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Variant 9.5.2. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 32^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees do... | Answer: (a) $106^{\circ}$. (b) $48^{\circ}$. | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.3. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 36^{\circ}$ and $\angle X D C = 52^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $108^{\circ}$. (b) $46^{\circ}$. | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Variant 9.5.4. Sides $A D$ and $D C$ of the inscribed quadrilateral $A B C D$ are equal. A point $X$ is marked on side $B C$ such that $A B = B X$. It is known that $\angle B = 38^{\circ}$ and $\angle X D C = 54^{\circ}$.
(a) (1 point) How many degrees does the angle $A X C$ measure?
(b) (3 points) How many degrees d... | Answer: (a) $109^{\circ}$. (b) $44^{\circ}$. | 109 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.
(A. Yakub... | Answer: $60^{\circ}$.
Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.
Perform a reflection about $\ell$. Since $\angle E D A = \angle F D C$, the ray $D C$ will be transformed into the ray $D A... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. The number 2019 is represented as the sum of different odd natural numbers. What is the maximum possible number of addends? | Answer: 43.
Solution. Evaluation. Let's calculate the sum of the 45 smallest odd natural numbers: 1 + 3 + ... + 87 + 89 = (1 + 89) / 2 * 45 = 2025 > 2019. Therefore, there are fewer than 45 addends, but the sum of 44 odd addends is an even number, so there are no more than 43 addends. Example. 2019 = 1 + 3 + ... + 81 ... | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.6. Find the largest $n$ such that the sum of the fourth powers of any $n$ prime numbers greater than 10 is divisible by $n$. | Answer: $n=240$.
Solution. In fact, it is required that the fourth powers of all prime numbers greater than 10 give the same remainder when divided by $n$. Indeed, if the fourth powers of some two numbers give different remainders, then one can take $n-1$ of the first number and one of the second, then the sum of the ... | 240 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A six-digit number $A$ is divisible by 19. The number obtained by removing its last digit is divisible by 17, and the number obtained by removing the last two digits of $A$ is divisible by 13. Find the largest $A$ that satisfies these conditions. | 2. Answer: 998412.
Solution. The largest four-digit number divisible by 13 is 9997. Among the numbers from 99970 to 99979, there is a number 99977 divisible by 17, but among the numbers from 999770 to 999779, there is no number divisible by 19. However, for the next number 9984, which is divisible by 13, the number 99... | 998412 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. A row of 50 people, all of different heights, stands. Exactly 15 of them are taller than their left neighbor. How many people can be taller than their right neighbor? (Provide all possible answers and prove that there are no others) | Answer: 34.
Solution. Let's call 15 people who are taller than their left neighbor - tall people.
If someone has a tall person to their right, then they are shorter than them, otherwise, they are taller.
There are a total of 49 pairs of adjacent people.
Tall people are on the right in exactly 15 pairs. In the remai... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.1. On 40 cells of an $8 \times 8$ chessboard, stones were placed. The product of the number of stones lying on white cells and the number of stones lying on black cells was calculated. Find the minimum possible value of this product. | Answer: 256
Solution. Let $b$ and $w=40-b$ be the number of stones on black and white cells, respectively. Without loss of generality, assume that $b \geqslant w$, so $b$ can take any integer value from 20 to 32. We are interested in the minimum value of the product $b(40-b)$. The quadratic function $x(40-x)=-x^{2}+40... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. Kolya had 10 sheets of paper. On the first step, he chooses one sheet and divides it into two parts. On the second step, he chooses one sheet from the available ones and divides it into 3 parts, on the third step, he chooses one sheet from the available ones and divides it into 4, and so on. After which step will ... | # Answer: 31
Solution. On the $k$-th step, the number of leaves increases by $k$, so after $k$ steps, the number of leaves will be $10+(1+2+\ldots+k)=10+\frac{k(k+1)}{2}$. When $k=30$, this number is less than 500 (it equals 475), and when $k=31$, it is already more than 500 (it equals 506), so the answer to the probl... | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.1. A two-digit natural number $\overline{a b}$ is randomly selected from 21 to 45 (the probability of selecting each number is the same). The probability that the number $\overline{a 8573 b}$ is divisible by 6 is $n$ percent. Find $n$. | Answer: 16
Solution. There are a total of 25 options. The number $\overline{a 8573 b}$ is divisible by 6 if and only if both conditions are met simultaneously: $b$ is even and the sum of the digits $a+8+5+7+3+b$ is divisible by 3. Among the numbers from 21 to 45, the numbers $22,28,34,40$ (every sixth number) fit - th... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. In a row, the numbers $\sqrt{7.301}, \sqrt{7.302}, \sqrt{7.303}, \ldots, \sqrt{16.002}, \sqrt{16.003}$ are written (under the square root - consecutive terms of an arithmetic progression with a common difference of 0.001). Find the number of rational numbers among the listed ones. | # Answer: 13
Solution. Multiply the numbers by 100, we get $\sqrt{73010}, \sqrt{73020}, \sqrt{73030}, \ldots, \sqrt{160030}$ (in this case, rational numbers will remain rational, and irrational numbers will remain irrational). The square root of a natural number $n$ is a rational number if and only if $n$ is a perfect... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. 72 vertices of a regular 3600-gon are painted red such that the painted vertices are the vertices of a regular 72-gon. In how many ways can 40 vertices of this 3600-gon be chosen so that they are the vertices of a regular 40-gon and none of them are red? | Answer: 81
Solution. Number the vertices in order $1,2, \ldots, 3600$, so that the painted vertices are those with numbers divisible by $3600 / 72=50$. 40 vertices form the vertices of a regular 40-gon if their numbers give the same remainder when divided by $3600 / 40=90$. There are 90 ways to choose a regular 40-gon... | 81 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 1. CONDITION
On a line, several points were marked. Then, between each pair of adjacent points, one more point was marked, and this operation was repeated once more. As a result, 101 points were obtained. How many points were marked initially? | Solution. Let there be $k$ points marked initially. Then $k-1$ more points were added to them (one between the first and second, second and third, $\ldots, (k-1)$-th and $k$-th marked points), and then another $(k+(k-1))-1=2k-2$ points. In total, the number of points became $4k-3$. Solving the equation $4k-3=101$, we f... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.2. If a class of 30 people is seated in a cinema hall, then in any case, at least two classmates will end up in the same row. If the same is done with a class of 26 people, then at least three rows will be empty. How many rows are there in the hall? | Answer: 29.
There are no more than 29 rows in the hall. Otherwise, a class of 30 students can be seated with one student per row. On the other hand, if a class of 26 students is seated with one student per row, at least three rows will be empty, which means there are no fewer than 29 rows.
Comment. An answer without ... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. In triangle $A B C$, points $E$ and $D$ are on sides $A B$ and $B C$ respectively, such that segments $A D$ and $C E$ are equal, $\angle B A D=\angle E C A$ and $\angle A D C=\angle B E C$. Find the angles of the triangle. | Answer: All angles are $60^{\circ}$.
Triangles $C E A$ and $A D B$ are equal (by side and adjacent angles). Therefore, sides $A C$ and $A B$ are equal, and angles $C A E$ and $A B D$ are equal, which means angles $C A B$ and $A B C$ are equal. This implies the equality of sides $A C$ and $B C$. Thus, triangle $A B C$ ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. 50 businessmen - Japanese, Koreans, and Chinese - are sitting at a round table. It is known that between any two nearest Japanese, there are as many Chinese as there are Koreans at the table. How many Chinese can there be at the table?
## Answer: 32. | Solution. Let there be $-x$ Koreans and $y$ Japanese at the table, then there are $-x y$ Chinese. According to the condition: $x+y+x y=50$. By adding 1 to both sides of the equation and factoring both sides, we get: $(x+1)(y+1)=17 \times 3$. Since $x \neq 0$ and $y \neq 0$, one of the factors is 3, and the other is 17.... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.5. In a row, there are 20 free chairs. From time to time, a person approaches and sits on one of the free chairs, at which point one of their neighbors, if any, immediately stands up and leaves (the two of them do not sit together). What is the maximum number of chairs that can be occupied? | Answer: 19.
Note that all chairs cannot be occupied, otherwise, when the 20th person sat down, no one would leave, and the neighboring chair would be occupied. We will show how 19 chairs can be occupied. The first person takes the first chair, the next person takes the third chair, the third person takes the second ch... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
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 poll was being 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 poll.
Then, 80 more people participated in the poll, voting only for October 22. After that, th... | 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 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), the right triangles $X A B$ and $Y D A$ are similar by the acut... | 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 $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). 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?
 | 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 $D Z$.
 standing consecutively. Find the number of chains, the sum of the numbers in which is positive. | Answer: 4951.
Solution. We will divide all chains (except the chain consisting of all numbers) into pairs that complement each other. If the sum of the numbers in one chain of the pair is $s$, and in the second is $t$, then $s+t=1$. Since $s$ and $t$ are integers, exactly one of them is positive. Therefore, exactly ha... | 4951 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. If a number is added to the sum of its digits, the result is 328. Find this number. | # 3. Answer: 317
Let's denote the digits of the desired number as a, b, c, and their sum as s. Then abc $+\mathrm{s}=328$. Since $s$ is the sum of three single-digit numbers, s does not exceed 27, so a $=3$. At the same time, b cannot be zero, because otherwise the maximum value of s would be $3+0+9=12$. If $\mathrm{b... | 317 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Petya wants to color several cells of a $6 \times 6$ square so that there are as many vertices as possible that belong to exactly three colored squares. What is the maximum number of such vertices he can achieve? | Answer: 25.
Solution. Each vertex of the grid belongs to one, two, or four squares, and the latter are 25. Therefore, the number of vertices in question is no more than 25. An example is shown in the figure.
Criteria. Example: 4 points. Evaluation: 3
points.
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. In the park, paths are laid out as shown in the figure. Two workers started to asphalt them, starting simultaneously from point $A$. They lay asphalt at constant speeds: the first on the section $A-B-C$, the second on the section $A-D-E-F-C$. In the end, they finished the work simultaneously, spending 9 ho... | Answer: 45.
Solution. Let the line $A D$ intersect the line $C F$ at point $G$, as shown in the figure below. Since $A B C G$ and $D E F G$ are rectangles, we have $A B=C G, B C=A G, E F=D G$ and $D E=F G$.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA + ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
=3(z-t)$. Find
$x / y$. | Answer: 25 (in the second variant 9). Note that $z-t=(\sqrt{} x-\sqrt{} y)^{2} / 2$. Denoting $a=\sqrt{} x, b=\sqrt{} y$, we get that $a^{2}-b^{2}=3(a-b)^{2} / 2$, which transforms into $\kappa(a-b)(2 a+2 b-3 a+3 b)=(a-b)(5 b-a)=0$. By the condition $a \neq b$, therefore $a=5 b$, i.e. $x / y=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. 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 2nd 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, and... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. The company specializes in manufacturing "boards with a hole": this is a $300 \times 300$ grid board, in which a hole in the form of a rectangle, not extending to the board's edge, is cut out by cells. Each such board comes with a tag indicating the maximum number of non-attacking rooks that can be placed on this bo... | Answer: 400 rooks (in the $2-nd$ variant 440 rooks).
Let the hole have dimensions $a \times b$. Note that all cells of the board with the hole are covered by $300-b$ columns and 300 - a rows. If $a, b \geqslant 100$, then all cells are covered by no more than 400 lines, and it is impossible to place more than 400 rook... | 400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Along the school corridor, there is a New Year's garland consisting of red and blue bulbs. Next to each red bulb, there is definitely a blue one. What is the maximum number of red bulbs that can be in this garland if there are 50 bulbs in total?
# | # Solution
Let's calculate the minimum number of blue bulbs that can be in the garland. We can assume that the first bulb is red. Since there must be a blue bulb next to each red bulb, three red bulbs cannot go in a row. Therefore, among any three consecutive bulbs, at least one bulb must be blue. Then, among the firs... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.6. Let $a, b, c$ be natural numbers, and the product $ab$ is divisible by $5c$, the product $c$ is divisible by $13a$, and the product $ca$ is divisible by $31b$. Find the smallest possible value of the product $abc$. Justify your answer. | Solution: From the condition that $ab$ is divisible by $5c$, it follows that at least one of the numbers $a$ and $b$ is divisible by 5. If this number is $a$, then from the condition that $bc$ is divisible by $13a$, it follows that one of the numbers $b$ or $c$ is also divisible by 5; if this number is $b$, then one of... | 4060225 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. What is the maximum number of cells that can be painted on a $6 \times 6$ board so that it is impossible to select four painted cells such that the centers of these cells form a rectangle with sides parallel to the sides of the board.
# | # Solution
Let $n$ be the number of cells that can be painted. Denote by $x_{1}, x_{2}, \ldots, x_{6}$ the number of painted cells in the 1st, 2nd, ..., 6th columns, respectively. Then $x_{1}+x_{2}+\ldots+x_{6}=n$. Let's count the number of pairs of cells in the columns that can form the sides of the desired rectangle... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Angle $A$ of the rhombus $A B C D$ is $60^{\circ}$. A line passing through point $C$ intersects segment $A B$ at point $M$ and line $A D$ - at point $N$. Prove that the angle between lines $M D$ and $N B$ is $60^{\circ}$. - | Solution. Let $K$ be the intersection point of the lines $M D$ and $N B$ (Fig. 4). Note that triangle $A B D$ is equilateral, $\angle B A D=\angle A B D=\angle B D A=60^{\circ}$. From the similarity of triangles $B M C$ and $A M N$, as well as $M A N$ and $C D N$, it follows that
$$
\frac{M B}{A B}=\frac{M C}{C N}, \q... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
1. On the board, three quadratic equations are written:
$$
\begin{gathered}
2020 x^{2}+b x+2021=0 \\
2019 x^{2}+b x+2020=0 \\
x^{2}+b x+2019=0
\end{gathered}
$$
Find the product of the roots of all the equations written on the board, given that each of them has two real roots. | Solution. According to the theorem converse to Vieta's theorem, the product of the roots of the first equation is $-\frac{2021}{2020}$, the product of the roots of the second equation is $\frac{2020}{2019}$, and the third is 2019. Therefore, the product of the roots of all equations is $\frac{2021}{2020} \cdot \frac{20... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The equation $x^{2}-a x+2022=0$ has 2 positive integer roots. What is the smallest value of $a$ for which this is possible? | Solution. According to the theorem converse to Vieta's theorem, we have: $x_{1}+x_{2}=a$, $x_{1} \cdot x_{2}=2022$. Note that the product of the roots can be factored into two factors in 4 ways: $1 \cdot 2022, 2 \cdot 1011, 3 \cdot 674, 6 \cdot 337$. The smallest sum of the factors is $6+337=343$. Answer: 343. | 343 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. How many even six-digit numbers exist, in the notation of which identical digits do not stand next to each other | Solution. Let $N_{k}(k>1)$ denote the number of even $k$-digit numbers in which identical digits do not stand next to each other. Similarly, define $N_{1}$, but exclude the number 0, so $N_{1}=4$. We can directly calculate $N_{2}=41$. Let $k>2$. Notice that from $k$-digit numbers satisfying the condition, we can form s... | 265721 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. From the set $\{10 ; 11 ; 12 ; \ldots ; 19\}$, 5 different numbers were chosen, and from the set $\{90 ; 91 ; 92 ; \ldots ; 99\}$, 5 different numbers were also chosen. It turned out that the difference between any two of the ten chosen numbers is not divisible by 10. Find the sum of all 10 chosen numbers. | Solution. Obviously, all the selected numbers have different last digits. Therefore, the sum can be found by separately adding the tens and units of these numbers. Among the tens, there are five 10s and five 90s, so the sum of the tens is $5 \cdot 10 + 5 \cdot 90 = 500$. And among the units, each digit appears exactly ... | 545 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10-6-1. In the figure, there are two circles with centers $A$ and $B$. Additionally, points $A, B$, and $C$ lie on the same line, and points $D, B$, and $E$ also lie on the same line. Find the degree measure of the angle marked with a “?”.
. Angles $ABE$ and $DBC$ are equal as vertical angles. Note that triangle $DBC$ is isosceles wi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10-8-1. There is a magical grid sheet of size $2000 \times 70$, initially all cells are gray. The painter stands on a certain cell and paints it red. Every second, the painter takes two steps: one cell to the left and one cell down, and paints the cell red where he ends up after the two steps. If the painter is in the ... | Answer: 14000.
Solution variant 1. We need to understand how many cells will be painted by the time the painter returns to the initial cell. Note that after every 2000 moves, the painter returns to the starting column, and after every 140 moves, he returns to the starting row. Therefore, he will return to the initial ... | 14000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The distance between cities $A$ and $B$ is 435 km. A train departed from $A$ at a speed of 45 km/h. After 40 minutes, another train departed from city $B$ towards it at a speed of 55 km/h. How far apart will they be one hour before they meet? | Answer: The trains are approaching each other at a speed of 100 km/h. Therefore, in the last hour, they will together cover 100 km. This is the distance that will be between them one hour before the meeting.
Evaluation. 7 points for the correct solution. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many ways are there to rearrange the letters of the word ГЕОМЕТРИЯ so that no two consonants stand next to each other?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Answer: 21600.
Solution. First, arrange the vowels, among which there are two letters E. They can be permuted in $5!/2=60$ ways. Now, 6 positions are formed where consonants can be placed (no more than one consonant per position) - these are the four gaps between the vowels, as well as the positions at the beginning a... | 21600 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
2. From one point on a straight highway, three cyclists start simultaneously (but possibly in different directions). Each of them rides at a constant speed without changing direction. An hour after the start, the distance between the first and second cyclist was 20 km, and the distance between the first and third - 5 k... | Answer: 25 km/h or 5 km/h.
Solution. Let's draw a numerical axis along the highway, taking the starting point of the cyclists as the origin and directing it in the direction of the second cyclist's movement. His speed is 10 km/h, so after an hour, he was at point B with a coordinate of 10. The distance from him to the... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. Petya wrote ten natural numbers on the board, none of which are equal. It is known that among these ten numbers, three can be chosen that are divisible by 5. It is also known that among the ten numbers written, four can be chosen that are divisible by 4. Can the sum of all the numbers written on the board be less... | Answer. It can.
Solution. Example: $1,2,3,4,5,6,8,10,12,20$. In this set, three numbers $(5,10,20)$ are divisible by 5, four numbers $(4,8,12,20)$ are divisible by 4, and the total sum is 71.
Remark. It can be proven (but, of course, this is not required in the problem), that in any example satisfying the problem's c... | 71 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Two vertices, the incenter of the inscribed circle, and the intersection point of the altitudes of an acute triangle lie on the same circle. Find the angle at the third vertex. | Answer: $60^{\circ}$.
Solution. Consider triangle $ABC$, in which altitudes $AA_1$ and $BB_1$ are drawn. Let point $H$ be the orthocenter, and point $I$ be the incenter. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.3 The number $\sqrt{1+2019^{2}+\frac{2019^{2}}{2020^{2}}}+\frac{2019}{2020}$ is an integer. Find it. | Solution. Let $a=2020$. Then the desired number is $\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+$ $\frac{a-1}{a}$. We will perform equivalent transformations:
$$
\begin{gathered}
\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+\frac{a-1}{a}=\frac{\sqrt{a^{2}(a-1)^{2}+(a-1)^{2}+a^{2}}}{a}+\frac{a-1}{a}= \\
=\frac{\sqrt{a^{4}... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.6 Each student in the eighth grade is friends with exactly two students in the seventh grade, and each student in the seventh grade is friends with exactly three students in the eighth grade. There are no more than 29 students in the eighth grade, and no fewer than 17 in the seventh grade. How many students are there... | Solution. Let there be $s$ students in the 8th grade and $-t$ in the 7th grade. Then the number of pairs of friends from different grades is $2s$ and it is also equal to $3t$. From the equality $2s = 3t$, we see that $s$ is a multiple of 3, and $t$ is a multiple of 2. Therefore, $2s = 3t$ is a multiple of 6. Additional... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.4. Ivan Ivanovich's age is 48 years 48 months 48 weeks 48 days 48 hours. How many full years old is Ivan Ivanovich? Don't forget to explain your answer. | Answer: 53 years.
48 months is 4 years, 48 weeks is 336 days, 48 days and 48 hours is 50 days, in total 53 years and 21 or 20 days, hence the answer.
Comment. Correct answer only - 3 points; answer with explanations or calculations leading to the answer - 7 points. | 53 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem №1
Misha suggested that Yulia move a chip from cell $A$ to cell $B$ along the shaded cells. In one step, you can move the chip to an adjacent cell by side or corner. To make it more interesting, Misha put 30 candies in the prize fund, but said he would take 2 candies for each horizontal or vertical move and ... | Answer: 14.
Solution. From $A$ to $B$, one can travel via the top or the bottom. If traveling via the top, the first 2 moves are diagonal (a diagonal move is more advantageous than 2 horizontal moves), and the next 5 moves are horizontal. Misha will collect $2 \cdot 3 + 5 \cdot 2 = 16$ candies, and Yulia will win 14. ... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem №3
A New Year's garland hanging along the school corridor consists of red and blue bulbs. Next to each red bulb, there is definitely a blue one. What is the maximum number of red bulbs that can be in this garland if there are 50 bulbs in total? | Answer: 33 bulbs
## Solution
Let's calculate the minimum number of blue bulbs that can be in the garland. We can assume that the first bulb is red. Since there must be a blue bulb next to each red bulb, three red bulbs cannot go in a row. Therefore, among any three consecutive bulbs, at least one bulb must be blue. T... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem №4
On a grid paper, a square with a side of 5 cells is drawn. It needs to be divided into 5 parts of equal area by drawing segments inside the square only along the grid lines. Can it be such that the total length of the drawn segments does not exceed 16 cells? | Answer: yes, it can
(the total length of the segments drawn is... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem №5
In a train, there are 18 identical cars. In some of the cars, exactly half of the seats are free, in some others, exactly one third of the seats are free, and in the rest, all seats are occupied. At the same time, in the entire train, exactly one ninth of all seats are free. In how many cars are all seats... | Answer: in 13 carriages.
Solution. Let's take the number of passengers in each carriage as a unit. We can reason in different ways.
First method. Since exactly one ninth of all seats in the train are free, this is equivalent to two carriages being completely free. The number 2 can be uniquely decomposed into the sum ... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Variant 1.
Write the smallest number with a digit sum of 62, in the notation of which at least three different digits are used. | Answer: 17999999.
Solution: $62=9 \cdot 6+8$. That is, the number is at least a seven-digit number. But if it is a seven-digit number, then in its decimal representation there are exactly 6 nines and 1 eight. According to the problem, the number must contain at least three different digits, so it is at least an eight-... | 17999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Variant 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 A... | 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 |
# 3. Option 1.
The Ivanov family consists of three people: dad, mom, and daughter. Today, on the daughter's birthday, the mother calculated the sum of the ages of all family members and got 74 years. It is known that 10 years ago, the total age of the Ivanov family members was 47 years. How old is the mother now, if s... | Answer: 33.
Solution: If the daughter had been born no less than 10 years ago, then 10 years ago the total age would have been $74-30=44$ years. But the total age is 3 years less, which means the daughter was born 7 years ago. The mother is now $26+7=33$ years old. | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Variant 1.
The figure shows a rectangle composed of twelve squares. The perimeter of this rectangle is 102 cm. What is its area? Express your answer in square centimeters.
 | Answer: 594.
Solution. We will call the squares large (one such), medium (three such), and small (eight such). Let's denote the side of the medium square as $4a$. Then the side of the large square is $12a$, and the side of the small squares is $12a: 4=3a$. Therefore, the sides of the rectangle are $12a$ and $4a+12a+3a... | 594 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 5. Variant 1.
The organizers of a ping-pong tournament have only one table. They call up two participants who have not yet played against each other. If, after the game, the losing participant has suffered their second defeat, they are eliminated from the tournament (there are no draws in tennis). After 29 games hav... | Answer: 16.
Solution: Each player 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, they have collectively suffered $2n$ losses, while the "finalists" could have $0 (0+0)$, $1 (0+1)$, or $2 (1... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 6. Variant 1.
A diagonal of a 20-gon divides it into a 14-gon and an 8-gon (see figure). How many of the remaining diagonals of the 20-gon intersect the highlighted diagonal? The vertex of the 14-gon is not considered an intersection.
, but it can consist of 2 and 3, and it equals 32. Arranging the remaining digits in descending ... | 8654232 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1.
On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, 2023 natives, among whom $N$ are liars, stood in a circle and each said: "Both of my neighbors are liars." How many different values can $N$ take? | Answer: 337.
Solution: Both neighbors of a knight must be liars, and the neighbors of a liar are either two knights or a knight and a liar. Therefore, three liars cannot stand in a row (since in this case, the middle liar would tell the truth). We can divide the entire circle into groups of consecutive liars/knights. ... | 337 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.1. The students of a school went on a trip in six buses. The number of students in the buses was not necessarily equal, but on average, there were 28 students per bus. When the first bus arrived at the destination, the average number of students in the buses that continued moving became 26. How many students were in ... | Answer: 38.
Solution: The initial total number of schoolchildren was $28 \cdot 6=168$. After the first bus finished its trip, there were $26 \cdot 5=130$ schoolchildren left. Therefore, there were $168-130=38$ schoolchildren in the first bus.
Comment: A correct answer without justification - 0 points. | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. On weekdays (from Monday to Friday), Petya worked out in the gym five times. It is known that in total he spent 135 minutes in the gym, and the time spent in the gym on any two different days differed by at least 7 minutes. What is the maximum duration that the shortest workout could have been? | Answer: 13 minutes.
Solution. Let the minimum training time be $x$ minutes, then the second (in terms of duration) is no less than $x+7$, the third is no less than $x+14$, the fourth is no less than $x+21$, and the fifth is no less than $x+28$. Therefore, the total duration of the trainings is no less than $5 x+70$ mi... | 13 | Other | math-word-problem | Yes | Yes | olympiads | false |
8.5. One hundred and one numbers are written in a circle. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 101 written numbers? | Answer: 41.
Solution. Consider any 5 consecutive numbers. Among them, there is a positive one. Fix it, and divide the remaining 100 into 20 sets of 5 consecutive numbers. In each such set, there will be at least two positive numbers. Thus, the total number of positive numbers is at least $1+2 \cdot 20=41$. Such a situ... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the board, three two-digit numbers are written, one of which starts with 5, the second with 6, and the third with 7. The teacher asked three students to each choose any two of these numbers and add them. The first student got 147, and the answers of the second and third students are different three-digit numbers ... | Answer: Only 78.
Solution: Let the number starting with 7 be denoted as $a$, the number starting with 6 as $b$, and the number starting with 5 as $c$. The sum $a+b \geqslant 70+60=130$, so it must be equal to 147. The maximum sum of numbers starting with 7 and 6 is $69+79=148$. The number 147 is only 1 less than 148, ... | 78 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a deck of 52 cards, each person makes one cut. A cut consists of taking the top $N$ cards and placing them at the bottom of the deck, without changing their order.
- First, Andrey cut 28 cards,
- then Boris cut 31 cards,
- then Vanya cut 2 cards,
- then Gena cut several cards,
- then Dima cut 21 cards.
The last... | Answer: 22.
Solution: Removing $N$ cards will result in the same outcome as moving $N$ cards one by one from the top to the bottom. We will consider that each of the boys moved one card several times.
After the last move, the order of the cards returned to the initial state, meaning the total number of card moves was... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Two equilateral triangles $C E F$ and $D I H$ are positioned as shown in the diagram. The diagram indicates the measures of some angles. Find the measure of angle $x$. Provide your answer in degrees.
- $\angle C E M=60^{\circ}$: this is the angle of the equilateral triangle $C E F$;
- $\angle ... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2. Solve the system of equations
$$
\left\{\begin{aligned}
10 x^{2}+5 y^{2}-2 x y-38 x-6 y+41 & =0 \\
3 x^{2}-2 y^{2}+5 x y-17 x-6 y+20 & =0
\end{aligned}\right.
$$ | Solution: We will eliminate the product $xy$. For example, multiply the first equation by 5, the second by 2, and add the left and right parts of the obtained equations. We get $56x^2 + 21y^2 - 224x - 42y + 245 = 0$. Divide the equation by 7, and then complete the squares for $x$ and $y$. We have $8(x-2)^2 + 3(y-1)^2 =... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. The distance between the midpoints of sides $AB$ and $CD$ of a convex quadrilateral $ABCD$ is equal to the distance between the midpoints of its diagonals. Find the angle formed by the lines $AD$ and $BC$ at their intersection. Justify your answer.
 $E M=A D / 2=F N ; 2) F M=B C / 2=E N$ (points $M, N, E, F-$ are the midpoints of segments $A B, C D, D B, A C$, respectively). Therefore, quadrilateral $E M F N$ is a parallelogram. According to the condition, its diagonals ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In the class, there are 30 students: excellent students, average students, and poor students. Excellent students always answer questions correctly, poor students always make mistakes, and average students answer the questions given to them strictly in turn, alternating between correct and incorrect answers. All stud... | Answer: 20 C-students
Solution. Let $a$ be the number of excellent students, $b$ be the number of poor students, $c$ be the number of C-students who answered the first question incorrectly, answered the second question correctly, and answered the third question incorrectly (we will call these C-students of the first t... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. What is the maximum number of natural numbers not exceeding 2016 that can be marked so that the product of any two marked numbers is a perfect square? | Answer: 44.
Solution. Let's find the number of natural numbers whose squares are no greater than 2016. There are 44 such numbers, since $44^{2}=1936 < 2016$. Since the product of two perfect squares is a perfect square, the numbers $1=1^{2}, 4=2^{2}, \ldots, 1936=44^{2}$ can be marked.
We will prove that it is imposs... | 44 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. If you add the last digit of a number to the number itself, you get 5574, and if you add the second-to-last digit, you get 557. What is this number? | Answer: 5567
Solution. Note that the guessed number differs from 5574 by one digit, so the guessed number is not less than $5574-9=5565$ and not more than 5574. Then the second to last digit of the number is 6 or 7, so our number is 5566 or 5567. The first one does not fit, but the second one does. | 5567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. In the table, there are numbers. It turned out that six sums: three sums by rows and three sums by columns are the same. The minimum of the numbers $a, b$, and $c$ is 101. Find the sum $a+b+c$.
| $\mathrm{a}$ | 1 | 4 |
| :---: | :---: | :---: |
| 2 | $\mathrm{~b}$ | 5 |
| 3 | 6 | $\mathrm{c}$ | | Answer: 309
Solution. Note that $a+5=b+7=c+9$, therefore, the number $c$ is the smallest, and the others are 2 and 4 more, that is, 103 and 105. The sum of all three numbers $101+103+105=309$ | 309 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. Alina and Masha wanted to create an interesting version of the school tour olympiad. Masha proposed several problems and rejected every second problem of Alina's (exactly half), Alina also proposed several problems and did not reject only every third problem of Masha's (exactly a third). In the end, there were 10 ... | Answer: 15
Solution. In total, 17 tasks were rejected. Note that among the tasks proposed by Alina, an equal number were included and rejected, and among the tasks proposed by Masha, the number of rejected tasks is twice the number of tasks that remained in the variant. Since the number of rejected tasks is 7 more tha... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4.1. On a grid sheet, a $1 \times 5$ rectangle was painted. Each minute, all those uncolored cells that have at least one side-adjacent cell already colored are colored. For example, after one minute, 17 cells will be colored. How many cells will be colored after 5 minutes? | Answer: 105
Solution 1. Note that after 5 minutes, we will get a stepped figure, in which the rows will have $5,7,9,11,13,15,13,11,9,7$ and 5 cells. In total, this is 105.
Solution 2. Note that if we shorten the strip to $1 \times 1$, 44 cells will disappear, which is $4 \cdot 11=44$. If there is only one cell at the... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. A four-digit number is called "beautiful" if it is impossible to append a digit to the right so that the resulting five-digit number is divisible by 11. How many beautiful numbers are there that are greater than 3100 and less than 3600? | Answer: 46
Solution. Note that the number 3101 is beautiful because 31009 and 31020 are divisible by 11, so 3101* cannot be divisible by 11. Suppose the number is beautiful, then we have found 10 consecutive numbers, none of which are divisible by 11, but among 11 consecutive numbers, at least one is divisible by 11, ... | 46 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the minimum value of the expression $\frac{25 x^{2} \sin ^{2} x+16}{x \sin x}$ for $0<x<\pi$. | Answer: 40.
Solution. Let $y = x \sin x$. Then the expression can be written as
$$
\frac{25 y^{2} + 16}{y} = 25 y + \frac{16}{y}.
$$
Note that $y > 0$ (since $x > 0$ and $\sin x > 0$), so we can apply the inequality between the means:
$$
25 y + \frac{16}{y} \geq 2 \sqrt{25 y \cdot \frac{16}{y}} = 40.
$$
Equality h... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a right-angled triangle, the lengths of the leg $a$ and the hypotenuse $c$ are expressed as two-digit integers, and both numbers use the same set of digits. The length of the leg $b$ is a positive rational number. What can the length of the hypotenuse be? (Provide all answers and prove that there are no others.) | Answer: 65.
Solution. Since $a \neq c$, in two-digit numbers, two different digits must be used, and they will stand in different places. Let $a=10 e+d, c=10 d+e$. Then $b=\sqrt{(10 d+e)^{2}-(10 e+d)^{2}}=3 \sqrt{11(d-e)(d+e)}$.
By the condition, $b$ is rational, so it cannot contain square roots. The number 11 is pr... | 65 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number $N$, which is divisible by $r$, ends in $r$, and has the sum of its digits equal to $p$, given that $p$ is a prime number and $2p+1$ is the cube of a natural number. | Answer: 11713.
Solution. Let $2 p+1=n^{3}$. Then $(n-1)\left(n^{2}+n+1\right)=2 p$. The number $2 p$ can only have the following positive divisors: $1,2, p, 2 p$. The number $n$ is obviously odd, so $n-1$ is divisible by 2. The number $n^{2}+n+1$ is greater than 1, so $n-1=2, n^{2}+n+1=p$. From this, $n=3, p=13$.
The... | 11713 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A paper triangle with sides $34, 30, 8 \sqrt{13}$ was folded along the midlines and formed into a triangular pyramid. Through the opposite edges of the pyramid, parallel planes were drawn (a total of 6 planes). Prove that the parallelepiped formed by the intersection of these planes is a rectangular parallelepiped, ... | Answer: 1224.
Solution: All four faces of the pyramid are equal, so the opposite edges of the pyramid are pairwise equal. On opposite faces of the parallelepiped, different diagonals are equal, as they are the opposite edges of the pyramid (see figure). Thus, in the parallelogram, which is a face of the parallelepiped... | 1224 | Geometry | proof | Yes | Yes | olympiads | false |
3. In the store, apples were sold. On the second day, they sold a quarter of the amount of apples sold on the first day, and an additional eight kilograms. On the third day, they sold a quarter of the amount of apples sold on the second day, and an additional eight kilograms. How many kilograms of apples were sold on t... | Solution. Let's reason from the end. On the third day, 18 kg of apples were sold. If we subtract 8 kilograms, the remaining 10 kg will be a quarter of the amount sold on the second day. Therefore, 40 kilograms of apples were sold on the second day. Of these, 32 kilograms are a quarter of the amount sold on the first da... | 128 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the swamp, there are 64 bumps, they are arranged in an $8 \times 8$ square. On each bump sits a frog or a toad. Frogs never lie, while toads only tell lies. Each of them, both frogs and toads, croaked: “At least one of the neighboring bumps has a toad.” What is the maximum number of toads that could sit on these ... | Solution. Note that frogs cannot sit on adjacent lily pads, otherwise they would be telling the truth. If we divide the lily pads into pairs, there will be no more than one frog in each pair, otherwise they would be telling the truth. Therefore, there can be no more than 32 frogs, as the lily pads can be divided into 3... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Vanya thought of a seven-digit number, and then subtracted from it the sum of all the digits except one. He got 9875352. What number would Vanya have gotten if he had subtracted the sum of all the digits except the second from the left? | Solution. Since the number is a seven-digit number, the sum of its digits is no more than 63. Adding 63 to 9875352, we get an upper estimate for the original number: 9875415. We see that in the intended number, after subtracting the sum of all digits, the first four digits do not change (the hundreds place initially ha... | 9875357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the largest natural number without repeating digits, in which the product of any two consecutive digits is divisible by 6. | 3. Answer: 894326705.
Among two adjacent digits, there is certainly one divisible by 3, and there are only four such digits. Therefore, a 10-digit number is impossible. Let's try to construct a 9-digit number. The digits $0,3,6,9$, which are multiples of three, must be in even positions. Moreover, next to 3 and 9, the... | 894326705 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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