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# Problem 3.
Find all functions $f(x)$ defined on the entire number line and satisfying the condition
$$
f(x-y)=f(x) \cdot f(y) \text { for all } x, y.
$$
# | # Solution.
One solution is obvious, which is $f(x) \equiv 0$.
Consider functions that are not identically zero over the entire number line.
By setting $t=x-y$, we get that $f(t)=f(t+y) f(y)$. Therefore, $f(y)$ does not equal 0 for any $y$, because otherwise the value of $f(t)$ would be 0 for all $t$, which contradi... | f(x)\equiv0,f(x)\equiv1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,851 |
# Problem 4.
The integer part $[x]$ of a real number $x$ is defined as the greatest integer $M$ such that $M \leq x$. For example, $[\sqrt{2}]=1,[2]=2,[\pi]=3$. Find all positive real numbers $x$ for which
$$
x[x[x[x]]]<2018
$$ | # Solution.
Note that for $x>0$, the left side of the equation is a monotonically non-decreasing function, and that for integer values of $x$ it equals $x^{4}$. Since $2401=7^{4}$, the inequality does not hold for $x \geq 7$.
We will show that it is satisfied for $0<x<7$. Indeed, in this case,
$$
\begin{aligned}
0 \... | 0<x<7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,852 |
# Task 1.
If Mary Ivanovna is on VKontakte at the pedcouncil, then Ivan Ilyich and Alexandra Varfolomeevna are also on VKontakte. The director has known this fact for a long time. He also knows the following. Only one of the two - Alexandra Varfolomeevna or Pyotr Petrovich - is on VKontakte. At least one of the other... | # Solution.
We will use the first letters of the names to denote the characters.
The first condition allows for two variants. Let's consider them in turn. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,853 |
# Task 2.
What is the last digit of the value of the sum $2019^{2020}+2020^{2019} ?$ | # Solution.
The number $2019^{n}$ for $n \in \mathbb{N}$ ends in 9 if $n$ is odd, and in 1 if $n$ is even. Therefore, $2019^{2020}$ ends in 1. The number $2020^{n}$ ends in 0 for any $n \in \mathbb{N}$, so $2020^{2019}$ ends in 0. Thus, the sum $2019^{2020} +$ $2020^{2019}$ ends in the digit 1.
Answer. The digit 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,855 |
# Problem 3.
On the coordinate plane, a square $K$ is marked with vertices at points $(0,0)$ and $(10,10)$. Inside this square, draw the set $M$ of points $(x, y)$ whose coordinates satisfy the equation
$$
[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}\}
$$
Inside the square $K$ specified in the problem, te... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,856 |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the code of the letter. Then, the weight of a word can be defined as the sum of the codes of all the letters in that ... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have
$k(C)+k(T)+k(O) \geq k(\amalg)+k(\mathrm{E})+k(C)+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T)$, which is equivalent to
$$
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 .
$$
The... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,857 |
# Problem 5.
In 10 minutes, Zhenya eats 5 cottage cheese vatrushki, while Sasha eats only 3. On Saturday, all the baked vatrushki went to Zhenya, and on Sunday, they went to Sasha. In total, exactly 70 vatrushki were eaten in three hours of pure time. How many cottage cheese vatrushki did each of the children get? | # Solution.
Let Zhene get $n$, and Sasha get $-k$ pancakes. We need to solve the system of equations in integers
$$
\left\{\begin{array}{l}
n+k=70 \\
2 n+\frac{10}{3} k=180
\end{array}\right.
$$
Substituting $n=70-k$ into the second equation, we get
$$
140-2 k+\frac{10}{3} k=180
$$
Or
$$
\frac{4}{3} k=40
$$
From... | Zhenegot40pancakes,Sashagot30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,858 |
# Problem 3.
On the coordinate plane, a square $K$ is marked with vertices at points $(0,0)$ and $(10,10)$. Inside this square, draw the set $M$ of points $(x, y)$,
the coordinates of which satisfy the equation
$$
[x]<[y]
$$
where $[a]$ denotes the integer part of the number $a$ (i.e., the greatest integer not excee... | Solution. Let $n \leq x < n+1$, where $n$ is an integer from 0 to 9. Then $[x]=n$ and $n < [y]$. The solution to the latter inequality is all $y \geq n+1$. Thus, the solution will be the union of strips
$$
\{n \leq x < n+1, n+1 \leq y, \quad n \in \mathbb{Z}\}
$$
Inside the square specified in the problem, part of ni... | \frac{45}{100} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,860 |
# 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 | 19,861 |
# Task 5.
A road paved with yellow bricks needs to be constructed. It will pass through an area where there is a straight section of a power line (PL) and a brick factory located at a distance $d$ from the PL $(d \neq 0)$. For rhythmic work, it is required that each point on the road being built is equally distant fro... | # Solution.
Let the power line run parallel to the coordinate axis $O X$. Then it will correspond to the line $y=d$. Note that the road being built will be located below this line, (in the half-plane $y \leq d$ ).
If $M(x, y)$ is the point on the road being built, then the distance to the factory $M O$ is $\sqrt{x^{2... | (-3,-4),(3,-4) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,862 |
Task 1. There are 3 types of installations totaling no less than 100. The number of installations of type 2 is 4 times the number of type 1, and the number of installations of type 3 is a multiple of the number of installations of type 1. If the number of installations of type 3 were 5 times more, then they would be 22... | Solution. If $x_{1}, x_{2}, x_{3}$ are the quantities of installations of types $1,2,3$, then the conditions are represented by the relations
$$
\begin{gathered}
x_{1}+x_{2}+x_{3} \geq 100 \\
x_{2}=4 x_{1} \\
x_{3}=k x_{1} \\
5 x_{3}=x_{2}+22 \\
k, x_{1}, x_{2}, x_{3} \in \mathbb{N}
\end{gathered}
$$
From (2)-(4) we ... | 22,88,22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,863 |
Task 2. A triangle was cut into two triangles. Find the greatest value of $N$ such that among the 6 angles of these two triangles, exactly $N$ are the same. | Solution. For $N=4$, an example is an isosceles right triangle divided into two isosceles right triangles: four angles of $45^{\circ}$. Suppose there are five equal angles. Then in one of the triangles, all three angles are equal, meaning all of them, and two angles of the other triangle are $60^{\circ}$. But then both... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,864 |
Problem 3. The set $M$ consists of $n$ numbers, $n$ is odd, $n>1$. It is such that when any of its elements is replaced by the sum of the other $n-1$ elements
from $M$, the sum of all $n$ elements does not change. Find the product of all $n$ elements of the set $M$. | # Solution. Let
$$
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 | 19,865 |
Problem 5. Mom put a vase with 15 tangerines on the table. One of the guests took two tangerines, and mom replaced them with one apple. Other guests also started taking two fruits each. Each time a guest took two of the same fruits (two tangerines or two apples), mom would put one apple back in the vase; if a guest too... | Solution. It can be noticed that the number of tangerines either remains the same or decreases by two. Since the number of tangerines was odd, one tangerine will definitely remain at the end.
Answer. A tangerine remained. | A\tangerine\remained | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,866 |
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
$$
\fr... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,867 |
2. A table of numbers with 20 rows and 15 columns, $A_{1}, \ldots, A_{20}$ are the sums of the numbers in the rows, $B_{1}, \ldots, B_{15}$ are the sums of the numbers in the columns.
a) Is it possible that $A_{1}=\cdots=A_{20}=B_{1}=\cdots=B_{15}$?
b) If the equalities in part a) are satisfied, what is the sum $A_{1... | Let $A_{i}=B_{j}=X$ for $i=1, \ldots 20$ and $j=1, \ldots, 15$. Consider the sum $S$ of all elements in the table. We have $S=20 X=15 X, X=0$ and $A_{1}+\cdots+A_{20}+B_{1}+\cdots+B_{15}=0$. An example of such a table is, for instance, a table consisting entirely of zeros. There is no need to consider other cases.
Ans... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,868 |
3. Is it possible to divide the numbers from 1 to 89 into groups such that each group contains at least four numbers, and one of the numbers in each group is equal to the sum of the other numbers in the same group? | Solution. Let $x_{4}=x_{1}+x_{2}+x_{3}$ in each group. Then the sum of all numbers in this group $x_{1}+x_{2}+x_{3}+x_{4}=2\left(x_{1}+x_{2}+x_{3}\right)$ is even. Thus, the sum of all numbers in the set, as the sum of even addends, is even.
On the other hand, it is equal to
$$
1+\cdots+89=\frac{1}{2}((1+89)+(2+88)+\... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,869 |
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 | 19,870 |
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
$$
\begin{aligned}
& {\left[\frac{... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,871 |
1. Find all roots of the equation
$$
a^{2} \cdot \frac{x-b}{a-b} \cdot \frac{x-c}{a-c}+b^{2} \cdot \frac{x-a}{b-a} \cdot \frac{x-c}{b-c}+c^{2} \cdot \frac{x-a}{c-a} \cdot \frac{x-b}{c-b}=x^{2}
$$
where $a \neq b \neq c$ are arbitrary given values. | # Solution
Note that the degree of the equation does not exceed two.
It is easy to verify that $a, b, c$ are roots of the equation:
$$
\begin{aligned}
& a^{2} \cdot \frac{a-b}{a-b} \cdot \frac{a-c}{a-c}+b^{2} \cdot \frac{a-a}{b-a} \cdot \frac{a-c}{b-c}+c^{2} \cdot \frac{a-a}{c-a} \cdot \frac{a-b}{c-b}=a^{2} \\
& a^{... | x\in\mathbb{R} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,872 |
2. At the food station, four track and field athletes drank all the cola-loc lemonade. If athlete Bystrov had drunk only half as much, a tenth of the lemonade would have remained. If, in addition, athlete Shustrov had also drunk half as much, an eighth of the lemonade would have remained. If, in addition to them both, ... | # Solution
Let $V$ be the amount of lemonade consumed in conditional units. If athlete Bystrov drank half as much, he would have consumed half of his share, which, according to the problem, is $\frac{1}{10} V$. Therefore, Bystrov drank $\frac{1}{5} V$.
If athletes Bystrov and Shustrov drank half as much, they would h... | \frac{1}{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,873 |
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
$$
k=7+... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,874 |
4. Through a point lying inside a triangle, three lines parallel to its sides are drawn, which divide the triangle into six parts: three triangles and three quadrilaterals. The areas of the three inner triangles are in the ratio $1: 4: 9$. Determine the range in which the ratio of the area of the largest of them to the... | # Solution
Draw $K F\|A C, L Q\| B C, P E \| A B$.
Triangles $K L O, O E F, P O Q$ are similar to each other and to triangle $A B C$ (their corresponding angles are equal as angles formed by... | \frac{1}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,875 |
5. Determine whether the number $\sqrt[2023]{3+\sqrt{8}}+\sqrt[2023]{3-\sqrt{8}}$ is greater or less than two.
# | # Solution
The given addends are reciprocals of each other, since
$$
\frac{1}{3+\sqrt{8}}=3-\sqrt{8}
$$
It is clear that they are not equal to each other.
Let $a=\sqrt[2023]{3+\sqrt{8}}$. Then the given expression is $a+\frac{1}{a}$, and the sum of two different positive reciprocals is strictly greater than two.
A... | \sqrt[2023]{3+\sqrt{8}}+\sqrt[2023]{3-\sqrt{8}}>2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,876 |
1. How many times is number $A$ greater or smaller than number $B$, if
$$
\begin{gathered}
A=\underbrace{1+\ldots+1}_{2022 \text { times }}+\underbrace{2+\ldots+2}_{2021 \text { times }}+\ldots+2021+2021+2022, \\
B=\underbrace{2023+\ldots+2023}_{2022 \text { times }}+\underbrace{2022+\ldots+2022}_{2021 \text { times }... | # Solution
The number $B$ will be calculated according to its representation as
$$
B=\sum_{m=1}^{2022} Q_{m}, \quad \text { where } \quad Q_{m}=(m+1) m
$$
The number $A$ can be rewritten as the sum of arithmetic progressions
$$
A=1+(1+2)+\ldots+(1+2+\ldots+2021)+(1+2+\ldots+2022)
$$
Then
$$
A=\sum_{m=1}^{2022} S_... | \frac{1}{2}B | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,877 |
2. Two circles touch each other externally and each touches a larger circle internally. The radius of one is half, and that of the other is one-third of the radius of the largest circle. Find the ratio of the length of the segment of the common internal tangent to the smaller circles, contained within the largest, to i... | # Solution
Let the radii of the smaller circles be $2r$ and $3r$. Then the radius of the largest (outer) circle is $6r$.
Consider $\triangle O_{1} O O_{2}$. Its sides are $3r, 4r$, and $5r$,... | \frac{2\sqrt{6}}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,878 |
3. Given a rectangular parallelepiped. The perimeters of each of its three mutually perpendicular faces are equal to the sides of a new rectangular parallelepiped. What can be the minimum ratio of the volume of the new parallelepiped to the volume of the original?
# | # Solution
Let $x, y, z$ be the sides of the original parallelepiped. Then the volume of the new one is
$$
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 | 19,879 |
4. Can the equation
$$
x^{2022}-2 x^{2021}-3 x^{2020}-\ldots-2022 x-2023=0
$$
have two positive roots? | # Solution 1
We will prove that this is impossible.
From the original equation, we move to an equation where the coefficients of the polynomial form an arbitrary permutation $(a_{2}, a_{3}, \ldots, a_{2023})$ of the numbers $\{2,3, \ldots, 2023\}:$
$$
x^{2022}-a_{2} x^{2021}-a_{3} x^{2020}-\ldots-a_{2022} x-a_{2023}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,880 |
5. Find the maximum value of the quantity $x^{2}+y^{2}$, given that
$$
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 | 19,881 |
# Task 1.
Hunter Pulkа received two bags of dog food for his dog Bulka from "AliExpress." In the morning, one bag was found empty. Nosy Nезнайка took up the investigation and identified three suspects, who made the following statements.
Siropchik said that he did not eat the dog food.
Toropyzhka claimed that the fo... | # Solution.
Let's consider the three cases one by one, in each of which one of the statements is false.
Case 1. Syropchik is lying.
In this case, Ponchik's statement would also be false, which contradicts the condition.
Case 2. Tropozhka is lying.
This means that neither Ponchik nor Syropchik ate the food. Then th... | Tropozhkaatetheentirebagofdogfood | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,882 |
# Problem 3.
A square sheet of paper was folded in half, then folded in half again, and for the third time in half. As a result, a triangle was obtained (see Fig. 1, where all fold lines are indicated by dashed lines). After that, a cut was made along the line $P Q$ and the sheet was unfolded. Will the shape of the re... | # Solution.
It is sufficient to draw two ways of unfolding (folding) and to make sure that there will be no differences. The result of the unfolding is presented on the right (Fig. 2).
Fig. ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,884 |
# 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
$$
Th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,885 |
# Problem 5.
In 10 minutes, Zhenya eats 5 cottage cheese vatrushki, while Sasha eats only 3. On Saturday, all the baked vatrushki went to Zhenya, and on Sunday, they went to Sasha. In total, 35 vatrushki were eaten in one and a half hours of pure time. How many cottage cheese vatrushki did each of the children get? | # Solution.
The problem can be solved by formulating a system of equations with two unknowns (similar to a 7th-grade problem). However, it is unlikely that 6th-grade students would take this approach.
An alternative method of reasoning might be as follows.
One and a half hours is 9 times 10 minutes. Suppose the chil... | Zhenyagot20pancakes,Sashagot15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,886 |
# Problem 2.
Solve the system of equations
$$
\begin{cases}2\left[x_{1}\right]+x_{2} & =3 / 2 \\ 3\left[x_{1}\right]-2 x_{2} & =4\end{cases}
$$
Here $[a]$ denotes the integer part of the number $a$.
# | # Solution.
Solving the system with respect to the unknowns $\left[x_{1}\right]$ and $x_{2}$, we find
$$
\left[x_{1}\right]=\frac{4+3}{7}=1, \quad x_{2}=\frac{9-16}{14}=-\frac{1}{2}
$$
Therefore, $x_{1}$ can be any number whose integer part is 1.
Answer: $x_{1} \in[1 ; 2), x_{2}=-1 / 2$. | x_{1}\in[1;2),x_{2}=-\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,887 |
# Problem 3.
Two circles intersect at points $P$ and $Q$. A line through $Q$, perpendicular to $P Q$, intersects the circles again at points $A$ and $B$ (with point $Q$ lying between $A$ and $B$), and the tangents to the circles at these points intersect at point $C$. Prove that the segments $A Q$ and $C B$ are seen f... | Solution. Case 1. Points $A$ and $B$ are on opposite sides of $Q$.
By the theorem on the angle between a tangent and a chord, $\angle B A C = \angle Q P A$.
Since the segment $PC$ subtends two right angles $\angle P A C$ and $\angle P B C$, the quadrilateral $A P B C$ is cyclic (Fig. 1).
Therefore, $\angle B P C = \... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,888 |
# 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 | 19,889 |
# Problem 5.
A road needs to be built, paved with concrete slabs. It will pass through an area where there is a straight section of a power line (PL) and a slab manufacturing plant located at a distance $d$ from the PL $(d \neq 0)$. For rhythmic operation, it is required that each point on the road being built is equa... | # Solution.
Let the power line run parallel to the coordinate axis $O X$. Then it will correspond to the line $y=d$. Note that the road being built will be located below this line, (in the half-plane $y \leq d$ ).
A) If $M(x, y)$ is the point on the road being built, then the distance to
the factory $M O$ is $\sqrt{... | A)(-3d,-4d),(3d,-4d)\\B)foranynaturaln | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,890 |
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 | 19,891 |
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
$$ | # Solution
## 1st Method
Introduce a Cartesian coordinate system and consider an arbitrary vector $\mathbf{a}$ with coordinates $(x, y, z)$ and a fixed vector $\mathbf{c}$ with coordinates $(3, 8, 1)$. Then, the left side of the condition represents the square of the length of vector $\mathbf{a}$, and the right side ... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,892 |
# 3. In the equation
$$
x^{2022}-2 x^{2021}-3 x^{2020}-\ldots-2022 x-2023=0
$$
the coefficients of all powers of $x$, except the highest, can be permuted in any way. Can such a permutation ensure that the equation has at least two positive roots? | # Solution 1
We will prove that this is impossible.
From the original equation, we move to an equation where the coefficients of the polynomial form an arbitrary permutation $(a_{2}, a_{3}, \ldots, a_{2023})$ of the numbers $\{2,3, \ldots, 2023\}:$
$$
x^{2022}-a_{2} x^{2021}-a_{3} x^{2020}-\ldots-a_{2022} x-a_{2023}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,893 |
4. Two spheres touch each other externally and each touches a larger sphere internally. The radius of one is half, and that of the other is one-third of the radius of the largest sphere. A tangent plane is constructed at the point of contact between the smaller spheres. Find the distance from this plane to the center o... | # Solution
Consider the section of the described composition by a plane passing through the centers of the three spheres. The required distance will be the length of the segment $O D$ on this plane.
^{2022}=\frac{2023}{2022^{2}} \cdot\left(1+\frac{1}{2022}\right)^{2022}
$$
We will use the fact that
$$
21
$$
Then
$$
\frac{2023}{2022^{2}} \cdot\left(1+\frac{1}{2022}\right)^{2022}... | 2023^{2023}<2022^{2024} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,895 |
# Task 1.
At the Faculty of Nuclear Airship Construction, it was calculated that the percentage of boys on the first year is greater than the percentage of boys on the entire faculty. Who is greater in percentage - first-year students among all boys in the faculty or all first-year students among all students in the f... | # Solution.
Let's introduce the following notations: $m$ - the number of boys on the 1st year, $M$ - the total number of boys on the faculty, $k$ - the number of students on the 1st year, $K$ - the total number of students on the faculty.
According to the condition,
$$
\frac{m}{k}>\frac{M}{K} .
$$
We need to compar... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,896 |
# Problem 2.
Solve the equation
$$
x^{2}-[x]=2019,
$$
where $[x]$ denotes the integer part of $x$.
# | # Solution.
If we represent an arbitrary number $x$ as $x=m+\alpha$, where $m$ is an integer, $\alpha \in(0,1)$, then $[x]=m=x-\alpha$. Thus, the original equation can be rewritten as
$$
x^{2}-x+\alpha=2019 \text {. }
$$
Since $\alpha \in(0,1)$, we have
$$
2018 \leq x^{2}-x \leq 2019
$$
Consider the function $f(x)... | x_{1}=-\sqrt{1974},x_{2}=\sqrt{2064} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,897 |
# 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 | 19,899 |
# Problem 5.
A circle of unit radius is divided into $2^{2019}$ equal parts. Prove that the distance from the center of the circle to the chord spanning one such part is exactly half of the value of
$$
\sqrt{\underbrace{2+\sqrt{2+\ldots+\sqrt{2}}}_{2018 \text { twos }}}
$$ | # Solution.
Let's depict (not to scale) $\angle A O B$, which is one $2^{2019}-$th part of the circle. Denote this angle as $\alpha$. Clearly,
$$
\alpha=\frac{2 \pi}{2^{2019}}=\frac{\pi}{2^{2018}}
$$
The problem gives a formula for $O H$. From the right triangle $O H B$ with hypotenuse $O B=1$ and acute angle $B O H... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,900 |
# Task 1.
Hunter Pulkа received two bags of dog food for his dog Bulka from "AliExpress." In the morning, one bag was found empty. Nosy Nезнайка took up the investigation and identified three suspects, who made the following statements.
Siropchik said that he did not eat the dog food.
Toropyzhka claimed that the fo... | # Solution.
Let's consider the three cases one by one, in each of which one of the statements is false.
Case 1. Syropchik is lying.
In this case, Ponchik's statement would also be false, which contradicts the condition.
Case 2. Tropozhka is lying.
This means that neither Ponchik nor Syropchik ate the food. Then th... | Tropozhka | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,903 |
# 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 | 19,904 |
# 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 | 19,905 |
# 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 | 19,906 |
# 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 | 19,907 |
# Task 2.
At a confectionery factory, they decided to develop a new type of candy. For technological reasons, the candy should have the shape of a cylinder with a volume $V$ and a total surface area $S$. Under what conditions on $V$ and $S$ will any two such cylinders be equal?
# | # Solution.
The solution to this problem will be divided into two parts: finding the extreme relationship between $\mathrm{S}$ and $\mathrm{V}$ and proving that in other cases, the cylinder may not be unique. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,908 |
# 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 | 19,910 |
# Problem 4.
On each face of a cube, a regular 4-sided pyramid is placed, with the base of the pyramid being this face of the cube. All pyramids are equal.
- 4A. Can the lateral edges of three pyramids, emanating from one vertex of the cube, lie in the same plane? If this is possible, find the heights of such pyramid... | # Solution.
Let's denote an arbitrary vertex of the cube as $A$, and the vertices of the pyramids connected to it by edges as $O_{1}, O_{2}, O_{3}$.
We introduce a coordinate system, placing its origin at point $A$ and directing the axes $A X, A Y$, and $A Z$ along the edges of the cube. Let the base of the pyramid w... | a | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,911 |
# Problem 5.
Solve the equation with three unknowns
$$
X^{Y}+Y^{Z}=X Y Z
$$
in natural numbers.
# | # Solution.
1) When $Y=1$, we get the equation $X+1=X Z$, hence $X(Z-1)=1$, i.e., $X=1, Z=2$.
2) When $Y=2$, the equation becomes
$$
(X-Z)^{2}+2^{Z}=Z^{2}
$$
For $Z=1$, it has no solutions. Substituting $Z=2,3,4$, we get the solutions $(2 ; 2 ; 2),(2 ; 2 ; 3)$ and $(4 ; 2 ; 3),(4 ; 2 ; 4)$, respectively. For $Z>4$, ... | (1;1;2),(2;2;2),(2;2;3),(4;2;3),(4;2;4) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,912 |
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 | 19,913 |
2. At the gathering of Madagascar legged-hand lovers, 23 people arrived, and some of them became friends with each other. Prove that there will be two participants at the gathering who have become friends with the same number of colleagues. | Solution. We can proceed by contradiction and assume that everyone has befriended a different number of like-minded individuals. Let's consider the options: a participant may not be friends with anyone, may befriend one, two, and so on, and the maximum number of friends they can have is 22. We get 23 possible options, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,914 |
5. On Monday, Donut started eating pastries with energy jam. Each day he ate the same number of them and one day he found that only a quarter of his daily portion was left from the initial stock of 340 pastries. On what day of the week did this happen? | # Solution
Let $n$ be the number of days with the same portion, and $s$ be the remainder (a quarter of the portion). Then
$$
2 \cdot 2 \cdot 5 \cdot 17=340=4 s \cdot n+s=s \cdot(4 n+1)
$$
Since the second factor is odd, we get three possible cases.
Either $s=4,4 n+1=85(n=21)$. If 21 days were eaten with a full port... | onMondayoronFriday | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,917 |
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 | 19,920 |
2. Find the solution to the system
$$
\left\{\begin{array}{l}
5 x^{7}+3 y^{2}+5 u+4 v^{4}=-2 \\
2 x^{7}+8 y^{2}+7 u+4 v^{4}=\frac{6^{5}}{3^{4} \cdot 4^{2}} \\
8 x^{7}+2 y^{2}+3 u+6 v^{4}=-6 \\
5 x^{7}+7 y^{2}+7 u+8 v^{4}=\frac{8^{3}}{2^{6} \cdot 4}
\end{array}\right.
$$ | # Solution.
Consider a system with the same coefficients but without the powers.
Add the first equation to the fourth, and the second to the third.
$$
\left\{\begin{array}{l}
10(x+y)+12(u+v)=0 \\
10(x+y)+10(u+v)=0
\end{array}\right.
$$
From which
$$
y=-x, \quad v=-u
$$
Substituting into the previous system, we ha... | {-1,\1,0,0} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,921 |
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 2, and the product of the numbers in any $3 \times 3$ square is 1. What number is in the center of the table? | # Solution
Consider the first 3 rows of the table. From the additional condition, it follows that if these rows are covered "butt-to-butt" with squares of size $3 \times 3$, moving simultaneously from left and right towards each other, then the 336-th square from the left and the 336-th square from the right will over... | 2^{-2017} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,922 |
4. The numbers $\sin \alpha, \cos \alpha, \operatorname{tg} \alpha, \sin 2 \alpha, \cos 2 \alpha$ are written in a row. The arithmetic means of any three consecutive numbers are equal. Find all values of $\alpha$ for which this is possible. | Solution. Let the 5 numbers be denoted by $x_{1}, x_{2}, \ldots, x_{5}$. According to the condition,
$$
\begin{aligned}
& x_{1}+x_{2}+x_{3}=x_{2}+x_{3}+x_{4} \quad \Rightarrow \quad x_{1}=x_{4}=a \\
& x_{2}+x_{3}+x_{4}=x_{3}+x_{4}+x_{5} \Rightarrow x_{2}=x_{5}=b
\end{aligned}
$$
Consider the pair of obtained equation... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,923 | |
5. Solve the equation $2^{[\sin x]}=3^{1-\cos x}$, where $[a]$ denotes the integer part of the number a. | Solution. The integer $[\sin x]$ can only take the values $0, \pm 1$.
1) If $[\sin x]=0$, then $2^{0}=1$. Therefore, $3^{1-\cos x}=1$, from which $\cos x=1$. The solution to this equation is $x=2 \pi n$. Note that $\sin x=0$ for such $x$.
2) If $[\sin x]=-1$, then $3^{1-\cos x}=2^{-1}=1 / 2$. But $1-\cos x \geqslant 0... | 2\pin,n\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,925 |
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 | 19,926 |
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:
$$
\frac{\text { Morning Exercise }}{\text { Walk }}=\frac{3}{2} \quad \frac{\text { Walk }}{\text { Jog }}=\frac{5}{3} \quad \frac{\text { Jog }}{\text { S... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,927 |
3. Represent the number $\frac{2}{7}$ as the sum of several different common fractions, the numerators of which are equal to one.
# | # Solution
We use the fact that
$$
\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}
$$
Then
$$
\frac{2}{7}=\frac{1}{7}+\frac{1}{7}=\frac{1}{7}+\frac{1}{8}+\frac{1}{56}
$$
Other variations are also possible.
Any other correct representation was also accepted as an answer. | \frac{1}{7}+\frac{1}{8}+\frac{1}{56} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,928 |
4. Two equal squares are superimposed on each other such that a vertex of the upper square coincides with the point $O$ where the diagonals of the lower square intersect. In this configuration, the upper square can freely rotate around point $O$. How should the upper square be positioned to cover the largest possible a... | # Solution
Let's illustrate an arbitrary position of the two squares.
Drop perpendiculars from the center of the stationary square to two sides (in the figure - the right and bottom sides). Then, the two shaded right triangles are equal by a leg and an acute angle.
Therefore, regardless of the angle of rotation of t... | anyposition | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,929 |
# Problem 2.
Can the number $n^{2}+n+17$ be divisible by 2019 for any natural $n$? Either find the smallest such $n$, or prove that it is impossible.
# | # Solution.
Let's factorize the divisor: $2019=3 \cdot 673$.
The number $n^{2}$ gives remainders of 0 and 1 when divided by 3 (which can be easily verified by constructing a table of remainders)
| n | $n$ mod | 3 $n^{2}$ | $\bmod 3$ |
| :---: | :---: | :---: | :---: |
| 0 | 0 | | 0 |
| 1 | 1 | | 1 |
| 2 | 2 | | 1... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,931 |
# 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 | 19,932 |
# Task 4.
In Donut's pantry and in Syrup's pantry, there is a total of 100 kg of jam. Each dwarf took the same amount of time to eat their own supplies, despite having different appetites. "If my supply were equal to yours, I would have eaten it in 45 days" - Donut told his friend. "And if my supply were equal to your... | # Solution.
Let Donut have $x$ kg of jam, and Syrup have $n$ times more, i.e., $n x$ kg. Then Donut's consumption rate is $\frac{n x}{45}$, and Syrup's consumption rate is $\frac{x}{20}$. The equality of the time it takes to consume their reserves gives the equation
$$
\frac{x}{n x / 45}=\frac{n x}{x / 20},
$$
from ... | Donut:40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,933 |
# Problem 5.
Is the inequality correct
$$
\sqrt{\underbrace{2019+\sqrt{2019+\ldots+\sqrt{2019}}}_{2019 \text { terms }}}<2019 ?
$$ | # Solution.
Consider the quantity
$$
x_{n}=\sqrt{\underbrace{2019+\sqrt{2019+\ldots+\sqrt{2019}}}_{n \text { terms }}}
$$
It is clear that $x_{n}>0$. Squaring the inequality $x_{n}<2019$, we get
$$
x_{n}^{2}=2019+x_{n-1}<2019^{2}
$$
or equivalently,
$$
x_{n-1}<2019^{2}-2019 \text { . }
$$
It is easy to verify th... | proof | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 19,934 |
Problem 1. In the country of "Energetika," there are 150 factories, and some of them are connected by bus routes that do not stop anywhere except at these factories. It turned out that any four factories can be divided into two pairs such that there is a bus route between the factories of each pair. Find the smallest n... | Solution. Suppose that some factory $X$ is connected by bus routes to no more than 146 factories. Then a quartet of factories, consisting of $X$ and any three with which it is not connected, does not satisfy the problem's condition, since $X$ cannot be paired with any of the three remaining factories. Therefore, each f... | 11025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,935 |
Problem 2. For the numerical sequence $x_{0}, x_{1}, \ldots, x_{n}, x_{n+1}, \ldots$ the relations $2 x_{n}=x_{0}+x_{1}+\cdots+x_{n-1}-x_{n}$ hold for all $n=0,1,2, \ldots$ Find each term $x_{n}$ of such a sequence and the values of the sums $S_{n}=$ $x_{0}+x_{1}+\cdots+x_{n}$. | Solution. We have
$$
x_{n}=(1 / 3)\left(x_{0}+x_{1}+\cdots+x_{n-1}\right) \text {. }
$$
Then $x_{0}-$ any, and by induction
$$
\begin{gathered}
x_{n}=x_{0} 4^{n-1} / 3^{n} \text { for } n \geq 1 \\
S_{n}=x_{0}(4 / 3)^{n}
\end{gathered}
$$
Answer: $x_{n}=x_{0} 4^{n-1} / 3^{n}$ for $n \geq 1, \quad S_{n}=x_{0}(4 / 3)... | x_{n}=x_{0}\frac{4^{n-1}}{3^{n}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,936 |
Problem 3. Six numbers are written in a row. It is known that among them there is a one and any three adjacent numbers have the same arithmetic mean. Find the maximum value of the geometric mean of any three adjacent numbers in this row, if the arithmetic mean of all 6 numbers is A. | Solution. Let's write in a row $a_{1}, \ldots, a_{6}$. From the conditions
\[
\begin{aligned}
& (1 / 3)\left(a_{1}+a_{2}+a_{3}\right)=(1 / 3)\left(a_{2}+a_{3}+a_{4}\right) \\
& (1 / 3)\left(a_{2}+a_{3}+a_{4}\right)=(1 / 3)\left(a_{3}+a_{4}+a_{5}\right) \\
& (1 / 3)\left(a_{3}+a_{4}+a_{5}\right)=(1 / 3)\left(a_{4}+a_{5... | \sqrt[3]{(3A-1)^{2}/4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,937 |
Problem 4. Given a quadratic trinomial $g(x)$, which has exactly one root. Find this root, if it is known that the polynomial $g(a x+b)+g(c x+d)$ $(a \neq c)$ also has exactly one root. | Solution. Let $x_{0}$ be the unique root of the polynomial $g(x)$. Then the function $g(x)$ maintains its sign at all points $x \neq x_{0}$ and only $g\left(x_{0}\right)=0$. Therefore, the root of the polynomial $f(x)=g(a x+b)+g(c x+d)$ is only such a point $x_{1}$ that $a x_{1}+b=c x_{1}+d=x_{0}$. We obtain the parame... | x_{0}=\frac{-}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,938 |
Problem 5. During the improvement of the city garden "Pythagoras," three alleys were initially laid out, forming a right-angled triangle with an acute angle $\alpha$. The next alleys were laid out as external squares on the sides of this triangle (resulting in a figure illustrating the Pythagorean theorem and called Py... | Solution. Let the original triangle be $ABC$ (angle $C$ is right), the centers of the squares be $O_{1}, O_{2}, O_{3}$, and one of the vertices of a square be $D$ (see figure). Let angle $A$ be $\alpha$.
Let the sides of the original triangle, opposite to vertices $A, B$, and $C$, have lengths $a, b, c$ respectively.
... | O_{1}O_{2}=CO_{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,939 |
Problem 4. Given a quadratic trinomial $g(x)=x^{2}+a x+b$, which has exactly one root. Find the coefficients $a$ and $b$, if it is known that the polynomial $g\left(x^{5}+2 x-1\right)+g\left(x^{5}+3 x+1\right)$ also has exactly one root. | Solution. Let $x_{0}$ be the unique root of the polynomial $g(x)$. Then the function $g(x)$ maintains its sign at all points $x \neq x_{0}$ and only $g\left(x_{0}\right)=0$. Therefore, the root of the polynomial $f(x)=g\left(x^{5}+2 x-1\right)+g\left(x^{5}+3 x+1\right)$ is only such a point $x_{1}$ that
$$
x_{1}^{5}+2... | =74,b=1369 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,940 |
Problem 5. There are 4 numbers, not all of which are the same. If you take any two of them, the ratio of the sum of these two numbers to the sum of the other two numbers will be equal to the same value k. Find the value of k. Provide at least one set of four numbers that satisfy the condition. Describe all possible set... | Solution. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be such numbers. Write the relations for the sums of pairs of numbers:
$$
\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 | 19,941 |
# Problem 1.
There are three electric generators with powers $x_{1}, x_{2}, x_{3}$ less than 1 MW. When analyzing the power system with such generators, it was found that for a certain process to be carried out, the condition
$$
2\left(x_{1}+x_{2}+x_{3}\right)+4 x_{1} x_{2} x_{3}=3\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} ... | # Solution.
Let's rewrite the equality from the condition as follows:
$$
\begin{gathered}
4 x_{1} x_{2} x_{3}-4\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)+4\left(x_{1} x_{2} x_{3}\right)-4= \\
\quad=-\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)+2\left(x_{1}+x_{2}+x_{3}\right)-3 \\
4\left(x_{1}-1\right)\left(... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,942 |
# Task 2.
In the country of Lemonia, there are only two denominations of money, 7 lemons and 9 lemons. Find all the ways to represent a sum of 997 lemons using these denominations and indicate their quantity.
# | # Solution.
The problem reduces to the equation
$$
7 x+9 y=997
$$
in non-negative integers $x, y$. Let's represent it as $7 x+7 y+2 y=$ $980+14+3$, from which $2 y-3=994-7 x-7 y$. Therefore, $2 y-3$ is divisible by $7,2 y=$ $7 n+3, n \in \mathbb{Z}$. Multiply the equality $2 y=7 n+3$ by 4 and transform the result: $... | 136-9k,7k+5\quadk=0,1,\ldots,15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,943 |
# Problem 3.
In the Kingdom of Enchanted Energy, an enchanted transformer booth stands on a flat plain: an observer looking parallel to the ground can see it only at an angle of $45^{\circ}$. In cross-section, the booth is square with a side length of $L$ cubits. Describe the geometric locus of points on the plain fro... | # Solution.
As is known, any inscribed angle in a circle is half the measure of the central angle that subtends the same arc.
Therefore, an arbitrary segment $A B$ will be seen at an angle of $45^{\circ}$ if the vertex of this angle $Q_{i}$ is located on the circle with the center at the vertex of a right isosceles t... | \rho_{\}=L,\quad\rho_{\max}=\frac{1+\sqrt{2}}{2}L | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,944 |
# 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 | 19,945 |
# 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 | 19,946 |
4. The Winter Palace (fourth) in St. Petersburg was erected by Francesco Bartolomeo Rastrelli from 1754 to 1762 for Empress Elizabeth Petrovna. The number of windows is 445 more than the number of rooms, the number of doors is 286 more than the number of rooms, and the total number of rooms and doors is 3731. Find the ... | 4. Taking into account (2), we get $N=1$ or $N=9$. If $N=9$, then $20 M^{2}=$ $=2017-17 \cdot 81=640, M^{2}=32$, which is impossible. Therefore, definitively $N=1$ and $M=10$. We have exactly 4 solutions to the equation (1$):(M ; N)=( \pm 10 ; \pm 1)$, from which, returning to the variables $X, Y$, we get
ANSWER. $(X ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,950 |
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 | 19,951 |
Problem 2. On each side of an equilateral triangle, a point is taken. Each side of the triangle with vertices at these points is perpendicular to one of the sides of the original triangle. In what ratio does each of the taken points divide the side of the original triangle? What is the ratio of the areas of the origina... | Solution. Let points $K, L, M$ lie on the sides $A B, B C$, and $A C$ of an equilateral triangle $A B C$, respectively, such that $K L \perp B C, L M \perp A C, M K \perp A B$.
1) Then $\angl... | 1:2,3:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,952 |
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:
$$
\begin{aligned}
& \frac{x_{1}+x_{2}}{x_{3}+x_{4}}=\frac{x_{3}+x_{4}}{x_{1}+x_{2}}=k \\
& \frac{x_{1}+x_{3}}{x_{2}+x_{4}}=\frac{x_{2}+x_{4}}{x_{1}+x_{3}}=k \\
& \frac{x_{1}+x_{4}}{x_{2}+x_{3}}=\frac{x_{2... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,953 |
# Task 1.
A road needs to be built, paved with concrete slabs. It will pass through an area where there is a straight section of a power line (PL) and a plant producing slabs, located at a distance $d$ from the PL $(d \neq 0)$. For rhythmic operation, it is required that each point of the road under construction be e... | # Solution.
Let's choose a rectangular coordinate system $X O Y$, placing the concrete slab factory at the origin $O$ of this coordinate system and directing the axis $O X$ along the power line. Let's assume for definiteness that the road passes through the half-plane $\{y \geq 0\}$. Then the power line will correspon... | \frac{}{2}-\frac{1}{2}\cdotx^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,954 |
# Problem 2.
Find all values of the real parameter $p$ for which the system of equations is solvable
$$
\left\{\begin{aligned}
2[x]+y & =3 / 2 \\
3[x]-2 y & =p
\end{aligned}\right.
$$
Here $[a]$ denotes the integer part of the number $a$. | # Solution.
Solving the system with respect to the unknowns $[x]$ and $y$, we find
$$
[x]=\frac{p+3}{7}, \quad y=\frac{9}{14}-\frac{2 p}{7}
$$
It remains to determine when the value $\frac{p+3}{7}$ is an integer.
Answer. $p=7 k-3, k \in \mathbb{Z}$. | 7k-3,k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,955 |
Problem 3.
Two circles intersect at points $P$ and $Q$. A line through $Q$, perpendicular to $P Q$, intersects the circles again at points $A$ and $B$, and the tangents to the circles at these points intersect at point $C$. Prove that the segments $A Q$ and $C B$ are seen from point $P$ at the same angle. | Solution. Case 1. Points $A$ and $B$ are on opposite sides of $Q$.
By the theorem on the angle between a tangent and a chord, $\angle B A C = \angle Q P A$.
Since the segment $\mathrm{PC}$ subtends two right angles $\angle P A C$ and $\angle P B C$, the quadrilateral $A P B C$ is cyclic (Fig. 1).
Therefore, $\angle ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,956 |
# Problem 4.
When designing a certain technical device, there arose a need to solve equations
$$
a \circ x=b,
$$
where the operation $\circ$ over two numbers is defined by the condition
$$
y \circ z=\frac{y+z+|y-z|}{2} \text {. }
$$
Find all numerical sets $X$ such that for any $a, b$ from $X$ the specified equati... | # Solution.
Notice that
$$
y \circ z=\max \{y, z\}
$$
Next, any one-element set fits
$$
X=\{a\}, \quad a \in(-\infty ; \infty)
$$
since $\max \{a, a\}=a$.
Suppose that the set $X$ contains at least two distinct elements $a$ and $b$. Without loss of generality, assume $a>b$. For such $a$ and $b$, equation (1) has ... | {},\in(-\infty;\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,957 |
# Task 1.
The Triassic Discoglossus tadpoles have five legs, while the saber-toothed frog tadpoles grow several tails (all have the same number of tails). An employee of the Jurassic Park scooped up several tadpoles with water. It turned out that the captured tadpoles had a total of 100 legs and 64 tails. How many tai... | # Solution.
Let $x$ be the number of tails of the saber-toothed frog's pollywog. Suppose $n$ five-legged and $k$ many-tailed pollywogs were caught. Counting the total number of legs and tails gives the equations
$$
\left\{\begin{aligned}
5 n+4 k & =100 \\
n+x k & =64
\end{aligned}\right.
$$
From the first equation, ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,958 |
# Problem 2.
Can the number $n^{2}+n+8$ be divisible by 2019 for any natural $n$? Either find the smallest such $n$, or prove that it is impossible.
# | # Solution.
Let's factorize the divisor: $2019=3 \cdot 673$.
The number $n^{2}$ gives remainders of 0 and 1 when divided by 3 (which is easy to check, for example, by constructing a table of remainders)
| n | $n \bmod 3$ | $\overline{n^{2}}$ | $\bmod 3$ |
| :---: | :---: | :---: | :---: |
| 0 | 0 | | 0 |
| 1 | 1 | ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,959 |
# Task 4.
In Donut's pantry and in Syrup's pantry, a total of 100 kg of jam is stored. Each dwarf took the same amount of time to eat their own supply, despite having different appetites. "If my supply were equal to yours, I would have eaten it in 45 days" - Donut told his friend. "And if my supply were equal to yours... | # Solution.
Let Donut have $x$ kg of jam, and Syrup have $n$ times more, i.e., $n x$ kg. Then Donut's consumption rate is $\frac{n x}{45}$, and Syrup's consumption rate is $\frac{x}{20}$. The equality of the time it takes to consume their reserves gives the equation
$$
\frac{x}{n x / 45}=\frac{n x}{x / 20}
$$
from w... | Donut:40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,960 |
# Problem 5.
Does the equation
$$
\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\cdots+\frac{1}{x^{2}-1}+\frac{1}{x^{2}}=1
$$
have a solution in natural numbers greater than one? | # Solution.
Let the integer $x=n \geq 2$ be a solution, and denote the expression on the left side of the equation as $S(n)$. In this case, there are exactly $n^{2}-n+1$ terms on the left side.
Each term except the first one will be replaced by a smaller positive number $1 / n^{2}$. Then
$$
S(n)>\frac{1}{n}+\left(n^... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,961 |
# 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.
$$
\begin{aligned}
& X \nabla 0=0, \\
& X \nabla 1=X \nabla(0+1)=(X \nabla 0)+2=X+2, \\
& X \nabla 2=X \nabla(1+1)=(X \nabla 1)+2=X+4, \\
& X \nabla 3=X \nabla(2+1)=(X \nabla 2)+2=X... | -673 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,962 |
# Problem 2.
What is greater: the sum of all odd numbers from 1 to 2019 (inclusive) or the sum of all even numbers from 2 to 2018 (inclusive)? | # Solution.
Let's write the specified sums one under the other
$$
\begin{aligned}
& N=1+3+5+\ldots+2017+2019 \\
& K=2+4+6+\ldots+2018
\end{aligned}
$$
The number of addends in the second sum is 2018/2 = 1009.
Olympiad for Schoolchildren "Hope of Energy". Final Stage. Solutions
It is clear that each addend in the s... | N>K | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,963 |
# Task 3.
Screwt and Pinch were designing a nanorover. Screwt drew a rectangle and marked twenty holes for wheels in it. Pinch divided the rectangle into sections by drawing two lines parallel to one side of the rectangle and two more parallel to the other. In doing so, none of Screwt's holes fell on Pinch's lines. Pr... | # Solution.
It is clear that 4 lines, each parallel to the sides of the rectangle, divide it into 9 parts.
If each part contains no more than two holes, then there will be no more than 18 holes in total, which contradicts the condition.
# | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,964 |
# Task 4.
In two departments of the "Phantasmagoria" laboratory, mobile applications for Android and iOS are being developed. On one of the working days, all employees of these departments exchanged a certain number of messages. Each developer from the Android department sent 7 and received 15 messages, while each dev... | # Solution.
Let $n$ be the number of people working in the Android department, and $m$ be the number of people working in the iOS department. We will calculate the total number of messages in two ways.
The number of messages sent is $7 n + 15 m$.
The number of messages received is $15 n + 9 m$.
Then,
$$
7 n + 15 m... | >n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,965 |
# 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 | 19,966 |
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