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2. (3 points) In a certain tetrahedron, the bases and midpoints of all four of its medians are marked. Each base of the median of the tetrahedron is connected to the midpoints of the other three. In the resulting polyhedron, all 12 edges turned out to be of equal length. Prove that the original tetrahedron is orthocent... | Solution: Let the position vectors of the vertices of the tetrahedron be $\vec{a}, \vec{b}, \vec{c}$, and $\vec{d}$. Then the bases of the medians are given by the formula $\frac{\vec{b}+\vec{c}+\vec{d}}{3}$ and similar ones, and the midpoints of the medians by the formula $\frac{\frac{\vec{a}+\vec{b}+\vec{c}}{3}+\vec{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 21,129 |
# 3. (3 points) Find the distance between the curves $y=e^{3 x+11}$ and $y=(\ln x-11) / 3$. | Solution: The distance between the graphs of the functions is equal to the distance between their nearest points. Note that the functions, between the graphs of which we need to find the distance, are inverses of each other, and their graphs are symmetric with respect to the line $y=x$, and, moreover, are on opposite s... | \sqrt{2}(4+\frac{\ln3}{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 21,130 |
4. (3 points) In a certain country, there are 800 cities and 8 airlines. Any two cities are connected by flights of one of the eight airlines. Can we assert that there will be an airline and more than 200 cities, between any two of which it is possible to travel using flights of this airline (possibly with layovers)? | Answer: No, this is not necessarily true.
Solution: Let's construct a counterexample. Divide the cities into 8 groups of 100 cities each. Let's call these groups $A, B, C, D, E, F, G$ and $H$.
Within each group, connect the cities with flights from company number 1.
Company number 2 will connect cities in group $A$ ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,131 |
5. (3 points) Find all functions continuous on the entire number line that satisfy the identity $2 f(x+y)=f(x) f(y)$ and the condition $f(1)=10$.
Answer: $f(x)=2 \cdot 5^{x}$. | Solution:
First, let's prove that \( f(x) \geqslant 0 \). Indeed, \( 2 f(x) = f\left(\frac{x}{2}\right)^{2} \geqslant 0 \).
Substituting \( x = y = 0 \) into the given condition, we get \( 2 f(0) = f(0) \cdot f(0) \), from which \( f(0) = 0 \) or \( f(0) = 2 \). If \( f(0) = 0 \), then \( 2 f(x + 0) = f(x) f(0) \), w... | f(x)=2\cdot5^{x} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,132 |
6. (3 points) Let $x, y, z$ and $t$ be non-negative numbers such that $x+y+z+t=2$. Prove the inequality
$$
\sqrt{x^{2}+z^{2}}+\sqrt{x^{2}+1}+\sqrt{z^{2}+y^{2}}+\sqrt{y^{2}+t^{2}}+\sqrt{t^{2}+4} \geqslant 5
$$ | Solution:
Consider the following points on the plane: $A(0,0) ; B(x, z) ; C(x+1, z+x) ; D(x+1+z, z+x+y) ; E(x+1+z+y, z+x+y+t) ; F(x+1+z+y+t, z+x+y+t+2)$. Then the length of the broken line $A B C D E F$ coincides with the expression that needs to be evaluated. By the triangle inequality, the length of the broken line ... | 5 | Inequalities | proof | Yes | Yes | olympiads | false | 21,133 |
7. (4 points) Solve the equation:
$$
\sin x + \sin 3x + \ldots + \sin 2015x = \cos x + \cos 3x + \ldots + \cos 2015x
$$ | Solution:
1) $x=\frac{\pi k}{1008}, k \in \mathbb{Z}, k$ does not divide 1008.
2) $x=\frac{\pi+4 \pi k}{4032}, k \in \mathbb{Z}$.
Proof:
First, let's prove the following formulas by induction in the case where $\sin x \neq 0$:
$$
\begin{gathered}
\sum_{k=1}^{n} \sin ((2 k-1) x)=\frac{\sin ^{2}(n x)}{\sin x} \\
\sum... | \frac{\pik}{1008},k\in\mathbb{Z},k | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,134 |
8. (4 points) With the number written on the board, the following operation is allowed: erase two adjacent digits whose sum does not exceed 9, and write this sum in their place. Initially, a 100-digit number $21162116 \ldots 2116$ was written. The operation was performed on the number on the board until it was no longe... | Answer:
$46 \ldots 46$ - a 50-digit number.
Solution:
The sum of the digits of the original number is $(2+1+1+6) \cdot 25=250$. The sum of the digits of the final number is the same.
We will divide the digits of the final number into pairs: the first with the second, the third with the fourth, and so on. The last d... | 46\ldots46 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,135 |
3. A four-digit number ABCD, consisting of different digits, is divisible by both the two-digit number CD and the two-digit number AB. Find the smallest possible value of ABCD. | Answer: 1248
## Examples of answer recording:
1234
# | 1248 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,136 |
3. Find all natural numbers $a, b, c$ satisfying the equations $a b c=80$, $(\mathrm{a}+1)(\mathrm{b}+1)(\mathrm{c}+1)=162$. In your answer, list all possible values of the number $a$ in ascending order separated by semicolons. | Answer: $2 ; 5 ; 8 \| 2,5,8$
# | 2;5;8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,142 |
1. Given a quadrilateral $A B C D$. It is known that $A B=3, B D=8, A C=4, C D=4$. It is also known that at least one of the two unknown sides of the quadrilateral $ABCD$ is also an integer. Find what it can be equal to. In your answer, list all suitable values in ascending order separated by semicolons. | Answer: $5,6,7 \| 5 ; 6 ; 7$ | 5;6;7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,143 |
2. Given a quadrilateral $A B C D$. It is known that $B C=4, B D=9, A C=5, A D=3$. It is also known that at least one of the two unknown sides of the quadrilateral $A B C D$ is also an integer. Find what it can be equal to. In your answer, list all suitable values in ascending order separated by a semicolon. | Answer: $6,7,8 \| 6 ; 7 ; 8$ | 6;7;8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,144 |
3. Given a quadrilateral $A B C D$. It is known that $A B=2, B C=4, A D=6, C D=3$. It is also known that at least one of the two diagonals has an integer length. Find what it can be equal to. In your answer, list all suitable values in ascending order separated by a semicolon. | Answer: $4,5,6 \| 4 ; 5 ; 6$ | 4;5;6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,145 |
3. In country Gamma, there are 83 cities. It is known that at least 2 roads lead out of each city. It is also known that if there is a road from city A to city B and from city B to city C, then there is also a road from city A to city C. What is the minimum number of roads that can be in the country? | Answer: 87
## Examples of answer notation:
17
# | 87 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,146 |
# Problem 1. (2 points)
The sum and product of three pairwise coprime numbers are divisible by 13. Can their sum of squares also be divisible by 13? | Answer: No.
## Solution:
Let these numbers be $x, y$, and $z$. Since the product of the numbers is divisible by 13, one of the numbers must be divisible by 13. Let this number be $x$, then from the condition of mutual simplicity, we get that neither $y$ nor $z$ is divisible by 13.
Next, $x^{2}+y^{2}+z^{2}=(x+y+z)^{2... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,147 |
# Problem 2. (2 points)
Does there exist a polynomial of the third degree such that all its roots are positive, and all the roots of its derivative are negative, given that both the polynomial and its derivative have at least one unique root? | Answer: Yes.
Solution:
For example, the polynomial $(x+1)^{3}-8$ works: its only root is 1, since $x+1=2$, and the only root of its derivative is -1. | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,148 |
# Problem 3. (3 points)
The sequence $\left\{a_{n}\right\}$ is defined by the conditions $a_{1}=-1$ and $a_{n+1}=\sum_{k=1}^{n}\left(a_{k}+1\right)$ for $n \geqslant 1$. Find the explicit formula for this sequence. | Answer: $a_{n}=2^{n-2}-1$ for $a>1, a_{1}=-1$
Solution:
By subtracting the formula for $a_{n}$ from that for $a_{n+1}$, we get $a_{n+1}-a_{n}=a_{n}+1$, that is, $a_{n+1}=2 a_{n}+1$ for $n>1$.
Now let's prove the formula $a_{n}=2^{n-2}-1$ by induction. The base case is $n=2$, and indeed, $a_{2}=a_{1}+1=0=2^{0}-1$.
N... | a_{n}=2^{n-2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,149 |
# Problem 4. (3 points)
Find $\operatorname{tg} x + \operatorname{ctg} x$, given that $\sec x - \operatorname{cosec} x = \sqrt{15}$.
Recall that $\sec x = 1 / \cos x$, and $\operatorname{cosec} x = 1 / \sin x$ | Answer: -3 or 5.
Solution:
We will use the following formulas: $\operatorname{tg} x+\operatorname{ctg} x=\frac{1}{\sin x \cos x} ; \sec x-\operatorname{cosec} x=$ $\frac{\sin x-\cos x}{\sin x \cos x}$.
Introduce the notations: $t=\sin x \cos x$ and $A=\sec x-\operatorname{cosec} x$.
Then $A^{2}=\frac{\cos ^{2} x-2 ... | -3or5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,150 |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $A B$ is equal to the diagonal $A C$. On the smaller arc $A D$ of the circumscribed circle of triangle $A B D$, a point $E$ is chosen such that $A B=A E$. Find the angle $\angle C E D$. | Answer: $90^{\circ}$.
Solution:
From the isosceles property of triangle $ABC$ and the parallelism of $BC$ and $AD$, we get $\angle ABC = \angle ACB = \angle CAD = \alpha$.
Let the line $BC$ intersect the circumcircle of triangle $ABD$ at point $F$. Then $ABFD$ is an inscribed, i.e., isosceles, trapezoid, from which ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,151 |
# Problem 6. (3 points)
In how many ways can natural numbers from 1 to 9 be arranged in a $3 \times 3$ square table so that the sum of the numbers in each row and each column is even? (Numbers can repeat) | Answer: $6 \cdot 5^{6} \cdot 4^{3}+9 \cdot 5^{4} \cdot 4^{5}+4^{9}$.
## Solution:
In each row and each column of the table, there can be either zero or two odd numbers, meaning either one or three even numbers. The total number of even numbers is an odd number from 3 to 9.
Let's consider several cases:
1) There are... | 6\cdot5^{6}\cdot4^{3}+9\cdot5^{4}\cdot4^{5}+4^{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,152 |
# Problem 7. (4 points)
A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $D$ and divides them in the ratio $5:1$ (not necessarily from vertex $D$). This plane also intersects the lines $AB$ and $AC$ at points $E$ and $F$. Find the ratio of the areas of triangles $AEF$ and $ABC$. | Answer: $1: 576$ or $625: 576$.
## Solution:
Let the points of intersection of the plane (which we will call $\alpha$) with the edges $D A, D B$, and $D C$ be $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. If $D A^{\prime}: A^{\prime} A = D B^{\prime}: B^{\prime} B$, the lines $A B$ and $A^{\prime} B^{\pri... | 1:576or625:576 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,153 |
# Problem 8. (5 points)
Girl Katya doesn't like the number 239. She listed several different numbers, none of which contain the sequence of digits 239 (in a row and exactly in that order). Prove that the sum of the reciprocals of these numbers is no more than 30000.
# | # Solution:
The number of suitable $3 n+1$-digit numbers is no more than $9 \cdot 999^{n}: 9$ options for the first digit and no more than 999 options for each subsequent triplet of digits. Each of them is no less than $10^{3 n}$.
The number of suitable $3 n+2$-digit numbers is no more than $90 \cdot 999^{n}$. Each o... | 30000 | Number Theory | proof | Yes | Yes | olympiads | false | 21,154 |
# Problem 1. (2 points)
The sum and product of three pairwise coprime numbers are divisible by 11. Can their sum of squares also be divisible by 11? | Answer: No.
Solution:
Let these numbers be $x, y$, and $z$. Since the product of the numbers is divisible by 11, one of the numbers must be divisible by 11. Let this number be $x$, then from the condition of mutual simplicity, we get that neither $y$ nor $z$ is divisible by 11.
Next, $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,155 |
# Problem 2. (2 points)
Does there exist a polynomial of the fifth degree such that all its roots are negative, and all the roots of its derivative are positive, given that both the polynomial and its derivative have at least one unique root? | Answer: Yes.
Solution:
For example, the polynomial $(x-1)^{5}+32$ works: its only root is -1, since $x-1=-2$, and the only root of its derivative is 1. | proof | Algebra | proof | Yes | Yes | olympiads | false | 21,156 |
# Problem 3. (3 points)
The sequence $\left\{a_{n}\right\}$ is defined by the conditions $a_{1}=1$ and $a_{n}=a_{1}+a_{2}+\ldots+a_{n-1}+n$ for $n \geqslant 2$. Find the explicit formula for this sequence. | Answer: $a_{n}=2^{n}-1$
Solution:
Subtracting the formulas for $a_{n+1}$ and $a_{n}$ from each other, we get $a_{n+1}-a_{n}=a_{n}+1$, that is, $a_{n+1}=2 a_{n}+1$ for $n>1$.
Now let's prove the formula $a_{n}=2^{n}-1$ by induction. The base case is $n=1$ and indeed, $a_{1}=1=2^{1}-1$.
Next, we will use the obtained... | a_{n}=2^{n}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,157 |
# Problem 4. (3 points)
Find $\operatorname{tg} x + \operatorname{ctg} x$, given that $\sec x - \operatorname{cosec} x = 4 \sqrt{3}$.
Recall that $\sec x = 1 / \cos x$, and $\operatorname{cosec} x = 1 / \sin x$ | Answer: -6 or 8.
Solution:
We will use the following formulas: $\operatorname{tg} x+\operatorname{ctg} x=\frac{1}{\sin x \cos x} ; \sec x-\operatorname{cosec} x=$ $\frac{\sin x-\cos x}{\sin x \cos x}$.
Introduce the notations: $t=\sin x \cos x$ and $A=\sec x-\operatorname{cosec} x$.
Then $A^{2}=\frac{\cos ^{2} x-2 ... | -6or8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,158 |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $B C$ is equal to the diagonal $B D$. On the smaller arc $A B$ of the circumscribed circle of triangle $A B C$, a point $E$ is chosen such that $B C=B E$. Find the angle $\angle A E D$. | Answer: $90^{\circ}$.
## Solution:
From the isosceles triangle $B C D$ and the parallelism of $A B$ and $C D$, we get $\angle B C D = \angle B D C = \angle D B A = \alpha$.
Let the line $C D$ intersect the circumcircle of triangle $A B C$ at point $F$. Then $B C F A$ is an inscribed, i.e., isosceles, trapezoid, from... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,159 |
# Problem 6. (3 points)
In how many ways can natural numbers from 1 to 7 be arranged in a $3 \times 3$ square table so that the sum of the numbers in each row and each column is odd? (Numbers can repeat) | Answer: $6 \cdot 3^{6} \cdot 4^{3}+9 \cdot 3^{4} \cdot 4^{5}+4^{9}$.
Solution:
In each row and each column of the table, there can be either one or three odd numbers. The total number of odd numbers is an odd number from 3 to 9.
Let's consider several cases:
1) There are a total of three odd numbers, one in each ro... | 6\cdot3^{6}\cdot4^{3}+9\cdot3^{4}\cdot4^{5}+4^{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,160 |
# Problem 7. (4 points)
A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $C$ and divides them in the ratio $4:1$ (not necessarily from vertex $C$). This plane also intersects the lines $AB$ and $BD$ at points $E$ and $F$. Find the ratio of the areas of triangles $BEF$ and $ABD$. | Answer: $1: 225$ or $256: 225$.
## Solution:
Let the points of intersection of the plane (which we will call $\alpha$) with the edges $C A, C B$, and $C D$ be $A^{\prime}, B^{\prime}$, and $D^{\prime}$, respectively. If $C A^{\prime}: A^{\prime} A = C B^{\prime}: B^{\prime} B$, the lines $A B$ and $A^{\prime} B^{\pri... | 1:225or256:225 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,161 |
# Problem 8. (5 points)
Boy Kolya doesn't like the number 1234. He listed several different numbers, none of which contain the sequence of digits 1234 (in a row and exactly in that order). Prove that the sum of the reciprocals of these numbers is no more than 400000. | Solution:
The number of suitable $4 n+1$-digit numbers is no more than $9 \cdot 9999^{n}: 9$ options for the first digit and no more than 9999 options for each subsequent quartet of digits. Each of them is no less than $10^{4 n}$.
The number of suitable $4 n+2$-digit numbers is no more than $90 \cdot 9999^{n}$. Each ... | 400000 | Number Theory | proof | Yes | Yes | olympiads | false | 21,162 |
# Problem 1. (2 points)
The sum and product of three pairwise coprime numbers are divisible by 7. Can their sum of squares also be divisible by 7? | Answer: No.
Solution:
Let these numbers be $x, y$, and $z$. Since the product of the numbers is divisible by 7, one of the numbers must be divisible by 7. Let this number be $x$, then from the condition of mutual simplicity, we get that neither $y$ nor $z$ is divisible by 7.
Next, $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,163 |
# Problem 3. (3 points)
The sequence $\left\{a_{n}\right\}$ is defined by the conditions $a_{1}=0 \quad$ and $a_{n+1}=\sum_{k=1}^{n}\left(a_{k}+1\right)$ for $n \geqslant 1$. Find the explicit formula for this sequence.
Answer: $a_{n}=2^{n-1}-1$ | Solution:
By subtracting the formulas for $a_{n+1}$ and $a_{n}$ from each other, we get $a_{n+1}-a_{n}=a_{n}+1$, that is, $a_{n+1}=2 a_{n}+1$ for $n>1$.
Now let's prove the formula $a_{n}=2^{n-1}-1$ by induction. The base case is $n=1$ and indeed, $a_{1}=0=2^{0}-1$.
Next, we will use the obtained recurrence formula ... | a_{n}=2^{n-1}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,164 |
# Problem 4. (3 points)
Find $\operatorname{tg} x + \operatorname{ctg} x$, given that $\sec x - \operatorname{cosec} x = \sqrt{35}$.
Recall that $\sec x = 1 / \cos x$, and $\operatorname{cosec} x = 1 / \sin x$ | Answer: -5 or 7.
Solution:
We will use the following formulas: $\operatorname{tg} x+\operatorname{ctg} x=\frac{1}{\sin x \cos x} ; \sec x-\operatorname{cosec} x=$ $\frac{\sin x-\cos x}{\sin x \cos x}$.
Let's introduce the notations: $t=\sin x \cos x$ and $A=\sec x-\operatorname{cosec} x$.
Then $A^{2}=\frac{\cos ^{2... | -5or7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,165 |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $C D$ is equal to the diagonal $A C$. On the smaller arc $B C$ of the circumscribed circle of triangle $B C D$, a point $E$ is chosen such that $C D=C E$. Find the angle $\angle A E B$. | Answer: $90^{\circ}$.
Solution:
From the isosceles triangle $A C D$ and the parallelism of $B C$ and $A D$, we get $\angle A D C = \angle C A D = \angle A C B = \alpha$.
Let the line $A D$ intersect the circumcircle of triangle $B C D$ at point $F$. Then $C D F B$ is an inscribed, i.e., isosceles, trapezoid, from wh... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,166 |
# Problem 6. (3 points)
In how many ways can natural numbers from 1 to 13 be arranged in a $3 \times 3$ square table so that the sum of the numbers in each row and each column is even? (Numbers can repeat) | Answer: $6 \cdot 7^{6} \cdot 6^{3}+9 \cdot 7^{4} \cdot 6^{2}+6^{9}$.
## Solution:
In each row and each column of the table, there can be either zero or two odd numbers, meaning either one or three even numbers. The total number of even numbers is an odd number from 3 to 9.
Let's consider several cases:
1) There are... | 6\cdot7^{6}\cdot6^{3}+9\cdot7^{4}\cdot6^{2}+6^{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,167 |
# Problem 7. (4 points)
A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $B$ and divides them in the ratio $2:1$ (not necessarily from vertex $B$). This plane also intersects the lines $CD$ and $AC$ at points $E$ and $F$. Find the ratio of the areas of triangles $EFC$ and $ACD$. | Answer: $1: 9$ or $16: 9$.
Solution:
Let the points of intersection of the plane (let's call it $\alpha$) with the edges $B A, B C$, and $B D$ be $A^{\prime}, C^{\prime}$, and $D^{\prime}$, respectively. If $B A^{\prime}: A^{\prime} A=B C^{\prime}: C^{\prime} C$, the lines $A C$ and $A^{\prime} C^{\prime}$ are parall... | 1:9or16:9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,168 |
# Problem 8. (5 points)
Girl Masha doesn't like the number 729. She listed several different numbers, none of which contain the sequence of digits 729 (in a row and exactly in that order). Prove that the sum of the reciprocals of these numbers is no more than 30000.
# | # Solution:
The number of suitable $3 n+1$-digit numbers is no more than $9 \cdot 999^{n}: 9$ options for the first digit and no more than 999 options for each subsequent triplet of digits. Each of them is no less than $10^{3 n}$.
The number of suitable $3 n+2$-digit numbers is no more than $90 \cdot 999^{n}$. Each o... | 30000 | Number Theory | proof | Yes | Yes | olympiads | false | 21,169 |
# Problem 1. (2 points)
The sum and product of three pairwise coprime numbers are divisible by 17. Can their sum of squares also be divisible by 17? | Answer: No.
Solution:
Let these numbers be $x, y$, and $z$. Since the product of the numbers is divisible by 17, one of the numbers must be divisible by 17. Let this number be $x$, then from the condition of mutual simplicity, we get that neither $y$ nor $z$ is divisible by 17.
Next, $x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,170 |
# Problem 3. (3 points)
The sequence $\left\{a_{n}\right\}$ is defined by the conditions $a_{1}=2$ and $a_{n}=a_{1}+a_{2}+\ldots+a_{n-1}+n$ for $n \geqslant 2$. Find the explicit formula for this sequence. | Answer: $a_{n}=5 \cdot 2^{n-2}-1$ for $a>1, a_{1}=2$
Solution:
By subtracting the formula for $a_{n}$ from the formula for $a_{n+1}$, we get $a_{n+1}-a_{n}=a_{n}+1$, that is, $a_{n+1}=2 a_{n}+1$ for $n>1$.
Now let's prove the formula $a_{n}=5 \cdot 2^{n-2}-1$ by induction. The base case is $n=2$, and indeed, $a_{2}=... | a_{n}=5\cdot2^{n-2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,171 |
# Problem 4. (3 points)
Find $\operatorname{tg} x + \operatorname{ctg} x$, given that $\sec x - \operatorname{cosec} x = 2 \sqrt{6}$.
Recall that $\sec x = 1 / \cos x$, and $\operatorname{cosec} x = 1 / \sin x$ | Answer: -4 or 6.
Solution:
We will use the following formulas: $\operatorname{tg} x+\operatorname{ctg} x=\frac{1}{\sin x \cos x} ; \sec x-\operatorname{cosec} x=$ $\frac{\sin x-\cos x}{\sin x \cos x}$.
Introduce the notations: $t=\sin x \cos x$ and $A=\sec x-\operatorname{cosec} x$.
Then $A^{2}=\frac{\cos ^{2} x-2 ... | -4or6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,172 |
# Problem 5. (3 points)
In trapezoid $A B C D$, the lateral side $A D$ is equal to the diagonal $B D$. On the smaller arc $C D$ of the circumscribed circle of triangle $A C D$, a point $E$ is chosen such that $A D=D E$. Find the angle $\angle B E C$. | Answer: $90^{\circ}$.
## Solution:
From the isosceles triangle $B A D$ and the parallelism of $A B$ and $C D$, we get $\angle B A D = \angle D B A = \angle B D C = \alpha$.
Let the line $A B$ intersect the circumcircle of triangle $A C D$ at point $F$. Then $D A F C$ is an inscribed, i.e., isosceles, trapezoid, from... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,173 |
# Problem 6. (3 points)
In how many ways can natural numbers from 1 to 9 be arranged in a $3 \times 3$ square table so that the sum of the numbers in each row and each column is odd? (Numbers can repeat) | Answer: $6 \cdot 4^{6} \cdot 5^{3}+9 \cdot 4^{4} \cdot 5^{2}+5^{2}$.
Solution:
In each row and each column of the table, there can be either one or three odd numbers. The total number of odd numbers is an odd number from 3 to 9.
Let's consider several cases:
1) There are a total of three odd numbers, one in each ro... | 6\cdot4^{6}\cdot5^{3}+9\cdot4^{4}\cdot5^{2}+5^{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,174 |
# Problem 7. (4 points)
A plane intersects the edges of the tetrahedron $ABCD$ emanating from vertex $A$ and divides them in the ratio $3:1$ (not necessarily from vertex $A$). This plane also intersects the lines $BC$ and $CD$ at points $E$ and $F$. Find the ratio of the areas of triangles $EFC$ and $BCD$. | Answer: $1: 64$ or $81: 64$.
## Solution:
Let the points of intersection of the plane (which we will call $\alpha$) with the edges $A B, A C$, and $A D$ be $B^{\prime}, C^{\prime}$, and $D^{\prime}$, respectively. If $A B^{\prime}: B^{\prime} B = A C^{\prime}: C^{\prime} C$, the lines $B C$ and $B^{\prime} C^{\prime}... | 1:64or81:64 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,175 |
# Problem 8. (5 points)
Boy Anton does not like the number 2048. He listed several different numbers, none of which contain the sequence of digits 2048 (in a row and exactly in this order). Prove that the sum of the reciprocals of these numbers is no more than 400000.
# | # Solution:
The number of suitable $4 n+1$-digit numbers is no more than $9 \cdot 9999^{n}: 9$ options for the first digit and no more than 9999 options for each subsequent quartet of digits. Each of them is no less than $10^{4 n}$.
The number of suitable $4 n+2$-digit numbers is no more than $90 \cdot 9999^{n}$. Eac... | 400000 | Number Theory | proof | Yes | Yes | olympiads | false | 21,176 |
3. $\mathrm{ABCD}$ is an isosceles trapezoid, $\mathrm{AB}=\mathrm{CD}=25, \mathrm{BC}=40, \mathrm{AD}=60$. $\mathrm{BCDE}$ is also an isosceles trapezoid. Find AE. (Points A and E do not coincide) If there are multiple possible values, list them in any order separated by a semicolon. | Answer: 44
## Examples of answer recording:
45
$45 ; 56$
Problem 10 (4 points). | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,180 |
1. (2 points) Does there exist a four-digit natural number with the sum of its digits being 21, which is divisible by $14 ?$ | Answer: Yes, for example 6384. | 6384 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,181 |
2. (2 points) Boys were collecting apples. Each collected either 10 apples or $10\%$ of the total number of apples collected, and there were both types. What is the smallest number of boys there could have been? | # Answer: 6
Solution: Example: one boy collected 10 apples, the rest collected 2 each, totaling 20.
Evaluation: Let's prove that there could not have been fewer boys. We know that $10\%$ of the total number of apples is an integer, let's denote it as $k$. Suppose $n$ people collected ten apples each, and $m$ people c... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,182 |
3. (3 points) Solve the system of equations: $\left\{\begin{array}{l}x y=6(x+y) \\ x z=4(x+z) \\ y z=2(y+z)\end{array}\right.$ | Answer: $x=y=z=0$ or $x=-24, y=\frac{24}{5}, y=\frac{24}{7}$
Solution: It is easy to verify that if one of the variables is 0, then the others are also 0. Otherwise, the system can be transformed into
$\left\{\begin{array}{l}\frac{1}{6}=\frac{1}{x}+\frac{1}{y} \\ \frac{1}{4}=\frac{1}{x}+\frac{1}{z} \\ \frac{1}{2}=\fr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,183 |
5. (3 points) Given the rebus: MIMIMI + NYANYANYA = OLALAOY. Identical letters represent identical digits, different letters represent different digits. Find MI + NY.
# | # Answer: 119
Solution: The rebus can be rewritten as OLAOLOY = (MI + NYA) $\cdot$10101. First, this means that the last digit of MI + NYA is Y. Second, if MI + NYA $<100$, the result will be a four-digit number. Let MI + NYA $=1 Z$ Y, where $Z$ is some digit.
Then OLAOLOY $=1 Z$ Y $0000+1 Z$ Y $00+1 Z$ Y. If Y $<9$,... | 119 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,185 |
6. (3 points) A quadrilateral is divided into 1000 triangles. What is the maximum number of different points at which the vertices of these triangles can be located? | Answer: 1002
Solution: The sum of the angles of a quadrilateral is $360^{\circ}$, and for thousands of triangles, it is $180000^{\circ}$. Where did the extra $1796400^{\circ}$ come from? Each internal vertex, where only the angles of triangles meet, adds $360^{\circ}$. Each vertex on the side of the original triangle ... | 1002 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,186 |
7. (4 points) Given a right triangle $A B C$, where angle $A$ is 60 degrees, and the hypotenuse $A B$ is $4+4 \sqrt{3}$. Through vertex $B$, a line $p$ parallel to $A C$ is drawn. Points $D$ and $E$ are placed on line $p$ such that $A B=B D, B C=B E$. $F$ is the intersection point of lines $A D$ and $C E$. Find what th... | Solution: First, we calculate the sides of triangle $ABC$: side $AC$ lies opposite the angle $30^{\circ}$, so it is half the hypotenuse, which is $2+2\sqrt{3}$. Now, using the Pythagorean theorem, we calculate $BC = \sqrt{(4+4\sqrt{3})^2 - (2+2\sqrt{3})^2} = \sqrt{16+32\sqrt{3}+48-4-8\sqrt{3}-12} = \sqrt{36+24\sqrt{3}+... | 4\sqrt{2}+2\sqrt{6}+18+10\sqrt{3} | Geometry | MCQ | Yes | Yes | olympiads | false | 21,187 |
8. (5 points) On the Island of Eye, there live knights who always tell the truth, and liars who always lie. In addition, all residents of the island are either blue-eyed or brown-eyed. One day, 100 residents of the island met, and each one said to each other one of two phrases: "You are a liar" or "You are blue-eyed," ... | Answer: The answer is from 46 to 54.
Solution: Note that at the meeting, there could not have been brown-eyed knights and blue-eyed liars, because no knight can say the given phrase, and no liar can say it either. Therefore, the islanders at the meeting are divided into blue-eyed knights and brown-eyed liars. Thus, th... | 46to54 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,188 |
1. (2 points) Does there exist a four-digit natural number with the sum of its digits being 14, which is divisible by $14 ?$ | Answer: Yes, for example 6314. | 6314 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,189 |
3. (3 points) Solve the system of equations: $\left\{\begin{array}{l}x y=5(x+y) \\ x z=4(x+z) \\ y z=2(y+z)\end{array}\right.$ | Answer: $x=y=z=0$ or $x=-40, y=\frac{40}{9}, y=\frac{40}{11}$
Solution: It is easy to verify that if one of the variables is 0, then the others are also 0. Otherwise, the system can be transformed into
$$
\left\{\begin{array}{l}
\frac{1}{5}=\frac{1}{x}+\frac{1}{y} \\
\frac{1}{4}=\frac{1}{x}+\frac{1}{2} \\
\frac{1}{2}... | -40,\frac{40}{9},\frac{40}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,191 |
5. (3 points) Given the puzzle: ЛЯЛЯЛЯ + ФУФУФУ = ГГЫГЫЫР. Identical letters represent identical digits, different letters represent different digits. Find ЛЯ + ФУ. | Answer: 109
Solution: The rebus can be rewritten as GGUGUUU $=(\text{LYA} + \text{FU}) 10101$. First, this means that the last digit of $\text{LYA} + \text{FU}$ is R. Second, if $\text{LYA} + \text{FU} < 100$, the result will be a four-digit number. Let $\text{LYA} + \text{FU} = 1ZP$, where $Z$ is some digit.
Then OL... | 109 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,193 |
7. (4 points) Given a right triangle $A B C$, where angle $A$ is 60 degrees, and the hypotenuse $A B$ is $2+2 \sqrt{3}$. Through vertex $B$, a line $p$ parallel to $A C$ is drawn. Points $D$ and $E$ are placed on line $p$ such that $A B=B D, B C=B E$. $F$ is the intersection point of lines $A D$ and $C E$. Find what th... | Solution: First, we calculate the sides of triangle $ABC$: side $AC$ lies opposite the angle $30^{\circ}$, so it is half the hypotenuse, which is $1+\sqrt{3}$. Now, using the Pythagorean theorem, we calculate $BC=\sqrt{(2+2\sqrt{3})^{2}-(1+1\sqrt{3})^{2}}=\sqrt{4+8\sqrt{3}+12-1}$ $\sqrt{9+6\sqrt{3}+3}=6+2\sqrt{3}$.
Tr... | 2\sqrt{2}+\sqrt{6}+9+5\sqrt{3},\sqrt{3}+1+\sqrt{2},9+5\sqrt{3}+2\sqrt{6}+3\sqrt{2},1+\sqrt{3}+\sqrt{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,194 |
8. (5 points) On the Hairy Island, there live knights who always tell the truth, and liars who always lie. In addition, all residents of the island are either blondes or brunettes. One day, 200 residents of the island met, and each one said to each other one of two phrases: "You are a liar" or "You are a blonde," and t... | Answer: The answer is from 93 to 107.
Solution: Note that at the meeting, there could not have been brown-eyed knights and blue-eyed liars, as no knight can say the given phrase, and no liar can say it either. Therefore, the islanders at the meeting are divided into blue-eyed knights and brown-eyed liars. Thus, the ph... | 93to107 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,195 |
3. Alice has six segments of different lengths. Five of them have lengths $6,7,8,9$ and 11. Any three of them can form a triangle. What integer values can the length of the fifth segment take? If there are multiple possible answers, write them in any order separated by a semicolon. | Answer: $12 ; 10\|12,10\| 10 ; 12 \| 10,12$
## Examples of answer notation:
5
## $5 ; 9$
# | 10;12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,198 |
1. In the room, there are knights who always tell the truth, and liars who always lie. 10 of them said: "In this room, there are more knights than liars." 12 said: "In this room, there are more liars than knights." The remaining 22 said: "In this room, there are an equal number of liars and knights." How many liars cou... | Answer: $22,32\|32,22\| 22 ; 32 \| 32 ; 22$ | 22;32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,199 |
2. In the room, there are knights who always tell the truth, and liars who always lie. 10 of them said: "In this room, there are more knights than liars." 15 said: "In this room, there are more liars than knights." The remaining 25 said: "In this room, there are an equal number of liars and knights." How many liars cou... | Answer: $25,35\|35,25\| 25 ; 35 \| 35 ; 25$ | 25;35 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,200 |
3. In the room, there are knights who always tell the truth, and liars who always lie. 15 of them said: "In this room, there are more knights than liars." 12 said: "In this room, there are more liars than knights." The remaining 27 said: "In this room, there are an equal number of liars and knights." How many liars cou... | Answer: $27,42\|42,27\| 27 ; 42 \| 42 ; 27$
## Examples of answer notation:
45
$45 ; 60$
# | 27;42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,201 |
3. Ani had two cups of tea with milk, with volumes $\mathrm{V}$ and $5 \mathrm{~V}$, and the percentage of milk in them was $5 \%$ and $2 \%$ of the volume, respectively. Ani poured both these cups into one large mug, adding 10 spoons of milk, resulting in a solution with a milk content of $10 \%$. What fraction of the... | Answer: $0.05\|0.05\| 5 \%\|1 / 20\| 5 / 100$
## Examples of answer notation:
0.01
$1 \%$
$1 / 100$ | 0.05 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,205 |
1. On the coordinate plane, lines of the form $y=a x+b$ are drawn, where $a$ and $b$ are natural numbers from 1 to 9. Among all the intersection points of these lines, select the point with the greatest sum of coordinates. Write the answer in the form ( $x ; y$ ). | Answer: $(8 ; 73)\|(8,73)\| 8 ; 73 \| 8,73$ | (8,73) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,208 |
3. Boy Zhenya lives in a building with a total of 100 floors. The elevator takes 1 second to travel between adjacent floors. Unfortunately, only two buttons in the elevator work: “+13 floors” and “-7 floors”. The buttons are pressed instantly.
How many seconds will it take Zhenya to get home from the 1st floor to the ... | Answer: 107
Examples of writing answers:
45 | 107 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,210 |
3. Two quadratic trinomials have a common root -3, and for one of them, it is the larger root, while for the other, it is the smaller root. The length of the segment cut off by the graphs of these trinomials on the y-axis is 12. Find the length of the segment cut off by the graphs of the trinomials on the x-axis. | Answer: 4
Allowed for input are digits, the division sign, and a period or comma as a decimal separator
# | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,211 |
# Problem 6. (3 points)
Quadrilateral $A B C D$ is inscribed in a circle with diameter $A C=65$. It is also known that $B C=$ $39, A D=25$. Find the distance between the feet of the perpendiculars dropped from $B$ and $D$ to $A C$. | Write the answer in the form of an improper fraction, do not round.
# | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,212 |
1. In a computer game, a turtle moves across a grid on the computer screen, which contains 5 columns and 7 rows. Initially, it is located at the bottom-left corner of the screen - on the cell with coordinates $(0,0)$. If the program instructs the turtle to move off the screen, it reappears on the opposite side - for ex... | Answer: $(2 ; 6)\|(2,6)\| 2 ; 6 \| 2,6$ | (2,6) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,213 |
3. 4. In a computer game, a turtle moves across a grid on the computer screen, which contains 5 columns and 9 rows. Initially, it is located at the bottom left corner of the screen - on the cell with coordinates $(0,0)$. If the program instructs the turtle to move off the screen, it reappears on the opposite side - for... | Answer: $(2 ; 1)\|(2,1)\| 2 ; 1 \| 2,1$
## Examples of answer notation:
$(1 ; 7)$
# | (2,1) | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,214 |
3. A 103-gon is inscribed in a circle with diameter $\mathrm{XY}=4$ and has an axis of symmetry perpendicular to this diameter. Find the sum of the squares of the distances from the vertices of the 103-gon to the point $X$. | Answer: 824
## Examples of how to write the answer:
1,7
$1 / 7$
17
# | 824 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,215 |
1. Given 50 quadratic trinomials. The graphs of these trinomials have their vertices on the $x$-axis. No three graphs intersect at the same point. What is the minimum number of intersection points they can form? | Answer: 600
## Examples of how to write the answer:
## 17
# | 600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,216 |
3. A circle of radius 60 touches three sides of quadrilateral $A B C D$: side $A B$ at point A, side $C D$ at point D, and side $B C$ at point $X$. It turns out that $\mathrm{BX}=20, \mathrm{CX}=45$. Find the area of the quadrilateral. | Answer: 2215.2
## Examples of how to write the answer:
# | 2215.2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,217 |
3. From the digits $1,2,3,4,5,6,7$, all possible seven-digit numbers with all different digits were formed. The obtained numbers were written in ascending order. Find which number will be in the 1972nd place. | Answer: 3641572
## Examples of how to write the answer:
1234567
# | 3641572 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,218 |
3. On the plane, points $A(-2,0), B(-1,4), C(1,3), D(0,-2)$ are given. What is the area of figure $A B C D$? | Answer: $10.5$.
## Examples of how to write answers:
17
$1 / 7$
1.7
# | 10.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,219 |
3. Given an equilateral triangle $A B C$. Inside it, a point $D$ is taken such that it divides triangle $A B C$ into three triangles $A B D, B C D, A C D$ such that one of them is a right triangle, and another is an isosceles triangle with a vertex angle of 172 degrees. Find all possible values of angle $B C D$ in degr... | Answer: $4,26,34,56 \| 4 ; 26 ; 34 ; 56$ | 4;26;34;56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 21,222 |
3. On Misfortune Island, a table tennis tournament was held in a round-robin format (i.e., everyone played against everyone else once). After each match, both participants separately approached the Chief Referee and reported the result. Among the participants, there were only three types of people: knights, who always ... | Answer: 16
## Examples of answer recording:
17
# | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,224 |
3. Two racers with speeds of $100 \mathrm{~m} / \mathrm{c}$ and $70 \mathrm{~m} / \mathrm{c}$ started simultaneously in the same direction from the same place on a circular track of length $800 \mathrm{m}$. How many times did the racers meet after the start if both drove for 320 seconds? | Answer: 12
## Examples of writing answers:
45
## Problem 4 (2 points). | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,226 |
3. Mitya wrote on the board the sum of the digits of each of the numbers from 1 to 800 inclusive. What is the sum of the numbers on the board | Answer: 10008
## Examples of answer notation:
12345
# | 10008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,228 |
3. Boy Zhenya lives in a building with a total of 100 floors. The elevator takes 1 second to travel between adjacent floors. Unfortunately, only two buttons in the elevator work: “+13 floors” and “-7 floors”. The buttons are pressed instantly.
How many seconds will it take Zhenya to get home from the 1st floor to the ... | Answer: 107
Examples of writing answers:
45 | 107 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,232 |
1. Seven dwarfs lined up by height, starting with the tallest. The first (i.e., the tallest) said: "My height is 60 cm." The second said: "My height is 61 cm." In order next: "My height is 62 cm," "My height is 63 cm," "My height is 64 cm," "My height is 65 cm," and finally the shortest said: "My height is 66 cm." What... | 1. Answer: 1.
No two dwarfs could tell the truth, because in this case the taller one would name the shorter height. However, one dwarf (any one) could tell the truth. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,233 |
2. Cat food is sold in large and small bags. There is more food in a large bag than in a small one, but less than in two small bags. One large and two small bags of cat food are enough for exactly two days. Will 4 large and 4 small bags of cat food be enough for six days? | 2. Answer: no.
4 large and 4 small bags contain less food than 3 large and 6 small bags, and they last exactly 6 days. Therefore, 4 large and 4 small bags would not be enough for the cat for 6 days. | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,234 |
3. Zhenya had 9 cards with numbers from 1 to 9. He lost the card with the number 7. Can the remaining 8 cards be arranged in a row so that any two adjacent cards form a number divisible by 7? | 3. Answer: no.
Note that after 5 there must be 6, after $6-3$, after $3-5$, after $5-6$. No other digits can precede these either. Therefore, it is impossible to arrange all these digits in one row. | no | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,235 |
5. A magician and an assistant perform the following trick. A spectator arranges 11 coins in a circle (each showing either heads or tails). The assistant then covers all the coins with circles, except for one. Then the magician enters and points to another (covered) coin that lies the same way as the open one. Explain ... | 5. The magician agrees with the assistant that he will point to the next (clockwise) coin after the uncovered one. Since there is an odd number of coins, some two consecutive coins will lie the same side up. The assistant leaves the first of them (if looking clockwise) open, and closes the rest. The magician points to ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,237 |
6. In the country, there are 100 cities, divided into three republics. Some cities are connected by air routes. The Minister of Communications found no fewer than 70 such cities, from each of which no fewer than 70 air routes depart. Prove that some air route connects two cities of the same republic. | 6. Assume the opposite: all airlines connect cities of different republics. Let's call millionaires those cities from which at least 70 airlines depart. If a republic has a millionaire, then such a republic has no more than 30 cities. Since there are 70 millionaires in total, they are present in all three republics, wh... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,238 |
9. Two hydras with 2016 and 2017 heads have contracted a strange disease. Each week, they lose 4 heads in total, and each one grows either 5 or 7 new heads. Death from the disease will occur if and only if the number of heads on the hydras becomes equal. The treating physician claims that the hydras will survive no mat... | 9. Answer: Yes.
Note that the total number of heads is always odd. Thus, the doctor is right, and the number of heads cannot be equal. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,241 |
10. The metro of city $\mathrm{N}$ consists of three lines, its scheme is shown in the figure. Initially, the trains are at the marked stations and start moving in the direction indicated by the arrow. Every minute, each train travels exactly one section between stations. Upon arriving at a terminal station, the train ... | 10. Note that the train takes exactly 7 minutes to travel the entire red branch, 8 minutes for the blue branch, and 9 minutes for the green branch. This means the trains return to their initial positions on the red branch every 14 minutes, on the blue branch every 16 minutes, and on the green branch every 9 minutes. Si... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,242 |
11. A lot of snow fell, and the children decided to build snowmen. For this, they rolled 99 snowballs with masses of 1 kg, 2 kg, 3 kg, ..., 99 kg. A snowman consists of three snowballs stacked on top of each other, and one snowball can be placed on another if and only if the mass of the first is at least twice as small... | 11. Answer: 24.
Two snowballs with masses from 50 to 99 kg cannot end up in the same snowman, so in each snowman, there will be at least two snowballs with masses from 1 to 49 kg. Therefore, there cannot be more than 24 snowmen.
An example can be constructed greedily: first, take three snowballs with masses, for exam... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,243 |
12. The king has 100 greyhounds, and he wants to select three for hunting. He gathered 100 courtiers and asked each one which three dogs they consider the fastest. It turned out that for any two courtiers, there are two dogs that both of them pointed out. Prove that the choice of some two courtiers completely coincided... | 12. Let's take a courtier, for example, he pointed out dogs 1, 2, and 3. Suppose also that no one else made the same choice. Then everyone is divided into 3 groups: those who chose dogs numbered 1 and 2, those who chose dogs 1 and 3, and those who chose the 2nd and 3rd dogs.
If at least two groups are non-empty, then ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 21,244 |
15. Cat food is sold in large and small bags (there is more food in a large bag than in a small one). One large and four small bags of cat food last exactly two weeks for a cat. Is it necessarily true that one large and three small bags will last for 11 days? | 15. Answer: enough.
Suppose it is not enough. Then, one small packet (which is the difference between these sets) lasts for more than three days. Therefore, five small packets would last for more than 15 days, and one large packet and four small packets even more so. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,247 |
17. Find the largest natural number that cannot be represented as the sum of two composite numbers. (Recall that a natural number is called composite if it is divisible by some natural number other than itself and one.) | 17. Answer: 11.
All large numbers can be represented in the required way: odd numbers as "9 + an even number greater than two", and even numbers as "8 + an even number greater than two". | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,249 |
18. Among 65 residents of the village, two are tricksters, and the rest are knights. A knight always tells the truth, while a trickster can either tell the truth or lie. You can show any resident a list of a group of residents (including a group of one person) and ask if they are all knights. How can you find both tric... | 18. In fact, 16 questions are enough (and this is not the minimum).
In the first four questions, we will find a knight. We will do this as follows: ask the first and second residents about the third. If both answered that he is a knight, then he is a knight (otherwise, all three are liars, which is impossible).
If no... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,250 |
19. Grandfather Frost has many identical dials in the form of regular 12-sided polygons, on which numbers from 1 to 12 are printed. He places these dials in a stack on top of each other (one by one, face up). In doing so, the vertices of the dials coincide, but the numbers in the coinciding vertices do not necessarily ... | 19. Answer: 12.
Let there be $k$ clock faces in the stack. Consider any two adjacent columns. The sum of the numbers in them differs by either $k$ or $k-12s$ (where $s$ is the number of clock faces for which one column has 12 and the next has 1). For $k=1,2, \ldots, 11$, all these differences are not divisible by 12, ... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,251 |
20. In the city, there are 10 parallel streets running from north to south, and another 10 streets running from west to east, all forming a $9 \times 9$ square. At each intersection, there is a square (a total of 100 squares). The distance from each square to the next along any street is 1 km. A pharmacy within walking... | 20. Answer. Not enough.
Solution. The position of each moving person is characterized by a segment of the street between squares (there are $2 \cdot 10 \cdot 9=180$ such segments) and the direction of movement - a total of 360 positions. Each pharmacy uses 36 positions (i.e., in 36 positions, this pharmacy will be wit... | Notenough | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,252 |
21. Baron Münchhausen told a story. "A whole crowd gathered. We reached a crossroads. Then half of our group turned left, a third - right, and a fifth - straight ahead." "But excuse me," the duke noticed, - the sum of a half, a third, and a fifth is not equal to one, so you are lying!" The baron objected: "I am not lyi... | 21. Answer: 37 (18 to the left, 12 to the right, 7 straight).
Evaluation: $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}=\frac{31}{30}$. If there were a total of people, then $\frac{1}{30} \cdot$ is the error. Let's calculate the maximum error (in fractions of a person): $\frac{1}{2}+\frac{1}{3}+\frac{2}{5}=\frac{37}{30}$. Ther... | 37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,253 |
24. Andryusha wrote a five-digit number. Then he wrote another five-digit number, differing from the first by the transposition of two digits. The sum of these numbers is 111111. Prove that the digit 0 appears in the number. | 24. Obviously, two five-digit numbers were added. Moreover, they changed the last digit and another one (let's call this place the lucky one). This means that the sum of the digits in the last and lucky places ends in 1. Let's assume the lucky place is the second to last. Then the sum of the digits in the last two plac... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 21,255 |
25. From 15 sticks, five triangles were formed. Is there necessarily another way to form five triangles from these sticks? | 25. Answer: no.
Example set of stick lengths:
$2,3,4$
$20,30,40$
$200,300,400$
$2000,3000,4000$
$20000,30000,40000$ | no | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,256 |
26. Point $M$ is the midpoint of side $A C$ of triangle $A B C$. Point $N$ on segment $A M$ is such that angle $M B N$ equals angle $C B M$. On the extension of segment $B N$ beyond point $N$, point $K$ is chosen such that angle $B M K$ is a right angle. Prove that $B C = A K + B K$. | 26. Let's construct the segment $K M$ and extend it until it intersects with $B C$ (point $X$). Notice that in triangle $K B X$, $B M$ is both an altitude and an angle bisector (since angle $B M K$ is a right angle, and angles $K B M$ and $M B C$ are equal). Therefore, triangle $K B X$ is isosceles, and $B K = B X$. Fr... | BC=AK+BK | Geometry | proof | Yes | Yes | olympiads | false | 21,257 |
27. Solve the equation $(a+b) c!=(a!+b!) c$ in natural numbers | 27. Answer: $\mathrm{c}=1, \mathrm{a}=1, \mathrm{~b}=2 ; \mathrm{c}=1, \mathrm{a}=2, \mathrm{~b}=1 ; \mathrm{c}=1, \mathrm{a}=2, \mathrm{~b}=2 ; \mathrm{c}=\mathrm{a}=\mathrm{b}-$
any.
Let's assume that $a \geqslant b$. Then there are three cases:
Case 1: $c>a$.
Then $(a+b) c!=(a+b) a! \cdot x \cdot c > 2 \cdot a! \... | =1,=1,b=2;=1,=2,b=1;=1,=2,b=2;==b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 21,258 |
28. A magician and an assistant perform the following trick. A spectator arranges 27 coins in a row (each showing either heads or tails). The assistant then covers all the coins except for five that are lying in the same way. Then the magician enters and points out another five coins lying in the same way as the uncove... | 28. We divide the coins into consecutive triples. In each triple, there are either more heads than tails or vice versa. Let's say there are more heads in five triples. The assistant covers one coin from these triples, and the magician guesses one more coin from these triples. The agreement can be as follows: if the two... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 21,259 |
30. Solve the equation $p^{2}-6 p q+q^{2}+3 q-1=0$ in prime numbers. | 30. Answer: $p=17, q=3$.
Let's look at the remainders when dividing by 3, we get $p^{2}+q^{2} \equiv 1(\bmod 3)$, this is only possible if one of the numbers is 3.
Let $q=3$, we get the quadratic equation: $p^{2}-18 p+17=0$. This gives us $p=1$ or 17. 1 is not a prime number, so $p=17$.
Let $p=3$, then $q^{2}-15 q+8... | p=17,q=3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 21,261 |
32. In the country, there are 1000000 people, each of whom is acquainted with at least one resident. After surveys, a paradoxical situation has emerged: exactly $90 \%$ of the population admitted that they believe in Santa Claus, however, each resident can claim that among their acquaintances, exactly $10 \%$ believe i... | 32. Each person knows at least 10 people. In total, we have 900,000 people who believe in Santa Claus, and they have at least 8,100,000 acquaintances with those who do not believe in him. In total, 100,000 people do not believe in Santa Claus, so among them, there will be someone who has at least 81 acquaintances who b... | 810 | Combinatorics | proof | Yes | Yes | olympiads | false | 21,263 |
33. Let there be a square grid $n \times n$. An original set of cells is a set of $n-1$ cells, no two of which are on the same vertical or horizontal line. For which $n$ can $n+1$ non-overlapping original sets be chosen from this square? | 33. Answer: for any.
Example: first, color the square in $n$ colors so that cells of the same color are located on the diagonals. Take the top left cell, it will be the first color. Repaint it to the $(n+1)$-th color, move 2 cells to the right and one down - we will land on a cell of the second color, repaint it, and ... | forany | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 21,264 |
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