problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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6. Winnie the Pooh ran down a moving escalator and counted 55 steps. Then he ran up the same escalator at the same speed relative to the escalator and counted 1155 steps. How many steps would he have counted if he had gone down a stationary escalator?
Answer: 105. | Let \( u \) be the speed of Winnie the Pooh, \( v \) be the speed of the escalator (both measured in steps per unit time), and \( L \) be the length of the escalator (in steps). Then, the time Winnie the Pooh spent descending the moving escalator is \( \frac{L}{u+v} \), and during this time, he counted \( \frac{L}{u+v}... | 105 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let $S(k)$ denote the sum of all the digits in the decimal representation of a positive integer $k$. Let $n$ be the smallest positive integer satisfying the condition $S(n)+S(n+1)=$ param1. As the answer to the problem, write down a five-digit number such that its first two digits coincide with the first two digits... | # Solution
Let $R$ be the radius of the circumcircle of triangle KBM and $r$ be the radius of the incircle of triangle ABC. First, we will prove that the centers of the circles mentioned in the problem statement coincide. Indeed, the center $I$ of the incircle of triangle $ABC$ lies on the bisector of angle $BAC$, whi... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given a strictly increasing function $f: \mathbb{N}_{0} \rightarrow \mathbb{N}_{0}$ (where $\mathbb{N}_{0}$ is the set of non-negative integers), which satisfies the relation $f(n+f(m))=f(n)+m+1$ for any $m, n \in \mathbb{N}_{0}$. Find all possible values that $f(2023)$ can take.
(T.A. Garmanova) | Solution. 1) Substituting $m=0, n=0$, we get $f(f(0))=f(0)+1$.
If $f(0)=0$, then we get $f(0)=f(0)+1$, which is impossible.
2) Let $f(0)=a$, then $a \in \mathbb{N}$. From the first point, we get that $f(a)=a+1$. If we substitute $m=0, n=a$, then we get that $f(2a)=f(a)+1=a+2$. Therefore, the values of the function at... | 2024 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number $N>9$, which is not divisible by 7, but if any of its digits is replaced by 7, the resulting number is divisible by 7. | (M. A. Evdokimov) Solution. Let the smallest such number be of the form $\overline{a_{1} a_{2} \ldots a_{n}}$. From the condition, it follows that among its digits there are no 0 and 7. If the number contains digits 8 or 9, then they can be replaced by 1 or 2, respectively, to obtain a smaller number with the same prop... | 13264513 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the relay race Moscow - Petushki, two teams of 20 people each participated. Each team divided the distance into 20 not necessarily equal segments and distributed them among the participants so that each ran exactly one segment (the speed of each participant is constant, but the speeds of different participants ma... | Answer: 38 overtakes.
Solution. First, let's prove that no more than 38 overtakes occurred. Note that between the start and the first overtake, and between two consecutive overtakes, at least one of the teams must have changed the runner. There were 19 changes of runners in each team, meaning a total of 38 changes, so... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given a polynomial of degree 2022 with integer coefficients and leading coefficient 1. What is the maximum number of roots it can have on the interval $(0,1)$?
(A. Kanel-Belov) | Answer: 2021
Solution. If the interval $(0,1)$ contains all 2022 roots of the polynomial, then by Vieta's theorem, the constant term of the polynomial must be equal to their product, and thus will also lie in the interval $(0,1)$ and will not be an integer. We will prove that the polynomial $P(x)=x^{2022}+(1-$ $4042 x... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. M. Evdokimov
}
A natural number is written on the board. If the last digit (in the units place) is erased, the remaining non-zero number will be divisible by 20, and if the first digit is erased, the remaining number will be divisible by 21. What is the smallest number that can be written on the board if its second... | Answer: 1609.
Solution.
The second to last digit of the number is 0, since the number without the last digit is divisible by 20. Therefore, the number is at least four digits. Note that the number remaining after erasing the last digit cannot be 100 according to the condition. Also, this number cannot be 120 or 140, ... | 1609 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. proposed by A. Shen
There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions are allowed:
- moving a stone from the bag to the first cell of the strip or back;
- if there is a stone in the ... | Answer: $\partial a$.
Solution. Note that for each action there is an inverse to it. Therefore, if we get from situation $A$ to situation $B$ by following the rules, we can also get from situation $B$ to situation $A$ by following the rules.
We will show by induction that if there is a reserve of $n$ stones, then, ac... | 1000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Sasha writes down the numbers 1, 2, 3, 4, 5 in some order, places the arithmetic operation signs "++", "-", "×" and parentheses, and looks at the result of the obtained expression. For example, he can get the number 8 using the expression $(4-3) \times (2+5)+1$. Can he get the number 123?
Forming numbers from ... | Answer: 2 yes
Solution. For example, $3 \times(2 \times 4 \times 5+1)=123$. | 123 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. In Alik's collection, there are two types of items: badges and bracelets. There are more badges than bracelets. Alik noticed that if he increases the number of bracelets by a certain (not necessarily integer) factor, without changing the number of badges, then his collection will have 100 items. And if, conversely, ... | Solution. Let Alik have $x$ badges and $y$ bracelets, and the increase occurs by a factor of $n$. Then we get the system
$$
x+n y=100, \quad n x+y=101
$$
By adding and subtracting the equations and eliminating $n$, we obtain
$$
\frac{201}{x+y}-\frac{1}{x-y}=2
$$
We transform this equation to the form
$$
(201-2 u)(... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The Sultan gathered 300 court wise men and proposed a test. There are 25 different colors of hats, known in advance to the wise men. The Sultan informed them that one of these hats would be placed on each of the wise men, and if the number of hats of each color is written down, all the numbers will be different. Eac... | (A. V. Grialko) Solution. Since $0+1+2+\ldots+24=300$, the quantities of caps of different colors take all values from 0 to 24.
Next, each sage counts the number of caps of each color. For two colors, the quantities of caps coincide, and the sage understands that he is wearing a cap of one of these two colors. It only... | 150 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Provide an example of a nine-digit natural number that is divisible by 2 if the second (from the left) digit is erased, by 3 if the third digit is erased in the original number, ..., and by 9 if the ninth digit is erased in the original number. | 1. Answer. For example, 900900000.
Note. In fact, there are 28573 numbers that satisfy the conditions of the problem, the smallest of which is 100006020, and the largest is 999993240. | 900900000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Tanya was sequentially writing down numbers of the form $n^{7}-1$ for natural numbers $n=2,3, \ldots$ and noticed that for $n=8$ the resulting number is divisible by 337. For what smallest $n>1$ will she get a number divisible by $2022?$ | (T. A. Garmanova) Solution. Let the natural number $n$ be such that $n^{7}-1$ is divisible by $2022=2 \cdot 3 \cdot 337$. Then $n^{7}-1$ is divisible by 2 and 3, so $n$ is an odd number, having a remainder of 1 when divided by 3. In addition, $n^{7}-1$ is divisible by 337. Note that if two numbers are congruent modulo ... | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. (A. Zaslavsky) In a certain state, there are 32 cities, each of which is connected by a road with one-way traffic. The Minister of Communications, a secret villain, decided to organize traffic in such a way that after leaving any city, it would be impossible to return to it. To achieve this, he can change th... | Solution. We will prove the general formula. Let there be $2^{n}$ cities in the state. Then the prime minister can achieve the desired result in no more than $2^{n-2}\left(2^{n}-n-1\right)$ days.
Lemma. Suppose there are $2k$ cities in the state, each two of which are connected by a one-way road. Choose half of them w... | 208 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. First solution. Let $a_{1}, a_{2}, \ldots, a_{15}$ be the days of the month that were sunny. Then Andrey Stepanovich will not drink a single drop from the 1st to the $a_{1}$-th day, he will drink one drop per day from the $a_{1}$-th day to the $a_{2}$-th day (including the $a_{1}$-th day but not the $a_{2}$-th day),... | The second solution. First, let's consider the situation where the first fifteen days of April were cloudy, and the last fifteen were sunny. It is easy to check that both characters in the problem will drink $1+2+3+\ldots+15=120$ drops of valerian - indeed, equally.
Let's swap some sunny day $s$ with some cloudy day $... | 120 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. At the vertices of the quadrilateral $A_{1} A_{2} A_{3} A_{4}$ (Fig. 62), the following masses are placed: $2, 6, 2, 3$. Calculate the moment of inertia of the resulting system of material points relative to the point $S$. (The side of each square on Fig. 62 is 1 unit.) (O. 117, P. 177.)
+6 \cdot 2^{2}+2\left(3^{2}+4^{2}\right)+3 \cdot 8^{2}=316
\end{aligned}
$$ | 316 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. There are two solutions of wine spirit in water ${ }^{1}$ ): The first solution weighs 4002 and contains $30 \%$ spirit, the second solution weighs 6002 and contains $80 \%$ spirit.
From these two solutions, one solution is made. What percentage of spirit will it contain? (O. 117, U. 126, P. 215.$)$ | 2. a) Let's represent the solutions using material points. For this, we will take a segment $AB$ of one unit length (Fig. 172). Let point $A$ represent pure water
[Boda] (cimupm.)
Fig. 172.... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. On the plane, there are a (generally non-convex) quadrilateral and a pentagon, and no vertex of one lies on the side of the other. What is the maximum possible number of intersection points of their sides? | 15. Note that no line can intersect the sides of a polygon in an odd number of points, since otherwise, moving along this line in a certain direction - after entering the polygon for the last time, we would not be able to leave its boundaries. Therefore, each side of the quadrilateral can have no more than four interse... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. Every two adjacent sides of a flat (self-intersecting!) 14-sided polygon are mutually perpendicular; no two sides of it lie on the same line. What is the maximum possible number of self-intersection points of the sides of such a polygon? | 16. Let's agree to consider that all sides of the 14-sided polygon are either "horizontal" or "vertical." Clearly, in this case, exactly 7 sides are horizontal (and 7 are vertical): after all, from each vertex, one horizontal (and one vertical) side extends. Summing over all 14 vertices, we count 14 horizontal sides; b... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18. a) Into how many parts can a plane be divided by two closed curves, one of which is a circle and the other is the boundary of a square?
b) $* *$ Into how many parts can space be divided by two closed surfaces, one of which is a sphere and the other is the surface of a cube? | 18. a) The boundary of a square and a circle can divide the plane into parts of the following types:
1) the part located outside the square and outside the circle; such a part is always one;
2) the part located inside the square and inside the circle; if the square does not intersect the circle, then there are no such ... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In how many different ways can the faces of a cube be painted using six given colors (each face must be painted entirely with one color), if only those colorings are considered different that cannot be made to coincide by rotating the cube? | 4. Suppose the faces of a cube are painted green, blue, red, yellow, white, and black. Let's place the cube so that the green face is at the bottom. In this case, the top face can be painted one of the five remaining colors. It is clear that no two colorings, in which the top (i.e., opposite the green) face is painted ... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In how many different ways can 30 workers be divided into 3 teams of 10 people each? | 5. 10 workers, forming the first brigade, can be chosen from thirty in $C_{30}^{10}=\frac{30 \cdot 29 \cdot \ldots \cdot 21}{10!}$ ways. After this, from the remaining 20 workers, 10 workers for the second brigade can be chosen in $C_{20}^{10}=\frac{20 \cdot 19 \cdots 11}{10!}$ ways. By combining each way of forming th... | 2775498395670 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. The commission consists of 11 people. The materials that the commission works on are stored in a safe. How many locks should the safe have, and how many keys should each member of the commission be provided with, so that access to the safe is possible when a majority of the commission members gather, but not possibl... | 7. According to the problem statement, for any five members of the committee, there should be a lock for which the key is absent from all these committee members (but is present with each of the six absent members, since the presence of any of the nine absent members already makes the meeting of the committee possible)... | 252 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. The numbers from 1 to 1000 are written in a circle. Starting from the first, every 15th number is crossed out (i.e., the numbers $1, 16, 31$, etc.), and during subsequent rounds, already crossed-out numbers are also taken into account. The crossing out continues until it turns out that all the numbers to be crossed ... | 8. After the first cycle, all numbers that give a remainder of 1 when divided by 15 will be crossed out; the last such number will be 991. The first number to be crossed out during the second cycle will be $991+15-1000=6$; subsequently, during the second cycle, all numbers that give a remainder of 6 when divided by 15 ... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. All numbers from 1 to 10000000000 are written in a row. Which numbers will be more - those in which the digit 1 appears, or those in which 1 does not appear? | 9. First solution. Let's calculate the number of numbers in our sequence that do not contain the digit 1. Add the number 0 (zero) at the very beginning of this sequence and omit the last number 10000000000; we will get a sequence of $10^{10}$ numbers, containing one more number, in the notation of which the digit 1 doe... | 6513215600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. All integers from 1 to 222222222 are written in sequence. How many times does the digit 0 appear in the recording of these numbers?
## 2. Factorization of numbers into products and their decomposition into sums
In solving some of the subsequent problems, the following notations prove useful.
The symbol $[x]$ (re... | 10. Obviously, among the first 222222222 integers, there will be 22222222 numbers ending in zero (numbers 10, $20,30, \ldots, 222222220$). Further, among them, there will be 2222222 numbers ending in the digits 00 (numbers $100,200,300, \ldots, 222222200$); 2222222 numbers ending in the digits 01 (numbers 101, 201, 301... | 175308642 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. a) How many integers less than 1000 are not divisible by 5 or 7?
b) How many of these numbers are not divisible by 3, 5, or 7? | 11. a) There are a total of 999 numbers less than 1000 (numbers $1,2,3, \ldots, 999$). Now, let's strike out those that are divisible by 5; there are $\left[\frac{999}{5}\right]=199$ such numbers (numbers $5,10,15,20, \ldots, 995=199.5$). Next, let's strike out all numbers divisible by 7; there are $\left[\frac{999}{7}... | 457 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12*. How many integers are there that are less than the number 56700000 and coprime with it? | 12. The problem is very close to the previous one. The factorization of the number 56700000 is $56700000=2^{5} \cdot 3^{4} \cdot 5^{5} \cdot 7$; thus, the problem reduces to determining how many numbers less than 56700000 are not divisible by 2, 3, 5, or 7.
Let's list all numbers from 1 to 56700000. We will strike out... | 12960000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. How many positive integers $x$, less than 10000, are there for which the difference $2^{x}-x^{2}$ is not divisible by 7? | 13. $2^{0}=1$ gives a remainder of $1$ when divided by 7, $2^{1}=2$ gives a remainder of $2$, $2^{2}=4$ gives a remainder of $4$, $2^{3}$ gives a remainder of $1$ again, $2^{4}$ gives a remainder of $2$ again, $2^{5}$ gives a remainder of $4$ again, $2^{6}$ gives a remainder of $1$ for the third time, and so on. Thus, ... | 7142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. How many different pairs of integers $x, y$, lying between 1 and 1000, are there such that $x^{2}+y^{2}$ is divisible by 49? | 14. If $x^{2}+y^{2}$ is divisible by 49, then $x^{2}+y^{2}$ is also divisible by 7. But $x^{2}$, when divided by 7, can only give remainders of $0, 1, 4$, or 2 (see the solution to problem 13). The remainder from dividing $x^{2}+y^{2}$ by 7 is equal to the sum of the remainders from dividing the numbers $x^{2}$ and $y^... | 10153 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
15. In how many ways can a million be factored into three factors? Factorizations that differ only in the order of the factors are considered the same. | 15. Since a million is equal to $2^{6} \cdot 5^{6}$, any of its divisors has the form $2^{\alpha} \cdot 5^{\beta}$, and the factorization of a million into 3 factors has the form
$$
1000000=\left(2^{\alpha_{1}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{2}} \cdot 5^{\beta_{2}}\right)\left(2^{\alpha_{3}} \cdot 5^{\beta... | 139 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
17. How many different pairs of integers $A, B$ exist for which the least common multiple is 59400000? | 17. The prime factorization of the number 59400000 is
$$
59400000=2^{6} \cdot 3^{3} \cdot 5^{5} \cdot 11
$$
Therefore, if the least common multiple of two numbers $A$ and $B$ is 59400000, then it must be
$$
A=2^{x_{1}} \cdot 3^{\beta_{1}} \cdot 5 \gamma_{1} \cdot 11^{\delta_{1}}, \quad B=2^{\alpha_{2}} \cdot 3^{\bet... | 1502 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18. Find the coefficients of $x^{17}$ and $x^{18}$ after expanding the brackets and combining like terms in the expression
$$
\left(1+x^{5}+x^{7}\right)^{20}
$$ | 18. Let's repeat 20 times the trinomial $1+x^{3}+x^{7}$ and multiply it according to the usual rules. Each term of the resulting sum will represent a product of 20 factors, each equal to either 1, or $x^{5}$, or $x^{7}$. It is essential that in these products of twenty factors, any of the three expressions can stand in... | 3420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
19. In how many ways can 20 kopecks be exchanged for coins worth 5 kopecks, 2 kopecks, and 1 kopeck? | 19. Let's list all the ways to change 20 kopecks. In such a change, we can use either four 5-kopeck coins, or three such coins, or two, or one, or none.
There is only one way to change 20 kopecks using four 5-kopeck coins (since four 5-kopeck coins already make 20 kopecks).
If we use three 5-kopeck coins, we need to ... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
22**. In how many ways can you make up a ruble using coins of 1, 2, 5, 10, 20, and 50 kopecks? | 22. Since from coins of 10, 20, and 50 kopecks, only a whole number of tens of kopecks can be formed, then from 1, 2, and 5 kopeck coins, a whole number of tens of kopecks must also be formed. Therefore, the following cases are possible:
1) The ruble is composed of coins worth 10, 20, and 50 kopecks,
2) Coins worth 10,... | 4562 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
24. How many integer solutions does the inequality
$$
|x|+|y|<100 ?
$$
have? Here, for $x \neq y$, the solutions $x, y$ and $y, x$ should be considered different. | 24. We need to find the number of pairs of integers \(x, y\) such that \(|x| + |y|\) is 0, 1, 2, 3, ..., or 99. Let's count the number of pairs \(x, y\) for which \(|x| + |y| = k\). Here, \(|x|\) can take \(k + 1\) different values, namely \(0, 1, 2, \ldots, k-1, k\); in this case, \(|y|\) will be \(k, k-1, k-2, \ldots... | 19801 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
32. a) What is the maximum number of rooks that can be placed on a chessboard of $n^{2}$ squares so that no two rooks threaten each other? In how many different ways can this be done?
b) What is the minimum number of rooks that can be placed on a chessboard of $n^{2}$ squares so that these rooks threaten all the squar... | 32. a) A chessboard consisting of $n^{2}$ squares (see Fig. 39, where the case $n=8$ is shown) contains $n$ rows and $n$ columns. To ensure that no two rooks placed on this board threaten each other, it is necessary that no two rooks stand on the same row or the same column. It is clear, therefore, that the total numbe... | 33514312 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
47. a) How many different squares, in terms of size or position, consisting of whole cells, can be drawn on a chessboard of 64 cells?
b) The same question for a chessboard of $n^{2}$ cells. | 47. a) This problem is close to the previous one. The number of all possible squares consisting of $k^{2}$ cells that can be chosen on a chessboard of 64 cells is $(9-k)^{2}$ (see the solution to problem 46 a)). From this, it follows that the number of all squares is $8^{2}+7^{2}+6^{2}+\ldots+1^{2}=$
$$
=64+49+36+25+1... | 204 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.2. Given a cube with an edge of 1. Find the angle and distance between the skew diagonals of two of its adjacent faces. | 1.2. Consider the diagonals $A B_{1}$ and $B D$ of the cube $A B C D A_{1} B_{1} C_{1} D_{1}$. Since $B_{1} D_{1} \| B D$, the angle between the diagonals $A B_{1}$ and $B D$ is equal to the angle $A B_{1} D_{1}$. But the triangle $A B_{1} D_{1}$ is equilateral, so $\angle A B_{1} D_{1}=60^{\circ}$.
It is easy to chec... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.10. At the base of a regular triangular prism lies a triangle $ABC$ with side $a$. Points $A_{1}, B_{1}$, and $C_{1}$ are taken on the lateral edges, the distances from which to the plane of the base are $a / 2$, $a$, and $3 a / 2$. Find the angle between the planes $ABC$ and $A_{1} B_{1} C_{1}$.
## § 3. Lines formi... | 1.10. Let $O$ be the point of intersection of the lines $A B$ and $A_{1} B_{1}$, $M$ be the point of intersection of the lines $A C$ and $A_{1} C_{1}$. First, we will prove that $M O \perp O A$. For this, we will take points $B_{2}$ and $C_{2}$ on the segments $B B_{1}$ and $C C_{1}$ such that $B B_{3}=C C_{2}=A A_{1}$... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.28. Through the vertex $A$ of a right circular cone, a section of maximum area is drawn. Its area is twice the area of the section passing through the axis of the cone. Find the angle at the vertex of the axial section of the cone. | 2.28. Consider an arbitrary section passing through the vertex $A$. This section is a triangle $A B C$, and its sides $A B$ and $A C$ are the generators of the cone, i.e., they have a constant length. Therefore, the area of the section is proportional to the sine of the angle $B A C$. The angle $B A C$ varies from $0^{... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.21. In the tetrahedron \(ABCD\), the plane angles at vertex \(A\) are right angles, and \(AB = AC + AD\). Prove that the sum of the plane angles at vertex \(B\) is \(90^\circ\). | 6.21. Let's take points $P$ and $R$ on rays $A C$ and $A D$ such that $A P = A R = A B$, and consider the square $A P Q R$. It is clear that $\triangle A B C = \triangle R Q D$ and $\triangle A B D = \triangle P Q C$, and therefore, $\triangle B C D = \triangle Q D C$. Thus, the sum of the planar angles at vertex $B$ i... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
11.18. The height of a regular quadrilateral prism $A B C D A_{1} B_{1} C_{1} D_{1}$ is half the length of the side of the base. Find the maximum value of the angle $A_{1} M C_{1}$, where $M$ is a point on the edge $A B$. | 11.18. Let $A A_{1}=1, A M=x$. Introduce a coordinate system, the axes of which are parallel to the edges of the prism. The vectors $\overrightarrow{M A}_{\text {}}$ and $\overrightarrow{M C}_{1}$ have coordinates $(0,1,-x)$ and ( $2,1,2-x$ ); their scalar product is equal to $1-2 x+x^{2}=(1-x)^{2} \geqslant 0$. Theref... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15.26. Into how many parts do the planes of the faces divide the space a) of a cube; b) of a tetrahedron? | 15.26. The planes of the faces of both polyhedra intersect only along the lines containing their edges. Therefore, the space divided into parts, into which the space is split, has a common point with the polyhedron. Moreover, to each vertex, each edge, and each face, one can correspond exactly one adjacent part to it, ... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
15.34. In space, 4 points are given, not lying in the same plane. How many different parallelepipeds exist for which these points serve as vertices? | 15.34. Consider a parallelepiped for which the given points are vertices, and mark its edges connecting these points. Let \( n \) be the maximum number of marked edges of this parallelepiped emanating from one vertex; the number \( n \) can vary from 0 to 3. A simple enumeration shows that only the variants depicted in... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
112. Make sure that the last digits of the numbers in the Fibonacci sequence $^{\circ}$ repeat periodically. What is the length of the period?
Ensure that the last digits of the numbers in the Fibonacci sequence $^{\circ}$ repeat periodically. What is the length of the period? | 112. The last digits of the Fibonacci sequence $\varnothing^{0}$ themselves form a Fibonacci sequence in the sense of 10-arithmetic (sequence $\Phi_{10}^{0}$).
Let's write out the terms of this sequence:
$0,1,1,2,3,5,8,3,1,4,5,9,4,3,7$, $0,7,7,4,1,5,6,1,7,8,5,3,8,1,9$, $0,9,9,8,7,5,2,7,9,6,5,1,6,7,3$, $0,3,3,6,9,5,4,... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In triangle $ABC$, the median $AD$ is drawn. $\widehat{D A C} + \widehat{A B C} = 90^{\circ}$. Find $\widehat{B A C}$, given that $|A B| = |A C|$. | 1. Take a point $A_{1}$ on the line $B A$ such that $\left|A_{1} B\right|=\left|A_{1} C\right|$. The points $A_{1}, A, D$ and $C$ lie on the same circle $\left(\widehat{D A_{1} C}=90^{\circ}-\widehat{A B C}=\widehat{D A C}\right)$. Therefore, $\widehat{A_{1} A C}=\widehat{A_{1} D C}=90^{\circ}$, which means $\widehat{B... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
58. Determine the measure of angle $\hat{A}$ of triangle $ABC$, if it is known that the bisector of this angle is perpendicular to the line passing through the point of intersection of the altitudes and the center of the circumscribed circle of this triangle. | 58. Let $O$ be the center of the circumscribed circle, $H$ the intersection point of the altitudes of $\triangle ABC$. Since the line $OH$ is perpendicular to the bisector of angle $A$, it intersects sides $AB$ and $AC$ at points $K$ and $M$ such that $|AK|=|AM|$. Thus, $|AO|=|OB|$ and $\widehat{AOB}=2 \hat{C}$ (assumi... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
61. In trapezoid $A B C D$, the lateral side $A B$ is perpendicular to $A D$ and $B C$, and $|A B|=\sqrt{|A D| \cdot|B C|}$. Let $E$ be the point of intersection of the non-parallel sides of the trapezoid, $O$ be the point of intersection of the diagonals, and $M$ be the midpoint of $A B$. Find $\widehat{E O M}$. | 61. Let $|A D|=a,|B C|=b$. Drop a perpendicular $O K$ from $O$ to $A B$. Now we can find $|B K|=\sqrt{\overline{a b}} \frac{b}{b+a},|B E|=\sqrt{\overline{a b}} \frac{b}{a-b}$, $|M K|=\frac{\sqrt{a b}}{2}-\sqrt{\overline{a b}} \frac{b}{b+a}=\sqrt{\overline{a b}} \frac{a-b}{2(a+b)},|E K|=|B E|+|B K|=$ $=\sqrt{a \bar{b}} ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
63. Two circles touch each other internally at point $A$. From the center of the larger circle, a radius $O B$ is drawn, touching the smaller circle at point $C$. Find $\widehat{B A C}$. | 63. If $O_{1}$ is the center of the smaller circle, and $\widehat{B O A}=\varphi$, then $\widehat{B A O}=90^{\circ}-\frac{\varphi}{2}$, $\widehat{C O_{1} A}=90^{\circ}+\varphi$, $\widehat{C A O_{1}}=45^{\circ}-\frac{\varphi}{2}$. Therefore,
$$
\begin{gathered}
\widehat{B A C}=\widehat{B A O}-\widehat{C A O_{1}}=45^{\c... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
64. Inside the square $A B C D$, a point $M$ is taken such that $\widehat{M A B}=60^{\circ}, \widehat{M C D}=15^{\circ}$. Find $\widehat{M B C}$. | 64. Construct an equilateral triangle $A B K$ on $A B$ inside the square. Then $\widehat{K A B}=60^{\circ}, \widehat{K C D}=15^{\circ}$, i.e., $K$ coincides with $M$. Answer: $30^{\circ}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
65. In triangle $ABC$ with angle $\widehat{ABC}=60^{\circ}$, the bisector of angle $A$ intersects $BC$ at point $M$. On side $AC$, a point $K$ is taken such that $\widehat{AM} K=30^{\circ}$. Find $\widehat{OKC}$, where $O$ is the center of the circumcircle of triangle $AMC$. | 65. If $\widehat{B A C}=2 \alpha$, it is easy to find that $\widehat{K M C}=\widehat{M K C}=$ $=30^{\circ}+\alpha$, i.e., $|M C|=|K C|$. Extend $M K$ until it intersects the circle at point $N$; $\triangle K M C$ is similar to $\triangle K A N$, so $|A N|=|K N|=R$ - the radius of the circle (since $\widehat{A M N}=30^{... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
66. Given a triangle $A B C$, where $|A B|=|A C|$, $\widehat{B A C}=80^{\circ}$. Inside the triangle, a point $M$ is taken such that $\widehat{M} \overrightarrow{B C}=30^{\circ}, \widehat{M C B}=10^{\circ}$. Find $\widehat{A M C}$. | 66. Let's draw (Fig. 12) the angle bisector of $\angle A$ and extend $B M$ until it intersects with the bisector at point N. Since $|B N|=|N C|$, then $\widehat{B N C}=120^{\circ} ;$ therefore, the angles $\widehat{B N A}, \widehat{C N A}$ are also $120^{\circ}$ each, $\widehat{N C A}=\widehat{N C M}=20^{\circ}, \quad$... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
67. In triangle $ABC$, given are $\widehat{ABC}=100^{\circ}, \widehat{ACB}=$ $=65^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=55^{\circ}$, and on $AC$ - a point $N$ such that $\widehat{NBC}=80^{\circ}$. Find $\widehat{NMC}$. | 67. Describe a circle around $\triangle M C B$ (Fig. 13) and extend $B N$ until it intersects with the circle
Fig. 12. at point $M_{1} ;\left|C M_{1}\right|=|C M|$, since the angles subtend... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
68. In triangle $ABC$, it is given that $|AB|=|BC|, \widehat{ABC}=$ $=20^{\circ}$; on $AB$ a point $M$ is taken such that $\widehat{MCA}=60^{\circ}$; on side $CB$ - a point $N$ such that $\widehat{NAC}=50^{\circ}$. Find $\widehat{NMA}$. | 68. Let's take a point $K$ on $BC$ (Fig. 14) such that $\widehat{K A C}=60^{\circ}$, $M K \| A C$. Let $L$ be the intersection point of $A K$ and $M C$; $\triangle A L C$ is equilateral, $\triangle A N C$ is isosceles (calculate the angles), so $\triangle L N C$ is also isosceles, $\widehat{L C N}=20^{\circ}$. Now let'... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
69. In triangle $ABC$, given are $\widehat{ABC}=70^{\circ}, \widehat{ACB}=$ $=50^{\circ}$. On $AB$, a point $M$ is taken such that $\widehat{MCB}=40^{\circ}$, and on $AC$-point $N$ such that $\widehat{NBC}=50^{\circ}$. Find $\widehat{NMC}$. | 69. Let's take point $K$ (Fig. 15) such that $\widehat{K B C}=\widehat{K C B}=30^{\circ}$, and denote by $L$ the intersection point of the lines $M C$ and $B K$.
Fig. 15. Since $\triangle ... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
75. In quadrilateral $A B C D$, it is given that $\widehat{D A B}=150^{\circ}$, $\widehat{D A C}+\widehat{A B D}=120^{\circ}, \widehat{D B C}-\widehat{A B \bar{D}}=60^{\circ}$. Find $\widehat{B D C}$. | 75. Let $\widehat{A B D}=\alpha, \widehat{B D C}=\varphi$. By the condition, $\widehat{D A C}=$ $=120^{\circ}-\alpha, \widehat{B A C}=30^{\circ}+\alpha, \widehat{A D B}=30^{\circ}-\alpha, \widehat{D B C}=60^{\circ}+\alpha$. Using the Law of Sines for triangles $A B C, B C D, A C D$, we get
$$
\begin{aligned}
& \frac{|... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
64*. Can 1965 points be placed in a square with a side of 1 so that any rectangle of area $1 / 200$ with sides parallel to the sides of the square contains at least one of these points? | 64. Answer: it is possible.
In the square $0 \leqslant x \leqslant 1, 0 \leqslant y \leqslant 1$ and on the line $y=1 / 2$, uniformly place $c_{0}=200$ points: ( $\left.k / 201 ; 1 / 2\right), k=1,2, \ldots$ $\ldots, 200$. Then on each of the lines $y=1 / 4$ and $y=3 / 4$, place $c_{1}=100$ points ( $\left.k / 101 ; 1... | 1704 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
90. In a sequence of integers (positive), each term, starting from the third, is equal to the absolute difference of the two preceding ones.
What is the maximum number of terms such a sequence can have if each of its terms does not exceed 1967? | 90. Answer: 2952. We will prove that the length (number of terms) of a sequence satisfying the condition of the problem, where the largest term is the second and equals $n$, does not exceed $d_{n} = [3(n+1) / 2]$, and for the sequence $n-1, n, 1, \ldots, \ldots, 1,1$ the length is exactly $d_{n}$.
We will reason by in... | 2952 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
126*. In the country's football championship, 20 teams are participating. What is the minimum number of games that must be played so that among any three teams, there are two that have already played against each other? | 126. Answer: 90 games.
Let among any three teams, there be two that have already played against each other. Choose a team $A$ that has played the least number of games $-k$. Each of the $k$ teams that have already played with $A$, as well as team $A$ itself, has played no fewer than $k$ games. From the $(19-k)$ teams ... | 90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
161. Find the greatest integer $x$ such that the number
$$
4^{27}+4^{1000}+4^{x}
$$
is a perfect square. | 161. Answer: $x=1972$. Since $4^{27}+4^{1000}+4^{x}=$ $=2^{54}\left(1+2 \cdot 2^{1945}+2^{2 x-54}\right)$, the expression in parentheses will be a perfect square when $2 x-54=2 \cdot 1945$, i.e., when $x=1972$. If $x>1972$, then $2^{2(x-27)}<1+2 \cdot 2^{1945}+2^{2(x-27)}<\left(2^{x-27}+1\right)^{2}$, i.e., the given n... | 1972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
179. The Tennis Federation has assigned qualification numbers to all its tennis players: the strongest player gets the first number, the next strongest gets the second number, and so on. It is known that in matches where the qualification numbers differ by more than 2, the player with the lower number always wins. A to... | 179. Answer! The highest possible number of the winner is 20.
Since a tennis player with number $k$ can lose (not counting stronger ones) only to the $(k+1)$-th and $(k+2)$-th tennis players, the number of the strongest winner after each round cannot increase by more than 2. Thus, the number of the winner of the entir... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
226. In a regular 1976-gon, the midpoints of all sides and the midpoints of all diagonals are marked. What is the greatest number of marked points that can lie on one circle? | 226. Answer: 1976. All marked points, except for the center $O$ of the 1976-gon, lie 1976 each on 987 circles with center $O$. Any other circle $\gamma$ intersects each of these 987 circles at two points; besides these intersection points, there can be only one more marked point on $\gamma$: $O$. Therefore, there are n... | 1976 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
252. Let $a_{n}$ denote the integer closest to $\sqrt{n}$. Find the sum
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{1980}}
$$ | 252. Answer: 88.
Each number $k=1,2,3, \ldots$ appears in the sequence $(a_{n})$ $2 k$ times, since the condition $a_{n}=k$ is equivalent to
$$
k-\frac{1}{2}<\sqrt{n}<k+\frac{1}{2}, \quad \text { or } \quad k^{2}-k<n \leqslant k^{2}+k
$$
Therefore, in the sum
$$
\begin{aligned}
&\left(\frac{1}{a_{1}}+\frac{1}{a_{2}... | 88 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
270. A kangaroo jumps in the corner $x \geqslant 0, y \geqslant 0$ of the coordinate plane $O x y$ as follows: from the point ( $x ; y$ ) the kangaroo can jump to the point ( $x-5$; $y+7)$ or to the point $(x+1 ; y-1)$, and jumping to points with a negative coordinate is not allowed. From which initial points $(x ; y)$... | 270. Answer: the set of points from which one cannot escape *to infinity* has an area of 15; this is the stepped figure $T$, shown in Fig. 112.
From any point outside $T$, one can reach the region $x \geq 5$ in several steps $(1; -1)$, and then make steps $(-5; 7) + 5(1; -1) = (0; 2)$
$\nabla$ Even more interesting s... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
281*. The final sequence $a_{1}, a_{2}, \ldots, a_{n}$ of numbers 0 and 1 must satisfy the following condition: for any integer $k$ from 0 to $n-1$, the sum
$$
a_{1} a_{k+1}+a_{2} a_{k+2}+\ldots+a_{n-k} a_{n}
$$
is an odd number.
a) Come up with such a sequence for $n=25$.
b) Prove that such a sequence exists for s... | 281. The value specified in the problem condition
$$
p_{k}=a_{1} a_{k+1}+a_{2} a_{k+2}+\ldots+a_{n-k} a_{n}
$$
is conveniently calculated as follows: the sequence $A_{n}=\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ is signed under itself with a shift of $k$ digits; in this case, $p_{k}=p_{k}\left(A_{n}\right)$ is the nu... | 1201 | Combinatorics | proof | Yes | Yes | olympiads | false |
322. Find at least one natural number $n$ such that each of the numbers $n, n+1, n+2, \ldots, n+20$ has a common divisor with the number $30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ that is greater than one. | 322. Answer: for example, 9440. Let $m=2 \cdot 3 \cdot 5 \cdot 7 \cdot k$. By choosing $k$ such that $m-1$ is divisible by 11, and $m+1$ is divisible by 13, we get that the number $n=m-10$ satisfies the condition of the problem (P2). | 9440 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
337. Natural numbers from 1 to 1982 are arranged in some order. A computer scans pairs of adjacent numbers (the first and second, the second and third, etc.) from left to right up to the last pair and swaps the numbers in the scanned pair if the larger number is to the left. Then it scans all pairs from right to left f... | 337. Answer: 100. Consider the number $a$-the largest of the first 99 numbers, and the number $b$-the smallest of the last 1882 numbers and ensure that $a<100<b$. | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.9. What is the maximum number of strings connecting adjacent nodes in a volleyball net with square cells that can be cut so that the net does not fall apart into separate pieces? The size of the net is $10 \times 100$ cells.
## PARALLEL SEGMENTS | 1.9. Answer. 1000.
Suppose that so many strings have been torn that the net has not yet split into pieces, but no more strings can be torn. This means that there are no closed loops of strings left in the net. Prove that in this case, the number of unbroken strings is one less than the total number of nodes (including... | 1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.1. The altitudes of triangle $ABC$ intersect at point $O$. It is known that $OC = AB$. Find the angle at vertex $C$. | ## 5.1. Answer. $45^{\circ}$.
Let $D$ be the foot of the perpendicular dropped from point $A$. Prove that triangles $C O D$ and $A B D$ are congruent. Then show that $\angle O B D=45^{\circ}$, and from this, deduce that the required angle is also $45^{\circ}$.
Another solution can be obtained by almost verbatim repea... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.3. Find the angle $C$ of triangle $ABC$, if the distance from vertex $C$ to the orthocenter of the triangle is equal to the radius of the circumscribed circle. | 5.3. Answer. $60^{\circ}$.
1st solution. Let $O$ be the center of the circumcircle of triangle $ABC$, $M$ be the orthocenter, and $CC'$ be the diameter of the circumcircle (see figure a). Inscribed angles $ABC$ and $AC'C$ are equal, so the acute angles $MCB$ and $ACO$, which complement them to $90^{\circ}$, are also e... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.5. In triangle $A B C$ with an angle of $120^{\circ}$ at vertex $A$, the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ are drawn. Find the angle $C_{1} A_{1} B_{1}$. | 5.5. Answer. $90^{\circ}$.
Let the length of side $AB$ be denoted by $c$, $AC$ by $b$, and $BC$ by $a$. First, we will prove that $AA_1 = \frac{bc}{b+c}$.
Fig. 5.5.
Extend $AA_1$ and draw... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.7. In an isosceles triangle \(ABC\) with lateral sides \(AB = BC\), the angle \(ABC\) is \(80^\circ\). Inside the triangle, a point \(O\) is taken such that the angle \(OAC\) is \(10^\circ\) and the angle \(OCA\) is \(30^\circ\). Find the angle \(AOB\). | 5.7. Answer. $70^{\circ}$.
Let $K$ be the point of intersection of the altitude dropped from vertex $B$ and the bisector of angle $OAB$ (see figure). First, show that point $K$ lies on the extension of $OC$. Then prove that triangles
$ denote the sum of the digits of a natural number $x$. Solve the equations:
a) $x+S(x)+S(S(x))=1993$
b)* $x+S(x)+S(S(x))+S(S(S(x)))=1993$.
2*. It is known that the number $n$ is the sum of the squares of three natural numbers. Show that the number $n^{2}$ is also the sum of the squares of three natural ... | 1. a) According to the divisibility rule for 3 (see fact 6), the numbers $x$ and $S(x)$ give the same remainder when divided by 3. The same remainder will also be given by the number $S(S(x))$. Therefore, the sum
$$
x + S(x) + S(S(x))
$$
is divisible by 3 (since it is the sum of three numbers with the same remainder ... | 1963 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Petya has a total of 28 classmates. Any two of the 28 have a different number of friends in this class. How many friends does Petya have? | 4. Petya's classmates can have $0,1,2, \ldots, 28$ friends - a total of 29 options. However, if someone is friends with everyone, then everyone has at least one friend. Therefore, either there is someone who is friends with everyone, or there is someone who is not friends with anyone. In both cases, there are 28 option... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Each pair of numbers $x$ and $y$ is assigned a number $x * y$. Find $1993 * 1935$, given that for any three numbers $x, y$ and $z$ the identities $x * x=0$ and $x *(y * z)=(x * y)+z$ are satisfied. | 5. Let's take $y=z$ in the second identity. Then we get
$$
(x * y)+y=x *(y * y)=x * 0
$$
Thus, $x * y=x * 0 - y$. It remains to compute $x * 0$. For this, take $x=y=z$ in the second identity:
$$
x * 0=x *(x * x)=x * x + x=0 + x=x.
$$
Thus, $x * y=x * 0 - y=x - y$. Therefore, $1993 * 1935=1993 - 1935=58$.
Comment. ... | 58 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A cooperative receives apple and grape juice in identical barrels and produces an apple-grape drink in identical cans. One barrel of apple juice is enough for exactly 6 cans of the drink, and one barrel of grape juice is enough for exactly 10 cans. When the recipe for the drink was changed, one barrel of apple juice... | 1. The first method. For one can of the drink, $\frac{1}{6}$ of a barrel of apple juice and $\frac{1}{10}$ of a barrel of grape juice are used, so the volume of the can is
$$
\frac{1}{6}+\frac{1}{10}=\frac{4}{15}
$$
of the barrel's volume.
After changing the recipe, for one can of the drink, $\frac{1}{5}$ of a barre... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A student did not notice the multiplication sign between two three-digit numbers and wrote a single six-digit number, which turned out to be seven times greater than their product. Find these numbers. | 2. First method. Let $x, y$ be the desired three-digit numbers. If we append three zeros to the number $x$, we get the number $1000 x$, and if we append $y$, we get $1000 x + y$ (see fact 11).
Thus, the student wrote the number $1000 x + y$. According to the problem, this number is seven times greater than $x \cdot y$... | 143 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5*. Find the largest natural number, not ending in zero, which, when one (not the first) digit is erased, decreases by an integer factor.
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。 | 5. Let $x$ be the deleted digit, $a$ be the part of the number to the left of $x$, and $c$ be the part of the number to the right of $x$. Then the number has the form $\overline{a x c}$, see fact 11. Suppose the digit $x$ is in the $(n+1)$-th place (counting from the right). Then
$$
\overline{a x c}=a \cdot 10^{n+1}+x... | 180625 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A rectangle of size $1 \times k$ for any natural number $k$ will be called a strip. For which natural numbers $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips? | 3. Idea of the solution: take the maximum strip (equal to the maximum side of the rectangle). The remaining strips will be combined in pairs, giving the sum of the maximum strip. If we have filled the rectangle, the problem is solved; otherwise, reasoning with areas shows that the rectangle cannot be cut into different... | 3989 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Is it enough to make a closed rectangular box from all sides, enclosing no less than 1995 unit cubes, a) 962; b) 960; c) 958 square units of material? | 4. It is sufficient to take a box of size $11 \times 13 \times 14$. Its volume is 2002, which is sufficient; and the total area of its walls is $2 \cdot(11 \cdot 13+11 \cdot 14+13 \cdot 14)=958$.
Comments. $1^{\circ}$. How can one guess such a solution? It is known that among all parallelepipeds with a given volume, t... | 958 | Geometry | MCQ | Yes | Yes | olympiads | false |
6. A line cuts off triangle $A K N$ from a regular hexagon $A B C D E F$ such that $A K+A N=A B$. Find the sum of the angles under which segment $K N$ is seen from the vertices of the hexagon ( $\angle K A N+\angle K B N+\angle K C N+\angle K D N+\angle K E N+$ $+\angle K F N$).
## 9 t h g r a d e | 6. Let's assume that $N$ lies on $A B$, and $K$ lies on $A F$ (Fig. 60). Note that $F K = A N$. We choose point $P$ on $B C$, point $R$ on $C D$, point $S$ on $D E$, and point $T$ on $E F$ such that the equalities $F K = A N = B P = C R = D S = E T$ hold. Then $\angle K B N = \angle T A K$, $\angle K C N = \angle S A T... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In a $10 \times 10$ grid, the centers of all unit squares are marked (a total of 100 points). What is the minimum number of lines, not parallel to the sides of the square, needed to cross out all the marked points? | 2. Let's draw all lines parallel to one of the diagonals of the square and containing more than one of the marked points - there are 17 such lines. The un-
Fig. 77 erased will be the two c... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given an equilateral triangle $A B C$. Side $B C$ is divided into three equal parts by points $K$ and $L$, and point $M$ divides side $A C$ in the ratio $1: 2$, starting from vertex $A$. Prove that the sum of angles $A K M$ and $A L M$ is $30^{\circ}$. | 4. Without loss of generality, we can assume that point $K$ is closer to point $B$ than point $L$ (Fig. 72). Then triangle $M K C$ is equilateral (since $M C = K C, \angle M C K = 60^{\circ}$). Therefore, $A B \| M K$ (since $\left.\angle M K C = \angle A B C = 60^{\circ}\right)$. This means that angles $A K M$ and $B ... | 30 | Geometry | proof | Yes | Yes | olympiads | false |
6. Ali-Baba and the bandit are dividing a treasure consisting of 100 gold coins, arranged in 10 piles of 10 coins each. Ali-Baba chooses 4 piles, places a cup next to each, and sets aside several coins from each pile (at least one, but not the entire pile). The bandit must then rearrange the cups, changing their initia... | 6. We will show that Ali-Baba can achieve no more than 4 coins in 7 piles, while the robber can ensure that there are no piles with fewer than 4 coins. Therefore, Ali-Baba will take $100 - 7 \cdot 4 = 72$ coins.
First, we will prove that the robber can act in such a way that there are no piles with fewer than 4 coins.... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. From the volcano station to the summit of Stromboli volcano, it takes 4 hours to walk along the road, and then 4 hours along the path. At the summit, there are two craters. The first crater erupts for 1 hour, then remains silent for 17 hours, then erupts again for 1 hour, and so on. The second crater erupts for 1 ho... | 2. The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the eruption of the first crater, it will not be dangerous.
Movement along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the eruption of the s... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5*. In rhombus $A B C D$ the measure of angle $B$ is $40^{\circ}, E-$ is the midpoint of $B C, F$ - is the foot of the perpendicular dropped from $A$ to $D E$. Find the measure of angle $D F C$. | 5. Let lines $D E$ and $A B$ intersect at point $G$ (Fig. 83). Then triangles $D E C$ and $B E G$ are congruent by the second criterion. Therefore, $B G=C D=B A$. Hence, points $A, G$, and $C$ lie on a circle with center at point $B$, and $A G$ is the diameter. Since $\angle A F G=90^{\circ}$, point $F$ lies on the sam... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. A road 1 km long is fully illuminated by street lamps, each of which illuminates a section of the road 1 m long. What is the maximum number of street lamps that can be on the road, given that after turning off any street lamp, the road will no longer be fully illuminated? | 3. Let's number the street lamps with natural numbers in the order of their placement along the road. If the segments illuminated by the $n$-th and $(n+2)$-th street lamps intersect (at least at one point), then the $(n+1)$-th street lamp can be turned off. Therefore, segments with different odd numbers do not intersec... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Pete's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300 or adding 198 dollars. What is the maximum amount Pete can withdraw from his account if he has no other money? | 4. Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars (see fact 5). The maximum number that is a multiple of 6 and does not exceed 500 is 498.
Let's show how to withdraw 498 dollars. We will perform the following operations: $500-300=200, 200+198=398, 39... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the elections to the 100-seat parliament, 12 parties participated. Parties that received strictly more than $5 \%$ of the voters' votes enter the parliament. Among the parties that entered the parliament, seats are distributed proportionally to the number of votes they received (i.e., if one party received $x$ ti... | 2. Idea of the solution: The Party of Mathematics Enthusiasts (PME) will receive the maximum number of seats in parliament if the total number of votes cast for non-qualifying (i.e., receiving no more than $5 \%$ of the votes) parties is maximized.
If 10 parties receive exactly $5 \%$ of the votes each, and two, inclu... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. What is the maximum number of knights that can be placed on a $5 \times 5$ board such that each one attacks exactly two others? (Provide an example and explain why it is not possible to place more knights.)
## 9 t h g r a d e | 6. Fig. 116 shows the arrangement of 16 knights that satisfies the problem's condition. We will show that it is impossible to place more knights. Let's color the cells of the board in black and white, as shown in Fig. 116. Note that the number of knights on black cells is equal to the number of knights on white cells. ... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Six words are given: ZANOZA, ZIPUNY, KAZINO, KEFAL', OTMEL', SHELEST. In one step, you can replace any letter in any of these words with any other letter (for example, in one step, you can get the word ZKNOZA from ZANOZA). How many steps are needed to make all the words the same (nonsense words are allowed)? Provide... | 3. Let's write the words in a column:
|  | | | |
| :---: | :---: | :---: | :---: |
After all the letter replacements in each column, the letters should become the same. The number of rep... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the city of Udoyev, mayoral elections proceed as follows. If in a given round of voting no candidate receives more than half of the votes, then a subsequent round is held with the participation of all candidates except the one who received the fewest votes. (No two candidates ever receive the same number of votes... | 5. a) Ostap could not take the last, 2002nd place in the first round, as otherwise he would have been immediately eliminated from the list of candidates. Therefore, $k \leqslant 2001$.
Suppose all candidates in the first round received almost the same number of votes, Ostap took the second-to-last place, and in each s... | 2001 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. There are 4 people in the family. If Masha's scholarship is doubled, the total income of the entire family will increase by $5 \%$, if instead, Mom's salary is doubled - then by $15 \%$, if Dad's salary is doubled - then by $25 \%$. By what percentage will the family's total income increase if Grandpa's pension is d... | 1. The first method. If Masha's scholarship is doubled, the family income will increase by the amount of this scholarship. Therefore, Masha's scholarship constitutes $5 \%$ of the total income. Similarly, Mom's salary is $15 \%$, and Dad's is $25 \%$. The remaining share $100 \% - 5 \% - 15 \% - 25 \% = 55 \%$ is attri... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In the country, there are 15 cities, some of which are connected by air routes belonging to three airlines. It is known that even if any one of the airlines ceases operations, it will still be possible to travel from any city to any other (possibly with layovers), using the flights of the remaining two airlines. Wha... | 5. We will prove that fewer than 21 airlines are not sufficient. First, note that if 15 cities are connected by airlines in such a way that one can travel from any city to any other, then there are no fewer than 14 airlines. Indeed, starting from an arbitrary city, we will try to visit all the others, and visiting each... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the shore of a round island called Somewhere, there are 20 villages, each inhabited by 20 wrestlers. A tournament was held where each wrestler faced all wrestlers from all other villages. Village A is considered stronger than Village B if at least $k$ matches between wrestlers from these villages end with a victo... | 6. Let's provide an example showing that the described situation is possible when $k \leqslant 290$. We will order all the wrestlers by strength and renumber them in ascending order of strength (the first being the weakest). We will call the 210 weakest wrestlers novices, and the 190 strongest - masters. In particular,... | 290 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. An extrasensory perception (ESP) practitioner has a deck of 36 cards face down in front of him (four suits, nine cards of each suit). He names the suit of the top card, after which the card is revealed to him. Then he names the suit of the next card, and so on. The task of the ESP practitioner is to guess the suit a... | 6. a) Note that guessing 18 cards is not difficult. Indeed, the first two backs can "encode" the suit of the second card (by associating each suit with one of the four possible arrangements of the two backs), the next two backs can encode the suit of the fourth card, and so on.
When only two cards remain in the deck, ... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. To some natural number, two two-digit numbers were appended sequentially to the right. The resulting number turned out to be equal to the cube of the sum of the three original numbers. Find all possible triples of the original numbers. | 5. Let $a$ denote the first natural number, and $b$ and $c$ the two-digit numbers written after it. Let $x=a+b+c$. According to the condition, the numbers $a, b, c$ and $x$ satisfy the equation $10^{4} a+100 b+c=x^{3}$ (see fact 11). Therefore,
$$
x^{3}=10^{4} a+100 b+c44$;
2) $x=45(x-1=44), 45^{3}=91125, a=9, b=11, c... | 91125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
42. Given three lines $l_{1}, l_{2}$ and $l_{3}$ and three points $A, B$ and $C$-one point on each of the lines. Draw a line $m$ intersecting lines $l_{1}, l_{2}$ and $l_{3}$ at points $X, Y$ and $Z$ such that $A X=B Y=C Z$. | 42. Let the lines $l_{1}, l_{2}$ and $l_{3}$ not all be parallel to each other, for example, $l_{3}$ is not parallel to either $l_{1}$ or $l_{2}$. Suppose the problem is solved (Fig. 155). By Theorem 1, there exists
![](https://cdn.mathpix.com/cropped/2024_05_21_42e14062cf36c74a4d8dg-194.jpg?height=442&width=739&top_l... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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