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4. At the beginning of the year, the US dollar was worth 80 European cents. An expert predicted that over the course of the year, the euro's exchange rate against the ruble would increase by 8% (i.e., 1 euro would be able to buy 8% more rubles than at the beginning of the year), and the US dollar's exchange rate agains... | Solution. Let's choose the amount in rubles that will correspond to 1 euro (i.e., 100 European cents) at the end of the year. At the beginning of the year, the same amount corresponded to 108 European cents or $108 \cdot 100 / 80=135$ American cents. Consequently, at the end of the year, it will correspond to $135 \cdo... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,797 |
1. Find the sum of all roots of the equation:
$$
\begin{gathered}
\sqrt{2 x^{2}-2024 x+1023131}+\sqrt{3 x^{2}-2025 x+1023132}+\sqrt{4 x^{2}-2026 x+1023133}= \\
=\sqrt{x^{2}-x+1}+\sqrt{2 x^{2}-2 x+2}+\sqrt{3 x^{2}-3 x+3}
\end{gathered}
$$
(L. S. Korechkova) | Solution. Note that the radicands in the left part are obtained from the corresponding radicands in the right part by adding $x^{2}-2023 x+1023130=$ $(x-1010)(x-1013)$. Since all radicands are positive (it is sufficient to check for $x^{2}-x+1$ and for $\left.2 x^{2}-2024 x+1023131=2(x-506)^{2}+511059\right)$, the left... | -2023 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,799 |
2. There are 8 white cubes of the same size. Marina needs to paint 24 faces of the cubes blue, and the remaining 24 faces - red. After this, Katya assembles them into a $2 \times 2 \times 2$ cube. If the surface of the cube has an equal number of blue and red squares, Katya wins. If not, Marina wins. Can Marina paint t... | Answer: No.
Solution. Suppose Marina painted the cubes in some way, and Katya assembled a cube from them in some way. Let there be $a$ blue and $24-a$ red faces on the surface of the cube. Using the idea of so-called discrete continuity, we will show that Katya can gradually bring the cube to the desired appearance. N... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,800 |
3. Pasha and Igor are flipping a coin. If it lands on heads, Pasha wins; if tails, Igor wins. The first time the loser pays the winner 1 ruble, the second time - 2 rubles, then - 4, and so on (each time the loser pays twice as much as the previous time). At the beginning of the game, Pasha had a single-digit amount of ... | Solution. Let $n$ be the amount of money Pasha has become richer (and Igor poorer). Note that Pasha won the last game (otherwise, he would have lost more money than he gained in all previous stages). Therefore, the sequence of games can be divided into series, in each of which Pasha won the last game and lost all the o... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,801 |
4. A regular pyramid $S A B C$ (with base $A B C$) and the height $A H$ of the face $S A B$ are shown in an orthogonal projection on a plane, as shown in the figure. How can the image of the center of the sphere circumscribed around the pyramid be constructed using a compass and a straightedge?
(A. A. Tesler)
. The projection of $M$ is constructed as the midpoint of the projection of $AC$, and the projection of $N$ is constructed as the point dividing the ... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,802 |
5. Solve the equation $a^{b}+a+b=b^{a}$ in natural numbers.
(O. A. Pyayve, E. Yu. Voronetsky) | Answer: $a=5, b=2$.
Solution. If $a=1$ or $b=1$, there are no solutions.
If $b=2$, we get $2^{a}=a^{2}+a+1$. For $ab^{a}>a^{b}$
taking the logarithm and dividing by $ab$, we get:
$$
\frac{\ln \left(a+\frac{1}{a}\right)}{a}>\frac{\ln b}{b}>\frac{\ln a}{a}
$$
Let $f(x)=\frac{\ln (x)}{x}$. Note that $f(a)$ decreases ... | =5,b=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,803 |
6. There are 28 candies on the table. Petya considers some of them to be delicious. Vasya can, in one move, point to any set of candies and ask Petya how many of them are delicious. How can Vasya guaranteedly find all the delicious candies... (a) in 21 moves; (b) in 20 moves?
(A. A. Tselery, E. Yu. Voronechik)
# | # Solution.
a) Let's divide the candies into 7 groups of 4 each. In 3 moves, we can learn everything about these 4 candies $a, b, c, d$, by asking, for example, about the sets $\{a, c, d\}, \{b, c, d\}, \{a, b, c\}$.
If the answers to the first two questions are different, we will learn about candies $a$ and $b$, and... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,804 |
1. Tom and Jerry are chasing each other on a track shaped like an eight (see figure).
They run in the same direction at constant speeds. Initially, Jerry was directly above Tom. After 20 minutes, Tom was directly above Jerry, and neither of them had completed a full lap. At the moment when Jerry had completed exactly ... | Solution. While Jerry runs the small loop, Tom runs the large one; while Jerry runs the large loop, Tom runs both the large and small loops together. Let's denote the lengths of the large and small loops by $L$ and $l$ respectively. Then we get that $L: l = (L + l): L$. If we denote $L: l$ by $x$, we get: $x = 1 + \fra... | 30+10\sqrt{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,805 |
3. Prove that a rectangle $1 \times 10$ can be cut into 5 pieces and these pieces can be rearranged to form a square. | Solution. Let's cut the rectangle as shown in the figure.
It is easy to see that $r=10-3 \sqrt{10}$. Furthermore, since the small triangle is similar to the large one, we have $r: s=\sqrt{1... | \sqrt{10} | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 11,807 |
4. On a plane, $2 n+1$ points are marked, and no three points lie on the same line, and no four points lie on the same circle. Prove that there exists a circle passing through three of these points, inside which there are $n-1$ points and outside - also $n-1$. | Solution. Lemma: We can choose points $A$ and $B$ such that all other points lie in one half-plane relative to the line $A B$.
Proof of the lemma. Take a line such that all points lie in one half-plane relative to it. We will move this line parallel to itself towards the points until one of the points (call it $A$) li... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,808 |
1. Petya divided a circle into 7 parts with three straight lines and wrote 7 different integers in them so that the sums of the numbers on either side of each line were the same. One of the numbers is zero. Prove that some number is negative. | Solution. Let's consider three cases of the zero's position (a picture for each case). In the pictures, the sum of the yellow sectors equals the sum of the pink ones (since yellow + orange = pink + orange = half the sum of all numbers). We see that the first two cases are impossible (there are duplicate numbers), and i... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,810 |
5. How many solutions in natural numbers does the equation $(a+1)(b+1)(c+1)=2 a b c$ have? | Solution. Rewrite the equation as $(1+1 / a)(1+1 / b)(1+1 / c)=2$. Due to symmetry, it is sufficient to find all solutions with $a \leqslant b \leqslant c$. Then $(1+1 / a)^{3} \geqslant 2$, which means $a \leqslant(\sqrt[3]{2}-1)^{-1}<4$ and $a \in\{1,2,3\}$. In the case $a=1$, the inequality $2(1+1 / b)^{2} \geqslant... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,812 |
7. $f(x)$ is a linear function, and the equation $f(f(x))=x+1$ has no solutions. Find all possible values of the quantity $f(f(f(f(f(2022)))))-f(f(f(2022)))-f(f(2022))$. | Solution. Let $f(x)=k x+b$, then $f(f(x))=k(k x+b)+b=k^{2} x+k b+b$. The equation can have no solutions only when $k^{2}=1$, that is, for functions $x+b$ or $-x+b$, so the answer is either $(2022+5 b)-(2022+3 b)-(2022+2 b)=-2022$, or $(-2022+b)-(-2022+b)-2022=-2022$.
Answer: -2022.
Criteria. Up to 2 points will be de... | -2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,813 |
8. Let's call the efficiency of a natural number $n$ the fraction of all natural numbers from 1 to $n$ inclusive that have a common divisor with $n$ greater than 1. For example, the efficiency of the number 6 is $\frac{2}{3}$.
a) Does there exist a number with an efficiency greater than $80\%$? If so, find the smalles... | Solution. Let's move on to studying inefficiency (1 minus efficiency). From the formula for Euler's function, it follows that it is equal to $\frac{p_{1}-1}{p_{1}} \cdot \ldots \frac{p_{k}-1}{p_{k}}$, where $p_{1}, \ldots, p_{k}$ are all distinct prime divisors of $n$. Then, by adding a new prime factor, we can increas... | 30030 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,814 |
3. Does there exist a five-segment non-planar closed broken line, all segments of which are equal, and each two adjacent segments are perpendicular? | Solution. Let the broken line be $A B C D E$. Without loss of generality, we can assume that the length of each segment is 1. Then we can introduce a coordinate system in which three vertices have coordinates $A(0,1,0), B(0,0,0), C(1,0,0)$. Then the coordinates of the other two vertices are $D(1, a, b)$ and $E(c, 1, d)... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,815 |
5. How many triples of natural numbers $(a, b, c)$, forming an arithmetic progression ( $a<b<c$ ), exist such that the numbers $a b+1, b c+1$ and $c a+1$ are perfect squares | Solution: Infinitely many.
Let $(2+\sqrt{3})^{n}=A_{n}+B_{n} \sqrt{3}\left(A_{n}, B_{n} \in \mathbb{N}\right)$. Then $(2-\sqrt{3})^{n}=A_{n}-B_{n} \sqrt{3}$ and $A_{n}^{2}-3 B_{n}^{2}=1$. Set $a=2 B_{n}-A_{n}, b=2 B_{n}, c=2 B_{n}+A_{n}$. Then $a, b, c$ form an arithmetic progression, $a b+1=\left(A_{n}-B_{n}\right)^{... | Infinitely\many | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,816 |
3. In a convex pentagon $A B C D E \angle A=60^{\circ}$, and the other angles are equal to each other. It is known that $A B=6, C D=4, E A=7$. Find the distance from point $A$ to line $C D$. | Solution. It is clear that all other angles are 120 degrees, so $A B \| D E$ and $B C \| A E$. Also, $D E=2$ and $B C=3$. The desired distance is the height of an equilateral triangle with side 9 (to which the original pentagon can be completed), which is $\frac{9 \sqrt{3}}{2}$. | \frac{9\sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,818 |
3. Calculate the area of the set of points on the coordinate plane that satisfy the inequality $(y+\sqrt{x})\left(y-x^{2}\right) \sqrt{1-x} \leqslant 0$. | Solution. The left side makes sense only for $0 \leqslant x \leqslant 1$. In this case, it is required that $y+\sqrt{x}$ and $y-x^{2}$ have different signs (or one of them equals zero), or that $x$ equals 1. If we exclude the case $x=1$ (which gives a zero area), we are left with a part of the plane bounded by the segm... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 11,819 |
4. $N$ children met. Some of them gave gifts to some others (one could not give more than one gift to another). It turned out that everyone received an equal number of gifts, although they all gave different numbers of gifts (including possibly someone giving no gifts at all). For which $N>1$ is this possible? | Solution. Everyone gave a different number of candies, and no one gave candies to themselves, so all quantities from 0 to $N-1$ were given. Therefore, the total number of candies given is $\begin{gathered}N(N-1) \\ 2\end{gathered}$, and this number must be divisible by $N$. This is only possible for odd $N \stackrel{2}... | N | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,820 |
5. A natural number $n$ is called cubish if $n^{3}+13 n-273$ is a cube of a natural number. Find the sum of all cubish numbers. | Solution. If $0<13n-273<13\cdot21$, so it remains to check all other numbers.
If $n=21$, then $13n-273=0$, so 21 is cubic. For $n-3n^{2}+3n-1$, the number $n$ will not be cubic (i.e., for $8<n<21$).
If $n=8$, then $13n-273=-169=-3\cdot8^{2}+3\cdot8-1$, so it is cubic. For $n \leqslant 5$, the expression $n^{3}+13n-27... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,821 |
2. $P$ Let $f(x)=x^{3}+9 x^{2}+27 x+24$. Solve the equation $f(f(f(f(x))))=0$.
---
Note: The translation preserves the original formatting and structure of the text. | Solution. Note that $f(x)=(x+3)^{3}-3$.
Therefore, $f(f(x))=f\left((x+3)^{3}-3\right)=\left(\left[(x+3)^{3}-3\right]+3\right)^{3}-3=(x+3)^{9}-3$;
similarly, we obtain that $f(f(f(x)))=(x+3)^{27}-3$ and $f(f(f(f(x))))=(x+3)^{81}-3$.
Thus, we need to solve the equation $(x+3)^{81}-3=0$; its root is $-3+{ }^{81} \sqrt{... | -3+{}^{81}\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,823 |
3. Prove that from a circle of radius 1, four parts can be cut out, from which a rectangle $1 \times 2.5$ can be formed. The parts can be rotated and flipped. | # Solution.
Method 1. A rectangle with sides 1 and $\sqrt{3}$ can be cut out from the circle, as well as three trapezoids (see figure). From these pieces, a rectangle with sides 1 and $\sqrt{3} + 1 / 2 + (\sqrt{3} / 2 - 1 / 2) = (3 \sqrt{3}) / 2 > 2.5$ can be formed.
. At least one of the lines containing the diagonals of the square with cent... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,825 |
6. Pavel came up with a new way of adding numbers: he calls the "pavlosum" of numbers $x$ and $y$ the value of the expression $x \# y=(x+y) /(1-x y)$, if it is defined. One day he "added" numbers $a$ and $b$ in his way, "added" $c$ to them, and then "subtracted" $d$ from the result. At the same time, his friend "added"... | Solution. We need to check that ((a\#b)\#c)\#d=a\#(b\#(c\#d)). This can be proven by direct computation, or one can notice that if $x$ and $y$ are the tangents of angles A and B, then $x \# y$ is the tangent of the angle $\mathrm{A}+\mathrm{B}$, and therefore the required property follows from the fact that $\operatorn... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,827 |
1. Pasha draws dots at the intersections of the lines on graph paper.
He likes it when four dots form a figure resembling a "kite," as shown on the right (the kite must be of this exact shape and size, but can be rotated). For example, the 10 dots shown in the second image form only two kites. Is it possible to draw a... | Solution. For example, like this. Here there are 21 points and 24 snakes (6 snakes in each direction).
 | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,828 |
3. Two players play a game. They take turns naming four-digit numbers that have no zeros in their notation, and the sum of the digits is divisible by 9. At the same time, each subsequent number must start with the same digit that the previous number ends with, for example: 3231 - 1539 - 9756 - $6561 \ldots$ Numbers can... | Solution. The first player wins. One possible strategy is as follows. He calls the number 9999, and then in response to any number $\overline{A B C D}$ called by the second player, he calls the number $\overline{D C B A}$ (the same number "backwards"). Note that after this, the second player will again have to call a n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,830 |
4. Flint has five sailors and 60 gold coins. He wants to distribute them into wallets, and then give the wallets to the sailors so that each gets an equal number of coins. But he doesn't know how many sailors will be alive by the time of the distribution. Therefore, he wants to distribute the coins in such a way that t... | Solution. Answer: 9 wallets. Example: $12,12,8,7,6,5,4,3,3$.
We will prove that 8 wallets are insufficient.
1) Note that each wallet should contain no more than 12 coins. Therefore, 15 coins must be made up of at least two wallets. This means that when we divide 8 wallets among four pirates, each should receive two w... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,831 |
1. In a certain triangle, the sum of the tangents of the angles turned out to be 2016. Estimate (at least to the nearest degree) the magnitude of the largest of its angles. | Solution. One of the tangents must exceed 600. This is only possible for an angle very close to $90^{\circ}$. We will prove that it exceeds $89.5^{\circ}$. This is equivalent to the statement that $\operatorname{tg} 0.5^{\circ}>$ $1 / 600$.
Let's start with the equality $\sin 30^{\circ}=1 / 2$. Note that $\sin 2 x=2 \... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,832 |
3. Find all natural numbers $n$ for which $2^{n}+n^{2016}$ is a prime number. | Solution. Let's consider three cases.
- If $n$ is even, then the given number is also even (and greater than two for $n>0$).
- If $n$ is odd and not divisible by 3, then $2^{n}$ gives a remainder of 2 when divided by 3, and $n^{2016}=\left(n^{504}\right)^{4}$ gives a remainder of 1 when divided by 3, so the sum is div... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,834 |
7. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit or all three numbers contain different digits.
The complexity of... | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,835 |
1. Do there exist three different digits $A, B, C$, such that $\overline{A B C}, \overline{C B A}, \overline{C A B}$ are squares of natural numbers? (The bar over the digits denotes the number formed by these digits in the given order.) | Solution. Yes. These are the digits $A=9, B=6, C=1$. In this case $961=31^{2}, 169=13^{2}, 196=14^{2}$. | A=9,B=6,C=1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,836 |
2. Each cell of a $100 \times 100$ board is painted blue or white. We will call a cell balanced if among its neighbors there are an equal number of blue and white cells. What is the maximum number of balanced cells that can be on the board? (Cells are considered neighbors if they share a side.) | Solution. Cells lying on the border of the board but not in the corner cannot be equilibrium cells, since they have an odd number of neighbors (three). There are $4 \cdot 98=392$ such cells.
All other cells can be made equilibrium, for example, with a striped coloring (the first row is blue, the second row is white, t... | 9608 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,837 |
3. On the sides $AB$ and $AC$ of triangle $ABC$, points $M$ and $N$ are marked respectively, such that $AM = AN$. Segments $CM$ and $BN$ intersect at point $O$, and $BO = CO$. Prove that $ABC$ is isosceles. | Solution. Suppose this is incorrect, for example, that $A B>A C$. Mark a point $D$ on side $A B$ such that $A D=A C$. By symmetry, segments $M C$ and $N D$ intersect at some point $P$, and $P M=P N$. From symmetry, it also follows that $C M=D N$.
 is a rectangle. Let's define the maximum possible lengths of its sides.
The vertical side of the first rectangle is shorter than the horizontal one, so it is less than $\sqrt{2020}$, which is less than 45; the same is true for the horizontal side o... | 1764 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,839 |
5. In the game "set," all possible four-digit numbers consisting of the digits $1,2,3$ (each digit appearing exactly once) are used. It is said that a triplet of numbers forms a set if, in each digit place, either all three numbers contain the same digit, or all three numbers contain different digits.
For example, the... | Solution. Note that for any two numbers, there exists exactly one set in which they occur. Indeed, the third number of this set is constructed as follows: in the positions where the first two numbers coincide, the third number has the same digit; in the position where the first two numbers differ, the third number gets... | 1080 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,840 |
2. On a sheet of lined paper, two rectangles are outlined. The first rectangle has a vertical side shorter than the horizontal side, while the second has the opposite. Find the maximum possible area of their common part, if the first rectangle contains 2015 cells, and the second - 2016. | Solution. The common part of these two rectangles (if it is non-empty) is a rectangle. Let's determine the maximum possible lengths of its sides.
The vertical side of the first rectangle is shorter than the horizontal one, so it is less than $\sqrt{2015}$, which is less than 45; the same is true for the horizontal sid... | 1302 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,841 |
5. Each cell of a $100 \times 100$ board is painted blue or white. We call a cell balanced if among its neighbors there are an equal number of blue and white cells. For which $n$ can the board be painted so that there are exactly $n$ balanced cells? (Cells are considered neighbors if they share a side.) | Solution. Note that the minimum number of equilibrium cells is 0 (for example, if all cells are white).
Let's find the maximum number of equilibrium cells. Cells lying on the edge of the board but not in a corner cannot be equilibrium, as they have an odd number of neighbors (three). There are $4 \cdot 98 = 392$ such ... | from0to9608inclusive | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,844 |
1. A motorcyclist set out from point $A$ with an initial speed of 90 km/h, uniformly increasing his speed (that is, over equal time intervals, his speed increases by the same amount). After three hours, the motorcyclist arrived at point $B$, passing through $C$ along the way. After that, he turned around and, still uni... | Solution. In 5 hours, the speed changed from 90 km/h to 110 km/h, so the acceleration is 4 km/h$^2$. From $A$ to $B$ the distance is
$$
90 \cdot 3+\frac{4}{2} \cdot 3^{2}=270+18=288(\text{km})
$$
from $B$ to $C-$
$$
110 \cdot 2-\frac{4}{2} \cdot 2^{2}=220-8=212(\text{km})
$$
And the required distance is 76 km. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,845 |
3. Prove that for all positive numbers $a$ and $b$, the inequality $\left(a^{2018}+b^{2018}\right)^{2019}>\left(a^{2019}+b^{2019}\right)^{2018}$ holds. | Solution. Without loss of generality, $a \geqslant b$. Let 2018 be denoted by $n$. Then the inequality can be rewritten as
$$
\begin{aligned}
a^{n(n+1)} & +\binom{n+1}{1} a^{n^{2}} b^{n}+\ldots+\binom{n+1}{k} a^{n(n+1-k)} b^{n k}+\ldots+b^{n(n+1)}> \\
& >a^{(n+1) n}+\binom{n}{1} a^{(n+1)(n-1)} b^{n+1}+\ldots+\binom{n}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 11,846 |
4. On a plane, five points are marked, any three of which form a triangle with an area of at least 2. Prove that there will be 3 points forming a triangle with an area of at least 3. | Solution. Let's denote the points as $A, B, C, X, Y$, and suppose that $\triangle ABC$ has the largest area $S$ among all triangles with vertices at these points. We need to prove that the areas of all other triangles cannot be greater than $\frac{2}{3} S$. Suppose this is not the case, then all areas are between $S$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,847 |
6. A natural number $n$ is called cubish if $n^{3}+13 n-273$ is a cube of a natural number. Find the sum of all cubish numbers. | Solution. If $021$, so it remains to check all other numbers.
If $n=21$, then $13 n-273=0$, so 21 is cubic. For $n-3 n^{2}+3 n-1$, the number $n$ will not be cubic (i.e., for $8<n<21)$
If $n=8$, then $13 n-273=-169=-3 \cdot 8^{2}+3 \cdot 8-1$, so it is cubic. For $n \leqslant 5$, the expression $n^{3}+13 n-273$ will ... | 29 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,848 |
4. Prove that from a circle of radius 1, three parts can be cut out, from which a rectangle $1 \times 2.4$ can be formed. The parts can be rotated and flipped. | Solution. First, place a rectangle in the circle with vertices $( \pm \sqrt{3} / 2, \pm 1 / 2)$. Its sides are 1 and $\sqrt{3}$. Next, draw two small rectangles: one with vertices ( $\pm 1 / 2, 1 / 2$ ) and ( $\pm 1 / 2, \sqrt{ } 3 / 2$ ), and the other, symmetric to the first. The sides of each of these rectangles are... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,851 |
5. Let $a$ and $n$ be natural numbers, and it is known that $a^{n}-2014$ is a number with $2014$ digits. Find the smallest natural number $k$ such that $a$ cannot be a $k$-digit number. | Solution. Let $a$ be a $k$-digit number, then $10^{k-1} \leq a2013$, i.e., $k>45$.
We will check for $k$ starting from 46, whether the condition of the absence of an integer in the specified interval is met. We will find that
(for $k=46$) $2013 / 462013$ );
(for $k=47$) $2013 / 472013$ );
(for $k=48$) $2013 / 48201... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,852 |
6. Pavel came up with a new way of adding numbers: he calls the "pavlosum" of numbers $x$ and $y$ the value of the expression $x \# y=(x+y) /(1-x y)$, if it is defined. One day, he "added" numbers $a$ and $b$ in his way and then "added" $c$ to them, while he asked a friend to "add" numbers $b$ and $c$ and then "add" $a... | Solution. We need to check that $a \#(b \# c)=(a \# b) \# c$. This can be proven by direct computation, or one can notice that if $x$ and $y$ are the tangents of angles A and B, then $x$ tuu - the tangent of the angle $\mathrm{A}+\mathrm{B}$, and therefore the required property follows from the fact that $\operatorname... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,853 |
1. In a certain language, there are 3 vowels and 5 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language? | 1. The language has $3 \cdot 5=15$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 30 different syllables. The total number of two-syllable words is $30 \cdot 30=900$ | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,854 |
3. Masha and Lena left home and went to the store for ice cream. Masha walked faster and got to the store in 12 minutes. Spending 2 minutes buying the ice cream, she headed back. After another 2 minutes, she met Lena. Walking a bit more, Masha finished her ice cream and, deciding to buy another one, turned around and w... | 3. Masha covers $1 / 6$ of the entire distance in 2 minutes, which means Lena covered $5 / 6$ of the distance by the time they met, and it took her 16 minutes. Therefore, $1 / 6$ of the distance takes Lena $16: 5=3 \frac{1}{5}$ minutes, i.e., 3 minutes and 12 seconds. For the entire distance, she would need 19 minutes ... | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,856 |
4. Six schoolchildren decided to plant 5 trees in the schoolyard. It is known that each tree was planted by a different number of schoolchildren and each schoolchild participated in planting the same number of trees. Could this have happened? | 4. Yes, this is possible. Here is an example of distributing trees among schoolchildren:
| | D1 | D2 | D3 | D4 | D5 |
| :--- | :---: | :---: | :---: | :---: | :---: |
| Schoolchild 1 | + | + | | | + |
| Schoolchild 2 | | + | | + | + |
| Schoolchild 3 | | | + | + | + |
| Schoolchild 4 | | | + | + | + |
| Schoo... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,857 |
5. In a flat world, there are two rectangular islands. The coastal waters of each island are considered to be the part of the sea that is no more than 50 km away from the shore. Can it happen that the area of the first island is larger than that of the second, while the area of the coastal waters of the second island i... | 5. Yes, this is possible. For example, the area of coastal waters around an island that is 1 km $\times 1000$ km is greater than that around an island that is 100 km $\times 100$ km. (In the first case, the coastal waters include strips $50 \times 1000$ km, with a total area of $100000 \mathrm{km}^{2}$. In the second c... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,858 |
6. Anya, Galia, Dasha, Sonya, and Lena arrived at the "Formula of Unity" camp from different cities: Kursk, Vologda, Novorossiysk, Petrozavodsk, and Cheboksary. Introducing themselves to other members of the team, they shared the following information. Sonya and Dasha have never been to Kursk. Galia and Sonya were in t... | 6. 7) Anya and Sonya are not from Cheboksary, Galia and Lena are also not from Cheboksary. Therefore, Dasha lives in Cheboksary.
2) Galia and Sonya are not from Novorossiysk, Anya is not from Novorossiysk, so Lena lives there.
3) Galia and Sonya are not from Vologda, so Anya lives there.
4) Sonya (and Dasha) are not fr... | Anya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,859 |
3. Three students wrote a two-digit number on the board, each of which is a perfect square. It turned out that if you "glue" them into one six-digit number, it is also a square of a natural number. Find all such six-digit numbers. | Solution. Let the two-digit squares be $x^{2}, y^{2}$, and $z^{2}$, then $x, y, z \in\{4,5,6,7,8,9\}$. The six-digit number is denoted by $t^{2}(t>0)$. Then
$$
t^{2}=10000 x^{2}+100 y^{2}+z^{2}=(100 x)^{2}+(10 y)^{2}+z^{2}
$$
Let $t=100 x+k$. Obviously, $k \in \mathbb{N}$, since $t^{2}>(100 x)^{2}$. Then
$$
\begin{a... | 166464=408^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,861 |
5. Two people play such a game. The first person thinks of 8 real numbers (not necessarily distinct) and writes down all their pairwise sums in any order (some of them may coincide). The second person has to determine the original numbers from the 28 sums provided. Can the second person always guarantee to do this? | Solution. No. For example, the following two sets of numbers cannot be distinguished: $1,5,7,9,12,14,16,20$ and $2,4,6,10,11,15,17,19$.
Or such: $-1,-1,-1,1,0,2,2,2$ and $-2,0,0,0,3,1,1,1$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,862 |
5. Let all angles of triangle $ABC$ be less than $120^{\circ}$ and $AB \neq AC$. Consider a point $T$ inside the triangle such that $\angle B T C=\angle C T A=\angle A T B=120^{\circ}$. Let the line $B T$ intersect side $A C$ at point $E$, and the line $C T$ intersect side $A B$ at point $F$. Prove that lines $E F$ and... | Solution. 1) Suppose that $E F \| B C$. Then, by Thales' theorem, $\frac{A F}{F B}=\frac{A E}{E C}$. Let $D$ be the intersection point of the lines $A T$ and $B C$, then $\angle B T D=180^{\circ}-120^{\circ}=\angle C T D$, which means $T D$ is the angle bisector in $\triangle T B C$. Similarly, $T E$ is the angle bisec... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,863 |
3. Prove that from a circle of radius 1, five pieces can be cut which can be used to form a rectangle $1 \times 2.7$. The pieces can be rotated and flipped. | # Solution.
Method 1. A hexagon and three trapezoids can be cut from the circle and arranged as shown in the figure. This results in a rectangle with sides 1 and $2+2 \cdot(\sqrt{3} / 2-1 / 2)=1+\sqrt{3}>2.7$
![](https://cdn.mathpix.com/cropped/2024_05_06_a5f438bf3e619f142877g-3.jpg?height=350&width=796&top_left_y=110... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,865 |
4. Does there exist a triangular pyramid with a height of 60, the height of each lateral face drawn to the side of the base equal to 61, and the perimeter of the base equal to 62? | Solution. Since the heights of the lateral faces are the same, the distances from the projection of the vertex to the sides are also the same and equal to 11, i.e., the radius of the inscribed circle in the base is 11.
By the known formula, the area of the triangle is $62 \cdot 11 / 2 = 341$. At the same time, the are... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,866 |
6. Pavel came up with a new way of adding numbers: he calls the "pavlosum" of numbers $a$ and $b$ the value of the expression $a \# b=(a+b) /(1-a b)$, if it is defined. Just like in ordinary arithmetic, multiplication by a natural number Pavel understands as adding the same number of addends: $a @ b=((a \# a) \# a)$...... | Solution. In Pavel's algebra:
$x @ y=x \# x \# \ldots \# x(y \text{ times})$,
$y @ x=y \# y \# \ldots \# y(x \text{ times})$.
Since $(A+B) /(1-AB)=\tan(\arctan(A)+\arctan(B))$, then $A \# B=\tan(\arctan(A)+\arctan(B))$.
Therefore, $x @ y=\tan(y \cdot \arctan(x))$,
$y @ x=\tan(x \cdot \arctan(y))$.
If $x @ y=y @ x... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,867 |
1. Petya prints five digits on the computer screen, none of which are zeros. Every second, the computer removes the initial digit and appends to the end the last digit of the sum of the remaining four digits. (For example, if Petya enters 12345, after one second it will become 23454, then 34546, and so on. However, he ... | Answer: 2.
Solution. The record 00000 cannot appear on the screen, as it can only result from 00000. A record with four zeros and one also cannot appear, since in that case, the last digit would not equal the remainder of the division of the sum of the first four by 10.
However, a sum of digits equal to 2 is possible... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,868 |
4. The territory of the Thirtieth Kingdom consists of all integers. A principality will be called a set of the form $\{a k+b \mid k \in \mathbb{Z}\}$, where $a \neq 0$ and $b$ are some integers (that is, an infinite arithmetic progression in both directions). The tsar wants to divide the entire territory of the kingdom... | Solution. Yes. We will separately partition even and odd numbers, then we need to split the progression without one point twice. We will show how to do this for odd numbers: place the odd number $x$ in the principality $s$ if $x-3$ is divisible by $2^s$, but not by $2^{s+1}$. Similarly for even numbers.
Criteria. If t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 11,869 |
5. A race on an unpredictable distance is held as follows. On a circular running track, two points $A$ and $B$ are randomly selected (using a spinning arrow), after which the athletes run from $A$ to $B$ along the shorter arc. A spectator has bought a ticket to the stadium and wants the athletes to run past his seat (t... | Solution. Identify each point on the path with its distance to the viewer clockwise. Then pairs $(A, B)$ can be identified with pairs of numbers from $[0,1)$ (when measured in kilometers). In this case, the probability that $(A, B)$ belongs to some subset of $[0,1) \times[0,1)$ is equal to the area of this subset. We a... | \frac{1}{4} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,870 |
6. On a plane, an equilateral triangle and three circles with centers at its vertices are drawn, and the radius of each circle is less than the height of the triangle. A point on the plane is painted yellow if it lies inside exactly one of the circles; green if inside exactly two; blue if inside all three. It turns out... | Solution. We will prove that the side is greater. Let $r_{1}, r_{2}, r_{3}$ be the radii of the circles, $a$ be the side of the triangle. We need to prove that $\left(r_{1}+r_{2}-a\right)+\left(r_{2}+r_{3}-a\right)+\left(r_{3}+r_{1}-a\right)<a$, that is, $r_{1}+r_{2}+r_{3}<2 a$. Then the sum of the areas of the circles... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,871 |
1. In a certain language, there are 5 vowels and 7 consonants. A syllable can consist of any vowel and any consonant in any order, and a word can consist of any two syllables. How many words are there in this language? | 1. The language has $5 \cdot 7=35$ syllables of the form "consonant+vowel" and the same number of syllables of the form "vowel+consonant", making a total of 70 different syllables. The total number of two-syllable words is $70 \cdot 70=4900$. | 4900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,872 |
3. On the table lie three candies. Anya and Lena have a bag with an unlimited number of candies, and they are playing a game. Each of them, on their turn, adds some number of candies from the bag to the table, but they cannot add more candies than are already on the table. The girls take turns, with Anya starting. The ... | 3. Note that the girl who, on her turn, has 1007 candies on the table will definitely lose. Indeed, after her move, there will be between 1008 and 2014 candies on the table, and her opponent will be able to take 2015 candies in any case. Therefore, the girl who leaves 1007 candies on the table after her move will win. ... | Anya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,873 |
5. In a flat world, there are two rectangular islands. The coastal waters of each island are considered to be the part of the sea that is no more than 50 km away from the shore. Can it happen that the perimeter of the first island is greater than that of the second, while the area of the coastal waters of the second is... | 5. No, this is impossible. The coastal waters consist of four rectangular strips along the shores and four quarters of a circle with a radius of 50 km. The area of the circle quarters is always the same, while the total area of the strips is equal to the perimeter of the rectangle multiplied by 50 km. Therefore, as the... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,874 |
1. In a math club, 20 people gathered, among them there are exactly 49 pairs of people who knew each other before the start of the classes. Prove that someone knew no more than 4 participants.
(S. S. Korechkova) | Solution. We will prove by contradiction: assume that everyone knew at least 5 people. Then, before the classes started, there were at least $\frac{20 \cdot 5}{2}=50$ pairs of acquaintances. Contradiction.
Criteria. The problem is solved under the assumption that each person knew exactly 5 people - 5 points. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,876 |
2. There are 81 squares of the same size. Arrange them into two rectangles so that their perimeters differ by a factor of 2. No squares should be left over.
(L. S. Korechkova) | Answer: $3 \times 19$ and $3 \times 8$ or $3 \times 23$ and $1 \times 12$.
Solution. The sum of their areas is 81 ( $57+24$ in the first option and $69+12$ in the second), and their perimeters indeed differ by a factor of 2: $2 \cdot(3+8)=22,2 \cdot(3+19)=44$ or $2 \cdot(1+12)=26,2 \cdot(3+23)=52$.
Criteria. A correc... | 3\times193\times8or3\times231\times12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,877 |
3. Two two-digit numbers are written on the board. Andrey multiplied them, getting a four-digit number with the first digit 2. Pasha added them and got a three-digit number. If the first digit is erased from Andrey's number, Pasha's number is obtained. What numbers were written?
(l. S. Korechkova) | Solution. If the numbers $x$ and $y$ were recorded, then by the condition $xy = 2000 + x + y$, which means $xy - x - y + 1 = 2001$, or $(x-1)(y-1) = 2001$. Since $2001 = 3 \cdot 23 \cdot 29$ and $x, y$ are two-digit numbers, then $x-1$ and $y-1$ can be $23, 29, 23 \cdot 3$ or $29 \cdot 3$. Therefore, $x, y \in \{24, 30... | {24,88} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,878 |
4. Given an isosceles triangle $A B C$, where $\angle A=30^{\circ}, A B=A C$. Point $D$ is the midpoint of $B C$. On segment $A D$, point $P$ is chosen, and on side $A B$, point $Q$ is chosen such that $P B=P Q$. What is the measure of angle $P Q C ?$ (S. S. Korechkova) | Answer: $15^{\circ}$.
Solution. Since $D$ is the midpoint of the base of the isosceles triangle, $A D$ is the median, bisector, and altitude of the triangle. Draw the segment $P C$. Since $\triangle P D B = \triangle P D C$ (by two sides and the right angle between them), $P C = P B = P Q$, meaning that all three tria... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,879 |
5. A few years ago, in the computer game "Minecraft," there were 9 different pictures (see the figure): one horizontal picture measuring $2 \times 1$ and $4 \times 2$, one square picture measuring $2 \times 2$, and two each of pictures measuring $1 \times 1$, $4 \times 3$ (horizontal), and $4 \times 4$. In how many way... | Answer: 896.
Solution. We will say that two paintings are in different columns if no block of the first painting is in the same column as any block of the second. It is clear that the $4 \times 4$ paintings are in different columns from each other and from the $4 \times 3$ paintings regardless of their placement. Thus... | 896 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,880 |
1. In each cell of a $100 \times 100$ table, a natural number was written. It turned out that each number is either greater than all its neighbors or less than all its neighbors. (Two numbers are called neighbors if they are in cells sharing a common side.) What is the smallest value that the sum of all the numbers can... | Solution. Let's divide the board into dominoes. In each domino, the numbers are different, so their sum is at least $1+2=3$. Then the total sum of the numbers on the board is at least 15000. This estimate is achievable if we alternate ones and twos in a checkerboard pattern. | 15000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,881 |
2. A natural number is written on the board. Every minute, the following operation is performed: if it contains two identical digits, one of them is erased; if all digits are different, the entire number is erased and replaced with a number three times larger. For example, from the number 57, you can get $57 \rightarro... | Solution. $25 \rightarrow 75 \rightarrow 225 \rightarrow 25$ or $75 \rightarrow 225 \rightarrow 25 \rightarrow 75$. | 25arrow75arrow225arrow25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,882 |
3. The pond has a square shape. On the first frosty day, the part of the pond that is no more than 10 meters away from the nearest point on the shore froze. On the second day, the part no more than 20 meters away froze, on the third day, the part no more than 30 meters away, and so on. On the first day, the area of ope... | Solution. Note that the larger the side of the pond, the smaller the percentage that will freeze on the first day. If the side of the pond is 100 m, then 36% will freeze on the first day, and if the side is 120 m, then on the first day, $1-\frac{100^{2}}{120^{2}}=11 / 36<1 / 3$ of the pond's area will freeze. Therefore... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,883 |
5. In 100 bags, there are 2018 candies, and no two bags contain the same number of candies, and there are no empty bags. Some bags may be inside other bags (in this case, a candy in an inner bag is also considered to be in the outer bag). Prove that there is a bag that contains a bag with a bag inside. | Solution. Let's estimate the total number of occurrences of all candies in all bags. Each of the 100 bags has a different number of occurrences, so there are at least $1+$ $2+\cdots+100=5050$. This means that for some candy, there are at least three occurrences, that is, it lies in at least three bags. This means that ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,884 |
1. Find all such numbers $k$ for which
$$
(k / 2)!(k / 4)=2016+k^{2}
$$
The symbol $n!$ denotes the factorial of the number $n$, which is the product of all integers from 1 to $n$ inclusive (defined only for non-negative integers; $0!=1$). | Solution. Note that the left side makes sense only for even values of $k$. We directly verify that $k=2,4,6,8,10$ do not work, while $k=12$ gives a correct equality.
With each further increase of $k$ by 2, the expression $(k / 2)!$ increases by at least 7 times, i.e., the left side grows by more than 7 times. At the s... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,885 |
4. On the coordinate plane, an isosceles triangle $A B C$ was drawn: $A B=2016, B C=$ $A C=1533$, with vertices $A$ and $B$ lying on nodes on the same horizontal line. Determine how many nodes lie within the triangle $A B C$ (including nodes lying on the sides). A node is a point on the coordinate plane where both coor... | Solution. Note that $1533^{2}-1008^{2}=(1533-1008)(1533+1008)=525 \cdot 2541=21 \cdot 25 \cdot 7 \cdot 363=$ $7 \cdot 3 \cdot 5^{2} \cdot 7 \cdot 3 \cdot 11^{2}=(7 \cdot 3 \cdot 5 \cdot 11)^{2}=1155^{2}$. Therefore, the height of the triangle is 1155.
We see that the GCD of 1155 and 1008 is 21. This means that there a... | 1165270 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,888 |
5. On a plane, there are 100 rectangles, the sides of which are parallel to the coordinate axes. Each intersects with at least 90 others. Prove that there exists a rectangle that intersects with all. | Solution. Project all rectangles onto the $x$-axis; as a result, each of them will turn into a segment $\left[a_{i}, b_{i}\right](i=1, \ldots, 100)$.
Let $A$ be the greatest starting coordinate, and $B$ be the smallest ending coordinate (i.e., $A=\max a_{i}, B=\min b_{i}$). It is not essential for the further solution... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 11,889 |
4. Let's call a rectangular parallelepiped typical if all its dimensions (length, width, and height) are different. What is the smallest number of typical parallelepipeds into which a cube can be cut? Don't forget to prove that this is indeed the smallest number. | Solution. A cube can be cut into four typical parallelepipeds. For example, a cube $5 \times 5 \times 5$ can be cut into parallelepipeds $5 \times 3 \times 1, 5 \times 3 \times 4, 5 \times 2 \times 1, 5 \times 2 \times 4$.
. At some point, it flies out in a random straight direction, and each time it reaches a wall, it turns $60^{\circ}$ and continues flying in a straight line (see figure). Can it happen tha... | Solution. Construct a triangle $A X C$ with $A C$ as the base and the opposite angle $60^{\circ}$ such that the fly initially flies along the ray $A X$. It is not difficult to show that the total path traveled by the fly is equal to $A X + X C$. Then, $X$ traces a certain arc of a circle, and the path length reaches it... | 5\sqrt{6}>12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,895 |
2. A company produces a certain product in the amount of $x_{1}$ in January, $x_{2}$ in February, $\ldots, x_{12}$ in December. The average production of the product from the beginning of the year is calculated as follows:
$x_{1}=x_{1}, \quad \bar{x}_{2}=\frac{1}{2}\left(x_{1}+x_{2}\right), \quad \bar{x}_{3}=\frac{1}{... | Solution. Note that
\[
\begin{aligned}
\bar{x}_{k+1} & =\frac{1}{k+1}\left(x_{1}+x_{2}+\ldots+x_{k+1}\right)=\frac{1}{k+1}\left(k \bar{x}_{k}+x_{k+1}\right) \\
x_{k+1}-\bar{x}_{k+1} & =k\left(\bar{x}_{k+1}-\bar{x}_{k}\right)
\end{aligned}
\]
In total, we have found that the average production increases (decreases) wh... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,896 |
5. On a plane, five points are marked, any three of which form a triangle with an area of at least 2. Prove that there will be 3 points forming a triangle with an area of at least 3. | Solution. Let's denote the points as $A, B, C, X, Y$, and suppose that $\triangle ABC$ has the largest area $S$ among all triangles with vertices at these points. We need to prove that the areas of all other triangles cannot be greater than $\frac{2}{3} S$. Suppose this is not the case, then all areas are between $S$ a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,897 |
5. Solve the system of equations:
$$
\left\{\begin{array}{l}
x y-2 y=x+106 \\
y z+3 y=z+39 \\
z x+3 x=2 z+438
\end{array}\right.
$$ | Solution. Note that $y \neq 1$ (since substituting into the first equation would then yield $x-2=x+106$). Then from the first two equations, we can express $x$ and $z$:
$$
\begin{aligned}
& x=\frac{106+2 y}{y-1} \\
& z=\frac{39-3 y}{y-1}
\end{aligned}
$$
Substitute into the last equation, multiply by $(y-1)^{2}$, and... | (38,4,9)(-34,-2,-15) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,898 |
2. There are 8 white cubes of the same size. Marina needs to paint 24 faces of the cubes blue, and the remaining 24 faces - red. After this, Katya assembles them into a $2 \times 2 \times 2$ cube. If the surface of the cube has an equal number of blue and red squares, Katya wins. If not, Marina wins. Can Marina paint t... | Answer: No.
Solution. Suppose Marina painted the cubes in some way, and Katya assembled a cube from them. Let there be $a$ blue and $24-a$ red faces on the surface of the cube. Using the idea of so-called discrete continuity, we will show that Katya can gradually bring the cube to the desired appearance. Note that eac... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,899 |
3. Petya's favorite TV game is called "Sofa Lottery." During the game, viewers can send SMS messages with three-digit numbers containing only the digits $1, 2, 3$, and 4. At the end of the game, the host announces a three-digit number, also consisting only of these digits. An SMS is considered a winning one if the numb... | Answer: 8.
Solution. An example of eight suitable SMS messages: 111, 122, 212, 221, 333, 344, 434, 443. Indeed, no matter what number the host names, it contains either at least two digits from the set $\{1,2\}$, or at least two from the set $\{3,4\}$. If the third digit is from the other set, we replace it with a dig... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,900 |
4. Given a cyclic quadrilateral $A B C D$ with a right angle $A D B$. Through point $C$, a line $l \| A D$ is drawn, and a point $F$ is marked on it such that the angle $B A F$ is equal to the acute angle between the diagonals $A C$ and $B D$, and $F$ and $C$ are on opposite sides of $A B$. Point $X$ is such that $F X ... | Solution. First, note that $F$ lies on the circumcircle of $A B C D$. Indeed, $F$ and $D$ lie on opposite sides of $A C$ and $\angle D A F + \angle D C F = \angle D A B + \angle B A F + \pi - \angle C D A = \angle D A B + \angle B A F + \frac{\pi}{2} - \angle C D B = \angle D A B + \angle B A F + \frac{\pi}{2} - \angle... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,901 |
5. Solve the equation $a^{b}+a+b=b^{a}$ in prime numbers. | Answer: $a=5, b=2$.
Solution. Move $b$ to the right side and group: $a\left(a^{b-1}+1\right)=b\left(b^{a-1}-1\right)$. Since numbers $a$ and $b$ are prime and, obviously, different, $a$ does not divide $b$, from which $a^{b-1}+1 \vdots b$, meaning $a^{b-1} \equiv-1(\bmod b)$. On the other hand, according to Fermat's l... | =5,b=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,902 |
6. There are 28 candies on the table. Petya considers some of them to be delicious. Vasya can, in one move, point to any set of candies and ask Petya how many of them are delicious. How can Vasya guaranteedly find all the delicious candies... (a) in 21 moves; (b) in 20 moves?
(A. A. Tselery, E. Yu. Voronetsky)
# | # Solution.
a) Let's divide the candies into 7 groups of 4 each. In 3 moves, we can learn everything about these 4 candies $a, b, c, d$, for example, by asking about the sets $\{a, c, d\}, \{b, c, d\}, \{a, b, c\}$.
If the answers to the first two questions are different, we will learn what candies $a$ and $b$ are, a... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 11,903 |
1. What is the maximum number of numbers that can be chosen from the set $\{1,2, \ldots, 12\}$ so that the product of no three chosen numbers is a perfect cube? | Solution. 9: all except $4,9,12$. Note that to remove the cubes, we need to remove at least one element from each of the sets $\{1,2,4\},\{3,6,12\},\{2,4,8\},\{1,3,9\},\{2,9,12\}$, $\{3,8,9\},\{4,6,9\}$. Note that all numbers, except 9, are in no more than three of these seven triples. Therefore, if we remove two numbe... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,904 |
3. Find the smallest possible value of the expression
$$
\left(\frac{x y}{z}+\frac{z x}{y}+\frac{y z}{x}\right)\left(\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y}\right)
$$
where $x, y, z$ are non-zero real numbers. | Solution. Note that the signs of all six numbers $\frac{x y}{z}, \frac{z x}{y}$, etc., are the same. If all of them are negative, then replace the numbers $x, y, z$ with their absolute values. As a result, each term ($\frac{x y}{z}$, etc.) will change its sign. The modulus of each bracket will remain the same, but the ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,905 |
4. All angles of a convex octagon are equal, and all sides have a rational length. Prove that it has a center of symmetry. | Solution. Let the octagon be denoted as $A B C D E F G H$. Extend its sides, taken one after the other, until they intersect (see the diagram), then the quadrilateral $P Q R S$ formed by them is a rectangle. Indeed, two external angles of $\triangle H A P$ each measure $135^{\circ}$, so the two internal angles each mea... | proof | Geometry | proof | Yes | Yes | olympiads | false | 11,906 |
5. In each cell of a $10 \times 10$ table, a natural number was written. Then, each cell was shaded if the number written in it was less than one of its neighbors but greater than another neighbor. (Two numbers are called neighbors if they are in cells sharing a common side.) As a result, only two cells remained unshad... | Solution. Answer: 20. The estimate matches the estimate in problem 7.5. One of the possible examples is shown below.
| 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 8 | 10 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 |
| 7 | 8 | 10 | 12 | 13 | 14 ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,907 |
5. The edge of a regular tetrahedron $A B C D$ is 1. Through a point $M$, lying on the face $A B C$ (but not on the edge), planes parallel to the other three faces are drawn. These planes divide the tetrahedron into parts. Find the sum of the lengths of the edges of the part that contains point $D$. | 5. See the figure at the end of the file.
Note that the part of the tetrahedron we are interested in is bounded by three of its faces containing point $D$, and by three planes parallel to the faces. Therefore, this is a parallelepiped.
Consider the faces $A B C$ and $A D C$. Align these faces so that vertex $B$ coinc... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,909 |
11.1. Petya wrote a five-digit number on the board, consisting of different even digits. Vasya erased one digit so that the resulting number is divisible by 18. Which digit did Vasya erase? | Answer: 2.
Solution: The sum of the digits initially equals $0+2+4+6+8=20$. By erasing a digit, Vasya subtracts 0, 2, 4, 6, or 8 from the sum of the digits. The resulting number is divisible by 18, i.e., it is divisible by 9, which means the sum of its digits is divisible by 9. Therefore, Vasya can only erase 2 (other... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,910 |
11.2. Two circles with radii 1 and 2 have a common center $O$. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.
 | Answer: $60^{\circ}$
Solution: Let $\angle A O B=\alpha$. Then the area of the shaded part is $\frac{360^{\circ}-\alpha}{360^{\circ}} \pi+$ $4 \frac{\alpha}{360^{\circ}} \pi-\frac{\alpha}{360^{\circ}} \pi=\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi$. The area of the large circl... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,911 |
11.3. Prove that $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$ is a rational number. | Solution: Let $x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$. Then $x^{3}=(\sqrt{5}+2)-(\sqrt{5}-2)-$ $3 \sqrt[3]{(\sqrt{5}+2)(\sqrt{5}-2)} \times(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2})=4-3 x ; x^{3}+3 x-4=0.0=x^{3}+$ $3 x-4=(x-1)\left(x^{2}+x+4\right), x-1=0$ or $x^{2}+x+4=0$. The second equation has no roots becau... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 11,912 |
11.4. In a row, $n$ integers are written such that the sum of any seven consecutive numbers is positive, and the sum of any eleven consecutive numbers is negative. For what largest $n$ is this possible? | Answer: 16
Solution: Let's provide an example for $n=16$:
$80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80, -31, -31, -31, -31, 80$. We will prove that for $n \geq 17$, it will not be possible to write down a sequence of numbers that satisfy the conditions of the problem. Let's construct a table for the first 17 n... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,913 |
11.5. Given a function $f(x)$, satisfying the condition
$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$
What is $f(2017)$, if it is known that $f(0)=1$? | Answer: 2018
Solution: $f(0 \cdot 0+1)=f(0) f(0)-f(0)-0+2, f(1)=1-1+2=2$,
$$
f(2017 \cdot 0+1)=f(2017) f(0)-f(0)-2017+2
$$
$$
\begin{gathered}
f(1)=f(2017)-1-2017+2=2 \\
f(2017)=2018
\end{gathered}
$$
Criteria: Only for the correct answer - 1 point. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,914 |
6. In space, there are two equal regular tetrahedra with side length $\sqrt{6}$. It is known that their centers coincide. Prove that the volume of their common part is greater than $\frac{1}{2}$. | Solution: It is not hard to notice that the inscribed spheres in these two regular tetrahedra coincide. Therefore, it is sufficient to prove that the volume of the inscribed sphere in the tetrahedron with side $\sqrt{6}$ is greater than $\frac{1}{2}$. The radius of the inscribed sphere is $r=$ $\frac{3 V_{\text {sphere... | \frac{\pi}{6}>\frac{1}{2} | Geometry | proof | Yes | Yes | olympiads | false | 11,915 |
7. Given an infinite sequence
$$
1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n}, \ldots
$$
Prove that for any natural number $k \geq 3$, from this sequence it is possible to select $k$ terms that are consecutive terms of some arithmetic progression. | Solution: Notice that the numbers $\frac{k}{k!}, \frac{k-1}{k!}, \ldots, \frac{1}{k!}$ are consecutive terms of an arithmetic progression and terms of the given sequence, because $k$ ! is divisible by all numbers from 1 to $k$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 11,916 |
1. Can the difference of squares of two natural numbers equal 2018? | Answer: No.
Solution: The difference of squares of natural numbers is the product of the difference and the sum of two natural numbers. The difference and the sum of natural numbers have the same parity. Therefore, their product is either divisible by 4 or odd. However, 2018 is neither. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,917 |
2. Prove that the graphs of the quadratic trinomials: $y=a x^{2}-b x+c, y=b x^{2}-$ $c x+a, y=c x^{2}-a x+b$ have a common point. | Solution: Notice that when $x=-1$, the value $y=a+b+c$ for all three quadratic polynomials, i.e., their graphs intersect at the point $(-1 ; a+b+c)$. | (-1;+b+) | Algebra | proof | Yes | Yes | olympiads | false | 11,918 |
3. The diagonals of the quadrilateral $ABCD$, inscribed in a circle with center at point $O$, intersect at point $P$. It is known that $OP \perp BC$. Prove that $AB = CD$. | Solution: Let $OP$ intersect $BC$ at point $K$. $OK$ is the height of the isosceles triangle $BOC$ ($BO=OC$ as radii), so $OK$ is also the median and $K$ is the midpoint of $BC$. $PK$ is the height and median in triangle $BPC$, so triangle $BPC$ is isosceles, $BP=PC$, and $\angle PBC = \angle PCB$. $\angle PAD = \angle... | AB=CD | Geometry | proof | Yes | Yes | olympiads | false | 11,919 |
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