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Find all natural numbers, when divided by 7, the quotient is the same as the remainder.
# | When dividing a number by 7, there can only be 7 different remainders.
## Solution
The remainder when dividing by 7 cannot exceed 6, so the numbers of interest can be represented as $7a + a = 8a$, where $a = 1, 2, \ldots, 6$. Thus, these numbers are: $8, 16, 24, 32, 40, 48$.
## Answer
$8, 16, 24, 32, 40, 48$. | 8,16,24,32,40,48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,859 |
Can a $5 \times 5$ table be filled with numbers so that the sum of the numbers in each row is positive, and the sum of the numbers in each column is negative?
# | Try to calculate the sum of all numbers in the table.
## Solution
If it were possible, the sum of all numbers in the table, calculated "by rows," would be positive, while "by columns" it would be negative, which is impossible.
## [ Motion Problems ] Problem 88124 Topics: [ Case Analysis $]
The distance between Atho... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,860 |
Rita, Luba, and Varya were solving problems. To speed things up, they bought candies and agreed that the girl who solved a problem first would get four candies, the second to solve it would get two, and the last to solve it would get one. The girls say that each of them solved all the problems and received 20 candies, ... | Notice that for solving each problem, the three girls together received 7 candies.
## Solution
If for solving each problem the three girls together received 7 candies (the first girl received 4, the second girl received 2, and the third girl received 1 candy), then the total number of candies they received should be ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,862 |
How many two-digit numbers are there where the digit in the tens place is greater than the digit in the units place?
# | The problem can be solved by simple enumeration.
1) If the units digit is 0, then the tens digit can take values from 1 to 9.
2) If the units digit is 1, then the tens digit can take values from 2 to 9.
3) If the units digit is 2, then the tens digit can take values from 3 to 9.
4) If the units digit is 8, then the te... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,863 |
The scent from a blooming lily-of-the-valley bush spreads in a radius of 20 m around it. How many blooming lily-of-the-valley bushes need to be planted along a straight 400-meter alley so that every point along it smells of lily-of-the-valley
# | Notice that to meet the conditions of the problem, the distance between adjacent lilies of the valley should not exceed $40 \mathrm{M}$.
## Solution
Let's mentally move the leftmost 20 m of the alley to the right end. Then we will have segments of the alley that are 40 m long, and to the right of each segment, a lily... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,865 |
Nazarov $\Phi$.
An automaton, when a five-kopeck coin is inserted, dispenses five two-kopeck coins, and when a two-kopeck coin is inserted, it dispenses five five-kopeck coins. Can Petya, approaching the automaton with one two-kopeck coin, end up with an equal number of two-kopeck and five-kopeck coins after several i... | Each operation does not change the parity of the number of coins Pete has.
## Solution
The operation does not change the parity of the number of coins Pete has, so he will always have an odd number of coins.
## Answer
It cannot. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,866 |
Kalinin D.A.
On a line, several points were marked. After that, between each pair of adjacent points, one more point was marked. This "densification" was repeated two more times (a total of three times). As a result, 113 points were marked on the line. How many points were initially marked? | Find how many points there were before the last compaction, i.e., solve the problem "from the end".
## Solution
If there were $n$ points marked (before compaction), then after compaction, there will be $2 n-1$ points (of which $n$ are old and $n-1$ are new). If after compaction, there are $k$ points, then $2 n-1=k$ o... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,868 |
a) Come up with three correct irreducible fractions, the sum of which is an integer, and if each of these fractions is "inverted" (i.e., replaced with its reciprocal), the sum of the resulting fractions will also be an integer.
b) The same, but the numerators of the fractions are unequal natural numbers. | a) Select three fractions with numerators equal to 1.
b) First, find three fractions with different denominators that sum up to 1.
## Answer
a) For example, $1 / 2, 1 / 2, 1 / 6$. b) For example, $2 / 11, 3 / 11, 6 / 11$. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,870 |
The digits of a three-digit number A were written in reverse order to obtain number B. Can the number equal to the sum of A and B consist only of odd digits?
# | Let, for example, $A=219$. Then $B=912, A+B=1131$.
## Answer
Yes, it can. | 1131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,871 |
For a two-digit number, the first digit is twice the second. If you add the square of its first digit to this number, you get the square of some integer. Find the original two-digit number.
# | The first digit is twice the second only in the following two-digit numbers: 21, 42, 63, and 84. By checking, we find that only the number 21 satisfies the condition of the problem.
## Answer
21.00 | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,872 |
Once Misha, Vitya, and Kolya noticed that they had brought the same toy cars to kindergarten. Misha has a car with a trailer, a small car, and a green car without a trailer. Vitya has a car without a trailer and a small green one with a trailer, and Kolya has a large car and a small blue one with a trailer. What type o... | If Kolya brought a small blue car with a trailer, then Vitya should have brought a car with a trailer, but Vitya's is green. This means Kolya brought a large car, and Vitya did not bring a small one, that is, a car without a trailer. Then Misha could not have brought a car with a trailer, and he could not have brought ... | largegreencarwithouttrailer | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,874 |
$\underline{\text { Folklore }}$
Is there a natural number that when divided by the sum of its digits gives a quotient and a remainder of 2011? | Suppose there exists a natural number $n$ with the sum of its digits $s$, for which $n=2011 s+2011$, hence $n-$ $s=2010 s+2011$. It is known that $n-s$ is divisible by 3. However, the number $2010 s+2011$ is not divisible by 3, since 2010s is a multiple of 3, but 2011 is not. Contradiction.
## Answer
It does not exis... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,875 |
$\underline{\text { Folklore }}$
Is it true that if $b>a+c>0$, then the quadratic equation $a x^{2}+b x+c=0$ has two roots? | The first method. The discriminant of the given equation $D=b^{2}-4 a c>(a+c)^{2}-4 a c=(a-c)^{2} \geq 0$.
The second method. Consider the function $f(x)=a x^{2}+b x+c$. From the condition, it follows that $f(-1)=a-b+c0$. Thus, the graph of the function must intersect the x-axis, and since it is a parabola, it interse... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,877 |
One hundred people are sitting at a round table, and more than half of them are men. Prove that there are two men sitting opposite each other.
# | Let's divide all people into 50 pairs: in each pair - two people sitting opposite each other. It is clear that in one of these pairs, both people are men. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,879 |
7,8
In the Russian Football Championship, 16 teams participate. Each team plays 2 matches against each of the others.
a) How many matches should "Uralan" play in a season?
b) How many matches are played in total in one season? | a) "Uralan" must play twice against each of the remaining 15 teams, therefore, over the season, they must play $2 \cdot 15=30$ matches.
b) First method. Each of the 16 teams must play 30 matches. Since each match is counted twice, the total number of matches played over the season is $16 \cdot 30: 2=240$.
Second meth... | )30;b)240 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,882 |
A country is called a "pentagonal" if every city in it is connected by air routes to exactly five other cities (there are no international flights).
a) Draw a diagram of air routes for a pentagonal country with 10 cities.
b) How many air routes are there in a pentagonal country with 50 cities?
c) Can there be a pent... | a) In a regular decagon, we draw all sides, all small diagonals (connecting vertices one apart), and all large diagonals (connecting opposite vertices).
b) $50 \cdot 5: 2=125$
c) If a country has $n$ cities, then $46=5 n / 2$. But 46 is not divisible by 5.
## Answer
b) 125 air routes; c) it cannot. | itcannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,883 |
Is the number $4^{9}+6^{10}+3^{20}$ prime?
# | This number is a perfect square.
## Solution
$4^{9}+6^{10}+3^{20}=\left(2^{9}\right)^{2}+2 \cdot 2^{9} \cdot 3^{10}+\left(3^{10}\right)^{2}=\left(2^{9}+3^{10}\right)^{2}$.
## Answer
It is not. | Itisnot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,885 |
In one glass there was milk, and in the other - the same amount of coffee. One spoonful was transferred from the glass of milk to the glass of coffee and mixed. Then the same spoonful of the mixture was transferred back to the glass of milk. What is now more: coffee in the glass of milk or milk in the glass of coffee?
... | From the vertex with the number 1 to the vertex with the number 8, you can get by moving along no more than three edges.
## Solution
Let's call the distance between two vertices of a cube the minimum number of edges that need to be traversed from one of these vertices to the other. It is easy to see that the distance... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,888 |
[ Semi-invariants $]$ [ $\left[\begin{array}{l}\text { Processes and operations }\end{array}\right]$
A chocolate bar has a size of $4 \times 10$ squares. In one move, it is allowed to break one of the existing pieces into two along a straight line. What is the minimum number of moves required to break the entire choco... | After each break, the number of pieces increases by 1.
## Solution
Notice that after each break, the number of pieces increases by 1. Initially, we have a whole chocolate bar, i.e., one piece, and in the end, we need to get 40 pieces of one tile each. Therefore, we need to perform exactly 39 breaks.
## Answer
In 39... | 39 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,891 |
[ Combinatorics (miscellaneous). ] [Principle of the extreme (miscellaneous) ]
How many ways are there to rearrange the numbers from 1 to 100 such that adjacent numbers differ by no more than 1?
# | Where can the number 1 be placed?
## Solution
Next to the number 1, only the number 2 can stand, so 1 must be at the edge. Suppose 1 is at the beginning. Then the next number is 2, the next is 3 (no other numbers can be next to 2), the next is 4, and so on. We get the arrangement $1, 2, \ldots, 99, 100$.
If 1 is at ... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,892 |
[ Weighing $\quad]$ [ Examples and Counterexamples. Constructions ]
What weights can three weights have so that with their help, it is possible to weigh any whole number of kilograms from 1 to 10 on balance scales (weights can be placed on both pans)? Provide an example. | Let's take, for example, weights of 3, 4, and 9 kg: $1=4-3, 2=9-3-4, 5=9-4, 6=9-3, 7=3+4, 8=3+9-4, 10=4+9-3$.
## Answer
For example, weights of 3, 4, and 9 kg.
In a corridor 100 m long, 20 r... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,893 |
[The median divides the area in half]
Prove that the medians divide the triangle into six equal-area triangles.
# | Triangles adjacent to one side have equal bases and a common height, so they are equal in area. Let $M$ be the point of intersection of the medians of triangle $ABC$. The line $BM$ cuts each of the triangles $ABC$ and $AMC$ into two equal-area triangles, so $S_{\mathrm{ABM}}=S_{\mathrm{BCM}}$. Similarly, $S_{\mathrm{BC... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,894 |
[Pairing and grouping; bijections] [Number and sum of divisors of a number]
Prove that a composite number $n$ always has a divisor greater than 1 but not greater than $\sqrt{n}$. | Each proper divisor $a$ of the number $n$ corresponds to a divisor $n / a$. One of these two numbers does not exceed $\sqrt{n}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,895 |
Does there exist a quadrilateral such that any diagonal divides it into two obtuse triangles?
# | For example, see Fig.
In each of the triangles of the partition, the obtuse angles are at vertices $B$ and $C$.
## Answer
It exists. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,896 |
[ Divisibility of numbers. General properties ] [Extracting a complete square. Sums of squares ]
Prove that for any natural $n$ the number $n^{2}+8 n+15$ is not divisible by $n+4$.
# | $n^{2}+8 n+15=(n+4)^{2}-1$.
$n^{2}+8n+15=(n+4)^{2}-1$.
(Note: The provided text is a mathematical equation, which is already in a universal format and does not change in translation. The equation is the same in English.) | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,897 |
7,8,9 | |
For what values of $m$ do the equations $m x-1000=1001$ and $1001 x=m-1000 x$ have a common root? | The roots of these equations are the numbers $2001 / m$ and $m / 2001$ respectively. ${ }^{2001} / m=m / 2001 \Leftrightarrow m^{2}=2001^{2}$ $m= \pm 2001$
## Answer
For $m= \pm 2001$.
Send a comment | \2001 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,898 |
Tokarev S.I.
Among any ten out of sixty schoolchildren, there will be three classmates. Is it necessarily true that among all sixty schoolchildren, there will be
a) 15 classmates;
b) 16 classmates? | a) Let's divide all 60 students into groups of classmates. If there are no 15 classmates among the students, then each group has no more than 14 students. Let $k$ be the number of groups consisting of two or more students. Such groups will be called large.
From the condition, it follows that $k \leq 4$ (otherwise, tak... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,899 |
Tomonvo A.K.
Petya's bank account contains 500 dollars. The bank allows only two types of transactions: withdrawing 300
dollars or adding 198 dollars.
What is the maximum amount Petya can withdraw from his account if he has no other money? | Since 300 and 198 are divisible by 6, Petya will only be able to withdraw an amount that is a multiple of 6 dollars. The maximum number that is a multiple of 6 and does not exceed 500 is 498.
Let's show how to withdraw 498 dollars. Perform the following operations: $500-300=200, 200+198=398, 398-300=98, 98+198=296, 29... | 498 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,900 |
[ Case Analysis ] [ Proof by Contradiction ]
In the cells of a $3 \times 3$ table, numbers are arranged such that the sum of the numbers in each column and each row is zero. What is the smallest number of non-zero numbers that can be in this table, given that this number is odd?
# | ## Example.
| 0 | -1 | 1 |
| :---: | :---: | :---: |
| -1 | 2 | -1 |
| 1 | -1 | 0 |
Evaluation. We will prove that it is impossible to use fewer non-zero numbers.
If the table contains exactly one non-zero number, then the sum of the numbers in the row containing this number is not zero. Suppose the table contains e... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,902 |
Vanya believes that fractions are "reduced" by crossing out identical digits in the numerator and denominator. Seryozha noticed that sometimes Vanya gets correct equalities, for example, $49 / 98 = 4 / 8$. Find all proper fractions with numerators and denominators consisting of two non-zero digits that can be "reduced"... | Let's consider all possible cases of cancellations.
1) $\frac{\overline{\beta a}}{\overline{\overline{b c}}}$. We get $(10 b + a) c = (10 b + c) a$, which simplifies to $b c = b a$. Since $b \neq 0$, it follows that $c = a$. However, by the condition, the fraction is proper. Therefore, there are no solutions.
2) $\fra... | 26/65,16/64,19/95,49/98 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,903 |
[ [ CaseAnalysis $\quad$]
Find all odd natural numbers greater than 500 but less than 1000, for each of which the sum of the last digits of all divisors (including 1 and the number itself) is 33.
# | For an odd number, all its divisors are odd. Since the sum of their last digits is odd, there must be an odd number of divisors. If a number has an odd number of divisors, then it is a square of a natural number (see problem $\underline{30365}$). Let's consider all the squares of odd numbers within the given range: $23... | 729 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,904 |
[ Arithmetic progression $]$ [ Extreme principle (other) . ] Author: Shapovalov $A . B$. In a $29 \times 29$ table, the numbers $1,2,3, \ldots, 29$ were written, each 29 times. It turned out that the sum of the numbers above the main diagonal is three times the sum of the numbers below this diagonal. Find the number... | Above (below) the diagonal there are $29 \cdot 14=406$ numbers. It is not difficult to check that the sum of the 406 largest numbers in the table (16, 17, ..., 29, taken 29 times each) is exactly three times the sum of the 406 smallest numbers (1, 2, ..., 14, taken 29 times each). Therefore, all the numbers on the diag... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,908 |
Zaslavsky A.A.
The midpoint of one of the sides of a triangle and the feet of the altitudes dropped to the other two sides form an equilateral triangle.
Is it true that the original triangle is also equilateral? | Actually, the same example can be constructed and justified in different ways.
First method. Consider an acute-angled triangle $ABC$, where $\angle B=60^{\circ}$. Let $AH$ and $CK$ be the altitudes, and $M$ be the midpoint of $AC$. In the right triangles $AHC$ and $AKC$, the medians $HM$ and $KM$ are equal to half the... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,909 |
Shapovalov A.V.
On the table, 28 coins of the same size are arranged in a triangular shape (see figure). It is known that the total mass of any three coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle.
First solution. Let's take a rhombus of 4 coins. As can be seen from the figure, the masses of two coins in it are equal. Cons... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,910 |
Do there exist such natural $x$ and $y$ that $x^{4}-y^{4}=x^{3}+y^{3}$? | After dividing by $x+y$, we get $(x-y)\left(x^{2}+y^{2}\right)=x^{2}-x y+y^{2}$. But the left side is no less than $x^{2}+y^{2}$ (it is clear from the condition that $x>y$), while the right side is smaller.
## Answer
They do not exist. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 54,911 |
$\underline{\text { Folklore }}$
Prove that for no natural values of $x$ and $y$ is the number $x^{8}-x^{7} y+x^{6} y^{2}-\ldots-x y^{7}+y^{8}$ a prime number. | $x^{8}-x^{7} y+x^{6} y^{2}-\ldots-x y^{7}+y^{8}=\frac{(x+y)\left(x^{8}-x^{7} y+\ldots-x y^{7}+y^{8}\right)}{x+y}=\frac{x^{9}+y^{9}}{x+y}=\frac{\left(x^{3}+y^{3}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right)}{x+y}=\left(x^{2}-x y\right.$ $\left.+y^{2}\right)\left(x^{6}-x^{3} y^{3}+y^{6}\right)$
The obtained factors have t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,912 |
Biinkov A. d.
Line $a$ intersects plane $\alpha$. It is known that in this plane there are 2011 lines equidistant from $a$ and not intersecting $a$.
Is it true that $a$ is perpendicular to $\alpha$? | Consider a cylindrical surface with axis $a$. Let's make a section of it with plane $\alpha$, which is not perpendicular to the axis (see figure). Then all tangents to the cylinder lying in plane $\alpha$ are equidistant from the cylinder's axis (the distance between the axis and the tangent is equal to the radius of t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,915 |
$\underline{\text { Frankin B.R. }}$
Petya marked several (more than two) points on a plane, all distances between which are distinct. A pair of marked points $(A, B)$ is called unusual if $A$ is the farthest marked point from $B$, and $B$ is the nearest marked point to $A$ (excluding the point $A$ itself). What is th... | Let $(A, B)$ be an unusual pair I, and $K$ be another marked point. Then $B K_{\mathrm{I}} B C>_{\mathrm{II}} C D>_{\mathrm{II}} A D>_{\mathrm{I}} A B$. Contradiction.
An example with one unusual pair $(A, B)$ - the vertices of triangle $A B C$, where $A C>A B>B C$.
## Answer
One pair. | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,917 |
Auto: :Shapovalov A.B.
a) There is a piece of cheese. It is allowed to choose any positive (possibly non-integer) number \( a \neq 1 \), and cut this piece in the ratio \( 1: a \) by weight, then cut any of the resulting pieces in the same ratio, and so on. Is it possible to act in such a way that after a finite numbe... | a) Cutting in the ratio $1: a$ is equivalent to cutting off a share of $\frac{a}{a+1}$. Choose such an $a$ that $\frac{a}{a+1}=\frac{1}{\sqrt{2}}$, and cut the largest of the pieces a second time. Then the weight of the largest of the resulting pieces is half the weight of the entire cheese.
b) Assume the opposite: af... | )Yes.b)No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,919 |
$\underline{\text { Folklore }}$
On a chessboard, $n$ white and $n$ black rooks are placed such that rooks of different colors do not attack each other. Find the maximum possible value of $n$.
# | Let's prove that for $n>16$, the specified arrangement is impossible. Note that on each row and each column, rooks can only be of one color (or the row/column can be free of rooks). We will denote a row (column) by the same color as the rooks standing on it.
Since there are more than 16 rooks, there are at least three... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,920 |
Golovanov A.S.
Given ten positive numbers, each pair of which is distinct. Prove that among them, there are either three numbers whose product is greater than the product of any two of the remaining, or three numbers whose product is greater than the product of any four of the remaining. | Let's take any five of the given numbers: $a, b, c, d, e$. If $a b c > d e$, then the statement is true. If, however, $d e > a b c$, let's take two more numbers $f$ and $g$. Suppose, for example, $f > g$. Then $d e f > a b c g$; thus, the statement is true in this case as well. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,921 |
$A . B$.
Four three-digit numbers are written on the board, which sum up to 2012. Only two different digits were used to write all of them.
Provide an example of such numbers. | The first method. $2012=503+503+503+503$. By replacing zero with five, we increase the sum by 200; to compensate, we replace one of the fives in the hundreds place with a three: $2012=353$ + 553 + 553 + 553.
The second method. Let's try to find two numbers, each written with two digits, whose sum is $2012: 2=$ 1006. A... | 2012=353+553+553+553=118+118+888+888=118+188+818+888=188+188+818+818 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,922 |
Pe
In the country of Distant, a province is called large if more than $7 \%$ of the country's population lives in it. It is known that for each large province, there are two provinces with a smaller population such that their combined population is greater than that of this large province. What is the smallest number ... | We will order the provinces by increasing population. The first and second provinces are not large, as for each of them, there will not be two provinces with a smaller population. In the third province, less than $14\%$ of the population lives, since in both provinces with a smaller population, the total is no more tha... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,923 |
[ Chessboards and chess pieces ] [ Examples and counterexamples. Constructions ]
Author: S. Preskova
Under one of the cells of an $8 \times 8$ board, a treasure is buried. Under each of the other cells, there is a sign indicating the minimum number of steps required to reach the treasure from that cell (one step all... | Let's dig up the corner cell $U$. Suppose there is a sign there. All cells at the specified distance from $U$ form a diagonal perpendicular to the main diagonal drawn from $U$. Let's dig up the corner cell $W$ on the same side as $U$. If there is also a sign there, then another diagonal perpendicular to the first one i... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,925 |
Gribalko A.V.
Given a strip of $2 n$ cells, numbered from left to right as follows:
|This strip is traversed by a token, which is moved each turn by the number of cells indicated in the current cell (to the right if the number is positive, and to the left if negative). It is known that the token, starting from any ce... | Suppose it is not prime. Then among the numbers $2,3, \ldots, n$ there is a divisor $d$ of the number $2 n+1$. We will show that starting from it, we will always land only on numbers divisible by $d$ (and thus not all numbers will be covered). For this, we will number the numbers from left to right. Note that the negat... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,927 |
Folklore
For numbers $a, b$ and $c$, different from zero, the equality holds: $a^{2}(b+c-a)=b^{2}(c+a-b)=c^{2}(a+b-c)$.
Does it follow from this that $a=b=c$? | $$
0=a^{2}(b+c-a)-b^{2}(c+a-b)=\left(a^{2}-b^{2}\right) c-\left(a^{2}+b^{2}\right)(a-b)=(a-b)\left(a c+b c-a^{2}-b^{2}\right)
$$
$$
\text { Similarly, }(b-c)\left(b a+c a-b^{2}-c^{2}\right)=0
$$
and $(c-a)\left(c b+a b-c^{2}-a^{2}\right)=0$
Let $a=b$. Then from equation (2), we get that $c(a-c)^{2}=0$, from which, c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 54,928 |
Gukhanov H.X.
Let $a_{1}, \ldots, a_{11}$ be distinct natural numbers, each not less than 2, with their sum equal to 407.
Can the sum of the remainders when some natural number $n$ is divided by the 22 numbers $a_{1}, \ldots, a_{11}, 4 a_{1}, 4 a_{2}, \ldots, 4 a_{11}$ equal 2012? | Suppose such a number $n$ exists.
The maximum possible remainder when dividing by a natural number $m$ is $m-1$. Therefore, the sum of the remainders when dividing an arbitrary number by $a_{1}, \ldots, a_{11}$ is no more than $407-11=396$, and the sum of the remainders when dividing it by $4 a_{1}, 4 a_{2}, \ldots, 4... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,929 |
[ Axial and sliding symmetries (other).] [ Examples and counterexamples. Constructions ]
Author: Yashchenko I.v.
On a sheet of graph paper, several cells were colored in such a way that the resulting figure had no axes of symmetry. Vanya colored one more cell. Could the resulting figure have four axes of symmetry? | To construct an example, one needs to take a figure with four axes of symmetry and remove a cell that does not lie on any of these axes. For example, a square has four axes of symmetry (see the left image). This results in the example in the center.
There are other examples (for example, on the right image).
, after which the upper half can be colored with one horizontal fold along line 3 - see the figure (the cells colored on the first, second, and third folds are marked with the numbers 1, 2, and 3, respectively).
... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,934 |
Khachaturyan A.V.
An astrologer believes that 2013 is a lucky year because 2013 is divisible by the sum of $20+13$.
Will there ever be two lucky years in a row? | For example, 2024 and 2025 are happy years: 2024 is divisible by $20+24=44$, and 2025 is divisible by $20+25=45$.
## Answer
It will be. | Itwillbe | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,935 |
$\underline{\text { Folklore }}$
Can some of the minuses in the expression $2013^{2}-2012^{2}-\ldots-2^{2}-1^{2}$ be replaced with pluses so that the value of the resulting expression equals 2013? | $(m+1)^{2}-m^{2}=2 m+1$. Therefore, $\left((n+3)^{2}-(n+2)^{2}\right)-\left((n+1)^{2}-n^{2}\right)=(2(n+2)+1)-(2 n+1)=2 n+5-2 n-1=4$. Thus, before the squares of any four consecutive natural numbers, signs can be placed in such a way that the value of the resulting expression is equal to 4.
We will divide the first 20... | 2013 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,936 |
A passenger left their belongings in an automated storage locker, and when they came to retrieve their items, they realized they had forgotten the number. They only remember that the numbers 23 and 37 were in the code. To open the locker, the correct five-digit number must be entered. What is the minimum number of code... | Let's consider several cases.
1) The number contains the combination 237. It can be placed in the number in three ways: **237, *237*, 237**. In each of these, each of the remaining digits can be chosen in 10 ways. In total, we get $3 \cdot 10^{2}=300$ numbers. 2) The number contains the combinations 23 and 37, with 23... | 356 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,938 |
[Permutations and Substitutions (other)] [ Invariants and Semi-invariants $]
Prove that if $a_{1} \geq a_{2} \geq \ldots \geq a_{n}, b_{1} \geq b_{2} \geq \ldots \geq b_{n}$, then the largest of the sums of the form $a_{1} b_{k_{1}}+a_{2} b_{k_{2}}+\ldots+a_{n} b_{k_{n}}$ $\left(k_{1}, k_{2}, \ldots, k_{n}-\right.$ pe... | Note that if $x \geq y, \quad z \geq w$, then $x z+y w \geq x w+y z$. Indeed, $(x z+y w)-(x w+y z)=(x-y)(z-w) \geq$ 0.
Consider in the sum $a_{1} b_{k_{1}}+a_{2} b_{k_{2}}+\ldots+a_{n} b_{k_{n}}$ the term $a_{j} b_{1}$, containing the factor $b_{1}$. If $j>1$, then by replacing $a_{j-1} b_{k_{j-1}}+a_{j} b_{1}$ with $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,943 |
Evenness and Oddness $\quad]$ Proof by Contradiction $\quad]$
In a country, 100 roads lead out of each city, and from each city, you can reach any other city. One road has been closed for repairs.
Prove that even now, you can still reach any city from any other city. | If the road $A B$ is closed, then it is sufficient to prove that it is still possible to reach $B$ from $A$ after this. If this is not the case, then in the connected component containing $A$, all vertices except $A$ are even. However, the presence of exactly one odd vertex contradicts problem 87972 b). | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,945 |
## [ Number of divisors and their sum ] [ Divisibility of numbers. General properties ] [ Partitioning into pairs and groups; bijections ] $[\quad \underline{\text { Arithmetic of residues (other). }}]$
Let a natural number $n$ be such that $n+1$ is divisible by 24. Prove that the sum of all natural divisors of $n$ ... | Prove that the sum of the divisors of $n$ is divisible by both 3 and 8.
## Solution
Since $n+1$ is divisible by 3 and 8, $n$ gives a remainder of 2 when divided by 3, and a remainder of 7 when divided by 8. We will divide the divisors into pairs of the form $\left\{d, \frac{n}{d}\right\}$. Note that if $d$ gives a re... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,946 |
[ Pigeonhole Principle (miscellaneous) $]$
$[$ Dirichlet's Principle (miscellaneous). ]
In a country, every two cities are connected by a road with one-way traffic.
Prove that there exists a city from which you can travel to any other city in no more than two roads.
# | This is the city from which the largest number of roads exit.
## Solution
Consider city $A$, from which the largest number of roads exit, and an arbitrary city $B$. If a road leads from $A$ to $B$, then everything is fine. If a road leads from $B$ to $A$, then, since the number of roads leading from $B$ is no greater... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,952 |
[ GCD and LCM. Mutual simplicity ] [ Proof by contradiction ]
Prove that the product of a hundred consecutive natural numbers cannot be the hundredth power of a natural number. | Suppose that for some natural $n$ and $k$ we have $n(n+1)(n+2) \ldots(n+99)=k^{100}$.
Then $n^{100}<k^{100}<(n+99)^{100}$, so $n<k<n+99$. Therefore, the number $k$ is a factor in the left-hand side of the original equation. The number $k+1$, also a factor in the left-hand side of this equation, is coprime with $k$. Th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,953 |
[ Chess coloring ] [ Examples and counterexamples. Constructions ]
a) What is the maximum number of bishops that can be placed on a 1000 by 1000 board so that they do not attack each other?
b) What is the maximum number of knights that can be placed on an $8 \times 8$ board so that they do not attack each other? | a) Since no more than one bishop can stand on one diagonal, and there are exactly 1999 diagonals running from the bottom-left to the top-right, and on the two outermost diagonals (each consisting of one cell) no more than one bishop can stand (they are located on the same perpendicular diagonal), no more than 1998 non-... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,955 |
Prove that for any natural $n$ there exist $n$ consecutive composite natural numbers.
# | Consider the numbers located "near" the number $(n+1)!$.
## Solution
Consider the next $n$ natural numbers: $(n+1)!+2, (n+1)!+3, \ldots, (n+1)!+(n+1)$. We will show that all these numbers are composite. Indeed, for each $k=2,3, \ldots, n+1$, the number $(n+1)!$ is divisible by $k$. Therefore, the number $(n+1)!+k$ is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,959 |
[ Dirichlet's Principle (Area and Volume) ] [ Partitions into Pairs and Groups; Bijections ]
In a square with side length 1, there are 100 figures, the total area of which is greater than 99. Prove that there is a point in the square that belongs to all these figures. | Consider the figures that complement the given 100 figures to form a square.
## Solution
Let the given 100 figures be $\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{100}$, and their areas be $\mathrm{S}_{1}, \mathrm{~S}_{2}, \ldots, \mathrm{S}_{100}$ respectively. According to the condition, $\mathrm{S}_{1}+\m... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,961 |
Kozhevnikov P.A.
What is the minimum number of connections required to organize a wired communication network of 10 nodes, so that the communication remains possible between any two remaining nodes (even through a chain of other nodes) in the event of any two nodes failing?
# | Evaluation. To ensure that the connection is maintained even when any two nodes fail, it is necessary that at least three lines of communication enter each node (if node $A$ is connected to two nodes $B$ and $C$, then when nodes $B$ and $C$ fail, node $A$ becomes inaccessible). Thus, the total number of communication l... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,963 |
[Parallel lines, properties and signs. Transversals]
$[$ Small movements $]$ On a plane, several points are drawn. Prove that a line can be drawn such that the distances from all points to it are different.
# | If the distances from two points to a line are the same, then this line is either parallel to the line connecting these two points or it passes through the midpoint of the segment connecting these points.
Consider two of the given points $A$ and $B$. Let some line $l$ have the property that the distances from points $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,964 |
In a communication system consisting of 2001 subscribers, each subscriber is connected to exactly $n$ others. Determine all possible values of $n$.
# | Calculate the number of pairs of connected subscribers and show that $n$ is even.
## Solution
Since each of the 2001 subscribers is connected to exactly $n$ others, the total number of connection directions is $2001n$. Therefore, the total number of connected pairs of subscribers is $2001n / 2$, since each connected ... | n=2tfor=0,1,\ldots,1000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,965 |
The sum of two letters is called the letter whose position in the alphabet has the same remainder when divided by the number of letters in the alphabet as the sum of the ordinal numbers of the original two letters. The sum of two letter sequences of the same length is called the letter sequence of the same length obtai... | Use the fact that the sum of all letters has an odd remainder when divided by 26.
## Solution
When adding two sequences, the sum of the ordinal numbers of all letters in the resulting sequence and the sum of the ordinal numbers of all letters in both addends have the same remainder when divided by 26. Therefore, the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,966 |
}
There are two glasses, the first glass is filled with a certain amount of water, and the second with the same amount of alcohol. It is allowed to pour some of the liquid from one glass to the other (at which point the solution is uniformly mixed). Is it possible, using such operations, to obtain a solution in the fi... | Prove that in the first glass, there will always be no less water than alcohol, and in the second glass, the opposite will be true.
## Solution
For definiteness, let's assume the amount of liquid poured into each glass is 100 g. We will prove that in the first glass, the percentage of alcohol will never exceed 50%, a... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,968 |
}
Find all finite sets of points on the plane that have the following property: no three points of the set lie on the same line, and together with any three points of the set, the orthocenter of the triangle formed by these points also belongs to the set. | The convex hull of this set cannot have an obtuse angle.
## Solution
Let $M$ be some such set of points. Consider the convex hull $V$ of this set. The polygon $V$ cannot have an obtuse angle. Indeed, if $A, B, C$ are three consecutive vertices of the polygon $V$, and the angle $ABC$ is obtuse, then the orthocenter of... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,969 |
On the plane, there is an angle of $60^{\circ}$. A circle touches one side of this angle, intersects the other side at points $A$ and $B$, and intersects the angle bisector at points $C$ and $D$. $A B = C D = \sqrt{6}$. Find the area of the circle bounded by this circle. | Let $K$ be the vertex of the angle, $O$ be the center of the circle, $R$ be its radius, $M$ be the point of tangency with the side of the angle, and $L$ be the midpoint of the chord $CD$.
From the condition, it is clear that $KO$ is the bisector of the angle $AOC$, so $\angle OKC = 15^\circ$. Therefore, $\angle OKM = ... | \pi\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,970 |
Is it necessarily true that a triangle is isosceles if the center of the inscribed circle is equidistant from the midpoints of two sides? | Consider a triangle with sides $6, 8, 10$.
## Solution
Consider a right triangle $ABC$ with legs $AB=6, BC=8$ and hypotenuse $AC=10$. Let $O$ be the center of its inscribed circle; $M, N$ be the midpoints of sides $AB$ and $AC$ respectively; $P, Q, F$ be the points of tangency of the inscribed circle with sides $AB, ... | Notnecessarily | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,971 |
[ Pythagorean Theorem (direct and inverse). Similarity criteria. In triangle $ABC$, a perpendicular line passing through the midpoint of side $AB$ intersects the extension of side $BC$ at point $M$, such that $MC: MB = 1: 5$. A perpendicular line passing through the midpoint of side $BC$ intersects side $AC$ at point $... | Continue the perpendicular bisector of side $B C$ until it intersects with line $A B$ and with the line drawn through vertex $C$ parallel to $A B$.
## Solution
Let $P$ and $Q$ be the midpoints of sides $B C$ and $A B$ respectively, and let $K$ be the point of intersection of lines $P N$ and $A B$. Draw a line through... | \angleA=\operatorname{arctg}2,\angleB=\operatorname{arctg}3,\angleC=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,972 |
In a certain kingdom, in a certain state, there are several cities, and the distances between them are all pairwise different. One fine morning, a plane takes off from each city and lands in the nearest neighboring city. Can more than five planes land in one city? | The largest angle of a triangle is not less than $60^{\circ}$.
## Solution
Suppose that in city $P$, 6 planes land, having departed from cities $A_{1}, A_{2}, \ldots, A_{6}$, and points $A_{1}, A_{2}, \ldots, A_{6}$ are consecutive vertices of a hexagon. Since the distance between cities $A_{1}$ and $A_{2}$ is greate... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,974 |
Prove that for any natural $a$ there exists such a natural $n$ that all numbers $n+1, n^{n}+1, n^{n^{n}}+1, \ldots$ are divisible by $a$.
# | $n=2 a-1$.
## Solution
If $n$ is odd, then all the specified numbers are divisible by $n+1$. Therefore, any odd number $n$ for which $n+1$ is a multiple of $a$ will work. For example,
$n=2 a-1$. | 2a-1 | Number Theory | proof | Yes | Yes | olympiads | false | 54,976 |
Prove the identities
a) $C_{r}^{m} C_{m}^{k}=C_{r}^{k} C_{r-k}^{m-k}(0 \leq k \leq m \leq r)$
b) $C_{n+1}^{m+1}=C_{n}^{m}+C_{n}^{m+1}$;
c) $C_{2 n}^{n}=\left(C_{n}^{0}\right)^{2}+\left(C_{n}^{1}\right)^{2}+\ldots+\left(C_{n}^{n}\right)^{2}$
d) $C_{m+n}^{k}=C_{n}^{0} C_{m}^{k}+C_{n}^{1} C_{m}^{k-1}+\ldots+C_{n}^{k} ... | a) Let us suppose we need to select a committee of $m$ people from $r$ people, and within it, a staff of $k$ people. If we first select the committee members and then choose the staff from them, the count of ways to do this will lead, by the rule of product, to the left side of the equation. However, we can proceed dif... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,977 |
In a company of 10 people, 14 pairwise arguments have occurred. Prove that it is still possible to form a group of three friends.
# | The total number of ways to choose a company of three people is $C_{10}^{3}=120$. Each quarrel destroys no more than eight such companies, so the number of destroyed companies is no more than $8 \cdot 14=112$. Therefore, at least 8 friendly companies remain. | 8 | Combinatorics | proof | Yes | Yes | olympiads | false | 54,978 |
[ [ Proof by contradiction ]
Prove that the set of prime numbers of the form $p=4k+3$ is infinite.
# | Let $p_{1}=3, p_{2}=7, \ldots, p_{n}$ be all such prime numbers. Consider the number $N=4 p_{2} \ldots p_{n}+3$. It does not divide any of the numbers $p_{1}, p_{2}, \ldots, p_{n}$, but it contains a prime divisor of the form $4 k+3$. Contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,979 |
For which integers $n$ is the number $n^{4}+4$ composite?
# | $n^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}$.
## Solution
$n^{4}+4=n^{4}+4 n^{2}+4-4 n^{2}=\left(n^{2}+2\right)^{2}-(2 n)^{2}=\left(n^{2}-2 n+2\right)\left(n^{2}+2 n+2\right)$.
The numbers $n^{2}+2 n+2=(n+1)^{2}+1$ and $n^{2}-2 n+2=(n-1)^{2}+1$ are greater than 1 for $n \neq \pm 1$. For $n= \pm 1 \quad n^{4}+4=5$.
## Answer
... | n\neq\1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 54,981 |
[ Euclid's Algorithm ] $[\quad \underline{\text { Induction (continued) }} \quad]$
a) Let $m_{0}$ and $m_{1}$ be integers, $01$. There exist integers $a_{0}, a_{1}, \ldots, a_{k}$ and $m_{2}, \ldots, m_{k}$ such that
$m_{1}>m_{2}>m_{3}>\ldots>m_{k}>0, a_{k}>1$,
$m_{0}=m_{1} a_{0}+m_{2}$
$m_{1}=m_{2} a_{1}+m_{3}$
$... | a) The numbers $a_{0}$ and $m_{2}$ are obtained as the quotient and remainder when $m_{0}$ is divided by $m_{1}$; the numbers $a_{1}$ and $m_{2}$ are obtained as the quotient and remainder when $m_{1}$ is divided by $m_{2}$, and so on. Since the numbers are always decreasing, the process will eventually terminate, that... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 54,982 |
Raskina I.V.
In the cells of an $m \times n$ table, numbers are placed. It turned out that in each cell, the number written is the count of adjacent cells (by side) that contain a one. At the same time, not all numbers are zeros. For which numbers $m$ and $n$, greater than 100, is this possible? | According to the condition, if a cell contains a one, then exactly one of the neighboring cells is also a one. This means that the ones form $1 \times 2$ rectangles. Moreover, no two rectangles share a side or overlap.
Furthermore, a cell that does not belong to such rectangles cannot border exactly one rectangle. Oth... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,984 |
Konoakov V.
The width of the river is one kilometer. By definition, this means that from any point on either bank, one can swim to the opposite bank by traveling no more than one kilometer. Can a boat sail along the river in such a way that at any moment the distance to either bank is no more than:
a) 700 m?
b) $800... | a) A counterexample is provided in the figure.
Here, \( AC = 1000 \, \text{m}, AB > 1400 \, \text{m}, CD = 1 \, \text{m} \). The ship's route must intersect segment \( AB \), but the distan... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 54,988 |
Ssiurov M.v.
On the plane, there is a finite number of strips, the sum of whose widths is 100, and a circle of radius 1.
Prove that each of the strips can be translated parallel to itself so that together they cover the circle.
# | Let the total width of the strips be no less than $2.5 \pi$.
We will assume that there are no horizontal or vertical strips. We will order the strips by the increasing angle of inclination relative to the positive $O x$-axis, and combine strips with the same angle of inclination into one strip. We can assume that the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 54,990 |
Laiko 0.
Around a round table, there is a company of thirty people. Each of them is either a fool or smart. Everyone sitting is asked: Is your right neighbor smart or a fool? In response, a smart person tells the truth, while a fool can say either the truth or a lie. It is known that the number of fools does not excee... | When $F=8$. If $F=0$, then one can point to any person sitting at the table. Let now $F \neq 0$. We divide all those sitting at the table into non-empty groups of consecutive smart and consecutive foolish people; the number of these groups is denoted by $2k$ ($k$ groups of smart people and $k$ groups of foolish people)... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,991 |
Tokarev S.I.
There are 8 coins, 7 of which are genuine and weigh the same, and one is counterfeit, differing in weight from the others. The balance scales without weights are such that if equal loads are placed on their pans, either pan may tip, but if the loads differ in mass, the pan with the heavier load will defin... | Let's denote the coins and their weights by the letters $A, B, C, D, E, F, G$ and $H$. It is clear that if 4 coins are placed on each pan of the balance, the balance cannot be in equilibrium. Also, note that if the coins are evenly distributed on the pans, the pan containing the counterfeit coin will always either be h... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 54,992 |
Authors: Chenokov G.R., Bogdanov I.I.
On the table lie banknotes of denominations $1, 2, ..., 2n$ tugriks. Two players take turns. Each move, a player picks up two banknotes, gives the larger one to the opponent, and keeps the smaller one for themselves. Each player aims to get as much money as possible. How many tugr... | $n^{2}=1+3+. .+(2 n-1)$ for odd $n, n(n+1)=2+4+. .+2 n$ for even $n$. Let's generalize the problem. Suppose at the beginning of the game, there are banknotes of denominations $M_{1}>M_{2}>. .>M 2 n$ on the table. We will show by induction on $n$ that the player making the last move (let's call this player the last one ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 54,993 |
Podlipsky o.k.
Prove that there does not exist a finite set containing more than $2 N (N>3)$ pairwise non-collinear vectors on a plane, possessing the following two properties.
1. For any $N$ vectors of this set, there is another $N-1$ vector from this set such that the sum of all $2 N-1$ vectors is zero;
2. for any ... | Assume the opposite.
Choose a line that is not orthogonal to any of the vectors in our set.
Introduce a direction on this line such that these vectors point in the negative direction, and select $N$ vectors $\vec{f}_{1}, \vec{f}_{N}$ such that the algebraic projection $s$ of their sum is maximal (it is clear from con... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 54,996 |
At a chemical conference, there were $k$ scientists, chemists, and alchemists, with chemists outnumbering the alchemists. It is known that chemists always tell the truth to any question, while alchemists sometimes tell the truth and sometimes lie. A mathematician attending the conference wants to determine whether each... | The provided solution allows us to determine who is a chemist and who is an alchemist in fewer than \(2k-2\) questions, specifically in \(q=3m\) questions if \(k=2m+1\) is odd, and in \(3m-2\) questions if \(k=2m\) is even.
First, we will consider the case when \(k\) is odd. The desired method will be defined by induc... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 55,000 |
Does there exist a polyhedron and a point outside it such that from this point none of its vertices are visible? | It is possible to create a spatial "cross" from 6 "pencils" - long, thin parallelepipeds, adjacent to the faces of a unit cube (the cube is only needed to explain the structure). The pencils lie symmetrically relative to the center of the cube, with the center of one face of a pencil coinciding with the center of one o... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,001 |
Avor: Bogosnov i.i.
On the plane, all points with integer coordinates $(x, y)$ such that $x^{2}+y^{2} \leq 1010$ are marked. Two players play a game (taking turns). On the first move, the first player places a token on some marked point and erases it. Then, on each subsequent move, a player moves the token to another ... | Let's prove a more general statement: Suppose the game with the same rules takes place on a finite set of points $S$, which contains the point $O(0,0)$ and is invariant under a $90^{\circ}$ rotation. Then the first player wins in this game. (It is clear that the set of points from the condition satisfies these conditio... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,003 |
6 $[$ Application of trigonometric formulas (geometry).]
Prove that in a regular triacontagon $A_{1} \ldots A_{30}$, the following triples of diagonals:
a) $A_{1} A_{7}, A_{2} A_{9}, A_{4} A_{23}$
b) $A_{1} A_{7}, A_{2} A_{15}, A_{4} A_{29}$
c) $A_{1} A_{13}, A_{2} A_{15}, A_{10} A_{29}$
intersect at a single poin... | In all cases, checking the condition of Ceva's theorem (see problem 35216) for the corresponding hexagon $\left(A_{1} A_{2} A_{4} A_{7} A_{9} A_{23}, A_{1} A_{2} A_{4} A_{7} A_{15} A_{29}, A_{1} A_{2} A_{10} A_{13} A_{15} A_{29}\right)$ reduces to checking the equality $\sin 6^{\circ} \sin 18^{\circ} \sin 84^{\circ}=$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 55,004 |
a) Can a $6 \times 6$ square be tiled with $1 \times 2$ dominoes such that there are no cracks, i.e., no straight line that does not cut through any dominoes?
b) Prove that any rectangle $m \times n$, where $m$ and $n$ are greater than 6 and $mn$ is even, can be tiled with $1 \times 2$ dominoes such that there are no ... | a) It is impossible. Suppose that a $6 \times 6$ square is tiled with $1 \times 2$ dominoes such that there are no gaps. Consider the 10 segments that divide the square into 36 cells (we do not consider the sides of the square itself). Each of these segments cuts through at least two dominoes. Indeed, if a segment cut ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,005 |
On the table, there are 15 magazines that completely cover it. Prove that you can remove seven magazines so that the remaining magazines cover at least 8/15 of the table's area. (No one solved this problem in the olympiad.)
# | Let's first prove that if $n$ magazines cover an area $S$, then one magazine can be removed so that the remaining magazines cover an area of at least $(n-1)S / n$. If after removing a certain magazine, the remaining magazines cover an area less than $(n-1)S / n$, then the area covered only by this magazine is greater t... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,007 |
Burr S.L.
Around a round table, there are 100 representatives from 25 countries, with 4 representatives from each. Prove that they can be divided into 4 groups in such a way that each group contains one representative from each country, and no two representatives in the same group sit next to each other at the table. | Lemma. Suppose there are $n$ non-intersecting pairs of acquaintances - representatives of $n$ countries, two from each country. Then it is possible to divide them into 2 groups, each containing one representative from each country and no acquaintances.
Choose any representative from country 1, place him in the first g... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 55,010 |
[Decimal numeral system]
[ $\underline{\text { Linguistics }}$ ]
$\left[\begin{array}{l}\text { Case enumeration }\end{array}\right]$
A three-digit number consists of different digits in ascending order, and all words in its name start with the same letter. Another three-digit number, on the contrary, consists of the ... | Notice, "the numbers are equal" and "the numbers start with the same letter" are two completely different statements.
## Solution
These numbers are 147 and 111, respectively. The problem is solved by simply enumerating the options, of which there are not so many. | 147111 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,017 |
[ Inequalities. Case Analysis ] [Examples and Counterexamples. Constructions]
Authors: Golenisheva-Kumuzova T.I., Gurvich V.M., Kozhevnikov P.A., Yatsenko I.V.
There was not enough space for building a typical house. The architect changed the project: removed two entrances and added three floors. As a result, the num... | Let's say the original project had 5 entrances, 2 floors, and one apartment on each floor. Then there were 10 apartments. After the project was changed, it became 3 entrances and 5 floors - 15 apartments, and after the second change - 1 entrance and 8 floors - 8 apartments.
## Answer
It could. | Itcould | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 55,018 |
A nut has the shape of a regular hexagonal prism. Each side face of the nut is painted in one of three colors: white, red, or blue, and adjacent faces are painted in different colors. How many differently colored nuts exist? (Not all three colors need to be used for the coloring of the nut.)
# | Faces of the same color cannot be adjacent, so there are no more than three of them. Let's consider three cases.
1) Only two colors are used - each three times. Then the colors must alternate, and the only coloring scheme is 121212. By substituting any pair of colors for 1 and 2, we get 3 types of nuts. Swapping the c... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,023 |
Prove that if the number $n!+1$ is divisible by $n+1$, then $n+1$ is a prime number.
# | Let $n+1$ be a composite number. If $p$ is some of its prime divisors, then $p \leq n$. Therefore, $n!$ is divisible by $p$, but $n! + 1$ is not. Contradiction. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 55,025 |
Does there exist a convex quadrilateral, each diagonal of which divides it into two acute triangles
# | From the condition, it follows that each angle of such a quadrilateral is acute. But then their sum is less than $360^{\circ}$.
## Answer
It does not exist. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,026 |
In a convex quadrilateral, the tangent of one of the angles is equal to the number $m$. Can the tangents of each of the three remaining angles also be equal to $m$? | From the condition, it follows that the quadrilateral has no right angles. Since the sum of its angles is $360^{\circ}$, there is at least one obtuse angle and at least one acute angle. However, the tangents of an obtuse and an acute angle have different signs.
## Answer
They cannot. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,027 |
[ Counting in Two Ways $]$ Vasya thought of three different non-zero digits. Petya wrote down all possible two-digit numbers, in the decimal representation of which only these digits were used. The sum of the written numbers is 231. Find the digits thought of by Vasya.} | Let a, $b$ and c be three digits thought of by Vasya. There are nine two-digit numbers, in the decimal representation of which only these digits are used: ; ; ; ; ; ; ; ; We will find their sum, by decomposing each number into a sum of place value addends: $(10 a+a)+(10 b+b)+(10 c+c)+(10 a+b)+(10 b+a)+$ $(10 a+c)+(10 c... | 1,2,4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 55,029 |
[Central Angle. Arc Length and Circumference Length]
Consider all moments of time when the hour and minute hands of a clock lie on the same straight line, forming a straight angle.
Will there be two such lines that are mutually perpendicular? | Suppose such straight lines do exist. Then at these moments, the hour hands are perpendicular. This means that a whole number of hours has passed between them (specifically $6n+3$, where $n$ is an integer). Then the positions of the minute hands at these moments are the same, and they cannot form an angle of $90^{\circ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 55,030 |
[ $\left.\begin{array}{c}\text { Work Problems } \\ \text { [ Processes and Operations }\end{array}\right]$
In a turning shop, parts are machined from steel blanks, one part from one blank. The chips left after processing three blanks can be melted down to make exactly one blank. How many parts can be made from nine b... | From nine blanks, on the first stage, you can obtain nine parts, and from the remaining swarf, you can make three blanks, from which (on the second stage) you can make three parts, and from the remaining swarf, you can make one blank and then one part. In total, 13 parts.
With 14 blanks, it is easier to proceed as fol... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 55,031 |
|
Nезнayka took a book from Pilulkin and counted how many digits were needed to number all the pages, starting from the first. He got 100 digits. Could this be possible, or did Nезнayka make a mistake? If it could, how many pages were there? | Notice, the number of the last page is a two-digit number.
## Solution
Under these conditions, the number of the last page is a two-digit number (the total number of digits in all two-digit and one-digit numbers is $9+90 \times 2>100$). However, all one-digit pages will contribute 9 digits, which is an odd number, an... | Nезнaykamademistake | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 55,032 |
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