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|---|---|---|---|---|---|---|---|---|---|
# Task 4. (12 points)
Solve the equation $x^{2}+y^{2}+1=x y+x+y$.
# | # Solution
$$
\begin{gathered}
x^{2}+y^{2}+1=x y+x+y \Leftrightarrow(x-y)^{2}+(x-1)^{2}+(y-1)^{2}=0 \Leftrightarrow \\
x=y=1 .
\end{gathered}
$$
Answer: $x=y=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,633 |
# Task 5. (12 points)
It is known that $5 a+3 b+2 c=0$ and $a \neq 0$. Prove that the equation $a x^{2}+b x+c=0$ has two distinct roots.
# | # Solution.
The given equality can be rewritten as:
$$
5 a+3 b+2 c=0 \Leftrightarrow a+b+c+4 a+2 b+c=0
$$
We can write this equality as $f(1)+f(2)=0$, where $f(x)=a x^{2}+b x+c$.
Thus, either $f(1)=f(2)=0$ and the numbers 1 and 2 are roots of the given equation, or these values have different signs, which also mean... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,634 |
# Task 6. (14 points)
Determine the sign of the number
$$
A=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\ldots+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2016}
$$
The signs are arranged as follows: “+" before the first fraction, then two “-” and two “... | # Solution
Let's divide all numbers into groups of four numbers:
$$
\left(\frac{1}{4 k+1}-\frac{1}{4 k+2}-\frac{1}{4 k+3}+\frac{1}{4 k+4}\right)
$$
The sum of the numbers in each group is positive:
$$
\begin{gathered}
\left(\frac{1}{4 k+1}-\frac{1}{4 k+2}-\frac{1}{4 k+3}+\frac{1}{4 k+4}\right)>0 \Leftrightarrow \\
... | A>0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,635 |
# Task 7. (14 points)
It was discovered that at the close of trading, the stock price of a certain company increased or decreased by exactly $n \%$ each time compared to the previous close over the course of a year. Does there exist a natural value of $n$ such that the stock price at the close of trading during the ye... | # Solution.
After $k$ price increases and $l$ price decreases by $n \%$, the price will be:
$$
P\left(1+\frac{n}{100}\right)^{k}\left(1-\frac{n}{100}\right)^{l}
$$
where $P$ is the initial stock price. If we assume that the price takes the same value twice, we get
$$
\left(1+\frac{n}{100}\right)^{k}\left(1-\frac{n}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,636 |
# Task 8. (16 points)
Let all companies in the country have a certain rank, which is a natural number. When two companies of ranks $m$ and $n$ merge, a new company of rank $(m+n)$ is formed. The profit of the resulting company will be $m \cdot n$ more than the sum of the profits of the companies forming it. The profit... | # Solution.
Let $p_{n}$ be the profit of the company of rank $n$. Then, according to the problem statement,
$$
p_{m+n}=p_{m}+p_{n}+m n
$$
Notice that for any $n$,
$$
p_{n}=p_{n-1}+p_{1}+(n-1) \cdot 1=p_{n-1}+1+n-1=p_{n-1}+n
$$
We will prove that $p_{n}=1+2+\ldots+n$.
We use mathematical induction.
1) Base case: ... | 63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,637 |
Task 2. (10 points)
Find the denominator of the fraction $\frac{100!}{28^{20}}$ after it has been reduced to its simplest form.
(The expression 100! is equal to the product of the first 100 natural numbers: $100!=1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100$.$) .$ | # Solution.
It can be noticed that among the numbers from 1 to 100, exactly 14 numbers are divisible by 7, and exactly two are divisible by 49. Therefore, in the prime factorization of $1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 100$, the number 7 will appear in the power of 16. Among the numbers from 1 to 100, exac... | 2401 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,639 |
# Task 3. (12 points)
The sequence $a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots$ is such that $a_{2 n}=\frac{1}{a_{2 n-1}}$, and $a_{2 n+1}=1-a_{2 n}$.
Find $a_{1}$, if $a_{2018}=2$. | # Solution.
Let's find several of the last values of the sequence, considering that $a_{2 n-1}=\frac{1}{a_{2 n}}$ and $a_{2 n}=1-a_{2 n+1}:$
$$
a_{2018}=2 ; a_{2017}=0.5 ; a_{2016}=0.5 ; a_{2015}=2 ; a_{2014}=-1 ; a_{2013}=-1 ; a_{2012}=2 ; a_{2011}=0.5 \ldots
$$
Thus, we have obtained a periodic sequence with a per... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,640 |
# Task 4. (12 points)
A shooting tournament involves several series of 10 shots each. In one series, Ivan scored 82 points, as a result of which the average number of points he scored per series increased from 75 to 76 points. How many points does Ivan need to score in the next series of shots to make the average numb... | # Solution.
Let $N$ be the points scored in the $n$ considered series, in the last of which Ivan scored 82 points. Then $N=76n$ and $N-82=75(n-1)$. Solving the obtained system, we find $n=7$ and $N=532$.
Let $x$ be the points Ivan needs to score to meet the condition of the problem. In this case, we get $77 \cdot 8 =... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,641 |
# Task 6. (14 points)
According to the regulations of the chess tournament, each participant must play one game with every other participant. After exactly 99 games were played, it turned out that the set of participants could be divided into two unequal groups such that all opponents belonging to the same group had a... | # Solution.
Let the number of participants in the tournament be $n$, and the number of those who fell into the 1st group be $k$. Then the number of matches played is:
$\frac{k(k-1)}{2}+\frac{(n-k)((n-k)-1)}{2}+m=99$, where $m$ is small, $\Rightarrow$
$k^{2}-k+(n-k)^{2}-(n-k)+2 m-198=0$
$k^{2}-k+k^{2}-2 k n+n^{2}-n+... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,642 |
# Task 7. (14 points)
In a company, there are 168 employees. Among any four people, at least one can be chosen who is acquainted with the other three. What is the minimum possible number of people who are acquainted with everyone? # | # Solution.
If there are no strangers, then the number of people who know everyone is 168. Let $A$ and $B$ not know each other, then all other people know each other (if $C$ does not know $D$, then in the group $A, B, C, D$ no one knows the other three). If $A$ and $B$ know all the others, then 166 people know everyon... | 165 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,643 |
# Task 8. (16 points)
Vova was playing with old domino tiles, on which all the dots had worn off, making them indistinguishable. Each tile is a $2 \times 1$ rectangle, and their total number is 24. Vova decided to arrange the tiles in a new pattern each day to form a $2 \times 12$ path, ensuring that the pattern of th... | # Solution.
We will assume that the paths are arranged horizontally, with the longer side along the x-axis. The dominoes in such paths can be of two types: vertical (placed across the path) and horizontal, placed along the path.
Each vertical domino "cuts" the path and divides it into two parts, to the right and left... | 233 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,644 |
# Assignment 1. (10 points)
Each of the 2017 middle school students studies English or German. English is studied by $70 \%$ to $85 \%$ of the total number of students, and both languages are studied by $5 \%$ to $8 \%$. What is the maximum number of students who can study German. | # Solution.
Let $A$ be the number of people who study English, $N$ be the number of people who study German, and $AN$ be the number of people who study both languages.
Then $N=2017-A+AN$.
It is known that $A \geq 2017 \cdot 0.7=1411.9$ and $AN \leq 0.08 \cdot 2017=161.36$.
Therefore, $N \leq 2017-1412+161=766$.
No... | 766 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,645 |
# Task 3. (12 points)
Find all pairs $(a, b)$ of real numbers $a$ and $b$ such that the equations $x^{2}+a x+b^{2}=0$ and $x^{2}+b x+a^{2}=0$ have at least one common root.
# | # Solution.
Let $x_{0}$ be the common root of the equations.
Substituting $x_{0}$ into the equations and subtracting one from the other, we get
$$
(a-b) x_{0}+b^{2}-a^{2}=0 \Leftrightarrow(a-b)\left(x_{0}-(a+b)\right)=0
$$
1) Suppose $a \neq b$, then $x_{0}=a+b$ and
$(a+b)^{2}+a(a+b)+b^{2}=0$ or $2 a^{2}+3 a b+2 b... | (0,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,648 |
# Task 4. (12 points)
Find the smallest positive integer in which the product of the digits is 5120. | # Solution.
Since $5120=2^{10} \cdot 5$, the main question is how many digits are in this number. Clearly, there are no 0s and there is a 5. From the ten factors of 2, we can form a minimum of 4 digits. This can be done in two ways: $2,8,8,8$ or $4,4,8,8$. Then, from the two sets of digits $5,2,8,8,8$ and $5,4,4,8,8$,... | 25888 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,649 |
# Task 5. (12 points)
Asha is learning to write and can write three letters A, S, and Y. Her mother suggested she write seven letters in a row. In the resulting "word," three consecutive letters formed the name "ASY". How many such different seven-letter "words" exist? | # Solution.
We can consider the word ASYA as the fourth letter, then the seven-letter words will turn into five-letter words. The "letter" ASYA can stand in one of 5 positions, and in the other positions, any of the 3 letters can stand, so we get \(5 \cdot 3^{4} = 405\) words. Among them, there are words where the "le... | 396 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,650 |
# Problem 6. (14 points)
On the chord $A B$ of a circle, a point $P$ is marked such that $A P=2 P B$. The chord $D E$ is perpendicular to $A B$ and passes through point $P$. Prove that the midpoint of segment $A P$ is the orthocenter of triangle $A E D$.
# | # Solution.
Let $M$ be the midpoint of segment $AP$, and line $EM$ intersects segment $AD$ at point $Q$. Triangles $MPE$ and $BPE$ are equal by two legs. Then the angles $\angle QAM = \angle DAB = \angle DEB = \angle PEB = \angle PEM$. Since $\angle QMA = \angle EMP$, triangles $QMA$ and $EMP$ have two equal angles. T... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,651 |
# Task 7. (14 points)
In a certain company, no two employees have jobs of the same difficulty, and no two employees receive the same salary. On April 1, each employee made two statements:
(a) There will not be 12 employees with more difficult jobs.
(b) At least 30 employees have a higher salary. How many employees a... | # Solution.
If on April 1st all employees of the company told the truth, then the second statement about the highest salary is false, which cannot be. If, however, all of them lied, then the first statement for the employee with the highest salary would be true, again leading to a contradiction.
Thus, there is at lea... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,652 |
# Task 8. (16 points)
In a class, there are 14 girls. Each of them found out how many girls in the class have the same first name as her, and how many have the same last name, and wrote down two numbers on the board. It turned out that among the numbers on the board, all numbers from 0 to 6 are present. Prove that the... | # Solution.
Consider sets of girls with the same name. These are subsets of the set of 14 girls. Also, consider sets of girls with the same last name, which are also subsets of the original set. Each girl belongs to two such sets, and according to the problem, these sets contain 1, 2, 3, 4, 5, 6, and 7 people, but the... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,653 |
# Task 6. (14 points)
A regular $n$-sided prism is described around a sphere of radius 1, all edges of which touch some other sphere. Prove that $n=4$ and find the volume of this prism. | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,654 | |
# Problem 6. (14 points)
All edges of a regular $n$-sided truncated pyramid circumscribed around a sphere $\sigma$ touch the sphere $\omega$ with its center at the center of the lower base of the pyramid. Prove that $n=3$ and find the angle of inclination of the lateral edge of the pyramid to the base plane.
# | # Solution.
Let points $O$ and $O_{1}$ be the centers of the lower and upper bases, respectively,
$A A_{1} B_{1} B$ - one of the lateral faces $\left(A_{1} B_{1} \| A B\right), M$ and $M_{1}$ - the midpoints of the edges $A B$ and $A_{1} B_{1}$ of this pyramid. Then $M M_{1}=O M+O_{1} M_{1}$ (as segments of two tange... | notfound | Geometry | proof | Yes | Yes | olympiads | false | 22,655 |
# Task 5. (12 points)
The product of two different integers $a$ and $b$ is a square of a natural number. Does there exist an integer $x$ such that $n=\frac{(a-x)(b+x)}{a-b}$? | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,656 | |
# Task 5. (12 points)
The product of two different integers $a$ and $b$ is a square of a natural number. Does there exist an integer $x$ such that $n=\frac{(a+x)(b-x)}{b-a}$? | # Solution.
Consider $x=n$. Then, indeed: $\frac{(a+n)(b-n)}{b-a}=\frac{a b-n^{2}+(-a+b) n}{b-a}=\frac{(b-a) n}{b-a}=n$. Answer: Yes, $x=n$.
| Criterion Content | Evaluation | Points |
| :---: | :---: | :---: |
| The problem is solved completely. | + | 12 |
| The solution contains a correct general scheme, but lacks ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,657 |
# Task 1. (10 points)
How many natural numbers $n$ exist such that the equation $n x-12=3 n$ has an integer solution
# | # Solution.
Since $n x-12=3 n \Rightarrow x=3+\frac{12}{n}$, the condition of the problem is satisfied if 12 is divisible by $n$. Therefore, $n$ can take the values $1,2,3,4,6,12$.
Answer. 6
| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 10 |
| All main logical steps of the ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,658 |
# Task 2. (10 points)
Below is a magic square with some numbers missing.
| 20 | | |
| :--- | :--- | :--- |
| | 17 | |
| 16 | | |
Restore the given magic square.
(In a magic square, the sums of the numbers in each row, each column, and on the diagonals are equal). | Solution.
| 20 | | 2) | 20 | | $x$ | 3) | 20 | 13 | $x$ | 4) | 20 | 13 | $x$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | 17 | | | 17 | | | $x-3$ | 17 | | | $x-3$ | 17 | 19 |
| 16 | | | 16 | | | | 16 | | | | 16 | $x+3$ | 14 |
... | \begin{pmatrix}\hline20&13&18\\\hline15&17&19\\\hline16&21&14\\\hline\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,659 |
Task 3. (12 points)
The arithmetic mean of ten different natural numbers is 20, and the arithmetic mean of any nine of these numbers is not less than 17. Find the maximum possible value of the largest of these numbers.
# | # Solution.
We will order the 10 given numbers in ascending order.
According to the problem, the sum of the first 9 numbers cannot be less than $9 \cdot 17=153$.
Therefore, the 10th largest number cannot be greater than $20 \cdot 10-153=47$.
At the same time, the set 17, 17, $, 17, 47$ satisfies the condition of th... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,660 |
# Task 4. (12 points)
A numerical sequence is such that $x_{n}=\frac{n+2}{n x_{n-1}}$ for all $n \geq 2$. Find the product $x_{1} x_{2} x_{3} \ldots x_{2016} x_{2017}$, if $x_{1}=1$.
# | # Solution.
Notice that $x_{n} x_{n-1}=\frac{n+2}{n}$, therefore,
$$
x_{1} x_{2} x_{3} \ldots x_{2016} x_{2017}=x_{1}\left(x_{2} x_{3}\right)\left(x_{4} x_{5}\right) \ldots\left(x_{2016} x_{2017}\right)=1 \frac{5}{3} \frac{7}{5} \ldots \frac{2017}{2015} \frac{2019}{2017}=673
$$
Answer. 673
| Criterion | Evaluation ... | 673 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,661 |
# Problem 6. (14 points)
The function $f(x)$ is such that $f(f(x))=x$ and $f(f(x+2)+2)=x$ for any $x$.
Find $f(2017)$, given that $f(0)=1$.
# | # Solution.
From the equality $f(x)=f(f(f(x+2)+2))=f(x+2)+2$, we obtain the formula $f(x+2)=f(x)-2$.
Moreover, $f(1)=f(f(0))=0$.
We will prove by induction that $f(x)=1-x$ for any integer $x$.
First, we will prove that the given equality holds for even $x$.
1) $f(0)=1$ - true.
2) Let $f(2n)=1-2n$.
3) We will prove... | -2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,663 |
# Task 7. (14 points)
Given an equilateral triangle $A B C$ with side length 2. Point $K$ lies on the extension of side $A C$ beyond point $A$, point $N$ lies on the line parallel to line $A C$ and passing through point $B$, such that $|A K|=2,|B N|=1$. Consider broken lines $K L M N$ such that point $L$ lies on side ... | # Solution.
Rotate segment $B N$ by $60^{\circ}$ around point $B$ so that point $N$ moves to the midpoint of side $B C$, which is point $N^{\prime}$ in the diagram. Then segment $M N$ will transform into segment $L N^{\prime}$.
Two players take turns laying out coins in a row. In one move, a player can place two or three coins. The player who places the 16th coin wins. Determine which player (the first or the second) has a strategy that will allow them to win regardless of the other player's moves. Describe this strateg... | # Solution.
The first player has a winning strategy. For this, the first player needs to place 2 coins on the first move, and in subsequent moves, place as many coins as necessary to make the sum of his coins and the coins placed by the second player before that equal to 5. In this case, after the third move of the fi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,665 |
# Task 1. (10 points)
Six consecutive natural numbers from 10 to 15 are written in circles on the sides of a triangle in such a way that the sums of the three numbers on each side are equal.
What is the maximum value that this sum can take?
Find the value of the parameter $p$ for which the equation $p x^{2}=|x-1|$ has exactly three solutions.
# | # Solution
Notice that for $p>0$.
The given equation will have three roots if the parabola $y=p x^{2}$ intersects the "corner" $y=|x-1|$ at exactly three points. Since the left branch of the "corner" intersects the parabola at exactly two points when $p>0$, the right branch must be tangent to the parabola.
This happ... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,667 |
# Task 3. (12 points)
Egorov decided to open a savings account to buy a car worth 900000 rubles. The initial deposit amount is 300000 rubles. After one month and subsequently every month, Egorov plans to top up his account by 15000 rubles. The bank accrues interest monthly at an annual rate of $12 \%$. Interest accrue... | # Solution
Let $S_{n}$ be the sum of the deposit after $n$ months, after interest accrual and after making additional contributions $D (15000$ rubles).
Since the bank accrues $1 \%$ per month, then
$$
\begin{gathered}
S_{1}=300000(1+0.01)+D \\
S_{2}=S_{1}(1+0.01)+D=\left(S_{0}(1+0.01)+D\right)(1+0.01)+D=S_{0}(1+0.01... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,668 |
# Task 4. (12 points)
Solve the equation $(x+2)^{4}+x^{4}=82$.
# | # Solution
Let $y=x+1$, then we get
$$
(y+1)^{4}+(y-1)^{4}=82 \Leftrightarrow y^{4}+6 y^{2}-40=0 \Leftrightarrow\left[\begin{array} { l }
{ y ^ { 2 } = - 1 0 } \\
{ y ^ { 2 } = 4 }
\end{array} \Rightarrow \left[\begin{array}{l}
x=-3 \\
x=1
\end{array}\right.\right.
$$
$\underline{\text { Answer: }} x_{1}=-3 ; x_{2}... | x_{1}=-3;x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,669 |
# Task 5. (12 points)
The safe code consists of five consecutive digits. Vasiliy Petrovich put money in the safe, and when he wanted to take it out, he found out that he had forgotten the code. He only remembered that the numbers 21 and 16 were in the code. What is the minimum number of five-digit numbers that need to... | # Solution
Let's consider several cases. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,670 |
# Problem 6. (14 points)
A finite sequence of numbers $x_{1}, x_{2}, \ldots, x_{N}$ has the following property:
$$
x_{n+2}=x_{n}-\frac{1}{x_{n+1}} \text { for all } 1 \leq n \leq N-2 \text {. }
$$
Find the maximum possible number of terms in this sequence if $x_{1}=20 ; x_{2}=16$. | # Solution
The sequence will have the maximum number of terms if its last term is equal to zero. Otherwise, this sequence can be continued.
For all $1 \leq n \leq N-2$ we have
$$
x_{n+2}=x_{n}-\frac{1}{x_{n+1}} \Leftrightarrow x_{n+2} \cdot x_{n+1}=x_{n+1} \cdot x_{n}-1 \Leftrightarrow x_{n+1} \cdot x_{n}=x_{n+2} \c... | 322 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,671 |
# Task 7. (14 points)
Several businessmen decided to open a company and divide all the profits into equal parts. One of the businessmen was appointed as the director. One day, this director of the company transferred part of the profit from the company's account to his own personal account. This amount of money was th... | # Solution.
Let $n$ be the number of businessmen and $d_{i}$ be the profit of the $i$-th director, $i=1, \ldots, n$.
By the condition $d_{i}=3 \frac{d_{i+1}+d_{i+2}+\ldots+d_{n}}{n-i}$. Then
$$
\begin{gathered}
d_{i-1}=3 \frac{d_{i}+d_{i+1}+\ldots+d_{n}}{n-i+1}=3 \frac{3 \frac{d_{i+1}+d_{i+2}+\ldots+d_{n}}{n-i}+d_{i... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,672 |
# Task 8. (16 points)
The tangent $l$ to the circle inscribed in a rhombus intersects its sides $A B$ and $B C$ at points $E$ and $F$ respectively. Prove that the product $A E \cdot F C$ does not depend on the choice of the tangent $l$.
# | # Solution.
Let
$$
q=B K, p=A K, x=E K, y=F L
$$
( $K$ and $L-$ points of tangency of the inscribed circle with sides $A B$ and $B C$ ) and $\alpha=\angle \mathrm{ABC}$.
We will prove that $\mathrm{AE} \cdot \mathrm{FC}=(\mathrm{p}+\mathrm{x})(\mathrm{p}+\mathrm{y})$ does not depend on $x$ and $y$.
Consider triang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,673 |
# Task 1. (10 points)
Five cards are lying on the table, as shown in the figure.
On each card, there is a letter on one side and a natural number on the other side. Peter said: "If there is ... | # Solution
To prove that Peter is wrong, Katya needs to find a card with an odd number on one side and a vowel on the other. The only card that could have this property is the fourth card, which has the number 7 written on it.
Answer: the fourth card. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,674 |
# Task 2. (10 points)
It is known that the graph of the function $f(x)=x^{2}-2016 x+2015$ passes through two different points with coordinates ( $a, c$ ) and ( $b, c$ ). Find the sum $a+b$.
# | # Solution
According to the problem,
$$
\begin{gathered}
f(a)=a^{2}-2016 a+2015=f(b)=b^{2}-2016 b+2015 \Leftrightarrow \\
\Leftrightarrow a^{2}-b^{2}=2016 a-2016 b \Leftrightarrow(a-b)(a+b)=2016(a-b) \Leftrightarrow a+b=2016
\end{gathered}
$$
Answer: 2016. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,675 |
# Task 3. (12 points)
In a school, there are 1200 students, each of whom has five lessons every day. Any teacher in this school gives 4 lessons per day. How many teachers work in the school if there are exactly 30 students in each class? | # Solution
Since each student has 5 lessons per day, if there was only one student in the class, the total number of lessons per day would be $5 \times 1200=6000$. Since there are 30 students in the class, the number of lessons conducted in the school each day is $\frac{6000}{30}=200$. Therefore, the number of teacher... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,676 |
# Task 4. (12 points)
Consider the sequence of numbers $x_{1}, x_{2}, \ldots, x_{2015}$. In this case,
$$
x_{n}= \begin{cases}7, & \text { if } n \text { is divisible by } 9 \text { and } 32 ; \\ 9, & \text { if } n \text { is divisible by } 7 \text { and } 32 ; \\ 32, & \text { if } n \text { is divisible by } 7 \te... | # Solution
Since $7 \cdot 9 \cdot 32=2016$, then
$$
x_{n}= \begin{cases}7, & \text { if } n=9 \cdot 32 \cdot k, \text { where } k=1, \ldots, 6 \\ 9, & \text { if } n=7 \cdot 32 \cdot k, \text { where } k=1, \ldots, 8 \\ 32, & \text { if } n=7 \cdot 9 \cdot k, \text { where } k=1, \ldots, 31 \\ & 0, \quad \text { in a... | 1106 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,677 |
# Task 5. (12 points)
On a plane, there are four different circles. We will call an intersection point a point where at least two circles intersect. Find the maximum possible number of intersection points of four circles.
# | # Solution
Any two circles can intersect at no more than two points. From four circles, we can choose 6 different pairs of circles. Therefore, the number of intersection points cannot exceed 12.
The figure below shows a case where there are exactly 12 intersection points.
It is known that $2016+a^{2}+a c<a b$. Prove that the equation $a x^{2}+b x+c=0$ has two distinct roots.
# | # Solution
From the condition, it follows that $a \neq 0$, otherwise we would get the incorrect inequality $20160$. If $a > 0$, then the branches of the parabola are directed upwards and $f(-1)<0$, and the given equation has two distinct roots. Similarly for $a<0$.
# | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,679 |
# Task 7. (14 points)
In the analysis of bank accounts, it was found that the remaining balance on each of them is more than 10 rubles. It also turned out that there is a group of clients, each of whom has the same amount of money on their account. This amount is a number consisting of only ones. If you add up all the... | # Solution
This problem is equivalent to the following.
Find the smallest natural number $m$, for which there exist natural numbers $n$ and $k$, such that $n>k>1$ and $\underbrace{11 \ldots 1}_{n}=\underbrace{11 \ldots 1}_{k} \cdot m$.
Obviously, $m>9$. If $m=\overline{a b}$, where $a \geq 1$, then the equality $\un... | 101 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,680 |
# Task 8. (16 points)
Several people arrived at a conference. Prove that they can be placed in two conference halls so that each of them has an even number of acquaintances in their hall. (One of the halls can be left empty.)
# | # Solution
We will solve the problem by induction on the number of conference participants. The base case of the induction is verified directly.
Assume that the statement of the problem is true for $n$ participants.
Suppose that $n+1$ participants have arrived at the conference. If each of them has an even number of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,681 |
# Task 1. (10 points)
Five cards are lying on the table, as shown in the figure.
On each card, there is a letter on one side and a natural number on the other side. Peter said: "If there is ... | # Solution
To prove that Peter is wrong, Katya needs to find a card with an odd number on one side and a vowel on the other. The only card that could have this property is the fourth card, which has the number 7 written on it.
Answer: the fourth card. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,682 |
# Task 1. (10 points)
How many natural numbers $n$ not exceeding 2017 exist such that the quadratic trinomial $x^{2}+x-n$ can be factored into linear factors with integer coefficients
# | # Solution.
According to the problem, \(x^{2}+x-n=(x-a)(x-b)\). Therefore, \(ab = -n\), which means the numbers \(a\) and \(b\) have different signs and are not zero. Without loss of generality, we assume that \(a \geq 0\). Since \(a + b = -1 \Rightarrow b = -1 - a\), we have
\[
ab = -n = a(-1 - a) \Rightarrow a^{2} ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,683 |
# Task 2. (10 points)
In the thirtieth kingdom, on February 29th of an olden year, a merchant was selling self-playing boots for 2000 altyn at a fair. According to trading rules, the price of goods is adjusted every morning before the market opens. The price can be increased by $10 \%$, decreased by $1 \%$ or $12 \%$ ... | # Solution.
The boyar is right.
If the price was $s$, then after the change, it should be $\frac{110}{100} s, \frac{99}{100} s$ or $\frac{88}{100} s$. With any such change, the new price must be divisible by 11. The number 9876543210 is not divisible by 11, and therefore it cannot be obtained through several such cha... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,684 |
# Task 3. (12 points)
A numerical sequence is such that $x_{n}=\frac{x_{n-1}}{5}+4$ for all $n \geq 2$. Find $x_{2017}$ if $x_{1}=6$.
# | # Solution.
$x_{2}=\frac{x_{1}}{5}+4=\frac{6}{5}+4=5+\frac{1}{5}$.
$x_{3}=\frac{x_{3}}{5}+4=\frac{5+\frac{1}{5}}{5}+4=5+\frac{1}{5^{2}}$.
We will prove by induction that $x_{n}=5+\frac{1}{5^{1-n}}$ for all natural $n$.
1) $n=1: \quad x_{1}=5+\frac{1}{5^{1-1}}=6$ - true.
2) Suppose the statement is true for $n=k: x_... | 5+\frac{1}{5^{-2016}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,685 |
# Task 4. (12 points)
Which of the numbers is greater $\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2016}\right)$ or $\frac{1}{2017}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2017}\right)$? | Solution.
Let
$$
x=\frac{1}{2016}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2016}\right), y=\frac{1}{2017}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2017}\right), a=\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2016}\right)
$$
Then $x=\frac{a}{2016}, y=\frac{1}{2017}\left(a+\frac{1}{2017}\right), x-y=\fr... | the\first\\is\greater | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,686 |
# Task 6. (14 points)
Natural numbers $a, b, c, d$, and $e$ are consecutive terms of an arithmetic progression. Find the smallest possible value of the number $c$, if the sum $b+c+d$ is a perfect square, and the sum $a+b+c+d+e$ is a perfect cube. | # Solution.
Since $b+d=2c$, then $3c=n^2$ for some natural number $n$.
Therefore, $n$ is divisible by 3 and $c=3l^2$ for some natural number $l$.
Since $a+b+d+e=4c$, then $5c=m^3$ for some natural number $m$.
Therefore, $m$ is divisible by 5 and $c=5^2 l^3$ for some natural number $l$.
The smallest number that sat... | 675 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,688 |
# Task 7. (14 points)
281 employees from 7 different branches of the company participated in the conference. In every group of six participants, at least two were of the same age. Prove that among all the participants, you can find five of the same age, of the same gender, and from the same branch of the company. | # Solution
By the Pigeonhole Principle, in at least one branch, there are at least 41 participants, and at least 21 of them are of the same gender. It is required to prove that at least 5 of these 21 participants are of the same age. If this is not the case, then there are no more than 4 participants from each age gro... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,689 |
# Task 8. (16 points)
Ten pirates are dividing gold and silver coins among themselves. There are twice as many silver coins as gold coins. They divided the gold coins such that the difference in the number of gold coins between any two pirates is not divisible by 10. Prove that they cannot divide the silver coins in a... | # Solution.
Let $a_{1}$ be the number of gold coins the first pirate has, $a_{2}$ for the second, and so on. Since $a_{i}-a_{j}$ is not divisible by 10 for any $i$ and $j$, the numbers $a_{1}, a_{2}, \ldots, a_{10}$ must give different remainders when divided by 10. We can write $a_{i}=10 k_{i}+l_{i}$, where $l_{i}$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 22,690 |
# Task 11.1. (10 points)
Given a sequence of numbers $x_{1}, x_{2}, \ldots$, such that $x_{1}=79$ and $x_{n}=\frac{n}{x_{n-1}}$ for all $n>1$. How many zeros does the number equal to the product $x_{1} x_{2} \ldots x_{2018}$ end with? | # Solution.
From the condition of the problem, it follows that $x_{n} x_{n-1}=n$. Therefore,
$$
x_{1} x_{2} \ldots x_{2018}=\left(x_{1} x_{2}\right)\left(x_{3} x_{4}\right) \ldots\left(x_{2017} x_{2018}\right)=2 \cdot 4 \cdot 6 \cdot \mathrm{K} \cdot 2018=2^{1009} \cdot 1009!
$$
In the obtained product, 201 numbers ... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,691 |
# Task 11.3. (12 points)
How many distinct roots does the equation $f(f(f(x)))=1$ have, if $f(x)=x-\frac{2}{x}$. | # Solution.
Let $f(x)=k=x-\frac{2}{x}, f(f(x))=f(k)=k-\frac{2}{k}, f(f(f(x)))=f(f(k))=f(k)-\frac{2}{f(k)}$. The equation $f(f(f(x)))=f(k)-\frac{2}{f(k)}=1 \Leftrightarrow f^{2}(k)-f(k)-2=0$ has two solutions $f_{1}(k)=-1$ and $f_{2}(k)=2$.
We obtain $\left[\begin{array}{l}f(k)=k-\frac{2}{k}=-1 \\ f(k)=k-\frac{2}{k}=2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,693 |
Task 11.4. (12 points)
Employees of the company are divided into hard workers and slackers. In 2016, the average salary of hard workers was twice the average salary of slackers. After improving their qualifications, hard workers in 2017 began to earn $50 \%$ more, while the salary of slackers remained unchanged. At th... | # Solution.
Let the number of hard workers in the company in 2016 be $x$, then the number of slackers was $9x$, and the total number of people in the company was $10x$. Let the average salary of a slacker be $s$, then the average hard worker received $2s$, and the average salary across the entire company was $\frac{2s... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,694 |
# Task 11.5. (12 points)
On the first step, a unit circle was drawn on a sheet of paper, and the circle bounded by it was colored black. On each subsequent step, for each of the circles drawn on the previous step, four new internally tangent circles of equal radii are drawn. These four circles are externally tangent t... | # Solution.
If at some step a circle of radius $R$ was drawn, then on the next step circles of radii $\frac{R}{1+\sqrt{2}}$ will be drawn. Thus, the area of the white region is the sum:
Thi... | \frac{4\pi}{(1+\sqrt{2})^{2}+4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,695 |
Task 11.6. (14 points)
Real numbers $a$ and $b$ are such that $\left(a+\sqrt{1+a^{2}}\right)\left(b+\sqrt{1+b^{2}}\right)=1$. Find the sum $a+b$.
# | # Solution.
$$
\begin{aligned}
& b+\sqrt{1+b^{2}}=\frac{1}{a+\sqrt{1+a^{2}}}=-a+\sqrt{1+a^{2}} \\
& a+\sqrt{1+a^{2}}=\frac{1}{b+\sqrt{1+b^{2}}}=-b+\sqrt{1+b^{2}}
\end{aligned}
$$
By adding the obtained equations, we get
$$
a+b+\sqrt{1+a^{2}}+\sqrt{1+b^{2}}=-a-b+\sqrt{1+a^{2}}+\sqrt{1+b^{2}} \Rightarrow a+b=0
$$
## ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,696 |
# Problem 11.7. (14 points)
We will call a positive number $a$ close above a positive number $b$ if $a$ exceeds $b$, but not by more than $1 \%$. Prove that if in a triangle the radian measure of one of the angles is close above the radian measure of another angle, then there will be two sides of this triangle such th... | # Solution.
Let the radian measure of angle $\alpha$ be close from above to the radian measure of angle $\beta$, that is,
$$
\beta\frac{\sin \beta}{b} \Rightarrow a rigorous justification of individual conclusions. | $\pm$ | 11 |
| The main idea of the solution is found, but it is not completed or contains errors. H... | proof | Geometry | proof | Yes | Yes | olympiads | false | 22,697 |
# Task 11.8. (16 points)
In the alphabet of the Alph language, there are three letters: $A$, Л, and $Ф$. All words of this language can be constructed by applying the following rules sequentially to any word in this language:
(1) reverse the order of the letters in the word
(2) replace two consecutive letters as fol... | # Solution.
Let's call the "value" of a word the number $n-m$, where $n$ and $m$ are the number of letters L and A, respectively, in the given word. Note that all allowed operations preserve the remainder of the word's "value" modulo 3. The "value" of the known word LLAFALAFALAFALAFALAFAL is $7-8=-1$, and the "value" ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,698 |
2. In the isosceles triangle $A N T$ with the angle at the vertex $90^{\circ}$, the angle at the base $\angle T A N = 45^{\circ}$ (similarly in triangle $B M K$). Then we have: $\angle T A K = \angle T A N - \angle M A N = 45^{\circ} - \angle M A N = 45^{\circ} - \angle M B N = \angle M B K - \angle M B N = \angle K B ... | # Solution.
Form two auxiliary vectors $(x ; 1)$ and $(\sqrt{1+x} ; \sqrt{3-x})$. Their dot product gives the left side of the equation, and the product of their lengths gives the right side of the equation. Therefore, these vectors are collinear, i.e., $x \sqrt{3-x}=\sqrt{1+x}$. Squaring the obtained equation, we not... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,701 |
2. In the isosceles triangle $A N T$ with the angle at the vertex $90^{\circ}$, the angle at the base $\angle T A N = 45^{\circ}$ (analogously in triangle $B M K$). Then we have: $\angle T A K = \angle M A N - \angle T A N = \angle M A N - 45^{\circ} = \angle M B N - 45^{\circ} = \angle M B N - \angle M B K = \angle K ... | # Solution.
Form two auxiliary vectors $(x ; 1)$ and $(\sqrt{x} ; \sqrt{4-x})$. Their dot product gives the left side of the equation, and the product of their lengths gives the right side of the equation. Therefore, these vectors are collinear, i.e., $x \sqrt{4-x}=\sqrt{x}$. Squaring the obtained equation and noting ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 22,702 |
# Task 1. (10 points)
Five different weights, each weighing a whole number of kilograms, were weighed in all possible groups of three weights. As a result, the following weights (in kilograms) of the ten weighed groups were obtained: $10,14,15,16,17,17,18$, $21,22,24$. Find the weights of these five weights. | Solution.
Let $a>b>c>d>e$ be the weights of the given weights. Since each weight is included in six out of ten weighed triples, the sum of the obtained weights is $6(a+b+c+d+e)$, therefore,
$$
a+b+c+d+e=\frac{10+14+15+16+17+17+18+21+22+24}{6}=29
$$
Meanwhile,
$$
\begin{aligned}
& a+b+c=24, c+d+e=10 \Rightarrow a+b+... | 10,9,5,3,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,703 |
# Task 2. (10 points)
Given 2018 numbers $x_{1}, x_{2}, \ldots, x_{2018}$, each of which is either $2-\sqrt{3}$ or $2+\sqrt{3}$. Find the greatest possible value of the sum $x_{1} x_{2}+x_{3} x_{4}+x_{5} x_{6}+\ldots+x_{2017} x_{2018}$, given that it is an integer. | # Solution.
Note that the product $x_{2 k-1} x_{2 k}$ can take one of three values:
$$
\begin{aligned}
& (2-\sqrt{3})(2-\sqrt{3})=7-2 \sqrt{3} \\
& (2+\sqrt{3})(2+\sqrt{3})=7+2 \sqrt{3}
\end{aligned}
$$
or
$$
(2-\sqrt{3})(2+\sqrt{3})=1
$$
Let $a$ be the number of times the number $2-\sqrt{3}$ appears in the consid... | 7057 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,704 |
# Task 3. (12 points)
Find the minimum value of the function $f(x)=|x|+|x+1|+K+|x+2022|$. | # Solution.
The numerical axis can be divided into 2024 subsets:
$(-\infty; -2022), (-2022; -2021), (-2021; -2020), \ldots, (-2; -1), (-1; 0), (0; +\infty)$. On each of these subsets, the function is linear and has the form $f(x) = k_i x + b_i$. Let $k_i$ be the slope coefficient on the $i$-th subset from the left ($... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 22,705 |
# Task 4. (12 points)
The angle bisectors of angles $A, B$, and $C$ of triangle $A B C$ intersect the circumcircle of this triangle at points $A_{1}, B_{1}$, and $C_{1}$, respectively. Find the distances between point $A_{1}$ and the center of the inscribed circle of triangle $A B C$, given that $\angle A_{1} B_{1} C_... | # Solution.
In the figure, identical numbers mark equal angles (this follows from the fact that $A A_{1}, B B_{1}, C C_{1}$ are the angle bisectors of triangle $ABC$, the angle marked "1+2" near point $O$ (which is the center of the inscribed circle) is equal to $\angle O A B + \angle O B A = \angle 1 + \angle 2$ by t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,706 |
# Task 7. (14 points)
Ivan and Petr are playing the following game. From a pile containing 2018 stones, they take some number of stones in turns. If there are $N$ stones in the pile before a move, the player can take $k$ stones only if $k$ is a divisor of the number $N$. The player who takes the last stone loses. Whic... | # Solution
We will show that Ivan has a winning strategy. For Ivan to win, it is sufficient for him to take one stone each move.
In this case, after his move, the number of stones will be odd. Since the divisors of an odd number are odd numbers, Peter will have to take an odd number of stones.
Since the number of st... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,707 |
# Task 8. (16 points)
How many solutions in integers does the equation $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}=2031$ have. | # Solution.
Note that if the number $n$ is even, then $n^{4}$ is divisible by 16.
If $n$ is odd, then the number
$$
n^{4}-1=(n-1)(n+1)\left(n^{2}+1\right)
$$
is divisible by 16.
Therefore, the remainder of the left-hand side of the equation $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}$ when divided by 16 is equal to the... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,708 |
# Problem 6. (14 points)
In a convex quadrilateral $A B C D \quad \angle A B C=70^{\circ}, \angle A D C=145^{\circ}$, $B C=B D=1$. Find the length of side $A B$.
# | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,709 | |
Problem 1. Anya calls a date beautiful if all 6 digits of its notation are different. For example, 19.04.23 is a beautiful date, while 19.02.23 and 01.06.23 are not.
a) How many beautiful dates will there be in April 2023?
b) How many beautiful dates are there in total in 2023? | Answer. a) 5; b) 30.
Solution. a) Note that in April, the date is written as **.04.23. Therefore, the first two digits can only be $1,5,6,7,8,9$. Of these digits, only 1 can be the first, so there are five dates in total: $15,16,17,18,19$ April.
b) See the solution to problem 1 for 7th grade (p. 9). | )5;b)30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,710 |
Problem 3. Vasya solved problems for 15 days - at least one problem each day. Each day (except the first), if the weather was cloudy, he solved one more problem than the previous day, and if it was sunny, one less problem. In the first 9 days, Vasya solved 13 problems.
a) What was the weather like on the 10th day? How... | Answer. a) The 10th day was cloudy, Vasya solved 2 problems; b) 7 problems.
Solution. a) See the solution to problem 2 for 6th grade (p. 4).
b) Since on the 10th day Vasya solved 2 problems (see part a), and in each subsequent day he can solve a maximum of 1 more problem, then in 5 days he can solve a maximum of $2+5... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,712 |
Problem 6. The tsar has 5 bags of gold coins, each containing 100 coins. The tsar remembers that in one bag all the coins weigh 10 g, in the second 11 g, in the third 12 g, in the fourth 13 g, and in the fifth 14 g, but he does not remember which is which. The tsar informed the court sage and pointed to one of the bags... | Answer. a, b) Yes, it can.
Solution. a) Let's take one coin from each of the bags, except for the specified one, and place them on the left pan, and put 5 coins from the specified bag on the right pan. If the specified bag indeed contains coins weighing 10 g each, the scales will show balance:
$$
11+12+13+14=10+10+10... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,713 |
Problem 1. On a sausage, thin rings are drawn across. If you cut along the red rings, you get 5 pieces; if along the yellow ones, you get 7 pieces; and if along the green ones, you get 11 pieces. How many pieces of sausage will you get if you cut along the rings of all three colors?
$[3$ points] (A. V. Shipovalov) | Answer: 21 pieces.
Solution: Note that the number of pieces is always one more than the number of cuts. Therefore, there are 4 red rings, 6 yellow ones, and 10 green ones. Thus, the total number of cuts is $4+6+10=20$, and the number of pieces is 21. | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,714 |
Problem 2. In Lesogoria, only elves and gnomes live. Gnomes lie when talking about their gold, but tell the truth in all other cases. Elves lie when talking about gnomes, but tell the truth in all other cases. One day, two Lesogorians said:
$A$ : All my gold I stole from the Dragon.
B: You are lying.
Determine wheth... | Answer: Both are gnomes.
Solution. Suppose $A$ is an elf. Then he told the truth, and $B$ lied. But neither gnomes nor elves lie when talking about elves. Therefore, $A$ is a gnome. Speaking about the gold, he lied. Therefore, $B$ told the truth about $A$. This could only be done by a gnome. | Bothgnomes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,715 |
Problem 3. Piglet Naaf-Naaf came up with a way to build a parallelepiped from identical cubes and cover it with three squares without gaps or overlaps. Do this yourself. $\quad[5$ points] (A. V. Shapovalov) | Answer. For example, he can stack a tower of four cubes, "fold" it into a $4 \times 4$ square, and glue the bottom and top with $1 \times 1$ squares.
 | 4\times4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,716 |
Problem 4. At the exchange office, there are two types of operations: 1) give 2 euros - get 3 dollars and a candy as a gift; 2) give 5 dollars - get 3 euros and a candy as a gift.
When the wealthy Pinocchio came to the exchange office, he only had dollars. When he left, he had fewer dollars, no euros appeared, but he ... | Answer: $10.
Solution: Since Buratino received 50 candies, he performed exactly 50 operations. At the same time, he exchanged all the euros he received back into dollars. Therefore, for every 3 operations of the first type, there were 2 operations of the second type. That is, Buratino received $3 30 times and gave awa... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,717 |
Problem 6. At the edge of a rotating round table, 30 cups of tea were placed at equal intervals. The March Hare and Alice sat down at the table and started drinking tea from two of the cups (not necessarily adjacent). When they finished their tea, the Hare turned the table so that each of them had a full cup in front o... | Solution. We will color the cups alternately in blue and red. Let the March Hare drink from a red cup at first. We will prove that Sonya drank from a blue cup at first. Indeed, if she drank from a red one, then after any rotation of the table, two cups of the same color would be emptied. Since there are 15 of each colo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,719 |
Problem 1. Yura has a calculator that allows multiplying a number by 3, adding 3 to a number, or (if the number is divisible by 3) dividing the number by 3. How can one use this calculator to get from the number 1 to the number 11? $\quad[3$ points] (T. I. Golenishcheva-Kutuzova) | Answer. For example, $((1 \cdot 3 \cdot 3 \cdot 3)+3+3): 3=11$ or $(1 \cdot 3+3): 3+$ $3+3+3=11$.
Comment. Note that on Yura's calculator, any number can be increased by $1:(x \cdot 3+3): 3=x+1$. Therefore, in principle, any natural number can be obtained from one on it. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,720 |
Problem 3. Little kids were eating candies. Each one ate 7 candies less than all the others together, but still more than one candy. How many candies were eaten in total?
[5 points] (A. V. Shapovalov) | Answer: 21 candies.
Solution. Let's choose one of the children, for example, Petya. If we take away 7 candies from all the others, there will be as many left as Petya has. This means that twice the number of candies Petya has equals the total number of candies minus seven. The same can be said about any of the childre... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,722 |
Problem 4. In a singing competition, a Rooster, a Crow, and a Cuckoo participated. Each member of the jury voted for one of the three performers. The Woodpecker calculated that there were 59 judges in total, and that the Rooster and the Crow received a total of 15 votes, the Crow and the Cuckoo received 18 votes, and t... | Answer: 13 judges.
Solution. The number of votes for the Rooster and the Raven cannot be more than $15+13=28$. Similarly, the number of votes for the Raven and the Cuckoo cannot exceed $18+13=31$, and the number of votes for the Cuckoo and the Rooster cannot exceed $20+13=33$. Adding these three quantities of votes, w... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,723 |
Problem 5. a) Piglet Naf-Naf came up with a way to build a parallelepiped from identical cubes and cover it with three squares without gaps or overlaps. Do this yourself.
b) Can Naf-Naf achieve that each pair of squares borders each other?
[7 points: a) 3 points, b) 4 points] (A.V. Shapovalov) | Solution. a) See the solution to problem 3 of the 6th grade.
b) The diagram shows how a parallelepiped $1 \times 4 \times 6$ can be covered with two $4 \times 4$ squares and one $6 \times 6$ square. The larger square covers three faces: the front, bottom, and back, while each of the smaller
![](https://cdn.mathpix.co... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,724 |
Problem 6. It is easy to place a set of ships for the game "Battleship" on a $10 \times 10$ board (see Fig. 4). On what smallest square board can this set be placed? (Let's recall that according to the rules, ships should not touch even at the corners.)
$[8$ points] ( $A$. D. Blinkov) | Answer. On a $7 \times 7$ board.
Fig. 4
Fig. 5
Solution. ... | 7\times7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,725 |
Task 1. In the rebus YAEMZMEYA $=2020$, replace each letter in the left part of the equation with a digit or an arithmetic operation symbol (the same letters are replaced by the same, different letters by different ones) so that the resulting equation is correct. It is sufficient to provide one example, no explanations... | Answer. $2 \times 505 \times 2=2020$. | 2\times505\times2=2020 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,726 |
Problem 3. On the table lie 6 apples (not necessarily of the same weight). Tanya divided them into 3 on each of the two pans of the scales, and the scales remained in balance. And Sasha divided the same apples differently: 2 apples on one pan and 4 on the other, and the scales were again in balance. Prove that it is po... | Solution. Let's look at the three apples that Tanya put on one of the pans. Let's call them green.
Could all of them have ended up on one pan with Sasha as well? No, because their weight is already half of the total weight of the apples. This means that two green apples will end up on one of the pans, and the third on... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 22,727 |
Problem 4. Three sides of a quadrilateral are equal, and the angles of the quadrilateral formed by these sides are $90^{\circ}$ and $150^{\circ}$. Find the other two angles of this quadrilateral. | Answer: $45^{\circ}$ and $75^{\circ}$.
Solution. Let's denote the vertices of the quadrilateral as shown in the diagram.
Extend $A B C$ to form a square $A B C X$. In triangle $X C D$, the angle $\angle X C D$ is equal to $\angle B C D - \angle B C X = 150^{\circ} - 90^{\circ} = 60^{\circ}$, and the sides $C X$ and $... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,728 |
Problem 6. Can this figure (the "camel") be divided
a) along the grid lines;
b) not necessarily along the grid lines into 3 parts, from which a square can be formed?
[a) 4 points; b) 6 poi... | Answer. a) No; b) Yes.
Solution. Note that the area of the camel is 25 cells. That is, we are supposed to assemble a square with a side of 5.
a) Let's look at the 4 cells marked in the picture. Any two of them are "far from each other": separated by at least 4 rows or columns. Therefore, when cutting, two marked cell... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,729 |
Problem 1. Little Know-Not has five cards with digits: $1,2,3,4$ and 5. Help him form two numbers - a three-digit number and a two-digit number - from these cards so that the first number is divisible by the second. $\quad[3$ points] (A.V. Shapovalov) | Answer. $5|3| 2$ and $1 \mid 4(532: 14=38)$ or $2|1| 5$ and $4 \mid 3$ $(215: 43=5)$.
Comment. Of course, this problem is solved by trial and error, but it is useful to use the properties of divisibility of numbers in the process. For example, when selecting a two-digit number, you don't need to consider:
a) neither ... | 532:14=38or215:43=5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,730 |
Problem 2. In a small town, there is only one tram line. It is circular, and trams run along it in both directions. There are stops on the loop: Circus, Park, and Zoo. The journey from Park to Zoo via Circus is three times longer than not via Circus. The journey from Circus to Zoo via Park is twice as short as not via ... | Answer: The path not through the Zoo is 11 times shorter.
Solution. Let's board the tram at the Zoo stop and travel through the Circus to the Park, and then, without leaving the tram, return to the Zoo. The second part of the journey is three times shorter than the first, meaning the first part takes up three quarters... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,731 |
Problem 3. An equilateral triangle with a side length of 8 was divided into smaller equilateral triangles with a side length of 1 (see figure). What is the minimum number of small triangles that need to be shaded so that all intersection points of the lines (including those on the edges) are vertices of at least one sh... | Answer: 15 small triangles. See the example in the figure.
Solution: The total number of intersection points of the lines is $1+2+3+\ldots+9=45$. Since a triangle has three vertices, at leas... | 15 | Combinatorics | proof | Yes | Yes | olympiads | false | 22,732 |
Problem 4. Anya wanted to write one digit in each cell of a $5 \times 8$ table in such a way that each digit appears exactly in four rows. (We consider both columns and rows as rows of the table.) Prove that she cannot achieve this. [6 points] (E.V. Bakayev) | Solution. Let's assume that the table has 5 rows and 8 columns, and suppose that Anya managed to arrange the digits. Note that each digit in the table can appear no more than 4 times. Indeed, if among the four rows where it appears, there are two vertical rows and two horizontal rows, then there are exactly four cells ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 22,733 |
Problem 5. The robot invented a cipher for writing words: he replaced some letters of the alphabet with single or double-digit numbers, using only the digits 1, 2, and 3 (different letters he replaced with different numbers). First, he wrote himself in code: РОБОТ $=3112131233$. After encrypting the words КРОКОДИЛ and ... | Answer: 2232331122323323132.
Solution. Consider the word ROBOT $=3112131233$. It contains 5 letters and 10 digits, so all codes are two-digit and can be determined without difficulty. Let's write down all twelve possible codes and the letters we definitely know:
$$
\begin{array}{llll}
1= & 11= & 21= & 31=P \\
2= & 12... | 2232331122323323132 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,734 |
Problem 6. Forty children were playing in a ring. Of them, 22 were holding hands with a boy and 30 were holding hands with a girl. How many girls were in the ring? [8 points] (E.V. Bakayev) | Answer: 24 girls.
Solution: $22+30=52$, so $52-40=12$ children held hands with both a boy and a girl. Therefore, $30-12=18$ children held hands only with girls. These 18 children held $18 \cdot 2=36$ girls' hands, and the other 12 held one girl's hand each, so the girls had a total of $36+12=48$ hands. Thus, there wer... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 22,735 |
Problem 1. On the surface of a planet shaped like a donut, two snails crawled, leaving trails behind: one along the outer equator, and the other along a spiral line (see figure). Into how many parts did the snails' trails divide the surface of the planet? (It is sufficient to write the answer.)
 }
$$ | Answer. One of the solutions is shown in the figure (the axis of symmetry is indicated by a dashed line).
 | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,737 |
Task 4. Fill in the asterisks with six different digits so that all fractions are irreducible and the equation is correct: $\frac{*}{*}+\frac{*}{*}=\frac{*}{*} . \quad[6$ points] (A.V. Lapovalov) | Answer. For example, $\frac{1}{6}+\frac{7}{3}=\frac{5}{2}$ (there are other examples).
Comment. Finding an example is simplified if you notice that no denominator can be equal to 1 (then the denominators of the remaining fractions would coincide), nor 5 or 7 (because if the denominators of two irreducible fractions ar... | \frac{1}{6}+\frac{7}{3}=\frac{5}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 22,738 |
Problem 5. One angle of a triangle is $60^{\circ}$, and the side opposite this angle is equal to one third of the perimeter of the triangle. Prove that the given triangle is equilateral. $[10$ points] (M.A. Volchkevich) | Solution. Suppose the opposite. Let in triangle $ABC$ angle $A$ be $60^{\circ}$, $BC=a$, $AB=a-x$, then $AC=$ $=p-AB-BC=3a-a-(a-x)=a+x$. Choose point $B'$ on ray $AB$ and point $C'$ on ray $AC$ such that $AB'=AC'=a$.
 differently, and a triangle remained. Draw an example of how this could have been. | Answer. One of the possible examples is shown in the figures.
 | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,741 |
Problem 2. Three frogs on a swamp jumped in turn. Each landed exactly in the middle of the segment between the other two. The length of the second frog's jump is 60 cm. Find the length of the third frog's jump.
$[4$ points]
(A. V. Shapovalov) | Answer: 30 cm.
Solution. Regardless of how the frogs were sitting initially, after the first jump, they will be on one straight line, with the first (A) in the middle.
B $\qquad$ A B
Now the second frog (B) jumps. It flies a distance to A and then half of this distance, which, according to the condition, is 60 cm. T... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 22,742 |
Problem 3. The digits from 0 to 9 are encrypted by the letters A, B, C, D, E, F, G, H, I, J in some order. In one question, you can find out the encrypted record of the sum of several different letters. For example, if you ask "A $+\mathrm{B}=$ ?", then in the case where $\mathrm{A}=9, \mathrm{~B}=1, \mathrm{C}=0$, the... | Solution. The sum of all ten digits is 45. Therefore, by naming all ten letters, we will learn which letters represent the digits 4 and 5. Excluding these letters and asking about the sum of the remaining eight, we will learn which letters represent the digits 3 and 6. In each subsequent question, we will similarly ask... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,743 |
Problem 4. The maze for mice (see the figure) is a $5 \times 5$ meter square, and the mice can only run along the paths. Two pieces of cheese (marked with crosses) are placed at two intersections. At another intersection, a mouse (marked with a circle) is sitting. She can smell the cheese, but she needs to run the same... | Answer. a)
b) The maximum number of places for thoughtful mice is 26:
![](https://cdn.mathpix.com/cropped/2024_05_06_51d7eed368663d90d59bg-3.jpg?height=285&width=291&top_left_y=487&top_left_... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 22,744 |
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