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Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.4. Misha made himself a homemade dartboard during the summer at the cottage. The round board is divided by circles into several sectors - darts can be thrown into it. Points are awarded for hitting a sector as indicated on the board.
Misha threw 8 darts 3 times. The second time he scored twice as many points... | Answer: 48.
Solution. The smallest possible score that can be achieved with eight darts is $3 \cdot 8=24$. Then, the second time, Misha scored no less than $24 \cdot 2=48$ points, and the third time, no less than $48 \cdot 1.5=72$.
On the other hand, $72=9 \cdot 8$ is the highest possible score that can be achieved w... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. In the chat of students from one of the schools, a vote was held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, given that $A K=4, B L=31, M C=3$.
. Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
. For what maximum number of pira... | Solution. We will show that if there are no more than 11 pirates, Silver can divide the coins as he wishes. Indeed, let the $i$-th pirate need to receive $S_{i}$ coins. $S_{i}=5 x_{i}+a_{i}$, for some integers $x_{i}$ and $a_{i}$ (where $a_{i}$ is the remainder of the division of $S_{i}$ by 5). Note that the sum of all... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.5. Uncle Chernomor assigns 9 or 10 of his thirty-three bogatyrs (knights) to duty each evening. What is the smallest number of days after which it can happen that all bogatyrs have been on duty the same number of times? | # 10.5. 7.
Let \( m \) and \( n \) be the number of days when 9 and 10 heroes were on duty, respectively. Let \( k \) be the number of days each hero was on duty. Then \( 9m + 10n = 33k \).... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.7. In a white $2016 \times 2016$ table, some cells are painted black. We will call a natural number $k$ successful if $k \leqslant 2016$, and in each of the grid squares with side $k$ located in the table, exactly $k$ cells are painted. (For example, if all cells are black, then only the number 1 is successful.) What... | Answer: 1008 numbers.
Solution. Consider an arbitrary coloring of the table. Let there be at least two successful numbers, and let $a$ be the smallest of them, and $b$ be the largest.
Divide $b$ by $a$ with a remainder: $b = qa + r$, where $0 \leq r < a$. If $r \neq 0$, then the number $qa$ is also successful, and $a... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.6. Ten-digit natural numbers $a, b, c$ are such that $a+b=c$. What is the maximum number of their 30 digits that can be odd?
(I. Bogdanov)
# | # Answer. 29.
Solution. Note that if $a+b=c$, then all three numbers $a, b$, $c$ cannot be odd simultaneously. Therefore, among them, there is at least one even number, and the last digit of this number will also be even. Thus, among the 30 digits, there is at least one even digit, and the number of odd digits is no m... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.8. Given a $1000 \times 1000$ chessboard. A figure called a cheetah, from any cell $x$, attacks all cells of a $19 \times 19$ square centered at $x$, except for the cells that are in the same row or column as $x$. What is the maximum number of non-attacking cheetahs that can be placed on the board?
(I. Bogdanov) | Answer: 100000.
Solution: Divide the board into $100^{2}$ squares of size $10 \times 10$. We will show that in each square, no more than 10 leopards can stand without attacking each other - from this it will follow that the total number of leopards cannot exceed $100^{2} \cdot 10=100000$.
Consider an arbitrary square... | 100000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. There are 30 logs, the lengths of which are 3 or 4 meters, and their total length is one hundred meters. How many cuts can be made to saw all these logs into pieces 1 meter long? (Each cut saws exactly one log). | Answer: 70.
Solution. First method The total length of the logs is 100 meters. If it were one log, 99 cuts would be needed. Since there are 30 logs, 29 cuts have already been made. Therefore, $99-29=70$ cuts remain to be made.
Second method Let's find the number of logs of each type. If all were 3-meter logs, their t... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. All natural numbers from 1 to 1000 inclusive are divided into two groups: even and odd. In which group is the sum of all digits used to write the numbers greater, and by how much? | Answer. The sum of the digits of odd numbers is 499 more.
Solution. The sum of the digits of the number 1 is equal to the sum of the digits of the number 1000; the other numbers can be divided into pairs: $2-3,4-5,6-7,8-9, \ldots, 998-999$. In each pair, the unit digit of the odd number is 1 more than that of the even... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. In the product of three natural numbers, each factor was decreased by 3. Could the product have increased by exactly 2022 in this case? | Answer. Yes, it could.
Solution. The product $1 \cdot 1 \cdot 678$ serves as an example. After the specified operation, it becomes $(-2) \cdot(-2) \cdot 675=2700=678+2022$.
Remark. The given example is the only one. Here is how to come up with it. Suppose two of the factors were 1, and the third was $a$. Their produc... | 678 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Daisies grew along the path. Between every two daisies, a cornflower grew, and then between each cornflower and daisy - a dandelion. It turned out that now 101 flowers grow along the path. How many daisies grow along the path? | Solution. If at some point there are $n$ flowers growing, then there are $n-1$ gaps between them, so $n-1$ flowers grow between them, and the total becomes $2n-1$ flowers. (The absence of this phrase does not result in a score reduction). Let's move to the solution from the end. After the dandelions appear, there are 1... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. Let all numbers $x, y, z$ be non-zero. Find all values that the expression
$$
\left(\frac{x}{|y|}-\frac{|x|}{y}\right) \cdot\left(\frac{y}{|z|}-\frac{|y|}{z}\right) \cdot\left(\frac{z}{|x|}-\frac{|z|}{x}\right)
$$
can take. | Solution: By the Pigeonhole Principle, among the numbers $x, y$, and $z$, there will be two numbers of the same sign. Then the corresponding bracket will be equal to 0, and the entire product will also be equal to 0.
Answer: Only the number 0.
Recommendations for checking:
| is in the work | points |
| :--- | :--- |... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. Petya and three of his classmates started a 100-meter race at the same time, and Petya came in first. After 12 seconds from the start of the race, no one had finished yet, and the four participants had run a total of 288 meters. When Petya finished the race, the other three participants had 40 meters left to run i... | Solution: The runners were supposed to run a total of 400 meters, and in 12 seconds they ran 288 meters, meaning that in 1 second the runners run $288: 12=$ 24 meters. When Petya finished, the runners had run $400-40=360$ meters, so from the start, $360: 24=15$ seconds had passed. Therefore, Petya runs at a speed of $1... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. All natural numbers from 1 to 20 were divided into pairs, and the numbers in each pair were added. What is the maximum number of the ten resulting sums that can be divisible by 11? Justify your answer. | Solution: The number 11 is the only number in the set that is divisible by 11, so adding it to any other number will disrupt divisibility by 11. Therefore, all 10 sums cannot be divisible by 11. One example where nine sums are divisible by 11 is as follows:
$(1,10),(2,20),(3,19),(4,18),(5,17),(6,16),(7,15),(8,14),(9,1... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Seven friends entered a cafe and ordered 3 small cups of coffee, 8 medium cups, and 10 large cups. The volume of a small cup is half the volume of a medium cup, and the volume of a large cup is three times the volume of a small cup. How should the friends divide the cups of coffee among themselves so that everyone d... | Solution: Let's call the amount of coffee in a small cup a "norm". Then, in total, we have $3+8 \times 2+10 \times 3=49$ norms. Since there are seven friends, each should get 7 norms.
We divide as follows: 1 small + 2 large - 3 people; 2 medium + 1 large - 4 people.
Criteria. A correct example - 7 points. No addition... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a football tournament, 12 teams participated. By September, they had played several games, and no two teams had played each other more than once. It is known that the first team played exactly 11 games. Three teams played 9 games each. One team played 5 games. Four teams played 4 games each. Two teams played only... | Answer: 5 games.
Solution. Let the first team K1 play 11 games - i.e., once with everyone. Teams K2 and K3 played 1 game each - these are games with team K1.
There are 9 teams left (K4-K12). Three of these teams (K4, K5, K6) played 9 games each. One of these games was with K1. And 8 with all teams K4-K12 (except them... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Given a convex quadrilateral ABCD. Point $M$ is the midpoint of side BC, and point $N$ is the midpoint of side CD. Segments AM, AN, and $MN$ divide the quadrilateral into four triangles, the areas of which, written in some order, are consecutive natural numbers. What is the maximum possible area of triangle $\mathrm... | Answer: 6.
Solution. Estimation. Let $n, n+1, n+2, n+3$ be the areas of the four triangles. Then the area of quadrilateral $ABCD$ is $4n+6$. $MN$ is the midline of triangle $BCD$, so $S_{BCD} = 4S_{MCN}$, but $S_{MCN} \geq n$, hence $S_{BCD} \geq 4n$. Then $S_{ABD} = S_{ABCD} - S_{BCD} \leq 6$.
Example. If $ABCD$ is ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Simplify the expression: $\frac{8}{1+a^{8}}+\frac{4}{1+a^{4}}+\frac{2}{1+a^{2}}+\frac{1}{1+a}+\frac{1}{1-a}$ and find its value at $a=2^{-\frac{1}{16}}$. | Solution. Add the last two terms, then add the obtained sum to the third term from the end, and so on. As a result, we get $\frac{16}{1-a^{16}}$. Substituting $a$ with the value $2^{-\frac{1}{16}}$, we get 32.
Answer: 32 | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the product of all roots of the equation $x^{4}+4 x^{3}-2015 x^{2}-4038 x+2018=0$. | Solution. Transform the equation to the form: $\left(x^{2}+2 x\right)^{2}-2019\left(x^{2}+2 x\right)+2018=0$. Introduce the substitution $\boldsymbol{t}=\boldsymbol{x}^{2}+2 \boldsymbol{x}$, we get the equation: $t^{2}-2019 t+2018=0$, the roots of which are $t_{1}=1, t_{2}=2018$. Then $x^{2}+2 x-1=0$ or $x^{2}+2 x-2018... | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Each participant in the school charity event brought either one encyclopedia, or three fiction books, or two reference books. In total, 150 encyclopedias were collected. After the event, two bookshelves in the library were filled, with an equal number of books on each. On the first shelf, there was one fifth of all ... | Solution. Let $x$ be the fifth part of the reference books, $y$ be the seventh part of all fiction books. Then on the first shelf, there are $-x+y+150$ books, and on the second shelf, there are $-4x+6y$ books. From the condition of the problem, the number of books on the shelves is equal, therefore, $x+y+150=4x+6y$, fr... | 267 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $L$ and $K$ are marked, respectively, and $M$ is the intersection point of segments $A K$ and $C L$. It is known that the area of triangle $A M C$ is equal to the area of quadrilateral $L B K M$. Find the angle $A M C$. | Solution. From the equality of the areas of triangle $A M C$ and quadrilateral $L B K M$, it follows that the areas of triangles $A C K$ and $C B L$ are equal (see the figure), because in this case, the same area of triangle $C M K$ is added to the equal areas. We have $S_{A C K}=\frac{1}{2} A C \cdot C K \cdot \sin \a... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 2. Clone 1
The teacher wanted to write an example for calculation on the board:
$$
1,05+1,15+1,25+1,4+1,5+1,6+1,75+1,85+1,95=?
$$
but accidentally forgot to write one comma. After this, Kolya went to the board and, correctly performing all the operations, obtained an integer result. What is it? | Answer: 27
## Solution
Let's calculate the sum
$$
1.05+1.15+1.25+1.4+1.5+1.6+1.75+1.85+1.95=13.5
$$
Therefore, in order to get an integer, the decimal point should be omitted in the number with the fractional part 0.5. As a result, instead of the number 1.5, the number 15 will appear on the board, and the result wi... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 3. Clone 1
On an island, there live knights who always tell the truth, and liars who always lie. Before a friendly match, 30 islanders gathered in T-shirts with numbers on them—arbitrary natural numbers. Each of them said: “I have a T-shirt with an odd number.” After that, they exchanged T-shirts, and each said: “I ... | # Answer: 15
## Solution
Notice that each knight exchanged a shirt with an odd number for a shirt with an even number, while each liar exchanged a shirt with an even number for a shirt with an odd number. This means that each knight exchanged with a liar, so there are no more knights than liars. On the other hand, ea... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 6. Clone 1
The figure shows a hexagon composed of identical equilateral triangles, each with an area of 10. Find the area of the shaded part.
# | # Answer: 110
## Solution
We will call the equilateral triangles that make up the original hexagon unit triangles. Consider each of the highlighted triangles inside the hexagon separately. The smallest one coincides with the unit triangle and has an area of 10. The medium triangle is inscribed in a hexagon made up of... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. When the passengers entered the empty tram, half of them took seats. How many passengers entered at the very beginning, if after the first stop their number increased by exactly $8 \%$ and it is known that the tram can accommodate no more than 70 people? | Answer: 50. From the condition, it follows that the number of passengers is divisible by 2 (since half took seats) and by 25 (8% is $2/25$ of the total number). Therefore, the initial number of passengers was divisible by 50, but it was less than 70, so it was 50.
Comment. Answer only - 0 points. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. The price of a ticket to the stadium was 25 rubles. After the ticket prices were reduced, the number of spectators at the stadium increased by $50 \%$, and the revenue from ticket sales increased by $14 \%$. What is the new price of a ticket to the stadium after the price reduction? | Answer: 19 rubles.
Let's denote: $a$ - the number of spectators coming to the stadium; $x$ - the ratio of the new ticket price to the price of 25 rubles. The new revenue is: on the one hand $(25 \cdot x) \cdot a \cdot 1.5$, on the other hand $25 \cdot 1.14 \cdot a$. From the equality $(25 \cdot x) \cdot a \cdot 1.5 = ... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. In triangle $A B C$, the bisectors $A D$ and $C E$ were drawn. It turned out that $A E + C D = A C$. Find the angle $B$.
---
The text has been translated while preserving the original formatting and line breaks. | Answer: angle $B$ is equal to $60^{\circ}$.
Let $\alpha$ and $\gamma$ be the measures of angles $A$ and $C$ of triangle $ABC$. On side $AC$, we mark segment $AF$ equal to $AE$. From the condition, we get the equality of segments $CD$ and $CF$. Let the bisectors $AD$ and $CE$ intersect segments $EF$ and $FD$
. After 30 exchanges, the number of stickers will increase by $30 * 4 = 120$. Initially, Petya had one sticker, so after 30 exchanges, he will have $1 + 120 = 121$. | 121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Three jumps of a two-headed dragon are equal to 5 jumps of a three-headed one. But in the time it takes for the two-headed dragon to make 4 jumps, the three-headed one makes 7 jumps. Which one runs faster? Justify your answer.
# | # Answer. Three-headed.
Solution. Consider the time it takes for a two-headed dragon to make 3*4=12 jumps. In this time, a three-headed dragon makes $3 * 7=21$ jumps. Since 12=4*3, 12 jumps of the two-headed dragon are equal to 4*5=20 jumps of the three-headed dragon. Thus, in the same amount of time, the three-headed... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.9. On a plane, $N$ points are marked. Any three of them form a triangle, the angles of which in degrees are expressed by natural numbers. For what largest $N$ is this possible
$$
\text { (E. Bakayev) }
$$ | Answer: 180.
Fig. 1
First Solution. Example. First, we will show that the required is possible when $N=180$. Mark 180 points on the circle, dividing it into 180 equal arcs, each $2^{\circ}$... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In the game "Sportlotto-Shish," the main prize is drawn according to the following rules. Each person in the studio independently writes down any number of different pairs of different integers from the set from 1 to 5. If some participants have written down the same pairs, these participants share the main prize. H... | Solution. A total of different pairs can be formed $5 * 4 / 2=10$. It is taken into account that the digits in the pair are different and the order of the digits within the pair does not matter. Each pair can be written down or not written down by the participant. A total of different sets can be formed $2^{10}=1024$. ... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In a row, 33 weights are arranged in ascending order. It is known that any four consecutive weights can be distributed on two scales so that equilibrium is achieved. The third weight weighs 9 g, and the ninth weighs 33 g. How much does the 33rd weight weigh? | Solution. Let the weights of the weights be $a_{1}<a_{2}<\cdots<a_{33}$. For all $k=1,2, \ldots, 30$ the equalities $a_{k}+a_{k+3}=a_{k+1}+a_{k+2}$, equivalent to $a_{k+3}-a_{k+2}=$ $a_{k+1}-a_{k}$, hold. Let $a_{4}-a_{3}=a_{2}-a_{1}=d$ and $a_{5}-a_{6}=a_{3}-a_{2}=\mathrm{c}$, then $a_{6}-$ $a_{5}=a_{4}-a_{3}=d, a_{7}... | 129 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The least common multiple of seven natural numbers is 2012. Find the smallest possible sum of these numbers. | Solution: Let's factorize the number 2012 into prime factors $2012=2^{2} \cdot 503$.
One of the numbers, the LCM of which is 2012, must be divisible by $2^{2}=4$, and one (possibly the same) - by 503. If the same number is divisible by both 4 and 503, then it is divisible by 2012, and the sum of seven numbers is not l... | 512 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Three lines intersect at one point 0. Outside these lines, a point M is taken and perpendiculars are dropped from it to them. The points $\mathrm{H}_{1}, \mathrm{H}_{2}$ and $\mathrm{H}_{3}$ are the bases of these perpendiculars. Find the ratio of the length of the segment OM to the radius of the circle circumscribe... | Solution
$2, \prime \prime$
First, consider two intersecting... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right ... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. On the board, there are 2017 digits. From these, several numbers were formed, the sums of the digits of these numbers were calculated, and then the sum of all the numbers was subtracted by the sum of the sums of their digits. The resulting number was broken down into digits, and the above operation was repea... | Solution. Since the difference between a number and the sum of its digits is divisible by 9, the first operation will result in a number that is a multiple of 9. Moreover, if we take the sum of several numbers and subtract the sum of the digits of these numbers, the result will also be a multiple of 9. Continuing the c... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. The numbers from 1 to 20 are arranged in a circle. We will paint a number blue if it is divisible without a remainder by the number to its left. Otherwise, we will paint it red. What is the maximum number of blue numbers that could be in the circle? | # Solution.
Evaluation. It is obvious that numbers cannot be blue if the number to their left is greater than or equal to 11. That is, no more than 10 numbers can be blue.
Example. As an example, both any correct arrangement and a correct algorithm are counted. An example of a correct algorithm.
1) write down the nu... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. A dandelion blooms in the morning, remains yellow for this and the next day, turns white on the third morning, and by evening of the third day, it has lost its petals. Yesterday afternoon, there were 20 yellow and 14 white dandelions on the meadow, and today there are 15 yellow and 11 white. How many yellow dandel... | Solution: All yellow dandelions the day before yesterday are the white dandelions of yesterday and the white dandelions of today. Therefore, the day before yesterday there were $14+11=25$ yellow dandelions.
Answer: 25 dandelions.
Note: The number of yellow dandelions yesterday and today is not needed for solving the ... | 25 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.4. It is known that $a b c=1$. Calculate the sum
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}
$$ | Solution: Note that
$$
\frac{1}{1+a+a b}=\frac{1}{a b c+a+a b}=\frac{1}{a(1+b+b c)}=\frac{a b c}{a(1+b+b c)}=\frac{b c}{1+b+b c}
$$
Similarly, by replacing 1 with the number $a b c$, we have
$$
\frac{1}{1+c+c a}=\frac{a b}{1+a+a b}=\frac{a b^{2} c}{1+b+b c}=\frac{b}{1+b+b c} .
$$
Then
$$
\frac{1}{1+a+a b}+\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6. On a certain segment, its endpoints and three internal points were marked. It turned out that all pairwise distances between the five marked points are different and are expressed in whole centimeters. What is the smallest possible length of the segment? Justify your answer. | Solution: There are 5 points, so there are 10 pairwise distances. If all of them are expressed as positive whole numbers of centimeters and are distinct, at least one of them is not less than 10. Therefore, the length of the segment is not less than 10. Suppose the length of the segment is exactly 10. Then the pairwise... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3.1. Condition:
The number 4597 is displayed on the computer screen. In one move, it is allowed to swap any two adjacent digits, but after this, 100 is subtracted from the resulting number. What is the largest number that can be obtained by making no more than two moves? | Answer: 8357
## Solution.
The first digit cannot exceed 8, since to obtain the other digits, you need to move the nine forward by two places and subtract one from it. Note that in any other example, the first digit will be less than 8, as we can only get an eight from a nine, and all other digits are less than 9. The... | 8357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 4.1. Condition:
In front of the elevator stand people weighing 50, 51, 55, 57, 58, 59, 60, 63, 75, and 140 kg. The elevator's load capacity is 180 kg. What is the minimum number of trips needed to get everyone up? | Answer: 4 (or 7)
## Solution.
In one trip, the elevator can move no more than three people, as the minimum possible weight of four people will be no less than $50+51+55+57=213>180$. Note that no one will be able to go up with the person weighing 140 kg, so a separate trip will be required for his ascent. For the rema... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4.2. Condition:
In front of the elevator stand people weighing 150, 60, 70, 71, 72, 100, 101, 102, and 103 kg. The elevator's load capacity is 200 kg. What is the minimum number of trips needed to get everyone up? | Answer: 5 (or 9)
## Solution
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+70+71=201>200$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remaining... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4.3. Condition:
In front of the elevator stand people weighing 150, 62, 63, 66, 70, 75, 79, 84, 95, 96, and 99 kg. The elevator's load capacity is 190 kg. What is the minimum number of trips needed to get everyone up? | # Answer: 6 (or 11)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $62+63+66=191>190$. Note that no one will be able to go up with the person weighing 150 kg, so a separate trip will be required for his ascent. For the remai... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 4.4. Condition:
In front of the elevator stand people weighing $130,60,61,65,68,70,79,81,83,87,90,91$ and 95 kg. The elevator's load capacity is 175 kg. What is the minimum number of trips needed to get everyone up? | # Answer: 7 (or 13)
## Solution.
In one trip, the elevator can move no more than two people, as the minimum possible weight of three people will be no less than $60+61+65=186>175$. Note that no one will be able to go up with the person weighing 135 kg, so a separate trip will be required for his ascent. Six trips wil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 5.1. Condition:
In the warehouse, there are 8 cabinets, each containing 4 boxes, each with 10 mobile phones. The warehouse, each cabinet, and each box are locked. The manager has been tasked with retrieving 52 mobile phones. What is the minimum number of keys the manager should take with them? | # Answer: 9
Solution. To retrieve 52 phones, at least 6 boxes need to be opened. To open 6 boxes, no fewer than 2 cabinets need to be opened. Additionally, 1 key to the warehouse is required. In total, $6+2+1=9$ keys need to be taken by the manager. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 8.2. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 2 meters away from me."
- The rightmost in the row said: "My fellow ... | # Answer:
The second islander -1 m
The third islander -1 m.
## Solution.
Let's number the islanders from left to right. Suppose the first one is a knight. Then from his statement, it follows that the third one is also a knight; by the principle of exclusion, the second and fourth must be liars. The fourth said that... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# 8.3. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Four islanders lined up, each 1 meter apart from each other.
- The leftmost in the row said: "My fellow tribesman in this row stands 1 meter away from me."
- The second from the left said: "My fellow t... | # Answer:
The third islander -1 m; 3 m; 4 m.
The fourth islander -2 m.
## Solution.
Let's number the islanders from right to left. Suppose the first one is a knight. Then, from his statement, it follows that the second one is also a knight. However, the second one said that his fellow tribesman is two meters away f... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9-1. Postman Pechkin calculated that he walked half the distance (at a speed of 5 km/h) and only a third of the time he was cycling (at a speed of 12 km/h). Did he make a mistake in his calculations? | Answer. Mistaken.
Solution. Let's denote the entire distance Pechkin traveled as $2 S$ km. Then, on foot, he covered a distance of $S$ km and spent $S / 5 = 0.2 S$ (hours) on it. According to the problem, this constituted $2 / 3$ of the total time spent, meaning the entire journey took $0.2 S : 2 / 3 = 0.3 S$ (hours),... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-2. The school volleyball team played several matches. After they won another match, the share of victories increased by $1 / 6$. To increase the share of victories by another 1/6, the volleyball players had to win two more consecutive matches. What is the minimum number of victories the team needs to achieve to incre... | Answer: 6.
Solution: Let the team initially play $n$ matches, of which $k$ were won. Then, after the next win, the share of victories increased by $\frac{k+1}{n+1}-\frac{k}{n}=\frac{1}{6}$. Similarly, after two more wins, the increase was $\frac{k+3}{n+3}-\frac{k+1}{n+1}=\frac{1}{6}$. Simplifying each equation, we get... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-5. Two three-digit numbers are written on the board in a multiplication example. If the multiplication sign is replaced with 0, a seven-digit number is obtained, which is an integer multiple of the product. By what factor exactly | Answer: 73.
Solution. Let the original numbers be denoted by $a$ and $b$. Then the specified seven-digit number will have the form $10000a + b$. According to the condition, $10000a + b = nab$, from which we get $b = \frac{10000}{na - 1}$.
Note that the numbers $a$ and $a-1$ do not have common divisors, so $na - 1 = p... | 73 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. From the digits $1,2,3,4,5,6,7,8,9$, nine (not necessarily distinct) nine-digit numbers are formed; each digit is used exactly once in each number. What is the maximum number of zeros that the sum of these nine numbers can end with?
(N. Agakhanov) | Answer: Up to 8 zeros.
Solution: We will show that the sum cannot end with 9 zeros. Each of the numbers formed is divisible by 9, since the sum of its digits is divisible by 9. Therefore, their sum is also divisible by 9. The smallest natural number divisible by 9 and ending with nine zeros is $9 \cdot 10^{9}$, so the... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. Polina has two closed boxes - a square one and a round one. She was told that the round one contains 4 white and 6 black balls, while the square one contains 10 black balls. In one move, Polina can take a ball from any box without looking and either throw it away or move it to the other box. Polina wants to make t... | Answer: 15
Solution. Suppose that in the end, both boxes still contain both white and black balls. Then the last action Polina took was to draw a ball of a certain color from one of the boxes, and there were still both black and white balls left. But she could not have done this with certainty, meaning she could have ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.2. In Olya's black box, there are 5 apples and 7 pears, and in the white box, there are 12 pears. In one move, Olya can blindly take a fruit from any box and either eat it or move it to the other box. Olya wants the contents of the boxes to be the same. What is the minimum number of moves Olya can guarantee to achiev... | # Answer: 18
Solution. Suppose that in the end both boxes still contain both apples and pears. Then the last action Olya took was to take some fruit from a box, and there were still both apples and pears left. Because of this, she could not guarantee that she would take the needed fruit, as she might have picked the w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.3. Inna has two closed boxes - a square one and a round one. She was told that the round box contains 3 white and 10 black balls, while the square box contains 8 black balls. In one move, Inna can, without looking, take a ball from any box and either throw it away or move it to the other box. Inna wants to make the c... | Answer: 17
Solution. Suppose that in both boxes, there are both white and black balls left in the end. Then the last action Inna took was to draw a ball of a certain color from one of the boxes, and there were still both black and white balls left. However, she could not have done this with certainty, i.e., she might ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.4. In Zhenya's black box, there are 8 bananas and 10 mangoes, and in the white box - 12 mangoes. In one move, Zhenya, without looking, can take a fruit from any box and either eat it or move it to the other box. Zhenya wants the contents of the boxes to be the same. What is the minimum number of moves Zhenya can guar... | # Answer: 24
Solution. Suppose that in the end, both boxes still contain both bananas and mangoes. Then the last action was Zhenya taking some fruit from a box, and both types of fruit remained. Because of this, she could not guarantee picking the right fruit, as she might have picked the wrong one.
If removing mango... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1. A chocolate bar weighed 250 g and cost 50 rubles. Recently, to save money, the manufacturer reduced the weight of the bar to 200 g, and increased its price to 52 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 30
Solution. We will calculate by what percentage the cost of one kilogram has increased. Before the increase, 1 kg cost 200 rubles, and after the increase, it costs $52 \cdot 5=260$ rubles. The increase is 60 rubles per kg, which originally cost 200 rubles. This is $\frac{60}{200} \cdot 100 \% = 30 \%$ | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.2. A chocolate bar weighed 400 g and cost 150 rubles. Recently, to save money, the manufacturer reduced the weight of the bar to 300 g, and increased its price to 180 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 60
Solution. We will calculate by what percentage the cost of 1 kg 200 g has increased. Before the increase, this amount cost 450 rubles, and after the increase, it costs $180 \cdot 4=720$ rubles. The increase is 270 rubles on 1.2 kg, which originally cost 450 rubles. This is $\frac{270}{450} \cdot 100 \%=$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.4. A package of milk with a volume of 1 liter cost 60 rubles. Recently, in order to save money, the manufacturer reduced the volume of the package to 0.9 liters, and increased its price to 81 rubles. By what percentage did the manufacturer's revenue increase? | # Answer: 50
Solution. We will calculate by what percentage the cost of 9 liters of milk has increased. Before the increase, 9 liters cost $9 \cdot 60=540$ rubles, and after the increase, $-81 \cdot 10=810$ rubles. The increase amounts to 270 rubles for 9 liters. This amount previously cost 540 rubles. This is $\frac{... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 2021 less than the originally thought number. What is N? | Answer: 2022
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numb... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.2. Galia thought of a number, multiplied it by $\mathrm{N}$, then added $\mathrm{N}$ to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 7729 less than the originally thought number. What is N? | Answer: 7730
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numb... | 7730 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.3. Galia thought of a number, multiplied it by N, then added N to the result, divided the resulting number by N, and subtracted N. In the end, she got a number that is 100 less than the originally thought number. What is N? | Answer: 101
Solution. Let the number she thought of be $\mathrm{k}$, then after two operations, she will have the number $\mathrm{kN}+\mathrm{N}$, and after division, she will have the number $\mathrm{k}+1$, which is 1 more than the number she thought of. And when she subtracts $\mathrm{N}$, the result will be a numbe... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 14 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 13 girls in thi... | # Answer: 15
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 7 pizzas. Then $m: 13=7: x$, from which $m x=91$. The number 91 has only one divisor greater than 13, which is 91. Therefore, $m=91, x=1$,... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.2. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 11 girls in thi... | # Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 11 = 5: x$, from which we get $m x = 55$. The number 55 has only one divisor greater than 11, which is 55. Therefore, $... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.3. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 22 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 13 girls in thi... | # Answer: 23
Solution. Let the number of boys be $m$, and the number of pizzas that the girls received be $x$. If each boy had eaten as much as each girl, the boys would have eaten 11 pizzas. Then $m: 13 = 11: x$, from which $m x = 143$. The number 143 has only one divisor greater than 13, which is 143. Therefore, $m ... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.4. For the celebration of the Name Day in the 5th grade parallel, several pizzas were ordered. 10 pizzas were ordered for all the boys, with each boy getting an equal share. Each girl also received an equal share, but half as much as each boy. How many pizzas were ordered if it is known that there are 17 girls in thi... | Answer: 11
Solution. Let the number of boys be $m$, and the number of pizzas that the girls got be $x$. If each boy had eaten as much as each girl, the boys would have eaten 5 pizzas. Then $m: 17 = 5: x$, from which $m x = 85$. The number 85 has only one divisor greater than 17, which is 85. Therefore, $m=85, x=1$, th... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. How many natural numbers $\mathrm{N}$ greater than 900 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $900, N+15,2 N$ exactly two are four-digit numbers? | Answer: 5069
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-900, \mathrm{~N}+15,2 \mathrm{~N}, 3 \mathrm{~N}$. Four-digit numbers can only be two consecutive ones. If the four-digit numbers are $\mathrm{N}-900$ and $\mathrm{N}+15$, then $2 \mathrm{... | 5069 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.2. How many natural numbers $\mathrm{N}$ greater than 300 exist such that among the numbers $4 \mathrm{~N}, \mathrm{~N}-$ $300, N+45,2 N$ exactly two are four-digit numbers? | Answer: 5410
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+15$, and if we write the numbers in ascending order, we get $\mathrm{N}-300, \mathrm{~N}+45,2 \mathrm{~N}, 4 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-300$ and $\mathrm{N}+15$, then $2 \mathrm{~N}$ ... | 5410 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.3. How many natural numbers $\mathrm{N}$ greater than 700 exist such that among the numbers $3 \mathrm{~N}, \mathrm{~N}-$ $700, N+35,2 N$ exactly two are four-digit numbers? | Answer: 5229
Solution. Note that $2 \mathrm{~N}>\mathrm{N}+35$, and if we write the numbers in ascending order, we get $\mathrm{N}-700, \mathrm{~N}+35,2 \mathrm{~N}, 3 \mathrm{~N}$. Only two consecutive numbers can be four-digit. If the four-digit numbers are $\mathrm{N}-700$ and $\mathrm{N}+35$, then $2 \mathrm{~N}$ ... | 5229 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved two seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Wh... | # Answer: 2
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 2, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decrease... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved three seats to the right, Galia had moved one seat to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. ... | Answer: 3
Solution. Let's see how the seat number changed for everyone except Anya. Varya's seat number increased by 3, Galia's decreased by 1, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have decreased by ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.3. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved one seat to the right, Galia had moved three seats to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. ... | # Answer: 3
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 1, Galia's decreased by 3, and the sum of Diana's and El's seat numbers did not change. At the same time, the total sum of the seats did not change, so Anya's seat number must have increased by 2. ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.4. In a cinema, five friends took seats numbered 1 to 5 (the leftmost seat is number 1). During the movie, Anya left to get popcorn. When she returned, she found that Varya had moved one seat to the right, Galia had moved two seats to the left, and Diana and Elia had swapped places, leaving the edge seat for Anya. Wh... | # Answer: 4
Solution. Let's see how the seat number of everyone except Anya has changed. Varya's seat number increased by 1, Galia's decreased by 2, and the sum of Diana's and Eli's seat numbers did not change. At the same time, the total sum of the seat numbers did not change, so Anya's seat number must have increase... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Misha calculated the products $1 \times 2, 2 \times 3$, $3 \times 4, \ldots, 2017 \times 2018$. For how many of them is the last digit zero? | Solution. The last digit of the product depends on the last digits of the factors. In the sequence of natural numbers, the last digits repeat every ten. In each ten, in the sequence of products, four products end in zero: ... $4 \times \ldots 5, \ldots 5 \times \ldots 6, \ldots 9 \times \ldots 0, \ldots 0 \times \ldots... | 806 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In the pantry, Winnie-the-Pooh keeps 11 pots, seven of which contain jam, and four contain honey. All the pots are lined up, and Winnie remembers that the pots with honey are standing together. What is the minimum number of pots Winnie-the-Pooh needs to check to find a pot with honey? | Answer: one.
Solution. Let's number the pots from 1 to 11 in the order of their arrangement in a row. Exactly one of the pots numbered 4 and 8 contains honey. Therefore, it is sufficient to check one of them. It is impossible to identify the pot with honey without checking, as any pot may contain either honey or jam.
... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. There is a $6 \times 6$ square, all cells of which are white. In one move, it is allowed to change the color of both cells in any domino (a rectangle of two cells) to the opposite. What is the minimum number of moves required to obtain a square with a checkerboard pattern? Don't forget to explain why a smaller numbe... | Answer: 18 moves.
Solution. Note that in the chessboard coloring of a $6 \times 6$ square, there are 18 black cells. At the same time, no two of them can be turned black in one move, because they are not covered by one domino. Therefore, at least 18 moves are required (at least one move per cell). This can be done in ... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. To the number $A$, consisting of eight non-zero digits, a seven-digit number, consisting of identical digits, was added, and the eight-digit number $B$ was obtained. It turned out that the number $B$ can be obtained from the number $A$ by rearranging some of the digits. What digit can the number $A$ start with if ... | Answer: 5.
Solution: Since the numbers $A$ and $B$ have the same sum of digits, their difference is divisible by 9. Therefore, the added seven-digit number with identical digits is divisible by 9. This means it consists of nines. That is, we can consider that $10^7$ was added to the number $A$ and 1 was subtracted. Th... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5. On a checkerboard of size $8 \times 8$, 8 checkerboard ships of size $1 \times 3$ are placed such that no two cells occupied by different ships share any points. One shot is allowed to pierce all 8 cells of one row or one column. What is the minimum number of shots needed to guarantee hitting at least one ship? | Answer: 2 shots.
Solution. We will make 2 shots as shown in Fig. 5. Suppose we did not hit any ship. Then there are no ships in area 1. In each of areas 2 and 3, there is no more than 1 ship. Therefore, there are at least 6 ships in area 4.
Area 4 is a $5 \times 5$ square. Then in this area, horizontally placed ships... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A tea set consists of six identical cups and six identical saucers. Six different sets were packed into two boxes, all saucers in one box, and all cups in the other. All items are wrapped in opaque paper and are indistinguishable by touch. Find the minimum number of items that need to be taken out of these boxes to ... | Solution. Explanation by example. 18 cups and 12 saucers may not be enough. Let's number the sets. By randomly picking 18 cups, we might get all the cups from sets №1, 2, 3.
By randomly picking 12 saucers, we might get all the saucers from sets №5, 6. It is impossible to form a cup-saucer pair. Adding one more saucer ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. From Central Square to the station, there is a straight street divided by 11 intersections into 12 equal blocks. At each intersection, there is a traffic light. All traffic lights simultaneously turn green for 3 minutes, then red for 1 minute. It takes the bus two minutes to travel one block (from intersection to in... | Solution. A car will travel three blocks without obstacles in the first three minutes. Upon approaching the intersection separating the third and fourth blocks, the car will stop at the traffic light for 1 minute. Thus, to travel three blocks and start moving on the fourth block, the car will need four minutes. The car... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In quadrilateral $A B C D$, points $X, Y, Z$ are the midpoints of segments $A B, A D, B C$ respectively. It is known that $X Y$ is perpendicular to $A B$, $Y Z$ is perpendicular to $B C$, and the measure of angle $A B C$ is $100^{\circ}$. Find the measure of angle $A C D$.
. We are given that $x+y=100^{\circ}$. We need to find $z+v$.
Th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. Three runners - Anton, Seryozha, and Tolya - are participating in a 100 m race. When Anton finished, Seryozha was 10 meters behind him, and when Seryozha finished, Tolya was 10 meters behind him. What distance was Tolya from Anton when Anton finished? (It is assumed that all boys run at constant, but of course, no... | Answer: 19 m.
Solution: Since Tolya's speed is $9 / 10$ of Seryozha's speed, by the time Anton finished, Tolya had run $9 / 10$ of the distance covered by Seryozha, which is $90 \cdot 9 / 10=81$ m. Thus, when Anton finished, Tolya and Anton were 19 m apart. | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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