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4. Some squares of the table are mined. Each number written in a square shows the number of mines in the squares adjacent to the given square. (See the figure. Adjacent are squares that have at least one common point; a square with a number is not mined). In how many ways can the mines be placed in the table? Justify y... | # Answer: 14.
Solution. The mine adjacent to the number 1 is in the second, third, or fourth column.
1 case. If the mine is in the second column, it has three possible positions, and it is adjacent to the number two in the first column, which means the second mine adjacent to this two is in the first column, and it h... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. A family of four octopuses came to a shoe store (each octopus has 8 legs). The father octopus already had half of his legs booted, the mother octopus had only 3 legs booted, and their two sons had 6 legs booted each. How many boots did they buy if they left the store fully booted? | Answer: 13.
Solution. The daddy octopus had half of his legs booted, that is, 4 legs. Thus, 4 legs were not booted.
The mommy octopus had 3 legs booted, meaning 5 legs were not booted.
Each of the two sons had 6 legs booted, meaning 2 legs were not booted.
Thus, a total of $4+5+2+2=13$ boots were bought.
## Criter... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task No. 1.1
## Condition:
On the potion-making exam, each student at Hogwarts School had to brew 4 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same s... | # Answer: 36
## Exact match of the answer -1 point
## Solution.
Let's calculate the potion-making speed of each of the three students, measured in potions per hour. Hermione made 4 potions in half an hour, so in one hour she would make twice as many potions, which is 8. Therefore, her speed is 8 potions per hour. Ha... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.3
## Condition:
At the Potions exam, each student of Hogwarts School had to brew 8 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron needed 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the same speed... | Answer: 72
Exact match of the answer -1 point
Solution by analogy with task №1.1
# | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.4
## Condition:
At the potion-making exam, each student from Hogwarts School had to brew 10 potions. Hermione managed to complete the task in half an hour, Harry in 40 minutes, and Ron took 1 hour. How many potions would Ron, Hermione, and Harry brew together in 2 hours if they continued working at the sam... | Answer: 90
Exact match of the answer -1 point
Solution by analogy with task №1.1
# | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.1
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 5 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables i... | Answer: 43750000
Exact match of the answer -1 point
## Solution.
Note that the number of vowels cannot be more than half of the total number of letters in the word, that is, more than 4, since in each syllable with a vowel there is at least one consonant. Also, the number of vowels cannot be less than $1 / 3$ of the... | 43750000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.2
## Condition:
In a certain language $\mathrm{N}$, letters denote only 10 consonants and 8 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into syllables i... | Answer: 194560000
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 194560000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.4
## Condition:
In a certain language $\mathrm{N}$, letters denote a total of 20 consonants and 3 vowels. Syllables in this language are of two types: either "consonant + vowel" or "consonant + vowel + consonant". A word in language $\mathrm{N}$ is any sequence of letters that can be broken down into sylla... | Answer: 272160000
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 272160000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 8.1
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=74^{\circ}$. On ray $\mathrm{BC}$, segment $\mathrm{CD}=\mathrm{CB}$ is laid out. Then, from point $\mathrm{D}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, dr... | Answer: 37
## Exact match of the answer -1 point
## Solution.
Let $\mathrm{F}^{\prime}$ be the midpoint of segment $\mathrm{AC}$. Then $\mathrm{F}^{\prime} \mathrm{C}=0.5 \mathrm{AC}=\mathrm{BC}=\mathrm{CD}=0.5 \mathrm{BD}$. From this, it follows that triangle BDF' is a right triangle, since its median is equal to h... | 37 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 8.2
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=106^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CX}=$ CB is laid out. Then, from point $\mathrm{X}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, draw... | Answer: 53
Exact match of the answer -1 point
Solution by analogy with task №8.1
# | 53 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8.3
## Condition:
Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=46^{\circ}$. On the ray $\mathrm{BC}$, segment $\mathrm{CM}=$ CB is marked. Then, from point M, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\... | Answer: 23
Exact match of the answer - 1 point
Solution by analogy with task №8.1
# | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1.
In the number, two digits were swapped, and as a result, it increased by more than 3 times. The resulting number is 8453719. Find the original number. | Answer: 1453789.
Solution. When the number 8453719 is reduced by at least 3 times, the result is a number less than 3000000. Therefore, 8 was swapped with a digit less than 3, and the only such digit is 1. | 1453789 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 5. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the café was open for 18 days" and "from April 10 to April 30, the café was also open for 18 days." It is known that he was wrong once. How many days was the café open from April 1 to ... | Answer: 23
Solution: In the period from April 10 to April 30, there are exactly 21 days. Dividing this period into three weeks: from April 10 to April 16, from April 17 to April 23, and from April 24 to April 30, we get exactly one weekend (Monday) in each of the three weeks. Therefore, in the second statement, Kolya ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
On a grid paper, a right triangle with sides 6 and 10 is drawn (see figure). Find the total length of the horizontal grid lines inside this triangle.
 | Answer: 27.
Solution. Extend the triangle to form a rectangle.
It is divided into two identical triangles, so the total length of the horizontal lines in the rectangle is twice the desired ... | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1.
101 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there will be at least two even numbers. What is the minimum number of even numbers that can be among the written numbers? | Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is an even number. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there are at least two even numbers. Thus, the total number of even numbers is no less than $1+2 \cdot 20=41$. Such a situation is... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. In the castle, there are 25 identical square rooms arranged in a $5 \times 5$ square. In these rooms, 25 people—liars and knights (liars always lie, knights always tell the truth)—have settled, one person per room. Each of these 25 people said: "At least one of the rooms adjacent to mine is occupied by a liar." Wh... | Answer: 13 liars.
Solution. Note that liars cannot live in adjacent rooms (otherwise, they would be telling the truth). Let's select one corner room, and divide the remaining rooms into 12 pairs of adjacent rooms. Then, in each pair of rooms, there can be no more than one liar. Therefore, the total number of liars can... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. The magpie-crow was cooking porridge and feeding her chicks. The third chick received as much porridge as the first two combined. The fourth one received as much as the second and third. The fifth one received as much as the third and fourth. The sixth one received as much as the fourth and fifth. And the seventh ... | Answer: 40 g.
Solution. Let the first chick receive $m$ g of porridge, and the second chick $-n$ g. Then the third chick received $m+n$ (g), the fourth chick $-m+2n$ (g), the fifth chick $-2m+3n$ (g), and the sixth chick $-3m+5n$ (g). Therefore, the total amount of porridge was: $m+n+(m+n)+(m+2n)+(2m+3n)+(3m+5n)=8m+12... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. $A B C D$ - a convex quadrilateral. It is known that $\angle C A D=\angle D B A=40^{\circ}, \angle C A B=60^{\circ}$, $\angle C B D=20^{\circ}$. Find the angle $C D B$. | Answer: $30^{\circ}$.
Solution. Since $\angle C A B=60^{\circ}, \angle A B C=\angle A B D+\angle D B C=60^{\circ}$, triangle $A B C$ is equilateral (see Fig. 8.3a). We can reason in different ways from here.
 In a three-digit number, the first digit (hundreds place) was increased by 3, the second digit by 2, and the third digit by 1. As a result, the number increased by 4 times. Provide an example of such an original number.
Answer: 107. | Solution. The answer can be found in the following way. Let $x$ be the desired number. Then the condition of the problem immediately leads to the equation $x+321=4x$, the only solution of which is $x=107$.
Criteria. Correct answer, even without any comments: 7 points.
Incorrect answer: 0 points. | 107 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) A movie ticket cost 300 rubles. When the price was reduced, the number of visitors increased by 50 percent, and the cinema's revenue increased by 35 percent. How many rubles does one ticket cost now? | Answer: 270.
Solution. Let the price of the new ticket be $s$ rubles. Suppose the initial number of visitors was $N$, and after increasing by $50 \%$ it became $1.5 N$. Then, according to the condition, the current revenue of the cinema $1.5 N \cdot s$ is $35 \%$ more than $N \cdot 300$, from which we have $1.5 N s = ... | 270 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Given an arithmetic progression. The sum of its first 10 terms is 60, and the sum of its first 20 terms is 320. What can the 15th term of this progression be? | Answer: 25.
Solution. Let the first term of the sequence be $a$, and the common difference be $b$. Then the sum of the first 10 terms is $a+(a+b)+\ldots+(a+9b)=$ $10a+45b$. The sum of the first 20 terms is $a+(a+b)+\ldots+(a+19b)=$ $20a+190b$. According to the problem, $10a+45b=60, 20a+190b=320$. Solving the system, w... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. On a circle, 2012 points are marked, dividing it into equal arcs. From these, $k$ points are chosen and a convex $k$-gon is constructed with vertices at the chosen points. What is the largest $k$ for which it could turn out that this polygon has no parallel sides? | Answer. For $k=1509$.
Solution. Let $A_{1}, A_{2}, \ldots, A_{2012}$ be the marked points in the order of traversal (we will assume that $A_{2013}=A_{1}, A_{2014}=$ $A_{2}$). We will divide them into quadruples of points $\left(A_{1}, A_{2}, A_{1007}, A_{1008}\right)$, $\left(A_{3}, A_{4}, A_{1009}, A_{1010}\right), \... | 1509 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the problem: octopuses with an even number of legs always lie, while octopuses with an odd number of legs always tell the truth. Five octopuses met, each having between 7 and 9 legs.
The first said: "We have 36 legs in total";
The second: "We have 37 legs in total";
The third: "We have 38 legs in total";
T... | Solution. All answers are different, so one is telling the truth or all are lying. If all were lying, they would have 8 legs, i.e., a total of 40, which matches the last answer, a contradiction.
Therefore, four are lying, one is telling the truth, and the one telling the truth has an odd number of legs since $4 \cdot ... | 39 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Two cars started from the same point on a circular track 150 km long, heading in opposite directions. After some time, they met and continued moving in the same directions. Two hours after the start, the cars met for the second time. Find the speed of the second car if the speed of the first car is 60 km/h. | Solution. Let the speed of the first car be $x$ km/h, then their closing speed will be $(x+60)$ km/h. Obviously, in two hours they traveled two laps, i.e., 300 km, hence $2 \cdot(x+60)=300$, from which we get that $x=90($ km/h).
Answer: 90 km/h | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. At the railway platform, many people gathered early in the morning waiting for the train. On the first train, one-tenth of all those waiting left, on the second train, one-seventh of the remaining left, and on the third train, one-fifth of the remaining left. How many passengers were initially on the platform if 216... | Solution. Solving from the end, we get that before the departure of the third train, there were $216: 6 \cdot 5=270$ passengers on the platform, before the departure of the second - $270: 6 \cdot 7=315$ passengers, before the departure of the first - $315: 9 \cdot 10=350$ passengers.
Answer: 350 passengers. | 350 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. On the board, there are $n$ different integers. The product of the two largest is 77. The product of the two smallest is also 77. For what largest $n$ is this possible?
(R. Zhenodarov, jury) | Answer. For $n=17$.
Solution. The numbers $-11, -7, -6, -5, \ldots, 6, 7, 11$ provide an example for $n=17$.
Assume that there are at least 18 such numbers. Then, at least 9 of them will have the same sign (all positive or all negative). Among these 9 numbers, the absolute values of the two largest will be at least 8... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. Find any solution to the puzzle
$$
\overline{A B}+A \cdot \overline{C C C}=247
$$
where $A, B, C$ are three different non-zero digits; the notation $\overline{A B}$ represents a two-digit number composed of the digits $A$ and $B$; the notation $\overline{C C C}$ represents a three-digit number consisting... | Answer: 251.
Solution. The number $\overline{C C C}$ is divisible by 111 and is less than 247, so $A \cdot \overline{C C C}$ is either 111 or 222. In the first case, we get that $\overline{A B}=247-111=136$, which is impossible. In the second case, $\overline{A B}=247-222=25$, that is, $A=2, B=5$, and therefore, $C=1$... | 251 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Does there exist a four-digit natural number with distinct non-zero digits that has the following property: if this number is added to the same number written in reverse order, the result is divisible by $101$? | 1. Answer. It exists.
For example, the number 1234 works. Indeed, $1234+4321=5555=101 \cdot 55$. | 1234 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. Given a rectangle $A B C D$. A line passing through vertex $A$ and point $K$ on side $B C$ divides the entire rectangle into two parts, the area of one of which is 5 times smaller than the area of the other. Find the length of segment $K C$, if $A D=60$.
Fig. 1: to the solution of problem 8.2
Solution. Draw a line through $K$ parallel to $AB$. Let it intersect side $AD$ at point $L$ (Fig. 1), then $ABKL$ and $DCKL$ are rectangl... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In a row, there are 127 balls, each of which is either red, green, or blue. It is known that
- there is at least one red, at least one green, and at least one blue ball;
- to the left of each blue ball, there is a red ball;
- to the right of each green ball, there is a red ball.
(a) (1 point) What is the... | # Answer:
(a) (1 point) 125.
(b) (3 points) 43.
Solution. (a) Among 127 balls, there is at least 1 green and at least 1 blue, so there are no more than 125 red balls. Note also that there can be exactly 125 if the leftmost ball is blue, the rightmost ball is green, and all 125 balls between them are red.
(b) Suppos... | 125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given a right isosceles triangle $ABC$ with a right angle at $A$. A square $KLMN$ is positioned as shown in the figure: points $K, L, N$ lie on sides $AB, BC, AC$ respectively, and point $M$ is located inside triangle $ABC$.
Find the length of segment $AC$, if it is known that $AK=7, AN=3$.
Fig. 2: to the solution of problem 8.7
Solution. Mark a point $H$ on the segment $B K$ such that $L H \perp B K$ (Fig. 2). Triangle $B H L$ is a right isosceles triangle, $H B=... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. A biologist sequentially placed 150 beetles into ten jars. Moreover, in each subsequent jar, he placed more beetles than in the previous one. The number of beetles in the first jar is no less than half the number of beetles in the tenth jar. How many beetles are in the sixth jar? | Answer: 16.
Solution. If the first jar contains no less than 11 beetles, then the second jar contains no less than 12, the third jar no less than 13, ..., and the tenth jar no less than 20. And in all jars, there are no less than $11+12+\ldots+20=155$ beetles. Contradiction. Therefore, the first jar contains no more t... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 8.2 (7 points)
The shares of the company "Nu-i-Nu" increase in price by 10 percent every day. Businessman Borya bought shares of the company for 1000 rubles every day for three days in a row, and on the fourth day, he sold them all. How much money did he make from this operation? | Solution: $1000 \cdot 1.1^{3}+1000 \cdot 1.1^{2}+1000 \cdot 1.1-3 \cdot 1000=1331+1210+1100-3000=641$.
| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Correct approach with arithmetic error | 4 |
| Correct answer without justification | 0 |
## Answer: 641 rubles
# | 641 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8.5 (7 points)
Find the smallest natural number that is divisible by $48^{2}$ and contains only the digits 0 and 1. | Solution:
$48^{2}=2^{8} \cdot 3^{2}$. For a number to be divisible by $2^{8}$, it must end with at least 8 zeros according to the divisibility rule, because otherwise, with fewer zeros ($n \leq 7$), it would have the form $1 \ldots .1 \cdot 10^{\text {n }}$ and would only be divisible by the n-th power of two, but we ... | 11111111100000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. In a store, there are 20 items, the costs of which are different natural numbers from 1 to 20 rubles. The store has decided to have a promotion: when buying any 5 items, one of them is given as a gift, and the customer chooses which item to receive for free. Vlad wants to buy all 20 items in this store, pa... | Answer: 136.
Solution. Vlad can take advantage of the offer no more than 4 times, so he will get no more than 4 items for free. The total cost of these 4 items does not exceed $17+18+$ $19+20$ rubles. Therefore, the rubles Vlad needs are not less than
$$
(1+2+3+\ldots+20)-(17+18+19+20)=1+2+3+\ldots+16=\frac{16 \cdot ... | 136 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. Vanya thought of two natural numbers, the product of which equals 7200. What is the greatest value that the GCD of these numbers can take? | Answer: 60.
Solution. Since each of these numbers is divisible by their GCD, their product is divisible by the square of this GCD. The greatest exact square that divides the number $7200=2^{5} \cdot 3^{2} \cdot 5^{2}$ is $3600=\left(2^{2} \cdot 3 \cdot 5\right)^{2}$, so the GCD of the two numbers in question does not ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are whole numbers of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
. Each of them said one of two phrases:
- "Among those gathered, at least 5 liar... | Answer: 70.
Solution. Suppose there are at least 11 liars. Arrange the numbers on their T-shirts in ascending order and select the liar with the 6th number. Then he must be telling the truth, as there are at least 5 liars with a smaller number and at least 5 liars with a larger number. Thus, there are no more than 10 ... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. Given an obtuse triangle $ABC$ with an obtuse angle $C$. On its sides $AB$ and $BC$, points $P$ and $Q$ are marked such that $\angle ACP = CPQ = 90^\circ$. Find the length of the segment $PQ$, if it is known that $AC = 25, CP = 20, \angle APC = \angle A + \angle B$.
$, whose leading coefficient is 1. On the graph of $y=P(x)$, two points with abscissas 10 and 30 are marked. It turns out that the bisector of the first quadrant of the coordinate plane intersects the segment between them at its midpoint. Find $P(20)$. | Answer: -80.
Solution. The midpoint of this segment has coordinates $\left(\frac{10+30}{2}, \frac{P(10)+P(30)}{2}\right)$. Since it lies on the bisector of the first quadrant, i.e., on the line $y=x$, these coordinates are equal. From this, we get $P(10)+P(30)=40$.
Since $P(x)$ is a monic polynomial, it can be writte... | -80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. In an $8 \times 12$ table, some $N$ cells are black, and the rest are white. In one operation, it is allowed to paint three cells forming a three-cell corner to white (some of them could already be white before repainting). It turned out that it is impossible to make the entire table white in fewer than 25... | Answer: 27.
Solution. Divide the $8 \times 12$ table into 24 squares of $2 \times 2$ (Fig. 9a).
(a)
 At a certain moment, Anya measured the angle between the hour and minute hands of her clock. Exactly one hour later, she measured the angle between the hands again. The angle turned out to be the same. What could this angle be? (Consider all cases.) | Answer: $15^{\circ}$ or $165^{\circ}$.
Solution. After 1 hour, the minute hand remains in its original position. During this time, the hour hand has turned $30^{\circ}$. Since the angle has not changed, the minute hand must bisect one of the angles between the positions of the hour hand (either the $30^{\circ}$ angle ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Determine the number of points at which 10 lines intersect, given that only two of them are parallel and exactly three of these lines intersect at one point. | Answer: 42.
Solution. Let's number the lines so that lines 1, 2, and 3 intersect at one point (denote this point as $X$). List all possible pairs of lines (1 and 2, 1 and 3, 1 and $4, \ldots, 8$ and 9, 8 and 10, 9 and 10) and their points of intersection. There are a total of 45 pairs of lines (there are 9 pairs of th... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Let's call a three-digit number interesting if at least one of its digits is divisible by 3. What is the maximum number of consecutive interesting numbers that can exist? (Provide an example and prove that it is impossible to have more consecutive numbers.)
| Answer: 122.
Solution. The numbers $289,290, \ldots, 299,300, \ldots, 399,400, \ldots, 409,410$ are interesting (recall that 0 is divisible by 3), and there are 122 of them in total. Let's prove that a larger number is not possible.
Suppose we managed to find a larger number of consecutive interesting numbers; choose... | 122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. In a country, there are 100 cities. Between any two cities, there is either no connection, or there is an air route, or there is a railway (both air routes and railways cannot exist simultaneously). It is known that if two cities are connected to a third city by railway, then there is an air route between th... | Solution. Note that no city is connected by railway to more than two other cities. Indeed, suppose some three cities are connected by railway to one. Then all of them are connected to each other by air routes, which is impossible by the condition of the problem. Therefore, each city is connected by railway to no more t... | 20 | Combinatorics | proof | Yes | Yes | olympiads | false |
Problem 9.3. Four cities and five roads are arranged as shown in the figure. The lengths of all roads are equal to an integer number of kilometers. The lengths of four roads are indicated in the figure. How many kilometers is the length of the remaining one?
 | 4. Answer: 25 numbers. Solution: let the record of a superclass number include digits a, b, c in some order. Then, two equalities are satisfied: $2 \mathrm{a}=\mathrm{bc} 2 \mathrm{~b}=\mathrm{ac}$. If the numbers a and b are different, then by swapping them, we will violate the existing equality. Therefore, $\mathrm{a... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1.
In the figure, an example is given of how 3 rays divide the plane into 3 parts. Into what maximum number of parts can 11 rays divide the plane?
 | Answer: 56.
## Solution.
If the $(n+1)$-th ray is drawn so that it intersects all $n$ previous rays, then $n$ intersection points on it divide it into $n+1$ segments. The segment closest to the vertex does not add new parts to the partition, while each of the other $n$ segments ( $n-1$ segments and 1 ray) divides som... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1.
In the "Triangle" cinema, the seats are arranged in a triangular shape: in the first row, there is one seat with number 1, in the second row - seats with numbers 2 and 3, in the third row - 4, 5, 6, and so on (the figure shows an example of such a triangular hall with 45 seats). The best seat in the cine... | Answer: 1035
Solution.
1st method.
Note that the number of rows in the cinema cannot be even, otherwise there would be no best seat. Let the total number of rows in the cinema be $2 n+1$, then the best seat is in the $n+1$ row. If we remove this row, the triangle can be divided into 4 parts, and the number of seats ... | 1035 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4. In triangle $A B C$, points $D$ and $F$ are marked on sides $A B$ and $A C$ respectively such that lines $D C$ and $B F$ are perpendicular to each other and intersect at point $E$ inside triangle $A B C$. It turns out that $A D=D C$ and $D E \cdot E C=F E \cdot E B$. What degree measure can angle $B A C$ have? (7... | # Solution:
$D E \cdot E C=F E \cdot E B \Rightarrow \frac{D E}{E F}=\frac{E B}{E C}, \angle D E B=\angle F E C=90^{\circ}$, so triangles $\triangle D E B$ and $\triangle F E C$ are similar.... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. On an $8 \times 8$ battleship game field, a "piglet" figure is placed. What is the minimum number of shots needed to definitely hit one "piglet"? (7 points)
# | # Solution:
## Evaluation:
Divide the board into rectangles $4 \times 2$. To ensure that a piglet cannot be placed inside one such figure, at least two shots are required. There are 8 such ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Let $O$ be the center of the circumcircle of triangle $ABC$, points $O$ and $B$ lie on opposite sides of line $AC$, $\angle AOC = 60^\circ$. Find the angle $AMC$, where $M$ is the center of the incircle of triangle $ABC$. | # Solution.
Since points $O$ and $B$ lie on opposite sides of line $A C$, the degree measure of arc $A C$, not containing point $B$, is $360^{\circ}-60^{\circ}=300^{\circ}$. Therefore,
$\angle A B C=(1 / 2) \cdot 300^{\circ}=150^{\circ}$. The sum of the angles at vertices $A$ and $C$ of triangle $A B C$ is $180^{\circ... | 165 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. On a chessboard of size $2015 \times 2015$, dominoes are placed. It is known that in every row and every column there is a cell covered by a domino. What is the minimum number of dominoes required for this? | Solution Let's start filling the 2015 × 2015 square with dominoes starting from the lower left corner (see figure). The first domino will be placed horizontally, covering two vertical lines and one horizontal line. The next domino will be placed vertically, covering two horizontal lines and one vertical line. In total,... | 1344 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Anya and Danya together weigh 82 kg, Danya and Tanya - 74 kg, Tanya and Vanya - 75 kg, Vanya and Manya - 65 kg, Manya and Anya - 62 kg. Who is the heaviest and how much does he/she weigh? | Answer. Vanya weighs 43 kg.
Solution. Adding the weights given in the condition: $82+74+75+65+62=358$, we get the doubled weight of all the children. That is, all the children together weigh $358 / 2=179$.
Anya, Danya, Tanya, and Vanya together weigh $82+75=157$, so Manya weighs 179 $157=22$.
Similarly, we find that... | 43 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.3
In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$.
## Number of points 7 | Answer:
$30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
Figure a
. How many minutes will it take him to trace all the lines of a checkered square $5 \times 5$? | Answer: 30 minutes.
Solution. Note that the grid rectangle $3 \times 7$ consists of four horizontal lines of length 7 and eight vertical lines of length 3. Then the total length of all lines is $4 \cdot 7 + 8 \cdot 3 = 52$. It turns out that Igor spends $26 : 52 = 1 / 2$ minutes on the side of one cell.
The rectangle... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.4. Karlson counts 200 buns baked by Fräulein Bock: «one, two, three, ..., one hundred and nine, one hundred and ten, two hundred». How many words will he say in total? (Each word is counted as many times as it was said.) | Answer. 443 words.
Solution. One word is required to pronounce 29 numbers: $1,2,3,4,5,6,7,8,9,10,11,12$, $13,14,15,16,17,18,19,20,30,40,50,60,70,80,90,100,200$.
Among the first 99 numbers, the number of those pronounced in two words: $99-27=72$, thus, the number of words required for their pronunciation is $2 \cdot 7... | 443 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Ivan and Petr are running in the same direction on circular tracks with a common center, and initially, they are at the minimum distance from each other. Ivan completes one full circle every 20 seconds, while Petr completes one full circle every 28 seconds. After what least amount of time will they be at the maximum... | Answer: 35 seconds.
Solution. Ivan and Petr will be at the minimum distance from each other at the starting points after the LCM $(20,28)=140$ seconds. In this time, Ivan will complete 7 laps, and Petr will complete 5 laps relative to the starting point. Consider this movement in a reference frame where Petr is statio... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. Find the number of three-digit numbers for which the second digit is less than the third by 3. | 7.1. The first digit of the number can be chosen in 9 ways (it can be any digit from 1 to 9), the second digit in 7 ways (it can be any digit from 0 to 6), and the third digit is uniquely determined. We get $9 \cdot 7 \cdot 1=63$ three-digit numbers that satisfy the condition of the problem.
Answer: 63 | 63 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. A plot of $80 \times 50$ meters is allocated for gardens and is fenced on the outside. How should 5 straight fences of the same length be installed inside the plot to divide it into 5 rectangular plots of equal area? | 7.3. One of the possible solutions is shown in the figure. The fences have a length of 40 m, their ends are marked with bold dots (the horizontal segment is composed of two fences).
 | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.4. The 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? | 7.4. Let's set the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed each monkey no more than one tangerine, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. We will show how to satisfy 45 monkeys:
5 monkeys eat a pear, a banana, ... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. Zya decided to buy a crumblik. In the store, they also sold kryambliks. Zya bought a kryamblik and received coupons worth $50\%$ of the cost of the purchased kryamblik. With these coupons, he was able to pay $20\%$ of the cost of the crumblik. After paying the remaining amount, he bought the crumblik as well. By w... | 7.5. From the condition of the problem, it follows that $50 \%$ of the cost of a kryamblik is equal to $20 \%$ of the cost of a krumblik. This means that the kryamblik constitutes $40 \%$ of the cost of the krumblik. Zya paid the full cost of the kryamblik and the remaining $80 \%$ of the cost of the krumblik. Thus, he... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. What is the sum of the digits of the number $A=100^{40}-100^{30}+100^{20}-100^{10}+1$? | # Answer: 361.
Solution. The number is the sum of three numbers: a number composed of 20 nines followed by 60 zeros, a number composed of 20 nines followed by 20 zeros, and finally the number 1. All nines and the one fall on the zeros of the other addends, so there is no carry-over, and the answer is $180+180+1=361$.
... | 361 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. Toshi is traveling from point A to point B via point C. From A to C, Toshi travels at an average speed of 75 km/h, and from C to B, Toshi travels at an average speed of 145 km/h. The entire journey from A to B took Toshi 4 hours and 48 minutes. The next day, Toshi travels back at an average speed of 100 km/h. The ... | Answer: 290 km.
Solution. Let x km be the distance between B and C, and y km be the distance between A and C. From the system of equations $x / 145 + y / 75 = 24 / 5$ and $(x + y) / 100 = 2 + y / 70$, we find: $x = 290$ and $y = 210$.
![](https://cdn.mathpix.com/cropped/2024_05_06_e20f1a290a060f9ea344g-1.jpg?height=4... | 290 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. Mother gives pocket money to her children: 1 ruble to Anya, 2 rubles to Borya, 3 rubles to Vitya, then 4 rubles to Anya, 5 rubles to Borya and so on until she gives 202 rubles to Anya, and 203 rubles to Borya. By how many rubles will Anya receive more than Vitya? | # Answer: 68
Solution. Note that Anya will receive money one more time than Vitya. If we remove the first ruble, then each subsequent time Anya receives one more ruble than Vitya. Thus, it remains to determine how many such times there were. Apart from 1 ruble and 2 rubles, all the "moves" by mom break down into tripl... | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. In a hotel, rooms are arranged in a row in order from 1 to 10000. Masha and Alina checked into the hotel in two different rooms. The sum of the room numbers they are staying in is 2022, and the sum of the room numbers of all the rooms between them is 3033. In which room is Masha staying, if her room has a lower nu... | Answer: 1009
Solution. If there is exactly one room between Masha's and Alina's rooms, then it is room number 3033, but then Masha's and Alina's rooms are 3032 and 3034, and their sum is not 2022.
Consider room B, whose number is 1 more than Masha's room. Also consider room V, whose number is 1 less than Alina's room... | 1009 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a batch, there are 90000 boxes of vaccine weighing 3300 grams each, and 5000 small boxes weighing 200 grams each. What is the minimum number of temperature-controlled containers needed if, according to the new rules, no more than 100 kilograms can be placed in each? | Answer: 3000
Solution. Note that no more than 30 boxes of vaccines can be placed in one container. Thus, less than $90000 / 30=3000$ containers would not be enough. Also, note that if a container holds 30 boxes, there is still room for 5 small boxes, so all these small boxes will fit into the free spaces. | 3000 | Other | math-word-problem | Yes | Yes | olympiads | false |
7.1. Ten guides were leading ten tour groups through the halls of the Hermitage. The groups did not necessarily have equal numbers, but on average, there were 9 people in each group. When one of the groups finished their tour, the average number of tourists per group decreased to 8 people. How many people were in the g... | Answer: 18.
Solution: The total number of excursionists initially was $9 \cdot 10=90$. After one of the groups finished the excursion, $8 \cdot 9=72$ excursionists remained. Therefore, the group that finished the excursion had $90-72=18$ excursionists.
Comment: A correct answer without justification - 0 points. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. In the school gym, there is one arm wrestling table. The physical education teacher organized a school tournament. He calls any two participants who have not yet met each other for a match. There are no draws. If a participant loses twice, they are eliminated from the tournament. After 29 matches were held, all pa... | Answer: 16.
Solution: Each participant is eliminated after exactly two losses. In the situation where two "finalists" remain, the total number of losses is 29. If $n$ people have been eliminated from the tournament, then they have collectively suffered $2 n$ losses, while the "finalists" could have $0(0+0), 1(0+1)$ or... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. Four pirates divided a treasure of 100 coins. It is known that among them, there are exactly two liars (who always lie) and exactly two knights (who always tell the truth).
They said:
First pirate: “We divided the coins equally.”
Second pirate: “Everyone has a different number of coins, but each got at least 15... | Answer: 40 coins
Solution. Note that the first and fourth pirates could not have told the truth simultaneously.
If the first pirate is a knight, then everyone received 25 coins (such a situation is possible). In this case, the maximum number is 25.
If the fourth pirate is a knight, then each received no more than 35... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. One hundred non-zero integers are written in a circle such that each number is greater than the product of the two numbers following it in a clockwise direction. What is the maximum number of positive numbers that can be among these 100 written numbers? | Answer: 50.
Solution: Note that two consecutive numbers cannot both be positive (i.e., natural numbers). Suppose the opposite. Then their product is positive, and the number before them (counterclockwise) is also a natural number. Since it is greater than the product of these two natural numbers, it is greater than ea... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Sergey wrote down the numbers from 500 to 1499 in a row in some order. Under each number, except the leftmost one, he wrote the GCD of that number and its left neighbor, obtaining a second row of 999 numbers. Then he applied the same rule to get a third row of 998 numbers, from it a fourth row of 997 numbers, and so... | Answer: 501 (in the second variant 10001).
Let's prove that all numbers in the 501st row are already equal to 1. Indeed, if there is a number $d \neq 1$ in it, then in the 500th row there are 2 numbers divisible by $d$, in the 449th row - 3 such numbers, ..., in the 1st row there are 501 such numbers. But no number $d... | 501 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a regular 20-gon, four consecutive vertices $A, B, C$ and $D$ are marked. Inside it, a point $E$ is chosen such that $A E=D E$ and $\angle B E C=2 \angle C E D$. Find the angle $A E B$. | Answer: $39^{\circ}$ (in the $2^{nd}$ variant: $36^{\circ}$).
Note that $ABCD$ is an isosceles trapezoid with angles $\angle ABC = \angle DBC = 180^{\circ} \cdot 18 / 20 = 162^{\circ}$. Point E lies on the perpendicular bisector of the base $AC$, and therefore, triangle $BEC$ is isosceles. Draw the height $EH$ in it, ... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The exam consists of $N \geqslant 3000$ questions. Each of the 31 students has learned exactly 3000 of them, and every question is known by at least 29 students. Before the exam, the teacher openly laid out all the question cards in a circle. He asked the students to point to one of the questions and explained that ... | Answer: $N=3100$.
Each student does not know $N-3000$ questions and thus can mentally mark exactly that many cards which do not suit them as the initial one. Together, they can indicate no more than $31(N-3000)$ different cards. If $31(N-3000)<N$, then the students can indicate a card that suits everyone. This inequal... | 3100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Ten different natural numbers are such that the product of any 5 of them is even, and the sum of all 10 numbers is odd. What is their smallest possible sum | Answer: 65.
Solution. The product of 5 odd numbers is odd $\Rightarrow$ among the given ten numbers, there are no more than 4 odd numbers. 4 odd numbers are also impossible, since then the sum of all numbers would be even, while according to the condition, it is odd. Therefore, there are a maximum of 3 odd numbers. If... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. There were 1009 gnomes with 2017 cards, numbered from 1 to 2017. Ori had one card, and each of the other gnomes had two. All the gnomes knew only the numbers on their own cards. Each gnome, except Ori, said: "I am sure that I cannot give Ori any of my cards so that the sum of the numbers on his two cards would be 20... | Answer: 1009
Solution: If one of the gnomes has a card with the number 1, then he must be sure that Ori does not have a card with the number 2017. This can only be certain if the card with the number 2017 is also in the hands of this gnome. Similarly, the gnome with the card that has the number 2 has another card with... | 1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Vanya wrote down a four-digit number, subtracted a two-digit number from it, multiplied the result by a two-digit number, divided by the sum of two single-digit numbers, added a single-digit number, and then divided the result by the sum of three single-digit numbers. To write all the numbers, he used only one digit... | Answer: 2017; any digit.
Solution. Let the digit be $a$. We get $\left(\frac{(\overline{a a a a}-\overline{a \bar{a}} \cdot \cdot \overline{a a}}{a+a}+a\right):(a+a+a)$. Then
$$
\overline{a a a a}-\overline{a a}=\overline{a a 00}=a \cdot 1100 \Rightarrow \frac{\bar{a}}{a a}=a \cdot 11
$$
The numerator of the fractio... | 2017 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The giants were prepared 813 burgers, among which are cheeseburgers, hamburgers, fishburgers, and chickenburgers. If three of them start eating cheeseburgers, then in that time two giants will eat all the hamburgers. If five take on eating hamburgers, then in that time six giants will eat all the fishburgers. If sev... | Answer: 252 fishburgers, 36 chickenburgers, 210 hamburgers, and 315 cheeseburgers.
Solution: Let $a, b, c$, and $d$ be the quantities of cheeseburgers, hamburgers, fishburgers, and chickenburgers, respectively. According to the problem, $a + b + c + d = 813$. The statement that while three people eat cheeseburgers, tw... | 252 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
![](https://cdn.mathpix.co... | Answer: 22.
Solution. Note that there are a total of 91 coins, so after all the moves, each box should have exactly 13 coins. At least 7 coins need to be moved from the box with 20 coins. Now consider the boxes adjacent to the box with 20 coins. Initially, they have a total of 25 coins, and at least 7 more coins will ... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $ABCD$, it is known that $\angle BAC = \angle CAD = 60^{\circ}$, $AB + AD = AC$. It is also known that $\angle ACD = 23^{\circ}$. How many degrees does the angle $ABC$ measure?
$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
 according to the following rules. In the top row, there should be 2019 real numbers, none of which are equal, and in the bottom row, there should be the same 2019 numbers, but in a different order. In each of the 20... | Answer: 2016.
Solution. Estimation. We will prove that in the first row of the table, where numbers are arranged according to the rules, there are no fewer than three rational numbers (and, accordingly, no more than 2016 irrational numbers). Each number appearing in the table is written in exactly two cells, one of wh... | 2016 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Petya was exchanging stickers. He trades one sticker for 5 others. At first, he had 1 sticker. How many stickers will he have after 50 exchanges? | Answer: 201.
Solution: After each exchange, the number of Petya's stickers increases by 4 (one sticker disappears and 5 new ones appear). After 50 exchanges, the number of stickers will increase by 50*4=200. Initially, Petya had one sticker, so after 50 exchanges, he will have $1+200=201$. | 201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Cut a square with a side of 4 into rectangles, the sum of the perimeters of which is 25. | For example, two rectangles $2 \times 0.5$ and one rectangle $3.5 \times 4-$ cm. The total perimeter is $2 * 2 * (2 + 0.5) + 2 * (3.5 + 4) = 25$.
 | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Someone wrote down two numbers $5^{2020}$ and $2^{2020}$ in a row. How many digits will the resulting number contain? | Solution. Let the number $2^{2020}$ contain $m$ digits, and the number $5^{2020}$ contain $n$ digits. Then the following inequalities hold: $10^{m-1}<2^{2020}<10^{m}, 10^{n-1}<5^{2020}<10^{n}$ (the inequalities are strict because the power of two or five is not equal to the power of ten). Multiplying these inequalities... | 2021 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The older brother noticed that in 10 years, the younger brother will be as old as he is now, and his age will be twice the current age of the younger brother. How old is the younger brother now? | Answer: 20 years.
Solution. Let the current age of the younger brother be $x$ years, and the older brother be $y$ years. In 10 years, the younger brother's age will be $y$ years, and the older brother's age will be $2x$ years. Since the age of each has changed by 10 years, we have the equations: $y+10=2x, x+10=y$. By ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A rectangle $10 \times 20$ is divided into unit squares. How many triangles are formed after drawing one diagonal?
 | Answer: 220.
Solution. The figure shows one of the obtainable triangles. All such triangles are right-angled, and the vertex of the right angle can be any lattice node,
except those lying on the diagonal. There are a total of $21 \times 11$ nodes, and 11 of them are on the
diagonal, so the number of triangles is $2... | 220 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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