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1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=4000$, if $f(0)=1$ and for any $x$ the equality $f(x+4)=f(x)+3 x+4$ holds. | Answer: 5998001
Solution: In the equation $f(x+4)-f(x)=3 x+4$, we will substitute for $x$ the numbers $0,4,8, \ldots, 3996$. We get:
$$
\begin{aligned}
& f(4)-f(0)=3 \cdot 0+4 \\
& f(8)-f(4)=3 \cdot 4+4
\end{aligned}
$$
$$
f(4000)-f(3996)=3 \cdot 3996+4
$$
Adding the equations, we get: $f(4000)-f(0)=3 \cdot(0+4+8+\... | 5998001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 36 positive and 36 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=36+36$, which means $x=9$.
2) Let there be $y$ people with "positive te... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 12 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-13.5,13.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 756
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-28 \leqslant y-1 \leqslant 26$ and $-25 \leqslant 2-x \leqslant 29$. Therefore, $(y-1)(2-x)+2 \leqslant 26 \cdot 29+2=756$. ... | 756 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) On the board, 25 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 25 minutes? | Answer: 300.
Solution: Let's represent 25 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=4 \\
z^{2}+x z+x^{2}=79
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 20
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=4, A C^{2}=79$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Given various natural numbers $a, b, c, d$, for which the following conditions are satisfied: $a>d, a b=c d$ and $a+b+c+d=a c$. Find the sum of all four numbers. | Answer: 12
Solution. From the relations $a>d$ and $ab=cd$, we have the inequality $ba+d$ and $\frac{ac}{2} \geqslant 2c > b+c$. Adding them, we get a contradiction with the condition.
Therefore, without loss of generality, assume that $a \geqslant 3$. Since $a$ is a natural number, then $a \in\{1,2,3\}$. Moreover, $d... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) When dividing the numbers 312837 and 310650 by a certain three-digit natural number, the remainders were the same. Find this remainder. | Answer: 96
Solution. $312837-310650=2187=3^{7}$. The three-digit divisors of this number are also powers of three, i.e., $3^{6}=729,3^{5}=243$. By dividing these numbers by 729 and 243 with a remainder, we find that the remainder is always 96. | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) The area of a sector of a circle is 100. For what value of the radius of the circle will the perimeter of this sector be minimal? If the answer is not an integer, round to the tenths.
Answer: 10 | Solution. The area of the sector corresponding to $\alpha$ radians $S_{\alpha}=\frac{1}{2} R^{2} \alpha=100$. The length of the perimeter is $L(R)=2 R+R \alpha=2 R+\frac{200}{R}$, which reaches a minimum at the point $R=10$. | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (13 points) On the coordinate plane, all points whose coordinates satisfy the conditions
$$
\left\{\begin{array}{l}
|2 x+3 y|+|3 x-2 y| \leq 13 \\
2 x^{2}-3 x y-2 y^{2} \leq 0
\end{array}\right.
$$
are shaded. Find the area of the resulting figure.
# | # Answer: 13
Solution. The solution to the first inequality defines the square $ABCD$, the solution to the second inequality is the shaded part of the plane between the perpendicular lines $2x + y = 0$ and $x - 2y = 0$.
 Find the smallest value of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2(x+7)^{2}+(y-4)^{2}=c \\
(x+4)^{2}+2(y-7)^{2}=c
\end{array}\right.
$$ | Answer: $c=6.0$.
Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have
$$
\begin{aligned}
& \left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right) \geqslant(|x+\alpha|+|y-\beta|)^{2} \\
& \left(1+\frac{1}{2}\right)\left((x+\beta)^{2}+2(y-\alpha)^{2}\right) \geqslant(|x+\beta|+|y-\alpha|)^{2}
\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (6 points) It is known that no digit of a three-digit number is zero and the sum of all possible two-digit numbers formed from the digits of this number is equal to this number. Find the largest such three-digit number. | # Answer: 396
Solution. If there are identical digits, then the sum of three two-digit numbers is less than 300. If the digits are different, then $\overline{a b c}=\overline{a b}+\overline{a c}+\overline{b a}+\overline{b c}+\overline{c a}+\overline{c b}=22(a+b+c)$. The sum of the digits gives the same remainder when... | 396 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) A square is divided into 2016 triangles, with the vertices of no triangle lying on the sides or inside another triangle. The sides of the square are sides of some triangles in the partition. How many points, which are vertices of the triangles, are located inside the square? | Answer: 1007
Solution. Let us have $k$ points inside the square. Then the sum of the angles of all triangles is $360 k+4 \cdot 90=180 \cdot 2016$ degrees. | 1007 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Anton, Boris, Vadim, Gena, Dima, and Egor gathered at the cinema. They bought 6 seats in a row. Anton and Boris want to sit next to each other, while Vadim and Gena do not want to. In how many ways can the boys sit in their seats considering these preferences?
# | # Answer: 144
Solution. The total number of seating arrangements where Anton and Boris sit next to each other is $2 \cdot 5!=240$. The number of seating arrangements where the pairs Anton-Boris and Vadim-Gena end up next to each other is $2 \cdot 2 \cdot 4!=96$. Subtracting the second set from the first gives the answ... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (12 points) On the coordinate plane, all points whose coordinates satisfy the condition
$$
|| x|-2|+|y-3| \leq 3
$$
are shaded. Find the area of the resulting figure. | Answer: 34
Solution. The polygon that results from solving the equation ||$x|-2|+$ $|y-3|=3$ consists of segments of straight lines, glued together at special points: $x=0, x=2, x=$ $-2, y=3$.
$, where $i=1,2, \ldots, n$ satisfy the following conditions:
1) $a_{i}+b_{i}+c_{i}=2017$ for all $i=1,2, \ldots, n$;
2) if $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$. What is the maximum possible value of $n$? (M. Popov) | Answer: 1343.
Solution: Note that
$$
\sum_{i=1}^{n} a_{i} \geqslant \sum_{i=1}^{n} i=\frac{n(n+1)}{2} .
$$
A similar inequality is written for the sums $b_{i}$ and $c_{i}$. Adding the three obtained inequalities, we get
$$
\begin{gathered}
3 \cdot \frac{n(n+1)}{2} \leqslant \sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}... | 1343 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Several people played a round-robin table tennis tournament. At the end of the tournament, it turned out that for any four participants, there would be two who scored the same number of points in the games between these four participants. What is the maximum number of tennis players that could have participated in t... | Answer: 7 tennis players.
Solution. Let $n \geqslant 8$ be the number of tennis players in the tournament. Then the total number of matches played is $\frac{1}{2} n(n-1)$, and thus the total number of victories is also $\frac{1}{2} n(n-1)$. Therefore, there must be a participant who has won at least $\frac{n-1}{2}$ ma... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Cinderella, together with her fairy godmother, released a collection of crystal slippers featuring 7 new models. The fairy tale heroines organized a presentation of the collection for some guests: the audience was supposed to say which slippers they liked. The guests wrote in a questionnaire which models they though... | Solution. Suppose that the guests were given a questionnaire where they could put a "1" next to each pair of shoes they liked and a "0" next to the pair they did not select as the best. Thus, the opinion of each guest can be recorded as a string of "1"s and "0"s. Then, the number of different sets of favorite shoes is ... | 128 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. On the side $AB$ of triangle $ABC$, a point $M$ is taken. It starts moving parallel to $BC$ until it intersects with $AC$, then it moves parallel to $AB$ until it intersects with $BC$, and so on. Is it true that after a certain number of such steps, point $M$ will return to its original position? If this is ... | Solution. Let the length of side $AB$ be 1, and let point $M$ be at a distance $a$ from point $B$. From the properties of the parallelogram, the small triangles are equal, so after 3 steps, point $M$ will be at a distance $a$ from point $A$, which is at a distance $1-a$ from point $B$. After another 3 steps, the point ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$. | Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, \q... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x+3}{2}\right]^{2}-x=1$. | # Solution.
From the equation, it follows that $x=\left[\frac{x+3}{2}\right]^{2}-1$ is an integer. Therefore, $x=n \in \mathbb{Z}$, but we need to consider the cases of even and odd $n$ separately.
First, let's move all terms to the left side of the equation.
1) If $x=2 k$.
$$
\left[\frac{2 k+3}{2}\right]^{2}-2 k-1... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 2.
What is the last digit of the value of the sum $2019^{2020}+2020^{2019} ?$ | # Solution.
The number $2019^{n}$ for $n \in \mathbb{N}$ ends in 9 if $n$ is odd, and in 1 if $n$ is even. Therefore, $2019^{2020}$ ends in 1. The number $2020^{n}$ ends in 0 for any $n \in \mathbb{N}$, so $2020^{2019}$ ends in 0. Thus, the sum $2019^{2020} +$ $2020^{2019}$ ends in the digit 1.
Answer. The digit 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
On the coordinate plane, a square $K$ is marked with vertices at points $(0,0)$ and $(10,10)$. Inside this square, draw the set $M$ of points $(x, y)$ whose coordinates satisfy the equation
$$
[x]=[y],
$$
where $[a]$ denotes the integer part of the number $a$ (i.e., the greatest integer not exceeding $a... | Solution. Let $n \leq x < n+1$, where $n$ is an integer from 0 to 9. Then $[x]=n$ and $[y]=n$. The solution to the latter equation is all $y \in [n, n+1)$. Thus, the solution will be the union of unit squares
$$
\{x \in [n, n+1), y \in [n, n+1), n \in \mathbb{Z}\}
$$
Inside the square $K$ specified in the problem, te... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the code of the letter. Then, the weight of a word can be defined as the sum of the codes of all the letters in that ... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have
$k(C)+k(T)+k(O) \geq k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T)$, which is equivalent to
$$
k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})=0 .
$$
Thus,
$$
k(\amalg)=k(\mathrm{E})=k(C)=k(T)=k(\mathrm{~b})=0 .
$$
The... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Over two days, 50 financiers raised funds to combat a new virus. Each of them made a one-time contribution of a whole number of thousands of rubles, not exceeding 100. Each contribution on the first day did not exceed 50 thousand, while on the second day, it was more than this amount; and no pair of the 50 c... | The solution significantly depends on whether all contributions were distinct or could repeat.
Let's first consider the case where all contributions are distinct. Any natural number greater than 50 but not exceeding 100 can be represented as $50+n$, where $n \in [1, 2, 3, \ldots, 50]$. According to the condition, ther... | 2525 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 2. A triangle was cut into two triangles. Find the greatest value of $N$ such that among the 6 angles of these two triangles, exactly $N$ are the same. | Solution. For $N=4$, an example is an isosceles right triangle divided into two isosceles right triangles: four angles of $45^{\circ}$. Suppose there are five equal angles. Then in one of the triangles, all three angles are equal, meaning all of them, and two angles of the other triangle are $60^{\circ}$. But then both... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements
from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$. | # Solution. Let
$$
M=\left\{x_{1}, \ldots, x_{n}\right\}, \quad x_{1}+\cdots+x_{n}=S
$$
Replace the element $x_{1}$ with the sum of the others. Then
$$
S=\left(S-x_{1}\right)+x_{2}+x_{3}+\cdots+x_{n}=\left(S-x_{1}\right)+\left(S-x_{1}\right)
$$
Reasoning similarly for the other elements, we get that
$$
2 x_{k}=S, ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. After another labor season, the electrified part of the Mediterranean Tundra doubled. At the same time, its non-electrified part decreased by $25 \%$. What fraction of the entire Tundra was not supplied with electricity at the beginning of the labor season? | # Solution
Let $x$ and $y$ be the fractions of the electrified and non-electrified parts, respectively. Clearly, $x+y=1$. According to the condition, $2x + 0.75y = 1$. We obtain the equation
$$
x+y=2x+0.75y
$$
from which we can find the ratio
$$
\frac{x}{y}=\frac{1}{4}
$$
Now we can find the required ratio
$$
\fr... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A table of numbers with 20 rows and 15 columns, $A_{1}, \ldots, A_{20}$ are the sums of the numbers in the rows, $B_{1}, \ldots, B_{15}$ are the sums of the numbers in the columns.
a) Is it possible that $A_{1}=\cdots=A_{20}=B_{1}=\cdots=B_{15}$?
b) If the equalities in part a) are satisfied, what is the sum $A_{1... | Let $A_{i}=B_{j}=X$ for $i=1, \ldots 20$ and $j=1, \ldots, 15$. Consider the sum $S$ of all elements in the table. We have $S=20 X=15 X, X=0$ and $A_{1}+\cdots+A_{20}+B_{1}+\cdots+B_{15}=0$. An example of such a table is, for instance, a table consisting entirely of zeros. There is no need to consider other cases.
Ans... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Early in the morning, the pump was turned on and they started filling the pool. At 10 am, another pump was connected and by 12 pm the pool was half full. By 5 pm, the pool was full. What could be the latest time the first pump was turned on? | Solution. Let the volume of the pool be $V$. Denote by $x$ and $y$ the capacities of the pumps, and by $t$ the time the first pump operates before the second pump is turned on.
Then $t x + 2 x + 2 y = V / 2.5 x + 5 y = V / 2$.
From this, $t x + 2 x + 2 y = 5 x + 5 y$ or $t x = 3 x + 3 y$.
In the end, $t = 3 + 3 y / ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The integer part $[x]$ of a number $x$ is defined as the greatest integer $n$ such that $n \leqslant x$, for example, $[10]=10,[9.93]=9,\left[\frac{1}{9}\right]=0,[-1.7]=-2$. Find all solutions to the equation $\left[\frac{x-1}{2}\right]^{2}+2 x+2=0$. | Solution. From the equation, it follows that $2 x=2-\left[\frac{x-1}{2}\right]^{2}-$ is an integer. Therefore, either $2 x=n \in \mathbb{Z}$, or $x=n+\frac{1}{2} \quad(n \in \mathbb{Z})$. In this case, we need to separately consider the cases of even and odd $n$.
1) If $x=2 k$, then
$$
\begin{aligned}
& {\left[\frac{... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The Atelier "Heavy Burden" purchased a large batch of cast iron buttons. If they sew two buttons on each coat or if they sew three buttons on each coat, in each case 1 piece will remain from the entire batch. If, however, they sew four buttons on each coat or if they sew five buttons on each coat, in each case 3 pie... | # Solution
Let $a$ be the desired number. From the condition, it follows that the number $a-1$ is divisible by 2 and 3. Therefore, $a=6k+1$. Also, the number $a-3$ is divisible by 4 and 5. Therefore, $a=20n+3$. We solve the equation
$$
6k+1=20n+3
$$
Or, equivalently,
$$
3k=10n+1
$$
Its general solution is
$$
k=7+... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangular parallelepiped. The perimeters of each of its three mutually perpendicular faces are equal to the sides of a new rectangular parallelepiped. What can be the minimum ratio of the volume of the new parallelepiped to the volume of the original?
# | # Solution
Let $x, y, z$ be the sides of the original parallelepiped. Then the volume of the new one is
$$
V_{2}=2 \cdot(x+y) \cdot 2 \cdot(y+z) \cdot 2 \cdot(z+x)
$$
The desired ratio of volumes is
$$
\begin{gathered}
\frac{V_{2}}{V_{1}}=\frac{8(x+y)(y+z)(z+x)}{x y z}=\frac{8\left(x y+y^{2}+x z+y z\right)(z+x)}{x ... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Find the maximum value of the quantity $x^{2}+y^{2}$, given that
$$
x^{2}+y^{2}=3 x+8 y
$$ | # Solution
## Method 1
Introduce a Cartesian coordinate system and consider an arbitrary vector $\mathbf{a}$ with coordinates $(x, y)$ and a fixed vector $\mathbf{c}$ with coordinates $(3, 8)$.
Then the left side of the condition represents the square of the length of vector $\mathbf{a}$, and the right side represen... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the code of the letter. Then, the weight of a word can be defined as the sum of the codes of all the letters in that ... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have:
$$
k(C)+k(T)+k(O) \geq k(\Pi)+k(\text { Ya) }+k(T)+k(\text{ b })+k(C)+k(O)+k(T)
$$
which is equivalent to
$$
k(\Pi)+k(T)+k(\text{ b })+k(\text { Ya) }=0
$$
from which it follows that
$$
k(\Pi)=k(T)=k(\text{ b })=k(\text { Ya) }=0
$$
Th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4.
Over two days, 100 bankers collected funds to fight a new virus. Each of them made a one-time contribution of a whole number of thousands of rubles, not exceeding 200. Each contribution on the first day did not exceed 100 thousand, while on the second day it was more than this amount; and no pair of all 1... | The solution significantly depends on whether all contributions were distinct or could repeat.
Let's first consider the case where all contributions are distinct. Any natural number greater than 100 but not exceeding 200 can be represented as $100+n$, where $n \in [1,2,3, \ldots, 100]$. According to the condition, the... | 10050 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the six houses on one side of the street is connected by cable lines to each of the eight houses on the opposite side. How many pairwise intersections do the shadows of these cables form on the surface of the street, if no three of them intersect at the same point? Assume that the light causing these shadows... | # Solution
Let's take an arbitrary pair of houses on one side of the street and an arbitrary pair on the other. They are the vertices of a convex quadrilateral (since two sides of the quadrilateral, coming from each chosen pair, lie on one side of the line, i.e., the angles do not exceed $180^{\circ}$), so its diagona... | 420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the maximum value of the quantity $x^{2}+y^{2}+z^{2}$, given that
$$
x^{2}+y^{2}+z^{2}=3 x+8 y+z
$$ | # Solution
## 1st Method
Introduce a Cartesian coordinate system and consider an arbitrary vector $\mathbf{a}$ with coordinates $(x, y, z)$ and a fixed vector $\mathbf{c}$ with coordinates $(3, 8, 1)$. Then, the left side of the condition represents the square of the length of vector $\mathbf{a}$, and the right side ... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Four brigades were developing an open coal deposit for three years, working with a constant productivity for each brigade. In the second year, due to weather conditions, work was not carried out for four months, and for the rest of the time, the brigades worked in rotation (one at a time). The ratio of the w... | Solution. Let the $i$-th brigade mine $x_{i}$ coal per month. Then we have the system
$$
\left\{\begin{array}{l}
4 x_{1}+x_{2}+2 x_{3}+5 x_{4}=10 \\
2 x_{1}+3 x_{2}+2 x_{3}+x_{4}=7 \\
5 x_{1}+2 x_{2}+x_{3}+4 x_{4}=14
\end{array}\right.
$$
By adding twice the first equation to thrice the second and subtracting the thi... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 2.
What is the last digit of the value of the sum $5^{2020}+6^{2019} ?$ | # Solution.
The number 5 to any power ends in 5.
The number 6 to any power ends in 6.
Therefore, the sum $5^{2020}+6^{2019}$ ends in the digit 1.
Answer. The digit 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the letter code. Then, the weight of a word can be defined as the sum of the codes of all the letters in that word. I... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have:
$$
k(C)+k(T)+k(O) \geq k(\Pi)+k(\text { ( })+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T)
$$
which is equivalent to
$$
k(\Pi)+k(T)+k(\mathrm{~b})+k(\text { Я) }=0
$$
from which it follows that
$$
k(\Pi)=k(T)=k(\mathrm{~b})=k(\text { Я })=0
$$
Th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5.
The carriages of the express train "Moscow - Yalta" must be numbered consecutively, starting from one. But in a hurry, two adjacent carriages received the same number. As a result, it turned out that the sum of the numbers of all carriages is 111. How many carriages are in the train and which number was u... | # Solution.
Let's start calculating the sums sequentially
$$
\begin{aligned}
& 1+2=3 \\
& 1+2+3=6 \\
& \cdots \\
& 1+2+\ldots+14=105 \\
& 1+2+\ldots+14+15=120
\end{aligned}
$$
From this, it is clear that the first 14 wagons had the first 14 sequential numbers, which sum up to 105, and one more wagon had a number equ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
Three electric generators have powers $x_{1}, x_{2}, x_{3}$, the total power of all three does not exceed 2 MW. In the power system with such generators, a certain process is described by the function
$$
f\left(x_{1}, x_{2}, x_{3}\right)=\sqrt{x_{1}^{2}+x_{2} x_{3}}+\sqrt{x_{2}^{2}+x_{1} x_{3}}+\sqrt{x_{... | # Solution.
It is clear that the minimum value of the function is zero (achieved when $\left.x_{1}=x_{2}=x_{3}=0\right)$.
Let's find the maximum. We can assume that $x_{1} \geq x_{2} \geq x_{3} \geq 0$. We will prove two inequalities:
$$
\begin{gathered}
\sqrt{x_{1}^{2}+x_{2} x_{3}} \leq x_{1}+\frac{x_{3}}{2} \\
\sq... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
A polynomial $P(x)$ with integer coefficients has the properties
$$
P(1)=2019, \quad P(2019)=1, \quad P(k)=k,
$$
where the number $k$ is an integer. Find this number $k$.
# | # Solution.
Since the polynomial $P(x)$ has integer coefficients, $P(a)-P(b)$ is divisible by $a-b$ for any integers $a$ and $b$.
We get that
$$
\begin{gathered}
P(k)-P(1)=(k-2019) \text { is divisible by }(k-1), \\
P(k)-P(2019)=(k-1) \text { is divisible by }(k-2019) .
\end{gathered}
$$
This can only be true if $|... | 1010 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that there are four ways to prepare magical pollen to create elixirs of kindness, joy, wisdom, luck, health, friendliness, and creativity. But elixirs of kindness, joy, and wisdom are made from fairy pollen, while elixirs of luck, health, friendliness, and creativity are made from elf pollen. Among the i... | Solution. Four methods of preparing pollen are represented by four branches of a tree. From two branches with methods of preparing fairy pollen, three branches of elixirs (of goodness, joy, and wisdom) branch off, and from two branches with methods of preparing elf pollen, four branches (of luck, health, friendliness, ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the free term $a_{0}$ of the polynomial $P(x)$ with integer coefficients is less than 100 in modulus, and $P(20)=P(16)=2016$. Find $a_{0}$. | Solution. We can write $P(x)-2016=(x-16)(x-20) Q(x)$, where $Q(x)$ is a polynomial with integer coefficients. The constant term of the right side is $320 k$, where $k$ is an integer. Thus, $a_{0}=2016-320 k$. The condition is satisfied only by the value $k=6, a_{0}=96$.
Answer. $a_{0}=96$. | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The council of the secret pipeline village is gathering around a round table, where each arriving member can sit in any free seat. How many different seating arrangements are possible if 7 participants will attend the council? (Two seating arrangements are considered the same if the same people sit to the left and r... | # Solution
Since empty seats are not taken into account, we can consider only the ways of arranging on seven seats. The first person can sit in any of the 7 seats, the next in any of the 6 remaining seats, and so on until the last. In total, there are $7 \cdot 6 \cdot \ldots \cdot 1=7!$ ways.
However, each arrangemen... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the summer, Ponchik eats honey cakes four times a day: instead of morning exercise, instead of a daytime walk, instead of an evening run, and instead of a nighttime swim. The quantities of cakes eaten instead of exercise and instead of a walk are in the ratio of $3: 2$; instead of a walk and instead of a run - as... | # Solution
According to the problem, we can establish the following ratios for the number of doughnuts eaten instead of engaging in a particular useful activity:
$$
\frac{\text { Morning Exercise }}{\text { Walk }}=\frac{3}{2} \quad \frac{\text { Walk }}{\text { Jog }}=\frac{5}{3} \quad \frac{\text { Jog }}{\text { S... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
Two swimmers are training in a rectangular quarry. The first swimmer finds it more convenient to exit at a corner of the quarry, so he swims along the diagonal to the opposite corner and back. The second swimmer finds it more convenient to start from a point that divides one of the quarry's shores in the ... | # Solution.
Let's draw a rectangular quarry $A D C D$ and the inscribed quadrilateral route of the second swimmer $N L M K$.
Reflect the drawing symmetrically first with respect to side $C D$, then with respect to side $C B^{\prime}$, and finally with respect to side $A^{\prime} B^{\prime}$.
![](https://cdn.mathpix.... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. In the country of "Energetika," there are 150 factories, and some of them are connected by bus routes that do not stop anywhere except at these factories. It turned out that any four factories can be divided into two pairs such that there is a bus route between the factories of each pair. Find the smallest n... | Solution. Suppose that some factory $X$ is connected by bus routes to no more than 146 factories. Then a quartet of factories, consisting of $X$ and any three with which it is not connected, does not satisfy the problem's condition, since $X$ cannot be paired with any of the three remaining factories. Therefore, each f... | 11025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be equal to the same value k. Find the value of k. Provide at least one set of four numbers that satisfy the condition. Describe all possible set... | Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:
$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4.
Find the number of numbers $N$ in the set $\{1,2, \ldots, 2018\}$ for which there exist positive solutions $x$ to the equation
$$
x^{[x]}=N
$$
( $[x]$ is the integer part of the real number $x$, i.e., the greatest integer not exceeding $x$) | # Solution.
Notice that suitable numbers $N$ for $x$ such that $[x]=n$ are the numbers from $n^{n}$ to $(n+1)^{n-1}$, that is, exactly the numbers for which $[\sqrt[n]{N}]=n$. Such numbers (among the numbers from 1 to 2018) are the number 1, the numbers from $2^{2}$ to $3^{2-1}$ (there are exactly 5 of them), the numb... | 412 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5.
An electric cable 21 meters long is cut into 21 pieces. For any two pieces, their lengths differ by no more than a factor of three. What is the smallest $m$ such that there will definitely be two pieces whose lengths differ by no more than a factor of $m$? | # Solution.
If among the pieces of cable there are at least two that differ in length, then by taking the ratio of the smaller to the larger, we get that $m \leq 1$.
However, the cable can be cut in such a way that all pieces are equal. This implies that if $m<1$, then a way of cutting has been found where no two pie... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Boys and girls formed a circle in such a way that the number of children whose right neighbor is of the same gender is equal to the number of children whose right neighbor is of a different gender. What could be the total number of children in the circle? | Solution. Let $n$ be the number of children next to whom stands a child of the opposite gender, and $m$ be the number of children next to whom stands a child of the same gender. Initially, $n=m$, i.e., the total number of children ($n+m$) is even. We will swap the positions of two adjacent children so that all boys gat... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be the same value $\mathrm{k}$. Find the value of $\mathrm{k}$. Provide at least one set of four numbers that satisfy the condition. Describe all... | Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:
$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1.
The Triassic Discoglossus tadpoles have five legs, while the saber-toothed frog tadpoles grow several tails (all have the same number of tails). An employee of the Jurassic Park scooped up several tadpoles with water. It turned out that the captured tadpoles had a total of 100 legs and 64 tails. How many tai... | # Solution.
Let $x$ be the number of tails of the saber-toothed frog's pollywog. Suppose $n$ five-legged and $k$ many-tailed pollywogs were caught. Counting the total number of legs and tails gives the equations
$$
\left\{\begin{aligned}
5 n+4 k & =100 \\
n+x k & =64
\end{aligned}\right.
$$
From the first equation, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
A rule is given by which each pair of integers $X$ and $Y$ is assigned a number $X \nabla Y$. (The symbol «»» means applying the rule to the numbers $X$ and $Y$.) It is known that for any integers $X, Y$ the following properties hold:
1) $X \nabla 0=X$
2) $X \nabla(Y-1)=(X \nabla Y)-2$
3) $X \nabla(Y+1)=... | # Solution.
Let's start writing down in order the result of applying the operation $X \nabla Y$ for $Y=0,1,2, \ldots$, using property 3.
$$
\begin{aligned}
& X \nabla 0=0, \\
& X \nabla 1=X \nabla(0+1)=(X \nabla 0)+2=X+2, \\
& X \nabla 2=X \nabla(1+1)=(X \nabla 1)+2=X+4, \\
& X \nabla 3=X \nabla(2+1)=(X \nabla 2)+2=X... | -673 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 5.
Once upon a time, Baba Yaga and Koschei the Deathless tried to divide a magical powder that turns everything into gold equally. Baba Yaga took out a scale and weighed all the powder. The scales showed 6 zolotniks. Then she started removing the powder until the scales showed 3 zolotniks. However, Koschei susp... | # Solution.
Let $A$ be the weight of the first part (the one that remained on the scales), $B$ be the weight of the second part (the one that was poured off), and $d$ be the error of the scales.
Then the result of the first weighing (of the entire powder) gives
$$
A+B+d=6,
$$
the result of the second weighing (afte... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Does there exist a convex polygon with 2015 diagonals?
# | # Solution.
The number of diagonals in a convex $n$-gon is
$$
\frac{n(n-1)}{2}-n=\frac{n^{2}-3 n}{2}
$$
Solve the equation
$$
\frac{n^{2}-3 n}{2}=2015 \quad \Longleftrightarrow \quad n^{2}-3 n-4030=0
$$
Its discriminant $D=9+4 \cdot 4030=16129$. Note that $120^{2}<16129<130^{2}$ and $D$ ends with the digit 9. If $... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2.
In a football tournament, each team is supposed to play one match against each of the others. But during the tournament, half of all the teams were disqualified and withdrew from further participation. As a result, 77 matches were played, and the teams that withdrew had managed to play all their matches a... | # Solution.
According to the condition, an even number of teams $2n$ started, of which $n$ were disqualified. The eliminated teams played $\frac{n(n-1)}{2}$ matches among themselves. The same number of matches were played by the teams that remained in the tournament.
Let each of the eliminated teams have played with ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4.
At 9:00 AM, ships "Anin" and "Vinin" departed from port O to port E. At the same moment, ship "Sanin" departed from port E to port O. All three vessels are traveling on the same course ("Sanin" heading towards "Anin" and "Vinin") at constant but different speeds. At 10:00 AM, "Vinin" was at the same dista... | # Solution.
For simplicity, let's assume the distance between the ports is 1. Let $x$ be the speed of "Anina", $y$ be the speed of "Vanina", and $z$ be the speed of "Sanina". Then at 10 o'clock (i.e., one hour after departure),
$$
2 y = x + 1 - z,
$$
and one and a half hours after departure,
$$
2(1 - 1.5 z) = 1.5 x... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a square table with 2015 rows and columns, positive numbers are arranged. The product of the numbers in each row and each column is 1, and the product of the numbers in any 1008 x 1008 square is 2. What number is in the center of the table?
# | # Solution
Consider the first 1008 rows of the table. From the additional condition, it follows that if these rows are covered by two squares of size $1008 \times 1008$, then these squares will overlap by one column. Denote the product of the numbers in this column (1008 numbers) by $M$.
Then the product of all the n... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Clever Dusya arranges six cheat sheets in four secret pockets so that the 1st and 2nd cheat sheets end up in the same pocket, the 4th and 5th cheat sheets also end up in the same pocket, but not in the same pocket as the 1st. The others can be placed anywhere, but only one pocket can remain empty (or all can be fill... | # Solution
Let's reason constructively and straightforwardly. Place the 1st and 2nd cheat sheets in any pocket. This can be done in 4 ways. Now place the 4th and 5th cheat sheets in any free pocket. This can be done in 3 ways. In total, there are 12 ways. There remain two cheat sheets (3rd and 6th) and two free pocket... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Four managers of responsibility shifting reported: "If they are arranged in pairs, there will be 1 left. If they are arranged in threes, there will also be 1 left. If they are arranged in fours, there will be 2 left, and if they are arranged in fives, there will also be 2 left." Should the head of the report recepti... | # Solution
Let $a$ be the desired quantity. If when dividing by 4, 2 remains, then $a$ is even. But then when dividing by two, 1 cannot remain. Therefore, these two statements are contradictory and the message as a whole is false.
Remove one of the contradictory statements.
1st var.: leave the division by 3, by 4, a... | 1042 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. (15 points) Real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ are such that
$$
\left\{\begin{array}{l}
x_{1}+x_{2} \geqslant 12 \\
x_{1}+x_{3} \geqslant 13 \\
x_{1}+x_{4} \geqslant 14 \\
x_{3}+x_{4} \geqslant 22 \\
x_{2}+x_{3} \geqslant 23 \\
x_{2}+x_{4} \geq 24
\end{array}\right.
$$
What is the smallest value t... | Answer: 37.
Solution. By adding the second equality to the last one, we get $x_{1}+x_{2}+x_{3}+x_{4} \geqslant 37$.
It is also worth noting that the value of the expression $x_{1}+x_{2}+x_{3}+x_{4}$ can be equal to 37, for example, when $x_{1}=1, x_{2}=11, x_{3}=12, x_{4}=13$. It is easy to verify that such numbers s... | 37 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. (15 points) In one move, you can choose a natural number $x$ and cross out all natural numbers $y$ such that $|x-y|$ is a natural composite number. At the same time, $x$ can be chosen from already crossed out numbers.
What is the minimum number of moves needed to cross out all numbers from the natural num... | Answer: 2.
Solution. First, note that one move is not enough, because if we choose some natural number $x$, then the number $y=x$ will remain unchecked. Let's prove that two moves are enough.
We will look for two suitable numbers $x_{1}$ and $x_{2}$ of different parity. Then one of the differences $\left|y-x_{1}\righ... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 7. Problem 7.1*
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha, in any order, tells the digits that stand in these places. The order in which to tell the digits is chosen by Misha. For examp... | # 7. Problem $7.1^{*}$
Misha thought of a five-digit number, all digits of which are different, and Igor is trying to guess it. In one move, Igor can choose several digits of the number, and Misha reports the digits in these positions in any order. The order in which to report the digits is chosen by Misha. For exampl... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 9. Problem 9.1*
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fin... | # 9. Problem 9.1*
Petya came up with four different natural numbers, wrote down all their pairwise sums on the board, and in the row below, all their sums of three. It turned out that the sum of the two largest numbers in the top row and the two smallest numbers in the bottom row (a total of four numbers) is 2017. Fin... | 806 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. (15 points) Do there exist polynomials $P(x), Q(x)$, and $R(x)$ with real coefficients such that the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have the same degree, while the polynomials $P(x)+Q(x), P(x)+R(x)$, and $Q(x)+R(x)$ have pairwise distinct degrees? (We consider that t... | Answer: $: 2$.
Solution. It is sufficient to take, for example, $P(x)=x^{2}, Q(x)=-x^{2}+1, R(x)=x^{2}+x$. Then the polynomials $P(x) \cdot Q(x), Q(x) \cdot R(x)$, and $P(x) \cdot R(x)$ have degree 4 (these are products of two polynomials of degree 2); the polynomial $P(x)+Q(x)$ equals 1 and has degree 0; the polynomi... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
Problem 10.4. (15 points) Once, 45 friends living in different parts of the world decided to exchange news with each other. To do this, they plan to arrange $k$ video meetings, at each of which every person will share their news, as well as all the news from other people they have learned previously.
For the video mee... | Answer: 5 days.
Solution. Let's provide an example of a situation where 4 days would not be enough. Suppose each of the 45 people has their own unique pair of days when they cannot participate in the meeting. Since the number of ways to choose a pair of days from the 10 offered is $\mathrm{C}_{10}^{2}=45$, for any pai... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 10.5. (20 points) Find all composite natural numbers $n$ that have the following property: each natural divisor of the number $n$ (including $n$ itself), decreased by 1, is a square of an integer. | Answer: 10.
Solution. Suppose that $n$ is divisible by the square of some prime number $p$. Then it has a divisor $p^{2}=b^{2}+1$; but two squares of integers can only differ by 1 if they are 0 and 1.
Let $n$ be divisible by some two prime numbers $p$ and $q$. Without loss of generality, we can assume that $p>q$. Fro... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 8. Problem $8^{*}$
A natural number $\mathrm{N}$ ends in 5. Ninth-grader Dima found all its divisors and discovered that the sum of the two largest proper divisors does not divide evenly by the sum of the two smallest proper divisors. Find the smallest possible value of the number N. A divisor of a natural number is... | # 8. Problem $\mathbf{8}^{2}$
A natural number $\mathrm{N}$ ends with 5. Ninth-grader Dima found all its divisors and noticed that the sum of the two largest proper divisors does not divide evenly by the sum of the two smallest proper divisors. Find the smallest possible value of the number N. A divisor of a natural n... | 725 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In an acute-angled triangle $A B C$, the altitude $A A_{1}$ is drawn. $H$ is the orthocenter of triangle $A B C$. It is known that $A H=3, A_{1} H=2$, and the radius of the circumcircle of triangle $A B C$ is 4. Find the distance from the center of this circle to $H$. | Solution. Draw the altitudes $B B_{1}$ and $C C_{1}$ in the triangle. Then the quadrilateral $A C_{1} H B_{1}$ is cyclic, since its opposite angles $C_{1}$ and $B_{1}$ are right angles. Therefore: $\angle B H C=$ $\angle C_{1} H B_{1}=180^{\circ}-\angle C_{1} A B_{1}$. Reflect point $H$ symmetrically with respect to th... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Let $x$ be a number from the interval $(\pi / 2, \pi)$ such that
$$
\frac{4}{3}\left(\frac{1}{\sin x}+\frac{1}{\cos x}\right)=1
$$
Prove that the number
$$
\left(\frac{4}{3}\right)^{4}\left(\frac{1}{\sin ^{4} x}+\frac{1}{\cos ^{4} x}\right)
$$
is an integer and find it. | Solution. For brevity, let $s=\sin x, c=\cos x$. We need to find the number $t=$ $1 / c s$. For this, we square the relation $\left(s^{-1}+c^{-1}\right)=3 / 4$. We get $s^{2}+c^{2}+2 c s=$ $9(c s)^{2} / 16$. Considering that $s^{2}+c^{2}=1$, we obtain the following quadratic equation for $t$:
$$
t^{2}+2 t-9 / 16=0
$$
... | 49 | Algebra | proof | Yes | Yes | olympiads | false |
4. On the board, the numbers $1 / 1, 1 / 2, 1 / 3, 1 / 4, \ldots 1 / 100$ are written.
It is allowed to erase any two numbers a, b and write $a b + a + b$ in their place,
then do the same with any two of the remaining numbers, and so on. What number can be left last? | Solution. If initially on the board there were not one hundred, but three numbers: a, b, c, then at the end on the board there would be the number abc + ab + bc + ac + a + b + c. That is, the sum of all possible monomials composed of the numbers a, b, c, taken no more than once each.
In the case when one hundred numbe... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 3. Problem 3
In an isosceles triangle $A B C$, the perpendicular bisector of the lateral side $B C$ intersects the base $A B$ at point $D$ such that $A C=A D$. Find the angle $A B C$.
Write the answer in degrees without the degree symbol. | # 3. Problem 3
In an isosceles triangle $ABC$, the perpendicular bisector of the lateral side $BC$ intersects the base $AB$ at point $D$ such that $AC = AD$. Find the angle $ABC$.
Write the answer in degrees without the degree symbol.
Answer: 36 | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9-1. Calculate the sum $1^{2}+2^{2}-3^{2}-4^{2}+5^{2}+6^{2}-7^{2}-8^{2}+9^{2}+10^{2}-\ldots+2017^{2}+2018^{2}$. | Answer: 4074341.
Solution. Note that for any $k$ the equality $k^{2}-(k+1)^{2}-(k+2)^{2}+(k+3)^{2}=4$ holds. Therefore, the entire sum is equal to $1+504 \cdot 4+2018^{2}=4074341$.
| Rating | Score | Content of the criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| $+/-$ | 16 | Correct solution wit... | 4074341 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9-3. In a right triangle $A B C$, angle $B$ is a right angle. On the leg $A B$, a point $M$ is chosen such that $A M=B C$, and on the leg $B C$, a point $N$ is chosen such that $C N=M B$. Find the acute angle between the lines $A N$ and $C M$.
Answer: $45^{\circ}$.
$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. On a plane, there are eight different points. A numbering of these points with numbers from 1 to 8 is called good if the following condition is met:
there exists a line such that all points lie on one side of it and at different distances from it, and the distances from the points to this line increase with the ... | Answer. $2 C_{8}^{2}=56$.
Solution. Note that when numbering points relative to a line, we are only interested in its direction and which side of it the points are on. All this information can be uniquely restored from a unit normal to it, directed towards the points. Therefore, the numbering of the octet is derived f... | 56 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. Along the shore of a circular lake with a perimeter of 1 km, two salmons are swimming - one at a constant speed of $500 \mathrm{m} /$ min clockwise, the other at a constant speed of 750 m/min counterclockwise. Along the edge of the shore, a bear is running, always moving along the shore at a speed of 200 m/min in ... | Solution. Note that the bear runs to the nearest salmon if and only if it runs from a point on the shore that is equidistant from the salmons and not separated from the bear by the salmons. Since both points equidistant from the salmons move counterclockwise at a speed of $|750-500| / 2=125<200$ m/min, they will never ... | 7 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively, such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30. | # 6. Problem 6
In triangle $A B C$, points $M$ and $N$ are chosen on sides $A B$ and $B C$ respectively such that $A M=2 M B$ and $B N=N C$. Segments $A N$ and $C M$ intersect at point $P$. Find the area of quadrilateral $M B N P$, given that the area of triangle $A B C$ is 30.
## Answer: 7
# | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10-1. For real numbers $a, b$, and $c$, it is known that $a b c + a + b + c = 10$, and $a b + b c + a c = 9$. For which numbers $x$ can it be asserted that at least one of the numbers $a, b, c$ is equal to $x$? (Find all such numbers $x$ and prove that there are no others.)
Answer: 1. | Solution. We will provide two solutions to the problem.
First Solution. Let $a+b+c=\lambda$. Vieta's theorem allows us to write a cubic equation depending on the parameter $\lambda$, whose roots are the set $a, b, c$ corresponding to the given $\lambda$:
$$
t^{3}-\lambda t^{2}+9 t-(10-\lambda)=0 \quad \Leftrightarrow... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10-5. Consider all reduced quadratic trinomials $x^{2}+p x+$ $q$ with integer coefficients $p$ and $q$. Let's call the range of such a trinomial the set of its values at all integer points $x=0, \pm 1, \pm 2, \ldots$ What is the maximum number of such trinomials that can be chosen so that their ranges do not in... | Solution. Note that the substitution of the variable $x \rightarrow x+k$ for any integer $k$ does not change the range of the polynomial. Then, by making the substitution $x \rightarrow x-\left[\frac{p}{2}\right]$ (square brackets denote the integer part), we can assume that any polynomial has one of two forms: $x^{2}+... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7-5. In the garden of the oracle, there live four turtles. A visitor can choose any subset of turtles in a move and ask the oracle how many of these turtles are males (the oracle's answers are always truthful). What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3. | Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question fo... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. (15 points) Natural numbers $a, b, c$ are such that $1 \leqslant a<b<c \leqslant 3000$. Find the greatest possible value of the quantity
$$
\text { GCD }(a, b)+\text { GCD }(b, c)+\text { GCD }(c, a)
$$ | Answer: 3000.
Solution. Note that GCD $(a, b)=$ GCD $(a, b-a) \leqslant b-a$, since the GCD of two natural numbers does not exceed either of them. Similarly, we obtain that GCD $(b, c) \leqslant c-b$, and also GCD $(c, a) \leqslant a$.
Adding these three inequalities, we get
$$
\text { GCD }(a, b)+\text { GCD }(b, c... | 3000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. (20 points) Quadratic trinomials $P(x)$ and $Q(x)$ with real coefficients are such that together they have 4 distinct real roots, and each of the polynomials $P(Q(x))$ and $Q(P(x))$ has 4 distinct real roots. What is the smallest number of distinct real numbers that can be among the roots of the polynomia... | # Answer: 6.
Solution. Note that if among the roots of the polynomial $P(Q(x))$ there is a root of $Q(x)$, say, the number $x_{0}$, then $P\left(Q\left(x_{0}\right)\right)=P(0)=0$, from which 0 is a root of $P(x)$. Similarly, if among the roots of $Q(P(x))$ there is a root of $P(x)$, then 0 is a root of $Q(x)$. Howeve... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8-2. Calculate the sum $1^{2}+2^{2}-3^{2}-4^{2}+5^{2}+6^{2}-7^{2}-8^{2}+9^{2}+10^{2}-\ldots+2017^{2}+2018^{2}$. | Solution. Note that for any $k$ the equality $k^{2}-(k+1)^{2}-(k+$ $2)^{2}+(k+3)^{2}=4$ holds. Therefore, the entire sum is equal to $1+504 \cdot 4+2018^{2}=4074341$.
| Grade | Score | Content criterion |
| :---: | :---: | :--- |
| + | 20 | Complete solution |
| .+ | 18 | Solution is correct, but there are minor flaws... | 4074341 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8-3. $\quad$ In a right triangle $A B C$, angle $B$ is a right angle. On the leg $A B$, a point $M$ is chosen such that $A M=B C$, and on the leg $B C$, a point $N$ is chosen such that $C N=M B$. Find the acute angle between the lines $A N$ and $C M$.
Answer: $45^{\circ}$.
. What is the minimum number of moves required to find out the gender of all the turtles?
Answer: 3. | Solution. In three questions, the answer can be obtained as follows. The first two questions are about turtles 1 and 2, and 2 and 3. If at least one of the answers is 0 or 2, we know who they are for the corresponding pair, and for the remaining one of the three, we know from the other question, leaving one question fo... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8-5. 8 8-5. A divisor of a natural number is called proper if it is different from 1 and the number itself. Find all natural numbers for which the difference between the sum of the two largest proper divisors and the sum of the two smallest proper divisors is a prime number. | Answer: $12(2,3,4,6)$.
Solution. One of two cases applies.
A) Suppose the two smallest divisors $p$ and $q$ are prime numbers. Then the number $r=(n / p+n / q)-(p+q)$ is also prime, and $p q r=(p+q)(n-p q)$. Since the numbers $p+q$ and $p q$ are coprime, we get $r=p+q$, from which $p=2$ and $n=4 q$. But then, due to ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.3. The sequence $a_{n}$ is constructed as follows: $a_{1}, a_{2}$ are arbitrary real numbers, and for $n \geq 3$, the number $a_{n}$ is the smallest of the numbers $\left|a_{i}-a_{j}\right|, 1 \leq$ $i<j \leq n-1$. For example, if $a_{1}=6, a_{2}=\frac{19}{2}$, then we get the sequence $6, \frac{19}{2}, \frac{7}{2},... | Solution. Let $S_{n}$ be the set of all numbers $\left|a_{i}-a_{j}\right|, 1 \leq i<j \leq n-1, \quad n \geq 3$. Then $a_{n}$ is the smallest number in the set $S_{n}$. Since $S_{n} \subset S_{n+1}$, then $a_{n} \geq a_{n+1}$, i.e., starting from $a_{3}$, the sequence is non-increasing. Therefore, $a_{9} \geq a_{10}=1$... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 2. Task 2
In a dumpling shop, you can order dumplings in portions of 6, 9, and 20 pieces. Thus, not every number of dumplings can be ordered with these sets, for example, 1, 2, 3, 4, 5, 7, and 8 cannot be bought. What is the largest number of dumplings that cannot be ordered in the dumpling shop? | # 2. Task 2
In a dumpling shop, you can order dumplings in portions of 6, 9, and 20 pieces. Thus, not every number of dumplings can be ordered with these sets, for example, 1, 2, 3, 4, 5, 7, and 8 cannot be bought. What is the largest number of dumplings that cannot be ordered in the dumpling shop?
## Answer: 43 | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Natural numbers $x$ and $y$ are such that the following equality holds:
$$
x^{2}-3 x=25 y^{2}-15 y
$$
How many times greater is the number $x$ than the number $y$? | Answer. 5
Solution. The equality $x^{2}-3 x=25 y^{2}-15 y$ is equivalent to the equality $(5 y-x)(5 y+x-3)=$ 0. Since $x$ and $y$ are natural numbers, $x, y \geq 1$. Therefore, the second bracket $5 y+x-3 \geq 3$, in particular, is non-zero. Therefore, the first bracket $5 y-x$ is zero, meaning $x$ is 5 times $y$.
Cr... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Pete, Sasha, and Misha are playing tennis in a knockout format. A knockout format means that in each match, two players compete while the third waits. The loser of the match gives up their place to the third player and becomes the waiting player in the next match. Pete played a total of 12 matches, Sasha played 7 ma... | Answer: 4
Solution. First, let's find the total number of games played. Petya, Pasha, and Misha participated in a total of $12+7+11=30$ games. Since each game involves two participants, the number of games is half of this: $30 / 2=15$.
Thus, Petya did not participate in $15-12=3$ games, Pasha in $15-7=8$ games, and M... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 1. Task 1.1
Timur and Alexander are counting the trees growing around their house. Both are moving in the same direction but start counting from different trees. What is the total number of trees growing around the house if the tree that Timur called the 12th, Alexander counted as $33-\mathrm{m}$, and the tree that... | # 1. Task 1.1
Timur and Alexander are counting the trees growing around their house. Both are moving in the same direction but start counting from different trees. What is the total number of trees growing around the house if the tree that Timur called the 12th, Alexander counted as $33-\mathrm{m}$, and the tree that... | 118 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 8. Problem 8.1
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without specifying the unit of measurement).
# | # 8. Problem 8.1
Pentagon $A B C D E$ is circumscribed around a circle.
Angles $\angle A B C, \angle B A E, \angle C D E$ are each $104^{\circ}$. Find $\angle A D B$. Write the answer in degrees (the number only, without specifying the unit of measurement).
## Answer: 38 | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. How many different (i.e., not equal to each other) acute triangles with integer side lengths and a perimeter of 24 exist? List the lengths of the three sides of all these triangles and prove that there are no others.
ANSWER: 6.
CRITERIA
b - correct criterion for acuteness and either a correct list of acute tria... | Solution. Note that a triangle with sides $a \geqslant b \geqslant c$ is obtuse if and only if $a^{2}>b^{2}+c^{2}$. Indeed, if $A B C$ is a triangle with sides $a \geqslant b \geqslant c$ opposite vertices $A, B$, and $C$ respectively, and $A B^{\prime} C$ is a triangle with a right angle $\angle A$ and $A B^{\prime}=A... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.5. (20 points) Real numbers $a, b, c, d$ are such that $a+b=\frac{9}{c-d}$ and $c+d=$ $=\frac{25}{a-b}$. What is the smallest value that the quantity $a^{2}+b^{2}+c^{2}+d^{2}$ can take? | # Answer: 34.
Solution. If the given equalities are multiplied by the denominators of the corresponding fractions and added, we get $2(a c-b d)=34$. We will prove that $a^{2}+b^{2}+c^{2}+d^{2} \geqslant 2(a c-b d)$. This follows from $a^{2}+c^{2} \geqslant 2 a c$ (equivalent to $(a-c)^{2} \geqslant 0$) and $b^{2}+d^{2... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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