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8. (4 points) For Valentine's Day, the boy Zhenya was given a chocolate bar of size $2 \times 4$, on each piece of which a unique picture is drawn. In each move, Zhenya can bite off one piece that has no more than two common sides with other pieces that have not yet been eaten. In how many ways can Zhenya eat his choco... | Answer: 6720.
## Solution:
The first cell to be eaten is necessarily a corner cell (4 ways). After this, there are 4 options:
1) The second cell to be eaten is the adjacent corner cell (1 way). A $2 \times 3$ rectangle remains. From this, the first cell to be eaten is again a corner cell (4 ways). After this, one ce... | 6720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. What is the greatest value that the sum $\sin ^{2} a+\sin ^{2}\left(a+60^{\circ}\right)+\sin ^{2}\left(a+120^{\circ}\right)+\ldots+\sin ^{2}\left(a+300^{\circ}\right)$ can take? | Answer: 3
## Examples of how to write the answer:
1,7
$1 / 7$
17
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. All seven-digit numbers consisting of different digits from 1 to 7 were listed in ascending order. What is the position of the number 3241765? | Answer: 1590
## Examples of how to write the answer:
1,7
# | 1590 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. What integer can be written in the form $\sqrt{12-\sqrt{12-\sqrt{12-\ldots}}}$ (the number of roots is infinite). | Answer: 3
## Examples of how to write the answer:
1,7
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. On a unit cube, a vector was drawn along each edge in one of the two possible directions. On each face, one diagonal was chosen and a vector was drawn along it in one of the two possible directions, and the 6 drawn diagonals turned out to be non-parallel. In total, 18 vectors were obtained. What is the maximum value... | Answer: 116
## Examples of how to write the answer:
## 1,7
$1 / 7$
17
# | 116 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a $13 \times 13$ grid, numbers are arranged such that the numbers in each row and each column form an arithmetic progression in the order they are written. The grid is colored in two colors in a checkerboard pattern. The numbers on the corner white cells of the grid are $1, 2, 3,$ and $6$. Find the sum of the num... | Answer: 252
## Examples of how to write the answer:
1,7
# | 252 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. At a rock festival, vocalists, guitarists, and drummers met, a total of 121 people. Each vocalist gave a slap on the back of each guitarist, each guitarist gave a slap on the back of each drummer, and each drummer gave a slap on the back of each vocalist. What is the maximum number of slaps on the back that the part... | Answer: 4880.
## Examples of how to write the answer:
17
# | 4880 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A 101-gon is inscribed in a circle with diameter $\mathrm{XY}=6$ and has an axis of symmetry perpendicular to this diameter. Find the sum of the squares of the distances from the vertices of the 101-gon to the point $\mathrm{X}$. | Answer: 1818
## Examples of how to write the answer:
# | 1818 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Around a circle, 300 (not necessarily integer) numbers from 20 to 400 inclusive were written. From each number, the logarithm to the base of the next number in the clockwise direction was taken, and then all the obtained logarithms were summed. What is the greatest value that the sum of these logarithms can take? | Answer: 375
## Examples of how to write the answer:
## Grade 9
# | 375 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Four different numbers $a, b, c, d$, greater than one and not divisible by 5, are such that $\operatorname{GCD}(a, b)=\operatorname{GCD}(c, d)$ and $\operatorname{LCM}(a, b)=\operatorname{LCM}(c, d)$. What is the smallest possible value of $a+b+c+d$? | Answer: 24
## Examples of how to write the answer:
17 | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The fox builds "pyramids" from 5 cubes in the following way: each "pyramid" consists of one or several levels; on each level, the number of cubes is strictly less than on the previous one; each new level consists of one or several consecutive cubes. You can see an example of a "pyramid" made of ten cubes in the pict... | Answer: 7
## Examples of answer notation:
12
# | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. $A B C D E$ - a pentagon inscribed in circle $S$. Lines $D E$ and $A B$ are parallel, $B C=C D$, $A D=D E=20, B C: B D=4: 5$. Find the radius of circle $S$. | Answer: 16
## Examples of how to write answers: $1 / 4$ 0.25
# | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. $A B D E, B C E F, C D F A$ - cyclic quadrilaterals with the intersection points of the diagonals $K, L$ and $M$ respectively. It is known that point $K$ lies on segments $B L$ and $A M$, point $M$ - on segment $C L$. Moreover, $E L=F L=K L=5, D M=4, A K=M K=6$. Find the length of segment $M C$. If there are multipl... | Answer: 4
## Examples of how to write answers:
$1 / 4$
0.25
$4 ; 10$
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given two quadratic trinomials with the leading coefficient $\frac{\sqrt{3}}{8}$. The vertices and the intersection point of their graphs form an equilateral triangle. Find the length of its side. If there are multiple possible answers, list them in any order separated by a semicolon. | Answer: 16
## Examples of answer notations:
$1 / 4$
$-0.25 ; 10$
8th grade. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9. (4 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | # Problem 9. (4 points)
$A B C D$ is a cyclic quadrilateral. The extension of side $A B$ beyond point $B$ and the extension of side $C D$ beyond point $C$ intersect at point $P$. The extension of side $A D$ beyond point $D$ and the extension of side $B C$ beyond point $C$ intersect at point $Q$. It turns out that angl... | 1040 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 1. (1 point)
Find the largest three-digit number ABC that is divisible by the two-digit numbers AB and BC. (Different letters do not necessarily represent different digits) | Answer: 990
Solution:
The number 990 clearly fits. A larger answer is not possible, as all three-digit numbers greater than 990 do not divide by 99: the next number divisible by 99 is 1089.
# | 990 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (2 points)
What is the maximum number of different reduced quadratic equations that can be written on the board, given that any two of them have a common root, but no four have a root common to all. | Answer: 3
Solution:
Consider three such equations. There are two cases:
1) These equations have a common root $a$. Then they can be represented in the form $(x-a)(x-b)=0, (x-a)(x-c)=0$ and $(x-a)(x-d)=0$, where $b, c$ and $d$ are their remaining roots (they are not equal to each other, otherwise the equations coinci... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In trapezoid $ABCD$, a point $X$ is taken on the base $AD$ such that segments $XB$ and $XC$ divide the trapezoid into three triangles that are similar to each other but pairwise unequal and non-isosceles. The side $AB$ has a length of 6. Find $AX \cdot DX$. | Answer: 36
## Solution:
The triangles are scalene, meaning all their angles are different.
$\alpha=\angle B X A=\angle X B C \neq \angle B X C=\beta$. The angle $\angle C X D$ together with these angles sums to $180^{\circ}$, so it is the third angle of the triangle and cannot be equal to either $\alpha$ or $\beta$.... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (2 points)
Find the smallest three-digit number ABC that is divisible by the numbers AB and BC (the digit A cannot be 0, and the digit B can be; different letters do not necessarily represent different digits) | Answer: 110
Solution:
The number 110 clearly fits. A smaller answer cannot be, as all three-digit numbers less than 110 start with 10. At the same time, none of them are divisible by 10, except for 100, and 100 is not divisible by 0. | 110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (2 points)
What is the maximum number of different reduced quadratic equations that can be written on the board, given that any two of them have a common root, but no five have a root common to all. | Answer: 4
## Solution:
Consider 4 such equations. There are two cases:
1) These equations have a common root $a$. Then they can be represented in the form $(x-a)(x-b)=0,(x-a)(x-c)=0,(x-a)(x-d)=0$ and $(x-a)(x-e)=0$, where $b, c, d$ and $e$ are their remaining roots (they are not equal to each other, otherwise the eq... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In trapezoid $ABCD$, a point $X$ is taken on the base $BC$ such that segments $XA$ and $XD$ divide the trapezoid into three triangles that are similar to each other but pairwise unequal and non-isosceles. The side $AB$ has a length of 5. Find $XC \cdot BX$.
Answer: 25 | Solution:
The triangles are scalene, meaning all their angles are different.
$\alpha=\angle B X A=\angle X A D \neq \angle A X D=\beta$. The angle $\angle C X D$ together with these angles sums up to $180^{\circ}$, so it is the third angle of the triangle and cannot be equal to either $\alpha$ or $\beta$.
If $\angle... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. How many pairs of natural numbers exist for which the number 189 is the LCM? (The numbers in the pair can be the same, the order of the numbers in the pair does not matter) | Answer: 11
Only digits are allowed as input
# | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the largest three-digit number that gives a remainder of 2 when divided by 13, and a remainder of 6 when divided by 15. | Answer: 951
2nd option. Find the largest three-digit number that gives a remainder of 3 when divided by 13, and a remainder of 2 when divided by 14.
Answer: 926
3rd option. Find the largest three-digit number that gives a remainder of 5 when divided by 14, and a remainder of 3 when divided by 15.
Answer: 873
Only ... | 951 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
$A B C$ is an equilateral triangle with a side length of 10. On side $A B$, a point $D$ is taken; on side $A C$, a point $E$ is taken; on side $B C$, points $F$ and $G$ are taken such that triangles $A D E$, $B D G$, and $C E F$ are also equilateral. $A D=3$. Find $F G$. | # Answer: 4
## Solution:
$B D=A B-A D=10-3=7$, so all sides of triangle $B D G$ are equal to 7.
$C E=A C-A E=A C-A D=10-3=7$, so all sides of triangle $C E F$ are also equal to 7.
Thus, $B G+C F=14$. This is greater than the length of segment $B C=10$, so $G F=14-10=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. (3 points)
Alice the Fox thought of a two-digit number and told Pinocchio that this number is divisible by $2, 3, 4, 5$, and $6$. However, Pinocchio found out that exactly two of these five statements are actually false. What numbers could Alice the Fox have thought of? In your answer, indicate the number o... | # Answer: 8
## Solution:
If a number is not divisible by 2, then it is not divisible by 4 or 6 either, and we already have three false statements. Therefore, the number must be divisible by 2. If an even number is divisible by 3, it is also divisible by 6, which means the statements about divisibility by 4 and 5 are ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (3 points)
All residents of the island are either blondes or brunettes with green or blue eyes. The proportion of brunettes among blue-eyed residents is $65 \%$. The proportion of blue-eyed residents among blondes is $70 \%$. Finally, the proportion of blondes among green-eyed residents is 10\%. What perc... | Answer: 54
## Solution:
Let the number of blue-eyed brunettes be $a$, blue-eyed blondes be $b$, green-eyed blondes be $c$, and green-eyed brunettes be $d$.
Then $\frac{a}{a+b}=0.65$, from which $\frac{a+b}{a}=\frac{20}{13}$ and $\frac{b}{a}=\frac{20}{13}-1=\frac{7}{13}$.
Similarly, $\frac{b}{b+c}=0.7$, from which $... | 54 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In the cells of a $7 \times 7$ table, pairwise distinct non-negative integers are written. It turns out that for any two numbers in the same row or column, the integer parts of their quotients when divided by 8 are different. What is the smallest value that the largest number in the table can t... | # Answer: 54
Solution: Each row must contain seven numbers that give different incomplete quotients when divided by 8. This means that in each row, there is at least one number with an incomplete quotient of 6 or more when divided by 8 (6, not 7, since the minimum value of the incomplete quotient is 0). Therefore, the... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (4 points)
In the Magic Country, there are 100 cities, some of which are connected by two-way air routes. Any two cities can be reached from each other with no more than 11 transfers, and there is a unique way to do so. If it is impossible to travel from city $A$ to city $B$ with 10 or fewer transfers, bo... | Answer: 89
## Solution:
Consider two peripheral cities and the route between them. All 11 cities where transfers are made on this route cannot be peripheral. All other cities can be peripheral if there are direct flights from them to the first or last of these 11 intermediate cities.
Thus, the maximum number of peri... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a cube and 12 colors. Find the number of ways to paint the faces of this cube using these colors (each face in one color) such that adjacent faces are of different colors. Colorings that differ by a rotation are considered different. | Answer: 987360
## Examples of writing the answer:
## 7th grade
# | 987360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In a four-digit number, the first digit was crossed out. The resulting number is 9 times smaller than the original. What is the largest value the original number could have had? | Answer: 7875
## Examples of how to write the answer:
1234 | 7875 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On a plane, a certain number of triangles are drawn, the lengths of whose sides are ten-digit natural numbers, containing only threes and eights in their decimal representation. No segment belongs to two triangles, and the sides of all triangles are distinct. What is the maximum number of triangles that can be drawn... | Answer: 341
## Examples of how to write the answer:
## 17
# | 341 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. At a round table sat raccoons, hedgehogs, and hamsters, a total of 134 animals. When asked: "Are any of your neighbors the same type of animal as you?", everyone answered "No." What is the maximum number of hedgehogs that could have been sitting at the table, given that hamsters and hedgehogs always tell the truth, ... | Answer: 44
## Examples of how to write the answer:
17
# | 44 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. The polynomial $P(x)=(x-4)^{4}\left(x^{2}+\frac{3}{2} x-1\right)^{5}$ is represented in the form $\sum_{k=0}^{14} a_{k}(x+1)^{k}$. Find $a_{0}+a_{2}+\ldots+a_{14}$. | Answer: -256
Allowed for input are digits, minus sign, and division sign, a dot or comma as a decimal separator
# | -256 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many pairs of natural numbers exist for which the number 392 is the LCM? (The numbers in the pair can be the same, the order of the numbers in the pair does not matter) | Answer: 18
Only digits are allowed as input
# | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The gnomes dug a system of tunnels that formed a rectangular grid 7 by 2. The main entrance is located at the bottom-left vertex $(0,0)$, and the main exit is at the top-right vertex $(7,2)$. The tunnels are numbered as follows: the tunnel connecting vertices $(x, k)$ and $(x+1, k)$ has a number equal to $7 k + x + ... | Answer: 8
Only digits are allowed as input | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
Let $p_{1}, p_{2}, \ldots, p_{97}$ be prime numbers (not necessarily distinct). What is the largest integer value that the expression
$$
\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1}=\frac{p_{1}}{p_{1}^{2}+1}+\frac{p_{2}}{p_{2}^{2}+1}+\ldots+\frac{p_{97}}{p_{97}^{2}+1}
$$
can take? | # Answer: 38
Solution: Note that the function $\frac{x}{x^{2}+1}$ decreases when $x \geqslant 2$, which means $\sum_{i=1}^{97} \frac{p_{i}}{p_{i}^{2}+1} \leqslant 97 \cdot \frac{2}{2^{2}+1}=$ $97 \cdot 0.4=38.8$. Therefore, the answer cannot be greater than 38.
At the same time, $\frac{3}{3^{2}+1}=0.3$, which is 0.1 ... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
On the side $B C$ of triangle $A B C$, points $A_{1}$ and $A_{2}$ are marked such that $B A_{1}=6, A_{1} A_{2}=8$, $C A_{2}=4$. On the side $A C$, points $B_{1}$ and $B_{2}$ are marked such that $A B_{1}=9, C B_{2}=6$. Segments $A A_{1}$ and $B B_{1}$ intersect at point $K$, and $A A_{2}$ and $... | # Answer: 12
## Solution:
Let $M$ be the point of intersection of line $KL$ and side $AB$. We write two Ceva's theorems, for point $K$ and for point $L$:
$$
\frac{AM}{MB} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}=1 ; \quad \frac{AM}{MB} \cdot \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A}=1
$$
From this, we get $\... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (3 points)
$4^{27000}-82$ is divisible by $3^n$. What is the greatest natural value that $n$ can take? | Answer: 5
Solution:
$4^{27000}=(1+3)^{27000}=1+27000 \cdot 3+\frac{27000 \cdot 26999 \cdot 3^{2}}{2}+\frac{27000 \cdot \ldots \cdot 26998 \cdot 3^{3}}{6}+$
$+\frac{27000 \cdot \ldots \cdot 26997 \cdot 3^{4}}{24}+\frac{27000 \cdot \ldots \cdot 26996 \cdot 3^{5}}{120} \ldots$
The last two terms listed are divisible by ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
Positive numbers $x, y$, and $z$ are such that $x+y+z=5$. What is the smallest value that the quantity $x^{2}+y^{2}+2 z^{2}-x^{2} y^{2} z$ can take? | # Answer: -6
## Solution:
Rewrite the condition as $\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z=5$ and write the inequality for the arithmetic mean and the quadratic mean of these five numbers:
$$
1=\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z}{5} \leqslant \sqrt{\frac{\frac{x^{2}}{4}+\frac{x^{2}}{4... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
In an $8 \times 8$ table, some 23 cells are black, and the rest are white. In each white cell, the sum of the number of black cells on the same row and the number of black cells on the same column is written; nothing is written in the black cells. What is the maximum value that the sum of the n... | Answer: 234
## Solution:
The number in the white cell consists of two addends: "horizontal" and "vertical". Consider the sum of all "horizontal" addends and the sum of all "vertical" addends separately across the entire table. If we maximize each of these two sums separately, the total sum will also be the largest.
... | 234 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Six natural numbers are written on the board, such that for any two $a$ and $b$ among them, $\log _{a} b$ or $\log _{b} a$ is an integer (the second logarithm does not necessarily exist). What is the smallest value that the maximum of these numbers can take? | The answer can be written in the form of a power of a number: $m^{n}$ is denoted as $\mathrm{m}^{\wedge} \mathrm{n}$.
Answer: 65536 || $2 \wedge 16\left\|4^{\wedge} 8\right\| 16^{\wedge} 4 \| 256^{\wedge} 2$ | 65536 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. $A B C A_{1} B_{1} C_{1}$ - a right triangular prism with a circumscribed sphere. The perimeter of the base $A B C$ is 32 units, and the product of the sides is 896 cubic units. The surface area of the prism is 192 square units. Find the square of the radius of its circumscribed sphere. | Answer: 53
## Examples of answer notations:
7.4
$7 / 14$
714
# | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) From the set of three-digit numbers that do not contain the digits $0,1,2$, $3,4,5$ in their notation, several numbers were written on paper in such a way that no two numbers can be obtained from each other by swapping two adjacent digits. What is the maximum number of such numbers that could have been wr... | Answer: 40
Solution:
Whether a number can be written or not depends only on numbers consisting of the same set of digits.
Suppose a number ABC, consisting of three different digits, is written. This means that the numbers BAC and ACB cannot be written. Among the numbers consisting of these same digits, BVA, VAB, and... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) On a $3 \times 3$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 5.
Solution: The people in the corner cells are obviously liars: they simply do not have three neighbors.
If the person in the center is a knight, then all the people in the side cells are also liars. But in this case, the knight has 4 liar neighbors, not three. This leads to a contradiction.
If the person i... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Does there exist a rectangular parallelepiped with integer sides, for which the surface area is numerically equal to the sum of the lengths of all twelve edges? | Answer: Yes, it exists.
Solution: For example, a $2 \times 2 \times 2$ cube fits. Both the surface area and the sum of the lengths of the edges are 24. | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) From the set of three-digit numbers that do not contain the digits 0, 6, 7, 8, 9 in their notation, several numbers were written on paper in such a way that no two numbers can be obtained from each other by swapping two adjacent digits. What is the maximum number of such numbers that could have been writt... | Answer: 75
Solution:
Whether a number can be written or not depends only on numbers consisting of the same set of digits.
Suppose a number ABC, consisting of three different digits, is written. This means that the numbers BAC and ACB cannot be written. Among the numbers consisting of these same digits, BVA, VAB, and... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) On a $2 \times 4$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 6.
Solution: On the corner cells, liars are obviously standing: they simply do not have three neighbors.
A situation where all people on the board are liars is impossible, as it would mean that four of these liars are telling the truth.
Therefore, there is at least one knight. Then all his neighbors are liar... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Given the function $f(x)=\frac{\sqrt{3} x-1}{x+\sqrt{3}}$. Find $\underbrace{f(f(\ldots(1) \ldots))}_{2013 \text { times }}$. | Answer: -1
## Examples of how to write answers:
$1 / 4$
0.25
10
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The polynomial $P(x)$ of the third degree, as well as its first, second, and third derivatives, take the value 1 at $x=-3$. Find $\mathrm{P}(0)$. | Answer: 13
## Examples of how to write answers:
$1 / 4$
0.25
$-10$
# | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In space, there are 37 different vectors with integer non-negative coordinates, starting from the point $(0 ; 0 ; 0)$. What is the smallest value that the sum of all their coordinates can take? | Answer: 115
## Examples of answer notation:
239
Answer: $[-1 ; 7]$ | 115 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Four spheres of radius $r$ touch each other externally. A sphere of radius $\sqrt{6}+2$ touches all of them internally. Find $r$. | Answer: 2
## Examples of how to write answers:
$1 / 4$
0.25
10
# | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In the city of Gentle-city, there live 19 gentlemen, any two of whom are either friends or enemies. At some point, each gentleman asked each of his friends to send a hate card to each of his enemies (gentleman A asks gentleman B to send a card to all enemies of gentleman B). Each gentleman fulfilled all the requests... | Answer: 1538.
## Examples of answer recording:
100
# | 1538 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is a trapezoid with bases $A D=15$ and $\mathrm{BC}=10$. $O$ is one of the intersection points of the circles constructed on the lateral sides of the trapezoid as diameters, and this point lies inside the trapezoid. Triangle $B C M$ is constructed on side $B C$ on the external side relative to the tr... | Answer: 4
## Examples of answer notations:
$1 / 4$
0.25
10 | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) In a row without spaces, all natural numbers are written in ascending order: $1234567891011 . .$. What digit stands at the 2017th place in the resulting long number? | Answer: 7.
Solution: The first 9 digits are contained in single-digit numbers, the next 180 - in two-digit numbers. $2017-180-9=1828$. Next, $1828: 3=609 \frac{1}{3}$. This means that the 2017-th digit is the first
 Does there exist a four-digit natural number with the sum of its digits being 23, which is divisible by 23? | Answer: Yes, for example $7682=23 \cdot 334$.
Solution: Generally speaking, the solution is contained in the answer, but let's explain how to find such a number. The sum of the digits of the product gives the same remainder when divided by 9 as the sum of the original numbers. Represent 23 as the difference between a ... | 7682 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Given the cryptarithm: ЖАЛО + ЛОЖА = ОСЕНЬ. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter А. | Answer: 8
Solution: The rebus can be rewritten as ОСЕНЬ $=($ ЖА + ЛО $) \cdot 101$. First, this means that the last digit of ЖА + ЛО is Ь. Second, if ЖА + ЛО $<100$, the result will be a four-digit number. Let $Ж А+Л О=1 Х Ь$, where $X$ is some digit.
Then ОСЕНЬ $=1 Х Ь 00+1 Х Ь$. If $\mathrm{b}<9$, then the second a... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. (4 points) A triangle is divided into 1000 triangles. What is the maximum number of different points at which the vertices of these triangles can be located? | Answer: 1002
Solution: The sum of the angles of a triangle is $180^{\circ}$, thousands of triangles $-180000^{\circ}$. Where did the extra $179820^{\circ}$ come from? Each internal vertex, where only the angles of triangles meet, adds $360^{\circ}$. Each vertex on the side of the original triangle or on the side of on... | 1002 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) In a row without spaces, all natural numbers from 999 to 1 are written in descending order: 999998 ...321. What digit is in the 2710th position of the resulting long number? | Answer: 9.
Solution: The first $3 \cdot 900=2700$ digits are contained in three-digit numbers. Therefore, we need to count another 10 digits in two-digit numbers. That is, we need the first digit of the fourth largest two-digit number. This number is 96, so we need the digit 9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Given the cryptarithm: RIVER + SQUARE = ABVAD. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter B. | Answer: 2
Solution: The rebus can be rewritten as ABVAD $=(\mathrm{KA}+\mathrm{PE}) \cdot 101$. First, this means that the last digit of $\mathrm{KA}+\mathrm{PE}$ is D. Second, if $\mathrm{KA}+\mathrm{PE}<100$, the result will be a four-digit number. Let $\mathrm{KA}+\mathrm{PE}=1X$D, where $X$ is some digit.
Then AB... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. (4 points) A triangle is divided into 1000 triangles. What is the minimum number of distinct points at which the vertices of these triangles can be located? | Answer: 503
Solution: The sum of the angles of a triangle is $180^{\circ}$, so for a thousand triangles, it is $180000^{\circ}$. Where did the extra $179820^{\circ}$ come from? Each internal vertex where only the angles of triangles meet adds $360^{\circ}$. Each vertex on the side of the original triangle or on the si... | 503 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
Three runners are moving along a circular track at constant equal speeds. When two runners meet, they instantly turn around and start running in opposite directions.
At some point, the first runner meets the second. Twenty minutes later, the second runner meets the third for the first time. An... | # Answer: 100
Solution: (in general form)
Let the first runner meet the second, then after $a$ minutes the second runner meets the third for the first time, and after another $b$ minutes the third runner meets the first for the first time.
Let the first and second runners meet at point $A$, the second and third at p... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In an isosceles trapezoid $A B C D$, the bisectors of angles $B$ and $C$ intersect on the base $A D$. $A B=50, B C=128$. Find the area of the trapezoid. | Answer: 5472
## Solution:
Let $K$ be the point of intersection of the angle bisectors. Angles $\angle B=\angle C$ are equal as the base angles of an isosceles trapezoid, so their halves are... | 5472 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (3 points)
Natural numbers $x, y, z$ are such that $\operatorname{GCD}(\operatorname{LCM}(x, y), z) \cdot \operatorname{LCM}(\operatorname{GCD}(x, y), z)=1400$.
What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$ can take?
# | # Answer: 10
## Solution:
Notice that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$1400=2^{3} \... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (3 points)
On an island, there live vegetarians who always tell the truth, and cannibals who always lie. 50 residents of the island, including both women and men, gathered around a campfire. Each of them said either "All men at this campfire are cannibals" or "All women at this campfire are vegetarians," ... | Answer: 48
## Solution:
We will prove that if we have $n$ people, the maximum number of vegetarian women is $n-2$.
First, note that if someone has called someone a cannibal, then we definitely have at least one cannibal.
Second, according to the condition, there is at least one man. Therefore, the only case where w... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. On a piece of paper, a square with a side length of 1 was drawn, next to it a square with a side length of 2, then a square with a side length of 3, and so on. It turned out that the area of the entire resulting figure is 42925. How many squares were drawn? | Answer: 50
## Examples of answer notation: 45
# | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is an isosceles trapezoid, $\mathrm{AB}=\mathrm{CD}=25, \mathrm{BC}=40, \mathrm{AD}=60$. $\mathrm{BCDE}$ is also an isosceles trapezoid. Find AE. (Points A and E do not coincide)
If there are multiple possible values, list them in any order separated by a semicolon. | Answer: 44.
## Examples of answer recording:
45
$45 ; 56$
## Problem 7 (3 points). | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let $f(x)=\frac{34^{x}}{4^{x}+2}$. Find the sum $f(0)+f\left(\frac{1}{2017}\right) \ldots f\left(\frac{2}{2017}\right) f(1)$. | Answer: 3027
## Examples of answer recording:
45
## Problem 9 (4 points). | 3027 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two tangents are drawn from point A to a circle. The distance from point A to the point of tangency is 10, and the distance between the points of tangency is 16. Find the greatest possible distance from point A to a point on the circle. | Answer: 30
## Examples of answer recording:
## Problem 10 (2 points). | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) From cards with letters, you can form the word KARAKATITSA. How many words (not necessarily meaningful) can be formed from these cards, in which the letters R and T are adjacent? | Answer: $\frac{9!}{4!}=9 \cdot 8 \cdot 7 \cdot 6 \cdot 5=15120$
## Solution:
Let's replace the adjacent letters R and T with one letter, for example, the letter Sh, since they stand together. Then we are left with 9 letters, and among them, 4 letters A and 2 letters K.
Thus, the number of ways to rearrange the lette... | 15120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Given a convex hexagon ABCDEF such that AB $\|\mathrm{CF}\| \mathrm{DE}, \mathrm{BC}\|\mathrm{AD}\| \mathrm{EF}$ and $\mathrm{CD} \| \mathrm{BE}$ $\| \mathrm{FA}$ and $\mathrm{AB}=\mathrm{DE}=\mathrm{CD}=\mathrm{AF}=13, \mathrm{BD}=24$. Find the area of the union of triangles $\mathrm{ACE}$ and BDF.
. What is the maximum number of chips that can be on the board? | Answer: 21
## Examples of answer recording:
14
## 9th grade.
# | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
Let $x, y, z$ be pairwise coprime three-digit natural numbers. What is the greatest value that the GCD $(x+y+z, x y z)$ can take? | Answer: 2994
Solution:
The GCD of two numbers cannot be greater than either of them. The maximum possible value of $x+$ $y+z=997+998+999=2994$ and for these numbers, $x y z$ is indeed divisible by $x+y+z=2994=3 \cdot 998$. | 2994 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. (2 points)
On a circle, 10 points are marked. Any three of them form three inscribed angles. Petya calculated the number of different values that these angles can take. What is the maximum number he could have obtained? | Answer: 80
Solution:
Any two points form two arcs. All inscribed angles subtending the same arc are equal. For two adjacent points, on one of the two arcs between them, no other point lies, meaning a pair of adjacent points gives us one possible angle value, while a pair of non-adjacent points gives two values.
In t... | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. (3 points)
Let $f(x)$ be a quadratic trinomial with integer coefficients. Given that $f(\sqrt{3}) - f(\sqrt{2}) = 4$. Find $f(\sqrt{10}) - f(\sqrt{7})$. | # Answer: 12
## Solution:
Let $f(x)=c x^{2}+d x+e$. Then $f(\sqrt{3})-f(\sqrt{2})=3 c+\sqrt{3} d+e-(2 c+\sqrt{2} d+e)=c+d(\sqrt{3}-\sqrt{2})$. This number can only be an integer if $d=0$. Therefore, $f(x)=c x^{2}+e$ and $f(\sqrt{3})-f(\sqrt{2})=c$.
Then $f(\sqrt{10})-f(\sqrt{7})=10 c+e-(7 c+e)=3 c=12$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
In triangle $A B C$, the midpoints of sides $A B=40$ and $B C=26$ are marked as points $K$ and $L$ respectively. It turns out that the quadrilateral $A K L C$ is a tangential quadrilateral. Find the area of triangle $A B C$. | # Answer: 264
## Solution:
According to the Midline Theorem, $K L=\frac{1}{2} A C$. In a cyclic quadrilateral, the sums of the opposite sides are equal, that is, $K L+A C=A K+C L=\frac{A B+B C}{2}$, so $\frac{3 A C}{2}=\frac{66}{2}$ and $A C=22$. Knowing the sides of the triangle, the area can be calculated using Her... | 264 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (3 points)
Positive numbers $x, y, z$ are such that $x y + y z + x z = 12$.
Find the smallest possible value of $x + y + z$.
# | # Answer: 6
## Solution:
By adding the inequalities $x^{2}+y^{2} \geqslant 2 x y, x^{2}+z^{2} \geqslant 2 x z$ and $y^{2}+z^{2} \geqslant 2 y z$ and dividing by 2, we get $x^{2}+y^{2}+z^{2} \geqslant x y+y z+x z$.
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z) \geqslant 3(x y+y z+x z)=36$, from which $x+y+z \geqslant... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (3 points)
From point $K$ on side $A C$ of triangle $A B C$, perpendiculars $K L_{1}$ and $K M_{1}$ were dropped to sides $A B$ and $B C$ respectively. From point $L_{1}$, a perpendicular $L_{1} L_{2}$ was dropped to $B C$, and from point $M_{1}$, a perpendicular $M_{1} M_{2}$ was dropped to $A B$.
It tu... | # Answer: 8
## Solution:
Notice that the quadrilateral $L_{1} M_{2} L_{2} M_{1}$ is cyclic, since $\angle L_{1} M_{2} M_{1}=\angle L_{1} L_{2} M_{1}=$ $90^{\circ}$. Therefore, $\angle B M_{2} L_{2}=180^{\circ}-\angle L_{1} M_{2} L_{2}=\angle L_{2} M_{1} L_{1}=\angle B M_{1} L_{1}$. Similarly, $\angle B L_{2} M_{2}=\a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
A cubic polynomial has three roots. The greatest value of the polynomial on the interval $[4 ; 9]$ is achieved at $x=5$, and the smallest value is achieved at $x=7$. Find the sum of the roots of the polynomial.
# | # Answer: 18
## Solution:
Since in all options the minimum and maximum on the interval are not reached at its ends, they are reached at the roots of the polynomial's derivative. Let the polynomial be of the form $a x^{3}+b x^{2}+c x+d$, and its derivative, respectively, $3 a x^{2}+2 b x+c$. In this case, the sum of t... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (3 points)
Find the sum of natural numbers from 1 to 3000 inclusive that have common divisors with the number 3000, greater than 1. | # Answer: 3301500
## Solution:
$3000=2^{3} \cdot 3 \cdot 5^{3}$, so we are interested in numbers divisible by 2, 3, or 5. First, let's find the number of such numbers. For this, we will use the principle of inclusion and exclusion. There are exactly $\frac{3000}{2}=1500$ even numbers from 1 to 3000, $\frac{3000}{3}=1... | 3301500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (4 points)
Quadrilateral $A B C D$ is circumscribed around a circle with center at point $O, K, L, M, N$ - the points of tangency of sides $A B, B C, C D$ and $A D$ respectively, $K P, L Q, M R$ and $N S$ - the altitudes in triangles $O K B, O L C, O M D, O N A . O P=15, O A=32, O B=64$.
Find the length ... | # Answer: 30
## Solution:
Triangles $O K A$ and $O N A$ are right triangles with a common hypotenuse and a leg equal to the radius of the circle, so they are congruent. Therefore, their altitudes fall on the same point of the common hypotenuse, meaning $K S$ is the altitude in triangle $O K A$. Thus, points $S$ and $... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
Two cubes with an edge of $12 \sqrt[4]{\frac{8}{11}}$ share a common face. A section of one of these cubes by a certain plane is a triangle with an area of 16. The section of the other by the same plane is a quadrilateral. What is the maximum value that its area can take? | # Answer: 128
## Solution:
Let our cubes be $A B C D A_{1} B_{1} C_{1} D_{1}$ and $A B C D A_{2} B_{2} C_{2} D_{2}$ with a common face $A B C D$. Let the triangular section of the first cube be $K L M$, where point $K$ lies on $A A_{1}$, point $L$ on $A B$, and point $M$ on $A D$. One side of the quadrilateral sectio... | 128 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (4 points)
Hansel and Gretel are playing a game, with Hansel going first. They take turns placing tokens on a $7 \times 8$ grid (7 rows and 8 columns). Each time Gretel places a token, she earns 4 points for each token already in the same row and 3 points for each token already in the same column.
Only o... | # Answer: 700
## Solution:
Let's say that Hansel also earns points according to the same principle as Gretel. In this case, each pair of cells in the same row will ultimately give one of the players 4 points, and each pair of cells in the same column will give 3 points. In one row, there are $\frac{8 \cdot 7}{2}=28$ ... | 700 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
The graphs of the quadratic trinomials $f(x)$ and $g(x)$ intersect at the point $(3 ; 8)$. The trinomial $f(x)+g(x)$ has a single root at 5. Find the leading coefficient of the trinomial $f(x)+g(x)$.
# | # Answer: 4
## Solution:
Since the quadratic polynomial $f(x)+g(x)$ has a unique root 5, it can be represented as $a(x-5)^{2}$, where $a$ is precisely the leading coefficient we are looking for. Additionally, the value of this quadratic polynomial at the point 3 is equal to the sum of the values of the polynomials $f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (2 points)
A natural number $n$ when divided by 12 gives a remainder $a$ and an incomplete quotient $b$, and when divided by 10, it gives a remainder $b$ and an incomplete quotient $a$. Find $n$. | Answer: 119
## Solution:
From the condition, it follows that $n=12 b+a=10 a+b$, from which $11 b=9 a$. Therefore, $b$ is divisible by 9, and $a$ is divisible by 11. On the other hand, $a$ and $b$ are remainders from division by 12 and 10, respectively, so $0 \leqslant a \leqslant 11$ and $0 \leqslant b \leqslant 9$. ... | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (3 points)
It is known that $a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c=30$ and $a^{2}+b^{2}+c^{2}=13$.
Find $a+b+c$. | Answer: 5
## Solution:
$(a+b+c)^{3}-(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)=2\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c\right)$. Let $a+b+c$ be $x$, substitute the known values of the expressions from the condition, and we get the equation $x^{3}-13 x-60=0$.
It is not hard to notice that the number ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
Circles $O_{1}$ and $O_{2}$ touch circle $O_{3}$ with radius 13 at points $A$ and $B$ respectively and pass through its center $O$. These circles intersect again at point $C$. It is known that $O C=12$. Find $A B$. | # Answer: 10
## Solution:
Since circles $O_{1}$ and $O_{2}$ touch circle $O_{3}$ at points $A$ and $B$ respectively and pass through its center $O$, $AO$ and $BO$ are their diameters. Therefore, angles $\angle OCA$ and $\angle OCB$ are right angles. These cannot be the same angle, as this would mean that circles $O_{... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (4 points)
Six positive numbers, not exceeding 3, satisfy the equations $a+b+c+d=6$ and $e+f=2$. What is the smallest value that the expression
$$
\left(\sqrt{a^{2}+4}+\sqrt{b^{2}+e^{2}}+\sqrt{c^{2}+f^{2}}+\sqrt{d^{2}+4}\right)^{2}
$$
can take? | Answer: 72
In the image, there are three rectangles $2 \times (a+b)$ and three rectangles $2 \times (c+d)$, forming a $6 \times 6$ square, since $a+b+c+d=6$. The segment of length 2 in the c... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
In an $8 \times 8$ table, some cells are black, and the rest are white. In each white cell, the total number of black cells on the same row or column is written; nothing is written in the black cells. What is the maximum value that the sum of the numbers in the entire table can take? | Answer: 256
## Solution:
The number in the white cell consists of two addends: a "horizontal" and a "vertical" one. Consider the sum of all "horizontal" addends and the sum of all "vertical" addends separately across the entire table. If we maximize each of these two sums separately, the total sum will also be the gr... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
32 volleyball teams participate in a tournament according to the following scheme. In each round, all remaining teams are randomly paired; if the number of teams is odd, one team skips this round. In each pair, one team wins and the other loses, as there are no draws in volleyball. After three ... | # Solution:
For all teams to be eliminated except one, they must suffer at least 93 losses, meaning at least 93 matches must be played.
We also note that after each round, the number of teams decreases by at most half, because no more than half of the teams lose. In particular, in the final round, only 2 teams could ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (2 points)
In a class, each student has either 5 or 6 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class? | Answer: 11
## Solution:
Let's look at some person. Suppose he has five friends. Then each of these five people has six friends. Similarly, there are at least another 6 people with five friends each. In total, there are 11 people.
It is quite easy to construct an example: two groups of 5 and 6 people, people from dif... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (2 points)
In a positive non-constant geometric progression, the arithmetic mean of the second, seventh, and ninth terms is equal to some term of this progression. What is the minimum possible number of this term? | Answer: 3
Solution:
The second element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first element is even less suitable.
To prove that the answer "3" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we need to solv... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (2 points)
How many negative numbers are there among the numbers of the form $\operatorname{tg}\left(\left(15^{n}\right)^{\circ}\right)$, where $\mathrm{n}$ is a natural number from 1 to 2019? | Answer: 1009
Solution:
$\operatorname{tg} 15^{\circ}>0$.
$15^{2}=225 ; \operatorname{tg} 225^{\circ}>0$.
Further, $225 \cdot 15=3375$, this number gives a remainder of 135 when divided by $360 . \operatorname{tg} 135^{\circ}<0$.
$135 \cdot 15=2025$, this number gives a remainder of 225 when divided by 360. The sequ... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
Can the numbers from 0 to 9999 (each used exactly once) be arranged in a $100 \times 100$ square table so that the sum of the numbers in each $2 \times 2$ square is the same? | Answer: Yes, it can.
Solution:
The desired table is obtained as the sum of two tables.
The first table looks like this:
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 98 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 9... | 19998 | Combinatorics | proof | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
In a class, each student has either 5 or 7 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class?
Answer: 12 | Solution:
Let's look at some person. Suppose he has five friends. Then each of these five people has seven friends. Similarly, there are at least seven more people with five friends each. In total, there are 12 people.
It is quite easy to construct an example: two groups of 5 and 7 people, people from different group... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (2 points)
In a positive non-constant geometric progression, the arithmetic mean of the third, fourth, and eighth terms is equal to some term of this progression. What is the minimum possible number of this term? | Answer: 4
Solution:
The third element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first and second elements are even less suitable.
To prove that the answer "4" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (2 points)
How many positive numbers are there among the numbers of the form $\operatorname{ctg}\left(\left(15^{n}\right)^{\circ}\right)$, where $\mathrm{n}$ is a natural number from 1 to 2019? | Answer: 1010
Solution:
$\operatorname{ctg} 15^{\circ}>0$.
$15^{2}=225 ; \operatorname{ctg} 225^{\circ}>0$.
Next, $225 \cdot 15=3375$, this number gives a remainder of 135 when divided by $360 . \operatorname{ctg} 135^{\circ}<0$.
$135 \cdot 15=2025$, this number gives a remainder of 225 when divided by 360. The seq... | 1010 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (4 points)
Can the numbers from 0 to 999 (each used exactly once) be arranged in a $100 \times 10$ rectangular table so that the sum of the numbers in each $2 \times 2$ square is the same? | Answer: Yes, it can.
Solution:
The desired table is obtained as the sum of two tables.
The first table looks like this:
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 99 | 0 | 99 | 0 | 99 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 98 | $\ldots$ |
| 1 | 98 | 1 | 98 | 1 | 9... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is a trapezoid with bases $\mathrm{AD}=6$ and $\mathrm{BC}=10$. It turns out that the midpoints of all four sides of the trapezoid lie on the same circle. Find its radius.
If there are multiple correct answers, list them in any order separated by a semicolon. | Answer: 4.
## Examples of answer notation:
45
$4 ; 5$
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given a regular tetrahedron with edge lengths that are integers. Two of these edges have lengths 9 and 11. What is the greatest possible value of the perimeter of the tetrahedron? | Answer: 68.
## Examples of answer recording: 45
# | 68 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given a cubic polynomial $p(x)$. It is known that $p(-6)=30, p(-3)=45, p(-1)=15, p(2)=30$.
Find the area of the figure bounded by the lines $x=-6, y=0, x=2$ and the graph of the given polynomial, if it is also known that on the interval from -6 to 2 the given polynomial takes only positive values. | Answer: 240.
## Examples of answer recording:
45 | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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