problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
132. Prove the inequality $(\sqrt{a}+\sqrt{b})^{8} \geqslant 64 a b(a+b)^{2}$, if $a \geqslant 0, b \geqslant 0$. | S o l u t i o n. If the given inequality is valid, then by taking the square root of both sides of the inequality twice, we will have:
$$
\begin{gathered}
(\sqrt{a}+\sqrt{b})^{4} \geqslant 8 \sqrt{a b}(a+b) \\
(\sqrt{a}+\sqrt{b})^{2} \geqslant 2 \sqrt{2 \sqrt{a b}(a+b)} \\
a+b+2 \sqrt{a b} \geqslant 2 \sqrt{2} \sqrt{\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,147 |
133. Find all arithmetic progressions for which the sums of the first $n_{1}$ and $n_{2}\left(n_{1} \neq n_{2}\right)$ terms are equal to $n_{1}^{2}$ and $n_{2}^{2}$, respectively. | S o l u t i o n. By the formula for the sum of the first $n$ terms of an arithmetic progression $S_{n}=\left(a_{1}+\frac{d}{2}(n-1)\right)$, we have:
$$
\begin{aligned}
& \left(a_{1}+\frac{d}{2}\left(n_{1}-1\right)\right) n_{1}=n_{1}^{2} \\
& \left(a_{1}+\frac{d}{2}\left(n_{2}-1\right)\right) n_{2}=n_{2}^{2}
\end{alig... | 1,3,5,7,9,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,148 |
134. Three numbers are consecutive terms of some geometric progression. If 2 is added to the second number, they will form an arithmetic progression. If after that 16 is added to the third number, they will again form a geometric progression. Find these numbers. | Solution. Let $b_{1}, b_{2}, b_{3}$ be consecutive terms of a geometric progression. Then, by the condition, the numbers $b_{1}, b_{2}+2, b_{3}$ form an arithmetic progression, and the numbers $b_{1}, b_{2}+2, b_{3}+16$ form a geometric progression. Based on the properties of arithmetic and geometric progressions, we c... | 1,3,9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,149 |
136. Prove that in a geometric progression, the ratio of the difference between the sum of its first $n$ terms and its $n$-th term to the difference between the same sum and its first term is equal to the number that is the reciprocal of the common ratio of this progression. | We have:
$$
\frac{S_{n}-b_{n}}{S_{n}-b_{1}}=\frac{\frac{b_{n} q-b_{1}}{q-1}-b_{n}}{\frac{b_{n} q-b_{1}}{q-1}-b_{1}}=\frac{b_{n} q-b_{1}-b_{n} q+b_{n}}{b_{n} q-b_{1}-b_{1} q+b_{1}}=\frac{b_{n}-b_{1}}{q\left(b_{n}-b_{1}\right)}=\frac{1}{q}
$$
if $q \neq 1$ (otherwise, dividing by $q-1$ is not possible).
If $q=1$, then... | \frac{1}{q} | Algebra | proof | Yes | Yes | olympiads | false | 36,151 |
137. Knowing that $\cos ^{6} x+\sin ^{6} x=a$, find $\cos ^{4} x+\sin ^{4} x$. | S o l u t i o n. Transform the expression $\cos ^{4} x+\sin ^{4} x$ :
$\cos ^{4} x+\sin ^{4} x=\left(\cos ^{2} x+\sin ^{2} x\right)^{2}-2 \cos ^{2} x \sin ^{2} x=1-2 \cos ^{2} x \sin ^{2} x$.
Find the value of the product $\cos ^{2} x \sin ^{2} x$ from the given equation, for which we transform its left side:
$$
\be... | \frac{1+2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,152 |
138. Calculate the sum for any $\alpha$
$$
\sin ^{2} \alpha+\sin ^{2}\left(\alpha+1^{\circ}\right)+\sin ^{2}\left(\alpha+2^{\circ}\right)+\ldots+\sin ^{2}\left(\alpha+179^{\circ}\right)
$$ | S o l u t i o n. By combining the first term with the ninety-first, the second with the ninety-second, and so on, and using the reduction formula, we get:
$$
\begin{gathered}
\left(\sin ^{2} \alpha+\sin ^{2}\left(\alpha+90^{\circ}\right)\right)+\left(\sin ^{2}\left(\alpha+1^{\circ}\right)+\sin ^{2}\left(\alpha+91^{\ci... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,153 |
140. Find the maximum value of the function
$$
y(x)=3 \sin x+4 \cos x
$$ | Solution. $y(x)=5\left(\frac{3}{5} \sin x+\frac{4}{5} \cos x\right)$. Since $\left(\frac{3}{5}\right)^{2}+$ $+\left(\frac{4}{5}\right)^{2}=1$, there exists an angle $x_{0}$ such that $\frac{3}{5}=\cos x_{0}, \frac{4}{5}=\sin x_{0}$. Then $y(x)=5 \sin \left(x+x_{0}\right)$. It is obvious that the maximum value of $y(x)$... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,154 |
142. Calculate, without using tables, the following products:
a) $A=\cos 4^{\circ} \cos 8^{\circ} \ldots \cos 84^{\circ} \cos 88^{\circ}$;
b) $B=\cos 12^{\circ} \cos 24^{\circ} \ldots \cos 72^{\circ} \cos 84^{\circ}$;
c) $C=\cos 1^{\circ} \cos 3^{\circ} \ldots \cos 87^{\circ} \cos 89^{\circ}$;
d) $D=\cos 5^{\circ} ... | Solution. a) Multiply and divide the given expression by $2 \sin 4^{\circ}:$
$$
\begin{gathered}
A=\frac{2 \sin 4^{\circ} \cos 4^{\circ} \cos 8^{\circ} \ldots \cos 84^{\circ} \cos 88^{\circ}}{2 \sin 4^{\circ}}= \\
=\frac{2 \sin 8^{\circ} \cos 8^{\circ} \cos 16^{\circ} \ldots \cos 88^{\circ}}{2 \cdot 2 \sin 4^{\circ}}=... | )(\frac{1}{2})^{22};b)(\frac{1}{2})^{7};)(\frac{1}{2})^{45}\sqrt{2};)(\frac{1}{2})^{9}\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,155 |
143. For which $n \in \boldsymbol{N}$ does the equality
$$
\sqrt[n]{17 \sqrt{5}+38}+\sqrt[n]{17 \sqrt{5}-38}=\sqrt{20} ?
$$ | S o l u t i o n. Denoting the first term by $a$, we will have $a+\frac{1}{a}=\sqrt{20}$, from which $a=\sqrt{5}+2$. Since $(\sqrt{5}+2)^{3}=17 \sqrt{5}+$ $+38=(\sqrt{5}+2)^{n}$, then $n=3$.
A n s w e r: $n=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,156 |
144. Extract the cube root $\sqrt[3]{2+\sqrt{5}}$. | Solution. We will look for rational numbers $a$ and $b$ such that $\sqrt[3]{2+\sqrt{5}}=a+b \sqrt{5}$.
Raising the obtained equality to the third power and equating the coefficients of $\sqrt{5}$ and the rational terms in both parts, we get the system:
$$
\left\{\begin{array}{l}
3 a^{2} b+5 b^{3}=1 \\
a^{3}+15 a b^{2... | \frac{1+\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,157 |
145. Prove that if the numbers $a, b, c$ are positive, then at least one of the numbers $(a+b+c)^{2}-8 a c,(a+b+c)^{2}-8 b c,(a+b+c)^{2}-8 a b$ is positive. | The sum of the given numbers is equal to
$$
\begin{gathered}
3\left(a^{2}+b^{2}+c^{2}\right)+6 a b+6 b c+6 a c-8 a c-8 a b-8 b c= \\
=3\left(a^{2}+b^{2}+c^{2}\right)-2(a b+a c+b c)=a^{2}+b^{2}+c^{2}+(a-b)^{2}+ \\
+(a-c)^{2}+(b-c)^{2}
\end{gathered}
$$
The value of this expression is positive, since by the condition $... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 36,158 |
146. Prove that if $x+y=z+t(x, y, z, t \in \boldsymbol{Z})$, then the number $x^{2}+y^{2}+z^{2}+t^{2}$ is the sum of the squares of three integers. | Solution. Indeed,
$x^{2}+y^{2}+z^{2}+t^{2}=x^{2}+y^{2}+z^{2}+t^{2}+2 x(x+y)-2 x(z+t)=$ $=3 x^{2}+y^{2}+z^{2}+t^{2}+2 x y-2 x z-2 x t=(x+y)^{2}+(x-z)^{2}+(x-t)^{2}$. | (x+y)^{2}+(x-z)^{2}+(x-)^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 36,159 |
147. What prime numbers can be divisors of numbers of the form $111 . .11$? | Solution. It is clear that numbers of the considered form are not divisible by 2 and 5. Let $p$ be a prime number different from 2 and 5. Consider $p+1$ numbers: $1, 11, 111, \ldots, 111 \ldots 11$.
At least two of them give the same remainder when divided by $p$. Then their difference $11 \ldots 1100 \ldots 0$ is div... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,160 |
148. Can the sum of the digits of a natural number that is a perfect square equal 1991? | Solution. If the sum of the digits of the number $n^{2}(n \in \boldsymbol{N})$ is 1991, then it does not divide by 3, hence the number $n$ does not divide by 3 either, i.e., when divided by 3, it gives a remainder of 1 or 2. But then $n^{2}$, when divided by 3, gives a remainder of 1. However, the remainder from dividi... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,161 |
151. Can the number $5^{n}+1$ be divisible by the number $5^{k}-1$ (n, k - natural numbers)? | Solution. For $n>1$ and $k>1$, the number $5^{n}+1$ ends in the number 26, i.e., it is not divisible by 4, whereas the number $5^{k}-1$ ends in the number 24 and is therefore divisible by 4. It follows that $5^{n}+1$ is not divisible by $5^{k}-1$.
For $n=1$ and $k=1$, we get the numbers 6 and 4, and 6 is not divisible... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,163 |
152. Can a number of the form $5^{n}-1$ be divisible by $4^{n}-1$? | Solution. If $n$ is even, then the number $4^{n}$ ends with the digit 6, so the number $4^{n}-1$ ends with the digit 5, i.e., it is divisible by 5.
If $n$ is odd, then $4^{n}-1$ is divisible by $3\left(3=4^{\prime}-1\right)$.
Since the number $5^{n}-1$ is not divisible by either 5 or 3, $5^{n}-1$ is not divisible by ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,164 |
161. Does there exist a natural number that is a perfect square and consists only of sixes and zeros? | Solution. If a natural number is a perfect square, then it ends with an even number of zeros. After discarding them, a perfect square remains, ending in 06 or 66, which in both cases is divisible by 2 but not by 4. Therefore, the required natural number does not exist.
$\triangle$ 162. 1989 people are lined up. Can th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,165 |
166. A rectangular picture is inserted into a rectangular frame made of strips of the same width. The area of the picture is equal to the area of the frame. What is the ratio of the length to the width of the picture in the frame, if the ratio of the length to the width of the picture without the frame is $2: 3$? | Solution. Let $x$ be the length of the picture without the frame. Then $\frac{3 x}{2}$ is its width. If $l$ is the width of the strips from which the frame is made, then the area of the picture with the frame is $(x+2 l)\left(\frac{3 x}{2}+2 l\right)$. By equating this area to twice the area of the picture without the ... | 3:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 36,166 |
167. To transport 60 tons of cargo from one place to another, a certain number of trucks were required. Due to the poor condition of the road, each truck had to carry 0.5 tons less than originally planned, which is why 4 additional trucks were required. How many trucks were initially required? | Solution. Let $x$ be the original number of trucks. Then, each truck was supposed to carry $\frac{60}{x}$ tons of cargo, but they carried $\left(\frac{60}{x}-0.5\right)$ tons each. Since all 60 tons were loaded with the help of four additional trucks, we have the equation
$$
\left(\frac{60}{x}-0.5\right)(x+4)=60, \tex... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,167 |
168. Two pedestrians set out towards each other at the same time: the first from point $A$, the second from point $B$. The first pedestrian walked 1 km more before the meeting than the second. The first pedestrian arrived at point $B$ 45 minutes after the meeting. The second pedestrian arrived at point $A$ 1 hour and 2... | S o l u t i o n. Let the speed of the first pedestrian who left point $A$ be $v_{1}$ kilometers per hour, and the speed of the second pedestrian be $v_{2}$ kilometers per hour. Since after meeting, the first pedestrian walked $\frac{3}{4} v_{1}$ kilometers, and the second $-\frac{4}{3} v_{2}$ kilometers, and the first ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,168 |
1. A person has 12 pints of wine in a barrel (pint - an old French unit of volume, 1 pint ≈ 0.568 liters) and wants to give away half of the wine, but does not have a 6-pint container. However, there are two empty containers with capacities of 8 pints and 5 pints. How can one use them to measure out exactly 6 pints of ... | $\triangle$ The solution to the problem can be written as follows:
| Vessel with a capacity of 8 pints | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Vessel with a capacity of 5 pints | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 |
This should be understood as follows. Initiall... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,169 |
5. Is it possible to measure out exactly 4 liters of water from a river using two empty buckets with capacities of 12 liters and 9 liters? | $\triangle$ Since the initial capacities of the buckets are divisible by 3, any pouring from one bucket to another or filling from the river will result in each bucket containing a volume of water that is divisible by 3. However, since 4 is not divisible by 3, it is impossible to obtain 4 liters of water.
Answer: It i... | Itisimpossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,170 |
8. There is no less than 10 liters of milk in the bucket. How can you pour exactly 6 liters of milk from it using an empty nine-liter bucket and a five-liter bucket? | $\triangle$ Let's denote the initial amount of milk in the first bucket as $a$ liters. Let's think about how to use the fact that the number $a$ is not less than 10. The difference $a-10$ can be used, but the difference $a-11$ cannot. The solution is written as follows:
| Bucket with volume $a$ l | $a$ | $a-5$ | $a-5$... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,171 |
12*. In how many ways can milk be transferred from a 12-liter barrel, filled with milk, to another empty barrel of the same volume using two empty cans of 1 liter and 2 liters? Transferring milk from one can to another is not allowed.
Note that the question in this problem is different from the previous problems. | Let's denote by $a_{n}$ the number of ways to transfer milk from a full container of $n$ liters (where $n \in \boldsymbol{N}$) using two empty cans of 1 liter and 2 liters.
If we first use the 1-liter can, there will be $n-1$ liters left in the first container, and it can be transferred in $a_{n-1}$ ways.
If we first... | 233 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,172 |
15. Three friends met - Belov, Serov, and Chernov. Chernov said to the friend dressed in a gray suit: "It's interesting that one of us is wearing a white suit, another a gray one, and the third a black one, but each suit is a color that does not match the surname." What color is each friend's suit? | $\triangle$ Let's take a $4 \times 4$ table. In the left column of the table, we will write the surnames of friends (denoting each by its first letter), and in the top row - the colors of their costumes. According to the condition, Belov does not wear a white costume, Serov does not wear a gray one, and Chernov does no... | Belov:gray,Serov:black,Chernov:white | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,173 |
20. In the volleyball championship draw, team A was three places behind team B, team E outperformed B but fell behind D, team V outperformed team G. What place did each of these six teams take? | $\triangle$ Consider the $7 \times 7$ table.
According to the first condition, team A could not have taken the first, second, or third place. Therefore, it took the fourth, fifth, or sixth place. Then team B took the first, second, or third place.
Since team B not only fell behind team E but also team D, it could onl... | D,E,B,V,G,A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,174 |
24*. The case of Brown, Jones, and Smith is being investigated. One of them committed a crime. During the investigation, each of them made two statements.
Brown: "I didn't do it. Jones didn't do it."
Smith: "I didn't do it. Brown did it."
Jones: "Brown didn't do it. Smith did it."
It was later found out that one of... | $\triangle$ The difficulty lies in the fact that we do not know who told the truth both times, who lied both times, and who lied only once. Let's consider three cases.
1) Suppose Brown told the truth both times.
Since, according to his statements, he and Jones did not commit the crime, the only one who could have com... | Brown | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,175 |
26. In five baskets A, B, V, G, and D, there are apples of five different varieties. In each of the baskets A and B, there are apples of the 3rd and 4th varieties, in basket V - 2nd and 3rd, in basket G - 4th and 5th, in basket D - 1st and 5th. Number the baskets so that basket No. 1 contains apples of the 1st variety ... | $\triangle$ Let's represent two sets - the set of baskets and the set of their numbers. Each of these sets has five elements; let's denote them with points (Fig. 1).
Fig. 1
We will estab... | D,B,A,B,GorD,B,B,A,G | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,176 |
30. Three friends were wearing white, red, and blue dresses. Their shoes were of the same three colors. Only Tamar had matching dress and shoe colors. Varya was wearing white shoes. Neither Lida's dress nor her shoes were red. Determine the color of the dress and shoes of each friend.
.
Draw solid and dashed lines on the diagram to match the conditions of the problem (similarly to how it was done in problem 26).
The answer should be in the form of three triangles with solid... | Tamara-inreddressredshoes,Valya-inbluedresswhiteshoes,Lida-inwhitedressblueshoes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,177 |
33. Is it possible to organize a football tournament with nine teams so that each team plays four matches?
In practice, there is a need to conduct competitions with a large number of participants in a short time; for example, when organizing football tournaments for the prize of the opening season and when organizing ... | $\triangle$ Let's represent each team as a point, and each match played by the team as a segment emanating from this point. Nine points are better arranged so that when they are sequentially connected by segments, they form a convex nonagon.
The problem boils down to the following: can nine points be connected by segm... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,178 |
45. In a round-robin school tournament, six chess players are participating: Alyosha, Borya, Vitya, Grisha, Dima, and Kostya. Every day, three games were played, and the entire tournament lasted five days. On the first day, Borya played against Alyosha, and on the second day, he played against Kostya. Vitya played agai... | Let's denote the chess players by points A, B, V, G, D, and K, and the games they played by segments connecting these points. The points should be arranged so that when connected sequentially, they form the vertices of a convex hexagon.
1) Let's start with the first day of the tournament. According to the conditions, ... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,180 |
48. Each pair of six computers is connected by a wire. Can all these wires be colored with one of five colors so that from each computer five wires of different colors come out? | $\triangle$ To solve this, let's draw a convex hexagon and draw all its diagonals (Fig. 13). Let each vertex of the hexagon represent one of the computers, and each segment represent a wire connecting two computers. We will number the different colors with natural numbers from 1 to 5 to distinguish them from each other... | itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,181 |
50. In Flower City, each dwarf is acquainted with 6 little girls, and each little girl is acquainted with 6 dwarfs. Prove that the number of dwarfs in this city is equal to the number of little girls. | $\triangle$ Let's represent two sets: the set of dwarfs $A$ and the set of little girls $B$ (Fig. 14). The relationship of acquaintance between a dwarf and a little girl will be denoted by a line connecting the corresponding points in sets $A$ and $B$. Let the number of dwarfs be denoted by $k$, and the number of littl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,182 |
55. In one institution, each employee subscribes to two newspapers, each newspaper is subscribed to by five people, and each pair of newspapers is subscribed to by only one person. How many people are in the institution and how many newspapers do they subscribe to?
$\mathbf{5 6}^{\circ}$. Six points, no three of which... | $\triangle$ Let's take one of the six points $A_{1}$. From it, there are five segments, colored either black or red (Fig. 22). Then among them, there will be three segments of the same color, for example, red: because if there were no more than two segments of each color, then from point $A_{1}$ there would be no more ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,185 |
60*. 17 scientists from different countries are corresponding with each other in one of three languages: English, French, or Russian. Prove that among them, there will be three who are corresponding with each other in the same language. | Let's denote each of the scientists by a point and connect these points with all possible segments. We will arrange the points so that no three of them lie on the same line. Since each scientist corresponds with 16 others, 16 segments will emanate from each point. Each segment representing correspondence in English wil... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,186 |
63. Find the sets $A$ and $B$, if
$$
A \cap B=\{1, 2, 3\}, A \cup B=\{1, 2, 3, 4, 5\}
$$ | $\triangle$ From the definitions of intersection and union of sets, it follows that elements 1, 2, and 3 are in both sets $A$ and $B$, while elements 4 and 5 are in only one of them. Hence, we get the answer by considering all cases.
Answer: $A=\{1 ; 2 ; 3\}, B=\{1 ; 2 ; 3 ; 4 ; 5\}$ (or vice versa, i.e., with $A$ and... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,187 | |
64. It is known that $A \cap B=\{1 ; 2\}, A \cap C=\{2 ; 5\}, A \cup B=$ $=\{1 ; 2 ; 5 ; 6 ; 7 ; 9\}, B \cup C=\{1 ; 2 ; 3 ; 4 ; 5 ; 7 ; 8\}$. Find the sets $A, B$ and $C$. | $\triangle$ When solving, we will create a table. It is more convenient to fill it in by columns rather than by rows. For example, if the element 5 belongs to set $B$, we will place a plus at the intersection of the corresponding row and column of the table; otherwise, we will place a minus.
1) Let's start with the nu... | A={1;2;5;6;9},B={1;2;7},C={2;3;4;5;7;8} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,188 |
66. In a Bashkir village, every resident speaks either Bashkir, or Russian, or both languages. 912 residents of the village speak Bashkir, 653 - Russian, and 435 people speak both languages. How many residents are there in this village? | $\triangle$ Let's apply Euler circles. Let $A$ denote the set of villagers who speak Bashkir, and $B$ the set of villagers who speak Russian (Fig. 26).
Fig. 26
=28, \quad n(B)=13, \quad n(C)=10, \quad n(A \cap B)=8 \\
n(A \cap C)=6, \quad n(B \cap C)=5, \quad n(A \cap B \cap C)=2
\end{gathered}
\]
First, we... | 75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,190 |
75. Among the applicants who passed the entrance exams to a technical university, "excellent" grades were received by: in mathematics - 48 people, in physics - 37, in literature - 42, in mathematics or physics - 75, in mathematics or literature - 76, in physics or literature - 66, in all three subjects - 4. How many ap... | $\triangle$ Let's apply Euler circles. Let M, Ph, and L denote the sets of applicants who have scored "excellently" in mathematics, physics, or literature, respectively; these sets have 48, 37, and 42 elements, respectively. The common part of all three sets has 4 elements. Let $a, b, c, x, y, z$ denote the number of a... | 65,25,94 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,191 |
77. Find the set $X$, if $X \cap A = X \cup A$, where $A$ is a given set. | $\triangle$ The given equality is an example of an equation with a set. In the future, we will encounter systems of such equations.
Based on property 2 of operations over sets (p. 29)
$$
X \cap A \subset X \subset X \cup A
$$
But since the extreme sets $X \cap A$ and $X \cup A$ coincide here, all three sets coincide... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,192 |
81. Find the sets $X$ and $Y$ from the system of equations
$$
\left\{\begin{array}{l}
X \cup Y=A \\
X \cap A=Y
\end{array}\right.
$$
where $A$ is a given set. | $\triangle$ From the first equation of the system, it follows that $X \subset A$. Then $X \cap A=X$, which means the second equation becomes $X=Y$. In this case, from the first equation we get: $X=Y=A$.
Verification shows that the values $X=A$ and $Y=A$ we found satisfy the given system of equations.
Answer: $X=Y=A$. | X=Y=A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,193 |
86. It is known that for sets $A, B$ and $C$ the inclusions
$$
A \cup B \subset C, A \cup C \subset B, B \cup C \subset A .
$$
Do these inclusions imply that $A=B=C$? | $\triangle$ Since $A \subset A \cup B, B \subset A \cup B, A \cup B \subset C$, it follows that $A \subset C$, $B \subset C$.
Since $A \subset A \cup C, C \subset A \cup C, A \cup C \subset B$, then $A \subset B, C \subset B$.
Since $B \subset B \cup C, C \subset B \cup C, B \cup C \subset A$, then $B \subset A, C \s... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,194 |
90. There are three coins that look identical, two of which are genuine and of the same weight, and one is counterfeit and lighter than the genuine ones. Can the counterfeit coin be found with a single weighing on correct balance scales without weights? | $\triangle$ Let's number the coins as $1, 2, 3$. Place the first coin on the left pan of the balance, and the second coin on the right pan. There are two possible outcomes.
38
The balanc... | itispossible | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,195 |
92. There are 9 coins, 8 of which are genuine and of the same weight, and one is counterfeit and heavier than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin? | $\triangle$ Let's number the coins with natural numbers from 1 to 9. Here, it is better to weigh the coins by placing three on each pan of the balance. Place coins numbered $1,2,3$ on the left pan, and coins numbered $4,5,6$ on the right pan. There are two possible outcomes, as in problem 90.
Suppose the balance is in... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,196 |
93. There are 10 coins, 9 of which are genuine and of the same weight, and one is counterfeit and lighter than the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the counterfeit coin? | $\triangle$ Let's number the coins. Place coins numbered $1,2,3$ on the left pan and coins numbered $4,5,6$ on the right pan. We can use Figure 30 for further steps.
If the scales balance, the counterfeit coin is among the remaining four. It takes two more weighings to determine the counterfeit coin from these four (s... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,197 |
97. There are 5 parts that are indistinguishable in appearance, 4 of which are standard and of the same mass, and one is defective, differing in mass from the others. What is the minimum number of weighings on a balance scale without weights that are needed to find the defective part? | $\triangle$ Let's number the parts. Now try to figure out the weighing scheme presented below (Fig. 32).
As can be seen from this, it took three weighings to find the defective part.
Answer: in three. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,198 |
99. There are 6 identical-looking coins, but 4 are genuine, of the same weight, while 2 are counterfeit, lighter, and also weigh the same. What is the minimum number of weighings on a balance scale without weights that are needed to find both counterfeit coins? | $\triangle$ As usual, let's number the coins. During the first weighing, we will place coins numbered $1,2,3$ on the left pan, and all the others on the right (Fig. 33).
The scales may balance. Then each of the triplets contains one counterfeit coin. To find them, we need two more weighings (see the solution to proble... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,199 |
102. The controller distributed 90 standard parts equally into 9 boxes and placed 10 defective parts in one box. He cannot remember which box contains the defective parts, but he knows that a standard part weighs 100 g, and a defective part weighs 101 g. How can he find the box with the defective parts in one weighing ... | $\triangle$ Let's number the boxes. From the first box, we take one part, from the second - two, and so on, from the tenth - 10 parts. In total, we have taken $1+2+3+\ldots+10=55$ parts.
We will weigh these parts together. If all of them were standard, the scales would show $55 \cdot 100=5500$ g. But in reality, they ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,200 |
106. Out of 11 balls, two are radioactive. In one check, you can determine if a group of balls contains a radioactive ball, but if there is a radioactive ball, you cannot determine how many there are - one or two. How can you find both radioactive balls in seven checks? | $\triangle$ Let's number the balls. We will divide them into three piles: the first pile will include balls $1-4$, the second pile will include balls $5-8$, and the third pile will include balls $9-11$. In three checks, we will determine if there is a radioactive ball in each of these piles; if it turns out that two of... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,202 |
111. How many five-digit natural numbers can be formed using the digits 1 and 0 if the digit 1 appears exactly three times in each number? | $\triangle$ We will search for the specified numbers by enumeration, ensuring that we do not miss any number. It is simpler to start by finding the positions for the two zeros, because if the positions for the zeros are determined, the three remaining positions will be filled with ones uniquely.
Let's fix one of the z... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,204 |
113. How many two-digit natural numbers are there in which the first digit is greater than the second? | $\triangle$ If the first digit of a two-digit number is 1, then there is only one such number - 10. If the first digit of the number is 2, then there are two such numbers - 20 and 21. If the first digit is 3, then there are already three such numbers - 30, 31, and 32. And so on. Finally, if the first digit is 9, then t... | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,205 |
124*. A checker can move in one direction along a strip divided into cells, moving either to the adjacent cell or skipping one cell in a single move. In how many ways can it move 10 cells? | $\triangle$ A checker can move one cell in 1 way, two cells in 2 ways, three cells in $-1+2=3$ ways, four cells in $-3+2=5$ ways, five cells in $-3+5=8$ ways, and so on. In fact, the Fibonacci sequence (see the solution to problem 12) is formed, which is defined by the recurrence relation
$$
a_{n}=a_{n-1}+a_{n-2}(n \i... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,208 |
128. The trainer brings three lions A, B, C and two tigers D and E onto the circus arena and seats them in a row on pedestals. At the same time, the tigers cannot be placed next to each other, otherwise a fight between them is inevitable. How many ways are there to arrange the animals? | $\triangle$ First, let's calculate the number of ways to arrange the tigers. If tiger G takes the first place, then tiger D can take the third, fourth, or fifth place. If G takes the second place, then D can only take the fourth or fifth place. If $\Gamma$ is placed on the third or fourth place, then in each of these c... | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,209 |
134. How many four-digit natural numbers without repeating digits can be written using only the digits $1,2,3,4,5$ and 6? | $\triangle$ A four-digit number can be represented as $\overline{x y z t}$, where $x, y, z, t$ are digits. Here, the digit $x$ can take 6 values, after which the digit $y-5$ (repetition of digits is not allowed), then the digit $z-4$, and the digit $t-3$. Therefore, the number of such four-digit numbers is $6 \cdot 5 \... | 360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,210 |
136. Prove that the product of the sum of $k$ terms by the sum of $n$ terms, before combining like terms, contains $k n$ terms. | $\triangle$ Any term of the product can be represented as $y \cdot z$, where $y$ is any term of the first sum, and $z$ is any term of the second. Then the variable $y$ takes $k$ values, the variable $z$ takes $n$ values. Therefore, the number of terms in the expansion is $\mathrm{kn} . \boldsymbol{\Delta}$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,211 |
142. On a plane, there are $n$ points ( $n \geq 3$ ), no three of which lie on the same line. How many lines can be drawn, each passing through two of the given points? | $\triangle$ By analogy with the solution of the previous problem, we get $n(n-1)$ lines. However, the point is that a line, unlike a vector, can be considered as an unordered pair of points - in the sense that line $A B$ coincides with line $B A$. Therefore, in such a calculation, each line is counted twice. Consequent... | \frac{n(n-1)}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,212 |
146. From a group of 15 people, four participants for the relay race $800+400+200+100$ are selected. In how many ways can the athletes be arranged for the relay stages? | $\triangle$ First, let's choose an athlete for the 800 m stage, then for the 400 m, followed by 200 m, and finally for the 100 m stage. Since the order in which the selected athletes compete is significant, we are dealing with permutations—permutations of 4 out of 15. We get:
$$
A_{15}^{4}=15 \cdot 14 \cdot 13 \cdot 1... | 32760 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,213 |
155. On a plane, there are 10 lines, and among them, there are no parallel lines and through each point of their intersection, exactly two lines pass. How many intersection points do they have? | $\triangle$ Each intersection point is determined by a pair of lines $a$ and $b$ that intersect at this point. Since the order of these lines does not matter, each pair of such lines is a combination - a combination of 10 elements taken 2 at a time. We get:
$$
C_{10}^{2}=\frac{10 \cdot 9}{1 \cdot 2}=45
$$
Answer $: 4... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,214 |
161. How many six-digit natural numbers exist, each of which has its digits arranged in ascending order? | $\triangle$ Let's take the nine-digit number 123456789, whose digits are in ascending order. To obtain any six-digit number with the same property, it is sufficient to erase any three digits from this nine-digit number. Since the order of the digits being erased does not matter, we are dealing with combinations. We hav... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,215 |
165. On a straight line, there are 6 points, and on a parallel line, there are 8 points. How many triangles exist with vertices at these points? | $\triangle$ First, let's calculate the number of triangles where two vertices lie on the first line, and one vertex lies on the second line:
$$
C_{6}^{2} \cdot C_{8}^{1}=15 \cdot 8=120
$$
Similarly, let's calculate the number of triangles where one vertex lies on the first line, and two vertices lie on the second lin... | 288 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,216 |
## KPOCC
$+$
KPOCC
SPORT,
where each letter represents a digit, and the same letters represent the same digits, different letters represent different digits. Find all solutions.
This is an example of a numerical riddle. | $\triangle$ Let's pay attention to the last two columns. Since the addition of the same digits C and C results in different digits, the sum in the fifth column must be at least 10, which means C is not less than 5.
Let's look at the third column: when can the sum $\mathrm{O}+\mathrm{O}$ end with the same digit O? This... | 35977+35977=71954 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,219 |
184. In the apartment, there are 13 people, cats, and flies. Together, they have 42 legs, with each fly having 6 legs. How many people, cats, and flies were there separately? List all possible answers. | Let's denote the number of people, cats, and flies by \( x, y, \) and \( z \) respectively. We obtain the system of equations:
\[
\left\{\begin{array}{l}
x+y+z=13 \\
2 x+4 y+6 z=42
\end{array}\right.
\]
We can simplify the second equation by dividing it by 2:
\[
x+2 y+3 z=21
\]
Subtracting the first equation from t... | (8,2,3),(7,4,2),(6,6,1) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,221 |
206. Find all two-digit numbers, each of which is 13 more than the sum of the squares of its digits. | $\triangle$ Let the desired number be denoted as $\overline{x y}$. We obtain:
$$
10 x + y = x^2 + y^2 + 13, \quad x^2 - 10 x + (y^2 - y + 13) = 0
$$
We will consider the last equation as a quadratic equation in terms of $x$. Its discriminant must be non-negative:
$$
\frac{1}{4} D = 25 - y^2 + y - 13 \geq 0, \quad -y... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,222 |
213. Solve the equation $x y z = x + y$ in natural numbers $x, y$ and $z$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | $\triangle$ Suppose that $x \leq y$. Such a restriction can be imposed mainly in cases where the equation is symmetric with respect to $x$ and $y$. It does not spoil the problem, as it can be removed during further solving.
We get:
$$
x y z = x + y \leq y + y = 2 y, \quad x y z \leq 2 y
$$
from which
$$
x z \leq 2 ... | (2;2;1),(1;1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,223 |
215. The sequence $\left(x_{n}\right)$ is defined by the recurrence relations
$$
x_{1}=1, x_{2}=2, x_{n}=\left|x_{n-1}-x_{n-2}\right|(n>2) .
$$
Find $x_{1994}$. | $\triangle$ First, let's compute the first few terms of the sequence: $x_{1}=1, x_{2}=2, x_{3}=|2-1|=1, x_{4}=|1-2|=1, x_{5}=|1-1|=0, x_{6}=$ $=|0-1|=1, x_{7}=1, x_{8}=0, x_{9}=1, x_{10}=1, x_{11}=0$.
Starting from the third term, the terms of this sequence are either 1 or 0. Only the terms with indices
$$
5,8,11, \l... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,224 |
218. Find the last two digits of the power $7^{1995}$. | $\triangle$ Let's find the last two digits of the first few terms of the sequence ($7^{n}$): $07,49,43,01,07,49,43,01,07,49, \ldots$.
From this, we can see that these digits repeat when the term number increases by 4, generally by $4k (k \in N)$. Since
$$
1995=4 \cdot 498 + 3
$$
the last two digits of $7^{1995}$ coi... | 43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,225 |
221. One person is a truth-teller, the other is a liar. Find at least one question that you need to ask each of them so that they give the same answer. | $\triangle$ Here the question should relate to the essence of these people. For example, you can ask both: "Are you a truth-teller or a liar?" Both the truth-teller and the liar will answer: “Truth-teller”.
Answer: for example, "Are you a truth-teller or a liar?".
| \ | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,227 |
224. On the island, there are two villages A and B. The residents of A are truth-tellers, and the residents of B are liars. Residents of A visit B, and residents of B visit A. A traveler met a person in one of these villages and wants to find out which village he is in. How can he find out from the islander he met:
a)... | $\triangle$ a) It is easy to do this in two questions. With the first question, the traveler should find out whether the person in front of them is a truth-teller or a liar (see problem 222). The second question can be straightforward: "Is this village A?" Depending on who this person turns out to be, the traveler shou... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,228 |
225. On the island of truth-lovers and liars, local resident $\mathrm{K}$ says about himself and another resident $\mathrm{M}$ of the island: "At least one of us is a liar". Who are K and M? | $\triangle$ Suppose that $\mathrm{K}$ is a liar. Then his statement is false, which means that K and M are truth-tellers. But this contradicts our assumption.
Therefore, $\mathrm{K}$ is a truth-teller. Then his statement is true, which means that M is a liar.
Answer: $\mathrm{K}$ is a truth-teller, $M$ is a liar. $\m... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,229 |
227. Of the three inhabitants K, M, and P of the island of truth-tellers and liars, two said:
K: We are all liars.
M: One of us is a truth-teller.
Who among the inhabitants K, M, and P is the truth-teller and who is the liar? | $\triangle$ If K were a truth-teller, then based on his statement, all three islanders, including K, would be liars, which contradicts our assumption. Therefore, K is a liar. Since his statement is false, among M and P there is a truth-teller.
Now let's consider M. If M were a liar, then based on the previous statemen... | K | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,230 |
229. Of the three residents K, M, and P of a remote area, one is a truth-teller, another is a liar, and the third is a trickster. They made the following statements:
K: I am a trickster.
M: That's true.
P: I am not a trickster.
Who are K, M, and P in reality? | $\triangle$ Let's start with K. He cannot be a truth-teller due to his statement. Let's consider two cases.
1) Suppose $\mathrm{K}$ is a trickster. Then $\mathrm{M}$ is a truth-teller based on his statement, which means $\mathrm{P}$ is a liar. But in this case, $\mathrm{P}$'s statement is true, which is impossible.
2)... | K | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,231 |
232. On the island of Buiane, there are three villages: Pravdino, Kryvdino, and Seredina-na-Polovine. The residents of Pravdino are truth-lovers, the residents of Kryvdino are liars, and the residents of Seredina-na-Polovine are tricksters, but with a quirk: one of any two consecutive statements they make is true, and ... | $\triangle$ Let's consider three possibilities.
1) Suppose the caller was a truth-teller. But this contradicts the fact that the caller referred to his village as Middle-on-Half.
2) Suppose the caller was a liar. Then his statement "There is a fire in our village" is false.
3) Suppose the caller was a trickster. In th... | continuereadingthenovel | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,232 |
234*. Once in a room, there were several inhabitants of an island where only truth-tellers and liars live. Three of them said the following.
- There are no more than three of us here. All of us are liars.
- There are no more than four of us here. Not all of us are liars.
- There are five of us. Three of us are liars.
... | $\triangle$ Let's consider the first of the three speakers. Suppose he is a truth-teller. Then both of his statements, including the second one, "All of us are liars," would be true, which means he is a liar. We have reached a contradiction. Therefore, he can only be a liar.
In this case, both statements of the first ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,233 |
241. The goddesses Hera, Aphrodite, and Athena came to the young Paris to have him determine which of them was the most beautiful. They made the following statements.
- Aphrodite: I am the most beautiful.
- Hera: I am the most beautiful.
- Athena: Aphrodite is not the most beautiful.
- Aphrodite: Hera is not the most ... | $\triangle$ Let's consider three possibilities.
1) Suppose the most beautiful of the goddesses is Hera. Then her statement is true, and all the statements of the other two goddesses are false. But, on the other hand, it turns out that Athena's first statement is true. We have reached a contradiction.
2) Suppose the mo... | Aphrodite | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,234 |
242. Once I found a curious notebook. In it, there were forty of the following statements.
- In this notebook, there is exactly one false statement.
- In this notebook, there are exactly two false statements.
- In this notebook, there are exactly three false statements.
- In this notebook, there are exactly forty fals... | Let $\triangle$ denote by $A$ the statement "Andrey will cope with the work," and by $B$ the statement "Boris will cope with the work."
Then Andrey said: "If $A$ is true, then $B$ is true." In short: "If $A$, then $B$," or in symbolic notation: $A \Rightarrow B$.
Similarly, Boris said: "If $B$ is false, then $A$ is f... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 36,235 |
247. Find the natural numbers $a$ and $b$, if out of the four statements:
1) $a-b$ is divisible by 3
2) $a+2b$ is a prime number;
3) $a=4b-1$
4) $a+7$ is divisible by $b$
three are true, and one is false. Find all solutions. | $\triangle$ We will look for a false statement among the given ones. For this, it would be good to find two incompatible statements.
The first and second statements cannot be true at the same time, because if the number $a-b$ is divisible by 3, then the number $a+2b = (a-b) + 3b$ is also divisible by 3, and therefore ... | (3;1),(7;2),(11;3) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,236 |
253. Given three statements:
1) the equation $x^{2}-a x+1=0$ has no roots (real);
2) the equality $|a-2|=2-a$ holds;
3) the system of equations
$$
\left\{\begin{array}{l}
x+y^{2}=a \\
x-\sin ^{2} y=-3
\end{array}\right.
$$
has a unique solution.
Find all values of $a$ for which two of these statements are true and o... | $\triangle$ The first statement is true if $a^{2}-4<0$, i.e., when $-2<a<2$. The second statement is true if $a \leq 2$.
To determine when the third statement is true, note that based on the form of the given system of equations, if ( $x_{0} ; y_{0}$ ) is a solution to the system, then ( $x_{0} ;-y_{0}$ ) is also a so... | -3,(-2;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 36,237 |
255. Given the following statements:
1) Joe is a trickster;
2) Joe is unlucky;
3) Joe is lucky, but he is not a trickster;
4) if Joe is a trickster, then he is unlucky;
5) Joe is a trickster if and only if he is lucky;
6) either Joe is a trickster or he is lucky, but not both at the same time.
What is the maximum numb... | $\triangle$ The third statement is incompatible with the first and the second. Therefore, among the first three statements, there is a false one. Since the number of true statements out of the given six should be the largest possible, the number of false statements should be the smallest. In this case, to have only one... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,238 |
257. In the class, there are 35 students. Prove that among them, there will definitely be at least two whose last names start with the same letter. | $\triangle$ In the Russian alphabet, there are 33 letters, while the number of students is greater - 35. Here, the letters play the role of boxes, and the students in the class play the role of objects being placed into these boxes. Since there are more objects than boxes, by the Pigeonhole Principle, there will be a b... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,239 |
265. Is it true that from any five natural numbers, one can choose two numbers, the difference of whose squares is divisible by 7? | $\triangle$ The square of a natural number, when divided by 7, can only give the following four remainders: $0, 1, 2$, and 4 (check this!). But there are five numbers. Based on the statement of problem 262, among the five squares
$$
a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, a_{4}^{2}, a_{5}^{2}
$$
there will be two whose diff... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,240 |
270. In a box, there are 105 apples of four varieties. Prove that among them, there will be at least 27 apples of one variety. | $\triangle$ We will provide two proofs of the statement.
1) Assume the opposite: 27 apples of the same variety cannot be found. Then the number of apples of each variety will not exceed 26. Therefore, the total number of apples in the box will not be more than $26 \cdot 4 = 104$, while according to the problem, there ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,241 |
277. Prove that among 82 cubes, each of which is painted in a certain color, there exist 10 cubes of different colors or 10 cubes of the same color. | $\triangle$ Consider two cases.
1) Suppose that at least 10 different colors were used to paint the cubes. Then there will be 10 cubes of different colors.
2) Suppose that no more than 9 different colors were used to paint the cubes. Divide 82 by 9 with a remainder: $82=9 \cdot 9+1$. Here $n=9, k=9$. Then by the gener... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 36,242 |
281. Is it true that among any 7 natural numbers, there will definitely be three numbers whose sum is divisible by 3? | $\triangle$ A natural number when divided by 3 can give one of three remainders: 0, 1, or 2. Since $7=3 \cdot 2+1$, there exist three numbers that give the same remainder when divided by 3. Their sum is divisible by 3.
Answer: correct. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,243 |
286*. Is it possible to number the vertices of a cube with natural numbers from 1 to 8 so that the sums of the numbers at the ends of each edge of the cube are different? | $\triangle$ The sums of the numbers at the ends of the edges of a cube can take values from $1+2=3$ to $7+8=15$ - a total of 13 different values.
Suppose among the sums there are such:
$$
1+2=3, \quad 1+3=4, \quad 1+4=5
$$
But then there cannot be a sum equal to 6, because the sum $1+5$ cannot be used - since the nu... | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,245 |
293*. Natural numbers $a$ and $b$, where $b \geq 2$, are coprime. Prove that there exists a natural number $n$, such that the number $pa$ when divided by $b$ gives a remainder of 1. | $\triangle$ Let's take $b-1$ numbers: $1 \cdot a, 2 \cdot a, 3 \cdot a, \ldots, (b-1) a$, i.e., numbers of the form $k a$, where $k$ is a natural number satisfying the inequality $1 \leq k \leq b-1$. Each of these numbers is not divisible by $b$, since $a$ and $b$ are coprime, and there are no $k_1$ and $k_2$ such that... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 36,246 |
295. There are two piles of balls on the table, with 30 balls in each pile. Two players take turns taking any number of balls from the table, but on one turn, from only one of the piles. The winner is the one who takes the last balls from the table. Who and how will win with the correct play? | $\triangle$ The second player wins by taking the same number of balls on each move as the player who starts, but from a different pile. For example, if the player who starts takes 5 balls from the first pile on their first move, the second player should take 5 balls from the second pile in response; if the first player... | theplayer | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,247 |
296. On the table lie three piles of pebbles, with 30 pebbles in each pile. Two players take turns taking any number of pebbles from the table, but in one move, only from one pile. The winner is the one who takes the last pebbles from the table. Who and how will win with the correct play? | $\triangle$ Let's compare the conditions of this problem with those of the previous one. There were two piles, not three, and as proven earlier, it was disadvantageous to start the game. This consideration leads to the following idea for the beginning player: he should take all 30 stones from one pile, for example, the... | thebeginningplayer | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,248 |
299. There are 8 balls: 2 red, 2 blue, 2 white, and 2 black. Players A and B take turns attaching one ball to one of the vertices of a cube. Player A aims to achieve a vertex such that this vertex and its three adjacent vertices each have a ball of a different color, while Player B aims to prevent this. Who will win wi... | $\triangle$ The problem does not specify who starts the game - A or B. Therefore, let's consider two cases.
1) Suppose A starts the game.
If A, for example, attaches a red ball to one of the vertices of the cube, then B's best response is to attach a red ball to one of the adjacent vertices, and so on (Fig. 36). Cons... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,249 |
301. There are $n$ minuses written on the board. Two players take turns changing one or two adjacent minuses to pluses. The player who changes the last minus wins. Who will win with the correct play, if: a) $n=9$; b) $n=10$? | $\triangle$ а) When $n=9$, the beginner should first move to change the central minus to a plus
Fig. 36
.
Fig. 38
1) Suppose the chip is initially on the cell numbered 0. Then the second player wins without making a single move.
2) Suppose ... | 4,8,12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,252 |
310. There are 60 coins on the table. Two players take turns picking up $1, 2, 3$ or 4 coins from the table. The player who picks up the last coins wins. Who will win with the correct play? | $\triangle$ Let's consider several cases, starting with a small number of coins on the table.
1) Suppose there are 1, 2, 3, or 4 coins on the table initially. Then the first player wins by taking all these coins on the first move.
2) Suppose there are 5 coins. Here, the second player wins by complementing to 5: if the... | theplayer | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,253 |
315*. In one box, there are 15 blue balls, and in another, 12 white balls. Two players are playing a game. In one move, each is allowed to take 3 blue balls or 2 white balls. The winner is the one who takes the last balls. How should the first player play to win? | $\triangle$ He should first take 2 white balls from the second box, because after this, the ratio of the number of blue balls to the number of white balls will become $15: 10=3: 2$.
If now the second player, in their first move, takes 3 blue balls from the first box or 2 white balls from the second, then the one who s... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 36,254 |
318. Out of 100 cubes, 80 have a red face, 85 have a blue face, and 75 have a green face. How many cubes have faces of all three colors? | $\triangle$ Compared to similar problems from § 4 "Operations on Sets," there is significantly less data here, so the answer will be ambiguous - "from" and "to." Therefore, Venn diagrams will be of little help here.
First, let's determine the maximum number of cubes that have faces of all three colors. It is 75 and is... | 40\leqn\leq75 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 36,255 |
321. On a line, there is a set $M$ of points such that each point in $M$ is the midpoint of a segment connecting two other points in $M$. Prove that the set $M$ is infinite. | $\triangle$ Suppose that the set $M$ is finite. Then among its points, there is the leftmost point $A$. Since it is the midpoint of the segment connecting two points from $M$, one of these two points will be located to the left of point $A$. We have reached a contradiction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,256 |
323. At each vertex of an octagon, a number is written, and each such number is equal to the arithmetic mean of the numbers written at the two adjacent vertices. Can the 8 written numbers be different? | $\triangle$ Suppose that among these 8 numbers there are different ones. Let $a$ be the smallest of them (if there are several, we take any of them), and let $b$ and $c$ be the numbers written in the two adjacent vertices. Then $a \leq b$ and $a \leq c$. Adding these inequalities term by term:
$$
2 a \leq b+c, \quad a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 36,257 |
326. Does the equation $x^{3}+2 y^{3}=4 z^{3}$ have solutions in natural numbers | $\triangle$ Suppose the equation has solutions in natural numbers. Consider the solution $(x_{0}; y_{0}; z_{0})$ for which the value $x=x_{0}$ is minimal. Later, we will see that if this equation has a solution in natural numbers, then it has another solution; if the equation has an infinite set of solutions in natural... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 36,258 |
329. Prove that the circles constructed on the sides of a convex quadrilateral $A B C K$ as diameters completely cover this quadrilateral. | $\triangle$ Let $M-$ be any point lying inside a quadrilateral. Consider the angles $A M B, B M C, C M K$ and $K M A$. Suppose the largest of them is
Fig. 39 angle $A M B$. Then (Fig. 39) ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 36,259 |
331. There is a $10 \times 10$ square table, in the cells of which natural numbers from 1 to 100 are written in sequence:
in the first row - numbers from 1 to 10, in the second row - from 11 to 20, and so on. Prove that the sum $S$ of any 10 numbers in the table, none of which are in the same row and none of which are ... | $\triangle$ Let's denote the term of the original sum $S$ from the first row as $a_{1}$, from the second row as $10+a_{2}$, from the third row as $20+a_{3}$, and so on, finally from the tenth row as $90+a_{10}$.
Here, each of the natural numbers $a_{1}, a_{2}, \ldots, a_{10}$ is within the range from 1 to 10, and thes... | 505 | Combinatorics | proof | Yes | Yes | olympiads | false | 36,260 |
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