problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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5. Lydia likes a five-digit number if none of the digits in its notation is divisible by 3. Find the total sum of the digits of all five-digit numbers that Lydia likes. | Solution. Lidia likes five-digit numbers composed only of the digits $1,2,4,5,7,8$. Note that all such numbers can be paired in the following way: each digit $a$ is replaced by $9-a$, that is, 1 is replaced by 8, 2 by 7, 4 by 5, and vice versa. For example, the number 42718 is paired with the number 57281. This partiti... | 174960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In each cell of a $10 \times 10$ table, a natural number was written. Then, each cell was shaded if the number written in it was less than one of its neighbors but greater than another neighbor. (Two numbers are called neighbors if they are in cells sharing a common side.) As a result, only two cells remained unshad... | Solution. Answer: 20. The estimate matches the estimate in problem 7.5. One of the possible examples is shown below.
| 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 8 | 10 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
| 7 | 8 | 10 | 12 | 13 | 14 ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. Two circles with radii 1 and 2 have a common center $O$. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.
 | Answer: $60^{\circ}$
Solution: Let $\angle A O B=\alpha$. Then the area of the shaded part is $\frac{360^{\circ}-\alpha}{360^{\circ}} \pi+$ $4 \frac{\alpha}{360^{\circ}} \pi-\frac{\alpha}{360^{\circ}} \pi=\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi$. The area of the large circl... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.4. In a row, $n$ integers are written such that the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative. For what largest $n$ is this possible? | Answer: 16
Solution: Let's provide an example for $n=16$:
$80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80$. We will prove that for $n \geq 17$, it will not be possible to write down a sequence of numbers that satisfy the conditions of the problem. Let's construct a table for the first 17 n... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.5. Given a function $f(x)$, satisfying the condition
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
What is $f(2017)$, if it is known that $f(0)=1$? | Answer: 2018
Solution: $f(0 \cdot 0+1)=f(0) f(0)-f(0)-0+2, f(1)=1-1+2=2$,
$$
f(2017 \cdot 0+1)=f(2017) f(0)-f(0)-2017+2
$$
$$
\begin{gathered}
f(1)=f(2017)-1-2017+2=2 \\
f(2017)=2018
\end{gathered}
$$
Criteria: Only for the correct answer - 1 point. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be?
Solution. Let the first number be $a$, and the second $b$. Then the equation $1996 a+\frac{b}{96}=a+b$ must hold, from which we find that $2016 a=b$. Therefore, their quotient... | Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 14 points; correct answer without solution - 2 points. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
untranslated text remains the same as requested. | 2. Find the sum of the digits of all numbers in the sequence $1,2,3, \ldots, 199,200$.
Answer: 1902 | 1902 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. A parallelepiped is composed of white and black unit cubes in a checkerboard pattern. It is known that the number of black cubes is $1 \frac{12}{13} \%$ more than the number of white cubes. Find the surface area of the parallelepiped, given that each side of the parallelepiped is greater than 1. | Solution. The number of black and white cubes can differ by only 1. Therefore, 1 cube is $1 \frac{12}{13} \%$ of the quantity of white cubes. Thus, there are 52 white cubes, 53 black cubes, and a total of 205 cubes. That is, our parallelepiped is $3 \times 5 \times 7$. The surface area is 142.
Answer: 142.
Criteria: ... | 142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5? | 2. How many three-digit numbers exist where all digits are odd numbers, and all two-digit numbers that can be obtained by erasing one of these digits are not divisible by 5?
Answer: 80 | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.2. Let $f(n)$ be equal to the product of the even digits of the natural number $\mathrm{n}$ or be zero if there are no even digits. Find the sum $f(1)+f(2)+\cdots+f(100)$. | Answer: 620
Solution: for a single-digit number $n$, $f(n)$ will be equal to $n$ itself if it is even and 0 if it is odd. For single-digit $\mathrm{n}$, we get the sum $2+4+6+8=20$. If $\mathrm{n}$ is a two-digit number, let's consider the cases:
1) If both digits are even. Then the first digit can be $2,4,6$ or 8, a... | 620 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students? | Problem 2. After the mathematics Olympiad, five students noticed that any two of them solved no more than 9 problems in total. What is the maximum number of problems that could have been solved by all the students?
| Solution | Criteria |
| :---: | :---: |
| Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \mathrm... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Solve the system of equations $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=10 \\ \sqrt[4]{x}+\sqrt[4]{y}=4\end{array}\right.$ and find the value of the product $x y$. | Answer: 81
Problem 5. From the vertex of the right angle $K$ of triangle $MNK$, a perpendicular $KL$ is drawn to the plane of the triangle, equal to 280. Find the distance from point $L$ to the line $MN$, given that the height of the triangle dropped from vertex $K$ is 96.
Answer: 296
Problem 6. The bank allocated a... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 12. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$ with side $3 \sqrt{2}$. Find the volume of a regular tetrahedron, one vertex of which coincides with point $A$, and the other three vertices belong to the plane $C M A_{1} N$, where $M$ and $N$ are the midpoints of edges $D D_{1}$ and $B B_{1}$. | Answer: 9
Problem 13. Solve the equations for all natural $n$:
$$
\cos ^{4} x+\sin x(\sin x+1)\left(\cos ^{2} x+\sin x-1\right)=n
$$
In the answer, write the number of roots in the interval $[0,2 \pi]$
Answer: 3
Problem 14. $M$ is the midpoint of the lateral side $A B$ of trapezoid $A B C D$, and $E$ is the inters... | 3986 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
 | Problem 4. A rectangle is divided into six squares (see figure). What is the side of the larger square if the side of the smaller one is 2.
Answer: 14 | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 6. In an acute-angled triangle $\mathrm{ABC}$, angle $\mathrm{B}$ is $30^{\circ}, \mathrm{BC}=12$. The altitude CD of triangle $\mathrm{ABC}$ and the altitude DE of triangle BDC are drawn. Find BE. | Answer: 9
Task 7.3a 2016 The number of books in the school library fund increased by $0.4 \%$, and in 2017 - by $0.8 \%$, remaining less than 50 thousand. By how many books did the library fund increase in 2017?
Answer: 251
Task 8. A rectangle is divided into six squares (see figure). What is the side of the larger ... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 7. On a circle, there are 25 non-overlapping arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the arc following it in a clockwise direction. What can the sum of all the numbers be? | Problem 7. On a circle, there are 25 non-intersecting arcs, and on each of them, two arbitrary prime numbers are written. The sum of the numbers on each arc is not less than the product of the numbers on the next arc clockwise. What can the sum of all the numbers be?
| Solution | Criteria |
| :---: | :---: |
| Let the... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In a convex quadrilateral $A B C D: A B=A C=A D=B D$ and $\angle B A C=\angle C B D$. Find $\angle A C D$. | Answer: $70^{\circ}$.
Solution: Triangle $A B D$ is equilateral, so the angles $\angle A B D=\angle B D A=$ $\angle D A B=60^{\circ}$. Let $\angle B A C=\angle C B D=\alpha$, then $\angle A B C=60^{\circ}+\alpha . A B=A C$, thus $\angle A C B=\angle A B C=60^{\circ}+\alpha$. The sum of the angles in triangle $A B C$ i... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In how many ways can a pile of 100 stones be divided into heaps so that the number of stones in any two heaps differs by no more than one? | Answer: 99.
Solution: We will prove that for any $k$ from 2 to 100, the pile can be divided into $\mathrm{k}$ such piles in a unique way. $100 = q k + r$, where $q$ and $r$ are the quotient and remainder of 100 when divided by $k$, respectively. Suppose there is a pile with no more than $q-1$ stones, then in the remai... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Find the remainder when the number 20172017... 2017 (2017 written 100 times) is divided by 8. | Answer: 1
8. MN and $\mathrm{PQ}$ are two parallel chords located on opposite sides of the center of a circle with radius $10 . \mathrm{MN}=12, \mathrm{PQ}=16$. Find the distance between the chords.
Answer: 14 | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one? | Task 5. In how many ways can a pile of 100 stones be divided into piles so that the number of stones in any two piles differs by no more than one?
## Solutions and Evaluation Criteria
| Task Number | Solution | Criteria |
| :---: | :---: | :---: |
| 1 | Answer: No. Suppose this is possible. No digit is 0, because the... | 99 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the largest natural number consisting of distinct digits such that the product of the digits of this number is 2016. | Answer: 876321.
Solution: Factorize the number 2016. $2016=2^{5} \cdot 3^{2} \cdot 7$. To make the number as large as possible, it should contain the maximum number of digits. Notice that the number must include the digit 1. Therefore, the number should consist of the digits $1,2,3,6,7,8$.
Grading Criteria. 20 points... | 876321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The parabolas in the figure are obtained by shifting the parabola $f(x)=x^{2}$ and are arranged such that the points of their intersections with the $O X$ axis respectively coincide in pairs, and all vertices lie on the same straight line. There are a total of 2020 parabolas. The length of the segment on the $O X$ a... | Solution. Note that the arrangement of parabolas can be changed by shifting the entire set to the left or right (the lengths of the segments on the $O X$ axis, enclosed between the roots of the parabolas, do not change with the shift). The intersection point of the line drawn through the vertices of the parabolas will ... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Three numbers are stored in a computer's memory. Every second, the following operation is performed: each number in this triplet is replaced by the sum of the other two. For example, the triplet $(1 ; 3 ; 7)$ turns into $(10 ; 8 ; 4)$. What will be the difference between the largest and smallest number in the triple... | 5. Answer: 19. Solution. Let the initial triplet be ( $a ; b ; c$). Since all numbers in the triplet are distinct, we will assume that $a<b<c$. In the next second, the triplet of numbers will look like this: $(b+c$; $a+c ; a+b)$. The largest number in this triplet is $b+c$, and the smallest is $a+b$. Their difference i... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. Find the largest four-digit number in which all digits are different and which is divisible by any of its digits (don't forget to explain why it is the largest). | Answer: 9864.
Firstly, the desired number cannot have the form $\overline{987 a}$, because divisibility by the digit 7 would mean that $a$ is 0 or 7. This means the desired number is smaller. Secondly, consider numbers of the form $\overline{986 a}$. Divisibility by the digit 9 would mean that $9+8+6+a=a+23$ is divisi... | 9864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. To some number $\kappa$, the sum of its digits was added and the result was 2014. Provide an example of such a number. | Answer: 1988 or 2006.
Grading Criteria:
+ a correct example is provided (one is sufficient)
$\pm$ a correct example is provided along with an incorrect one
- the problem is not solved or solved incorrectly | 1988 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.2. The Wolf, the Hedgehog, the Chizh, and the Beaver were dividing an orange. The Hedgehog got twice as many segments as the Chizh, the Chizh got five times fewer segments than the Beaver, and the Beaver got 8 more segments than the Chizh. Find out how many segments were in the orange, if the Wolf only got the peel. | Answer: 16 segments.
Solution. First method. Let the number of orange segments given to Chizh be $x$, then Hedgehog received $2x$ segments, and Beaver received $5x$ segments (Wolf - 0 segments). Knowing that Beaver received 8 more segments than Chizh, we set up the equation: $5x - x = 8$. The solution to this equation... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Yura was walking down the road and met a tractor pulling a long pipe. Yura decided to measure the length of the pipe. For this, he walked along it "against the direction of the tractor" and counted 20 steps. After that, he walked along the pipe "in the direction of the tractor" and counted 140 steps. Knowing that ... | Answer: 35 m.
Let's say that in the time it takes Yura to make 20 steps, the train travels $x$ meters. Denoting the length of the train by $L$, we get: $20=L-x, 140=L+7 x$. From this, $L=(140+20 \cdot 7): 8=35$. | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.2. In the quarry, there are 120 granite slabs weighing 7 tons each and 80 slabs weighing 9 tons each. A railway platform can load up to 40 tons. What is the minimum number of platforms required to transport all the slabs? | Answer: 40 platforms
It is impossible to load 6 slabs onto one platform, even if they weigh 7 tons each. Therefore, at least $200 / 5=40$ platforms are needed. Forty platforms are sufficient: on each platform, you can load 3 slabs weighing 7 tons and 2 slabs weighing 9 tons.
Comment. The necessity of at least 40 plat... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. All seats at King Arthur's round table are numbered clockwise. The distances between adjacent seats are the same.
One day, King Arthur sat at seat number 10, and Sir Lancelot sat directly opposite him at seat number 29. How many seats are there in total at the round table? | Answer: 38.
Solution. Along one side of the table between Arthur and Lancelot are seats numbered $11,12, \ldots, 28$ - a total of exactly 18 seats. Since these two are sitting directly opposite each other, there are also 18 seats on the other side of the table. Therefore, the total number of seats at the table is $18+... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.7. At the festival of namesakes, 45 Alexanders, 122 Borises, 27 Vasily, and several Gennady arrived. At the beginning of the festival, all of them stood in a row so that no two people with the same name stood next to each other. What is the minimum number of Gennady that could have arrived at the festival? | Answer: 49.
Solution. Since there are a total of 122 Boris, and between any two of them stands at least one non-Boris (Alexander/Vasily/Gennady), there are at least 121 non-Boris. Since there are a total of 45 Alexanders and 27 Vasilies, there are at least $121-45-27=49$ Gennadys.
Note that there could have been exac... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. Dasha calls a natural number special if four different digits are used to write it. For example, the number 3429 is special, while the number 3430 is not special.
What is the smallest special number greater than 3429? | Answer: 3450.
Solution. Note that all numbers of the form $343 \star$ and $344 \star$ are not special. And the next number after them, 3450, is special. | 3450 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.2. At first, the magical island was divided into three counties: in the first county lived only elves, in the second - only dwarves, in the third - only centaurs.
- During the first year, each county where non-elves lived was divided into three counties.
- During the second year, each county where non-dwarve... | Answer: 54.
Solution. Initially, there was 1 county of each kind.
After the first year, there was 1 elven county, 3 dwarf counties, and 3 centaur counties.
After the second year, there were 4 elven counties, 3 dwarf counties, and 12 centaur counties.
After the third year, there were 24 elven counties, 18 dwarf coun... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. A large rectangle consists of three identical squares and three identical small rectangles. The perimeter of the square is 24, and the perimeter of the small rectangle is 16. What is the perimeter of the large rectangle?
The perimeter of a figure is the sum of the lengths of all its sides.
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.6. Ksyusha runs twice as fast as she walks (both speeds are constant).
On Tuesday, when she left home for school, she first walked, and then, when she realized she was running late, she started running. The distance she walked was twice the distance she ran. In the end, it took her exactly 30 minutes to get ... | # Answer: 24.
Solution. Let the distance from home to school be $3 S$, Ksyusha's walking speed be $v$, and her running speed be $-2 v$ (distance is measured in meters, and speed in meters per minute). Then on Tuesday, Ksyusha walked a distance of $2 S$ and ran a distance of $S$. And on Wednesday, she walked a distance... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.8. In the country of Dragonia, there live red, green, and blue dragons. Each dragon has three heads, each of which always tells the truth or always lies. Moreover, each dragon has at least one head that tells the truth. One day, 530 dragons sat around a round table, and each of them said:
- 1st head: “To my ... | Answer: 176.
Solution. Consider an arbitrary red dragon. To the right of this dragon, at least one head must tell the truth. Note that the 1st and 3rd heads cannot tell the truth (since there is a red dragon to the left), so the 2nd head must tell the truth, and to the right of this dragon, there must be a blue dragon... | 176 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Seven boxes are arranged in a circle, each containing several coins. The diagram shows how many coins are in each box.
In one move, it is allowed to move one coin to a neighboring box. What is the minimum number of moves required to equalize the number of coins in all the boxes?
. On each tree, there is a sign that reads: "Two different trees grow next to me." It is known that among all the trees, the statement is false on all lindens and on exactly one birch. How many birches could have been pla... | Answer: 87.
Solution. Let's divide all the trees into alternating groups of consecutive birches and consecutive lindens (by the condition, there are groups of both types).
Suppose there exists a group of at least 2 lindens. Then the truth would be written on the outermost of them (since it is between a birch and a li... | 87 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In front of a pessimist and an optimist, there are glasses (the glasses are identical). Each of them was given water in their glass such that the pessimist's glass turned out to be $60\%$ empty, while the optimist's glass, on the contrary, was $60\%$ full. It turned out that the amount of water in the pess... | Answer: 230.
Solution. The pessimist's glass is $40 \%$ full, while the optimist's is $60 \%$ full. The pessimist has $20 \%$ less water than the optimist, which is $\frac{1}{5}$ of the total volume of the glass. Since this difference is 46 milliliters according to the problem, the total volume of the glass is $46 \cd... | 230 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. On the board, 23 signs are drawn - some pluses and some minuses. If any 10 of them are chosen, there will definitely be at least one plus among them. If any 15 of them are chosen, there will definitely be at least one minus among them. How many pluses are there in total? | Answer: 14.
Solution. Since among any 10 signs there is a plus, the number of minuses on the board is no more than 9 (otherwise, we could choose 10 minuses).
Since among any 15 signs there is a minus, the number of pluses on the board is no more than 14 (otherwise, we could choose 15 pluses).
Then the total number o... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In the grove, there are 140 chameleons - blue and red. One day, several blue chameleons changed their color to red. As a result, the number of blue chameleons decreased by 5 times, and the number of red chameleons increased by 3 times. How many chameleons changed their color? | Answer: 80.
Solution. Let the number of blue chameleons become $x$. Then initially, there were $5 x$ blue chameleons. Accordingly, the number of red chameleons initially was $140-5 x$. Then the number of red chameleons became $3 \cdot(140-5 x)$. Since the total number of chameleons remained the same, we get the equati... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.
These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued fac... | Answer: 75.
Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.
Consider one such pair of faces: on one of them, ... | 75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. For quadrilateral $A B C D$, it is known that $\angle B A C=\angle C A D=60^{\circ}, A B+A D=$ $A C$. It is also known that $\angle A C D=23^{\circ}$. How many degrees does the angle $A B C$ measure?
, so at least $42-1=41$ photos of the first monument were taken. Similarly, at le... | 123 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.5. On the base $AC$ of isosceles triangle $ABC (AB = BC)$, a point $M$ is marked. It is known that $AM = 7, MB = 3, \angle BMC = 60^\circ$. Find the length of segment $AC$.
 | Answer: 17.
Fig. 3: to the solution of problem 9.5
Solution. In the isosceles triangle \(ABC\), draw the height and median \(BH\) (Fig. 3). Note that in the right triangle \(BHM\), the angl... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.8. On the side $CD$ of trapezoid $ABCD (AD \| BC)$, a point $M$ is marked. A perpendicular $AH$ is dropped from vertex $A$ to segment $BM$. It turns out that $AD = HD$. Find the length of segment $AD$, given that $BC = 16$, $CM = 8$, and $MD = 9$.
. Since $B C \| A D$, triangles $B C M$ and $K D M$ are similar by angles, from which we obtain $D K = B C \cdot \frac{D M}{C M} = 16 \cdot \frac{9}{8} = 18$.
Fig. 6: to the solution of problem 10.3
F... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. Greedy Vovochka has 25 classmates. For his birthday, he brought 200 candies to the class. Vovochka's mother, to prevent him from eating them all himself, told him to distribute the candies so that any 16 of his classmates would collectively have at least 100 candies. What is the maximum number of candies ... | Answer: 37.
Solution. Among all 25 classmates, select 16 people with the smallest number of candies given. Note that among them, there is a person who received no less than 7 candies (otherwise, if they all received no more than 6 candies, then in total they received no more than $16 \cdot 6=96$ candies, which is less... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.6. In a convex quadrilateral $A B C D$, the midpoint of side $A D$ is marked as point $M$. Segments $B M$ and $A C$ intersect at point $O$. It is known that $\angle A B M=55^{\circ}, \angle A M B=$ $70^{\circ}, \angle B O C=80^{\circ}, \angle A D C=60^{\circ}$. How many degrees does the angle $B C A$ measure... | Answer: 35.
Solution. Since
$$
\angle B A M=180^{\circ}-\angle A B M-\angle A M B=180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}=\angle A B M
$$
triangle $A B M$ is isosceles, and $A M=B M$.
Notice that $\angle O A M=180^{\circ}-\angle A O M-\angle A M O=180^{\circ}-80^{\circ}-70^{\circ}=30^{\circ}$, so $\angle A C D... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. A square board $30 \times 30$ was cut along the grid lines into 225 parts of equal area. Find the maximum possible value of the total length of the cuts. | Answer: 1065.
Solution. The total length of the cuts is equal to the sum of the perimeters of all figures, minus the perimeter of the square, divided by 2 (each cut is adjacent to exactly two figures). Therefore, to get the maximum length of the cuts, the perimeters of the figures should be as large as possible.
The ... | 1065 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.8. A natural number $1 \leqslant n \leqslant 221$ is called lucky if the remainder of 221
divided by $n$ is divisible by the quotient (in this case, the remainder can be equal to 0). How many lucky numbers are there? | Answer: 115.
Solution. Let for some successful number $n$ the quotient be $k$, and the remainder be $k s$, by definition it is non-negative and less than the divisor $n$ (from the condition it follows that $k$ is a natural number, and $s$ is a non-negative integer.) Then
$$
221=n k+k s=k(n+s)
$$
Since $221=13 \cdot ... | 115 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.2. In the store, there are 9 headphones, 13 computer mice, and 5 keyboards. In addition, there are 4 sets of "keyboard and mouse" and 5 sets of "headphones and mouse". In how many ways can you buy three items: headphones, a keyboard, and a mouse? Answer: 646. | Solution. Let's consider the cases of whether any set was purchased.
- Suppose the set "keyboard and mouse" was purchased, then headphones were added to it. This results in exactly $4 \cdot 9=36$ ways to make the purchase.
- Suppose the set "headphones and mouse" was purchased, then a keyboard was added to it. This re... | 646 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.5. Quadrilateral $ABCD$ is inscribed in a circle. It is known that $BC=CD, \angle BCA=$ $64^{\circ}, \angle ACD=70^{\circ}$. A point $O$ is marked on segment $AC$ such that $\angle ADO=32^{\circ}$. How many degrees does the angle $BOC$ measure?
Fig. 10: to the solution of problem 11.8
Draw the height $S H$ to the base ... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers from 1 to 8 are arranged in a circle. A number is called large if it is greater than its neighbors, and small if it is less than its neighbors. Each number in the arrangement is either large or small. What is the smallest possible sum of the large numbers? | Answer: 23.
Instructions. Adjacent numbers cannot be of the same type, so larger and smaller numbers alternate, and there are four of each. 8 is large. 7 is also large, since a small number must be less than two numbers, and seven is less than only one. 1 and 2 are small. 3 and 4 cannot both be large, as there are onl... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The circle inscribed in the right triangle ABC touches the legs CA and CB at points P and Q, respectively. The line PQ intersects the line passing through the center of the inscribed circle and parallel to the hypotenuse at point N. M is the midpoint of the hypotenuse. Find the measure of angle MCN. | Answer: $90^{\circ}$.
Instructions. Let the center of the inscribed circle be denoted by I. The points of intersection of the line,
parallel to the hypotenuse and passing through I, with th... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The function $y=f(x)$ is such that for all values of $x$, the equality $f(x+1)=f(x)+2x+3$ holds. It is known that $f(0)=1$. Find $f(2018)$. | Solution. Rewrite the condition of the problem as $f(x+1)-f(x)=2 x+3$. Substituting sequentially instead of $x$ the numbers $0,1,2, \ldots, 2017$, we get the following equalities
$$
\begin{aligned}
& f(1)-f(0)=2 \cdot 0+3 \\
& f(2)-f(1)=2 \cdot 1+3 \\
& f(3)-f(2)=2 \cdot 2+3
\end{aligned}
$$
$$
f(2018)-f(2017)=2 \cdo... | 4076361 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.1. In a basket, there are 41 apples: 10 green, 13 yellow, and 18 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What is... | Answer: (a) 13 apples. (b) 39 apples.
Solution. (a) Alyona can take all 13 yellow apples from the basket, for example, if the first 13 apples turn out to be yellow. Since at no point do the yellow apples become fewer than the red ones, Alyona can indeed do this.
(b) We will show that Alyona can take all the apples ex... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.4. On a horizontal floor, there are three volleyball balls with a radius of 18, each touching the other two. Above them, a tennis ball with a radius of 6 is placed, touching all three volleyball balls. Find the distance from the top point of the tennis ball to the floor. (All balls are spherical.)
Fig. 9: to the solution of problem 11.4
Solution. Let the centers of the volleyballs be denoted by $A, B, C$, and the center of the tennis ball by $D$; the top point of the tenn... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.6. Point $M$ is the midpoint of the base $B C$ of trapezoid $A B C D$. A point $P$ is chosen on the base $A D$. Ray $P M$ intersects ray $D C$ at point $Q$. The perpendicular to base $A D$, drawn through point $P$, intersects segment $B Q$ at point $K$.
It is known that $\angle K Q D=64^{\circ}$ and $\angle... | Answer: $39^{2}$.
Fig. 11: to the solution of problem 11.6
Solution. From the condition, it follows that $\angle B K D=\angle K Q D+\angle K D Q=64^{\circ}+38^{\circ}=102^{\circ}$.
Let the... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 11.8. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
a^{2}+a b+b^{2}=11 \\
b^{2}+b c+c^{2}=11
\end{array}\right.
$$
(a) (1 point) What is the smallest value that the expression $c^{2}+c a+a^{2}$ can take?
(b) (3 points) What is the largest value that the expression $c^{2}+c a+a^{2}$ can take... | Answer: (a) 0; (b) 44.
Solution. (a) Note that if $a=c=0$, and $b=\sqrt{11}$, then the given system of equalities is satisfied, and $c^{2}+c a+a^{2}$ equals 0. On the other hand,
$$
c^{2}+c a+a^{2}=\frac{c^{2}}{2}+\frac{a^{2}}{2}+\frac{(c+a)^{2}}{2} \geqslant 0
$$
Thus, the smallest value of the expression $c^{2}+c ... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.1. In a basket, there are 41 apples: 10 green, 13 yellow, and 18 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What ... | Answer: (a) 13 apples. (b) 39 apples. | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.2. In a basket, there are 38 apples: 9 green, 12 yellow, and 17 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What i... | Answer: (a) 12 apples. (b) 36 apples. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.3. In a basket, there are 35 apples: 8 green, 11 yellow, and 16 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What i... | Answer: (a) 11 apples. (b) 33 apples. | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.1.4. In a basket, there are 44 apples: 11 green, 14 yellow, and 19 red. Alyona sequentially takes one apple at a time from the basket. If at any point she has pulled out fewer green apples than yellow ones, and fewer yellow ones than red ones, she will stop taking apples from the basket.
(a) (1 point) What ... | Answer: (a) 14 apples. (b) 42 apples. | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Variant 11.3.3. The polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-7$.
(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.
(b) (2 points) What is the smal... | Answer: (a) -21 . (b) 6. | -21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. $\left(225-4209520: \frac{1000795+(250+x) \cdot 50}{27}\right)=113$
4209520: $\frac{1000795+(250+x) \cdot 50}{27}=112$
$\frac{1000795+(250+x) \cdot 50}{27}=37585$
$1000795+(250+x) \cdot 50=1014795$
$250+x=280$
$x=30$ | Answer: $\boldsymbol{x}=30$.
Criteria. Correct solution - 7 points. $3 a$ for each computational error minus 1 point. Error in the algorithm for solving the equation - 0 points | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. 11088 is $110\%$ of the containers in December compared to November. This means that in November, $11088: 1.10=10080$ containers were manufactured, which is $105\%$ compared to October. This means that in October, $10080: 1.05=9600$ containers were manufactured, which is $120\%$ compared to September. Therefore, in ... | Answer: in September 8000, in October 9600, in November 10080 containers. | 8000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In one lyceum, $76 \%$ of the students have at least once not done their homework, and $\frac{5}{37}$ sometimes forget their second pair of shoes. Find the number of students in the lyceum, if it is more than 1000 but less than 2000. | Solution. Since $76 \%=\frac{76}{100}=\frac{19}{25}$, and the numbers 25 and 37 are coprime, the number of students is a multiple of $25 \cdot 37$, i.e., $925 \mathrm{k}$, where k is a natural number. Since $1000<925 k<2000$, then $\mathrm{k}=2$, and the number of students $925 \cdot 2$ $=1850$.
Answer. 1850
Recommen... | 1850 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Denis housed chameleons that can change color only to two colors: red and brown. Initially, the number of red chameleons was five times the number of brown chameleons. After two brown chameleons turned red, the number of red chameleons became eight times the number of brown chameleons. Find out how many chameleons D... | Solution. Let $t$ be the number of brown chameleons Denis had. Then the number of red chameleons was $5t$. From the problem statement, we get the equation $5 t+2=8(t-2)$. Solving this, we find $t=6$. Therefore, the total number of chameleons is $6 t$, which is 36.
Answer. 36
Recommendations for checking. Only the cor... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. An odd six-digit number is called "simply cool" if it consists of digits that are prime numbers, and no two identical digits stand next to each other. How many "simply cool" numbers exist? | Solution. There are four single-digit prime numbers in total - 2, 3, 5, 7. We will place the digits starting from the least significant digit. In the units place, three of them can stand $-3, 5, 7$. In the tens place, there are also three out of the four (all except the one placed in the units place). In the hundreds p... | 729 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Three boys and 20 girls stood in a row. Each child counted the number of girls who were to the left of them, the number of boys who were to the right of them, and added the results. What is the maximum number of different sums that the children could have obtained? (Provide an example of how such a number could be o... | Solution. Let's consider how the number changes when moving from left to right by one person. If the adjacent children are of different genders, the number does not change. If we move from a girl to a girl, the number increases by 1, and if from a boy to a boy, it decreases by 1. Thus, the smallest number in the row co... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. 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 |
9.4. In the company, some pairs of people are friends (if $A$ is friends with $B$, then $B$ is also friends with $A$). It turned out that among any 100 people in the company, the number of pairs of friends is odd. Find the largest possible number of people in such a company.
(E. Bakayev) | Answer: 101.
Solution: In all solutions below, we consider the friendship graph, where vertices are people in the company, and two people are connected by an edge if they are friends.
If the graph is a cycle containing 101 vertices, then any 100 vertices will have exactly 99 edges, so such a company satisfies the con... | 101 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The numbers $1,2,3, \ldots, 29,30$ were written in a row in a random order, and partial sums were calculated: the first sum $S_{1}$ equals the first number, the second sum $S_{2}$ equals the sum of the first and second numbers, $S_{3}$ equals the sum of the first, second, and third numbers, and so on. The last sum $... | Answer: 23.
Solution: Evaluation: adding an odd number changes the parity of the sum, there are 15 odd numbers, so the parity of the sums changes at least 14 times. Therefore, there will be at least 7 even sums, and thus no more than 23 odd sums.
Implementation: arrange the numbers as follows: 1, then all even number... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
113. In the forest, there grew pines, cedars, and larches, and there were cones on all the trees, and they were equal in number. A gentle breeze blew, and several cones fell to the ground. It turned out that $11\%$ of the cones fell from each pine, $54\%$ from each cedar, and $97\%$ from each larch. At the same time, e... | 11.3. Let $m$ be the number of cones on each tree, $s, k, l-$ the number of pines, cedars, and deciduous trees in the forest. Then the condition of the problem corresponds to the equation $0.3 m(s+k+l)=0.11 m s+0.54 m k+0.97 m l$. This is equivalent to $19(s+k+l)=43 k+86 l$. The right side of this expression is divisib... | 43 | Number Theory | proof | Yes | Yes | olympiads | false |
115. In how many non-empty subsets of the set $\{1,2,3, \ldots, 10\}$ will there be no two consecutive numbers? | 115. 143.
Let $A_{n}$ denote the set of non-empty subsets of the set $\{1,2,3, \ldots, n\}$ that do not contain two consecutive numbers, and let $a_{n}$ be the number of such subsets. Clearly, the set $A_{1}$ consists of the subset $\{1\}$, and the set $A_{2}$ consists of the subsets $\{1\}$ and $\{2\}$. Thus, $a_{1}... | 143 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.3. Lёsha colors cells inside a $6 \times 6$ square drawn on graph paper. Then he marks the nodes (intersections of the grid lines) to which the same number of colored and uncolored squares are adjacent. What is the maximum number of nodes that can be marked? | Answer: 45.
## Reasoning.
Evaluation. Each grid node belongs to one, two, or four squares.
The corner vertices of the original square are adjacent to only one small square each, so Lёsha will not be able to mark them. Therefore, the maximum number of marked nodes does not exceed $7 \cdot 7-4=45$.
, where \( N > 5 \). What is the smallest value that \( N \) can have?
(O. Dmitriev)
# | # Answer: 14.
Solution. The number $N$ can equal 14, as shown, for example, by the quartet of numbers $4, 15, 70, 84$. It remains to show that $N \geqslant 14$.
Lemma. Among the pairwise GCDs of four numbers, there cannot be exactly two numbers divisible by some natural number $k$.
Proof. If among the original four ... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. On the board, the expression $\frac{a}{b} \cdot \frac{c}{d} \cdot \frac{e}{f}$ is written, where $a, b, c, d, e, f$ are natural numbers. If the number $a$ is increased by 1, then the value of this expression increases by 3. If in the original expression the number $c$ is increased by 1, then its value increases b... | # Answer: 60.
First solution. Let the value of the original expression be $A$. Then, as a result of the first operation, the product will take the value $\frac{a+1}{a} \cdot A=A+3$, from which $A=3a$. This means that $A$ is a natural number. Moreover, from this equality, it follows that it is divisible by 3. Similarly... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. $\mathrm{ABCD}$ is a rhombus. Points $E$ and $F$ lie on sides $AB$ and $BC$ respectively, such that $\frac{AE}{BE}=\frac{BF}{CF}=5$. Triangle $DEF$ is equilateral. Find the angles of the rhombus. Solution.
The notation is shown in the figure. The situation where angle $D$ of the rhombus is acute is impossible (see ... | Answer. The angles of the rhombus are $60^{\circ}$ and $120^{\circ}$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.4. In pentagon $M N P Q S: \quad M N=N P=P Q=Q S=S M$ and $\angle M N P=2 \angle Q N S$. Find the measure of angle $M N P$. | Answer: $60^{\circ}$
Solution. Since $\angle S N Q=\angle M N S+\angle P N Q$, we can take a point $T$ on the side $S Q$ such that $\angle S N T=\angle M N S=\angle M S N$, i.e., $N T \| M S$.
Then $\angle T N Q=\angle S N Q-\angle S N T=\angle P N Q=\angle N Q P$, i.e., $N T \| P Q$. Therefore, $M S \| P Q$, and sin... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. The teams participating in the quiz need to answer 50 questions. The cost (in integer points) of a correct answer to each question was determined by experts after the quiz, the cost of an incorrect answer - 0 points. The final score of the team was determined by the sum of points received for correct answers. Whe... | Answer: 50.
Solution: We will prove that with 50 teams, such a distribution of points can exist. The example is obvious - let the $k$-th team answer only the $k$-th question. Then, by assigning the costs of the questions as $a_{1}, a_{2}, \ldots, a_{50}$, where $\left\{a_{1}, a_{2}, \ldots, a_{50}\right\}=\{1,2, \ldot... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9-3-1. The numbers from 1 to 217 are divided into two groups: one group has 10 numbers, and the other has 207. It turns out that the arithmetic means of the numbers in the two groups are equal. Find the sum of the numbers in the group of 10 numbers. | Answer: 1090.
Solution Variant 1. The shortest solution to this problem is based on the following statement:
If the arithmetic means of the numbers in two groups are equal, then this number is also equal to the arithmetic mean of all the numbers.
A formal proof of this is not difficult: it is sufficient to denote th... | 1090 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
 | Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii... | 54 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9-6-1. Six pirates - a captain and five members of his crew - are sitting around a campfire facing the center. They need to divide a treasure: 180 gold coins. The captain proposes a way to divide the treasure (i.e., how many coins each pirate should receive: each pirate will receive a non-negative integer number of coi... | Answer: 59.
Solution variant 1. Let's number the pirates clockwise, starting from the captain: First, Second, ..., Fifth. The main observation in this problem is:
two pirates sitting next to each other cannot both vote "for": one of them will not vote "for" because the captain offers no more to him than to his neighb... | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9-8-1. In Midcity, houses stand along one side of a street, each house can have $1,2,3, \ldots, 9$ floors. According to an ancient law of Midcity, if two houses on one side of the street have the same number of floors, then no matter how far apart they are from each other, there must be a house with more floors between... | Answer: 511.
Solution Variant 1. Estimation. Let $a_{k}$ be the maximum number of houses if all of them have no more than $k$ floors. Clearly, $a_{1}=1$. Let's find $a_{k}$. Consider the tallest house with $k$ floors, which is only one by the condition. Both to the left and to the right of it stand houses with $1,2,3,... | 511 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a finite set of cards. On each of them, either the number 1 or the number -1 is written (exactly one number on each card), and there are 100 more cards with -1 than cards with 1. If for each pair of different cards, the product of the numbers on them is found, and all these products are summed, the result is 1... | Answer: 3950.
Solution: Let the number of cards with 1 be $m$, and the number of cards with -1 be $k$. Then, among all pairs, there are $\frac{m(m-1)}{2}$ pairs of two 1s, $\frac{k(k-1)}{2}$ pairs of two -1s, and $m k$ pairs of 1 and -1. Therefore, the sum in the condition is $\frac{m(m-1)}{2}+\frac{k(k-1)}{2}-m k$, f... | 3950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The sides of a rectangle were reduced: the length - by $10 \%$, the width - by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$? | Answer: by $18 \%$.
Solution. Let $a$ and $b$ be the length and width of the rectangle. After decreasing the length by $10 \%$ and the width by $20 \%$, we get a rectangle with sides $0.9 a$ and $0.8 b$, the perimeter of which is $0.88$ of the perimeter of the original rectangle. Therefore, $2(0.9 a + 0.8 b) = 0.88 \c... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a six-digit number, one digit was crossed out to obtain a five-digit number. The five-digit number was subtracted from the original number, and the result was 654321. Find the original number. | Answer: 727023.
Solution. Note that the last digit was crossed out, as otherwise the last digit of the number after subtraction would have been zero. Let $y$ be the last digit of the original number, and $x$ be the five-digit number after crossing out. Then the resulting number is $10 x + y - x = 9 x + y = 654321$. Di... | 727023 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the three friends either always tells the truth or always lies. They were asked the question: "Is there at least one liar among the other two?" The first answered: "No," the second answered: "Yes." What did the third answer? | Solution Since the first and second friends gave different answers, one of them is a liar, and the other is a knight. Moreover, the knight could not have answered “No” to the question posed, as in that case, he would have lied (there is definitely a liar among the two remaining). Therefore, the first one is a liar. He ... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Tourists arrived at the campsite. For lunch, each of them ate half a can of soup, a third of a can of stew, and a quarter of a can of beans. In total, they ate 39 cans of food. How many tourists were there? Answer: 36. | Solution. Note that 12 tourists ate 6 cans of soup, 4 cans of stew, and 3 cans of beans - that is, a total of 13 cans of food.
39 cans is 3 sets of 13 cans. One set is eaten by 12 people. Therefore, the total number of tourists is 36. | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. Given nine cards with the numbers $5,5,6,6,6,7,8,8,9$ written on them. From these cards, three three-digit numbers $A, B, C$ were formed, each with all three digits being different. What is the smallest value that the expression $A+B-C$ can have? | # Answer: 149.
Solution. By forming the smallest sum of numbers $A$ and $B$, and the largest number $C$, we get the smallest value of the expression $A+B-C$. This is $566+567-988=145$. However, this partition is not valid: two numbers have the same digits. By swapping the digits 6 and 8 in the units place, we get the ... | 149 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. Little kids were eating candies. Each one ate 7 candies less than all the others together, but still more than one candy. How many candies were eaten in total? | Answer: 21 candies.
Solution: Let $S$ denote the total number of candies eaten by the children. If one of the children ate $a$ candies, then according to the condition, all the others ate $a+7$ candies, and thus together they ate $S=a+(a+7)=2a+7$ candies. This reasoning is valid for each child, so all the children ate... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-02.jpg?height=369&width=366&top_left_y=1409&to... | Answer: 57.
Solution. The number of options along the long side is $1+2+3+4+5+6=21$. Along each of the other two directions, it is $-4+4+4+3+2+1=18$. The total number of options is $21+18 \cdot 2=57$ | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-8. In a chess club, 90 children attend. During the session, they divided into 30 groups of 3 people, and in each group, everyone played one game with each other. No other games were played. In total, there were 30 games of "boy+boy" and 14 games of "girl+girl". How many "mixed" groups were there, that is, groups wher... | Answer: 23.
Solution: There were a total of 90 games, so the number of games "boy+girl" was $90-30-14=46$. In each mixed group, two "boy+girl" games are played, while in non-mixed groups, there are no such games. In total, there were exactly $46 / 2=23$ mixed groups.
## Grade 5 | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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