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7.3. Find a six-digit number that, when multiplied by 9, is written with the same digits as the original number but in reverse order? How many such six-digit numbers are there? | Answer: 109989 is the only number. Solution. Let $\overline{a b c d e f}$ be the desired number, i.e., $\overline{a b c d e f} \cdot 9=\overline{f e d c b a}$. Then it is obvious that $a=1, b=0$ (otherwise, multiplying by 9 would result in a seven-digit number). Therefore, $f=9$, and the second-to-last digit $e=8$ (whi... | 109989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. Find the natural number $x$ that satisfies the equation
$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3} .
$$ | Answer. $\quad 2012$.
Solution. See problem 8.4. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. On the side $AC$ of triangle $ABC$, a point $M$ is taken. It turns out that $AM = BM + MC$ and $\angle BMA = \angle MBC + \angle BAC$. Find $\angle BMA$. | Answer. $\quad 60^{\circ}$. Solution. First, we will show that triangle $A B C$ is isosceles. Indeed, this follows from the condition $\angle B M A=\angle M B C+\angle B A C$ and the property of the exterior angle: $\angle B M A=\angle M B C+\angle B C A$. From these two equalities, we have $\angle B C A=\angle B A C$.... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. A father had three sons, and he left them 9 ares of land as an inheritance - a rectangle measuring 25 m $\times 36$ m. The brothers decided to divide the land into three rectangular plots - three ares for each brother. How many options are there for the division (in terms of the length and width of the plots), and... | Answer: 4 options; the smallest length is 49 m in the option of dividing into a plot of $25 \times 12$ and two plots of $12.5 \times 24$. Solution. Let $ABCD$ be the original rectangle; $AB=25, BC=36$. Since it has 4 vertices and 3 plots, two vertices must belong to one plot. First, consider the case where the vertices... | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.4. How many zeros does the product $s(1) \cdot s(2) \cdot \ldots \cdot s(100)$ end with, where $s(n)$ denotes the sum of the digits of the natural number $n$? | Answer: .19 zeros. Solution. Consider the numbers from the first hundred for which the sum of the digits is divisible by 5. Such numbers have a digit sum of either 5, 10, or 15. There are 6 numbers with a digit sum of 5: these are $5, 14, 23, 32, 41, 50$. There are 9 numbers with a digit sum of 10: these are $19, 28, \... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. A father had three sons, and he left them 9 ares of land as an inheritance - a rectangle measuring 25 m $\times 36$ m. The brothers decided to divide the land into three rectangular plots - three ares for each brother. How many options are there for the division (in terms of the length and width of the plots), and... | Answer: 4 options; the shortest length is 49 m in the option of dividing into a plot of $25 \times 12$ and two plots of $12.5 \times 24$. Solution. See problem 7.2 | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Let $s(n)$ denote the sum of the digits (in decimal notation) of a natural number $n$. Find all natural $n$ for which $n+s(n)=2011$. | Answer: 1991. Hint. Since $n<2011$, then $s(n) \leq 2+9+9+9=29$. Therefore, $n=2011-$ $s(n) \geq 1982$. Since the numbers $n=2011$ and $n=2010$ obviously do not fit, the first three digits of the number $n$ can be one of three possibilities: 198, or 199, or 200. Let the last (fourth) digit of the number $n$ be $x$. The... | 1991 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The unit square in the first quadrant of the coordinate plane ( $0 \leq x, y \leq 1$ ) is divided into squares with side length $2 \cdot 10^{-4}$. How many nodes of this partition (inside the unit square) lie on the parabola $y=x^{2}$? | Answer: 49. Note. The nodes of the partition have coordinates of the form ( $i / 5000, j / 5000$ ), where $i$, $j=1,2, \ldots, 4999$. The condition that a given node lies on the parabola is $\frac{j}{5000}=\left(\frac{i}{5000}\right)^{2}$, i.e., $i^{2}=j \cdot 5^{4} \cdot 2^{3}$. Therefore, the number $i$ must have the... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. The bisectors of the external angles $B$ and $C$ of triangle $ABC$ intersect at point $M$. a) Can angle $BMC$ be obtuse? b) Find angle $BAC$ if it is known that $\angle BMC = \frac{\angle BAM}{2}$. | Answer: a) cannot; b) $120^{\circ}$. Hint. Let $\angle A=\alpha, \angle B=\beta, \angle C=\gamma$. Then $\angle B M C=180^{\circ}-\left(\frac{180^{\circ}-\beta}{2}+\frac{180^{\circ}-\gamma}{2}\right)=\frac{\beta+\gamma}{2}=90^{\circ}-\frac{\alpha}{2}$. a) Therefore, angle $B M C$ cannot be obtuse. b) from the equation ... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.5. Let $s(n)$ denote the sum of the digits of a natural number $n$. How many zeros does the number equal to the product $s(1) \cdot s(2) \cdot \ldots \cdot s(100)$ end with? | Answer: 19 zeros. Hint. Consider the numbers from the first hundred for which the sum of the digits is divisible by 5. Such numbers have a sum of digits of either 5, 10, or 15. There are 6 numbers with a sum of 5: these are $5,14,23,32,41,50$. There are 9 numbers with a sum of digits 10: these are $19,28, \ldots, 91$. ... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. A rectangular grid with a cell side of 1 cm and an area of $2021 \mathrm{~cm}^{2}$ is cut into four rectangular pieces by two perpendicular cuts along the grid lines. Prove that at least one of the pieces has an area of at least $528 \mathrm{~cm}^{2}$. | Solution. The prime divisors of the number 2021 are 43 and 47, and $2021=43 \cdot 47$. Therefore, the integer sides $a$ and $b$ of the original rectangle can be either 1) $a=2021, b=1$, or 2) $a=47, b=43$. However, it is obvious that case 1) $a=2021, b=1$ is impossible, since such a rectangle cannot be cut into square ... | 528 | Number Theory | proof | Yes | Yes | olympiads | false |
8.3. From points $A$ and $B$, two cyclists set off towards each other at the same time. They traveled at constant speeds. After meeting, the first cyclist took 40 minutes to reach point $B$, and the second cyclist took one and a half hours to reach point $A$. Find the time from the start of the journey until they met a... | Answer. Time until the meeting - 1 hour. The speed of the first cyclist is 1.5 times greater than the speed of the second. Solution. Let $v_{1}, v_{2}$ be the speeds of the cyclists, $t$ - the time until the meeting. Then the first cyclist traveled the distance $\mathrm{v}_{1} t$ before the meeting, and the second - th... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. On a chessboard, the centers of some cells are marked in such a way that no triangle with marked vertices is a right triangle. What is the maximum number of points that could have been marked? | Answer: 14 points. Solution. Let's call a marked point vertical if there are no other marked points on its vertical line. Similarly, we define a horizontal point. Note that any marked point is either vertical or horizontal (or both at the same time). Indeed, if the marked point $A$ had another point $B$ on its vertical... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 2. (10 points)
In a right triangle $A B C$ (angle $C$ is right), medians $A M$ and $B N$ are drawn, the lengths of which are 19 and 22, respectively. Find the length of the hypotenuse of this triangle. | # Solution.
Let $AC = 2x$ and $BC = 2y$. By the Pythagorean theorem, we have
$$
\left\{\begin{array}{c}
(2 x)^{2}+y^{2}=19^{2} \\
x^{2}+(2 y)^{2}=22^{2}
\end{array} \Rightarrow 5 x^{2}+5 y^... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 8. (16 points)
Let all companies in the country have a certain rank, which is a natural number. When two companies of ranks $m$ and $n$ merge, a new company of rank $(m+n)$ is formed. The profit of the resulting company will be $m \cdot n$ more than the sum of the profits of the companies forming it. The profit... | # Solution.
Let $p_{n}$ be the profit of the company of rank $n$. Then, according to the problem statement,
$$
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 |
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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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 |
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 |
# 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
$$
Answer. 673
| Criterion | Evaluation ... | 673 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (14 points)
The function $f(x)$ is such that $f(f(x))=x$ and $f(f(x+2)+2)=x$ for any $x$.
Find $f(2017)$, given that $f(0)=1$.
# | # Solution.
From the equality $f(x)=f(f(f(x+2)+2))=f(x+2)+2$, we obtain the formula $f(x+2)=f(x)-2$.
Moreover, $f(1)=f(f(0))=0$.
We will prove by induction that $f(x)=1-x$ for any integer $x$.
First, we will prove that the given equality holds for even $x$.
1) $f(0)=1$ - true.
2) Let $f(2n)=1-2n$.
3) We will prove... | -2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (10 points)
Six consecutive natural numbers from 10 to 15 are written in circles on the sides of a triangle in such a way that the sums of the three numbers on each side are equal.
What is the maximum value that this sum can take?
Egorov decided to open a savings account to buy a car worth 900000 rubles. The initial deposit amount is 300000 rubles. After one month and subsequently every month, Egorov plans to top up his account by 15000 rubles. The bank accrues interest monthly at an annual rate of $12 \%$. Interest accrue... | # Solution
Let $S_{n}$ be the sum of the deposit after $n$ months, after interest accrual and after making additional contributions $D (15000$ rubles).
Since the bank accrues $1 \%$ per month, then
$$
\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 |
# 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
$$
x_{n+2}=x_{n}-\frac{1}{x_{n+1}} \Leftrightarrow x_{n+2} \cdot x_{n+1}=x_{n+1} \cdot x_{n}-1 \Leftrightarrow x_{n+1} \cdot x_{n}=x_{n+2} \c... | 322 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 7. (14 points)
Several businessmen decided to open a company and divide all the profits into equal parts. One of the businessmen was appointed as the director. One day, this director of the company transferred part of the profit from the company's account to his own personal account. This amount of money was th... | # Solution.
Let $n$ be the number of businessmen and $d_{i}$ be the profit of the $i$-th director, $i=1, \ldots, n$.
By the condition $d_{i}=3 \frac{d_{i+1}+d_{i+2}+\ldots+d_{n}}{n-i}$. Then
$$
\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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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 |
# 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 |
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 |
# 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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
Problem 6. A row of new recruits stood facing the sergeant. On the command "left," some turned left, while the rest turned right. It turned out that six times more soldiers were looking at the back of their neighbor than in the face. Then, on the command "about face," everyone turned in the opposite direction. Now, sev... | Answer: 98.
Solution. Let's assume that the sergeant lined up the soldiers between two posts. After the first command, each recruit either looks at the back of the neck of the neighbor or at the face, except for the two soldiers at the ends, who can look at the posts.
If a soldier is looking at the back of the neck o... | 98 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 1. A number was multiplied by the sum of its digits and the result was 2008. Find this number.
$[4$ points] (I. V. Yashchenko) | Answer: 251.
Solution: The desired number is a divisor of the number 2008. Let's factorize the number 2008 into prime factors: $2008=2 \cdot 2 \cdot 2 \cdot 251$. List all divisors of the number $2008: 1,2,4,8,251,502,1004,2008$. By finding the sum of the digits of each of them, we notice that the condition of the pro... | 251 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. In quadrilateral $A B C D$, it is known that $A B=$ $=B C=C D, \angle A=70^{\circ}$ and $\angle B=100^{\circ}$. What can the angles $C$ and $D$ be equal to? $\quad[8$ points] (M.A.Volchkevich) | Answer: $60^{\circ}$ and $130^{\circ}$ or $140^{\circ}$ and $50^{\circ}$.
First Solution. Draw segment $B E$ such that point $E$ lies on $A D$, and angle $A B E$ is $40^{\circ}$. Then $\angle A E B = 180^{\circ} - 70^{\circ} - 40^{\circ} = 70^{\circ}$, hence triangle $A B E$ is isosceles, $A B = B E$. Consider triangl... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 1. The year 2009 has the following property: by swapping the digits of the number 2009, it is impossible to obtain a smaller four-digit number (numbers do not start with zero). In which year will this property reoccur for the first time? | Answer. In 2022.
Solution. In the years $2010, 2011, \ldots, 2019$ and in 2021, the year number contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be decreased by rearranging the digits. | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 5. A curious tourist wants to stroll through the streets of the Old Town from the station (point $A$ on the map) to their hotel (point $B$). The tourist wants their route to be as long as possible, but it is not interesting for them to visit the same intersection twice, and they do not do so. Draw the longest poss... | Solution. One of the possible routes of the tourist is shown in the figure. By following this path, the tourist will walk 34 streets (a street is defined as a segment between two adjacent intersections). We-
.
$[4$ points] (G. Galperin) | Answer: The sums are equal.
Solution. Let's write both sums in a column, and for better clarity, the second one in reverse order. In both sums, the digits from 1 to 9 will be added in the units place, the digits from 2 to 9 in the tens place, the digits from 3 to 9 in the hundreds place, and so on. The digit obtained ... | 1097393685 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The figure "violinist" attacks the cell to the left on the side (with the elbow) and the cell to the upper right diagonally (with the bow), if he is right-handed, and, conversely, the right cell on the side and the upper left cell diagonally, if he is left-handed (all violinists are facing us). Place as many... | Solution. Placing 32 violinists is not difficult: for example, you can fill four columns every other one (it doesn't matter whether they are right-handed or left-handed). However, it is possible to place more. An example with 34 violinists is shown in the figure.
| $\mathrm{p}$ | $\Lambda$ | | | $\mathrm{p}$ | $\Lam... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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. How many beautiful dates are there in 2023? 3 $[4$ points] (M. Evdokimov) | Answer: 30.
Solution: The digits 2 and 3 are already used in the year number, so we need to consider only the months 01, 04, 05, 06, 07, 08, 09, and 10. Each of these month numbers contains a 0, so in a beautiful date, there will be no day number starting with 0, 2, or 3, and there will also be no days 10, 11, 12, and... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In the morning, a dandelion blooms, it flowers yellow for two days, on the third day in the morning it turns white, and by evening it sheds its seeds. Yesterday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and today there are 15 yellow and 11 white.
a) How many yellow dandelions wer... | Answer. a) 25 yellow dandelions; b) 9 white dandelions.
Solution. a) All dandelions that were yellow the day before yesterday have turned white yesterday or today. Therefore, there were $14+11=25$.
b) Out of the yellow dandelions from yesterday, 11 turned white today, and the remaining $20-11=9$ will turn white tomor... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. An isosceles triangle $ABC (AB = BC)$ and a square $AKLM$ are positioned as shown in the figure. Point $S$ on $AB$ is such that $AS = SL$. Find the measure of angle $SLB$.
[8 points] (L. Ponov) | Answer: $90^{\circ}$.
Solution. Consider triangles $A K S$ and $L K S$. They are equal by three sides.
Therefore, angles $K A S$ and $K L S$ are equal.
In the isosceles triangle $A B C$, ang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. A monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines? Justify your answer. $[8$ points] (A.V. Shapovalov) | Answer: 45.
Solution. Let's set the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed the monkeys no more than one tangerine each, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. Let's show how 45 monkeys can be made happy:
5 mo... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. There is a set of two cards: 1 and 2. In one operation, it is allowed to form an expression using the numbers on the cards, arithmetic operations, and parentheses. If its value is a non-negative integer, it is issued on a new card. (For example, having cards 3, 5, and 7, you can form the expression 7 [ 5/3 a... | Answer. a) For example
$[1+2=3 ; \quad 3+2=5 ; \quad 3-2-11=0 ; \quad 201[5=2015$
or
$[1+2=3 ; \quad[1]=[13 ; \quad 3[1=31 ; \quad([2+3) \cdot 13 \cdot 31=2015$.
b) $[1+2=3 ; \quad 3 \cdot 2-1=63 ; \quad(63+2) \cdot 31=2015$.
Comments. 1. To solve the problem, it is useful to first factorize 2015 into prime factor... | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. A $3 \times 3$ square is filled with digits as shown in the figure on the left. It is allowed to move along the cells of this square, transitioning from one cell to an adjacent one (by side), but it is not allowed to visit any cell more than once.
| 1 | 8 | 4 |
| :--- | :--- | :--- |
| 6 | 3 | 9 |
| 5 | 7 | 2 ... | Answer. The largest number that can be obtained is -573618492 (see fig.).
Comments. 1. Let's explain how the problem could be solved (this was not required of the participants).
Notice that a number larger than the one given
| 1 | 8 | 4 |
| :---: | :---: | :---: |
| 6 | 3 | 9 |
| 5 | 7 | 2 |
in the problem can be o... | 573618492 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Having defeated Koschei, Ivan demanded gold to ransom Vasilisa from the bandits. Koschei led him to a cave and said:
"In the chest lie gold ingots. But you cannot simply take them: they are enchanted. Put one or several into your bag. Then I will put one or several from the bag back into the chest, but it m... | Answer. a) 13; b) 13.
Solution. Ivan will act in such a way that each time Kashchey's move will be the only one possible: all other numbers have either already appeared in previous moves or are too large - Ivan does not have that many ingots at that moment. We will record the moves of the game as follows: the number o... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 1. "This is for you to see a bit too early,"- said Baba Yaga to her 33 students and commanded: "Close your eyes!" The right eye was closed by all the boys and a third of the girls. The left eye was closed by all the girls and a third of the boys. How many students still saw what they were not supposed to see yet? ... | Answer: 22 students.
Solution: What is seen is still too early, two-thirds of the girls saw with their right eye, and two-thirds of the boys saw with their left eye. In total, then, one eye was not closed by two-thirds of all students - 22 people. | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. A wooden block was cut into eight smaller blocks with three cuts. On Fig. 5, the surface area of seven blocks is indicated. What is the surface area of the invisible block?
$[8$ points] (A. V. Shapovalov) | Answer: 22.
Solution: For each small block, the surface of the cuts constitutes half of its entire surface. We will only consider this. We will color the small blocks in black and white as shown in Fig. 6 (the invisible block is black). Then, every two identical rectangles touching on the cut are of different colors. ... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The magician taught Kashtanka to bark as many times as he showed her secretly. When Kashtanka correctly answered how much two times two is in this way, he hid a delicious cake in a suitcase with a combination lock and said:
- The eight-digit code for the suitcase is the solution to the puzzle УЧУЙ $=\kappa ... | Answer. a) No. b) Yes, УЧУЙ $=2021$.
Solution. a) Note that КЕ and КС represent different numbers, but swapping them does not change the product УЧУЙ. Therefore, for each solution to the puzzle, there is a paired solution where the digits corresponding to $\mathrm{E}$ and $\mathrm{C}$ are swapped. Thus, it is impossib... | 2021 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Inside a rectangular grid with a perimeter of 50 cells, a rectangular hole with a perimeter of 32 cells is cut out along the cell boundaries (the hole does not contain any boundary cells). If the figure is cut along all horizontal grid lines, 20 strips 1 cell wide will be obtained. How many strips will be ob... | Answer: 21.
First solution. Let the rectangle occupy $a$ cells vertically and $b$ horizontally, $a+b=50: 2=25$. Similarly, let the dimensions of the hole be $x$ cells vertically and $y$ horizontally, $x+y=32: 2=16$.
 | Answer: 33.
Solution: Among the numbers from 1 to 100, the digit 1 appears exactly twenty times: the number 1 itself, ten numbers from 10 to 19, the numbers $21, 31, \ldots, 91$ (eight of them), and the number 100. Therefore, none of these numbers were erased. Similarly, the digit 2 appears exactly nineteen times: the... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. A group of tourists is dividing cookies. If they evenly distribute two identical packs, one extra cookie will remain. But if they evenly distribute three such packs, 13 extra cookies will remain. How many tourists are in the group? [7 points] (I.V. Raskina) | Answer: 23.
First solution. Distribute three times two packs, 3 * 1 = 3 cookies will remain. But the same six packs of cookies can be distributed differently - three and another three, and then 2 * 13 = 26 cookies will remain. Therefore, 26 - 3 = 23 cookies can be divided equally among the tourists. Since the number 2... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Given a square $ABCD$. On the extension of the diagonal $AC$ beyond point $C$, a point $K$ is marked such that $BK=AC$. Find the angle $BKC$. $[6$ points] ( | Answer: $30^{\circ}$.
Solution: Since the picture is symmetric with respect to the line $A C$, we have $D K=B K$. By the condition, $B K=A C$.
And since the diagonals in a square are equal, $A C=B D$. Thus, in triangle $B K D$ all sides are equal, i.e., it is equilateral, and $\angle B K D=60^{\circ}$. Again, due to ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Find a six-digit number where the first digit is 6 times less than the sum of all digits to the right of it and the second digit is 6 times less than the sum of all digits to the right of it.
$[4$ points] (A. V. Shapovalov) | Answer: 769999.
First solution. The sum of the last five digits of the number, by condition, is divisible by 6 (the quotient is equal to the first digit). The sum of the last four digits of the number is also divisible by 6 (the quotient is equal to the second digit). Therefore, the second digit, as the difference of ... | 769999 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2.
Here is a fairly simple rebus: EKH is four times greater than OI. AI is four times greater than OH. Find the sum of all four.
[4 points | Answer: 150.
Solution: The digits Й and Х are even because ЭХ and АЙ are divisible by 4. The numbers ОЙ and ОХ are less than 25, otherwise, when multiplied by 4, they would no longer be two-digit numbers. Therefore, $\mathrm{O}=2$ or $\mathrm{O}=1$. Let's consider the first case. The number 20 as $\mathrm{OX}$ or ОЙ i... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A dog and a cat simultaneously grabbed a sausage loaf with their teeth from different sides. If the dog bites off its piece and runs away, the cat will get 300 g more than the dog. If the cat bites off its piece and runs away, the dog will get 500 g more than the cat. How much sausage will be left if both bi... | Answer: 400 g.
Solution: If two identical dogs grab the sausage from both sides, there will be a piece of 300 g between them. If two identical cats grab the sausage from both sides, there will be a piece of 500 g between them (see the figure).
 were hired to guard Lake Luka for 240 coins. The cunning Chernomor can divide the bogatyrs into squads of any size (or record all in one squad), and then distribute the entire salary among the squads. Each squad divides its coins equally, and the remainder goes to Cher... | Answer. a) 31 coins; b) 30 coins.
Solution. From each detachment of $N$ bogatyrs, Chernomor will receive at most $N-1$ coins in the best case, since the remainder is less than the divisor. Therefore, he will receive no more than $33-K$ coins in total, where $K$ is the number of detachments. Can Chernomor get 32 coins ... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
1. In Fedia the gardener's garden, there is a wonder-tree with seven branches. On each branch, either 6 apples, 5 pears, or 3 oranges can grow. Fedia noticed that there are fruits of all types on the tree, and the most pears have grown, while the fewest apples have grown.
How many fruits have grown on the wonder-tree?... | # Problem 1.
Answer: 30.
Since the tree ended up with fruits of all kinds, there must be at least one branch with apples, meaning there are no fewer than 6 apples. Since apples are the fewest, oranges must have grown on at least three branches, meaning there are at least 9.
There are three branches left. Suppose tha... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. On the side $AB$ of an equilateral triangle $ABC$, a point $K$ is chosen, and on the side $BC$, points $L$ and $M$ are chosen such that $KL = KM$, with point $L$ being closer to $B$ than $M$.
a) Find the angle $MKA$ if it is known that $\angle BKL = 10^{\circ}$.
b) Find $MC$ if $BL = 2$ and $KA = 3$.
Justify your... | # Problem 6.
Answer: a) $130^{\circ} ;$ b) 5 .
First solution. Mark a point $N$ on the segment $MC$ such that $NM = BL$. Since triangle $LKM$ is isosceles, the angles $KLM$ and $KML$ at its base are equal. Therefore, the adjacent angles $KLB$ and $KMN$ are also equal. This implies the equality of triangles $BKL$ and ... | 130 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A phone charges fully in 1 hour 20 minutes on fast charging, and in 4 hours on regular charging. Fedya first plugged in a completely discharged phone to regular charging, and then, when he found the necessary adapter, switched it to fast charging until it was fully charged. Find the total charging time of th... | Answer: 144 minutes = 2 hours 24 minutes.
Solution. Since the fast charging lasts 1 hour 20 minutes, which is 80 minutes, the phone charges by $1 / 80$ of the full charge per minute. For the regular charging, which lasts 4 hours, or 240 minutes, the phone charges by $1 / 240$ of the full charge per minute.
Let's deno... | 144 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2.
Here is a rather simple rebus: EKH is four times greater than OY. AY is four times greater than OH. Find the sum of all four.
$[4 points] | Answer: 150.
Solution: The digits Y and X are even because EH and AI are divisible by 4. The numbers OY and OX are less than 25, otherwise, when multiplied by 4, they will no longer be two-digit numbers. Therefore, O = 2 or O = 1. Let's consider the first case. The number 20 as OX or OY is not suitable because 80 ends... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. In the forest, there live 40 animals - foxes, wolves, hares, and badgers. Every year they organize a masquerade ball: each one wears a mask of another animal type, and they do not wear the same mask for two consecutive years. Two years ago, at the ball, there were 12 "foxes" and 28 "wolves", last year there ... | Answer. The most numerous are the badgers.
Solution. Let's record the data from the problem in a table.
| | "Wolves" | "Foxes" | "Rabbits" | "Badgers" |
| :--- | :---: | :---: | :---: | :---: |
| Two years ago | 28 | 12 | | |
| Last year | | 10 | 15 | 15 |
| This year | | 25 | 15 | |
Let's look at the "rabbits... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Vanya is coming up with a number consisting of non-repeating digits without zeros - a password for his phone. The password works as follows: if, without lifting his finger from the screen, he sequentially connects the points corresponding to the digits of the password with line segments, the phone will unloc... | Answer. For example, 12769. See the figure.
This password meets Vanya's requirements. Let's see how we can connect its digits without any intersections. The digit 7 must be connected to some digit, which can be either 2 or 6. Suppose, for example, we draw the segment $7-6$. Now 9 can only be connected to 6. Next, it i... | 12769 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. In a multicolored family, there were an equal number of white, blue, and striped octopus children. When some blue octopus children became striped, the father decided to count the children. There were 10 blue and white children in total, while white and striped children together amounted to 18. How many child... | Answer: 21.
First solution. Note that the white octopuses were one third of the total number, and they did not change color. If we add 10 and 18, we get the total number of all children, plus the number of white ones, which is $4 / 3$ of the total number of all children. Thus, $4 / 3$ of the number of children in the ... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. The figure "violinist" attacks the cell to the left on the side (with the elbow) and the cell to the upper right diagonally (with the bow), if he is right-handed, and, conversely, the right cell on the side and the upper left cell diagonally, if he is left-handed (all violinists are facing us). Place as many... | Solution. Placing 32 violinists is not difficult: for example, you can fill four columns every other one (it doesn't matter whether they are right-handed or left-handed). However, it is possible to place more. An example with 34 violinists is shown in the figure.
| $\mathrm{p}$ | $\Lambda$ | | | $\mathrm{p}$ | $\Lam... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Along the path between the houses of Nезнайка (Nезнayka) and Синеглазка (Sineglazka), there were 15 peonies and 15 tulips growing in a row, mixed together.
Setting out from home to visit Nезнайка, Синеглазка watered all the flowers in a row. After the 10th tulip, the water ran out, and 10 flowers remained u... | Answer: 19 flowers.
Solution: 10 flowers were left unwatered, which means $30-10=20$ flowers were watered. Consider the last flower watered by Blue-Eyes, which is a tulip. Since there are 15 tulips in total, there are $15-10=5$ tulips after this tulip.
Therefore, Nезнайка will pick these 5 tulips and finish picking f... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The inhabitants of the Island of Misfortune, like us, divide the day into several hours, an hour into several minutes, and a minute into several seconds. But they have 77 minutes in a day and 91 seconds in a minute. How many seconds are there in a day on the Island of Misfortune?
[5 points] (I. V. Raskina) | Answer: 1001.
Solution: If you divide 77 by the number of minutes in an hour, you get the number of hours in a day. If you divide 91 by the number of minutes in an hour, you get the number of seconds in a minute. Therefore, both 77 and 91 are divisible by the number of minutes in an hour. Since there are obviously mor... | 1001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. The cake is packed in a box with a square base. The height of the box is half the side of this square. A ribbon of length 156 cm can be used to tie the box and make a bow on top (as shown in the left image). To tie it with the same bow on the side (as shown in the right image), a ribbon of length 178 cm is n... | Answer: 22 cm $\times$ 22 cm $\times$ 11 cm.
Solution: In the first method of tying, the ribbon encircles the box twice along the length, twice along the width, and four times along the height, meaning its length is equal to six sides of the base plus the bow. In the second method of tying, the ribbon encircles the bo... | 22 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. First-year students admitted to the university were distributed into study groups so that each group had the same number of students. Due to a reduction in the number of specialties, the number of groups decreased by 9, and all first-year students were redistributed into groups; as a result, the groups were again eq... | Let the new number of groups be $n$, then initially there were ( $n+9$ ) groups. In each group, there was an equal number of students, so $2376: n$ and $2376:(n+9)$. We factorize 2376 into prime factors ( $2376=2^{3} \cdot 3^{3} \cdot 11$ ) and list all divisors: $1,2,4,8,3,6,12,24,9,18,36,72,27,54$, $108,216,11,22,44,... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the first thirty-three terms of an arithmetic progression, given that the sum of the seventh, eleventh, twentieth, and thirtieth terms of this progression is 28. | Answer. $S_{33}=231$.
Solution Let $a_{k}$ be the $k$-th term of the arithmetic progression, and $d$ be its common difference. Then, according to the condition, $a_{7}+a_{11}+a_{20}+a_{30}=28$, from which we have $a_{7}+\left(a_{7}+4 d\right)+\left(a_{7}+13 d\right)+\left(a_{7}+23 d\right)=28 \Leftrightarrow a_{7}+10 ... | 231 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. An infinite geometric progression consists of natural numbers. It turned out that the product of the first four terms equals param1. Find the number of such progressions.
The infinite geometric progression consists of positive integers. It turned out that the product of the first four terms equals param1. Find the ... | Solution
If $b_{1}$ is the first term of the progression and $q$ is its common ratio, then the product of the first four terms of the progression is $b_{1}^{4} q^{6}$. Therefore, $b_{1}^{2} q^{3}=2^{100} \cdot 3^{150}$. Hence, $b_{1}=2^{a} 3^{b}, q=2^{c} 3^{d}$, and we obtain the system: $2 a+3 c=100, 2 b+3 d=150$. Th... | 442 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a function $f: \sqcup \rightarrow \sqcup$ such that $f(1)=1$, and for any $x \in \sqcup, y \in \sqcup$ the equality $f(x)+f(y)+x y+1=f(x+y)$ holds. Find all integers $n$ for which the equality $f(n)=$ param 1 holds. In the answer, write down the sum of cubes of all such values of $n$.
Function $f: \sqcup \rig... | # Solution
Substituting $y=1$ into the given functional equation, we get: $f(x)+f(1)+x+1=f(x+1)$, which means $f(k)-f(k-1)=k+1$. Therefore, $f(n)-f(1)=(f(n)-f(n-1))+(f(n-1)-f(n-2))+\ldots$ $+f(2)-f(1)=n+1+n+\ldots+2$. Hence, $f(n)=\frac{n^{2}+3 n-2}{2}$. Reasoning similarly,
we get $f(n)=\frac{n^{2}+3 n-2}{2}$ for $n ... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. What is the smallest possible area of the pentagon $A B C D E$, given that the quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and... | # Solution
Let the areas of parallelograms ABPM and DCPN, triangle BCP, and quadrilateral MPNE be $S_{1}, S_{2}, S_{3}$, and $S_{4}$, respectively. According to the problem, $A M \sqcup B \mathcal{B}$ and $P N \sqcup C \mathcal{D}$. Therefore, $\mathrm{AE} \sqcup \mathcal{B} N \cup C D$. Similarly, $D E \sqcup C M \sq... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The clock hand points to 12. Jack writes a sequence consisting of param 1 symbols, each symbol being plus or minus. After that he gives this sequence to a robot. The robot reads it from right to left. If he sees a plus he turns the clock hand $120^{\circ}$ clockwise and if he sees a minus he turns it $120^{\circ}$ coun... | # Solution
Let $a_{n}$ be the number of sequences of length $n$ that result in the arrow pointing at 12 o'clock, and $b_{n}$ be the number of sequences of length $n$ that result in the arrow pointing at 4 or 8 o'clock. It is not difficult to understand that $a_{n+1}=2 b_{n}, b_{n+1}=a_{n}+b_{n}$. From this, we get tha... | 682 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits... | # Solution
Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, whi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
14. In a convex pentagon $A B C D E$, a point $M$ is taken on side $A E$, and a point $N$ is taken on side $D E$. Segments $C M$ and $B N$ intersect at point $P$. Find the area of pentagon $A B C D E$ if it is known that quadrilaterals $A B P M$ and $D C P N$ are parallelograms with areas param1 and param2, respectivel... | # Solution
Let the areas of parallelograms ABP M and DCPN, triangle BCP, and quadrilateral
$A E \sqcup B N N C D$. Similarly, $D E \sqcup C M \sqcup\{B$. Then M PNE is also a parallelogram,... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
17. The difference of the squares of two different real numbers is param 1 times greater than the difference of these numbers, and the difference of the cubes of these numbers is param 2 times greater than the difference of these numbers. By how many times is the difference of the fourth powers of these numbers greater... | # Solution
From the equalities $a^{2}-b^{2}=k(a-b)$ and $a^{3}-b^{3}=m(a-b)$ (in the problem $k=37, m=1069$), it follows that $a+b=k$ and $a^{2}+a b+b^{2}=m$. Squaring the first equality and subtracting the second from it, we get: $a b=k^{2}-m$. Then the ratio $\left(a^{4}-b^{4}\right):\left(a^{2}-b^{2}\right)=a^{2}+b... | 769 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Out of two hundred ninth-grade students, 80% received excellent grades on the first exam, 70% on the second exam, and 59% on the third exam. What is the smallest number of participants who could have received excellent grades on all three exams?
Answer: 18. | Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the problem,
$$
\left\{\begin... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Piglet ran down a moving escalator and counted 66 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 198 steps. How many steps would he have counted if he had gone down a stationary escalator?
Answer. 99. | Let \( u \) be the speed of Piglet, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Piglet spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v} \cdot u \) steps.... | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The remainder of dividing a certain natural number $n$ by 22 is 7, and the remainder of dividing $n$ by 33 is 18. Find the remainder of dividing $n$ by 66. | Answer: 51.
Solution: According to the condition $n=22l+7, \quad l \in Z$ and $n=33m+18, m \in Z$. By equating these two expressions, we get $22l+7=33m+18, 2l=3m+1$. Since the left side of the equation is even, the right side must also be divisible by 2, so $m$ is an odd number, i.e., $m=2q+1, q \in Z$. Then $n=33(2q+... | 51 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Out of three hundred eleventh-grade students, excellent and good grades were received by $77 \%$ on the first exam, $71 \%$ on the second exam, and $61 \%$ on the third exam. What is the smallest number of participants who could have received excellent and good grades on all three exams?
Answer: 27. | Let $M_{i}$ be the number of students who received excellent grades only on the $i$-th exam; $M_{ij}$ be the number of students who received excellent grades only on exams $i$ and $j$; $M_{123}$ be the number of students who received excellent grades on all three exams. Then, according to the conditions,
\[
\left\{\be... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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