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9.3. Natural numbers $a, x$ and $y$, greater than 100, are such that $y^{2}-1=$ $=a^{2}\left(x^{2}-1\right)$. What is the smallest value that the fraction $a / x$ can take? | # Answer. 2.
First solution. Rewrite the condition of the problem as $y^{2}=a^{2} x^{2}-a^{2}+1$. Notice that $y100$, if we set $a=2 x, y=a x-1=2 x^{2}-1$.
Second solution. We provide another proof of the estimate $a / x \geqslant 2$. Rewrite the equality from the condition as
$$
(a x-y)(a x+y)=a^{2} x^{2}-y^{2}=a^{... | 2 | Number Theory | 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 |
5.1. $B$ of trapezoid $A B C D: \angle A=\angle B=90^{\circ}, A D=2 \sqrt{7}, A B=\sqrt{21}, B C=2$. What is the minimum value that the sum of the lengths $X A+X B+X C+X D$ can take, where $X-$ is an arbitrary point in the plane?
, B\left(x_{b}, y_{b}\right), C\left(x_{c}, y_{c}\right)$ are taken. It is known that $x_{c}=5$ and $y_{a}=y_{b}$. Find the abscissa of the point of intersection of the medians of triangle $A B C$.
. 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 |
7.4. Little One gave Karlson 111 candies. They immediately ate some of them together, $45\%$ of the remaining candies went to Karlson for lunch, and a third of the candies left after lunch were found by Fräulein Bok during cleaning. How many candies did she find? | Answer: 11.
First method. Let Carlson have had $n$ candies before lunch. Then, after lunch, there were $\frac{55}{100} n$ left, and Fräulein Bock found $\frac{1}{3} \cdot \frac{55}{100} n=\frac{11 n}{60}$ candies. Since the number of candies must be an integer, $11 n$ is divisible by 60. Since the numbers 11 and 60 ar... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) A seller bought a batch of pens and sold them. Some customers bought one pen for 10 rubles, while others bought 3 pens for 20 rubles. It turned out that the seller made the same profit from each sale. Find the price at which the seller bought the pens. | Answer: 5 rubles.
Solution. Let the purchase price of a pen be $x$. Then the profit from one pen is $10-x$, and from 3 pens is $20-3x$. Solving the equation $10-x=20-3x$, we get $x=5$.
Criteria. Correct solution by any method: 7 points.
If it is not justified that the purchase price of the pen should be 5 rubles, bu... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11-5. Santa Claus is preparing gifts. He has distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest - 50. How many bags are the candies distributed into? How many candies are in the smallest gift? | Answer: 10 packages, 5 candies.
Solution. Let's number the gifts from the smallest to the largest, from 1 to n. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, the ... | 10 | Logic and Puzzles | 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 |
8.5. The numbers from 1 to 10 were written in some order and resulted in the numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{10}\), and then the sums \(S_{1}=a_{1}\), \(S_{2}=a_{1}+a_{2}\), \(S_{3}=a_{1}+a_{2}+a_{3}\), \ldots, \(S_{10}=a_{1}+a_{2}+a_{3}+\ldots+a_{10}\) were calculated. What is the maximum number of prime num... | Answer: 7.
Among the numbers from 1 to 10, there are five odd numbers. Adding an odd number changes the parity of the sum. Let $y_{1}^{\prime}, y_{2}, y_{3}, y_{4}, y_{5}$ be the odd numbers, in the order they appear on the board. After adding $y_{2}$, the sum will become even and greater than 2, as it will after addi... | 7 | Number Theory | 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).
, 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 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 8.7. In the figure, two equal triangles are depicted: $A B C$ 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), 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?
=\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 11.1. The product of nine consecutive natural numbers is divisible by 1111. What is the smallest possible value that the arithmetic mean of these nine numbers can take? | Answer: 97.
Solution. Let these nine numbers be $-n, n+1, \ldots, n+8$ for some natural number $n$. It is clear that their arithmetic mean is $n+4$.
For the product to be divisible by $1111=11 \cdot 101$, it is necessary and sufficient that at least one of the factors is divisible by 11, and at least one of the facto... | 97 | Number Theory | 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 $D Z$.
 \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Around a circle, 100 integers are written, the sum of which is 1. A chain is defined as several numbers (possibly one) 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 |
1. A 2019-digit number written on the board is such that any number formed by any two adjacent digits (in the order they follow) is divisible by 13. Find the last digit of this number, given that the first digit is 6. | 1. Answer: 2.
Two-digit numbers divisible by 13: $13,26,39,52,65,78,91$. If the first digit of the number is 6, then the second digit must be 5 (forming the number 65, which is divisible by 13), the third digit is 2 (forming 52), and the fourth digit is again 6 (forming 26). Thus, the digits will be arranged in triple... | 2 | Number Theory | 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
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. In the park, paths are laid out as shown in the figure. Two workers started to asphalt them, starting simultaneously from point $A$. They lay asphalt at constant speeds: the first on the section $A-B-C$, the second on the section $A-D-E-F-C$. In the end, they finished the work simultaneously, spending 9 ho... | Answer: 45.
Solution. Let the line $A D$ intersect the line $C F$ at point $G$, as shown in the figure below. Since $A B C G$ and $D E F G$ are rectangles, we have $A B=C G, B C=A G, E F=D G$ and $D E=F G$.
.
For convenience, let's introduce some notations. Let the top-left corner cell of the $5 \times 5$ table be called $A$, and the bottom-right corner ce... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\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 $B C=9$, $C D=30$.
^{2}+1=25 m^{2}+40 m+17$ is not divisible by 5. | Answer: $\left\{\begin{array}{c}x=5 m+2, \\ y=5 m^{2}+4 m+1 .\end{array}\right.$ or $\left\{\begin{array}{c}x=5 m+3, \\ y=5 m^{2}+6 m+2 .\end{array}\right.$ where $m \in \mathbb{Z}$.
5 points - the solution is correct, complete, and contains no errors.
4 points - if the answer is incorrect, the solution method is cor... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given positive numbers $x>y$. Let $z=\frac{x+y}{2}$ and $t=\sqrt{x y}$. It turns out that $(x-y)=3(z-t)$. 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 |
4. In a school chess tournament, boys and girls competed, with the number of boys being five times the number of girls. According to the tournament rules, each chess player played against every other player twice. How many players in total participated if it is known that the boys scored exactly twice as many points in... | Answer: 6 players.
Solution. Let $d$ girls and $5d$ boys participate in the tournament. Then the total number of players was $d + 5d = 6d$; playing two matches each with every other, they played a total of $2 \cdot \frac{1}{2} \cdot 6d(6d-1) = 6d(6d-1)$ matches. Since each match awards one point, the total number of p... | 6 | 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
$$
\frac{M B}{A B}=\frac{M C}{C N}, \q... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
10.7. For natural numbers $a>b>1$, define the sequence $x_{1}, x_{2}, \ldots$ by the formula $x_{n}=\frac{a^{n}-1}{b^{n}-1}$. Find the smallest $d$ such that this sequence does not contain $d$ consecutive terms that are prime numbers, for any $a$ and $b$.
(V. Senderov) | 10.7. Answer. 2.
For $a=4, b=2$ we have $\frac{a^{1}-1}{b^{1}-1}=3, \frac{a^{2}-1}{b^{2}-1}=5$. It remains to show that more than two consecutive prime numbers will not occur.
We will prove a stronger statement than required: for $n \geqslant 2$ at least one of the numbers $\frac{a^{n}-1}{b^{n}-1}, \frac{a^{n+1}-1}{b... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On the board, three quadratic equations are written:
$$
\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 |
2. There are 25 coins, 12 of which are counterfeit and differ in weight by exactly 1 g from the genuine ones. All coins weigh an integer number of grams. Some may be lighter than the genuine ones, while others may be heavier. There are balance scales without weights, with a needle that shows the difference in weight. W... | Solution. We will prove that one weighing is sufficient. Set aside the coin under investigation, and place the rest on the scales, 12 coins on each side. If the scales show a difference in an even number of grams, then the coin is genuine; if in an odd number, then it is counterfeit. Indeed, if an odd number of counter... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the hypotenuse of a right triangle if the height drawn to it is 1 cm, and one of the angles of the triangle is $15^{\circ}$. If the answer is not an integer, round it to the tenths. | Solution. Let's call the original triangle $ABC$. Let $CH$ be the height drawn to the hypotenuse; $\angle C=90^{\circ}$, and $\angle A=15^{\circ}$. Draw the median $CM$. It is clear that $CM=MA=MB$, so triangle $CMA$ is isosceles $\left(CM=MA \text{ and } \angle MCA=\angle MAC=15^{\circ}\right)$. Note that $\angle BMC=... | 4 | Geometry | 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 |
9.2 In the fishing, 11 experienced fishermen and $n$ children participated. Together they caught $n^{2}+$ $5 n+22$ fish, with all experienced fishermen catching the same amount, and all children catching the same amount, but each 10 less than an experienced fisherman. Who was there more of at the fishing - experienced ... | Solution: Let each child catch $m$ fish. Then $n m+11(m+10)=n^{2}+5 n+22$. From this, $(n+11) m=n^{2}+3 n-88$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+5 n-88=$ $(n+11)(n-6)-22$, so 22 is divisible by $n+11$. The only divisor of 22 greater than 11 is 22 itself, so $n+11=22, n=11$. Therefore, the... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4 Is there a rectangular box, all three dimensions of which (height, width, and depth) are expressed as irrational numbers, while the surface area and volume are integers? | Solution 1: Consider a box with sides $\sqrt{2}-1, \sqrt{2}-1,3+2 \sqrt{2}$. Its volume is $(\sqrt{2}-1)^{2}(3+2 \sqrt{2})=(3-2 \sqrt{2})(3+2 \sqrt{2})=1$, and the surface area is $2(\sqrt{2}-1)^{2}+4(\sqrt{2}-1)(3+2 \sqrt{2})=2(3-2 \sqrt{2})+4(\sqrt{2}+1)=10$.
Solution 2: Consider the polynomial $f(x)=x^{3}-30 x^{2}+... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5 What is the largest number of non-overlapping groups into which all integers from 1 to 20 can be divided so that the sum of the numbers in each group is a perfect square? | Solution: A group consisting of a single number can only be formed by 4 squares. The remaining 16 numbers must be divided into groups of at least two. Therefore, there will be no more than 12 groups in total. Let's check that exactly 12 groups are not possible. Indeed, in such a case, the numbers 1, 4, 9, 16 would form... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the numerical value of the expression
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{2}{a b+1}
$$
if it is known that $a$ is not equal to $b$ and the sum of the first two terms is equal to the third. | Answer: 2.
Solution. Let's bring the condition to a common denominator
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}=\frac{2}{a b+1}
$$
we get
$$
\frac{\left(a^{2}+b^{2}+2\right)(a b+1)-2\left(a^{2}+1\right)\left(b^{2}+1\right)}{\left(a^{2}+1\right)\left(b^{2}+1\right)(a b+1)}=0
$$
expand all brackets in the numerator, c... | 2 | Algebra | 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 |
1. Find the number of integer points $(x, y)$ satisfying the equation $\frac{1}{|x|}+\frac{1}{|y|}=\frac{1}{2017}$. | Solution. Transform the original equation $\frac{1}{|x|}+\frac{1}{|y|}=\frac{1}{2017} \Rightarrow$
$2017(|x|+|y|)-|x||y|=0 \Rightarrow 2017 \cdot|x|+2017 \cdot|y|-|x||y|-2017^{2}=-2017^{2}$, from which it follows that $(|x|-2017)(|y|-2017)=2017^{2}$. Since 2017 is a prime number and $|x|$ and $|y|$ are natural numbers... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The brother left the house 6 minutes after his sister, following her, and caught up with her after 12 minutes. How many minutes would it have taken him to catch up if he had walked twice as fast? Both the brother and the sister walk at a constant speed. | # Answer: 3 minutes.
Solution. Since the brother walked for 12 minutes before meeting his sister, and the sister walked for 18 minutes, the brother's speed was $3 / 2$ times the sister's speed. If the brother's speed is 3 times the sister's speed, which is 2 times faster than before, then the difference of 6 minutes w... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the neighboring houses on Happy Street in Sunny Village, two families of 10 people each lived, with the average weight of the members of one family being 1 kg more than the average weight of the members of the other family. After the eldest sons from both families left to study in the city, it turned out that the... | Answer: by 1 kg or by 19 kg.
Solution. The total weight of the members of one of the families before the departure of the elder sons was 10 kg more than the total weight of the members of the other family. If the heavier family remained heavier, then its total weight became 9 kg more than the total weight of the other... | 1 | Algebra | math-word-problem | 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 |
9.5 The numbers $x$, $y$, and $z$ satisfy the equations
$$
x y + y z + z x = x y z, \quad x + y + z = 1
$$
What values can the sum $x^{3} + y^{3} + z^{3}$ take? | Solution 1: Let $x y z=p$. Then, from the condition $x y+y z+z x$ is also equal to $p$. Therefore, by Vieta's theorem, the numbers $x, y$, and $z$ are the roots of the polynomial $t^{3}-t^{2}+p t-p$. However, the number 1 is a root of such a polynomial, so one of the numbers is equal to 1. Then the other two numbers ar... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (7 points) Two vertices, the incenter of the inscribed circle, and the intersection point of the altitudes of an acute triangle lie on the same circle. Find the angle at the third vertex. | Answer: $60^{\circ}$.
Solution. Consider triangle $ABC$, in which altitudes $AA_1$ and $BB_1$ are drawn. Let point $H$ be the orthocenter, and point $I$ be the incenter. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.3 The number $\sqrt{1+2019^{2}+\frac{2019^{2}}{2020^{2}}}+\frac{2019}{2020}$ is an integer. Find it. | Solution. Let $a=2020$. Then the desired number is $\sqrt{1+(a-1)^{2}+\frac{(a-1)^{2}}{a^{2}}}+$ $\frac{a-1}{a}$. We will perform equivalent transformations:
$$
\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 |
8.5 On an island, there live 25 people: knights, liars, and tricksters. Knights always tell the truth, liars always lie, and tricksters answer the questions posed to them in turn, alternating between truth and lies. All the islanders were asked three questions: "Are you a knight?", "Are you a trickster?", "Are you a li... | Solution. Each knight will answer "yes" to the first question and "no" to the other two. Each liar will answer "yes" to the first two questions and "no" to the last one. The tricksters can be divided into those who answered the first question truthfully (tricksters of the first type) and those who answered the first qu... | 4 | Logic and Puzzles | 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 |
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