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[ Quadratic equations and systems of equations] Principle of the extreme (further). ]
Find all positive solutions of the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=x_{3}^{2} \\
x_{2}+x_{3}=x_{4}^{2} \\
x_{3}+x_{4}=x_{5}^{2} \\
x_{4}+x_{5}=x_{1}^{2} \\
x_{5}+x_{1}=x_{2}^{2}
\end{array}\right.
$$ | Let $a$ be the largest of the numbers $x_{1}, \ldots, x_{5}$, and $b$ be the smallest of these numbers. Then $a^{2} \leq 2 a$ and $b^{2} \geq 2 b$. By the condition, the numbers $a$ and $b$ are positive, so $a \leq 2$ and $b \geq 2$, therefore, $a=b=2$.
## Answer
$(2,2,2,2,2)$. | (2,2,2,2,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,330 |
[ Convex polygons ] [ Pairings and groupings; bijections ]
At each vertex of a convex $k$-gon, there is a hunter armed with a laser gun. All hunters simultaneously shot at a rabbit sitting at point $O$ inside this $k$-gon. At the moment of the shot, the rabbit crouched, and all the hunters died. Prove that there is no... | According to the condition, the vertices of the $k$-gon are paired $\left\{A_{\mathrm{i}}, A_{\mathrm{j}}\right\}$ such that the point $O$ belongs to each segment $A_{\mathrm{i}} A_{\mathrm{j}}$. Moreover, for any other pair $\left\{A_{\mathrm{p}}, A_{\mathrm{q}}\right\}$, the points $A_{\mathrm{p}}$ and $A_{\mathrm{q}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,331 |
$$
\begin{aligned}
& [\text { Theory of Algorithms (other) }] \\
& {[\underline{\text { Backward Pass }}]}
\end{aligned}
$$
Petya bought a computational machine from a store that multiplies any number entered into it by 3 for 5 k, and adds 4 to any number for 2 k. Petya wants to start with one, which can be entered fo... | First, note that an even number cannot be entered on the machine. Therefore, Petya will not be able to enter the number 1982.
Lemma. After the first multiplication by 3, it is not advantageous to add 4 more than twice in a row.
Proof of the lemma. By replacing the sequence of actions $\times 3,+4,+4,+4$ with $+4, \ti... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,332 |
3 $[\quad$ Case Analysis $\quad]$
Given five different positive numbers that can be divided into two groups such that the sums of the numbers in these groups are equal. In how many ways can this be done?
# | Let's take some partition that satisfies the condition of the problem and prove that there are no other partitions. If one of the groups contains only one number, then this number is equal to the sum of all the others, which means that in any other partition (specifically, in the group containing this number), the sum ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,334 |
Solve the equation $\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}$. | Let $u=\frac{x-49}{50}, v=\frac{x-50}{49}$. Then the equation becomes $u+v=1 / u+1 / v$. Multiplying both sides by $uv$, moving all terms to one side, and factoring, we get $(u+v)(uv-1)=0$, from which it follows that either $u+v=0$ or $uv=1$. Substituting the expressions for $u$ and $v$, in the first case we get $x=490... | 0,\frac{4901}{99},99 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,335 |
$\begin{aligned} & {\left[\begin{array}{l}\text { Irrational Equations } \\ \text { [ Completing the Square. Sums of Squares }\end{array}\right]}\end{aligned}$
Solve the equation
$$
\left(x^{2}+x\right)^{2}+\sqrt{x^{2}-1}=0
$$ | Since the numbers $\left(x^{2}+x\right)^{2}$ and $\sqrt{x^{2}-1}$ are non-negative, and their sum is zero, then both these numbers are equal to zero. On the other hand, if both these numbers are equal to zero, then their sum is zero. Therefore, the original equation is equivalent to the following system:
$$
\left\{\be... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,336 |
3 [ Completing the square. Sums of squares ]
Prove that for a natural number $n$, the number $n m+1$ will be composite for at least one natural number $m$.
# | If $m=n+2$ then $n m+1=n^{2}+2 n+1=(n+1)^{2}$ - a composite number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,338 |
[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameter $]
For the quadratic trinomial $f(x)=a x^{2}-a x+1$, it is known that $|f(x)| \leq 1$ for $0 \leq x \leq 1$. Find the greatest possible value of $a$. | Since $f(0)=f(1)=1$, the graph of the quadratic function is a parabola symmetric about the line $x=0.5$. From the condition $|f(x)| \leq 1$ for
$0 \leq x \leq 1$, it follows that the branches of the parabola are directed upwards. The minimum value of $f(x)$ is $f(0.5)=1-\frac{a}{4}$.
The maximum possible value of $a$... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,339 |
Compose the equation of the plane passing through the midpoint of the segment with endpoints at points $P(-1 ; 2 ; 5)$ and $Q(3 ;-4 ; 1)$, perpendicular to the line passing through points $A(0 ;-2 ;-1)$ and $B(3 ; 2 ;-1)$. | The coordinates of the midpoint $M$ of the segment $P Q$ are equal to the arithmetic means of the coordinates of its ends, i.e., ( $1 ; -1 ; 3$ ).
The desired plane is perpendicular to the vector $\overrightarrow{A B}=(3-0 ; 2-(-2) ;-1-(-1))=(3 ; 4 ; 0)$, hence, its equation has the form
$$
3(x-1)+4(y+1)+0(z-3)=0, \t... | 3x+4y+1=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,340 |
[ Tournaments and tournament tables ] Counting in two ways
[ Examples and counterexamples. Constructions ]
In the competition, 10 figure skaters participate. The competition is judged by three judges in the following way: each judge independently distributes places (from first to tenth) among the skaters, after which ... | Evaluation. Since each of the three judges distributed a set of places from first to tenth, the sum of the places awarded by all judges to all participants in the competition is $3 \cdot (1 + 2 + \ldots + 10) = 165$.
On the other hand, if the winner received a sum of no less than 16, then all the others received a sum... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,342 |
$\left[\begin{array}{c}\text { Absolute value of a number }\end{array}\right]$
$\left[\begin{array}{c}\text { Evenness and oddness }\end{array}\right]$
$\left[\begin{array}{l}\text { Partitions into pairs and groups; bijections }\end{array}\right]$
[ Examples and counterexamples. Constructions ]
Find the maximum value... | The modulus of the difference of two non-negative numbers is not greater than their maximum.
## Solution
Estimate. Note that the modulus of the difference of two non-negative numbers is not greater than their maximum. Therefore, $\left|x_{1}-x_{2}\right|$ $\leq \max \left\{x_{1}, x_{2}\right\}$
||$x_{1}-x_{2}\left|-... | 1989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,343 |
Given $n$ pairwise coprime numbers, greater than 1 and less than $(2 n-1)^{2}$. Prove that among them there must be a prime number.
# | Assume the opposite and consider the smallest prime divisors of the given numbers.
## Solution
Let $a_{1}, a_{2}, \ldots, a_{n}$ be pairwise coprime composite numbers. Denote by $p_{i}$ the smallest prime divisor of the number $a_{i}$. Then $a_{i}=p_{i} q_{i}$, where $p_{i} \leq q_{i}$. Since $a_{i}<(2 n-1)^{2}$, it ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,344 |
[ Similar triangles (further). Application of trigonometric formulas (geometry). The area of a figure is equal to the sum of the areas of the figures into which it is divided.
The radius of the circle inscribed in triangle $ABC$ is 4, and $AC = BC$. A point $D$ is taken on the line $AB$, which is at a distance of 11 a... | Let $CM$ be the altitude of triangle $ABC$, and $P$ and $Q$ be the projections of point $D$ onto lines $AC$ and $BC$, respectively. Clearly, point $D$ cannot lie on the extension of $AB$ beyond point $A$.
If point $D$ lies on the extension of side $AB$ beyond point $B$, then the difference $DQ - DP = 8$ equals the hei... | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,347 |
In a triangle, two angles $\beta$ and $\gamma$ and the radius $R$ of the circumscribed circle are given. Find the radius of the inscribed circle.
# | Let in triangle $ABC$, $\angle B = \beta$, $\angle C = \gamma$, $r$ - the radius, $O$ - the center of the inscribed circle, $K$ - the point of tangency of this circle with side $AB$.
Method 1. $AK = \frac{1}{2}(AB + AC - BC) = R(\sin \beta + \sin \gamma - \sin (\beta + \gamma)) = R(\sin \beta (1 - \cos \gamma) + \sin ... | 4R\sin\beta/2\sin\gamma/2\cos(\beta+\gamma/2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,351 |
8,9 [ Examples and counterexamples. Constructions ]
Is there a figure that has neither axes of symmetry nor centers of symmetry, but maps onto itself under some rotation? | Construct such a figure from segments.
## Solution
From point $O$, lay off three equal segments $O A, O B$, and $O C$ at angles of $120^{\circ}$ to each other. Then, at points $A, B$, and $C$, erect equal perpendiculars to them (in the same direction). We will obtain the required figure.
## Answer
It exists. | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 53,352 |
Folklore $^{2}$
In the square $A B C D$, points $K$ and $M$ lie on sides $B C$ and $C D$ respectively, such that $A M$ is the bisector of angle $K A D$.
Prove that $A K=D M+B K$. | Let's rotate triangle $A D M$ by $90^{\circ}$ around point $A$ (see the figure). In this case, it will transform into an equal triangle $A B L$, where $B L$ is the extension of segment $C B$. Note that $\angle K A L = \angle K A B + \angle L A B = \angle K A B + \angle K A M = \angle B A M = \angle D M A = \angle K L A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,353 |
[ Examples and counterexamples. Constructions ]
Prove that for any natural $n$ there exist $n$ consecutive natural numbers, among which there is exactly one prime. | According to problem $\underline{34990}$, there will be $n$ consecutive composite numbers. Consider the smallest prime number $p$ greater than all these numbers (such a number exists according to problem $\underline{30410}$). Then the numbers $p-n+1, p-n+2, \ldots, p-1, p$ are the desired ones.
Submit a comment | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,354 |
10,11 [ Methods for solving problems with parameters ]
For all values of the parameter $a$, find the number of real roots of the equation $x^{3}-x-a=0$.
# | Setting the derivative $3 x^{2}-1$ to zero, we find that the function $x^{3}-x$ has a maximum of $\frac{2 \sqrt{3}}{9}$ at the point $x_{1}=-\frac{\sqrt{3}}{3}$ and a minimum of $-\frac{2 \sqrt{3}}{9}$ at the point $x_{2}=\frac{\sqrt{3}}{3}$. From this, we get the answer.
## Answer
Three roots when $|a|<\frac{2 \sqrt... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,356 |
8,9,10 |
| | [ Proof by contradiction | |
On the board, a pentagon inscribed in a circle was drawn. Masha measured its angles and got that they are $80^{\circ}, 90^{\circ}, 100^{\circ}, 130^{\circ}$, and $140^{\circ}$ (in that exact order). Did Masha make a mistake? | Let $A B C D E$ be the given pentagon, in which angles $A, B, C, D$ and $E$ are respectively $80^{\circ}, 90^{\circ}, 100^{\circ}, 130^{\circ}$ and $140^{\circ}$ (see figure). We can reason in different ways.
First method. Draw the diagonal $A D$, then the quadrilateral $A B C D$ is also inscribed, so $\angle B A D = ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,361 |
9,10 $[\quad$ The Pigeonhole Principle (etc.).
The sum of ten natural numbers is 1001. What is the greatest value that the GCD (greatest common divisor) of these numbers can take? | Example. Consider nine numbers equal to 91 and the number 182. Their sum is 1001.
Estimate. We will prove that the GCD cannot take a value greater than 91. Note that $1001=7 \cdot 11 \cdot 13$. Since each term in this sum is divisible by the GCD, the GCD is a divisor of the number 1001. On the other hand, the smallest... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,362 |
Auto: :Shapovalov A.B.
a) Fill in each circle with a non-zero digit so that the sum of the digits in the two top circles is 7 times less than the sum of the other digits, and the sum of the ... | b) From the condition, it is clear that the sum of all digits is divisible by both 8 and 6. The smallest such natural number is 24. The next number is 48, but the sum of all five digits cannot exceed \(5 \cdot 9 = 45\).
Thus, the sum of all digits is 24, the sum of the two top digits is 3, and the sum of the two left ... | Foroddn | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,363 |
[ Pigeonhole Principle (continued). ] [ Proof by contradiction ]
Each of 102 students in one school is acquainted with at least 68 others.
Prove that among them, there will be four who have the same number of acquaintances.
# | Assume the opposite. Then for each number from 68 to 101, there are exactly three people who have such a number of acquaintances.
## [ directed graphs ]
Problem | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,394 |
8, |
| :---: | :---: | :---: |
| | [ Isosceles, Inscribed, and Circumscribed Trapezoids ] | |
| | $\left[\begin{array}{l}\text { [ Area of a Trapezoid } \\ \text { [Application of Trigonometric Formulas (Geometry) }\end{array}\right.$ | |
Vertex $D$ of square $A B C D$ lies on side $E F$ of isosceles trapezoid $B ... | ## Answer
$\angle C B F=\pi / 2-1 / 2 \operatorname{arctg} 3 ; \sqrt{5}: \sqrt{2}$. | \sqrt{5}:\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,397 |
Shen A.H.
There is an infinite one-way strip of cells, numbered with natural numbers, and a bag with ten stones. Initially, there are no stones in the cells of the strip. The following actions can be performed:
- move a stone from the bag to the first cell of the strip or vice versa;
- if there is a stone in the cell... | Note that for each action, there is an inverse action. Therefore, if we move from situation $A$ to situation $B$ by following the rules, we can move from situation $B$ back to situation $A$ by following the rules.
We will show by induction that if there is a reserve of $n$ stones, then, acting according to the above r... | 197 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,404 |
Author: Belukhov $H$.
In a certain state, addition and subtraction are denoted by the signs "!" and "?", but you do not know which sign corresponds to which operation. Each operation is applied to two numbers, but you do not know whether subtraction involves subtracting the left number from the right or the right from... | Note that both operations are sums, with only one of the addends possibly being multiplied by -1. Therefore, when calculating, we will not swap the addends. There are four possible combinations of operations. For example, the variant "?L" means that when performing "?", the Left addend is multiplied by -1.
Consider th... | 20a-18b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,405 |
Novikov V. V.
There are 100 indistinguishable coins of three types: gold, silver, and copper (each type appears at least once). It is known that gold coins weigh 3 g each, silver coins weigh 2 g each, and copper coins weigh 1 g each.
How can you determine the type of all the coins using a balance scale without weight... | Author: Ryabichev A.
Even 100 weighings are enough.
Lemma 1. Let there be $k$ coins, among which all three types are represented, and for a pair of coins $(A, a)$, it is already known that $A > a$. Then it is possible to determine which of the $k$ coins belong to which type in $k-1$ weighings.
Proof. Induction. Base... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,406 |
$:$ Markelov Y.
Let's call a pair ( $\$ m, n \$$ ) of distinct natural numbers $\$ m \$$ and $n$ good if $\$ m n \$$ and $\$(m+1)(n+1) \$$ are perfect squares. Prove that for each natural $\$m\$ there exists at least one $\$ n>m \$$ such that the pair (\$m, n\$) is good. | The pair $\$(m, m(4 m+3) \wedge 2) \$$ is good. Indeed, $\$(m+1)(m(4 m+3) \wedge 2+1)=(m+1)(16 m \wedge 3+24 m \wedge 2+9 m+1)$ $=(m+1)^{\wedge} 2(16 m \wedge 2+8 m+1)=((m+1)(4 m+1))^{\wedge} 2 \$$. Leave a comment
## Problem | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,407 |
Robapko $A . B$. In a checkered wooden square, 102 cells are painted black. Petya, using the square as a stamp, applied it 100 times to a white sheet of paper, and each time these 102 cells (and only they) left a black mark on the paper. Could it be that in the end, a $101 \times 101$ square was formed on the sheet, al... | Any square \$(2N+1) \times (2N+1) \$$ without a corner cell can be obtained by applying a stamp of \$2N+2 \$$ cells \$2N \$$ times. To illustrate, we will provide a diagram for \$N=4\$.
A s... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,408 |
Gribalko A.V.
On a line, 2022 points are marked such that every two adjacent points are at the same distance. Half of the points are painted red, and the other half are painted blue. Can the sum of the lengths of all possible segments where the left end is red and the right end is blue equal the sum of the lengths of ... | We can consider the marked points to be integers from 1 to 2022. It is sufficient to show that the sum S of the lengths of all segments with endpoints of different colors is odd.
Method 1. Let the number of red-even points be x, then the number of red-odd and blue-even points is y = 1011 - x, and the number of blue-od... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,412 |
$\underline{\text { Markelov S.V. }}$
For each of the numbers $1, 19, 199, 1999$, etc., a separate card was made and the number was written on it.
a) Is it possible to choose at least three cards such that the sum of the numbers on them equals a number, all of whose digits, except one, are twos?
b) Suppose several ca... | a) For example, $19+199+1999+\ldots+199999999=222222212$.
b) Example. Example with zero: $1+19=20$; example with 1 is given in part a).
In the selected \$k cards, the sum \$S satisfies the ine... | 0or1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,413 |
Vazicina T.V.
Baron Munchausen claims that he drew a polygon and a point inside it such that any line passing through this point divides the polygon into three polygons. Can the baron be right?
# | See the examples in the figures. In the second figure, it is essential that the triple of points $\mathrm{A}, \mathrm{O}, \mathrm{A}^{\prime}$, the triple B, O, B' and the triple C, O, C lie on one straight line.
, and in the meantime, each of the remaining ones moves half of the garbage from their desk to the desk of the one who went out.... | a) An example fits when the first person leaving has 2 g of garbage, and the second has 4 g.
b) Let's number the slobs in the order they leave for lunch. Suppose on the table of the і-th slob, there are \$ 2 \wedge 1 \$$ g of garbage. After the first one leaves, there will be \$ 2+2+4+8+\ldots+2 \wedge 9=2 \wedge\{10\... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,415 |
Prove that the number of all digits in the sequence $1,2,3, \ldots, 10^{\mathrm{k}}$ is equal to the number of all zeros in the sequence $1,2,3, \ldots, 10^{\mathrm{k}+1}$
# | We will match a digit from the first sequence to a zero from the second sequence as follows. We write a zero after the given digit. As a result, we get a number from the second sequence with a marked zero. We match our digit precisely to this zero. Conversely, to a zero from the number of the second sequence, we will m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,420 |
On the plane, there are four lines, none of which are parallel, and no three intersect at the same point. A pedestrian is walking along each line at a constant speed. It is known that the first pedestrian meets the second, the third, and the fourth, and the second pedestrian meets the third and the fourth. Prove that t... | Let's introduce, along with the coordinates in the plane where pedestrians move, a third coordinate axis - the time axis. Consider the graphs of pedestrian movements. It is clear that pedestrians meet when their movement graphs intersect. From the condition, it follows that the graphs of the third and fourth pedestrian... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,421 |
$[$ Extreme Principle (Miscellaneous) ]
Prove that no rectangular chessboard 4 cells wide can be toured by a knight's move, visiting each square exactly once and returning to the starting square on the last move.
# | We imagine a board $n \times 4$, having 4 verticals and $n$ horizontals. We will call the cells located on the first and fourth verticals "edge" cells; the rest of the cells will be called "middle" cells. Note that from any of the edge verticals, a knight can move to only a middle cell in one move. Therefore, if the kn... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,422 |
$\left[\begin{array}{ll}\text { Combinations and Permutations }\end{array}\right]$ $\left[\begin{array}{l}\text { Quadratic Equations. Vieta's Theorem } \\ {[\text { Graph Theory (other) }}\end{array}\right]$
$n$ people gathered. Some of them are acquainted with each other, and every two unacquainted people have exact... | Let one of the gathered people - let's call him $X$ - have $m$ acquaintances: $a_{1}, a_{2}, \ldots, a_{m}$. By the condition, no two people from $a_{1}, a_{2}, \ldots, a_{m}$ are acquainted with each other. Therefore, for any two people $a_{i}, a_{j}$, there must be another common acquaintance besides $X$. This person... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,423 |
[ Least or greatest distance (length) ]
There are 100 points on a plane, and the distance between any two of them does not exceed 1, and if $A, B$, $C$ are any three points from the given ones, then triangle $A B C$ is obtuse. Prove that one can draw a circle of radius $1 / 2$ such that all the given points lie inside... | Let's choose two points $A$ and $B$ among the given points such that the distance between them is the greatest (if there are several such pairs of points, we can take any one of them). If $C$ is any of the given points, then in the triangle $A B C$, the obtuse angle can only be at vertex $C$. Therefore, point $C$ lies ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,424 |
10,11
$k$ people were traveling in a bus without a conductor, and all of them had coins only of 10, 15, 20 kopecks. It is known that each of them paid for the fare and received change. Prove that the smallest number of coins they could have had is $k+\left[\frac{k+3}{4}\right]$, where the symbol [a] denotes the greate... | For each used coin, we will draw an arrow from the person who had it before the payment to the person who ended up with it after the payment (some arrows will lead to the payment machine). The number of resulting arrows is equal to the number of used coins, so it is enough to prove that at least $k+\left[\frac{k+3}{4}\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,425 |
In the cells of an $m \times n$ table, some numbers are written. It is allowed to simultaneously change the sign of all numbers in a certain column or a certain row. Prove that by repeatedly performing this operation, the given table can be transformed into one where the sums of the numbers in each column and each row ... | By changing the signs of numbers in some column or some row of the table, we can obviously obtain no more than $2^{m n}$ different tables; ($2^{m n}$ is the number of ways to choose signs for the n numbers). Since there is a finite number of such tables, among them there exists (possibly more than one) table whose sum ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,426 |
9,10
What is the maximum number of self-intersection points that a closed 14-segment broken line can have, passing along the lines of a grid paper so that no line contains more than one segment of the broken line? | Let's first show that any closed 14-segment broken line has 7 horizontal and 7 vertical segments. Indeed, from each vertex of the broken line, one vertical and one horizontal segment emerge. Counting along all 14 vertices, we will thus count 14 horizontal and 14 vertical segments. But in this counting, each segment is ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,427 |
[Processes and Operations] $[$ Motion Problems ]
From point $A$, 100 planes (1 flagship and 99 additional) take off simultaneously. With a full tank of fuel, a plane can fly 1000 km. In flight, planes can transfer fuel to each other. A plane that has transferred fuel to others makes a gliding landing. How should the f... | Let's describe the optimal fuel exchange procedure. Initially, 100 aircraft take off with full tanks. As soon as possible, one of the aircraft distributes its fuel to the others, after which 99 aircraft have full tanks, and the freed-up aircraft lands. By this point, the aircraft will have flown \(1000 \cdot \frac{1}{1... | 5187 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,429 |
Given two sets of numbers: $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{n}$. Arrange the numbers $a_{k}$ in ascending order, and the numbers $b_{k}$ in descending order. The resulting sets will be
$A_{1} \leq \ldots \leq A_{n}, B_{1} \geq \ldots \geq B_{n}$. Prove that $\max \left\{a_{1}+b_{1}, \ldots, a_{n}+b_{n}\ri... | Let us renumber the numbers in the original sets $\left\{a_{i}\right\},\left\{b_{i}\right\}$ such that $b_{i}=B_{\mathrm{i}}$ for all $i$. Suppose for some indices $k<l$ the inequality $a_{l}<a_{k}$ holds. We will prove that if we swap $a_{k}$ and $a_{l}$, the number $\max _{k}\left\{a_{k}+\right.$ $\left.b_{k}\right\}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,430 |
A white sphere has 12% of its surface area painted red. Prove that a parallelepiped can be inscribed in the sphere, with all its vertices being white.
# | Let's pass through the center of the sphere three mutually perpendicular planes and for each point of the sphere consider its images under the symmetries with respect to these planes and under the compositions of these symmetries. Each point not lying on these planes has exactly 8 images. Therefore, the red points and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,431 |
## [ Coordinate method in space $\quad]$ [ Distance between skew lines ]
In a rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$, it is known that $A B=3, B C=2, C C 1=4$. On the edge $A B$, a point $M$ is taken such that $A M: M B=1: 2 ; K-$ is the point of intersection of the diagonals of the face $C C 1 D 1 D$. ... | Let's choose a coordinate system with the origin at point $D 1$. The $x$-axis is directed along the ray $D 1 C 1$, the $y$-axis along the ray $D 1 A 1$, and the $z$-axis along the ray $D 1 D$. Then the coordinates of the endpoints of segments $D 1 M$ and $B 1 K$ are:
$$
D 1(0 ; 0 ; 0), M(1 ; 2 ; 4), B 1(3 ; 2 ; 0), K\... | \arccos\frac{5}{\sqrt{861}},\frac{20}{\sqrt{209}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,433 |
4 [ Distance between skew lines ]
The base of the pyramid $A B C S$ is an equilateral triangle $A B C$ with a side length of $4 \sqrt{2}$. The lateral edge $S C$ is perpendicular to the base plane and equals 2. Find the angle and the distance between the skew lines, one of which passes through point $S$ and the midpoi... | Let $M$ and $K$ be the midpoints of edges $B C$ and $A B$ respectively, $P$ be the projection of point $M$ onto the line passing through vertex $C$ and parallel to $A B$. We choose a coordinate system with the origin at point $C$. The $x$-axis is directed along the ray $C P$,
the $y$-axis is directed along the ray $C K... | 45,\frac{2}{\sqrt{3}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,434 |
Anozhans A.
A square is divided into $n^{2}$ equal smaller squares. It is known that a certain broken line passes through the centers of all the smaller squares (the broken line can intersect itself). What is the minimum number of segments in this broken line?
# | In the figure, an example of such an eight-segment broken line for $n=5$ is shown. It consists of a known traversal of nine points by a four-segment broken line and an unwinding spiral.
Obv... | 2n-2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,436 |
Anjans.
Several guys are standing in a circle. Each has a certain number of candies. Initially, each has an even number of candies. On command, each gives half of their candies to the person on their right. If after this, someone ends up with an odd number of candies, one candy is added from outside. This is repeated ... | Let $2 m$ be the largest, and $2 n$ be the smallest number of candies a person has. After one round of exchange and, possibly, adding candies from outside, $m$ will not increase, and the number of people having $2 n$ candies will decrease. (Indeed, each person keeps no more than $m$ candies and receives no more than $m... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,437 |
Folklore
In a regular decagon, all diagonals are drawn. The number +1 is placed at each vertex and at each point of intersection of the diagonals (only the diagonals themselves, not their extensions are considered). It is allowed to simultaneously change the signs of all numbers lying on one side or one diagonal. Is i... | Let's consider three diagonals of the decagon, as shown in the figure.
Notice that no other diagonals pass through the three marked points. On each step, the sign of the even number of units... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,440 |
Kurrmynoik d.
Given a non-increasing sequence of non-negative numbers \(a_{1} \geq a_{2} \geq a_{3} \geq \ldots \geq a_{2 k+1} \geq 0\). Prove the inequality: \(a_{1}^{2}-a_{2}^{2}+a_{3}^{2}-\ldots+a_{2 k+1}^{2} \geq\left(a_{1}-a_{2}+a_{3}-\ldots+a_{2 k+1}\right)^{2}\). | Consider a trapezoid bounded by the lines $x=a_{2 k}, x=a_{2 k-1}, y=0, y=2 x$ (see figure).
Its area is equal to $a_{2 k-1}^{2}-a_{2 k}^{2}$, therefore, the expression on the left side of t... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,441 |
Proizvolov V.V.
Prove that for any natural $n \geq 2$ the inequality holds:
$$
\sqrt{2 \sqrt{3 \cdots \sqrt{(n-1) \sqrt{n}}}}<3
$$ | $$
\begin{aligned}
& k^{2}>(k-1)(k+1)=(k-1) \sqrt{(k+1)^{2}}. \text{ Hence } \\
& \quad 3=\sqrt{3^{2}}>\sqrt{2 \sqrt{4^{2}}}>\sqrt{2 \sqrt{3 \sqrt{5^{2}}}}>\ldots> \\
& \sqrt{2 \sqrt{3 \ldots \sqrt{(n-1) \sqrt{(n+1)^{2}}}}}>\sqrt{2 \sqrt{3 \ldots \sqrt{(n-1) \sqrt{n}}}}
\end{aligned}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 53,442 |
Imre Barany
Three triangles - white, green, and red - have a common interior point $M$. Prove that it is possible to choose one vertex from each triangle such that point $M$ is inside or on the boundary of the triangle formed by the chosen vertices.
# | Consider all possible angles formed by segments going from $M$ to some two vertices of different colors. Let the largest of these angles be $-\angle A M B \leq 180^{\circ}$, where $A$ is a red point and $B$ is a white point. Then, within the angles $A M B^{\prime}$ and $B M A^{\prime}$, adjacent to angle $A M B$, there... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,443 |
On an infinite chessboard, pawns are placed every three squares, forming a square grid.
Prove that a knight cannot visit all free squares, visiting each square exactly once.
# | Consider a "sub-board" of size $61 \times 61$ on an infinite board, where $16^{2}=256$ pawns are placed (we will assume they are on black squares). Then, the total number of black squares on the considered $61 \times 61$ board is 1861, and white squares are 1860. Let's calculate the number of black squares a knight can... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,444 |
8,9 Avor: Konievizi $M$. In quadrilateral $A B C D$, $A B=B C=C D=1$, $A D$ is not equal to 1. The positions of points $B$ and $C$ are fixed, while points $A$ and $D$ undergo transformations that preserve the lengths of segments $A B, C D$, and $A D$. The new position of point $A$ is obtained from the old one by refl... | The broken line $A B C D$ is defined by specifying two angles: $\angle B C A=\beta$ and $\angle C D B=\gamma$; in this case, if we denote by $X$ and $Y$ the points on the extensions of segment $B C$, then (since triangles $A B C$ and $B C D$ are isosceles, see the figure) $\angle X B A=2 \beta$, and $\angle Y C D=2 \ga... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,446 |
Shapovalov A.V.
At a round table, 12 places were prepared for the jury with each place labeled with a name. Nikolai Nikolaevich, who arrived first, absent-mindedly sat not in his own place, but in the next one clockwise. Each subsequent jury member, approaching the table, would take their own place or, if it was alrea... | Let's consider a certain way of seating the jury members. We will call a jury member lucky if they are sitting in their own seat. The first of the unlucky ones (excluding Nikolai Nikolaevich) to approach the table is the one whose seat is taken by Nikolai Nikolaevich (another unlucky one would sit in their still free s... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,447 |
[ Examples and counterexamples. Constructions ]
A first-grader has a hundred cards with natural numbers from 1 to 100 written on them, as well as a large supply of "+" and "=" signs. What is the maximum number of correct equations he can form? (Each card can be used no more than once, each equation can contain only on... | In the equality, no less than three numbers are involved, so there cannot be more than 33 equalities. Let's show how to form 33 equalities.
25 equalities: $49+51=100, 47+52=99, \ldots, 1+75=76$. The remaining even numbers are from 2 to 50. Another 5 equalities: $18+30=48, 14+32=46, \ldots, 2+38=40$. The remaining numb... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,448 |
[ [principle of the extreme]
Natural numbers are placed at the vertices of a cube $A B C D E F G H$ such that the numbers in adjacent (by edge) vertices differ by no more than one. Prove that ... | Take the vertex where the smallest of these numbers is located, and look at the adjacent vertices.
## Solution
Let the numbers in the vertices $A, B, C, D, E, F$ and $H$ of the cube be denoted by the corresponding lowercase Latin letters: $a, b, c, d, e, f, g$ and $h$. Take one of the vertices where the smallest numb... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,449 |
Konstantinov N.
In a country where residents' homes are represented by points on a plane, two laws are in effect:
1. A person can play basketball only if they are taller than the majority of their neighbors.
2. A person is entitled to free public transportation only if they are shorter than the majority of their neig... | Let's consider an example. Suppose in a country there are 10 people, and all their houses are located on a straight line, in the order of increasing height of these people. Let the distances between them be: 1 km, 2 km, 3 km, 4 km, 5 km, 4 km, 3 km, 2 km, 1 km (see the figure).
^{2}-1$ into two groups $A_{n}=\left\{n^{2}, n^{2}+1, \ldots, n^{2}+n\right\}$ and $B_{n}=\left\{n^{2}+n+1, n^{2}+n+2, \ldots\right.$, $\left.n^{2}+2 n\right\}$.
For the numbers in group $A_{n}$, the nearest square is $n^{2}$, and for $B_{n}$, the nearest square is $(n+1)... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,455 |
Berlov S.L.
At the alumni meeting, 45 people attended. It turned out that any two of them who had the same number of acquaintances among those present were not acquainted with each other. What is the maximum number of pairs of acquaintances that could have been among those who attended the meeting? | Since $45=1+2+3+\ldots+9$, we can divide 45 people into groups of 1, 2, ... 9 people. Let people in the same group not know each other, while people in different groups know each other. Then each person in the $k$-th group has $45-k$ acquaintances.
Under this condition, the problem is solved, and the total number of p... | 870 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,460 |
A machine produces parts of two types. On the conveyor belt, 75 parts are laid out in a single line. While the conveyor belt is moving, the machine prepares a part of the type that is less represented on the belt. Every minute, the next part falls off the belt, and the prepared part is placed at its end. After a certai... | Let after $t$ minutes, there are $m(t)$ parts of type $A$ and $k(t)$ parts of type $B$ on the conveyor belt. Since $m(t)$ and $k(t)$ have different parities, $|m(t)-k(t)|$ can only take values $1, 3, 5, \ldots$ Since each minute a part of the type that is less on the conveyor belt is added, and a random part is removed... | 2,4,38,76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,462 |
Berpovo S.L. The given quadratic trinomials $f(x)$ and $g(x)$ are such that the equations $f(g(x))=0$ and $g(f(x))=0$ have no real roots.
Prove that at least one of the equations $f(f(x))=0$ and $g(g(x))=0$ also has no real roots. | Without loss of generality, we can assume that the minimum value of $f(x)$ does not exceed the minimum value of $g(x)$. If the quadratic polynomial $g(x)$ has no roots, then $g(x)>0$ for all $x$, and therefore $g(g(x))>0$ for all $x$, which proves the statement.
If $g(x)$ does have roots, then the minimum value of $f(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,463 |
Authors: Agakhanov N.Kh., $\underline{\text { Bogdanov I.I. }}$
Find all triples of real numbers $x, y, z$ such that $1+x^{4} \leq 2(y-z)^{2}, 1+y^{4} \leq 2(z-x)^{2}, 1+z^{4} \leq 2(x-y)^{2}$. | Let for definiteness $x \geq y \geq z$. Since $0 \leq\left(x^{2}-1\right)^{2}=\left(x^{4}+1\right)-2 x^{2}$, we have $2 x^{2} \leq 1+x^{4} \leq 2(y-z)^{2}$, from which $|x| \leq y-z$. Similarly,
$|z| \leq x-y$. Therefore, $|z|+|x| \leq(x-y)+(y-z)=x-z$. This is possible only if $x \geq 0, z \leq 0$, in which case
the ... | {-1,0,1} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 53,464 |
Berrov S.L.
On some cells of a $10 \times 10$ board, $k$ rooks were placed, and then all cells that are attacked by at least one rook were marked (it is assumed that a rook attacks the cell it stands on). For what largest $k$ can it happen that after removing any rook from the board, at least one marked cell will no l... | Let's consider the placement of $k$ rooks satisfying the condition. There are two possible cases.
1. Suppose there is at least one rook in each column. Then the entire board is under attack, and a rook can be removed from any column that has at least two rooks. Therefore, in this case, there is exactly one rook in eac... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,466 |
Frankin B.R.
At some point on a circular island with a radius of 1 km, a treasure is buried. On the shore of the island stands a mathematician with a device that indicates the direction to the treasure when the distance to the treasure does not exceed 500 m. In addition, the mathematician has a map of the island, on w... | Let $O$ - the center of the island, $A$ - the starting point, $w$ - a circle with center at point $O$ and radius $500 \, \text{m}, l_{1}$ and $l_{2}$ - tangents to $w$, parallel to $OA$ (see figure).
 Let the lengths of the blue sticks be $12, 17, 20$, and the red ones be $2, 23, 24$. Since the only pair with a difference less than 2 is (23, 24), and after recoloring, the stick of length 2 will end up in a different composition triplet, the difference between the largest sides in it will be greater than 2, and it... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,469 |
[ Auxiliary coloring (other) ]
Can a $10 \times 10$ checkerboard be tiled with $1 \times 4$ tiles? | The first method. We will paint some cells black and the rest white (see the figure). Each tile can contain either 0 or 2 black cells, that is, an even number. But there are 25 black cells in total. Therefore, the board 10×10 cannot be tiled with $1 \times 4$ tiles, as otherwise the number of black cells would be even.... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,476 |
$\underline{\text { Tokarev S.I. }}$
Prove that there are infinitely many pairs of integers such that in the decimal representation of each number all digits are at least 7 and the product of the numbers in each pair is also a number where all digits are at least 7. | Consider the product of 3n-digit numbers 887887...887887$\cdot$999999...999877. 998
First, note that $887 \cdot 123=109101$. Therefore, $887887 \ldots 887887 \cdot 123=109210210 \ldots 210101$ (3n + 3 digits). Finally, 887887...887887$\cdot$999999... $999877=887887 \ldots 887887 \cdot\left(10^{3 n}-123\right)=887887 \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,478 |
A.K.
Twelve grasshoppers sit at various points on a circle. These points divide the circle into 12 arcs. We mark the midpoints of these 12 arcs. At a signal, the grasshoppers simultaneously jump, each to the nearest marked point in the clockwise direction. Again, 12 arcs are formed, and the jumps to the midpoints of t... | a) First method. After each "move," the grasshopper does not jump to the point where the next grasshopper was previously. Therefore, the next one will not reach this point in two consecutive jumps. Continuing this, we find that after the 12th jump, the first one will not return to its initial position.
Second method. ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,479 |
9In a scalene triangle, a circle is inscribed, the points of tangency of which with the sides are taken as the vertices of a second triangle. In this second triangle, another circle is inscribed, the points of tangency of which are the vertices of a third triangle; a third circle is inscribed in it, and so on. Prove th... | Let $\alpha \leq \beta \leq \gamma$ be the angles of a given triangle. By the condition, this triangle is scalene, so $\gamma-\alpha>0$. As seen from the solution of problem $\underline{78113}$, the angles of the second obtained triangle are $\frac{\beta+\gamma}{2} \geq \frac{\alpha+\gamma}{2} \geq \frac{\alpha+\beta}{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 53,481 |
Functions $f(x)$ and $g(x)$ are defined on the set of integers not exceeding 1000 in absolute value. Let $m$ be the number of pairs $(x, y)$ for which
$f(x)=g(y)$, $n$ be the number of pairs for which $f(x)=f(y)$, and $k$ be the number of pairs for which $g(x)=g(y)$. Prove that $2 m \leq n+k$.
# | Let $a$ be one of the values taken by the function $f(x)$, and let $n_{a}$ and $k_{a}$ be the number of those $x$ for which $f(x)=a$ and $g(x)=a$ respectively (it is possible that $k_{a}=0$). Then $n_{a} k_{a}$ pairs of numbers $(x, y)$ will satisfy the equations $f(x)=g(y)=a$, $n_{a}^{2}$ pairs will satisfy the equati... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,482 |
In a "tic-tac-toe" tournament held on a "lose and you're out" system, 18 schoolchildren participated. Each day, one game was played, with participants chosen by lot from those who had not yet been eliminated. Each of the six schoolchildren claims that they played exactly four games. Is anyone of them mistaken? | A total of 17 games were played in the tournament, as each loser was eliminated. If each of the six schoolchildren played 4 games, then each won no less than three. Since there is only one winner in a game, there were no fewer than $6 \cdot 3=18$ games. Contradiction.
## Answer
Incorrect.
, he called it unexpected if, up to that point, it had occurred less frequently than each of the other possible grades. (For example, if he had received the grades 3, 4, 2, 5, 5, 5, 2, ... | The first unexpected grade will be the last one received for the first time. The second unexpected grade will be the last one received for the second time, and so on. Therefore, there will be a total of 10 unexpected grades.
## Answer
It can be. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,488 |
[ Pairing and grouping; bijections ] [ Examples and counterexamples. Constructions ]
There are 64 weights laid out in a row, and the mass of any two adjacent weights differs by 1 gram. It is required to divide the weights into two piles with equal masses and an equal number of weights. Is this always possible? | Let's divide all the weights into groups of four consecutive weights. Consider any of these groups of four. Place the first two weights in different piles (the larger one in the first pile, and the smaller one in the second pile), and the next two weights also in different piles, but in reverse order (the smaller one i... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,489 |
Frankin 5.P.
An inscribed $n$-gon ( $n>3$ ) is divided by non-intersecting (in interior points) diagonals into triangles. Each of the resulting triangles is similar to at least one of the others. For which $n$ is the described situation possible? | For even $n$, a regular $n$-sided polygon can be divided into two equal polygons by a diagonal passing through its center, and then these two polygons can be cut in the same way. In addition, equal triangles with vertices on the circumscribed circle can be constructed on three sides of a regular $2k$-sided polygon. The... | 4forn>5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,490 |
[Mathematical logic (miscellaneous).] Case analysis $\quad]$
After a hockey game, Anton said that he scored 3 goals, and Ilya only one. Ilya said that he scored 4 goals, and Seryozha as many as 5. Seryozha said that he scored 6 goals, and Anton only two. Could it be that together they scored 10 goals, given that each ... | There are two cases.
1) Anton told the truth about himself, that is, he scored 3 goals. Then Seryozha lied about Anton, so he told the truth about himself, that is, he scored 6 goals. Therefore, Ilya lied about Seryozha and told the truth about himself, that is, he scored 4 goals. In this case, the boys scored a total... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,491 |
In one of the vertices of a hexagon, there is a gold coin, and there is nothing in the others. Koschei the Deathless is pining over the gold and every morning takes an arbitrary number of coins from one vertex, after which he immediately places six times more coins on the adjacent vertex. If by the end of some day all ... | Number the vertices of the hexagon, starting from the one where the coin lies, with consecutive natural numbers from 1 to 6. According to the condition, the difference between the number of coins in the odd vertices and the number of coins in the even vertices changes by a multiple of 7 each day.
At the initial moment... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,493 |
A set of several numbers, none of which are the same, has the following property: the arithmetic mean of some two numbers from this set is equal to the arithmetic mean of some three numbers from the set and is equal to the arithmetic mean of some four numbers from the set. What is the smallest possible number of number... | Let $C\left(a_{1}, \ldots, a_{k}\right)$ be the arithmetic mean of the numbers $\left(a_{1}, \ldots, a_{k}\right)$. Note that adding a number different from the arithmetic mean of a set changes the original arithmetic mean of the set.
Suppose that $(a, b, c, d)$ is a set of four numbers satisfying the condition, and $... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,496 |
The numbers $1, 2, \ldots, 100$ are written on a board. At each step, all numbers that do not have any divisors among the non-erased numbers, except for themselves, are erased simultaneously. For example, on the first step, only the number 1 is erased. Which numbers will be erased on the last step? | On the next step, after erasing the unit, all prime numbers (which have no divisors other than 1 and themselves) will be erased. Next, the numbers that have exactly two prime factors (not necessarily distinct) in their factorization will be erased, followed by the numbers that have exactly three prime factors, and so o... | 6496 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,497 |
In three cells of a grid sheet, numbers are written, while the other cells are empty. It is allowed to choose two numbers from different non-empty cells and write their sum in an empty cell; also, one can choose numbers $a, b, c$ from three different non-empty cells and write the number $a b+c^{2}$ in an empty cell. Pr... | Let the numbers $a, b, c$ be written down. Sequentially form $a+b, b+c, a+c,(a+b) c+a^{2},(b+c) a+b^{2},(c+$ $a) b+c^{2},(a+b) c+a^{2}+(b+c) a+b^{2}$ and $(a+b) c+a^{2}+(b+c) a+b^{2}+(c+a) b+c^{2}=(a+b+c)^{2}$.
Send a comment | (+b+)^2 | Algebra | proof | Yes | Yes | olympiads | false | 53,499 |
Commenting P...
Given natural numbers $a$ and $b$, where $a<1000$. Prove that if $a^{21}$ is divisible by $b^{10}$, then $a^{2}$ is divisible by $b$.
# | Suppose there exists a prime number $p$, which appears in the prime factorization of $a^2$ with an exponent smaller than in the factorization of $b$. That is, if $a$ is divisible by $p^k$ but not by $p^{k+1}$, and $ab$ is divisible by $p^m$ but not by $p^{m+1}$, then $m > 2k$, and thus $m \geq 2k + 1$. However, from th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 53,500 |
Vomienkov $C$.
In a company of six people, any five can sit at a round table such that any two neighbors are acquainted.
Prove that the entire company can also be seated at a round table such that any two neighbors are acquainted.
# | Note that each person in the company has at least three acquaintances. Indeed, if someone $X$ were acquainted with fewer than three, then by excluding one of his acquaintances from the company, we would end up with a group of five people in which $X$ has no more than one acquaintance, making it impossible to seat them ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,501 |
Shapovalov A.V.
A monkey becomes happy when it eats three different fruits. What is the maximum number of monkeys that can be made happy with 20 pears, 30 bananas, 40 peaches, and 50 tangerines? | Let's put the tangerines aside for now. There are $20+30+40=90$ fruits left. Since we feed no more than one tangerine to each monkey, each monkey will eat at least two of these 90 fruits. Therefore, there can be no more than $90: 2=45$ monkeys. Here's how we can satisfy 45 monkeys:
5 monkeys eat a pear, a banana, and ... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,502 |
Senderov B.A.
Integers $a, x_{1}, x_{2}, \ldots, x_{13}$ are such that $a=\left(1+x_{1}\right)\left(1+x_{2}\right) \ldots\left(1+x_{13}\right)=\left(1-x_{1}\right)\left(1-x_{2}\right) \ldots\left(1-x_{13}\right)$. Prove that $a x_{1} x_{2} \ldots x_{13}=0$. | If one of the numbers $x_{i}$ is equal to 0, the statement is obvious. If one of the $x_{i}$ is equal to $\pm 1$, then $a=0$, and the statement is also true. Otherwise, each product $\left(1+x_{i}\right)\left(1-x_{i}\right)=1-x_{i}^{2}$ is negative. Therefore, $a^{2}=(1$ $\left.+x_{1}\right)\left(1+x_{2}\right) \ldots\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 53,503 |
[ Examples and counterexamples. Constructions ] [ Regular polygons ]
Does there exist a non-right triangle inscribed in a circle of radius 1, for which the sum of the squares of the lengths of two sides is 4? | Consider the quadrilateral $ABCD$, which is half of a regular hexagon inscribed in a circle of radius 1 (see figure).
We have $AB=1, AC=\sqrt{3}, AB^{2}+AC^{2}=4$.
## Answer
It exists.
#... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,506 |
Berroov S.L.
One hundred integers are written in a circle. Each number is greater than the sum of the two numbers following it in a clockwise direction.
What is the maximum number of positive numbers that can be among those written?
# | Evaluation. Suppose two non-negative numbers stand next to each other. Then the number preceding them is greater than their sum, meaning it is positive. Similarly, the number before it is also positive, and so on. In the end, we get that all numbers are non-negative; but then the smallest of them cannot be greater than... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,508 |
Shapovalov A.V.
a) There are three identical large vessels. One contains 3 liters of syrup, another contains 20 liters of water, and the third is empty. You can pour all the liquid from one vessel into another or into the sink. You can choose two vessels and pour from the third into one of them until the levels of liq... | Let's denote the vessels with syrup, water, and the empty one as $A, B$, and $C$ respectively.
a) Pour 3 liters of water from $B$ to $C$, and then pour these 3 liters into the sink; repeat this operation several times until 5 liters of water remain in $B$, and in the last operation, pour the water from $C$ not into th... | N\not\equiv0\pmod{3}N>6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,510 |
Bakayev E.V.
We will call a grid polygon outstanding if it is not a rectangle and several copies of it can be used to form a similar polygon. For example, a corner made of three cells is an outstanding polygon (see figure).
 For example, a corner made of four cells (see fig.).
b) Consider such a corner made of $n$ cells, such that two copies of it can form a rectangle $2 \times n$. From such rectangles, a squ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 53,512 |
[ Divisibility of numbers. General properties ] [ Examples and counterexamples. Constructions ]
Natural numbers $A$ and $B$ are divisible by all natural numbers from 1 to 65. What is the smallest natural number that $A+B$ may not be divisible by? | From the condition, it follows that $A+B$ is divisible by all numbers from 1 to 65. This sum is also divisible by $66=2 \cdot 33$. However, $A+B$ is not necessarily divisible by 67. For example,
$A=65!, B=2 \cdot 65!, A+B=3 \cdot 65!$ which is not divisible by 67.
## Answer
Not divisible by 67. | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,513 |
The sum of nine different natural numbers is 200. Can we always choose four of them so that their sum is greater than 100?
# | Consider nine numbers: $28,25,24,23,22,21,20,19,18$. Their sum is 200, but the sum of the four largest of them is $28+25+24+23=100$. Therefore, the sum of any four numbers from these nine is no more than 100. A similar example: $27,26,24,23,22,21,20,19,18$.
## Problem | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,514 |
Can the products of all non-zero digits of two consecutive natural numbers differ by exactly 54 times?
# | For example, the numbers 299 and 300 have this property. Indeed, $2 \cdot 9 \cdot 9=54 \cdot 3$.
## Answer
They can. | 2\cdot9\cdot9=54\cdot3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 53,515 |
P.G. Zhenoarove
In a box, there are 111 balls: red, blue, green, and white. It is known that if you pull out 100 balls from the box without looking, there will definitely be four balls of different colors among them. What is the smallest number of balls that need to be pulled out without looking to ensure that there a... | There are no less than 12 balls of each color (otherwise, they could all end up among the 11 remaining in the box). Therefore, the number of balls of two colors is no more than
$111-24=87$. Hence, among any 88 balls, there will be balls of at least three colors.
87 balls are insufficient, for example, for the color ... | 88 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,516 |
Zhendarov R.G.
In a box, there are 100 balls: white, blue, and red. It is known that if 26 balls are drawn from the box without looking, there will definitely be 10 balls of the same color among them. What is the smallest number of balls that need to be drawn from the box, without looking, to ensure that there are 30 ... | Let's prove that 66 balls are sufficient. Suppose among them there are no more than 29 of each color. Then there are no less than $66-2 \cdot 29=$ 8 balls of each color and no less than $66-29=17$ of any two colors. But then there is a set of 8, 9, 9, which contradicts the condition.
65 balls are insufficient, for exa... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,517 |
Bakayev E.V.
Forty children were holding hands in a circle. Out of them, 22 were holding hands with a boy and 30 were holding hands with a girl. How many girls were in the circle?
# | $22+30=52$, so $52-40=12$ children held hands with both a boy and a girl. Therefore, $30-12=18$ children held hands only with girls. These 18 children held $18 \cdot 2=36$ girls' hands, and the other 12 held one girl's hand each, so the girls had a total of $36+12=48$ hands. Therefore, there were $48: 2=24$ girls.
## ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 53,518 |
Does there exist a function $f(x)$, defined for all real numbers, such that $f(\sin x)+f(\cos x)=\sin x$? | Let such a function exist. Then $f(\sin 0) + f(\cos 0) = \sin 0$, that is, $f(0) + f(1) = 0$. But $f(\sin \pi / 2) + f(\cos \pi / 2)$
$= \sin \pi / 2$, that is, $f(0) + f(1) = 1$. Contradiction.
## Answer
It does not exist. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 53,519 |
Bogdanov I.I.
There are 23 students in a class. During the year, each student in this class celebrated their birthday once, to which some (at least one, but not all) of their classmates came. Could it be that every two students in this class met at such celebrations the same number of times? (It is assumed that at eac... | Let's provide examples of how this could happen.
First example. Arrange the students in a circle. Suppose that on each student's birthday, all classmates except the next one in the clockwise direction came. Then, every two students $A$ and $B$ met at all celebrations except two: the one where $A$ did not come and the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 53,520 |
ranуoov G.
At the market, there is a man with many carpets. He agrees to exchange a carpet of size $a \times b$ for either a carpet of size $1 / a^{1} \times \frac{1}{b}$, or two carpets of sizes $c \times b$ and $a / c \times b$ (for each such exchange, the customer can choose the number $c$ themselves). The traveler... | Let's call a carpet big if all its sides are greater than 1, and small if all its sides are less than 1. Thus, initially, the traveler had one big carpet. We will prove that the total number of big and small carpets does not decrease; from this it follows that the described situation is impossible. For this, it is suff... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,521 |
The Unified State Exam (USE) in mathematics in the magical country of Oz is organized as follows. Each paper is independently checked by three teachers, and each gives 0 or 1 point for each problem. Then the computer finds the arithmetic mean of the scores for that problem and rounds it to the nearest integer. The poin... | It's easy to come up with a case where you get 4 points. To get 5 or more, there must be at least 10 units in total, but the teachers have only given 9 units in total. Therefore, it's impossible to get 5.
## Answer
4 points. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 53,522 |
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