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4. Vladimir has 100 boxes. The first box contains one stone, the second contains two, and so on. Vladimir can move stones from one box to another if the total number of stones in these boxes is 101. Can he, using these operations, achieve that the seventieth box contains 69 stones, the fiftieth box contains 51 stones, ... | Answer: No, it is impossible.
Solution. Initially, there are 50 boxes with even numbers, each containing an even number of stones. When rearranging, two boxes are taken, the total number of stones in which is 101, meaning the number of stones in them differs in parity. After any rearrangement, there will again be one ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,982 |
5. Each of two students wrote down three consecutive natural numbers, the average of which is the cube of a natural number. Then they multiplied all six numbers. Prove that the resulting product is divisible by 5184. | Solution. Consider the product $N$ of three numbers written by one student, and prove that it is divisible by 9 and 8, and therefore by 72.
By the condition $N=\left(k^{3}-1\right) k^{3}\left(k^{3}+1\right)$, where $k$ is some natural number. In the case of even $k$, $k^{3}$ is divisible by 8. If $k$ is odd, then the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,983 |
1. Does there exist an integer value of $a$ for which the equation
$$
x^{2}-2023 x+2022 a+1=0
$$
has an integer root?
(Find all such integer values of $a$ or prove that they do not exist.) | Solution. The product of the roots of the equation is $x_{1} \cdot x_{2}=2022a+1$, if they exist. If one root is an integer, then the second root is also an integer. The sum of the roots is 2023. Since the product of the roots is odd, both roots must be odd. Therefore, their sum should be even. We conclude that the equ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,984 |
2. For all non-negative $a, b$ prove the inequality
$$
a^{2}+a b+b^{2} \leq 3(a-\sqrt{a b}+b)^{2}
$$ | Solution. Let's represent the left side as the difference of squares.
$$
a^{2}+a b+b^{2}=a^{2}+2 a b+b^{2}-a b=(a+b)^{2}-(\sqrt{a b})^{2}=(a+b-\sqrt{a b})(a+b+\sqrt{a b})
$$
The factor $(a+b-\sqrt{a b})$ is non-negative, it can be represented as $\left((\sqrt{a})^{2}+(\sqrt{b})^{2}-\sqrt{a} \sqrt{b}\right)$, and it i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,985 |
3. Points on a line are colored in two colors (each color is present, each point has exactly one color). Prove that there will be three points of the same color, such that one lies exactly in the middle between the other two. | Solution. We will assume that the points are colored red and blue. If there is exactly one red point on the line, and all the others are blue, the problem has an obvious solution - we take two blue points on one side of the red point, and the midpoint of the segment between the blue points will also be blue. Now let's ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,986 |
4. Two circles intersect at points $M$ and $N$. A line through point $M$ intersects the first circle at point $A$ and the second circle at point $C$. A line through point $N$ intersects the first circle at point $B$ and the second circle at point $D$. Prove that
$$
\smile A M+\smile M C=\smile B N+\smile N D
$$
where... | Solution. 1st method. Quadrilaterals $A M N B$ and $C M N D$ are cyclic, therefore $\angle M N B=$ $180^{\circ}-\angle M A B, \angle M N D=\angle M A B, \angle M C D=180^{\circ}-\angle M A B, \angle D C K=\angle M A B$. From this, lines $A B$ and $C D$ are parallel. Draw line $M N$ until it intersects lines $A B$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,987 |
5. A flock of 64 sparrows is perched on trees. Then, through a series of flights, the flock changes its arrangement on the trees. If at least half of all the birds end up on one tree, then from it to each of the other trees, exactly as many sparrows fly as are already sitting on that tree (and this counts as one flight... | Solution. After the first flight, on each tree to which the sparrows have flown, there will be a number of sparrows divisible by 2. The total number of sparrows is also divisible by 2, so the number of sparrows remaining on the first tree, from which they flew, is also divisible by 2. Similarly, after the second flight... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 15,988 |
3. Boy Seryozha saw two two-headed dragonets, whose heads were tangled. Dragons are either truthful, i.e., both heads always tell the truth, or lying, i.e., both heads always lie. Seryozha decided to help the dragonets untangle their heads. But for this, he needed to know which head belonged to which dragon. He asked t... | The second and third heads contradict each other, which means they are not siblings (sibling heads would either both say they are siblings or both say they are not). Therefore, the third head told the truth (i.e., it is truthful), and the second head lied (i.e., it is deceitful). This means the fourth head lied when it... | Thethirdfirstheadsfromthesame(truthful)dragon,whilethefourthheadsfromanother(deceitful)dragon | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,990 |
5. On a certain island, the climate is unusually regular: it always rains on Mondays and Fridays, and there is fog on Saturdays, while the rest of the days are sunny. On which day of the week should a group of tourists start their vacation if they want to stay for 30 days and catch as many sunny days as possible? | Solution: Let's find out how many full weeks are in 30 days. We get 4 weeks. During these weeks, the number of sunny days does not depend on when the vacation starts. For the remaining two days, we choose Thursday and Friday - sunny days. Therefore, we send the tourists in the morning on Thursday.
Answer: on Thursday. | Thursday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,991 |
1. Eight cards have the numbers $1,1,2,2,3,3,4,4$ written on them. Can these cards be placed in a row so that there is one card between the ones, two cards between the twos, three cards between the threes, and exactly four cards between the fours? | Answer: Yes.
Solution: For example, this: 41312432.
Instructions for checking:
The score can only be one of two: 0 points or 7 points. | 41312432 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,992 |
3. Dima and Vova were solving a math olympiad consisting of two problems: one in geometry and the other in algebra. Dima solved the algebra problem twice as fast as Vova, while he took twice as long to solve the geometry problem compared to Vova, but still finished the olympiad earlier. Who spent more time: Dima on the... | Answer: Dima spent more time solving an algebra problem than Vova spent on a geometry problem.
## Solution:
Let $D$ be the time Dima spent solving the algebra problem, and $V$ be the time Vova spent solving the geometry problem.
According to the condition, $D + 2V < 2D + V$.
From this, $V < D$.
## Instructions for... | V<D | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,994 |
5. We have coins worth 1, 2, 3, and 5 cruzeros, with one of each denomination. If they are genuine, they weigh 1, 2, 3, and 5 grams, respectively. It turns out that one of them is counterfeit and differs in weight from the normal one (but it is not known whether it is heavier or lighter than the genuine one). How can w... | # Solution:
It is impossible to solve this with one weighing. Indeed, if one of the pans outweighs, it could mean either that a heavier fake coin is on that pan, or that a lighter fake coin is on the other pan.
Two weighings are sufficient. For example, we can do the following. First, place the coins worth 2 and 3 cr... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,995 |
1. Solve the system $\left\{\begin{array}{l}\sqrt{2 x^{2}+2}=y-1 ; \\ \sqrt{2 y^{2}+2}=z-1 ; \\ \sqrt{2 z^{2}+2}=x-1 .\end{array}\right.$ | Answer: The system has no solutions.
Sketch of the solution. Squaring each of the equations and adding them, we move all the terms to the left side. By completing the squares, we get $(x+1)^{2}+(y+1)^{2}+(z+1)^{2}=0$. From this, $x=y=z=-1$. Checking shows that this triplet does not work, as in this case the left side ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,996 |
4. Find the largest $\mathrm{C}$ such that for all $\mathrm{y} \geq 4 \mathrm{x}>0$ the inequality $x^{2}+y^{2} \geq$ C $x y$ holds. | Answer: $17 / 4$.
Sketch of the solution. Let $y=4, x=1.1+16 \geq 4$. So, $\mathrm{C} \leq 17 / 4$.
We will prove that for $\mathrm{C}=17 / 4$ and all $y \geq 4 x>0$, the inequality $x^{2}+y^{2} \geq С x y$ holds. Divide the inequality by $x^{2}>0$ and introduce a new variable $z=y / x \geq 4$. Then, $z^{2}+1 \geq C ... | \frac{17}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,998 |
6. For a natural number $n$, $G(n)$ denotes the number of natural numbers $m$ for which $m+n$ divides $m n$. Find $G\left(10^{\mathrm{k}}\right)$. | Answer: $G\left(10^{k}\right)=2 k^{2}+2 k$.
Sketch of the solution. Since $m n=n(m+n)-n^{2}$, if $m+n$ divides $m n$, then $m+n$ divides $n^{2}$. Therefore, $G(n)$ is equal to the number of divisors of $n^{2}$ that are greater than $n$. For each divisor $d > n$ of $n^{2}$, we can associate a divisor $n^{2} / d < n$, w... | 2k^{2}+2k | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,999 |
2. Three runners start simultaneously on a circular track at equal distances from each other (they are at the vertices of an inscribed equilateral triangle). The speeds of the first, second, and third runners are in the ratio $1: 2: 4$ respectively (at the start, the second runner is behind the first, the third is behi... | # Solution
Let \( a \) be one-third of the length of the circular track, \( x, 2x, 4x \) be the speeds of the runners, and \( n \) and \( k \) be the number of complete laps. We obtain the equation
\[
\frac{a + 3an}{2x - x} = \frac{2a + 3ak}{4x - x} \quad(*)
\]
which simplifies to
\[
3k - 9n = 1 \quad(* *)
\]
For ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,000 |
3. Inside a convex quadrilateral, find a point for which the sum of the distances from it to the vertices of the quadrilateral is minimal.
# | # Solution
This is the point of intersection of the diagonals of this quadrilateral.
Let's prove that the sum of the distances from any other point to the vertices of the quadrilateral will be greater.
Let M be the point of intersection of the diagonals AC and BD of the quadrilateral ABCD, and N be a point inside th... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,001 |
4. Prove that $1+2+3+\cdots+(2 n-1)$ is divisible by $(2 n-1)$ for any natural $n$.
# | # Solution
The expression represents the sum of $2 n-1$ consecutive natural numbers, with the last number being odd.
Without the last number ($2 n-1$) in the sum, there is an even number of addends. We can divide them into pairs of the extremes, that is, 1 and $(2 n-2)$, 2 and $(2 n-3)$, and so on. The sum in each pa... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,002 |
4. Express the sum $8+88+888+8888+\ldots+8 \ldots 8$ in terms of n, if the last term in its notation contains n eights | # Solution
$8+88+888+8888+\ldots+8 \ldots 8=8(1+11+111+\ldots+1 \ldots 1)$
Notice that
$$
\begin{gathered}
10-1=9 * 1 \\
100-1=9 * 11 \\
\ldots \ldots \ldots \ldots \\
10^{n}-1=9 * 1 \ldots 1
\end{gathered}
$$
Then $\frac{8}{9}\left(\frac{10 \times 10^{n}-10}{9}-n\right)=\frac{8}{81}\left(10^{n+1}-10-9 n\right)$.
... | \frac{8}{81}(10^{n+1}-10-9n) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,004 |
5. From vertex O, three rays $\mathrm{OA}, \mathrm{OB}$, and $\mathrm{OC}$ emerge, not lying in the same plane, such that angles $\mathrm{AOB}$ and $\mathrm{BOC}$ are equal and acute. Prove that if angle $\mathrm{AOC}$ is obtuse and line $\mathrm{OB}$ is not perpendicular to the plane $\mathrm{AOC}$, then the dihedral ... | # Solution
Let $\angle \mathrm{AOB} = \angle \mathrm{BOC} = \alpha, \angle \mathrm{AOC} = \beta, OB = a$.
We construct $\mathrm{AB}$ perpendicular to $\mathrm{OB}$ and $\mathrm{BC}$ perpend... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,005 |
Task 1. Lёnya can multiply numbers by 7, Gleb can add 3, Sasha can divide by 4, and Andrey can subtract 5. In what order should they perform their operations (each exactly once) to get the number 30 from the number 8? | Answer: $(8: 4+3) \cdot 7-5$.
Remark. This is the only possible solution.
## Criteria
4 p. The correct answer is provided.
It is sufficient to write down the arithmetic example. | (8:4+3)\cdot7-5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,006 |
Problem 3. On an island, there live knights and liars. Knights always tell the truth, while liars always lie. One day, 6 inhabitants of the island gathered together and each said: "Among the other five, there are exactly four liars!". How many knights could there be among them? | Answer: 0 or 2.
Solution. Let's consider two cases: either there is a knight among the attendees, or there is not. If there is at least one knight, then he told the truth, and there are four liars among the attendees. In this case, the second knight also told the truth, while the liars lied, so this case is possible.
... | 0or2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,007 |
Task 4. Masha wrote a three-digit number on the board, and Vera wrote the same number next to it, but she swapped the last two digits. After that, Polina added the obtained numbers and got a four-digit sum, the first three digits of which are 195. What is the last digit of this sum? (The answer needs to be justified.)
... | Solution. Let Masha write the number $100 x+10 y+z$. Then Vera wrote the number $100 x+10 z+y$, and the sum of these numbers is $200 x+11 y+11 z$. For $x \leqslant 8$ this expression does not exceed 1798, and therefore cannot start with 195. Thus, $x=9$. Then $11(y+z)$ is a three-digit number starting with 15. Among th... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,008 |
Problem 5. Vasya and Petya live in the mountains and love to visit each other. When climbing up the mountain, they walk at a speed of 3 km/h, and when descending, they walk at a speed of 6 km/h (there are no horizontal road sections). Vasya calculated that it takes him 2 hours and 30 minutes to get to Petya's, and 3 ho... | Answer: 12 km.
Solution. The journey from Petya to Vasya and back takes 6 hours, during which, as it is twice as slow to go uphill as downhill, the boys spend twice as much time on all ascents as on descents.
Thus, if traveling from Petya to Vasya and back, 2 hours will be spent on descents, and 4 hours on ascents, m... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,009 |
3. In an equilateral triangle $ABC$, segments $AE, BF, CD$ are drawn as shown in the figure. The areas of the shaded triangles are $S_{0}, S_{1}, S_{2}, S_{3}$, and it is given that $S_{0}=S_{1}+S_{2}+S_{3}, S=5 S_{0}$, where $S$ is the area of triangle $ABC$. Prove that $BC=BE+CF+AD$. | Solution. Let the areas of quadrilaterals $C H I E, D G I B$, and $G H F A$ be denoted as $a, c$, and $b$ respectively. Then
$$
\begin{aligned}
& \frac{S_{A B E}}{S}=\frac{B E}{B C} \Rightarro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,012 |
4. The function $f(x)$ is such that for all values of $x$, the equality $f(x+1)-f(x)=x+1$ holds. It is known that $f(0)=4$. Find $f(62)$.
# | # Solution.
$$
f(62)-f(61)=61+1=62 \rightarrow f(62)-f(0)=31 * 63=1953
$$
Therefore, $f(62)=1957$. | 1957 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,013 |
5. The device must transform the polynomial $2020 x^{4}+x+1$ into the polynomial $x^{4}+2020 x+1$ by changing its coefficients in several steps, ensuring that at no step does a polynomial with integer roots result. Can this device perform the transformations if it can only do one of two operations in a single step: 1) ... | Solution. Let $P(x)=2020 x^{4}+x+1$ and $Q(x)=x^{4}+2020 x+1$. We have, $P(-1)=2020>0$ and $Q(-1)=-2020<0$. Applying operation $\mathbf{1})$, the device changes the value of the polynomial at point -1 by $\pm 1$, so one of the intermediate polynomials will have a root at -1. Applying operation 2), the device either doe... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,014 |
Problem 7.1.1. Let $A, B, C, D, E, F, G, H$ - be different digits from 0 to 7 - satisfy the equation
$$
\overline{A B C}+\overline{D E}=\overline{F G H}
$$
Find $\overline{D E}$, if $\overline{A B C}=146$.
(The notation $\overline{A B C}$ represents a three-digit number consisting of digits $A, B, C$, similarly cons... | Answer: 57.
Solution. Substitute 146 for $\overline{A B C}$ and write the example in a column:
$$
\begin{array}{r}
146 \\
+\quad D E \\
\hline F G H
\end{array}
$$
$D, E, F, G, H$ can only be the digits $0,2,3,5,7$. Let's consider $E$.
- If $E=0$, then $H=6$ - but this contradicts the fact that $C=6$.
- If $E=2$, t... | 57 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,015 |
Problem 7.2.1. In the figure, rays $O A, O B, O C, O D, O E, O F$ are such that:
- $O B$ is the bisector of angle $A O C$
- $O E$ is the bisector of angle $D O F$;
- $\angle A O F=146^{\circ}, \angle C O D=42^{\circ}$.
How many degrees does angle $B O E$ measure?
, there are no fewer balls than in the previous one. It is also known that ther... | Answer: 1 red ball, 3 blue balls.
Solution. Among all the boxes, there can be:
- a maximum of 2 boxes with one ball (one with a red ball, the other with a blue ball),
- a maximum of 3 boxes with two balls (one with two red balls, another with one red and one blue, and the third with two blue balls),
- a maximum of 4 ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,023 |
Problem 7.7.1. A seven-digit natural number $N$ is called interesting if:
- it consists of non-zero digits;
- it is divisible by 4;
- any number obtained from the number $N$ by permuting its digits is divisible by 4.
How many interesting numbers exist? | Answer: 128.
Solution. We will use the divisibility rule for 4: a number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4.
In an interesting number, the digits can be rearranged in any way, and the divisibility by 4 should remain unchanged. This implies that there are no odd... | 128 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,027 |
Problem 7.8.1. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
\frac{a b}{a+b}=2 \\
\frac{b c}{b+c}=5 \\
\frac{c a}{c+a}=9
\end{array}\right.
$$
Find the value of the expression $\frac{a b c}{a b+b c+c a}$. | Answer: $\frac{180}{73}$.
Solution. Note that
$$
\begin{gathered}
\frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}, \quad \frac{1}{b}+\frac{1}{c}=\frac{b+c}{b c}, \quad \frac{1}{c}+\frac{1}{a}=\frac{c+a}{c a} \\
\text { and } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+c a}{a b c}
\end{gathered}
$$
This means that if w... | \frac{180}{73} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,028 |
Variant 7.8.2. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
\frac{a b}{a+b}=3 \\
\frac{b c}{b+c}=5 \\
\frac{c a}{c+a}=8
\end{array}\right.
$$
Find the value of the expression $\frac{a b c}{a b+b c+c a}$. | Answer: $\frac{240}{79}$. | \frac{240}{79} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,029 |
Variant 7.8.3. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
\frac{a b}{a+b}=2 \\
\frac{b c}{b+c}=5 \\
\frac{c a}{c+a}=7
\end{array}\right.
$$
Find the value of the expression $\frac{a b c}{a b+b c+c a}$. | Answer: $\frac{140}{59}$. | \frac{140}{59} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,030 |
Variant 7.8.4. Real numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
\frac{a b}{a+b}=4 \\
\frac{b c}{b+c}=5 \\
\frac{c a}{c+a}=7
\end{array}\right.
$$
Find the value of the expression $\frac{a b c}{a b+b c+c a}$. | Answer: $\frac{280}{83}$. | \frac{280}{83} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,031 |
10.1. Let's call the main divisors of a composite number $n$ the two largest natural divisors of $n$ different from $n$. Composite natural numbers $a$ and $b$ are such that the main divisors of $a$ coincide with the main divisors of $b$. Prove that $a=b$.
(A. S. Golyanov) | Solution. Let $n>k$ be the main divisors of the number $a$; then $a / n$ and $a / k$ are the two smallest divisors of the number $a$, greater than one. Let $p$ be the smallest prime divisor of the number $a$, and $q$ be the smallest prime divisor of $a$ other than $p$ (if such exists). Then $a / n = p$. Next, $a / k$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,032 |
10.2. On the side $B C$ of an acute-angled triangle $A B C$, points $D$ and $E$ are marked such that $B D=C E$. On the arc $D E$ of the circumcircle of triangle $A D E$, not containing point $A$, points $P$ and $Q$ are found such that $A B=P C$ and $A C=B Q$. Prove that $A P=A Q$.
(A. Kuznetsov) | The first solution. Without loss of generality, we assume that point $D$ lies on the segment $B E$ and $A D \leqslant A E$. Let $O$ be the center of the circle $(A D E)$. Let point $A^{\prime}$ be symmetric to $A$ with respect to the perpendicular bisector of the segment $D E$ (see Fig. 5). By symmetry, $A^{\prime} B =... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,033 |
10.3. Initially, the pair of numbers $(1,1)$ is written on the board. If for some $x$ and $y$ one of the pairs $(x, y-1)$ and $(x+y, y+1)$ is written on the board, then the other can be added. Similarly, if one of the pairs $(x, x y)$ and $\left(\frac{1}{x}, y\right)$ is written on the board, then the other can be adde... | The first solution. Let's call the discriminant of a pair of numbers $(a, b)$ the quantity $D(a, b)=b^{2}-4 a$. We will prove that the discriminant of all pairs of numbers written on the board is always negative. Indeed, the discriminant of the pair of numbers initially written is $D(1,1)=-30$, and for any pair of numb... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,034 |
10.4. Given a natural number $n>4$. On the plane, $n$ points are marked, no three of which lie on the same line. Vasily draws all segments connecting pairs of marked points, one by one. At each step, while drawing the next segment $S$, Vasily marks it with the smallest natural number that has not yet been used to mark ... | Answer. $k=2 n-3$ for odd $n$, and $k=2 n-4$ for even $n>4$.
Solution. Estimation. Consider the step at which Vasily marks some segment $A B$. Before this step, from each of the points $A$ and $B$, there are at most $n-2$ segments, and they contain at most $2 n-4$ different marks. Therefore, Vasily can definitely mark... | k=2n-3foroddn,k=2n-4forevenn>4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,035 |
10.3. Two ships are sailing on the sea on perpendicular courses such that both are to pass through a fixed point 0 (each at its own time. After passing point 0, the ships, without stopping, will continue their straight-line motion). The speeds of the ships are the same and constant. At noon, the ships had not yet passe... | 10.3. Two ships are sailing on the sea along perpendicular courses such that both are to pass through a fixed point 0 (each at its own time. After passing point 0, the ships, without stopping, will continue their straight-line motion). The speeds of the ships are the same and constant. At noon, the ships have not yet p... | yes,theywill | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,037 |
10.4 Given a triangle $A B C$. Point $P$ is the incenter. Find the angle $B$, if it is known that $R_{A B C}=R_{A P C}$, where $R_{A B C}, R_{A P C}$ are the radii of the circumcircles of triangles $ABC$ and $APC$ respectively. | 10.4. Given a triangle $A B C$. Point $P$ is the incenter. Find the angle $B$, if it is known that $R_{A B C}=R_{A P C}$, where $R_{A B C}, R_{A P C}$ are the radii of the circumcircles of triangles $ABC$ and $APC$ respectively.
Answer: $60^{\circ}$. Hint: Let $\alpha=\angle A B C$. Then $\angle A P C=90^{\circ}+\frac... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,038 |
10.5. a) Prove the inequality $\sqrt{n+1}+2 \sqrt{n}<\sqrt{9 n+3}$ for all natural numbers $n$; b) does there exist a natural number $n$ such that $[\sqrt{n+1}+2 \sqrt{n}]<[\sqrt{9 n+3}]$, where $[a]$ denotes the integer part of the number $a$? | 10.5. a) Prove the inequality $\sqrt{n+1}+2 \sqrt{n}<\sqrt{9 n+3}$ for all natural numbers $n$; b) Does there exist a natural number $n$ for which $[\sqrt{n+1}+2 \sqrt{n}]<[\sqrt{9 n+3}]$, where $[a]$ denotes the integer part of the number $a$?
Answer: b) No such $n$ exists. Hint a) After squaring both sides and isola... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,039 |
1. The roots of the quadratic equation $a^{2}+b x+c=0$ are 2011 times the roots of the quadratic equation $\mathrm{cx}^{2}+\mathrm{dx}+\mathrm{a}=0$. Prove that $\mathrm{b}^{2}=\mathrm{d}^{2}$. | 1. Let $x_{1}, x_{2}$ be the roots of the equation $c^{2} + d x + a = 0$, then $n x_{1}$ and $n x_{2}$ are the roots of the equation $a x^{2} + b x + c = 0$, where $n = 2011$. Then $x_{1} + x_{2} = -\frac{d}{c}, x_{1} x_{2} = \frac{a}{c}$,
 \cdot 80=40$ km ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,041 |
3. Each cell of a $5 \times 10$ board contains one checker. In one move, you can choose any two checkers and move each of them to an adjacent cell. If at least two checkers end up on the same cell, you can remove exactly two of these checkers from that cell. Is it possible to remove all the checkers from the board usin... | 3. It is impossible. Initially, there were 25 checkers on black squares (an odd number). After each move, the number of checkers on black squares will remain odd (after a move, either two checkers leave black squares, or two checkers arrive on black squares, or one arrives and one leaves; if pairs of checkers are remov... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,042 |
4. On the base AC of the isosceles triangle $\mathrm{ABC}$, a point $\mathrm{E}$ is taken, and on the lateral sides $\mathrm{AB}$ and $\mathrm{BC}$ - points K and M, such that KE is parallel to BC and EM is parallel to AB. What part of the area of triangle ABC does the area of triangle $\mathrm{KEM}$ occupy, if $\mathr... | 4. Quadrilateral VKEM is a parallelogram, triangles AKE and EMC are similar to triangle ABC with similarity coefficients equal to
$$
\begin{aligned}
& \frac{\mathrm{KE}}{\mathrm{BC}}=\frac{2 \mathrm{x}}{2 \mathrm{x}+3 \mathrm{x}}=\frac{2}{5} \text { and } \frac{\mathrm{MC}}{\mathrm{BC}}=\frac{3 \mathrm{x}}{2 \mathrm{x... | \frac{6}{25}\mathrm{~S}_{\mathrm{ABC}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,043 |
5. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $\mathrm{k}$ transfers. At the same time, the number of air routes should not exceed four. What is the smallest $\mathrm{k}$ for which this is... | 5. $\mathrm{k}=2$. At least two transfers will be required. From an arbitrary city A, one can reach no more than four cities without a transfer, and with one transfer - no more than $4 \times 3=12$ cities. That is, if using no more than one transfer, one can fly to no more than 16 other cities, but 19 are required. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,044 |
1.1. On September 1, 2021, Vasya deposited 100,000 rubles in the bank. Exactly one year later, the bank accrues 10% annual interest (that is, it increases the amount by 10 percent of what was on the account at that moment, for example, on September 2, 2022, Vasya will have 110,000 rubles on his account). Find the small... | Answer: 2026
Solution. Note that on September 2nd of the $(2021+n)$-th year, the account will have $100000 \cdot 1.1^{n}$ rubles. Since $100000 \cdot 1.1^{4}=146410<150100<161051=100000 \cdot 1.1^{5}$, the minimum $n$ for which the amount will exceed 150,100 rubles is 5. Therefore, the answer is: $2021+5=2026$. | 2026 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,045 |
3.1. Semyon was solving the quadratic equation $4 x^{2}+b x+c=0$ and found that its two roots are the numbers $\operatorname{tg} \alpha$ and $3 \operatorname{ctg} \alpha$ for some $\alpha$. Find $c$. | Answer: 12
Solution. By Vieta's theorem, $c / 4$ equals the product of the roots. Considering that $\operatorname{tg} \alpha \cdot \operatorname{ctg} \alpha=1$, we get $c / 4=3$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,047 |
4.1. Given an arithmetic progression $a_{1}, a_{2}, \ldots, a_{100}$. It is known that $a_{3}=9.5$, and the common difference of the progression $d=0.6$. Find the sum $\left\{a_{1}\right\}+\left\{a_{2}\right\}+\ldots+\left\{a_{100}\right\}$. The notation $\{x\}$ represents the fractional part of the number $x$, i.e., t... | Answer: 50
Solution. Let's look at the first digits after the decimal point in the decimal representation of the terms of the progression. Since $d=0.6$ and $a_{3}=9.5$, the sequence of these digits is: $3,9,5,1,7,3,9,5,1,7, \ldots$ It is periodic with a period of 5. Then the sum of the fractional parts of any five co... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,048 |
5.1. A random number sensor outputs a number $a$ - one of the natural numbers $1,2, \ldots, 100$ (with equal probability). For this value of $a$, we find the maximum possible value $M$ of the function
$$
f(x)=\frac{700}{x^{2}-2 x+2 a}
$$
The probability that $M>10$ is $n$ percent. What is $n$? | Answer: 35
Solution. The minimum value of the quadratic trinomial $x^{2}-2 x+2 a$ is $2 a-1$. Since for the considered values of $a$ we have $2 a-1>0$ (which is true for $a>1 / 2$), the maximum value of $f(x)$ is $M=\frac{700}{2 a-1}$. Next, the condition $M>10$ (for $2 a-1>0$) is equivalent to the following condition... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,049 |
6.1. In a row, the numbers are written: $100^{100}, 101^{101}, 102^{102}, \ldots, 876^{876}$ (i.e., the numbers of the form $n^{n}$ for natural numbers n from 100 to 876.) How many of the written numbers are perfect cubes? (A perfect cube is the cube of an integer.) | # Answer: 262
Solution. Consider a number of the form $m^{k}$, where $m$ and $k$ are natural numbers. If $k$ is divisible by 3, then $m^{k}$ is a perfect cube. Otherwise, $m^{k}$ is a perfect cube if and only if $m$ is a perfect cube. Thus, the answer to our problem is the total number of numbers that are divisible by... | 262 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,050 |
7.1. In a triangular pyramid $A B C D$, it is known that: $A B=C D=6, A D=B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R$ is the radius of the smallest sphere that can contain such a pyramid. | Answer: 49
Solution. Since the segment $AC$ fits inside the sphere, $2R \geqslant AC$. On the other hand, the sphere constructed with $AC$ as its diameter covers both the triangle $ABC$, since $\angle ABC > 90^{\circ}$, and the congruent (by three sides) triangle $BAD$, and thus covers the entire tetrahedron. Therefor... | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,051 |
8.1. On the coordinate plane, a triangle $O A B$ is drawn, where the point of intersection of the medians is at $\left(\frac{19}{3}, \frac{11}{3}\right)$, and points $A$ and $B$ have natural coordinates. Find the number of such triangles. (Here, $O$ denotes the origin - the point $(0,0)$; two triangles with the same se... | Answer: 90.
Let $M$ be the midpoint of $AB$. Then, by the property of the median, $OM = \frac{3}{2} OG$, where $G$ is the centroid. Therefore, $M$ has coordinates $\left(\frac{19}{2}, \frac{11}{2}\right)$.
:
- X202020;
- 2 X 02020;
- $20 \times 2020$
- $202 \times 020$
- $2020 \times 20$
- $20202 \mathrm{X} 0$
- $202020 \mathrm{X}$.
Case 1. If the first... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,056 |
Problem 6.5. On a line, 5 points $P, Q, R, S, T$ are marked, exactly in that order. It is known that the sum of the distances from $P$ to the other 4 points is 67, and the sum of the distances from $Q$ to the other 4 points is 34. Find the length of the segment $P Q$. | Answer: 11.
Solution. From the condition of the problem, it is known that
$$
P Q+P R+P S+P T=67 \quad \text { and } \quad Q P+Q R+Q S+Q T=34
$$
Let's find the difference of these quantities:
$$
\begin{aligned}
33 & =67-34=(P Q+P R+P S+P T)-(Q P+Q R+Q S+Q T)= \\
& =(P Q-Q P)+(P R-Q R)+(P S-Q S)+(P T-Q T)=0+P Q+P Q+P... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,057 |
Problem 6.6. Zhenya painted three faces of a white cube $6 \times 6 \times 6$ red. Then he sawed it into 216 identical small cubes $1 \times 1 \times 1$. How many of the small cubes could have no red faces? List all possible options. | Answer: $120,125$.
Solution. There are essentially two cases for painting the faces of the large cube:
- three painted faces form a "P" shape;
- three painted faces share a common vertex.
In ... | 120,125 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,058 |
Problem 6.7. The Amur and Bengal tigers started running in a circle at 12:00, each at their own constant speed. By 14:00, the Amur tiger had run 6 more laps than the Bengal tiger. Then the Amur tiger increased its speed by 10 km/h, and by 15:00, it had run a total of 17 more laps than the Bengal tiger. How many meters ... | Answer: 1250.
Solution. In the first 2 hours, the Amur tiger ran 6 more laps, i.e., in 1 hour, it ran 3 more laps. If it had not increased its speed, in the first 3 hours, it would have run 9 more laps. However, the increase in speed resulted in an additional $17-9=8$ laps in the third hour. Since it increased its spe... | 1250 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,059 |
Problem 6.8. In class 6 "A", there are several boys and girls. It is known that in 6 "A"
- girl Tanya is friends with 12 boys;
- girl Dasha is friends with 12 boys;
- girl Katya is friends with 13 boys;
- any girl will have a friend among any three boys.
How many boys can there be in 6 "A"? List all possible options. | Answer: 13, 14.
Solution. From the condition, it follows that there are at least 13 boys in the class. If there were at least 15, then among them we could choose three who do not befriend Tanya. But then Tanya would not have a friend among them, which is a contradiction. Therefore, there are 13 or 14 boys in total. Le... | 13,14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,060 |
1. (7 points) On the first day, Masha collected $25 \%$ fewer mushrooms than Vasya, and on the second day, she collected $20 \%$ more than Vasya. Over the two days, Masha collected $10 \%$ more mushrooms than Vasya. What is the smallest number of mushrooms they could have collected together? | Solution. Let Vasya collected $x$ mushrooms on the first day, and $-y$ mushrooms on the second day, then Masha collected $\frac{3}{4} x$ and $\frac{6}{5} y$ mushrooms respectively. According to the condition, $\frac{11}{10}(x+y)=\frac{3}{4} x+\frac{6}{5} y$
Solving this equation, we get $22 x+22 y=15 x+24 y \Leftright... | 189 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,061 |
2. (7 points) Seven dwarfs are sitting around a round table. In front of each of them is a mug. Some of these mugs contain milk. One of the dwarfs pours all his milk equally into the mugs of the others. Then his right neighbor does the same. Then the next right neighbor does the same, and so on. After the last, seventh... | # Solution.
The point is that after the first gnome pours the milk (giving $\frac{1}{7}$ to each of the others), the distribution remains exactly the same but shifted by one gnome, and the sum $(1+2+\ldots+6): 7$ is exactly 3. It remains to prove that there are no other solutions. Let $x$ be the maximum amount of milk... | \frac{6}{7},\frac{5}{7},\frac{4}{7},\frac{3}{7},\frac{2}{7},\frac{1}{7},0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,062 |
3. (7 points) Let $f(x)=x^{2}+12 x+30$. Solve the equation $f(f(f(f(f(x)))))=0$.
# | # Solution.
Rewrite the expression for $f(x)$ as $(x+6)^{2}-6$.
Then
$$
\begin{gathered}
f(f(x))=(\underbrace{(x+6)^{2}-6}_{f(x)}+6)^{2}-6=(x+6)^{4}-6 \\
f(f(f(x)))=(\underbrace{(x+6)^{4}-6}_{f(f(x))}+6)^{2}-6=(x+6)^{8}-6
\end{gathered}
$$
Thus, we have:
$$
f(f(f(f(f(x)))))=(x+6)^{32}-6
$$
Therefore, the original... | -6\\sqrt[32]{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,063 |
5. (7 points) Solve the equation $(\sin x)^{2018}+(\cos x)^{2018}=1$. | Solution.
Since $\sin ^{2} x+\cos ^{2} x=1$, we obtain the equation:
$$
(\sin x)^{2018}+(\cos x)^{2018}=\sin ^{2} x+\cos ^{2} x
$$
from which we have:
$$
\sin ^{2} x\left(1-\sin ^{2016} x\right)+\cos ^{2} x\left(1-\cos ^{2016} x\right)=0
$$
Since $|\sin x| \leq 1,|\cos x| \leq 1$, the equality holds only when both... | \frac{\pin}{2},n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,065 |
10.5. Given an infinite grid plane. A teacher and a class of 30 students play a game, taking turns - first the teacher, then each of the students in turn, then the teacher again, and so on. In one move, a player can paint a unit segment that is a boundary between two adjacent cells. Segments cannot be painted more than... | Solution. The teacher will choose a square $K$ of size $100 \times 100$ and will color segments of its boundary if possible. Let's say that before her $n$-th move, all these segments are colored. Then $n \leqslant 401$, since there are 400 segments on the boundary. By this point, no more than $30 \cdot 400$ segments ha... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 16,066 |
10.6. Given a polynomial $P(x)$ of degree $n>1$ with real coefficients. It is known that the equation $P(P(P(x)))=P(x)$ has exactly $n^{3}$ distinct real roots. Prove that these $n^{3}$ roots can be divided into two groups with equal arithmetic means. | Solution. Let $t \in \mathbb{R}$. Note that $P(P(P(t)))=P(t)$ if and only if $P(t)$ is a root of the polynomial $P(P(x))-x$. In particular, this polynomial has roots, denote them by $x_{1}, x_{2}, \ldots, x_{k}$. Since $n>1$, the degree of the polynomial $P(P(x))-x$ is $n^{2}$, so $k \leqslant n^{2}$. Therefore, all ro... | proof | Algebra | proof | Yes | Yes | olympiads | false | 16,067 |
10.7. Natural numbers $n>20$ and $k>1$ are such that $n$ is divisible by $k^{2}$. Prove that there exist natural numbers $a, b$, and $c$ such that $n=a b+b c+c a$.
(A. Khryabrov) | Solution. Note that from the equality $n+a^{2}=(a+b)(a+c)$ follows the equality $n=a b+b c+c a$. Therefore, to solve the problem, it is sufficient to find a natural number $a$ such that the number $n+a^{2}$ can be factored into the product of two natural numbers $x$ and $y$, both greater than $a$ (then we can set $b=x-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,068 |
10.8. A pentagon $ABCDE$ is inscribed in a circle $\omega$. The line $CD$ intersects the rays $AB$ and $AE$ at points $X$ and $Y$ respectively. Segments $EX$ and $BY$ intersect at point $P$ and intersect the circle $\omega$ again at points $Q$ and $R$. The point $A'$ is symmetric to point $A$ with respect to the line $... | Solution. Note that point $P$ lies inside the circle $(Q D R)$, and quadrilateral $P Q D R$ is convex. Therefore, point $D$ lies inside the circle $(P Q R)$. Meanwhile, point $Y$ lies outside the circle $(P Q R)$. Consequently, the circle $(P Q R)$ intersects the circle $(D R Y)$ again at some point $N_{1}$, which lies... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,069 |
# Problem 10.6
## General Part
[+1 pt.] It is proven that $P(P(x))-x$ has $n^{2}$ roots, or that $P(P(x))-x$ divides $P(P(P(x)))-P(x)$
[ 4 pts.] something from the previous is used, but not proven - no more than 4 points.
[ $\leq 3$ pts.] the solution relies on the values of the two leading coefficients of the poly... | # Solution 1 - by grouping the roots $P(x)=t$, where $P(P(t))=t$
This part of the criteria applies if it is specified that the roots $P(x)=t$, where $P(P(t))=t$, are grouped, and it is specified that the sum or average of the roots in such a group depends only on the coefficients of $P(x)$. The solution is considered ... | notfound | Algebra | proof | Yes | Yes | olympiads | false | 16,070 |
1. Provide an example of a natural number that is divisible by 2019, and the sum of its digits is also divisible by 2019. Don't forget to show that your example meets the condition. | 1. Answer. For example, $20192019 \ldots 2019$ (2019 times).
Grading criteria. Any correct example with verification - 7 points. Just an example without verification - 2 points. Incorrect example - $\mathbf{0}$ points. | 20192019\ldots2019 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,071 |
2. Uncle bought a New Year's gift for each of his nephews, consisting of a candy, an orange, a pastry, a chocolate bar, and a book. If he had bought only candies with the same amount of money, he would have bought 224. He could have bought 112 oranges, 56 pastries, 32 chocolate bars, and 16 books with the same amount o... | 2. Answer. 8. Solution. Let's express the prices of all items in terms of the price of a candy. An orange costs as much as two candies, a pastry - as 4 candies, a chocolate bar - as 224:32 = 7 candies, a book - as 14 candies. The total price of the gift is equal to the price of $1+2+4+7+14=28$ candies, and the number o... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,072 |
3. In square $A B C D$, diagonals $A C$ and $B D$ intersect at point $O$. Point $K$ is the midpoint of side $A B$. Points $M$ and $N$ are chosen on sides $A D$ and $B C$ respectively such that rays $O K, O M$, and $O N$ divide the square into three parts of equal area. In what ratio does point $M$ divide side $A D$? | 3. Answer. 5:1 Solution. Let the areas of ACO and BCO be $S$ (obviously, they are the same). Then, from the equality of the areas BLOK and AKOM, it follows that the areas of triangles BLO and AOM are equal, denote their areas by $\mathrm{S}_{1}$. Since the area of BCO is $2 \mathrm{~S}$, the area of OLC is $2 S-S_{1}$ ... | 5:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,073 |
4. On the board, the numbers $a, b$, and $c$ were written. They were erased, and in their place, the numbers $a^{4}-2 b^{2}, b^{4}-2 c^{2}, c^{4}-2 a^{2}$ were written. After this, it turned out that the numbers on the board were the same as those at the beginning (possibly in a different order). Find the numbers $a, b... | 4. Answer: $a=b=c=-1$. Solution. From the condition, it follows that $a^{4}-2 b^{2}+b^{4}-$ $2 c^{2}+c^{4}-2 a^{2}=a+b+c=-3 \Leftrightarrow a^{4}-2 b^{2}+b^{4}-2 c^{2}+c^{4}-2 a^{2}+3=\left(a^{2}-1\right)^{2}+\left(b^{2}-\right.$ $1)^{2}+\left(c^{2}-1\right)^{2}=0$. Therefore, $a^{2}=b^{2}=c^{2}=1$, meaning each of the... | =b==-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,074 |
5. The inscribed circle touches the sides $A B, B C, A C$ of the right triangle $A B C\left(\angle C=90^{\circ}\right)$ at points $C_{0}, A_{0}, B_{0}$ respectively. $C H-$ is the altitude of triangle $A B C$. Point $M$ is the midpoint of segment $A_{0} B_{0}$. Prove that $M C_{0}=M H$. | 5. Solution. Let point $O$ be the center of the inscribed circle of the triangle. Then its radii $O A_{0}, O B_{0}, O C_{0}$ are perpendicular to the corresponding sides. Therefore, the quadrilateral $O A_{0} C B_{0}$ is a square, and the midpoint of its diagonal $A_{0} B_{0}$ is also the midpoint of the diagonal $O C$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,075 |
6. Two brothers sold a flock of sheep, receiving as many rubles for each sheep as there were sheep in the flock. Wishing to divide the proceeds equally, they began to take 10 rubles from the total sum in turns, starting with the older brother. After the older brother took 10 rubles again, the younger brother was left w... | 6. Answer: 2 rubles. Solution. Let there be $n$ sheep in the flock. Then the brothers earned $n^{2}$ rubles. From the condition, it follows that the number of tens in the number $n^{2}$ is odd. Represent the number $n$ as $10 k+m$, where $k-$ is the number of tens, and $m-$ is the number of units in it. Then $n^{2}=100... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,076 |
# 4. CONDITION
In a circle of radius 3, several circles are placed arbitrarily, the sum of whose radii is 25. Prove that there exists a straight line that intersects at least nine of these circles. | Solution. Let's project all circles onto an arbitrary diameter of the large circle. The sum of the lengths of the projections is equal to the sum of the diameters of the circles, i.e., 50. If each point on the large diameter is covered by projections no more than eight times, the sum of their lengths does not exceed $6... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,077 |
# 5. CONDITION
The sequence of natural numbers $\mathrm{q}_{1}<\mathrm{q}_{2}<\mathrm{q}_{3}<\ldots$ is such that $\mathrm{q}_{\mathrm{n}}<2 \mathrm{n}$ for any index $\mathrm{n}$. Prove that any natural number can be represented as the difference of two numbers from this sequence or as a number from the sequence itse... | Solution. Let $\mathrm{m}$ be an arbitrary natural number. Consider the first $\mathrm{m}$ terms of the sequence: $\mathrm{q}_{1}<\mathrm{q}_{2}<\ldots<\mathrm{q}_{\mathrm{m}}<2 \mathrm{~m}$. Then, if $\mathrm{q}_{\mathrm{k}}$ is divisible by $\mathrm{m}$, due to the inequality $\mathrm{q}_{\mathrm{k}}<2 \mathrm{~m}$, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 16,078 |
7.1 In the expression $5 * 4 * 3 * 2 * 1=0$ replace the asterisks with arithmetic operation signs $+ - \times \div$, using each sign exactly once, so that the equality becomes true (note that $2+2 \times 2=6$ ). | 7.1 $5-4 \times 3: 2+1=0$.
7.1 $5-4 \times 3: 2+1=0$.
(Note: The mathematical expression is kept as is, since it is a universal notation and does not require translation.) | 5-4\times3:2+1=0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,079 |
7.2 There were 25 sparrows sitting on two bushes. After 5 sparrows flew from the first bush to the second, and 7 sparrows flew away from the second bush, it turned out that there were twice as many sparrows left on the first bush as on the second. How many sparrows were there on each bush initially? | 7.2 On the first bush, there were 17 sparrows, and on the second, there were 8.
After 7 sparrows flew away, 18 remained. At this point, there were twice as many sparrows on the first bush as on the second. This means there were 12 sparrows on the first bush and 6 on the second. If we return 7 sparrows to the second bu... | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,080 |
7.4. What angle do the hour and minute hands form at twenty minutes past one? (Don't forget to justify your answer). | 7.4 $110^{\circ}$.
At 12:00, the angle between the hour and minute hands is $0^{\circ}$. The minute hand completes a full circle of $360^{\circ}$ in 60 minutes, which is $6^{\circ}$ per minute, and in 20 minutes it will cover $120^{\circ}$. The hour hand moves 12 times slower than the minute hand. Therefore, in 20 min... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,081 |
7.5. Find all three-digit numbers $\mathrm{N}$ such that the sum of the digits of the number $\mathrm{N}$ is 11 times smaller than the number $\mathrm{N}$ itself (do not forget to justify your answer). | 7.5. The number 198 is unique.
From the condition, we get the relation $11 \cdot(a+b+c)=100 a+10 b+c$, or $10 c+b=89 a$. In this relation, the left side is a number less than 100. If $a$ is greater than 1, then the right side will be a number greater than 100. Therefore, $a=1$, $c=8$, $b=9$. All-Russian School Olympia... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 16,082 |
9.6. Petya and Misha start on a circular track from the same point in the counterclockwise direction. Both run at constant speeds, with Misha's speed being $2\%$ greater than Petya's. Petya always runs counterclockwise, while Misha can change his direction at any moment, immediately before which he has run half a lap o... | Solution. Let Misha, after running half a lap, turn around and run back. While he runs half a lap back, he will meet Petya. When Misha reaches the starting point, Petya will not have reached it yet. Therefore, if Misha continues to run in the same direction, he will meet Petya at some point at a distance $d$ from the s... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 16,083 |
9.7. A green chameleon always tells the truth, while a brown chameleon lies and immediately turns green after lying. In a company of 2019 chameleons (green and brown), each in turn answered the question of how many of them are currently green. The answers were the numbers $1,2,3, \ldots, 2019$ (in some order, not neces... | Answer: 1010.
Solution. Consider two chameleons who spoke in a row. One of them was brown at the moment of speaking; indeed, if both were green, the number of green chameleons would not have changed after the first one spoke, and the second one would have named the same number as the first. We can divide all the chame... | 1010 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,084 |
9.8. In an acute-angled triangle $A B C$, the bisector $B L$ is drawn. The circumcircle of triangle $A B L$ intersects side $B C$ at point $D$. It turns out that point $S$, symmetric to point $C$ with respect to the line $D L$, lies on side $A B$ and does not coincide with its endpoints. What values can $\angle A B C$ ... | Answer: $60^{\circ}$.
First solution. From the symmetry, triangles $C L D$ and $S L D$ are equal, so $D S = D C$, $\angle C D L = \angle S D L$, and $\angle D L C = \angle D L S$. Since the quadrilateral $A L D B$ is inscribed in a circle, we have $\angle B A L = \angle L D C$ (see Fig. 1). The chords $A L$ and $D L$ ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,085 |
9.9. Let's call a polygon good if it has a pair of parallel sides. A certain regular polygon was cut by non-intersecting (at interior points) diagonals into several polygons, each having the same odd number of sides. Can it happen that at least one of these polygons is good?
(I. Bogdanov) | Answer. No, it cannot.
Solution. We will prove the following simple lemma.
Lemma. If a p-gon is divided by non-intersecting diagonals into $(d+2)$-gons, the number of which is $t$, then $n=td+2$.
Proof. Induction on $t$; the base case for $t=1$ is obvious.
For the inductive step. Assuming that the statement is true... | proof | Geometry | proof | Yes | Yes | olympiads | false | 16,086 |
9.10. Prove that for any positive $x_{1}, x_{2}, \ldots, x_{9}$, the inequality
$$
\frac{x_{1}-x_{3}}{x_{1} x_{3}+2 x_{2} x_{3}+x_{2}^{2}}+\frac{x_{2}-x_{4}}{x_{2} x_{4}+2 x_{3} x_{4}+x_{3}^{2}}+\ldots+
$$
$$
+\frac{x_{8}-x_{1}}{x_{8} x_{1}+2 x_{9} x_{1}+x_{9}^{2}}+\frac{x_{9}-x_{2}}{x_{9} x_{2}+2 x_{1} x_{2}+x_{1}^{... | Solution. Let's continue the numbering cyclically: we will assume that $x_{i+9}=x_{i}$.
Fix the index $i$ and for simplicity denote $a=x_{i}, b=$ $=x_{i+1}, c=x_{i+2}$. Notice that when $a \geqslant c$, the inequality
$$
a c+2 b c+b^{2} \leqslant a c+a b+b c+b^{2}=(a+b)(b+c)
$$
holds, and when $a \leqslant c$, the i... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 16,087 |
# 1. Option 1.
Vasya strikes the strings of a 6-string guitar from 1 to 6 and back. Each subsequent strike hits the adjacent string. On which string number will the 2000th strike fall? (The order of striking the strings: $1-2-3-4-5-6-5-4-3-2-1-2-\ldots$) | Answer. 2.
Solution. The first string receives the strikes $1,11,21,31, \ldots, 2001$. Therefore, the 2000-th strike will fall on the 2-nd string.
# | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,088 |
# 2. Option 1.
Tourists Vitya and Pasha are walking from city A to city B at equal speeds, while tourists Katya and Masha are walking from city B to city A at equal speeds. Vitya met Masha at 12:00, Pasha met Masha at 15:00, and Vitya met Katya at 14:00. How many hours after noon did Pasha meet Katya? | Answer: 5.
Solution: The distance between Masha and Katya and their speeds do not change, and the speeds of Vitya and Pasha are equal. Vitya met Katya 2 hours after Masha, so Pasha will also meet Katya 2 hours after Masha, i.e., at 5:00 PM - 5 hours after noon. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,089 |
6. Increase Grisha's catch by $40 \%$ and Vasya's catch by $20 \%$.
Grisha, the most resourceful of them, calculated that their total catch in each case would increase by 5, 10, and 9 kg respectively. What was the total catch of the friends (in kilograms) before meeting with Shapka? | Answer: 40.
Solution. Let the harvest of Vasya, Misha, and Grisha be denoted by $x, y, z$ respectively. Then, $0.4 x + 0.2 y = 5 ; 0.4 y + 0.2 z = 10 ; 0.4 z + 0.2 x = 9$. Adding the equations, we get $0.6(x + y + z) = 24$, from which $x + y + z = 40$. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,090 |
# 4. Variant 1
If from the discriminant of the quadratic polynomial $f(x)=a x^{2}+2 b x+c$ we subtract the discriminant of the quadratic polynomial $g(x)=$ $(a+1) x^{2}+2(b+2) x+c+4$, the result is 24. Find $f(-2)$. | Answer: 6.
Solution. We have: $D_{1}-D_{2}=4\left(b^{2}-a c-(b+2)^{2}+(a+1)(c+4)\right)=4(-4 b+4 a+c)=4 f(-2)$.
## Variant 2
If the discriminant of the quadratic trinomial $f(x)=a x^{2}+2 b x+c$ is subtracted from the discriminant of the quadratic trinomial $g(x)=$ $(a+1) x^{2}+2(b+2) x+c+4$, the result is 28. Find ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,091 |
5. Variant 1 Petya wrote 9 different positive integers on the board. It turned out that the arithmetic mean of these numbers is 16. What is the greatest value that the largest number on the board can take | Answer: 108.
Solution: The sum of the given numbers is $9 \cdot 16=144$. Since all the numbers are distinct, the sum of the 8 smallest of them is no less than $1+2+\cdots+8=36$. Therefore, the largest number cannot be greater than $144-36=108$. This is possible: $(1+2+\cdots+8+108): 9=16$.
Variant 2 Petya wrote 9 dif... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 16,092 |
# 6. Variant 1.
Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1, 2, and 3. It turned out that if Vasya's triangle is rotated 15 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, the triangle will return to its original position. If Vasya's t... | Answer: 5.
Solution. Let the angle at the first vertex be $\alpha$, at the second $\beta$, and at the third $\gamma$. When the triangle is rotated around its vertices, it may complete several full rotations before returning to its initial position. Suppose that during the rotation around the first vertex, it makes $k$... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,093 |
# 7. Option 1.
Unit cubes were used to assemble a large parallelepiped with sides greater than 3. Two cubes will be called adjacent if they touch by faces. Thus, one cube can have up to 6 neighbors. It is known that the number of cubes that have exactly 6 neighbors is 429. Find the number of cubes that have exactly 4 ... | Answer: 108.
Solution: Let $a, b$ and $c$ be the lengths of the sides of the large parallelepiped. Then, the number of cubes with exactly 6 neighbors is: $(a-2)(b-2)(c-2)$. Since each of the factors $a-2, b-2$, and $c-2$ is greater than 1 and their product equals the product of the three prime numbers 3, 11, and 13, t... | 108 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 16,094 |
8. Variant 1. Point $I$ is the center of the circle inscribed in triangle $A B C$ with sides $B C=6$, $C A=8, A B=10$. Line $B I$ intersects side $A C$ at point $K$. Let $K H$ be the perpendicular dropped from point $K$ to side $A B$. Find the distance from point $I$ to line $K H$. | Answer: 2.
Solution.
By the converse of the Pythagorean theorem, angle $C$ is a right angle. Then, triangles $B K C$ and $B K H$ are congruent by the hypotenuse and an acute angle. Therefore... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 16,095 |
Problem 4.1. Four arithmetic examples were written on the board. Vera erased one "plus" sign, one "minus" sign, one "multiply" sign, one "divide" sign, and four "equals" signs.
Instead of the same signs, she wrote the same letters, and instead of different signs, different letters. Restore the examples.
![](https://c... | Answer: $\mathrm{a} 4 \mathrm{~b} 5 \mathrm{c} 2 \mathrm{~d} 1 \mathrm{e} 3$.
Solution. Notice that the letter $B$ appears in all examples. It turns out that the letter $B$ must be replaced by the equals sign:
$$
\begin{aligned}
& 4 A 2=2 \\
& 8=4 C 2 \\
& 2 D 3=5 \\
& 4=5 E 1
\end{aligned}
$$
Now we will look at th... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,096 | |
Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,098 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 16,099 |
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