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# 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 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.
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 11.3. (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 |
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 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 |
# 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 |
# 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 |
# 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 |
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 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 |
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 |
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 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 |
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 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: 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 |
Problem 2. The she-rabbit bought seven drums of different sizes and seven pairs of sticks of different lengths for her seven bunnies. If a bunny sees that both its drum is larger and its sticks are longer than those of one of its brothers, it starts to drum loudly. What is the maximum number of bunnies that can start d... | Answer: 6 baby rabbits.
Solution: Not all baby rabbits can play the drum, as the baby rabbit that gets the smallest drum will not play it. On the other hand, if the same baby rabbit is also given the shortest drumsticks, then all the other baby rabbits will play the drum. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. There were several whole cheese heads on the warehouse. At night, rats came and ate 10 heads, and everyone ate equally. Several rats got stomachaches from overeating. The remaining 7 rats the next night finished off the remaining cheese, but each rat could eat only half as much cheese as the night before. Ho... | Answer: 11 cheese heads
Solution: Let the total number of rats be $k$ ( $k>7$ ), then each rat ate $10 / k$ cheese heads on the first night. On the second night, each rat ate half as much, that is, $5 / k$ cheese heads. Thus, the 7 rats ate a total of $35 / k$ cheese heads. This is an integer. The only divisor of the ... | 11 | 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 3. Dima lives in a nine-story building. He descends from his floor to the first floor by elevator in 1 minute. Due to his small stature, Dima cannot reach the button for his floor. Therefore, when going up, he presses the button he can reach, and then walks the rest of the way. The entire journey upwards takes ... | Answer. Dima lives on the seventh floor.
First solution. Consider the part of the journey that Dima travels down by elevator and up on foot. On the one hand, the walk takes twice as long, and on the other, it is 10 seconds longer. Therefore, he traveled this part by elevator in 10 seconds and walked it in 20 seconds. ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Bus stop $B$ is located on a straight highway between stops $A$ and $C$. After some time following departure from $A$, the bus found itself at a point on the highway such that the distance from this point to one of the three stops is equal to the sum of the distances to the other two. After the same amount o... | Answer: 3 hours.
Solution. At both moments in time mentioned in the problem, the sum of distances will obviously be the distance from the bus to the farthest stop from it. This cannot be $B$, as it is closer than $C$. Therefore, these were $C$ (before the bus had traveled halfway from $A$ to $C$) and $A$ (after this m... | 3 | Algebra | 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 |
Problem 3. In the park, there were lindens and maples. Maples among them were $60 \%$. In spring, lindens were planted in the park, after which maples became $20 \%$. In autumn, maples were planted, and maples became $60 \%$ again. By what factor did the number of trees in the park increase over the year?
[6 points] (... | Answer: 6 times.
First solution. Before the planting, lindens constituted $2 / 5$, and maples $-3 / 5$ of all the trees in the park. By summer, the number of maples did not change, but they began to constitute $1 / 5$ of all the trees. Therefore, the total number of trees in the park increased threefold. At the same t... | 6 | Algebra | 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 1. Children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will have as many mushrooms as all the others combined. How many children went mushroom picking? | Answer: 6 children.
Solution: Let Anya give half of her mushrooms to Vitya. Now all the children have the same number of mushrooms (this means that Vitya did not have any mushrooms of his own). For Sanya to now get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushr... | 6 | Logic and Puzzles | 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. Mom baked pies - three with rice, three with cabbage, and one with cherries - and laid them out in a circle on a plate (see fig.). Then she put the plate in the microwave to warm them up. On the outside, all the pies look the same.
 were either seals or otters, and Igor remembered that 5 were either otters or sea l... | 5 | 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 2. In an aquarium, there are three types of fish: gold, silver, and red. If the cat eats all the gold fish, the number of fish will be 1 less than $2 / 3$ of the original number. If the cat eats all the red fish, the number of fish will be 4 more than $2 / 3$ of the original number. Which fish - gold or silver ... | Answer: There are 2 more silver fish.
Solution: From the first condition, there is 1 more goldfish than a third. From the second condition, there are 4 fewer red fish than a third. Therefore, there are 3 more silver fish than a third. | 2 | Algebra | 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 5. Given an equilateral triangle $ABC$. On side $AB$, point $K$ is marked, and on side $BC$ - points $L$ and $M$ ( $L$ lies on segment $BM$ ) such that $KL=KM, BL=2, AK=3$. Find $CM$.
[7 points]
. 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 3. Senya cannot write some letters and always makes mistakes in them. In the word TETRAHEDRON he would make five mistakes, in the word DODECAHEDRON - six, and in the word ICOSAHEDRON - seven. How many mistakes would he make in the word OCTAHEDRON?
. Assign (draw with arrows) several more one-way flights so that from any city to any other, one could get there with no more than two transfers. Try to make the number of additional flights as small as possible.
 | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Each face of a cube $6 \times 6 \times 6$ is divided into $1 \times 1$ cells. The cube is covered with $2 \times 2$ squares such that each square covers exactly four cells, no squares overlap, and each cell is covered by the same number of squares. What is the maximum value that this identical number can tak... | Answer: 3.
Solution: Estimation. A cell in the corner of a face can be covered in three ways (entirely within the face, with a fold over one edge of the corner, with a fold over the other edge of the corner). Therefore, each cell is covered by no more than three squares.
Example. Consider the usual covering of a cube... | 3 | Combinatorics | 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 |
4. Petya observes two ants crawling along a straight path at constant speeds. After 3 minutes of observation, the distance between the ants was 9 meters, after 5 minutes - 5 meters, after 9 minutes - 3 meters. What was the distance between the ants after 8 minutes of observation?
$[5$ points]
.
![](https://cdn.mathpix.com/cropped/2024_05_06_031fac0a0a7981eac22cg-04.jpg... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. 13 children sat at a round table and agreed that boys would lie to girls, but tell the truth to each other, and girls, on the contrary, would lie to boys, but tell the truth to each other. One of the children said to their right neighbor: "The majority of us are boys." The latter said to their right neighbor... | # Answer: 7.
Solution. It is clear that there were both boys and girls at the table. Let's see how the children were seated. After a group of boys sitting next to each other comes a group of girls, then boys again, then girls, and so on (a group can consist of just one person). Groups of boys and girls alternate, so t... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Small Island and Big Island have a rectangular shape and are divided into rectangular counties. In each county, a road is laid along one of the diagonals. On each island, these roads form a closed path that does not pass through any point more than once. Here is how Small Island is organized, with a total of... | Answer. Figure 2 provides an example for 9 counties.
Comment. We will show that examples do not exist for 7 counties (or fewer), while at the same time pointing out a property characteristic of all such examples.
All roads can be divided into two types: some roads connect the top-left corner of a county with the bott... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Thirty-three bogatyrs (Russian knights) 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 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:
a) in 5 or fewer;
b) in 4 or fewer;
.
Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)
Comment. It is impossible to paint all cells ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. A bag of sunflower seeds was passed around a table. The first person took 1 seed, the second took 2, the third took 3, and so on: each subsequent person took one more seed than the previous one. It is known that in the second round, the total number of seeds taken was 100 more than in the first round. How ma... | Answer: 10 people.
Solution: Let there be $n$ people sitting at the table. Then on the second round, the first person took the $n+1$-th sunflower seed, the second person took the $n+2$-th - and generally, each person took $n$ more seeds than on the first round. Altogether, on the second round, they took $n \cdot n=n^{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Three square paths with a common center are 1 m apart from each other (see figure). Three ants start simultaneously from the lower left corners of the paths and run at the same speed: Mu and Ra counterclockwise, and Wei clockwise. When Mu reaches the lower right corner of the largest path, the other two, who... | Answer: 4 m, 6 m, 8 m.
Solution: The lengths of the sides of two adjacent paths differ by 2 m (Fig. 3). Therefore, at the moment when Mu reached the corner, Ra had run 2 m along the right side of the path and was at a distance of $2+1=3$ m from the "lower" side of the outer path. Since $\mathrm{Pa}$ is halfway between... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Fox Alice and Cat Basil have grown 20 fake banknotes on a tree and are now filling in seven-digit numbers on them. Each banknote has 7 empty cells for digits. Basil calls out one digit at a time, either "1" or "2" (he doesn't know any others), and Alice writes the called digit in any free cell of any banknot... | Answer: 2.
Solution: Basil can always get two banknotes: he knows the place where the last digit should be written and names it so that it differs from the digit in the same place on some other banknote. Then the numbers on these two banknotes will be different, and the cat can take them.
We will show how Alice can e... | 2 | Combinatorics | 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 |
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.)
 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 6. On the sides \(AB\) and \(BC\) of square \(ABCD\) with side length 10, points \(K\) and \(L\) are marked such that \(AK = CL = 3\). On segment \(KL\), a point \(P\) is chosen, and on the extension of segment \(AB\) beyond point \(B\), a point \(Q\) is chosen such that \(AP = PQ = QL\) (see figure).
a) Prove... | Solution. Drop a perpendicular $PH$ from point $P$ to side $AB$. Since triangle $KBL$ is isosceles and right-angled, $\angle PKH = 45^\circ$. Therefore, triangle $KPH$ is also isosceles and $KH = HP$.
, while at the same time pointing out a property characteristic of all such examples.
All roads can be divided into two types: some roads connect the top-left corner of a county with the bott... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In the square, some cells are shaded as shown in the figure. It is allowed to fold the square along any grid line and then unfold it back. Cells that coincide with shaded cells when folded are also shaded. Can the entire square be shaded:
a) in 5 or fewer;
b) in 4 or fewer;
.
Solution. For example, it is possible to paint the entire lower half of the board with two vertical bends, after which the upper half can be painted with one horizontal bend - see the figure. (There are other solutions as well.)
Comment. It is impossible to paint all cells ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. The numbers $2,3,4, \ldots, 29,30$ are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? [6 points]
(I.V. Yashch... | Answer. For 5 rubles.
Solution. Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$),... | 5 | Number Theory | 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 4. A rectangular sheet of paper was folded, aligning a vertex with the midpoint of the opposite shorter side (Fig. 12). It turned out that triangles I and II are equal. Find the longer side of the rectangle if the shorter side is 8.
[6 points] (A. V. Khachatryan) | Answer: 12.
Fig. 12
Solution. Let's mark the equal segments (Fig. 13 - here we used the fact that in congruent triangles, sides opposite equal angles are equal). We see that the length of t... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Pete liked the puzzle, he decided to glue it together and hang it on the wall. In one minute, he glued together two pieces (initial or previously glued). As a result, the entire puzzle was joined into one complete picture in 2 hours. How long would it have taken to assemble the picture if Pete had glued togethe... | Answer: In one hour.
Solution 1. Each gluing reduces the number of pieces on the table by 1. Since after 120 gluings one piece (the complete puzzle) was obtained, there were 121 pieces at the beginning. Now, if three pieces are glued together per minute (i.e., the number of pieces is reduced by 2), one piece will rema... | 1 | 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 |
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