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Place different positive integers, not greater than 25, in the cells of a $3 \times 3$ square, such that in any pair of adjacent cells (sharing a side), one number is divisible by the other. | The central cell has the most neighbors - let's put 1 there. In the cells adjacent to the central one, we will place smaller numbers - 2, 3, 4, 5. Every number is divisible by the central number (any number is divisible by 1). To ensure the condition is met for the corner cells as well, we will place in each corner the... | \begin{pmatrix}\hline10&2&8\\\hline5&1&4\\\hline15&3&12\\\hline\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,062 |
There are 6 locked suitcases and 6 keys to them. However, it is unknown which key fits which suitcase. What is the minimum number of attempts needed to definitely open all the suitcases? And how many attempts would be needed if there were not 6, but 10 suitcases and keys?
# | Try to determine which of the 6 suitcases the first key fits in five attempts.
## Solution
Standard incorrect solution: "Each of the six suitcases is tried with each of the six keys, totaling $66=36$ attempts." However, it is possible to find the correspondence between the keys and the suitcases with fewer attempts. ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,064 |
[ Cube
Can the numbers from 1 to 12 be placed on the edges of a cube so that the sums of the numbers on all faces are the same?
# | On the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, we will place numbers as follows:
$A B-10, B C-5, C D-7, A D-4$,
$A_{1} B_{1}-3, B_{1} C_{1}-9, C_{1} D_{1}-6, A_{1} D_{1}-8$
$A A_{1}-2, B B_{1}-11, C C_{1}-1, D D_{1}-12$
On each face of the cube, the sum of the numbers is 26.
## Answer
It is possible. | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,065 |
Along the right side of the road, 100 cars are parked. Among them, there are 30 red, 20 yellow, and 20 pink Mercedes. It is known that no two Mercedes of different colors are parked next to each other. Prove that in this case, there must be three Mercedes of the same color parked consecutively. | We have 70 Mercedes and 30 "other" cars. According to the condition, next to a Mercedes can only be another Mercedes of the same color or an "other" car. The more Mercedes are parked in pairs, the fewer "other" cars will be needed. But there are 35 pairs of Mercedes, and 33 "other" cars will be needed for their "surrou... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,066 |
Blinkov A.d:
The square of the sum of the digits of the number $A$ is equal to the sum of the digits of the number $A^{2}$. Find all such two-digit numbers $A$. | Answer: $10,11,12,13,20,21,22,30,31$.
Solution 1. Note that $A^{2} \leq 99^{2}=9801<9999$. Therefore, the sum of the digits of $A^{2}$ is less than $9 * 4=36$. Since it is equal to the square of the sum of the digits of $A$, the sum of the digits of $A$ is less than $36^{1 / 2}=6$, that is, less than or equal to 5. Th... | 10,11,12,13,20,21,22,30,31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,068 |
Galochkin A.i.
Prove that if for numbers $a, b$ and $c$ the inequalities $|a-b| \geq|c|,|b-c| \geq|a|,|c-a| \geq|b|$ hold, then one of these numbers is equal to the sum of the other two. | The first method. Suppose, first, that one of the numbers is zero. Let, for example, $a=0$ (other cases are analogous). Then we obtain the inequalities: $|b| \geq|c|$ and $|c| \geq|b|$, from which $|b|=|c|$, i.e., $b=c$ or $b=-c$.
In the first case, $b=a+c$, in the second case, $a=b+c$. Everything is proven.
Now let ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,069 |
[ Decimal numeral system ]
A natural number had the sum of its digits subtracted from it, resulting in 2007. What could the original number have been? | It is not difficult to verify that the desired number must be a four-digit number. Let it be $\overline{a b c d}$ $=1000 a+100 b+10 c+d$. Then $1000 a+100 b+10 c+d-(a+b+c+d)=999 a+99 b+9 c=9(111 a+11 b+c)$. We obtain the equation: $9(111 a+11 b+c)=2007 \Leftrightarrow 111 a+11 b+c=223$. By enumeration, we can see that ... | 2010to2019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,070 |
In a $10 \times 10$ table, integers are arranged such that any two numbers in adjacent cells differ by no more than 5.
Prove that among these numbers, there are two that are equal.
# | Since from each cell to any other can be reached by no more than 19 moves to an adjacent cell, all numbers are between
$a$ and
$a+95$, where $a$ is the minimum of all the numbers placed. Therefore, there are no more than 96 distinct numbers among these. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,071 |
From the set of dominoes, all tiles with sixes were removed. Can the remaining tiles be laid out in a row?
# | Suppose we managed to do it. Now the number five appears 7 times. Inside the chain, it appears an even number of times. Therefore, a five must be at one of the ends. Similarly, we can prove that all other "numbers" of the dominoes are at the ends. But there are six numbers, and only two ends. Contradiction. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,072 |
The product of 22 integers is 1. Prove that their sum is not equal to zero.
# | Clearly, all these numbers are equal to $\pm 1$. The sum can be zero only if there are an equal number of 1s and -1s, each 11. But then their product would be -1. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,073 |
[ Pairing and grouping; bijections $]$ [ Decimal number system ]
Find the last digit of the number $1^{2}+2^{2}+\ldots+99^{2}$. | $1^{2}+2^{2}+\ldots+99^{2} \equiv 10\left(1^{2}+2^{2}+\ldots+9^{2}\right)=0(\bmod 10)$.
## Answer
0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,074 |
There were seven boxes. In some of them, seven more boxes (not nested within each other) were placed, and so on. In the end, there were 10 non-empty boxes.
How many boxes are there in total? | At each operation, one empty box is filled. Since there are now 10 non-empty boxes, 10 operations have been performed. With each operation, seven boxes were added. Therefore, in the end, there are $7 + 10 \cdot 7 = 77$ boxes.
The Russian national football team won against the Tunisian national team with a score of $9:... | 77 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,076 |
Ten people are sitting at a round table. A sum of ten dollars must be distributed among them so that each person receives half of the sum that their two neighbors received together. Does this rule uniquely determine the distribution of money? | Let's consider the person who received the largest amount (if there are several, we will consider any one of them). Then both of his neighbors received the same amount. The same is true for the neighbors of these neighbors. Continuing, we will be convinced that everyone received an equal amount.
## Answer
Definitely. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,077 |
$[$ P Sequences (other) ]
Given 20 different natural numbers, each less than 70. Prove that among their pairwise differences, there will be four identical ones.
# | Arrange the numbers in ascending order and assume the opposite - among the differences between consecutive numbers, there are no four different ones.
## Solution
Let the numbers be denoted by $\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{20}$ in ascending order, so $\mathrm{a}_{1}<\mathrm{a}_{2}<\ldots<\mathrm... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,078 |
[ Methods for solving problems with parameters ] [ [examples and GMP on the coordinate plane]
Consider the quadratic functions $y=x^{2}+p x+q$, for which $p+q=2002$.
Show that the parabolas, which are the graphs of these functions, intersect at one point.
# | Find the values of these functions at the point $x=1$.
## Solution
$y(1)=1+p+q=2003$. This means that each of the graphs passes through the point $(1,2003)$ on the coordinate plane. | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,079 |
[ Properties of symmetries and axes of symmetry ] [Examples and counterexamples. Constructions]
Is the following statement true: "If a quadrilateral has an axis of symmetry, then it is either an isosceles trapezoid, or a rectangle, or a rhombus"? | Provide a counterexample.
## Solution
Quadrilateral $ABCD$, in which $AB=AD$, $BC=CD$, and $AB \neq BC$, has an axis of symmetry but does not have parallel sides.
## Answer
No. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,080 |
For numbers $a, b$ and $c$, the equality $\frac{a^{2}+b^{2}}{b^{2}+c^{2}}=\frac{a}{c}$ holds. Does it follow that $\frac{a}{b}=\frac{b}{c}$? | For example, when $a=c=1, b=2$ the first equality holds, while the second does not.
Answer
It does not follow.
Send comment | Itdoesnotfollow | Algebra | proof | Yes | Yes | olympiads | false | 53,082 |
[ Algebraic inequalities (miscellaneous).] $[$ Factorization $]$
Prove that the polynomial $x^{12}-x^{9}+x^{4}-x+1$ is positive for all values of $x$.
# | If $x \leq 0$, then $x^{12}-x^{9}+x^{4}-x \geq 0$. If $0 < x < 1$, then $x^{12}-x^{9}+x^{4}-x > 0$. If $x \geq 1$, then $x^{9}\left(x^{3}-1\right)+$ $x\left(x^{3}-1\right)+1>0$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,084 |
On 99 cards, the numbers $1,2,3, \ldots, 99$ are written. Then the cards are shuffled, laid out with their clean sides up, and the numbers $1,2,3,4, \ldots, 99$ are written again on the clean sides. For each card, the numbers on it are added, and the 99 resulting sums are multiplied. Prove that the result will be an ev... | Among the numbers $1,2, \ldots, 99$ there are 50 odd and 49 even. Consider 50 cards, on which the odd numbers are written. On the reverse side of at least one of these cards, an odd number is written, so the sum of the numbers on it is even. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,086 |
[ Investigation of a quadratic trinomial ]
A quadratic trinomial $y=a x^{2}+b x+c$ has no roots and $a+b+c>0$. Determine the sign of the coefficient $c$.
# | 
The quadratic polynomial has no roots, which means its graph does not intersect the x-axis. Since \( y(1) = a + b + c > 0 \), the graph is located in the upper half-plane (see the figure), th... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,088 |
[ Invariants ] [ Algorithm Theory (other) $)]
On a magic apple tree, 15 bananas and 20 oranges have grown. It is allowed to pick one or two fruits at a time. If you pick one fruit, the same one will grow back. If you pick two identical fruits, an orange will grow back. If you pick two different fruits, a banana will ... | Can the fruits be picked in such a way that the number of bananas on the apple tree becomes even?
## Solution
b), c) Note that the number of bananas on the apple tree will always be odd.
a) For example, we can pick 7 times two bananas (leaving one banana and 27 oranges), and then 27 times pick one banana and one ora... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,090 |
Hooligans Vasya and Petya tore up the wall newspaper, with Petya tearing each piece into 5 parts, and Vasya into 9. When trying to collect the wall newspaper, 1988 scraps were found. Prove that not all pieces were found.
# | Both Petya and Vasya added an even number of pieces at each "step". Therefore, the total number of pieces was
odd. But 1988 is an even number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,091 |
$:$ Folkiore
Given two two-digit numbers $-X$ and $Y$. It is known that $X$ is twice as large as $Y$, one digit of the number $Y$ is equal to the sum, and the other digit is equal to the difference of the digits of the number $X$.
Find these numbers. | Let $a$ be the first digit of the number $X$, and $b$ be the second. The first digit of the number $Y$ is less than $a$, so it is equal to $|a-b|$, and the second digit is $a+b$. We can reason in different ways.
First method 1. Let's consider 2 cases.
1) $10 a+b=20(a-b)+2(a+b)$. Then $12 a=19 b$, which is impossible:... | 3417 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,092 |
$\underline{\text { Nazarov } \Phi}$.
Positive numbers $a, b, c, d$ are such that $a \leq b \leq c \leq d$ and $a+b+c+d \geq 1$. Prove that $a^{2}+3 b^{2}+5 c^{2}+7 d^{2} \geq 1$.
# | $a^{2}+3 b^{2}+5 c^{2}+7 d^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2\left(b^{2}+2 c^{2}+3 d^{2}\right) \geq a^{2}+b^{2}+c^{2}+d^{2}+2(a b+(a+b) c+(a+b+c) d)=(a+b+c+d)^{2} \geq 1$
$$
\begin{aligned} &\({\left[\begin{array}{l}\text { Decimal number system } \\ \text { Problem } \underline{98333} \text { Topics: }\end{array}\right] ... | 399 | Inequalities | proof | Yes | Yes | olympiads | false | 53,093 |
[ $\left[\begin{array}{ll}\text { Motion Problems } & \text { Processes and Operations }\end{array}\right]$ [ Divisibility of Numbers. General Properties ]
Two frogs, Kva and Kvi, are participating in a "race" - 20 meters forward in a straight line and back. Kva covers 6 dm in one jump, while Kvi only 4, but Kvi makes... | Think about how many jumps the second frog will have to make.
## Solution
The distance of 12 dm is covered simultaneously by both frogs. However, here they need to cover the distance of 20 m twice. 20 m = 200 dm is divisible by 4, so the second frog will jump exactly 40 m. But 200 is not divisible by 6, which means t... | The\\frog\will\win | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,094 |
Ivan the Tsarevich fought with the three-headed and three-tailed Zmey Gorynych. With one blow, he could cut off either one head, one tail, two heads, or two tails. But if he cut off one tail, two would grow back; if he cut off two tails, a head would grow back; if he cut off a head, a new head would grow back, and if h... | We need to ensure that Zmey Gorynych has an even number of heads and no tails. To achieve this, we first need to reach a certain number of tails.
## Solution
Notice that cutting off one head is meaningless. There are three types of strikes left: with the first, the number of tails increases by 1; with the second, the... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,096 |
On the plane, several points are colored blue and red such that no three points of the same color lie on the same line (there are at least three points of each color). Prove that some three points of the same color form a triangle, on the three sides of which there are no more than two points of the other color. | ## Solution
Consider a triangle $T$ of the smallest area $S$ with vertices at three points of the same color. Suppose that on its sides there are at least three points of another color. Then these three points are the vertices of a triangle with an area smaller than $S$, which contradicts the choice of triangle $T$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,097 |
On a circle, 2000 blue and one red point are marked. All possible convex polygons with vertices at these points are considered. Which are there more of - those that have a red vertex, or those that do not? | To each polygon with vertices only at blue points, we will correspond a polygon that, in addition to these blue vertices, has a red vertex.
## Solution
Let's agree to call a polygon with vertices only at blue points a polygon of type $A$, and a polygon, among whose vertices there is a red one, a polygon of type $B$. ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,100 |
3 [ Numerical tables and their properties ]
In the sky, there are an infinite number of stars. An astronomer assigned each star a pair of natural numbers representing its brightness and size. Each pair of stars differs in at least one parameter. Prove that there will be two stars, the first of which is not less than t... | Since the number of stars is infinite, at least one of the parameters takes an infinite number of values. Let this be the size. Then let's choose a star $A$ with the least brightness. Let the size of star $A$ be expressed by the number $n$. Since size takes an infinite number of values, there will be a star $B$ of size... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,101 |
[ Least or greatest distance (length).]
There are coins on the table without any overlaps. Prove that one of them can be pushed out without touching the others.
# | Select the coin with the farthest center relative to a certain direction.
## Solution
Introduce a rectangular coordinate system on the plane. Consider the coin whose center has the greatest ordinate. Denote its center and radius by $C$ and $r$. We will prove that this coin can be pushed upwards without touching the r... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,102 |
In each of $n$ countries, either the right party or the left party is in power. Each year, in one of the countries $A$, the government may change. This can happen if the majority of the countries bordering country $A$ are ruled by the party that is not in power in country $A$. Prove that the changes of government canno... | How does the number of pairs of neighboring countries with the same party in power change?
## Solution
Suppose the power has changed in some country $A$, where the left (for definiteness) are in power. Let there be $k$ neighboring countries where the left are in power and $m$ countries where the right are in power, w... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,103 |
[ Dirichlet's Principle (continued).]
Prove that in any company there will be two people who have the same number of friends (from this company).
# | In the company, there cannot be a person who is friends with everyone and a person who is not friends with anyone at the same time.
## Solution
Let there be $n$ people in the company. Then each person can have from 0 to $n-1$ friends. Thus, the number of friends can take $n$ different values: $0,1,2, \ldots, n-1$. Th... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,104 |
7,8,9 |
What is the maximum number of kings that can be placed on a chessboard so that no two of them attack each other
# | If in a square of four cells there are two kings, then they attack each other.
## Solution
We will divide the board into 16 squares $2 \times 2$.
Estimate. In each of these 16 squares, there can be no more than one king.
Example. Place a king in the lower left corner of each of the 16 squares.
## Answer
16 kings. | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,105 |
Can seven non-negative integers be arranged in a circle so that the sum of some three consecutive numbers is equal to 1, the sum of some three consecutive numbers is 2, ..., and the sum of some three consecutive numbers is 7? | If it were possible, then in the sum $1+2+\ldots+7$ each of the seven numbers, arranged in a circle, would appear three times.
## Solution
Suppose this is possible. There are a total of seven different sums of three consecutive numbers. Then among these seven sums, each of the numbers $1,2, \ldots, 7$ must appear onc... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,106 |
A piece of cheese has the shape of a $3 \times 3 \times 3$ cube, from which the central cube has been removed. A mouse starts gnawing at this piece of cheese. First, it eats some $1 \times 1 \times 1$ cube. After the mouse eats another $1 \times 1 \times 1$ cube, it proceeds to eat one of the adjacent (by face) cubes t... | Paint the $1 * 1$ cubes in a checkerboard pattern. After eating a black cube, the mouse can only eat a white cube, and vice versa.
## Solution
We will paint the $1 * 1$ cubes in a checkerboard pattern. Notice that there will be two more cubes of one color than the other (12 of one color, 14 of the other). But after a... | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,107 |
When setting up a code lock, each of the 26 Latin letters on its keyboard is assigned a natural number known only to the owner of the lock. Different letters may be assigned the same number. After entering any combination of pairwise distinct letters, the numerical values corresponding to the entered letters are summed... | Prove that the desired combination can be found as several consecutive letters of the alphabet.
## Solution
Denote by $S(n)$ the remainder of the division by 26 of the sum of the numbers corresponding to the first $n$ letters of the alphabet $(n=1,2, \ldots, 26)$. If among the numbers $S(1), S(2), \ldots, S(26)$ ther... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,108 |
$[$ Case Analysis] [ Cryptography ]
Letters of the Russian alphabet are numbered according to the table:
А Б В Г Д Е Ж З И К. Ф Х Ц Ч Ш Щ Ъ Ы Э Ю Я
$\begin{array}{llllllllllllllllllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \ldots & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 \\ \text { For encryption }\... | There are three possible decodings for each letter. Decode each letter in three ways, and then search for a meaningful decoding.
## Solution
We will decode each letter of the encrypted message in three ways, assuming sequentially that the corresponding letter of the encrypting sequence is the letter A, B, or V:
encr... | NAHSKORRESPONDENT | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,109 |
Two numbers $a$ and $b(a>b)$ are written on the board. They are erased and replaced by the numbers ${ }^{a+b} / 2$ and ${ }^{a-b} / 2$. The same procedure is applied to the newly written numbers. Is it true that after several erasures, the difference between the numbers written on the board will become less than $\frac... | Perform the operation twice and see what happens to the numbers.
## Solution
By performing the specified operation twice, we get the numbers $\frac{a}{2}, \frac{b}{2}$. This means that after $2 k$ operations, the numbers will be $2^{-k} a, 2^{-k} b$. But for sufficiently large $k$
$2^{-k} a-2^{-k} b<1 / 2002$
## An... | 2^{-k}-2^{-k}b<1/2002 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,110 |
[ Continuity Considerations ]
A row of recruits stood facing the sergeant. On the command "left," some turned left, some turned right, and the rest turned around.
Can the sergeant always find a place in the row so that there are an equal number of recruits facing him on both sides? | Move the sergeant along the formation. The number of faces turned towards the sergeant changes by no more than 1 in the process.
## Solution
For each position of the sergeant in the formation, we calculate the difference $d$ between the number of people standing to the left of the sergeant facing him and the number o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,111 |
8,9
The four-digit number written on the board can be replaced by another by adding one to each of its two adjacent digits, provided that neither of these digits is 9; or by subtracting one from each of the two adjacent digits, provided that neither of them is 0. Is it possible to obtain the number $2002$ from the num... | The sum of the digits in even positions and the sum of the digits in odd positions change in the same way.
## Solution
Let the number on the board be abcd. Then the operations under consideration do not change the number $\mathrm{M}=(\mathrm{d}+\mathrm{b})-(\mathrm{a}+\mathrm{c})$, since they increase (decrease) by o... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 53,112 |
Prove that no line can intersect all three sides of a triangle (at points other than the vertices).
# | A straight line divides the plane into two half-planes. By the Dirichlet principle, there will be two vertices lying in one half-plane. The side connecting these vertices does not intersect the line. | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,113 |
[ [Dirichlet's principle (pigeonhole)]
Prove that in every nonagon there is a pair of diagonals, the angle between which is less than $7^{\circ}$.
# | Through an arbitrary point, draw 27 lines, each parallel to the diagonals of a given nonagon.
## Solution
A nonagon has 9$\cdot$6:2 27 diagonals. Through an arbitrary point, draw 27 lines, each parallel to these diagonals. They will divide the full angle into 54 angles. Therefore, one of them is no more than ${ }^{36... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,114 |
[ Inscribed and Circumscribed Circles $]$ [ Rational and Irrational Numbers ]
The coordinates of the vertices of a triangle are rational. Prove that the coordinates of the center of its circumscribed circle are also rational
# | Let $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)$ and $\left(a_{3}, b_{3}\right)$ be the coordinates of the vertices of a triangle. The coordinates of the center of its circumscribed circle are given by the system of equations
$$
\begin{aligned}
& \left(x-a_{1}\right)^{2}+\left(y-b_{1}\right)^{2}=\left(x-a_{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,116 |
[ Smallest or Largest Angle ]
Prove that if the lengths of all sides of a triangle are less than 1, then its area is less than $\sqrt{3} / 4$.
# | Let $\alpha$ be the smallest angle of the triangle. Then $\alpha \leq 60^{\circ}$. Therefore $S=(b c \sin \alpha) / 2 \leq\left(\sin 60^{\circ}\right) / 2=\sqrt{3} / 4$. | \frac{\sqrt{3}}{4} | Geometry | proof | Yes | Yes | olympiads | false | 53,117 |
[ Smallest or largest distance (length). ]
On a plane, there are $n \geq 3$ points, and not all of them lie on the same line. Prove that there exists a circle passing through three of these points and not containing any of the remaining points inside it.
# | Let $A$ and $B$ be the points among the given points such that the distance between them is minimal. Then, inside the circle with diameter $A B$, there are no given points. Let $C$ be the point among the remaining points from which the segment $A B$ is seen at the largest angle. Then, inside the circle passing through ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,118 |
Several points are located on a plane, all pairwise distances between which are distinct. Each of these points is connected to the nearest one. Can a closed broken line be obtained in this case?
# | Suppose we have a closed broken line. Let $A B$ be the largest link of this broken line, and $A C$ and $B D$ be the links adjacent to it. Then $A C < A B$, i.e., $B$ is not the nearest point to $A$, and $B D < A B$, i.e., $A$ is not the nearest point to $B$. Therefore, points $A$ and $B$ cannot be connected. This leads... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,119 |
7,8 Nodes of an infinite grid paper are colored in two colors. Prove that there exist two horizontal and two vertical lines, at the intersections of which lie points of the same color.
# | Let's take three vertical lines and nine horizontal ones. We will only consider the intersection points of these lines. Since there are only $2^{3}=8$ ways to color three points in two colors, there will be two horizontal lines on which there are identically colored triples of points. Among the three points colored in ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,122 |
[ $\underline{\text { Invariants }}$ ]
Given a chessboard. It is allowed to repaint all the cells located inside a $2 \times 2$ square to another color at once. Can there be exactly one black cell left on the board in this case?
# | When repainting a $2 \times 2$ square containing $k$ black and 4 - $k$ white cells, it will result in 4 - $k$ black and $k$ white cells. Therefore, the number of black cells will change by ( $4-k$ ) - $k=4$ - $2k$, i.e., by an even number. Since
the parity of the number of black cells is preserved, from the initial 32 ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,123 |
[Chessboard coloring]
In each cell of a $5 \times 5$ board, there is a beetle. At some moment, all the beetles crawl to adjacent (horizontally or vertically) cells. Is it necessarily true that there will be an empty cell? | Since the total number of cells on a $5 \times 5$ chessboard is odd, the number of black and white cells cannot be equal. Let's assume, for definiteness, that there are more black cells. Then the number of beetles sitting on white cells is less than the number of black cells. Therefore, at least one of the black cells ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,124 |
8,9
The plane is colored in three colors. Prove that there are two points of the same color, the distance between which is 1.
# | Suppose any two points lying at a distance of 1 are colored in different colors. Consider an equilateral triangle $ABC$ with side length 1; all its vertices are of different colors. Let point $A_1$ be symmetric to $A$ with respect to the line $BC$. Since $A_1B = A_1C = 1$, the color of point $A_1$ is different from the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,126 |
7,8,9 |
| Calculate the product | $\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \ldots \cdot \frac{n^{3}-1}{n^{3}+1}$ | | | Notice that $n^{2}-n+1=(n-1)^{2}+(n-1)+1$. Therefore,
$$
\begin{gathered}
\frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \ldots \cdot \frac{n^{3}-1}{n^{3}+1}=\frac{(2-1)\left(2^{2}+2+1\right)}{(2+1)\left(2^{2}-2+1\right)} \cdot \frac{(3-1)\left(3^{2}+3+1\right)}{(3+1)\left(3^{2}-3+1\right)} \cdot \ldots \... | \frac{2}{3}(1+\frac{1}{n(n+1)}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,127 |
60433 [underachievers] $]$ topics: [
Counting in two ways
Combinatorics (miscellaneous).
[Summation of numerical sequences and difference series]
In a class, there are $a_{1}$ students who received at least one two during the year, $a_{2}$ students who received at least two twos, ..., $a_{k}$ students who received at ... | The number of students who received exactly one two is $a_{1}-a_{2}$, exactly two twos is $-a_{2}-a_{3}$, and so on. Therefore, the total number of twos is
$\left(a_{1}-a_{2}\right)+2\left(a_{2}-a_{3}\right)+3\left(a_{3}-a_{4}\right)+\ldots+(k-1)\left(a_{k-1}-a_{k}\right)+k a_{k}=a_{1}+\left(2 a_{2}-a_{2}\right)+\left... | a_{1}+a_{2}+\ldots+a_{k} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,129 |
[ Quadratic equations. Formula for roots ] [ Quadratic equations. Vieta's theorem ]
For what values of the parameter $a$ is one of the roots of the equation $x^{2}-15 / 4 x+a^{3}=0$ the square of the other? | According to Vieta's formula, the free term is equal to the cube of the specified root. Therefore, the roots of the equation are $a$ and $a^{2}$.
Thus, $a+a^{2}=15 / 4$, from which
$a=\frac{3}{2}$ or $a=-\frac{5}{2}$.
## Answer
When $a=1.5$ and $a=-2.5$. | =1.5=-2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,130 |
[ Sorting in ascending (descending) order $]$
## Authors: Folklore, Folklore
Long, long ago, the country of Tarnia was ruled by King Yatianr. To make the Tarnians talk less, he invented a simple language for them. Its alphabet consisted of only six letters: A, I, N, R, T, Y, but the order of these letters differed fr... | In the first word, the letters are arranged in alphabetical order: T, A, R, N, I, Y. For convenience, let's number the letters in alphabetical order: $\mathrm{T}=1, \mathrm{~A}=2, \mathrm{P}=3$,
$\mathrm{H}=4, \mathrm{I}=5$, Y $=6$. We will replace each word with the corresponding six-digit number. If the words are ar... | Ytiran | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,131 |
In a class of 33 students, the total age of all of them is 430 years.
Prove that if you select the 20 oldest among them, their combined age will be no less than 260 years. (The age of any student is an integer.)
# | The total age of the "older" group is not less than $20 / 33 \cdot 430=260.6 \ldots$ years. | 260.6 | Inequalities | proof | Yes | Yes | olympiads | false | 53,132 |
Shapovalov A.v.
There are 100 numbers. When each of them was increased by 1, the sum of their squares did not change. Each number was increased by 1 again.
Will the sum of the squares change this time, and if so, by how much? | $(a+2)^{2}-(a+1)^{2}=(a+1)^{2}-a^{2}+2$. Therefore, the second change is 2$\cdot$100 greater than the first.
## Answer
It will increase by 200.
Problem | 200 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,134 |
Kanel-Belov A.Y.
A mushroom is called bad if it contains no less than 10 worms. In the basket, there are 90 bad and 10 good mushrooms. Can all the mushrooms become good after some worms crawl from the bad mushrooms to the good ones? | Let there be exactly 10 worms in each bad mushroom, and no worms in good mushrooms. Further, let one worm from each bad mushroom crawl into the good ones, with 9 worms crawling into each good mushroom. As a result, there will be 9 worms in each mushroom, and all mushrooms will be good.
## Answer
They can. | Theycan | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,135 |
}
Petya cut a rectangular sheet of paper along a straight line into two parts. Then he cut one of the parts again along a straight line into two. Then he cut one of the resulting parts again into two, and so on, a total of one hundred times. Then Petya counted the total number of vertices of all the resulting polygons... | After a hundred cuts, 101 polygons were obtained. And each has no less than 3 vertices.
## Answer
It couldn't. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,136 |
Bakayev E.V.
In a circle, there are boys and girls (both are present), a total of 20 children. It is known that for each boy, the neighbor in the clockwise direction is a child in a blue T-shirt, and for each girl, the neighbor in the counterclockwise direction is a child in a red T-shirt. Can the number of boys in th... | The MHD (clockwise) arrangement cannot be due to the color of child X's T-shirt. Therefore, one boy should stand clockwise from the boy, one boy from him, and so on. This means that there are no fewer than half of all the children in the circle who are boys. By similar considerations, there are no fewer than half of th... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,137 |
[ Examples and counterexamples. Constructions ]
Can all natural numbers from 1 to 100 be arranged in a row so that any two adjacent numbers differ either by 2 or by a factor of two? | For example: 99, 97, ..., 3, 1, 2, 4, ..., 98, 100 (first odd numbers in descending order, then even numbers in ascending order).
## Answer
Yes, it can be done. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,138 |
[ Examples and counterexamples. Constructions ]
Seven mushroom pickers collected a total of 100 mushrooms. Is it necessarily true that there will be two mushroom pickers who together collected at least 36 mushrooms, if the number of mushrooms collected by each is pairwise distinct? | Let's say, for example, the mushroom pickers collected 10, 12, 13, 14, 16, 17, and 18 mushrooms respectively. Then any pair together collected no more than 35 mushrooms, and
$10+12+13+14+16+17+18=100$
## Answer
Not necessarily. | Notnecessarily | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,139 |
[ Examples and counterexamples. Constructions ]
Can 40 candies be distributed among six children so that each has a different number of candies and any two together have less than half of all the candies? | For example, $10+9+8+7+5+1=40$, while the largest sum of any two is $10+9=19<20$.
There are other examples as well.
## Answer
it can be done. | itcanbedone | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,140 |
[ Examples and counterexamples. Constructions ]
It is known about a quadrilateral that there are two lines, each of which divides it into two isosceles right triangles. Is it necessarily a square?
# | Consider a non-convex quadrilateral $ABCD$ where $\angle A = \angle B = \angle D = 45^{\circ}$ (see the figure). Then each of the lines $BC$ and $DC$ divides it into two isosceles right triangles.
. The ten triangles mentioned in the condition can be divided into two groups of five: 1) triangles formed by two adjacent sides and a diagonal of the pentagon; 2) triangles formed by a side and two diagonals of the pentagon.
![](https://cdn.mathpix.com/croppe... | Itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,142 |
Belukhov H.
In a certain state, addition and subtraction are denoted by the signs "!" and "?", but it is unknown to you which sign corresponds to which operation. Each operation is applied to two numbers, but it is unknown to you whether subtraction involves subtracting the left number from the right or the right from... | First, note that the expression ( $a$ ? $a$ ) ! ( $a$ ? $a$ ) is always equal to zero. In the future, we can use 0, implying that this exact expression should be written instead.
The expression ( $x$ ? 0) ? (0 ? y) is always equal to $x+y$. Similarly, we can now use the + operation with two arguments.
Finally, the ex... | 4:2+3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,143 |
Authors: Yatsenko I.V., Shapovelov A.V.
All taverns in the kingdom belong to three firms. In an effort to combat monopolies, King Pea issued the following decree: every day, if a certain firm has more than half of all taverns and the number of its taverns is divisible by 5, then this firm retains only one-fifth of its... | Let's consider an example. Suppose initially the firms have 60, 35, 20 taverns respectively. Then on the first day, the first firm will lose 48 taverns, leaving 12, 35, 20 taverns. On the second day, the second firm will lose 28 taverns, leaving 12, 7, 20 taverns. Finally, on the third day, the third firm will lose 16 ... | Yes,itcould | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,145 |
[ Processes and operations]
Given a natural number \$N\$. Vera performs the following operations with it: first, she adds 3 until the resulting number is divisible by 5 (if \$N\$ is initially divisible by 5, nothing needs to be added). The resulting number is then divided by 5. She then performs these same operations ... | Indeed, the number $1$ cannot be obtained from numbers that are multiples of $3$, because if a number $N$ is a multiple of $3$, then $N + 3$ is also a multiple of $3$, and if $N = 5k$ is a multiple of $3$, then $k$ is a multiple of $3$, since $3$ and all numbers resulting from these operations will be multiples of $3$ ... | 3k,k\in\mathbb{N} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,146 |
Shapovalov A.V. A sequence of several natural numbers is written, with a sum of 20. No number and no sum of several consecutive numbers equals 3. Could there be more than 10 numbers written? | Example with 11 numbers: $1,1,4,1,1,4,1,1,4,1,1$.
## Answer
It could.
Send a comment | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,147 |
Bakayev E.B.
In each cell of a strip of length 100, there is a chip. For 1 ruble, you can swap any two adjacent chips, and you can also swap any two chips that have exactly three chips between them for free. What is the minimum number of rubles needed to rearrange the chips in reverse order? | Evaluation. Each chip must change the parity of its number. A free operation does not change parity, while a paid operation changes the parity of two chips. Therefore, at least 50 rubles will be required.
Algorithm. We will number the chips in order from 0 to 99. We will color the cells in four colors: $a b c d a b c ... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,148 |
Bakayev E.B.
What is the maximum number of different integers that can be written in a row so that the sum of any 11 consecutive numbers is 100 or 101? | Evaluation. Suppose we have managed to list such distinct numbers $x_{1}, ..., x_{23}$, that the sum of any 11 consecutive numbers equals $A$ or $B$ (otherwise $x_k = x_{k+11}$). Thus, $S_k = S_{k+2}$. Since $x_{1} + S_{2} + S_{13} = S_{1} + S_{12} + x_{23}$, then $x_{1}= x_{23}$. Contradiction.
Example. Let's choose ... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,149 |
\section*{[ Examples and Counterexamples. Constructions ]
$[\quad$ Triangles (miscellaneous). $\quad]$
Auto: $\underline{\text { Koppu }}$ Tririar Gorodovov
In triangle $\$ \mathrm{ABC} \$$, the altitudes $\$ \mathrm{~A} X$ and $\$ \mathrm{~B} Z \$$, as well as the angle bisectors $\$ \mathrm{AY} \$$ and $\$ \mathrm... | For example, it is not difficult to check that in a triangle with angles $\angle A = 40^\circ$, $\angle B = 80^\circ$, $\angle C = 60^\circ$, both specified angles are $10^\circ$, and in a triangle with angles $\angle A = 30^\circ$, $\angle B = 90^\circ$, $\angle C = 60^\circ$, both specified angles are $15^\circ$.
##... | Notnecessarily | Geometry | proof | Yes | Yes | olympiads | false | 53,151 |
Mitranov I.V.
In front of Sharik lies an infinite number of cutlets, each with a fly sitting on it. On each move, Sharik sequentially performs two operations:
1) eats one cutlet along with all the flies sitting on it;
2) moves one fly from one cutlet to another (there can be any number of flies on a cutlet).
Sharik ... | Assume the opposite. Let's number the cutlets in the order in which Sharik will eat them. In the first million steps, Sharik will move some flies to some cutlets. Let \( n \) be the largest among the numbers of these cutlets. Consider the first \( n \) cutlets. In the first million steps, the total number of flies on t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,153 |
Authors: Gooovanov A.S., Sokolov A.
Sasha writes down the numbers 1, 2, 3, 4, 5 in some order, places arithmetic operation signs "\$\$\$", "\$-\$", "\$|times\$", and parentheses, and looks at the result of the obtained expression. For example, he can get the number 8 using the expression $\$(4-3) \backslash t i m e s(... | For example, $\$ 3 \backslash$ times $(2 \backslash$ times 4 \times $5+1)=123 . \$$
## Answer
Yes, it can. | 123 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,154 |
9 | |
On a circle, points $A_{1}, A_{2}, \ldots, A_{16}$ are given. We will construct all possible convex polygons whose vertices are among the points $A_{1}, A_{2}, \ldots, A_{16}$. We will divide these polygons into two groups. The first group will include all polygons where $A_{1}$ is a vertex. The second group wi... | To each polygon that does not have a vertex $A_{1}$, we associate a polygon with the vertex $A_{1}$ by simply adding $A_{1}$ to its vertices. The reverse operation (removing the vertex $A_{1}$) is impossible for triangles.
## Answer
In the first.
Five people play several domino games (two on two) such that each play... | In\the\first | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,155 |
Let's draw some diagonals in a convex polygon so that no two of them intersect (several diagonals can originate from one vertex). Prove that there will be at least two vertices of the polygon from which no diagonal is drawn.
# | Let $n$ be the number of vertices of a polygon. We will prove by induction on $n$ that there will be at least two non-adjacent vertices from which no diagonals are drawn. For $n=4$, this is obvious. We will prove the induction step. Suppose a diagonal is drawn through vertices $M$ and $N$ in the polygon. This diagonal ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,157 |
There is a labyrinth consisting of $n$ circles, each touching the line $A B$ at point $M$. All circles are located on one side of the line, and their lengths form a geometric progression with a common ratio of 2. Two people start walking through this labyrinth at different times. Their speeds are the same, but their di... | Let's prove that they will definitely meet on the largest circle. For this, it is enough to prove that at some point in time, one of them will enter this circle while the other is walking on it. If at the moment when the first person, having walked the second largest circle, enters the largest circle, the second is alr... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,158 |
9,10
In a chess tournament, 12 people participated. After the tournament, each participant made 12 lists. The first list includes only themselves, the second list includes them and those they defeated, the third list includes everyone from the second list and those they defeated, and so on. The 12th list includes ever... | Answer: 54. If the $(k+1)$-th list is the same as the $k$-th, then the lists numbered $k+2, \ldots, 11, 12$ will also be exactly the same. But according to the condition, the 11-th list and the 12-th are different. Therefore, each participant's $k$-th list contains exactly $k$ people. In particular, the 2-nd list conta... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,159 |
10,11
The mass of each of the 19 weights does not exceed 70 grams and is an integer number of grams. Prove that it is impossible to form more than 1230 different mass sets from these weights.
# | Let's divide all sets into two parts: those with no more than seventeen weights and all the others. The mass of each set from the first part does not exceed $17 \cdot 70=1190$, so among them, there are no more than 1190 different by mass. On the other hand, the number of sets in the second part is $19+1=20$. Therefore,... | 1210 | Combinatorics | proof | Yes | Yes | olympiads | false | 53,163 |
8
At the congress, 1000 delegates from different countries arrived. Each delegate knows several languages. It is known that any three of them can communicate with each other without the help of others. (At the same time, it is possible that one of them will have to translate the conversation of the other two.) Prove t... | Let's choose three delegates of the congress; among them, there will be two who know one language - we will place them in the same hotel room. From the remaining 998 delegates, we will again select three, among whom there will again be two who can be placed in the same room - and so on, until we are left with only 4 de... | proof | Other | proof | Yes | Yes | olympiads | false | 53,164 |
[ Dirichlet's Principle (finite number of points, lines, etc.).]
In square $ABCD$, there are 5 points. Prove that the distance between some two of them does not exceed $\frac{1}{2}$
AC. | By drawing lines through the center of the square parallel to its sides, we will divide the square into 4 equal squares. Some 2 of the 5 points lie in one of these squares, and the distance between them does not exceed the length of the diagonal of this square. | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,166 |
[ Counting in Two Ways ]
Each participant of the two-day olympiad solved as many problems on the first day as all the others combined did on the second day.
Prove that all participants solved the same number of problems.
# | Let on the second day all participants in total solved $N$ problems. Consider any participant; let him solve $m$ problems on the second day. Then on the second day, all the others in total solved $N-m$ problems. Therefore, on the first day, this participant solved $N-m$ problems, and thus, in total, he solved
$m+(N-m)... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,167 |
[Coordinate method in space]
Find the angle between the line passing through points $A(-3 ; 0 ; 1)$ and $B(2 ; 1 ;-1)$, and the line passing through points $C(-2 ; 2 ; 0)$ and $D(1 ; 3 ; 2)$. | Let's find the coordinates of the vectors $\overrightarrow{A B}$ and $\overrightarrow{C D}$:
$$
\begin{aligned}
& \overrightarrow{A B}=(2-(-3) ; 1-0 ;-1-1)=(5 ; 1 ;-2) \\
& \overrightarrow{C D}=(1-(-2) ; 3-2 ; 2-0)=(3 ; 1 ; 2)
\end{aligned}
$$
Let $\phi$ be the angle between the vectors $\overrightarrow{A B}$ and $\o... | \arccos\frac{2\sqrt{105}}{35} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,168 |
[ Coordinate method in space ] [ Rectangular parallelepipeds ]
The non-intersecting diagonals of two adjacent faces of a rectangular parallelepiped are inclined to the base plane at angles $\alpha$ and $\beta$. Find the angle between these diagonals. | Consider a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$. Let $A D 1$ and $D C 1$ be the specified diagonals. Then $\angle A D 1 A 1=\alpha, \angle D C D 1=\beta$. Choose a coordinate system with the origin at point $A$. The $x$-axis is directed along the ray $A B$, the $y$-axis along the ray $A D$, and the $z$-... | \arccos(\sin\alpha\sin\beta) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,169 |
$\left[\begin{array}{l}{[\text { Counting in two ways }]} \\ {[\quad \underline{\text { Degree of a vertex }}]}\end{array}\right.$
At a cat show, each visitor pet exactly three cats. It turned out that each cat was petted by exactly three visitors.
Prove that the number of visitors was exactly the same as the number ... | If we mentally draw threads between each cat and the visitor who petted it, then from each cat, 3 threads will be stretched, and from each visitor, also 3. This means that the number of threads is simultaneously three times the number of visitors and three times the number of cats. Therefore, the number of cats is equa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,170 |
.
#
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | The invariant is the sum of the numbers. Indeed,
$(a+b-c)+(b+c-a)+(c+a-b)=a+b+c$.
No matter how many times the sets of numbers change, it is impossible to obtain the specified set from the original one, since their sums differ by 1. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,171 |
Let $A$ be a pyramid.
We will say that two pyramids touch by faces if these pyramids do not have common internal points and some face of one pyramid intersects with some face of the other pyramid along a polygon. Is it possible to arrange eight pyramids in space so that every two touch by faces? | Consider eight triangles located in the horizontal plane $\alpha$ as shown in the figure. They serve as the bases of eight sought pyramids: four of them, whose bases are triangles bounded by black lines, have a common vertex located above the plane $\alpha$; the other four, whose bases are triangles bounded by red line... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,172 |
[ Rectangles and Squares. Properties and Characteristics ] [ Area of a Parallelogram $\quad$]
## Author: Professor V.B.
A square is divided into five rectangles such that the four corners of the square are the corners of four rectangles, the areas of which are equal to each other, and the fifth rectangle does not sha... | 
Consider the smallest side of the "corner" rectangles. Then its second side is the largest. But the largest side is also the side of the adjacent rectangle (complementing the smallest side t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,173 |
V.V. Proizvolov
A square was cut by 18 lines, nine of which are parallel to one side of the square, and nine to the other, into 100 rectangles. It turned out that exactly nine of them are squares. Prove that among these squares, there are two that are equal to each other.
# | If among these nine squares there are no two squares of the same size, then they all lie in different columns and different rows into which the original square is divided by straight lines. But then the rectangle lying at the intersection of the tenth row and the tenth column (not containing these nine squares) is also... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,175 |
3 $[$ Equations in integers $]$
On the table, there is a pile of 637 shells. One shell is removed from it and the pile is divided into two (not necessarily equal). Then, from any pile containing more than one shell, one shell is removed again and the pile is divided into two. And so on. Is it possible to leave only pi... | Each time after removing a pebble and splitting a pile, the number of pebbles decreases by 1, and the number of piles increases by 1.
## Solution
After each procedure (removing a pebble and splitting a pile), the number of pebbles decreases by 1, and the number of piles increases by 1. Since initially there were 637 ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,176 |
Electrician I.V. was called to repair a string of four lamps connected in series, one of which had burned out. It takes 10 seconds to unscrew any lamp from the string, and 10 seconds to screw it back in. The time spent on other actions is negligible. What is the minimum time the electrician can definitely find the burn... | If after replacing one bulb the garland does not light up, then we replaced a working bulb.
## Solution
Suppose we did not replace some two bulbs. Then, if one of them is burnt out, we will not be able to determine which one. Therefore, to definitively identify the burnt-out bulb, we need to unscrew at least three of... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,177 |
Fedorov R.M.
In the Banana Republic, parliamentary elections were held in which all residents participated. All those who voted for the "Mandarin" party love mandarins. Among those who voted for other parties, 90% do not like mandarins. What percentage of the votes did the "Mandarin" party receive in the elections, if... | Find in two ways how many people love tangerines.
## Solution
Let the entire population of the republic be $-N$ people, of which $M$ people voted for "Tangerine." Then, on one hand, tangerines are loved by $0.46 N$ people, and on the other hand, this number is equal to the number of people who voted for "Tangerine" p... | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,178 |
[ Coordinate method on the plane ]
On the number line, two points are marked. Where on this line is the point corresponding to their arithmetic mean located?
# | The desired point is exactly in the middle of the two marked ones, since the arithmetic mean of any two numbers is as much less than the larger number as it is greater than the smaller one.
## Answer
In the middle between the points.
## Problem | In\the\middle\between\the\points | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,180 |
Several points were placed on a line. Then, between each pair of adjacent points, another point was placed, and this was done several times. Prove that after each such operation, the total number of points will be odd. | Note: Each time, the number of points added is one less than the number of those that were.
## Solution
Each time, the number of points added is one less than the number of those that were. Therefore, the total number of points will be odd.
## Answer
The number of points added is one less than the number of those t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,181 |
Around each vertex of the triangle, write some numbers, and next to each side of the triangle, write the sum of the numbers at the ends of that side. Now add each number at a vertex to the number next to the opposite side. Do you think why the resulting sums are the same? | The sum of the two numbers at the vertex and the opposite side is equal to the sum of the three numbers at the three vertices. Since this sum is constant, the sum of the number at the vertex and the number at the opposite side will be constant for any vertex of the triangle.
## Answer
Each of these three sums is equa... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,182 |
[ [Reverse Process ]
Lёnya thought of a number. He added 5 to it, then divided the sum by 3, multiplied the result by 4, subtracted 6, divided by 7, and got 2. What number did Lёnya think of?
Try to form an equation to determine the unknown number.
# | Let's denote the number thought of by Lёna as $x$. Then we can form the equation
$$
\{[((x+5): 3)-4]-6\}: 7=2
$$
By sequentially moving all numbers from the left side to the right, we get a new equation
$$
x=\{[((27)+6): 4] 3\}-5
$$
from which it is easy to determine that $x=10$. From this, it is also clear that to... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,183 |
[ Dirichlet's Principle (miscellaneous).]
Five young workers received a total salary of 1500 rubles. Each of them wants to buy a tape recorder priced at 320 rubles. Prove that someone among them will have to wait until the next salary to make the purchase.
# | If each of the workers could buy a tape recorder, then in total they would have no less than $5 \cdot 320=1600$ rubles. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,184 |
In the lake, lotuses grow. Over the course of a day, each lotus splits in half, and instead of one lotus, two appear. After another day, each of the resulting lotuses splits in half, and so on. After 30 days, the lake is completely covered with lotuses. How long did it take for the lake to be half full?
# | Note that the number of lotus flowers doubles in one day.
## Solution
If you read the condition of the problem carefully, you will understand that the lake was half full after 29 days. The day before the lake is completely filled, it will be exactly half full.
## Answer
After 29 days | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,185 |
[ Dirichlet's Principle (continued).]
Is it necessary that among twenty-five "copper" coins (i.e., coins of 1, 2, 3.5 kopecks) there will be seven coins of the same denomination? | Think about how many coins there would be if there were no more than six of each of the four types of coins.
## Solution
Sure. If there were no more than 6 coins of each of the four types, then there would be no more than 64 $=24$ coins in total, but there are 25.
## Answer
Yes. If there were no more than 6 coins o... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,186 |
[ Dirichlet's Principle (continued).]
a) Prove that in any football team there are two players who were born on the same day of the week. b) Prove that among the residents of Moscow, there are ten thousand who celebrate their birthday on the same day.
# | a) Let football players be rabbits, and days of the week be cells. We get 7 cells into which we need to place at least 11 rabbits, which means that by the Pigeonhole Principle, at least one cell will contain at least two rabbits.
b) Suppose that fewer than 10000 Muscovites were born on each of the 366 days of the year... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,187 |
Bootin D.A.
Find in the sequence $2,6,12,20,30, \ldots$ the number standing a) at the 6th; b) at the 1994th position. Explain your answer. | $2=1 \cdot 2 ; 6=2 \cdot 3$
## Solution
We can notice that $2=1 \cdot 2, 6=2 \cdot 3, 12=3 \cdot 4$, and assume that the $n$-th term of the sequence is $n \cdot (n+1)$. Checking on the 4th ($20=4 \cdot 5$) and 5th ($30=5 \cdot 6$) terms of the sequence shows that we are correct. Therefore, the number at the 6th posit... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,188 |
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