5 + 1 = 13$, which is 5. For even $m$, the remainder of $q$ when divided by 8 is also 5. But the remainders of $11^s$ when divided by 8 are only 3 and 1, so $q$ cannot be a power of 11.
|
3124
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In the black cells of an $8 \times 8$ chessboard, the numbers $1,2,3, \ldots, 32$ are placed such that the sum of the numbers in any $2 \times 2$ square does not exceed $S$. Find the smallest possible value of $S$.
|
Answer: 33.
Solution. Divide the chessboard into 16 squares of $2 \times 2$. Since the sum of the numbers in all the black cells is
$$
1+2+\cdots+32=\frac{32 \cdot 33}{2}=16 \cdot 33
$$
the arithmetic mean of the sums of the numbers in these 16 squares is 33. Therefore, in at least one square, the sum of the numbers is not less than 33, i.e., $S \geqslant 33$. An example of an arrangement where $S=33$ is achieved is shown in the figure.
| 32 | | 31 | | 30 | | 29 | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 1 | | 2 | | 3 | | 4 |
| 28 | | 27 | | 26 | | 25 | |
| | 5 | | 6 | | 7 | | 8 |
| 24 | | 23 | | 22 | | 21 | |
| | 9 | | 10 | | 11 | | 12 |
| 20 | | 19 | | 18 | | 17 | |
| | 13 | | 14 | | 15 | | 16 |
|
33
|
Combinatorics
|
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: $90^{\circ}$
|
Solution. Let $O$ be the center of $\omega$, $M$ be the intersection point of lines $A B$ and $C D$. Note that
$$
\angle A O C=2\left(180^{\circ}-\angle A B C\right)=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secant theorem,
$$
D M^{2}=M C^{2}=B M \cdot A M, \quad \text { that is } \quad \frac{D M}{A M}=\frac{B M}{D M} .
$$
Therefore, triangles $D B M$ and $A D M$ are similar, from which
$$
\angle A B D=180^{\circ}-\angle M B D=180^{\circ}-\angle A D M=90^{\circ}
$$
|
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 transportation of all goods?
|
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 required.
We will show that if the weight of each container does not exceed 26 tons, then it is always possible to transport them from the warehouse. Imagine that the containers store boxes, each weighing one ton, and the containers themselves are weightless. We will sequentially load the platforms with these boxes: as soon as the current platform is full, we move to the next one. We will load the boxes starting with all the boxes from the first container, then from the second, and so on. As a result, the boxes will fit on 19 platforms, since $19 \cdot 80=1520>1500$. With such placement, boxes from no more than 18 containers can end up on two platforms. To accommodate these containers, 6 platforms are sufficient, since any three containers can fit on one platform.
|
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 $a$ and $b$ cannot both be divisible by 7, and $y$ will not be a power of 7.
The remainders of the powers of 8 when divided by 31 are $8,2,16,4,1$ and then repeat cyclically. Therefore, the divisibility of $x$ by 31 means that $n$ is a multiple of 5, i.e., $x=8^{5 m}-1$. Note also that $m$ is not divisible by 4, otherwise the divisors of $x$ will include 3, 5, and 7. For $m=1$, we get $x=32767=7 \cdot 31 \cdot 151$, which fits. Let's show that there are no solutions for $m>1$. By the formula for the sum of a geometric progression,
$$
x=p \cdot q, \quad \text{where } p=8^{m}-1, q=1+8^{m}+8^{2 m}+8^{3 m}+8^{4 m}
$$
Let's prove that the numbers $p$ and $q$ are coprime. Indeed, let the number $r$ divide $p$ and $q$. Then
$$
5=q-\left(8^{m}-1\right)-\left(8^{2 m}-1\right)-\left(8^{3 m}-1\right)-\left(8^{4 m}-1\right)
$$
Differences of the form $8^{k m}-1$ are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But the remainders of $8^{m}$ when divided by 5 are $3,4,2,1$ and then repeat cyclically. Since $m$ is not a multiple of 4, the number $p$ is not divisible by 5, hence $r=1$. This means that $q$ is not divisible by 7, since $p \vdots$. Consider two cases.
1) $p$ is a multiple of 31. Then $m \vdots 5$, i.e., $m=5 k$. If $k=1$, then $p$ is divisible by 31 and 151. For $k>1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will decompose into the product of two coprime numbers, neither of which are powers of 7. Taking into account the divisibility of $x$ by 7, we get that $x$ has at least four different prime divisors.
2) $q$ is a multiple of 31. Note that $p$ is divisible by 7, not being a power of 7, and $q$ is coprime with $p$. Then $q$ must be a power of 31, i.e., $q=31^{s}$. Transitioning to remainders when divided by 8, we get $1=(-1)^{s}$, hence $s$ is even. Moreover, the remainders of $31^{s}$ when divided by 7 for even $s$ are $2,4,1$ and then repeat cyclically. But the remainder of $q$ when divided by 7 is 5. Thus, $q$ cannot be a power of 31.
|
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 before and after the intermission?
|
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 that there are no more than three in each row. Thus, $k=16$ works.
Now, we will show that with 17 rows, the audience can be seated in such a way that the required four spectators do not exist. Let the $n$-th column be the seats in the hall with number $n$, cyclically ordered by rows:
$$
1,2, \ldots, 17,1,2, \ldots, 17, \ldots
$$
A cyclic shift of a column by $m$ rows is a permutation of the column where the new row number of each spectator is obtained from the old one by shifting $m$ positions to the right in the sequence (*). Fill the columns numbered from 1 to 45 with spectators, as well as the first 5 seats of the 46th column. Thus, there will be $17 \cdot 45 + 5 = 770$ people in the hall. After the intermission, we will seat the spectators as follows. The spectators in columns $1,2,3$ sit in their original seats; in columns $4,5,6$, the spectators are cyclically shifted by one row, in columns $7,8,9$ by two rows, and so on. As a result, we will have a situation where there are no four spectators who sat in the same row both before and after the intermission.
|
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 when $s>2$. Indeed, for $s>1$, the number $3^{s}+1$ is greater than 8 and is not divisible by 8, as its remainder when divided by 8 is 4 or 2. Now let $s>2$. If $s$ is odd, the number $3^{s}-1$ is greater than 8 and is not divisible by 8, as its remainder when divided by 8 is 2. If $s=2k$, then $3^{s}-1$ is divisible by the number $3^{k}+1$, which is not a power of two. From the above, it follows, in particular, that $9^{s}-1$ will be a power of two only when $s=1$.
The remainders of the powers of 9 when divided by 11 are $9, 4, 3, 5, 1$ and then repeat cyclically. Therefore, the divisibility of $x$ by 11 means that $n$ is a multiple of 5, i.e., $x=9^{5m}-1$. If $m=1$, we get $x=9^{5}-1=59048=2^{3} \cdot 11^{2} \cdot 61$, which fits our criteria. Let's show that there are no solutions for $m>1$. Consider two cases.
1) $m$ is odd. By the formula for the sum of a geometric progression,
$$
x=p \cdot q, \quad \text{where } p=9^{m}-1, \quad q=1+9^{m}+9^{2m}+9^{3m}+9^{4m}
$$
Let's prove that the numbers $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then
$$
5=q-(9^{m}-1)-(9^{2m}-1)-(9^{3m}-1)-(9^{4m}-1)
$$
Differences of the form $9^{km}-1$ are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But $p$ is not divisible by 5 when $m$ is odd, so $r=1$.
Suppose $p$ is divisible by 11. Then $m \vdots \leq 5$, i.e., $m=5k$. If $k=1$, then $p$ is divisible by 11 and 61. For $k>1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will decompose into two coprime factors that are not powers of two. Therefore, $x$ has at least three different odd prime divisors, which is impossible.
Now suppose $q$ is divisible by 11. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 11, i.e., $q=11^{s}$. But the remainder of $11^{s}$ when divided by 8 is 3 for odd $s$ and 1 for even $s$, while the remainder of $q$ when divided by 8 is 5. Therefore, the number $q$ cannot be a power of 11.
2) $m$ is even. Then $n=2k$ for $k \geq 5$, and
$$
x=(9^{k}-1)(9^{k}+1)=(3^{k}-1)(3^{k}+1)(9^{k}+1)
$$
The factors on the right side are coprime, since the pairs of numbers $3^{k} \pm 1$ and $9^{k} \pm 1$ are odd and differ by 2. Moreover, none of the factors is a power of two. Indeed, we have already proven this for the first two, and $9^{k}+1$ is greater than 8 and gives a remainder of 2 when divided by 8. Thus, $x$ has at least three different odd prime divisors, which is impossible.
|
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 $x$ by 61 means that $n$ is a multiple of 5, i.e., $x = 9^{5m} - 1$, where $m$ is odd. By the formula for the sum of a geometric progression,
$$
x = p \cdot q, \quad \text{where } p = 9^m - 1, \quad q = 1 + 9^m + 9^{2m} + 9^{3m} + 9^{4m}
$$
We will prove that the numbers $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then
$$
5 = q - (9^m - 1) - (9^{2m} - 1) - (9^{3m} - 1) - (9^{4m} - 1)
$$
Differences of the form $9^{km} - 1$ are divisible by $p$ and, therefore, by $r$. Hence, $r$ is a divisor of 5. But $p$ is not divisible by 5 when $m$ is odd, so $r = 1$.
We will prove that the number $p$ is a power of two exactly when $m = 1$. Indeed, for $m > 1$, we can write $p = (3^m - 1)(3^m + 1)$. In the right-hand side, we have the product of two consecutive even numbers, both greater than 2. Therefore, at least one of them is not divisible by 4 and, hence, is not a power of 2.
If $m = 1$, we get $x = 9^5 - 1 = 59048 = 2^3 \cdot 11^2 \cdot 61$, which fits our criteria. We will show that there are no solutions for $m > 1$. Consider two cases.
1) $p$ is divisible by 61. Then $m \vdots 5$, i.e., $m = 5k$. If $k = 1$, then $p$ is divisible by 11 and 61. For $k > 1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will decompose into two coprime factors, neither of which is a power of 2. Therefore, $x$ has at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 61. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 61, i.e., $q = 61^s$. Note that $s \vdots 3$, since $q - 1$ is divisible by 9, and the remainders of $61^s$ when divided by 9 are $7, 4, 1$ and then repeat cyclically. Additionally, when divided by 8, the number $q$ gives a remainder of 5, while the remainder of $61^s$ is 5 for odd $s$ and 1 for even $s$. Therefore, $s$ is odd, so $s$ has the form $6k + 3$ for some natural number $k$.
Now let's compare the remainders of $q$ and $61^s$ when divided by 7. The remainders of $61^s$ are $5, 4, 6, 2, 3, 1$ and then repeat cyclically. For $s = 6k + 3$, all of them coincide with 6. The remainders of $9^m$ are $2, 4, 1$ and then repeat cyclically. Therefore, it is sufficient to find the remainders of $q$ for $m = 1, 2, 3$. They are
$$
\begin{aligned}
& (1 + 2 + 4 + 1 + 2) \bmod 7 = 3, \quad \text{if } m = 1, \\
& (1 + 4 + 1 + 2 + 4) \bmod 7 = 5, \quad \text{if } m = 2, \\
& (1 + 1 + 2 + 4 + 1) \bmod 7 = 2, \quad \text{if } m = 3.
\end{aligned}
$$
Since none of these numbers coincide with 6, $q$ cannot be a power of 61.
|
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 $\angle DEF = \angle ABC$. Find the angle $\angle CME$.
|
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 $B C \parallel A F$. Then the line $\ell$, passing through point $M$ and perpendicular to the tangent $B C$, contains the diameter of the circle $\omega$ and is perpendicular to its chord $A F$. Thus, points $A$ and $F$ are symmetric with respect to $\ell$, meaning that $\ell$ is the axis of symmetry of the trapezoid $B C A F$. This implies that $A C = B F$ and $A M = F M$. Therefore, triangles $B F M$ and $C A M$ are congruent by the three sides, so $\angle B F M = \angle E A M$. Additionally, $\angle E F M = \angle E A M$ as inscribed angles subtending the same chord. Thus, $F M$ is the angle bisector of $\angle B F E$, and
$$
\angle E A M = \angle E F M = \frac{1}{2} \angle B F E = 36^{\circ}.
$$
Notice now that the angle between the chord $E M$ and the tangent $B C$ is equal to the inscribed angle subtending $E M$. Therefore, $\angle C M E = \angle E A M = 36^{\circ}$.
|
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 the formula for the sum of a geometric progression,
$$
x=p \cdot q, \quad \text{where } p=7^{m}+1, q=1-7^{m}+7^{2m}-7^{3m}+7^{4m}
$$
Note that $p$ and $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then
$$
5=q-(7^{m}+1)-(7^{2m}-1)-(7^{3m}+1)-(7^{4m}-1)
$$
The expressions in parentheses on the right-hand side are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But the remainders of the powers of 7 when divided by 5 are 2, 4, 3, 1, and then they cyclically repeat. Therefore, $p$ is not divisible by 5 for odd $m$, hence $r=1$.
We will prove that the number $p$ is a power of two only when $m=1$. Indeed, let $m>1$ and $7^{m}+1=2^{s}$. Then $s \geqslant 4$ and the number $7^{m}+1$ is divisible by 16. But this is impossible, since the remainders of $7^{m}$ when divided by 16 are only 7 and 1.
If $m=1$, we get $x=16808=2^{3} \cdot 11 \cdot 191$, which fits our criteria. We will show that there are no solutions for $m \geqslant 3$. Consider two cases.
1) $p$ is divisible by 11. Then $m \vdots 5$, that is, $m=5k$. If $k=1$, then $p$ is divisible by 11 and 191. For $k>1$, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ will decompose into two coprime factors that are not powers of two. Therefore, $x$ will have at least three different odd prime divisors, which is impossible.
2) $q$ is divisible by 11. Note that $p$ has an odd divisor, and $q$ is odd and coprime with $p$. Then $q$ must be a power of 11. The remainder of $7^{m}$ when divided by 8 is -1 due to the oddness of $m$. Therefore, the number $q$ gives a remainder of 5 when divided by 8. But the remainders of $11^{s}$ when divided by 8 are only 3 and 1, so $q$ cannot be a power of 11.
|
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 circle $\omega$, a point $L$ is marked such that $\angle LAK = \angle CQB$. Find the angle $\angle PCA$, given that $\angle ALQ = 60^{\circ}$.
|
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 $B L \| P Q$. Then the line $\ell$, passing through point $C$ and perpendicular to the tangent $P Q$, contains the diameter of the circle $\omega$ and is perpendicular to its chord $B L$. Therefore, points $B$ and $L$ are symmetric with respect to $\ell$, meaning that $\ell$ is the axis of symmetry of the trapezoid $P Q L B$. From this, it follows that $B P = L Q$ and $B C = L C$. Hence, triangles $Q L C$ and $P B C$ are congruent by the three sides, so $\angle Q L C = \angle A B C$. Additionally, $\angle A L C = \angle A B C$ as inscribed angles subtending the same chord. Therefore, $L C$ is the angle bisector of $\angle A L Q$, and
$$
\angle A B C = \angle A L C = \frac{1}{2} \angle A L Q = 30^{\circ}.
$$
Note that the angle between the chord $A C$ and the tangent $C P$ is equal to the inscribed angle subtending $A C$. Therefore, $\angle P C A = \angle A B C = 30^{\circ}$.
|
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 marked. Therefore, no more than \(1007 + 2 = 1009\) vertices can be marked.
On the other hand, number the vertices of the 2016-gon in order from 1 to 2016 and mark the first 1009 vertices. Only the vertices numbered 1 and 1009 will be diametrically opposite. Therefore, 1009 satisfies the condition of the problem.
|
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 $q$ are coprime. Indeed, let the number $r$ divide both $p$ and $q$. Then
$$
5=q-(6^{m}+1)-(6^{2m}-1)-(6^{3m}+1)-(6^{4m}-1)
$$
The expressions in parentheses on the right-hand side are divisible by $p$ and, therefore, by $r$. Thus, $r$ is a divisor of 5. But the remainder of the division of $p$ by 5 is 2 for any $m$. Therefore, $p$ is not divisible by 5, which means $r=1$.
Since the number $m$ is odd, one of the prime divisors of $p$ is 7. We will prove that $p$ is a power of 7 only when $m=1$. Suppose $m \geqslant 3$ and $6^{m}+1=7^{s}$. Then the number $7^{s}-1$ is divisible by $2^{m}$ and, therefore, by 8. This implies that $s$ is even, that is, $s=2k$. Therefore, $2^{m} \cdot 3^{m}=(7^{k}-1)(7^{k}+1)$. Since $7^{k}+1$ is not divisible by 3, the number $7^{k}-1$ is divisible by $3^{m}$ and, therefore, $7^{k}-1 \geqslant 3^{m}$. But then
$$
7^{k}+1 \leqslant 2^{m} \cdot 1
$$
1) $p$ is divisible by 11. If $p$ is not a power of 7, we can apply the same reasoning to $p$ as to $x$. In both cases, $p$ is the product of 7 and two coprime factors that are not powers of 7. Therefore, $x$ has at least four different prime divisors, which is impossible.
2) $q$ is divisible by 11. Note that $p$ is divisible by 7 and is not a power of 7, and $q$ is coprime with $p$. Then $q$ must be a power of 11. The remainder of the division of $6^{m}$ by 7 is -1 due to the oddness of $m$. Therefore, the number $q$ gives a remainder of 5 when divided by 8. But the remainders of the division of $11^{s}$ by 7 can only be 4, 2, and 1, so $q$ cannot be a power of 11.
|
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 move loses. Sets of items are considered different if they differ by at least one item or if they contain a different number of items. In the game, Vasily and Fyodor are participating; each player has the opportunity to make at least one move. It is known that: a) Vasily is 2 years older than Fyodor; b) Fyodor is at least 5 years old; c) Fyodor always wins. What is the minimum number of years Vasily can be?
|
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., $C_{64}^{x}$, will always win. Let $V$ be Basil's age in years, and $F$ be Fyodor's age in years. According to the conditions: $5 \leq F, F+2=V, V \leq 64$ (since each player can make a move, i.e., at least one combination can be chosen). It is also known that $C_{64}^{F}>C_{64}^{V}$ always (in the case of equality, Fyodor would lose if he went first). We need to find the minimum $V$ that satisfies all the listed conditions.
Let Fyodor's age be denoted by $x$, and Basil's age by $x+2$. Consider the integer function $x \mapsto C_{64}^{x}$. It increases from 0 to 32 and decreases from 32 to 64. Additionally, $C_{64}^{31}=C_{64}^{33}$. Therefore, the inequality $C_{64}^{x}>C_{64}^{x+2}$ holds if and only if $x \geq 32$. Thus, the minimum value of $x+2$ is 34.
|
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, there are always at least two identical ones?
|
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 of units. Since along with the number $\overline{x y z}$, the number $\overline{z y x}$ is also considered, then $x>0$ and $z>0$. According to the condition, $100 x+10 y+z+297=100 z+10 y+x$. Let's transform this:
$$
99 z-99 x=297 \Leftrightarrow x=z-3
$$
Since $0<x=z-3 \leq 9$ and $z \leq 9$, then $0<x \leq 6$. Thus, we have 6 suitable values for $x$. Each $x$ corresponds to one value of $z$, and $y$ is any number from 1 to 9 for which $\overline{x y z}$ is divisible by 3. Note that
$$
0=(x+y+z) \quad \bmod 3=(2 x+y) \quad \bmod 3
$$
which means any pair $(x, z)$ uniquely determines $y \bmod 3$. Among the numbers from 1 to 9, there are exactly 3 numbers with a given remainder when divided by 3. Therefore, the condition of the problem is satisfied by $6 \cdot 3=18$ numbers, and the minimum number of gnomes is 19.
|
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
\[
\begin{gathered}
\frac{1}{m}+\frac{1}{n}=\frac{1}{2020} \Leftrightarrow \frac{n+m}{m \cdot n}=\frac{1}{2020} \Leftrightarrow m n=2020(n+m) \Leftrightarrow \\
\Leftrightarrow m n-2020 n-2020 m+2020^{2}-2020^{2}=0 \Leftrightarrow(n-2020)(m-2020)=2020^{2}
\end{gathered}
\]
The solutions to this equation are \( m = d + 2020 \) and \( n = \frac{2020^{2}}{d} + 2020 \), where \( d \) is any divisor of \( 2020^{2} \). It remains to find the number of such divisors. Note that \( 2020^{2} = 2^{4} \cdot 5^{2} \cdot 101^{2} \). If \( d \) is a divisor of \( 2020^{2} \), then 2 can appear in its prime factorization in powers \( 0, \ldots, 4 \), and 5 and 101 can appear in powers \( 0, 1, 2 \). Therefore, the total number of such factorizations is \( 5 \cdot 3 \cdot 3 = 45 \).
|
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 considered to be laying without gaps, without overlaps, and not extending beyond the boundaries of the area.
|
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 room with length 12, an integer number of tiles with side length 1.5 can be placed; and, obviously, along the side of the room with length 8, an integer number of tiles with side length 2 can also be placed). The area of the room is $8 \times 12 = 96$, and the area of one tile is $1.5 \times 2 = 3$. Therefore, we get that a total of $96 \div 3 = 32$ tiles are required.
|
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 accidentally told the truth. How many liars can sit at this table?
|
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, but this is impossible since the knight would then be lying. Therefore, no two knights sit next to each other, and the neighbors of each knight are liars.
We will call those liars who did not make a mistake, true liars. According to the condition, the neighbors of each true liar are from different tribes. Therefore, the neighbors of true liars must be a knight and a liar. This means the sequence "LRLRL...LRL" repeats until it encounters a false liar. If a knight sits before him, then a knight also sits after him. If a liar sits before him, then a liar also sits after him. Therefore, the seating arrangement with false liars can be obtained from the seating arrangement "LRLRL...LRL" by either seating a false liar between two liars or by a true liar leaving the table (and then his neighbor liar becomes a false liar). The first action increases the remainder of the total number of people sitting at the table divided by 1, the second decreases it by 1. Since in the seating arrangement "LRLRL...LRL" the number of islanders is divisible by 3, and 2017 gives a remainder of 1 when divided by 3, we need to reduce the remainder 0 by 1 twice. Thus, we need to remove two liars from the seating arrangement of 2019 islanders. Therefore, the number of liars is $2019 \cdot \frac{2}{3}-2=1344$.
|
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 $\left(a_{1}-a_{2}\right)^{4}+\left(a_{2}-a_{3}\right)^{4}+\left(a_{3}-a_{4}\right)^{4}+\left(a_{4}-a_{5}\right)^{4}+\left(a_{5}-a_{6}\right)^{4}+\left(a_{6}-\right.$ $\left.a_{7}\right)^{4}+\left(a_{7}-a_{1}\right)^{4} \geqslant 82$.
|
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 the smallest of them, and $a_{k}$ be the largest. Then $\left|a_{i}-a_{i+1}\right|$ is the length of the segment connecting points $a_{i}$ and $a_{i+1}$. Let's assume for definiteness that $k>j$. Then $\left|a_{k}-a_{j}\right| \leqslant\left|a_{j+1}-a_{j}\right|+\ldots+\left|a_{k}-a_{k-1}\right|$ and $\left|a_{j}-a_{k}\right| \leqslant\left|a_{k+1}-a_{k}\right|+\ldots+\left|a_{7}-a_{6}\right|+\left|a_{1}-a_{7}\right|+\ldots+\left|a_{j}-a_{j-1}\right|$. Adding these inequalities, we get the inequality
$$
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\ldots+\left|a_{7}-a_{1}\right| \geqslant 2\left(a_{k}-a_{j}\right) \geqslant 12
$$
Thus, the sum of the absolute values is at least 12, which means we have either at least six twos, or five twos and two ones. The sum of the fourth powers of the differences in both cases is no less than 82.
|
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 attack each other, so there are no rooks in the rows where they are located. Thus, rooks can only be in $300-2k$ rows. Since there are no more than two rooks in each of these rows, the total number of rooks is no more than $2(300-2k) + 2k = 600 - 2k$. On the other hand, in $k$ columns, there are two rooks each, and in the remaining $300-k$ columns, there is no more than one rook, so the total number of rooks is no more than $(300-k) + 2k = 300 + k$. Therefore, the total number of rooks is no more than $\frac{1}{3}((600-2k) + 2(300+k)) = 400$.
Next, we will show how to place 400 rooks. On a $3 \times 3$ board, 4 rooks can be placed as shown in the figure, and then 100 such squares can be placed along the diagonal.
Second solution. On the board, there can be rooks that do not attack any other rooks, as well as pairs of rooks that attack each other. Let there be $k$ single rooks and $n$ pairs of rooks that attack each other. Then the total number of rooks on the board is $k + 2n$. We will count the total number of occupied rows and columns (for convenience, we will call them lines). Each single rook occupies its own row and its own column, i.e., two lines. Each pair of rooks occupies three lines. Therefore, the total number of lines occupied is $2k + 3n$. Thus, $2k + 3n \leq 600$. Therefore, $k + 2n = \frac{2}{3}(2k + 3n) - \frac{k}{3} \leq \frac{2}{3}(2k + 3n) \leq \frac{2}{3} \cdot 600 = 400$.
The example of placement is the same as in the first solution.
|
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. Therefore, a total of no more than $2 \cdot 2 + 5 \cdot 3 = 19$ chips can be placed on the black cells.
chips. Similarly, no more than 19 chips can be placed on the white cells. Therefore, a total of no more than 38 chips can be placed on the board.
An example of placing 38 chips is shown in the figure.
|
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 this table?
|
# 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 $2 k \leqslant 50$. Moreover, to the left of this pair there is definitely a liar, and all these liars are different. Let's call such two knights and a liar a triplet. In triplets, $3 k$ islanders are involved. Consider the remaining $100-3 k$ islanders. No two knights among them sit next to each other, nor next to knights from the triplets (otherwise, three knights in a row would form, which is impossible). Then to the left of each knight, there must be a liar, all these liars are different and distinct from the liars included in the triplets. Thus, the number of knights is no more than $2 k+\frac{1}{2}(100-3 k)=50+\frac{k}{2} \leqslant 50+\frac{25}{2}=67 \frac{1}{2}$. Therefore, the number of knights is no more than 67.
Here is an example showing that 67 knights can be present:
$$
\text { LKL LKL LKL ... LKL LK LK LK ... LK L }
$$
(the block "LKL" is repeated 25 times, the block "LK" - 12 times).
|
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 purchase, 75% of the remaining respondents bought the product under the influence of advertising, and the number of customers who chose the product based on the advice of a sales consultant is one-third of the number who chose the product under the influence of advertising.
|
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 integer, $x-7$ must be divisible by 4. The maximum suitable number $x$, less than 50, is 47.
|
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 considered to be laying without gaps, overlaps, and not exceeding the boundaries of the area.
|
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 4. Therefore, to tile the hall, $4 \cdot 4^{2}=64$ tiles will be required.
|
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 pairs of acquainted students that could be among those who arrived at the base?
|
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 is a student $A$ who has three unfamiliar students $B_{1}, B_{2}$, and $B_{3}$. Consider the group of six students: $A, B_{1}$, $B_{2}$, $B_{3}$, and two other arbitrary students $C$ and $D$. They can be seated in two three-person rooms, so $A$ must be in the same room with $C$ and $D$, which means $C$ and $D$ are familiar with each other and with $A$. Therefore, any two students, different from $B_{1}$, $B_{2}$, and $B_{3}$, are familiar with each other. In particular, if some student, different from $B_{1}$, $B_{2}$, and $B_{3}$, is unfamiliar with someone, then this someone is one of $B_{1}$, $B_{2}$, and $B_{3}$. But each $B_{i}$ is unfamiliar with no more than three students, so the number of pairs of unfamiliar students is no more than nine. Therefore, in this case, the number of pairs of familiar students is no less than $175 \cdot 174 / 2 - 9 = 15216 > 15050$. We will now show that 15050 pairs of students can exist. Seat all students at a large round table and make everyone familiar except those sitting next to each other. It is not difficult to verify that this construction satisfies the condition of the problem.
|
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, $x = 11 k + 8$ for some integer $k$, from which we have
$$
(11 k + 8) \cdot 100 + 77 y = 5882 \Longleftrightarrow 1100 k = 5082 - 77 y = 77(66 - y)
$$
The right side is divisible by 100, but 77 and 100 are coprime. Therefore, the number $66 - y$, which lies between 0 and 65, must be divisible by 100. This is only possible when $y = 66$. Therefore, $x = 0.01 \cdot (5882 - 76 \cdot 66) = 8$ and $x + y = 74$.
|
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, we get $30+60=90$.
|
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. Suppose there is a third number $c$ in this group. Then it lies in both $G_{k}$ and $G_{6-k}$. This is possible only for $k=0$ and $k=3$. Numbers not in $G_{0}$ and $G_{3}$ can only be included in two-element groups, and there are 400 such numbers. Therefore, they form at least 200 groups. The sets $G_{0}$ and $G_{3}$ can form groups, but they cannot be combined into one. Thus, at least 202 groups are required in total.
We will provide an implementation of the partition into 202 groups. One of them will be $G_{0}$, another will be $G_{3}$, and we will also choose groups of the form $\{6 n+k, 6 n+6-k\}$, where $n=0,1, \ldots, 99$ and $k=1,2$.
|
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 than $\left[\frac{25}{2}\right]=12$ of them, and the total number of words does not exceed $10+12=22$.
Example of the distribution of repeated words $\left(s_{i}\right)$ :
| | $s_{1}$ | $s_{2}$ | $s_{3}$ | $s_{4}$ | $s_{5}$ | $s_{6}$ | $s_{7}$ | $s_{8}$ | $s_{9}$ | $s_{10}$ | $s_{11}$ | $s_{12}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $M_{1}$ | + | + | + | + | + | | | | | | | |
| $M_{2}$ | + | | | | + | + | + | + | | | | |
| $M_{3}$ | | + | | | | + | | | + | + | | + |
| $M_{4}$ | | | + | | | | + | | + | | + | + |
| $M_{5}$ | | | | + | | | | + | | + | + | + |
|
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 less than $N$ rubles receives a $20\%$ discount on the next purchase (of any number of items).
A customer, visiting the store for the first time, wants to purchase exactly four items with a total cost of 1000 rubles, the cheapest of which costs no less than 99 rubles. Determine the largest $N$, for which the second offer is more beneficial for him.
|
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 $s_{1}, s_{2}, s_{3}, s_{4}$ rubles, where $s_{1} \geqslant s_{2} \geqslant s_{3} \geqslant s_{4} \geqslant X$. In the case of the first special offer, the customer can only receive the item costing $s_{4}$ for free; thus, they will spend $S-s_{4}$ rubles. In the case of the second special offer, the customer will spend
$$
s_{i}+0.8\left(S-s_{i}\right)=0.2 s_{i}+0.8 S
$$
where $s_{i}$ is the minimum of the costs that are not less than $N$ (it exists, otherwise the second offer could not be used).
The second special offer is more beneficial to the customer if
$$
0.2 s_{i}+0.8 S<S-s_{4} \Leftrightarrow 0.2 s_{i}<0.2 S-s_{4} \Leftrightarrow s_{i}<S-5 s_{4}
$$
from which $N \leqslant s_{i}<S-5 s_{4} \leqslant S-5 X$, that is, $N \leqslant S-1-5 X=1000-1-5 \cdot 99=504$.
Let's check that the value 504 is achievable for $N$. Suppose the costs of the items are 99, 198, 199, and 504 rubles. And suppose the customer buys the item costing 504 rubles to use the second special offer. Then they will pay the amount $504+$ $+0.8 \cdot 496=900.8$, which is less than the amount spent in the case of using the first special offer $-1000-99=901$.
|
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: $90^{\circ}$
|
Solution. Let $O$ be the center of $\omega$, $M$ be the intersection point of lines $A B$ and $C D$. Note that
$$
\angle A O C=2\left(180^{\circ}-\angle A B C\right)=2\left(180^{\circ}-135^{\circ}\right)=90^{\circ} .
$$
Then quadrilateral $A O C D$ is a square, and thus $\angle A D C=90^{\circ}$. By the tangent-secant theorem,
$$
D M^{2}=M C^{2}=B M \cdot A M, \quad \text { that is } \quad \frac{D M}{A M}=\frac{B M}{D M} .
$$
Therefore, triangles $D B M$ and $A D M$ are similar, from which
$$
\angle A B D=180^{\circ}-\angle M B D=180^{\circ}-\angle A D M=90^{\circ}
$$
|
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 $\overline{x y z}$ and the number $\overline{z y x}$ are considered together, then $x>0$ and $z>0$. According to the condition, $100 x + 10 y + z + 198 = 100 z + 10 y + x$. Transforming this, we get:
$$
99 z - 99 x = 198 \Leftrightarrow x = z - 2.
$$
Since $0 < x = z - 2 \leq 9$ and $z \leq 9$, then $0 < x \leq 7$. Therefore, there are 7 possible values for $x$. Each $x$ corresponds to one value of $z$, while $y$ can be any digit. Thus, the number of numbers that satisfy the condition of the problem is $7 \cdot 10 = 70$.
|
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 of his hat to be sure of what it is?
|
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 to survey no fewer than 31 people to ensure that the opinion of the knights prevails.
|
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 of people sitting at the table. There is a rule at the court - the lunch of a knight together with the lunch of a courtier equals the lunch of the king. Determine the maximum possible number of knights the king can invite to lunch on this day. How many courtiers will sit at their table in this case?
|
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 value of $b$ is equivalent to minimizing the value of $\frac{1}{b}$, which, from the dining rule, is equivalent to maximizing the value of $\frac{1}{a}$, which is equivalent to minimizing the value of $a$. Therefore, the options should be tried starting with the minimum possible number of courtiers.
$$
\begin{aligned}
& a=12 \Rightarrow b=\frac{7 a}{a-7}=\frac{7 \cdot 12}{5} \notin \mathbb{N} \\
& a=13 \Rightarrow b=\frac{7 a}{a-7}=\frac{7 \cdot 13}{6} \notin \mathbb{N} \\
& a=14 \Rightarrow b=\frac{7 a}{a-7}=\frac{7 \cdot 14}{7}=14
\end{aligned}
$$
Second 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 $\frac{1}{a} + \frac{1}{b} = \frac{1}{7} \Leftrightarrow 7(a+b) = ab$. Thus, the value of $ab$ must be divisible by 7, i.e., one of the numbers $a$ and $b$ must be divisible by 7. If the number of knights is divisible by 7, then it is 14, as among the numbers from 10 to 20, only 14 is divisible by 7. If the number of courtiers is divisible by 7, then it is also 14, as it lies in the range from 12 to 18. Thus, in any case, one of the numbers $a$ and $b$ is 14, and then the other is also 14.
|
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 other hand, this sum is also equal to $1703 = 8 \cdot 213 - 1$. Therefore,
$$
8 \cdot 213 - 1 = 8a - b, \quad \text{from which} \quad 8(a - 213) = b - 1
$$
Then the number $b-1$ is divisible by 8 and lies in the interval $[-5, 3]$. Thus, $b=1, a=213$, and the number that was erased is $213 + 1 = 214$.
|
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 points. The points earned for the number are summed. Which number should be chosen in this game to score the maximum number of points?
|
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 $11+7+5+3=26$ points. If the number is divisible by both 11 and 9, then it is a multiple of 99. In the given range, the only such number is 2079. Since 2079 is also divisible by 7, it gives $11+9+7+3=30$ points.
It remains to note that more than 30 points cannot be scored. Indeed, such a result can only be achieved by a number that is simultaneously divisible by $3, 5, 7, 9$, and 11. The smallest such number is $11 \cdot 9 \cdot 7 \cdot 5=3465$, but it is greater than 2117.
|
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, $\triangle K I A$ is isosceles with base $K I$, hence $A$ lies on the perpendicular bisector of $K I$, from which we have that $A H=a$.
It is easy to notice that $\triangle A H N$ is a right triangle, in which the leg $A H$ is half the hypotenuse $A N$. Therefore, $\angle H A N=60^{\circ}$. Since $A H \| K I$, then $\angle K A H = \angle A K I$. From this, we get that, since point $N$ can be located on different sides of the line $K I$, either $\angle K A N = \angle H A N + \angle H A K = 60^{\circ} + 30^{\circ}$, or $\angle K A N = \angle H A N - \angle H A K = 60^{\circ} - 30^{\circ}$.
|
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 cities connected by express trains?
|
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 remaining cities. Therefore, each city is connected to at least 147 cities. Consequently, the total number of pairs of cities connected by express trains is no less than $\frac{147 \cdot 150}{2}=11025$.
Now, let's show that there can be exactly 11025 pairs of cities. Number the cities from 1 to 150 and connect by express trains all cities except the first and the 150th, as well as cities whose numbers differ by one. Let's check that this construction satisfies the problem's condition. Since each city is connected by express trains to 147 cities, the total number of connected city pairs is exactly $\frac{147 \cdot 150}{2}=11025$. Now, take any quartet of cities. There are two cases.
1) There is a city not connected to two of the three remaining cities. Let city $A$ not be connected to cities $B$ and $C$, but be connected to city $D$. Then cities $B$ and $C$ must be connected to each other, since the remainders of their numbers when divided by 150 differ by 2. Therefore, the pairs $(A, D)$ and $(B, C)$ are suitable.
2) All cities are connected to at least two of the three remaining cities. Let city $A$ be connected to cities $B$ and $C$. By assumption, city $D$ must be connected to $B$ or $C$. If it is connected to $B$, then the pairs $(A, C)$ and $(B, D)$ are suitable, and if to $C$, then the pairs $(A, B)$ and $(C, D)$.
|
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 to at least two cities: one to arrive from, and another to depart to. Therefore, each city is connected to at least 123 cities. Consequently, the total number of pairs of cities connected by express trains is no less than $\frac{123 \cdot 125}{2}=7687 \frac{1}{2}$. But the number of pairs must be an integer, so it is at least 7688.
Now, let's show that the number of pairs of cities can be exactly 7688. Number the cities from 1 to 125 and connect by express trains all pairs of cities, except for pairs with numbers $2k-1$ and $2k$ for $k=1,2, \ldots, 62$. Let's verify that this construction satisfies the problem's condition. Since each city is connected by express trains to 123 cities, and the 125th city is connected to 124 cities, the total number of pairs of connected cities is $\frac{1}{2}(124 \cdot 123 + 124) = 7688$. Now, take any quartet of cities $A, B, C, D$. In this quartet, each city is connected to at least two of the others. Suppose that $B$ is connected to $A$ and $C$. If $D$ is also connected to $A$ and $C$, then the route $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ works. Otherwise, $D$ is not connected to exactly one of the cities $A$ and $C$ (let it be $C$). Then $C$ is connected to $A$, and $D$ is connected to $B$, and the cities can be traveled in a circle $A \rightarrow C \rightarrow B \rightarrow D \rightarrow A$.
|
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 exist in Sosnovka?
|
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, since next to Petya there will inevitably be a person he is not acquainted with. Therefore, each resident of the village cannot have more than two unfamiliar people. Hence, each is acquainted with at least 237 residents, and the total number of pairs of acquaintances is no less than $\frac{240 \cdot 237}{2}=28440$.
Now, let's show that the number of pairs of acquaintances can be exactly 28440. Seat all residents around a large round table and introduce everyone except those sitting next to each other. Verify that this construction satisfies the problem's condition. Since each resident is not acquainted with exactly two others, he is acquainted with exactly 237 people, and the total number of pairs of acquaintances is $\frac{240 \cdot 237}{2}=28440$. Now, take any five residents and seat them in the required manner. Remove all others from the large round table and denote the remaining in cyclic order as $A, B, C, D, E$. Then the seating arrangement $A, C, E, B, D$ will work. Indeed, any two new neighbors did not sit next to each other at the large round table and therefore know each other.
|
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, a difference of 1 point between opponents is allowed. The tournament ends as soon as a sole leader is determined. How many schoolchildren will finish the tournament with 5 points?
|
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 if athletes with $n$ and $n+1$ points meet, they will end up with $n$ and $n+2$ points, respectively, regardless of the outcome. We will prove two statements.
1) In a tournament with $2^n$ participants, after $m$ rounds, $2^{n-m} \cdot C_m^k$ athletes will have $k$ points, where $m \leq n$ and $k \in \{0, \ldots, m\}$. Let $f(m, k)$ be the number of participants who have $k$ points after $m$ rounds. We will use induction on $m$. If $m=0$, then $k=0$, and $f(0,0)=2^n$. For the inductive step, assume that the required equality holds for some $m < n$. Then each group with the same number of points contains an even number of participants, so all participants can be paired. After the $(m+1)$-th round, the groups of athletes with no wins and no losses will be halved. Therefore,
$$
f(m+1,0)=\frac{1}{2} f(m, 0)=\frac{1}{2} \cdot 2^{n-m}=2^{n-(m+1)} \cdot C_{m+1}^{0} \quad \text{and} \quad f(m+1, m+1)=\frac{1}{2} f(m, m)=2^{n-(m+1)} \cdot C_{m+1}^{m+1}
$$
Now, let $k \in \{1, \ldots, m\}$. After the $(m+1)$-th round, $k$ points will be held by those who had $k$ points and lost, as well as those who had $k-1$ points and won. Therefore,
$$
f(m+1, k)=\frac{1}{2} \cdot f(m, k)+\frac{1}{2} \cdot f(m, k-1)=2^{n-m-1}\left(C_m^k + C_m^{k-1}\right)=2^{n-(m+1)} \cdot C_{m+1}^k
$$
(in the last transition, we used the main identity of Pascal's triangle). Thus, the inductive step is complete.
2) In a tournament with $2^n - 2$ participants and fewer than $n$ rounds, the numbers $f(2m, k)$ are even for any $k$ from 0 to $n$, and the numbers $f(2m-1, k)$ are odd only for $k=m-1$ and $k=m$. We will use induction on $n$. For $n=2$, everything is obvious. Assume that the statement is true for some $n$. Then, in a tournament with $2^{n+1}$ participants, after the first round, $2^n - 1$ participants will have 0 points and 1 point. After the second round, $2^{n-1}$ participants will have 0 points and 2 points, and $2^n - 2$ participants will have 1 point. For the first two groups, we use 1), and for the third group, we use the inductive hypothesis.
Now, introduce fictitious participants $A$ and $B$. Let them play against each other in odd rounds, with $A$ winning; this does not affect the other participants. Additionally, let $A$ play against some $A'$ and lose, and $B$ win against some $B'$ in the $(2m)$-th round. Then $A'$ must have $m$ points, and $B'$ must have $(m-1)$ points. By 2), the numbers $f(2m-1, m-1)$ and $f(2m-1, m)$ are odd, and by 1), after adding $A$ and $B$, they become even. Therefore, in the absence of $A$ and $B$, participants $A'$ and $B'$ must play against each other; as a result, they will have $m-1$ and $m+1$ points, respectively, just as after playing against $A$ and $B$. Thus, adding $A$ and $B$ does not affect the distribution of points among the other participants.
Now, apply statement 1) for $n=8$. The tournament will end after 8 rounds, and $C_8^5 = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$ participants will have 5 points. Note that $A$ and $B$ will have 4 points each and, therefore, will not be included in this number.
|
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 Beryozovka?
|
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, since there will inevitably be an unfamiliar person sitting next to Vasya. Therefore, each resident of the village cannot have more than three unfamiliar people. Hence, each is acquainted with at least 196 residents, and the number of pairs of acquaintances is no less than $\frac{200 \cdot 196}{2}=19600$.
Now, let's show that the number of pairs of acquaintances can be exactly 19600. Seat all residents around a large round table. We will call two people almost opposite if there are at least 98 people sitting between them in both directions. Now, let's make all residents acquainted with each other except those sitting almost opposite. Since each resident is not acquainted with exactly three others, each is acquainted with exactly 196 people, and the total number of pairs of acquaintances is $\frac{200 \cdot 196}{2}=19600$. Now, take any six residents and denote them (in the order of their seating at the table) as $A_{1}, \ldots, A_{6}$. If in this cycle all neighbors are acquainted with each other, the desired seating arrangement is found. Otherwise, there exist two unfamiliar neighbors (let these be $A_{1}$ and $A_{2}$). Number all people at the large table in the same cyclic order from 1 to 200, starting with $A_{1}$. Note that the numbers of two unfamiliar residents can differ only by 99, 100, or 101. Let $N_{1}, \ldots, N_{6}$ be the numbers of $A_{1}, \ldots, A_{6}$, respectively. Then $N_{1}=1$ and $N_{2} \geqslant N_{1}+99=100$. Since the numbers $N_{i}$ are natural and strictly increasing, for $i \geqslant 2$ the inequalities $98+i \leqslant N_{i} \leqslant 194+i$ hold. Let's make two simple observations.
1) Resident $A_{1}$ is acquainted with $A_{5}$ and $A_{6}$. Indeed, for $i=5$ or $i=6$,
$$
N_{i}-N_{1}=N_{i}-1 \geqslant N_{5}-1 \geqslant 103-1=102>101
$$
2) For $2 \leqslant j < i \leqslant 6$, $A_{i}$ is acquainted with $A_{j}$. Indeed, $N_{i} > N_{j}$ and
$$
N_{i}-N_{j} \leqslant 194+i-(98+j)=96+(i-j) \leqslant 98<99
$$
Now, from 1) and 2) it follows that the cyclic seating arrangement $A_{1} \rightarrow A_{5} \rightarrow A_{3} \rightarrow A_{2} \rightarrow A_{4} \rightarrow A_{6} \rightarrow A_{1}$ satisfies the condition of the problem.
|
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 the second loss. The tournament continues as long as at least one pair of opponents can be formed. How many rounds will have to be played?
|
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 $m$. For $m=1$, everything is obvious. Suppose for some $m<n$ the statement is true. Then before the $(m+1)$-th round, all participants will be divided into pairs. After the round, the number of participants without any losses will be halved, i.e., it will become $2^{n-m-1}$. One loss will be incurred by those who lost in the group of leaders (there are $2^{n-m-1}$ such people), and by those who already had one loss and won (by the induction hypothesis, there are $m \cdot 2^{n-m-1}$ such people). In total, we will get $(m+1) \cdot 2^{n-m-1}$. The rest of the participants will leave the tournament. Thus, the inductive step is completed.
2) We can divide the participants into two groups of 1024 and 128 participants and assume that for the first 10 rounds they play separate tournaments. Indeed, in each round, we will only pair participants from the same group. By 1), all participants in the first group can be paired. Therefore, if someone is left without a pair in the overall tournament, we can assume that this participant is from the second group. After 10 rounds, the results of the two groups are simply combined.
Now let's count the number of rounds. We will record the remaining participants as pairs $(a, b)$, where $b$ is the number of leaders, and $a$ is the number of participants with one loss. By 1), in the first group, after the tenth round, there will be $(10,1)$ participants. In the second group, after the seventh round, there will be $(7,1)$ participants, and their number will change as follows:
$$
(7,1) \rightarrow(4,1) \rightarrow(2,1) \rightarrow(1,1)
$$
In total, after the tenth round, there will be $(11,2)$ participants remaining in the tournament. Their further dynamics up to the fourteenth round are as follows:
$$
(11,2) \rightarrow(7,1) \rightarrow(4,1) \rightarrow(2,1) \rightarrow(1,1)
$$
After this, no more pairs can be formed, and the tournament ends.
|
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 this perpendicular (see the left figure). By symmetry, $\angle C^{\prime} B A = \angle C A B = 2 \angle A B C$, hence $\angle C^{\prime} B C = \angle A B C$. On the other hand, lines $C C^{\prime}$ and $A B$ are parallel, so $\angle A B C = \angle B C C^{\prime}$. Therefore, $\angle B C C^{\prime} = \angle C^{\prime} B C$, triangle $B C^{\prime} C$ is isosceles, and $B C^{\prime} = C^{\prime} C$. By symmetry, $B C^{\prime} = A C$ and $C P = C^{\prime} P$, and by the condition $A C = C P$. Thus,
$$
C^{\prime} P = C P = A C = B C^{\prime} = C^{\prime} C
$$
and triangle $P C C^{\prime}$ is equilateral. Therefore, $\angle C C^{\prime} P = 60^{\circ}$. Note that
$$
180^{\circ} - 2 \angle P B C^{\prime} = \angle P C^{\prime} B = \angle B C^{\prime} C - 60^{\circ} = 120^{\circ} - 2 \angle C B C^{\prime}
$$
Therefore,
$$
\angle C B P = \angle P B C^{\prime} - \angle C B C^{\prime} = \frac{1}{2} \left(180^{\circ} - 120^{\circ}\right) = 30^{\circ}.
$$
Solution 2. Let $\alpha = \angle P A B, \beta = \angle P A C, \gamma = \angle P B C, \varphi = \alpha + \gamma$ (see the right figure). By the condition, $\angle P B A = \alpha, \angle C P A = \beta$, and
$$
\alpha + \beta = 2 \varphi = 2(\alpha + \gamma) \Longleftrightarrow 2 \gamma = \beta - \alpha
$$
By the Law of Sines,
$$
\frac{A C}{\sin \varphi} = \frac{A B}{\sin \left(180^{\circ} - 3 \varphi\right)} \Longleftrightarrow \frac{\sin 3 \varphi}{\sin \varphi} = \frac{A B}{A C}
$$
Note that
$$
A B = 2 A P \cos \alpha = 4 A C \cos \alpha \cos \beta \Longleftrightarrow \frac{A B}{A C} = 2(\cos (\alpha + \beta) + \cos (\beta - \alpha)) = 2 \cos 2 \varphi + 2 \cos 2 \gamma
$$
Moreover,
$$
\frac{\sin 3 \varphi}{\sin \varphi} = 3 - 4 \sin^2 \varphi = 2 \cos 2 \varphi + 1
$$
Therefore,
$$
2 \cos 2 \varphi + 2 \cos 2 \gamma = 2 \cos 2 \varphi + 1 \Longleftrightarrow \cos 2 \gamma = \frac{1}{2} \Longleftrightarrow 2 \gamma = 60^{\circ} \Longleftrightarrow \gamma = 30^{\circ}
$$
|
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 could have come to the camp knowing each other?
|
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 acquainted with no more than one person. Therefore, each student has no more than three unknowns. Let's consider two cases.
1) Each student is not acquainted with no more than two children. Then any student has at least 172 acquaintances, and the total number of pairs of acquaintances is no less than $\frac{175 \cdot 172}{2}=15050$.
2) There exists a student who is not acquainted with three children. Let's denote such a student by $A$, and the unknown people by $B_{1}, B_{2}$, and $B_{3}$. We will select the following six children: $A, B_{1}, B_{2}, B_{3}$, and two more arbitrary students $C$ and $D$. Since they can be seated in two three-person rooms, $A$ must end up in a room with $C$ and $D$. But then students $C$ and $D$ must be acquainted with each other and with $A$. Thus, we have established that any students, other than $B_{1}, B_{2}$, and $B_{3}$, are acquainted with each other. In particular, this means that any student, other than $B_{1}, B_{2}$, and $B_{3}$, can be not acquainted only with someone from $B_{i}$. But each $B_{i}$ is not acquainted with no more than three students, so the number of pairs of unknowns does not exceed nine. Therefore, the number of pairs of acquaintances is no less than $\frac{175 \cdot 174}{2}-9=15216>15050$.
Considering 1) and 2), it remains to show that the number of pairs of acquaintances can be exactly 15050. Let's seat all the students at a large round table and make everyone acquainted except those sitting next to each other. Let's check that this construction satisfies the problem's condition. Since each student is acquainted with 172 students, the total number of pairs of acquaintances is exactly $\frac{175 \cdot 172}{2}=15050$. Take any six students and seat them in two rooms. Remove all the others from the large round table and denote the remaining ones in a cyclic order as $A, B, C, D, E, F$. Now seat students $A, C, E$ in one room, and $B, D, F$ in the other. The inhabitants of each room are not sitting next to each other at the table, and, in particular, they were not sitting next to each other at the large table. Therefore, they all know each other.
|
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 $A P=L Q$ by the condition and $A L=L B$ as chords corresponding to equal angles. Moreover,
$$
\angle A P L=\angle A_{1} P C=90^{\circ}-\alpha=\angle B_{1} Q C=\angle B Q L
$$
Then, by the Law of Sines,
$$
\frac{A P}{\sin \angle A L P}=\frac{A L}{\sin \angle A P L}=\frac{L B}{\sin \angle B Q L}=\frac{L Q}{\sin \angle L B Q}, \quad \text { hence } \quad \sin \angle A L P=\sin \angle L B Q
$$
But $\angle A L P=\angle A B C<90^{\circ}$ and
$$
\angle L B Q=180^{\circ}-\angle B L C-\angle B Q L=90^{\circ}-\angle B A C+\alpha<90^{\circ}-\angle B A C+\angle A C B \leqslant 90^{\circ}
$$
(the last inequality is true because $B C \geqslant A B$). Therefore, the angles $A L P$ and $L B Q$ are acute, and they are equal due to the equality of their sines. Thus, triangles $A L P$ and $B L Q$ are equal by a side and two angles. Therefore, $\angle L A P=\angle Q L B=\gamma$. Now from triangle $A L P$
$$
\beta+\gamma+90^{\circ}-\alpha=180^{\circ} \Longrightarrow \beta+\gamma=90^{\circ}+\alpha
$$
and from quadrilateral $A L B C$
$$
\beta+\gamma=180^{\circ}-2 \alpha \Longrightarrow 180^{\circ}-2 \alpha=90^{\circ}+\alpha \Longrightarrow \alpha=30^{\circ} \Longrightarrow \angle A C B=60^{\circ}
$$
|
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. Thus, each student is acquainted with at least 220 students, and the number of pairs of acquaintances is no less than $\frac{225 \cdot 220}{2}=24750$.
Now, let's show that the number of pairs of acquaintances can be exactly 24750. Seat all the students around a large round table and introduce everyone except those sitting next to each other and those sitting one apart. Let's check that this construction satisfies the condition of the problem. Since each student is not acquainted with exactly four other students, each student is acquainted with exactly 220 students, and the total number of pairs of acquaintances is $\frac{225 \cdot 220}{2}=24750$.
Now consider any group of six students. Leave only them at the large round table. Notice that each of them is acquainted with the one sitting "opposite" him (i.e., the one sitting two seats away), since at the large table, he could not be either his neighbor or the one sitting one seat away. Thus, we have selected three non-overlapping pairs of acquaintances.
|
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 sole leader is determined. How many schoolchildren will finish the tournament with 6 points?
|
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 statements are true. Then, in a tournament with $2^{n+1}$ participants, after the first round, $2^{n}$ students will have 0 points and $2^{n}$ students will have 1 point. We will continue to conduct the draw so that participants from these groups only meet each other. By the inductive hypothesis, this is always possible, and such an assumption clearly does not affect the distribution of points. In other words, we are conducting two independent tournaments with $2^{n}$ participants each. By the inductive hypothesis, in the first tournament, $C_{n}^{k}$ students will score $k$ points, and in the second tournament, $-C_{n}^{k-1}$ students will score $k$ points, since all participants started with one point. In the end, $C_{n}^{k}+C_{n}^{k-1}=C_{n+1}^{k}$ students will score $k$ points (we used the main identity of Pascal's triangle). Thus, the inductive step is completed. It remains to note that $C_{9}^{6}=\frac{9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3}=84$.
|
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 $C P$ as $D$ (see the left figure). Triangles $B P D$ and $A P C$ are similar with a coefficient of $\frac{1}{2}$, hence $D B = \frac{1}{2} A C = A H$. Therefore, $A D B H$ is a rectangle, which means $\angle A D B = 90^{\circ}$. Note that triangles $A B Q$ and $C A P$ are congruent by two sides and an angle. Then
$$
\angle B D K = \angle D C A = \angle B A K.
$$
Thus, quadrilateral $A D B K$ is cyclic, from which $\angle A K B = 180^{\circ} - \angle A D B = 90^{\circ}$.
Solution 2. Draw the height $C H$ in triangle $A B C$ (see the right figure). Since $B H = \frac{1}{2} A B$, we get $\frac{B P}{B H} = \frac{2}{3} = \frac{B Q}{B C}$. Therefore, triangles $B P Q$ and $B H C$ are similar, from which $\angle B P Q = \angle B H C = 90^{\circ}$. Note now that $B Q = A P$, $A B = C A$, and $\angle A B Q = \angle C A P = 60^{\circ}$. Then triangles $A B Q$ and $C A P$ are congruent by two sides and an angle. Since $\angle A Q B = \angle C P A = 180^{\circ} - \angle C P B$, quadrilateral $B P K Q$ is cyclic, from which
$$
\angle A K B = 180^{\circ} - \angle B K Q = 180^{\circ} - \angle B P Q = 90^{\circ}
$$
|
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 several layovers). What is the maximum number of companies that can buy the airlines?
|
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 101, then the number $bk - ak = (b-a)k$ is divisible by 101, which is impossible because $1 \leqslant k \leqslant 100$ and $0 < b-a < 101$. Now consider the numerical sequence
$$
\mathcal{P}_{k}: \quad 0, k, 2k, 3k, \ldots, 99k, 100k, 100k+101, 99k+101, \ldots, 3k+101, 2k+101, k+101, 101
$$
Notice that all the remainders of the numbers in $\mathcal{P}_{k}$ when divided by 202 are distinct. Indeed, if two different numbers in $\mathcal{P}_{k}$ give the same remainder when divided by 202, then they also give the same remainder when divided by 101. Then these numbers are of the form $ak$ and $ak + 101$, which is impossible because they have different parities. Since $\mathcal{P}_{k}$ contains exactly 202 numbers, their remainders take all possible values from 0 to 201.
Now let's move on to solving the problem. Since each airline must ensure communication between all cities, it must have at least 201 air routes. The total number of air routes is $\frac{202 \cdot 201}{2}$, so the number of airlines does not exceed 101.
We will now show that the number of airlines participating in the privatization of air routes can be exactly 101. Let's agree to denote the remainder of the division of an integer $a$ by 202 by $\langle a \rangle$. We will number the cities from 0 to 201. For any $k = 1, \ldots, 100$, we will assign to the $k$-th airline the air routes connecting cities with numbers $\langle a \rangle$ and $\langle b \rangle$, where $a$ and $b$ are adjacent numbers in the sequence $\mathcal{P}_{k}$. Since the remainders of the numbers in $\mathcal{P}_{k}$ run through (in a certain order) all values from 0 to 201, the airline can ensure travel from any city to any other. Note also that no air route will be assigned to two airlines at the same time. Indeed, let $k$ and $m$ be different numbers from 1 to 100, and $(a, b)$ and $(c, d)$ be pairs of adjacent numbers from the sequences $\mathcal{P}_{k}$ and $\mathcal{P}_{m}$ respectively, such that $\{\langle a \rangle, \langle b \rangle\} = \{\langle c \rangle, \langle d \rangle\}$. Then, by the above, $\{a, b\} = \{c, d\}$, which is impossible since $b - a = k \neq m = d - c$.
We will sell all the remaining air routes from the first hundred airlines to the 101st airline. We will prove that it can also ensure travel from any city to any other. Obviously, it owns the air route $(0, 101)$, since 0 and 101 do not neighbor in the sequence $\mathcal{P}_{k}$ for any $k$. Now we will show that for any $m \in \{1, \ldots, 100\}$, the air routes $(0, 202 - m)$ and $(101, 101 - m)$ also belong to the 101st airline. Indeed, let the route $(0, 202 - m)$ belong to the airline with number $k \leqslant 100$. Then $202 - m = k$, which is impossible since $k + m \leqslant 200$. The second statement is verified similarly. Thus, the 101st airline connects any two cities via the cities 0 and 101.
|
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,
$$
A C \cdot B C=6+3 \sqrt{3}=3(2+\sqrt{3})=A L \cdot K B, \quad \text { or } \quad \frac{K B}{B C}=\frac{C A}{A L} .
$$
Since $\angle K B C=45^{\circ}=\angle C A L$, triangles $C B K$ and $L A C$ are similar. Therefore, $\angle B C K=\angle A L C$ and
$$
\angle A L C=\angle B C L+\angle L B C=\angle B C L+45^{\circ}=\angle B C K-\angle K C L+45^{\circ}=\angle A L C-\angle K C L+45^{\circ},
$$
from which $\angle K C L=45^{\circ}$.
Solution 2. Extend triangle $A B C$ to form a square $A B C D$, and let $M$ be the intersection point of lines $B D$ and $C L$ (see the figure). Let $a=A C$. Triangles $A L C$ and $B L M$ are similar with a ratio of $\sqrt{3}$. Therefore,
$$
B M=\frac{a}{\sqrt{3}}, \quad C M=\sqrt{a^{2}+\frac{a^{2}}{3}}=\frac{2 a}{\sqrt{3}}, \quad \frac{A B}{C M}=\sqrt{\frac{3}{2}}
$$
Moreover,
$$
B L=\frac{A B}{\sqrt{3}+1}, \quad C L=\frac{C M \cdot \sqrt{3}}{\sqrt{3}+1}, \quad \frac{C L}{B L}=\sqrt{3} \cdot \frac{C M}{A B}=\sqrt{2}
$$
and also
$$
K L=\frac{2 A B}{\sqrt{3}+3}, \quad M L=\frac{C M}{\sqrt{3}+1}, \quad \frac{K L}{M L}=\frac{2}{\sqrt{3}} \cdot \frac{A B}{C M}=\sqrt{2}
$$
Thus, $\frac{C L}{B L}=\frac{K L}{M L}$, and triangles $B L M$ and $C L K$ are similar. Therefore, $\angle K C L=\angle M B L=45^{\circ}$.
|
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 with several layovers). What is the maximum number of companies that can buy the air routes?
|
# 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 $a k$ and $b k$ give the same remainder when divided by 127, then the number $b k - a k = (b - a) k$ is divisible by 127, which is impossible since $0 < b - a < 127$. Therefore, the numbers $N_{k, 0}, \ldots, N_{k, 126}$ take all values from 0 to 126, and $N_{k, 127}$ is zero.
Now let's proceed to solving the problem. Take an arbitrary airline participating in the privatization. Since it must ensure communication between all cities on its own, for any city, it must have an airline leading to this city and an airline leading out of it. Thus, each city must be connected by at least two airlines belonging to this company. Therefore, the company must have at least 127 airlines. The total number of airlines is $\frac{127 \cdot 126}{2} = 127 \cdot 63$, so the number of airlines does not exceed 63.
Now we show that exactly 63 airlines could have participated in the privatization. Number the cities from 0 to 126. For any $k = 1, \ldots, 63$, assign to the $k$-th airline the airlines leading from the city with number $N_{k, n}$ to the city with number $N_{k, n+1}$, where $n \in \{0, \ldots, 126\}$. By the above, such airlines, starting and ending in city 0, cyclically pass through (in a certain order) all other cities. Thus, the $k$-th airline can ensure a flight from any city to any other. It remains to check that no airline is assigned to two companies at the same time. Indeed, let $k, m \in \{1, \ldots, 63\}$ and some airline is assigned to both the $k$-th and $m$-th companies. Then there exist $i, j \in \{1, \ldots, 126\}$ such that $N_{k, i} = N_{m, j}$ and $N_{k, i+1} = N_{m, j+1}$. We get $N_{k, i+1} - N_{k, i} = N_{m, j+1} - N_{m, j}$, which implies $k = m$.
|
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 \( n \), while if they are connected, then the numbers \( a + b \) and \( n \) must have a common natural divisor greater than 1. For what smallest \( n \) does such an arrangement exist?
|
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 among four consecutively connected numbers, there exists a pair of adjacent numbers whose sum is not divisible by $q$. Consider a chain $(a, b, c, d)$ of consecutively connected numbers. By the condition,
$$
a+d=(a+b)-(b+c)+(c+d) \vdots p
$$
Then the numbers $a$ and $d$ are also connected, which means there is a cycle of length 4 in the picture, which does not exist.
From 1) and 3), it follows that the number $n$ has at least two distinct odd prime divisors. Let there be exactly two (say, $p$ and $q$). We will show that they are different from 3. Suppose, for example, that $p=3$. No more than two numbers are divisible by 3 (if there are three, they form a cycle). The remaining numbers can be divided into two groups, giving remainders 1 and 2 when divided by 3. One of these groups is empty, otherwise any number from the smaller group will be connected to at least three numbers from the other group, which is impossible. The sum of the numbers in one group is not divisible by 3. Therefore, there exists a three-link chain in which the sum of any pair of connected numbers is not divisible by 3 and, hence, is divisible by $q$. But this contradicts 2).
Thus, if $n$ has exactly two distinct odd prime divisors, then $n \geqslant 5 \cdot 7=35$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>35$. The arrangement for $n=35$ is shown in the figure.
|
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 left). Find the sum of the $r$-ary digits of the number $x^{2}$.
|
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
$$
s^{2}\left(2 r^{2}+5\right)^{2}\left(r^{2}+2 r+1\right)=a\left(1+r^{6}\right)+b\left(r+r^{5}\right)+c\left(r^{2}+r^{4}\right)
$$
where $a, b, c$ are some $r$-ary digits. Let's make two observations.
1) For any natural $n$
$$
r^{n}=(1+r-1)^{n} \equiv(-1)^{n}(1-n(1+r))\left(\bmod (1+r)^{2}\right)
$$
The right side of $(*)$ is divisible by $(1+r)^{2}$, from which
$$
0 \equiv a(2-6(1+r))-b(2-6(1+r))+c(2-6(1+r))=2(1-3(1+r))(a-b+c)\left(\bmod (1+r)^{2}\right)
$$
Since $1-3(1+r)$ is coprime with $(1+r)^{2}$, $2(a-b+c)$ is divisible by $(1+r)^{2}$. But this number lies in the interval $(-2 r, 4 r) \subset\left(-(1+r)^{2},(1+r)^{2}\right)$, hence $b=a+c$.
2) Equate the remainders of the left and right sides of $(*)$ when divided by $1+r^{2}$:
$$
18 s^{2} r \equiv b r\left(1+r^{4}\right) \equiv 2 b r\left(\bmod \left(1+r^{2}\right)\right)
$$
Since $r$ is coprime with $1+r^{2}$, $2\left(9 s^{2}-b\right)$ is divisible by $1+r^{2}$. Note that $4 s^{2} \leqslant r-1$, otherwise the number $x^{2}$ will be eight-digit. In addition, $r \geqslant 5 s+1 \geqslant 6$. Therefore
$$
2\left(9 s^{2}-b\right)-2 r>-1-r^{2}
$$
Thus, $b=9 s^{2}$.
Since $b$ is an $r$-ary digit, from 2) it follows that $9 s^{2} \leqslant r-1$. Since $s>0$, we get $s=1$ and $b=9$. By 1), the sum of the digits of $x^{2}$ is $2(a+b+c)=4 b=36$.
Remark. It can be verified by direct calculation that $2255_{21}^{2}=4950594_{21}$. Thus, the situation described in the problem is realized.
|
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 connected, then the numbers $a+b$ and $n$ must have a common natural divisor greater than 1. For what smallest $n$ does such an arrangement exist?
|
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 among four consecutively connected numbers, there exists a pair of adjacent numbers whose sum is not divisible by $q$. Consider a chain $(a, b, c, d)$ of consecutively connected numbers. By the condition,
$$
a+d=(a+b)-(b+c)+(c+d) \vdots p
$$
Then the numbers $a$ and $d$ are also connected, which means there is a cycle of length 4 in the picture, which does not exist.
From 1) and 3), it follows that the number $n$ has at least two distinct odd prime divisors. Let there be exactly two (say, $p$ and $q$). We will show that they are different from 3. Suppose, for example, that $p=3$. No more than two numbers are divisible by 3 (if there are three, they form a cycle). The remaining numbers (at least 6) can be divided into two groups, giving remainders 1 and 2 when divided by 3. We will show that one of these groups is empty. If this is not the case, then each number from one group is connected to any number from the other. But then in one group there are no more than three numbers, and in the other no more than two, which is impossible. The sum of the numbers in one group is not divisible by 3. Therefore, there exists a three-link chain in which the sum of any pair of connected numbers is not divisible by 3 and, therefore, is divisible by $q$. But this contradicts 2).
Thus, if $n$ has exactly two distinct odd prime divisors, then $n \geqslant 5 \cdot 7=35$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>35$. The arrangement for $n=35$ is shown in the figure.
|
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 right to left). Find the sum of the $r$-ary digits of the number $x^{2}$.
|
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
$$
s^{2}\left(7 r^{2}+17\right)^{2}\left(r^{2}+2 r+1\right)=a\left(1+r^{6}\right)+b\left(r+r^{5}\right)+c\left(r^{2}+r^{4}\right)
$$
where $a, b, c$ are some $r$-ary digits. Let's make two observations.
1) For any natural $n$
$$
r^{n}=(1+r-1)^{n} \equiv(-1)^{n}(1-n(1+r))\left(\bmod (1+r)^{2}\right)
$$
The right side of $(*)$ is divisible by $(1+r)^{2}$, from which
$$
0 \equiv a(2-6(1+r))-b(2-6(1+r))+c(2-6(1+r))=2(1-3(1+r))(a-b+c)\left(\bmod (1+r)^{2}\right)
$$
Since $1-3(1+r)$ is coprime with $(1+r)^{2}$, $2(a-b+c)$ is divisible by $(1+r)^{2}$. But this number lies in the interval $(-2 r, 4 r) \subset\left(-(1+r)^{2},(1+r)^{2}\right)$, hence $b=a+c$.
2) Equate the remainders of the left and right sides of $(*)$ when divided by $1+r^{2}$:
$$
200 s^{2} r \equiv b r\left(1+r^{4}\right) \equiv 2 b r\left(\bmod \left(1+r^{2}\right)\right)
$$
Since $r$ is coprime with $1+r^{2}$, $2\left(100 s^{2}-b\right)$ is divisible by $1+r^{2}$. Note that $49 s^{2} \leqslant r-1$, otherwise the number $x^{2}$ will be an eight-digit number. In addition, $r \geqslant 17 s+1 \geqslant 18$. Therefore
$$
2\left(100 s^{2}-b\right)-2 r>-1-r^{2}
$$
Thus, $b=100 s^{2}$.
Since $b$ is an $r$-ary digit, from 2) it follows that $100 s^{2}0$, we get $s=1$ and $b=100$. By 1$)$, the sum of the digits of $x^{2}$ is $2(a+b+c)=4 b=100$.
Remark. Direct calculation shows that $(7,7,17,17)_{120}^{2}=(49,100,51,0,51,100,49)_{120}$. Thus, the situation described in the condition is realized.
|
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 connected, then the numbers $a^{2}+b^{2}$ and n must have a common natural divisor greater than 1. For what smallest n does such an arrangement exist?
|
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 $n$, then no more than two numbers are divisible by $d$. Suppose there are three such numbers. Then they must be pairwise connected, and thus form a three-link cycle.
3) If $q$ is a prime divisor of $n$, then among four sequentially connected numbers, there exists a pair of adjacent numbers whose sum of squares is not divisible by $q$. Suppose there is a chain $(a, b, c, d)$ of sequentially connected numbers such that the sum of squares of any pair of adjacent numbers is divisible by $q$. Then
$$
a^{2}+d^{2}=\left(a^{2}+b^{2}\right)-\left(b^{2}+c^{2}\right)+\left(c^{2}+d^{2}\right) \vdots q
$$
This means that the numbers $a$ and $d$ are connected, so there is a cycle of length 4 in the picture, which does not exist.
From 1) and 3), it follows that the number $n$ has at least two different odd prime divisors. Let there be exactly two (say, $p$ and $q$). We will show that they are different from $3, 7, 11$. Suppose, for example, that $p \in \{3, 7, 11\}$. Note that for any natural number $a$
$$
a^{2} \bmod 3 \in \{0,1\}, \quad a^{2} \bmod 7 \in \{0,1,2,4\}, \quad a^{2} \bmod 11 \in \{0,1,3,4,5,9\}
$$
Therefore, $a^{2}+b^{2}$ is divisible by $p$ if and only if $a^{2}$ and $b^{2}$ are divisible by $p$, and thus $a$ and $b$ are divisible by $p$. By 2), this condition can be satisfied by only one pair. From the remaining four pairs of connected numbers, a three-link chain can be formed in which the sum of squares of each pair of adjacent numbers is divisible by $q$. But this contradicts 3).
Thus, if $n$ has exactly two different odd prime divisors, then $n \geqslant 5 \cdot 13=65$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>65$. The arrangement for $n=65$ is shown in the figure.
|
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 this is possible?
|
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, for the leftmost of them, the condition of the problem is violated). Therefore, there are no more than 2021 blue cells with a zero. In addition, the number of blue cells with a cross does not exceed $3 \cdot 2021$. Thus, the total number of blue cells does not exceed $2021 + 3 \cdot 2021 = 4 \cdot 2021$. These arguments are valid for any color. Therefore, the total number of colors is not less than
$$
\frac{2021^{2}}{4 \cdot 2021} = \frac{2021}{4} = 505.25
$$
Thus, $n \geqslant 506$.
We will show that $n=506$ is achievable. Sequentially number all diagonals from bottom to top, running from northwest to southeast, with numbers from 1 to 4041. Paint the first four diagonals in the first color, the next four in the second color, and so on in a cycle (so after the 506th color, the first color will follow again). We will check that this coloring satisfies the condition of the problem. Note that no color is used more than twice, since $506 \cdot 4 \cdot 2 = 4048 > 4041$. Let the $k$-th color appear twice. In the first block, the topmost cell of this color will be in the row with number $4k$. In the second block, the bottommost cell of this color is in the row with number
$$
4(506+k-1)+1-2020 = 4k+1
$$
Thus, cells of the same color from different blocks cannot be in the same row.
## Variant 4
|
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 $a$ and $b$ are connected, then the numbers $a^{2}+b^{2}$ and $n$ must have a common natural divisor greater than 1. For what smallest $n$ does such an arrangement exist?
|
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 $n$, then no more than two numbers are divisible by $d$. Suppose there are three such numbers. Then they must be pairwise connected and, therefore, form a three-link cycle.
3) If $q$ is a prime divisor of $n$, and in the chain $(a, b, c, d)$ of sequentially connected numbers, the sum of the squares of any pair of adjacent numbers is divisible by $q$, then $(a, b, c, d)$ forms a cycle. Indeed,
$$
a^{2}+d^{2}=\left(a^{2}+b^{2}\right)-\left(b^{2}+c^{2}\right)+\left(c^{2}+d^{2}\right) \vdots q
$$
that is, the numbers $a$ and $d$ are also connected.
From 1) and 3), it follows that the number $n$ has at least two distinct odd prime divisors. Let there be exactly two (say, $p$ and $q$). We will show that they are different from $3, 7, 11$. Suppose, for example, that $p \in\{3,7,11\}$. Note that for any natural number $a$
$$
a^{2} \bmod 3 \in\{0,1\}, \quad a^{2} \bmod 7 \in\{0,1,2,4\}, \quad a^{2} \bmod 11 \in\{0,1,3,4,5,9\}
$$
Therefore, $a^{2}+b^{2}$ is divisible by $p$ if and only if $a^{2}$ and $b^{2}$ are divisible by $p$, and hence $a$ and $b$ are divisible by $p$. By 2), this condition can be satisfied by only one pair. From the remaining six pairs of connected numbers, a non-cyclic three-link chain can be formed (for example, passing along the left and top or right and bottom sides of the large rectangle). In this chain, the sum of the squares of each pair of adjacent numbers is divisible by $q$, which contradicts 3).
Thus, if $n$ has exactly two distinct odd prime divisors, then $n \geqslant 5 \cdot 13=65$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>65$. The arrangement for $n=65$ is shown in the figure.
|
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). For which $r$ is this possible?
|
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
$$
25(r+1)^{2}\left(r^{2}+1\right)^{2}=a\left(r^{7}+1\right)+b\left(r^{6}+r\right)+c\left(r^{5}+r^{2}\right)+d\left(r^{4}+r^{3}\right)
$$
Let's make some observations.
1) Since $r^{6} \equiv-1\left(\bmod \left(r^{2}+1\right)\right)$, the following equalities hold:
$$
r^{7}+1 \equiv 1-r\left(\bmod \left(r^{2}+1\right)\right) \quad \text { and } \quad r^{6}+r \equiv r-1\left(\bmod \left(r^{2}+1\right)\right)
$$
Moreover, $r^{4} \equiv 1\left(\bmod \left(r^{2}+1\right)\right)$, hence
$$
r^{5}+r^{2} \equiv r-1\left(\bmod \left(r^{2}+1\right)\right) \quad \text { and } \quad r^{4}+r^{3} \equiv 1-r\left(\bmod \left(r^{2}+1\right)\right)
$$
Then, by $(*)$
$$
0 \equiv(a-b-c+d)(1-r)\left(\bmod \left(r^{2}+1\right)\right)
$$
Note that $r^{2}+1+(1-r)(r+1)=2$, and the numbers $1-r$ and $r^{2}+1$ are odd. Therefore, they are coprime, which implies $a-b-c+d \vdots\left(r^{2}+1\right)$. But
$$
|a-b-c+d| \leqslant 2(r-1)<2 r<r^{2}+1
$$
Thus, $b-a=d-c=2$.
2) Since $r^{n} \equiv 1(\bmod (r-1))$ for any natural $n$, from $(*)$ and 1$)$ it follows that
$$
25 \cdot 16 \equiv 2(a+b+c+d)=4(b+c)(\bmod (r-1))
$$
Since $r-1$ is odd, $r-1$ divides the number $100-(b+c)$.
3) For any natural $n$
$$
r^{n}=(r+1-1)^{n} \equiv(-1)^{n}(1-n(r+1))\left(\bmod (r+1)^{2}\right)
$$
The right-hand side of $(*)$ is divisible by $(r+1)^{2}$, hence
$$
0 \equiv(7 a-5 b+3 c-d)(r+1)\left(\bmod (r+1)^{2}\right)
$$
Thus, the number
$$
7 a-5 b+3 c-d=7(b-2)-5 b+3 c-(c+2)=2(b+c)-16
$$
is divisible by $r+1$. Therefore, $b+c-8 \vdots(r+1)$ due to the oddness of $r+1$. Hence, $b+c=8$ or $b+c=8+r+1$. In the second case,
$$
0 \equiv 92-(r+1) \equiv 90(\bmod (r-1))
$$
Since $r$ is even and $r-1 \geqslant 17$, we get $r=46$. Then $a=25$ and $b=27$, which is impossible since $b=50 \bmod 46=4$. Therefore, $b+c=8$. By 2$) 92 \vdots(r-1)$, hence $r=24$.
Remark. If $b+c=8$ and $r=24$, then
$$
a=25 \bmod 24=1, \quad b=a+2=3, \quad c=8-b=5, \quad d=c+2=7
$$
The equality $5555_{24}^{2}=13577531_{24}$ is verified by direct computation.
|
24
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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