problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
3. The AC-2016 calculator can perform two operations: taking the cube root and calculating the tangent. Initially, the number $2^{-243}$ was entered into the calculator. What is the minimum number of operations required to obtain a number greater than 1? | Answer: 7
## Examples of answer notation:
## 10 | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Dima took the fractional-linear function $\frac{a x+2 b}{c x+2 d}$, where $a, b, c, d-$ are positive numbers, and added it to all the remaining 23 functions that result from it by permuting the numbers $a, b, c, d$. Find the root of the sum of all these functions, independent of the numbers $a, b, c, d$. | Answer: -1
## Examples of answer notation:
14
$1 / 4$
$-1.4$
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the last digit of the integer part of the number $(\sqrt{37}+\sqrt{35})^{2016}$. | Answer: 1
## Examples of answer recording:
## 0
# | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
Circles $O_{1}, O_{2}$, and $O_{3}$ are located inside circle $O_{4}$ with radius 6, touching it internally, and touching each other externally. Moreover, circles $O_{1}$ and $O_{2}$ pass through the center of circle $O_{4}$. Find the radius of circle $O_{3}$.
# | # Answer: 2
## Solution:
We will solve the problem in general for all cases, specifically proving that if the radius of circle $O_{4}$ is $R$, then the radius of circle $O_{3}$ is $\frac{R}{3}$.
Since circles $O_{1}$ and $O_{2}$ pass through the center of circle $O_{4}$ and are internally tangent to it, for each of ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given triangle $\mathrm{ABC}: \mathrm{BK}, \mathrm{CL}$ - angle bisectors, M - the point of their intersection. It turns out that triangle $\mathrm{AMC}$ is isosceles, one of whose angles is 150 degrees. Find what the perimeter of triangle $\mathrm{ABC}$ can be, if it is known that $\mathrm{BK}=4-2 \sqrt{3}$. | Answer: 4
## Examples of how to write answers:
14
$1 / 4$
0.25
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of solutions in natural numbers for the equation $(x-4)^{2}-35=(y-3)^{2}$. | Answer: 3
## Examples of answer notation:
14
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. At the hitmen convention, 1000 participants gathered, each receiving a registration number from 1 to 1000. By the end of the convention, it turned out that all hitmen, except number 1, were killed. It is known that each hitman could only kill hitmen with higher numbers, and the number of his victims could not exceed... | Answer: 10
## Examples of answer notation:
5
# | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (3 points)
Natural numbers $a$ and $b$ are such that $2a + 3b = \operatorname{LCM}(a, b)$. What values can the number $\frac{\operatorname{LCM}(a, b)}{a}$ take? List all possible options in ascending or descending order, separated by commas. If there are no solutions, write the number 0. | Answer: 0 (no solutions)
## Solution:
Let $d=\operatorname{GCD}(a, b)$. Then $a=x d, b=y d, \operatorname{LCM}(a, b)=x y d$, where $x$ and $y$ are coprime. We get $2 x d+3 y d=x y d$. Dividing by $d$ yields $2 x+3 y=x y$.
The right side of this equation is divisible by $x$, so $2 x+3 y$ is divisible by $x$, which me... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) Find the minimum value of the expression $x^{2}+4 x \sin y-4 \cos ^{2} y$. | Answer: -4.
Solution: Add and subtract $4 \sin ^{2} y$. We get the expression $x^{2}+4 x \sin y+4 \sin ^{2} y-4=(x+2 \sin y)^{2}-4$. It is clear that this expression has a minimum value of -4.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) Find the minimum value of the expression $x^{2}-6 x \sin y-9 \cos ^{2} y$. | Answer: -9.
Solution: Add and subtract $9 \sin ^{2} y$. We get the expression $x^{2}-6 x \sin y+9 \sin ^{2} y-9=(x-3 \sin y)^{2}-9$. It is clear that this expression has a minimum value of -9.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (2 points) Find the minimum value of the expression $4 x^{2}+4 x \sin y-\cos ^{2} y$. | Answer: -1.
Solution: Add and subtract $\sin ^{2} y$. We get the expression $4 x^{2}+4 x \sin y+\sin ^{2} y-1=(x+2 \sin y)^{2}-1$. It is clear that this expression has a minimum value of -1.
Idea of Solution 2: To find the minimum value, we can take the derivatives of this expression with respect to $x$ and $y$ and f... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Among six different quadratic trinomials that differ by the permutation of coefficients, what is the maximum number that can have no roots? | Answer: 6
Solution:
Consider, for example, the coefficients 1000, 1001, 1002. The square of any of them is obviously less than the quadruple product of the other two, so all discriminants will be negative. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) Among six different quadratic trinomials that differ by the permutation of coefficients, what is the maximum number that can have two distinct roots | Answer: 6
Solution: Let's take, for example, the coefficients, $-5,1,2$.
If the number -5 is the leading coefficient or the constant term, the equation obviously has two roots of different signs.
For the case when -5 is the second coefficient, let's calculate the discriminant: $5^{2}-1 \cdot 2=23>0$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the greatest value that the sum $\sin ^{2} a+\sin ^{2}\left(a+60^{\circ}\right)+\sin ^{2}\left(a+120^{\circ}\right)+\ldots+\sin ^{2}\left(a+300^{\circ}\right)$ can take? | Answer: 3
## Examples of how to write the answer:
1,7
$1 / 7$
17
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What integer can be written in the form $\sqrt{12-\sqrt{12-\sqrt{12-\ldots}}}$ (the number of roots is infinite). | Answer: 3
## Examples of how to write the answer:
1,7
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. The fox builds "pyramids" from 5 cubes in the following way: each "pyramid" consists of one or several levels; on each level, the number of cubes is strictly less than on the previous one; each new level consists of one or several consecutive cubes. You can see an example of a "pyramid" made of ten cubes in the pict... | Answer: 7
## Examples of answer notation:
12
# | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. $A B D E, B C E F, C D F A$ - cyclic quadrilaterals with the intersection points of the diagonals $K, L$ and $M$ respectively. It is known that point $K$ lies on segments $B L$ and $A M$, point $M$ - on segment $C L$. Moreover, $E L=F L=K L=5, D M=4, A K=M K=6$. Find the length of segment $M C$. If there are multipl... | Answer: 4
## Examples of how to write answers:
$1 / 4$
0.25
$4 ; 10$
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (2 points)
What is the maximum number of different reduced quadratic equations that can be written on the board, given that any two of them have a common root, but no four have a root common to all. | Answer: 3
Solution:
Consider three such equations. There are two cases:
1) These equations have a common root $a$. Then they can be represented in the form $(x-a)(x-b)=0, (x-a)(x-c)=0$ and $(x-a)(x-d)=0$, where $b, c$ and $d$ are their remaining roots (they are not equal to each other, otherwise the equations coinci... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (2 points)
What is the maximum number of different reduced quadratic equations that can be written on the board, given that any two of them have a common root, but no five have a root common to all. | Answer: 4
## Solution:
Consider 4 such equations. There are two cases:
1) These equations have a common root $a$. Then they can be represented in the form $(x-a)(x-b)=0,(x-a)(x-c)=0,(x-a)(x-d)=0$ and $(x-a)(x-e)=0$, where $b, c, d$ and $e$ are their remaining roots (they are not equal to each other, otherwise the eq... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. How many pairs of natural numbers exist for which the number 189 is the LCM? (The numbers in the pair can be the same, the order of the numbers in the pair does not matter) | Answer: 11
Only digits are allowed as input
# | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
$A B C$ is an equilateral triangle with a side length of 10. On side $A B$, a point $D$ is taken; on side $A C$, a point $E$ is taken; on side $B C$, points $F$ and $G$ are taken such that triangles $A D E$, $B D G$, and $C E F$ are also equilateral. $A D=3$. Find $F G$. | # Answer: 4
## Solution:
$B D=A B-A D=10-3=7$, so all sides of triangle $B D G$ are equal to 7.
$C E=A C-A E=A C-A D=10-3=7$, so all sides of triangle $C E F$ are also equal to 7.
Thus, $B G+C F=14$. This is greater than the length of segment $B C=10$, so $G F=14-10=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. (3 points)
Alice the Fox thought of a two-digit number and told Pinocchio that this number is divisible by $2, 3, 4, 5$, and $6$. However, Pinocchio found out that exactly two of these five statements are actually false. What numbers could Alice the Fox have thought of? In your answer, indicate the number o... | # Answer: 8
## Solution:
If a number is not divisible by 2, then it is not divisible by 4 or 6 either, and we already have three false statements. Therefore, the number must be divisible by 2. If an even number is divisible by 3, it is also divisible by 6, which means the statements about divisibility by 4 and 5 are ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The gnomes dug a system of tunnels that formed a rectangular grid 7 by 2. The main entrance is located at the bottom-left vertex $(0,0)$, and the main exit is at the top-right vertex $(7,2)$. The tunnels are numbered as follows: the tunnel connecting vertices $(x, k)$ and $(x+1, k)$ has a number equal to $7 k + x + ... | Answer: 8
Only digits are allowed as input | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
On the side $B C$ of triangle $A B C$, points $A_{1}$ and $A_{2}$ are marked such that $B A_{1}=6, A_{1} A_{2}=8$, $C A_{2}=4$. On the side $A C$, points $B_{1}$ and $B_{2}$ are marked such that $A B_{1}=9, C B_{2}=6$. Segments $A A_{1}$ and $B B_{1}$ intersect at point $K$, and $A A_{2}$ and $... | # Answer: 12
## Solution:
Let $M$ be the point of intersection of line $KL$ and side $AB$. We write two Ceva's theorems, for point $K$ and for point $L$:
$$
\frac{AM}{MB} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}=1 ; \quad \frac{AM}{MB} \cdot \frac{BA_2}{A_2C} \cdot \frac{CB_2}{B_2A}=1
$$
From this, we get $\... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (3 points)
$4^{27000}-82$ is divisible by $3^n$. What is the greatest natural value that $n$ can take? | Answer: 5
Solution:
$4^{27000}=(1+3)^{27000}=1+27000 \cdot 3+\frac{27000 \cdot 26999 \cdot 3^{2}}{2}+\frac{27000 \cdot \ldots \cdot 26998 \cdot 3^{3}}{6}+$
$+\frac{27000 \cdot \ldots \cdot 26997 \cdot 3^{4}}{24}+\frac{27000 \cdot \ldots \cdot 26996 \cdot 3^{5}}{120} \ldots$
The last two terms listed are divisible by ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. (3 points)
Positive numbers $x, y$, and $z$ are such that $x+y+z=5$. What is the smallest value that the quantity $x^{2}+y^{2}+2 z^{2}-x^{2} y^{2} z$ can take? | # Answer: -6
## Solution:
Rewrite the condition as $\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z=5$ and write the inequality for the arithmetic mean and the quadratic mean of these five numbers:
$$
1=\frac{\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+z}{5} \leqslant \sqrt{\frac{\frac{x^{2}}{4}+\frac{x^{2}}{4... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) On a $3 \times 3$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 5.
Solution: The people in the corner cells are obviously liars: they simply do not have three neighbors.
If the person in the center is a knight, then all the people in the side cells are also liars. But in this case, the knight has 4 liar neighbors, not three. This leads to a contradiction.
If the person i... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) On a $2 \times 4$ chessboard, there are knights, who always tell the truth, and liars, who always lie. Each of them said: "Among my neighbors, there are exactly three liars." How many liars are on the board?
Neighbors are considered to be people on cells that share a common side. | Answer: 6.
Solution: On the corner cells, liars are obviously standing: they simply do not have three neighbors.
A situation where all people on the board are liars is impossible, as it would mean that four of these liars are telling the truth.
Therefore, there is at least one knight. Then all his neighbors are liar... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Given the function $f(x)=\frac{\sqrt{3} x-1}{x+\sqrt{3}}$. Find $\underbrace{f(f(\ldots(1) \ldots))}_{2013 \text { times }}$. | Answer: -1
## Examples of how to write answers:
$1 / 4$
0.25
10
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Four spheres of radius $r$ touch each other externally. A sphere of radius $\sqrt{6}+2$ touches all of them internally. Find $r$. | Answer: 2
## Examples of how to write answers:
$1 / 4$
0.25
10
# | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is a trapezoid with bases $A D=15$ and $\mathrm{BC}=10$. $O$ is one of the intersection points of the circles constructed on the lateral sides of the trapezoid as diameters, and this point lies inside the trapezoid. Triangle $B C M$ is constructed on side $B C$ on the external side relative to the tr... | Answer: 4
## Examples of answer notations:
$1 / 4$
0.25
10 | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) In a row without spaces, all natural numbers are written in ascending order: $1234567891011 . .$. What digit stands at the 2017th place in the resulting long number? | Answer: 7.
Solution: The first 9 digits are contained in single-digit numbers, the next 180 - in two-digit numbers. $2017-180-9=1828$. Next, $1828: 3=609 \frac{1}{3}$. This means that the 2017-th digit is the first
 Given the cryptarithm: ЖАЛО + ЛОЖА = ОСЕНЬ. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter А. | Answer: 8
Solution: The rebus can be rewritten as ОСЕНЬ $=($ ЖА + ЛО $) \cdot 101$. First, this means that the last digit of ЖА + ЛО is Ь. Second, if ЖА + ЛО $<100$, the result will be a four-digit number. Let $Ж А+Л О=1 Х Ь$, where $X$ is some digit.
Then ОСЕНЬ $=1 Х Ь 00+1 Х Ь$. If $\mathrm{b}<9$, then the second a... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. (2 points) In a row without spaces, all natural numbers from 999 to 1 are written in descending order: 999998 ...321. What digit is in the 2710th position of the resulting long number? | Answer: 9.
Solution: The first $3 \cdot 900=2700$ digits are contained in three-digit numbers. Therefore, we need to count another 10 digits in two-digit numbers. That is, we need the first digit of the fourth largest two-digit number. This number is 96, so we need the digit 9. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Given the cryptarithm: RIVER + SQUARE = ABVAD. Identical letters represent identical digits, different letters represent different digits. Find the value of the letter B. | Answer: 2
Solution: The rebus can be rewritten as ABVAD $=(\mathrm{KA}+\mathrm{PE}) \cdot 101$. First, this means that the last digit of $\mathrm{KA}+\mathrm{PE}$ is D. Second, if $\mathrm{KA}+\mathrm{PE}<100$, the result will be a four-digit number. Let $\mathrm{KA}+\mathrm{PE}=1X$D, where $X$ is some digit.
Then AB... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (3 points)
Natural numbers $x, y, z$ are such that $\operatorname{GCD}(\operatorname{LCM}(x, y), z) \cdot \operatorname{LCM}(\operatorname{GCD}(x, y), z)=1400$.
What is the greatest value that $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$ can take?
# | # Answer: 10
## Solution:
Notice that $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $z$, and $z$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$, so $\operatorname{LCM}(\operatorname{GCD}(x, y), z)$ is divisible by $\operatorname{GCD}(\operatorname{LCM}(x, y), z)$.
$1400=2^{3} \... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. (3 points)
Let $f(x)$ be a quadratic trinomial with integer coefficients. Given that $f(\sqrt{3}) - f(\sqrt{2}) = 4$. Find $f(\sqrt{10}) - f(\sqrt{7})$. | # Answer: 12
## Solution:
Let $f(x)=c x^{2}+d x+e$. Then $f(\sqrt{3})-f(\sqrt{2})=3 c+\sqrt{3} d+e-(2 c+\sqrt{2} d+e)=c+d(\sqrt{3}-\sqrt{2})$. This number can only be an integer if $d=0$. Therefore, $f(x)=c x^{2}+e$ and $f(\sqrt{3})-f(\sqrt{2})=c$.
Then $f(\sqrt{10})-f(\sqrt{7})=10 c+e-(7 c+e)=3 c=12$. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (3 points)
Positive numbers $x, y, z$ are such that $x y + y z + x z = 12$.
Find the smallest possible value of $x + y + z$.
# | # Answer: 6
## Solution:
By adding the inequalities $x^{2}+y^{2} \geqslant 2 x y, x^{2}+z^{2} \geqslant 2 x z$ and $y^{2}+z^{2} \geqslant 2 y z$ and dividing by 2, we get $x^{2}+y^{2}+z^{2} \geqslant x y+y z+x z$.
$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z) \geqslant 3(x y+y z+x z)=36$, from which $x+y+z \geqslant... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7. (3 points)
From point $K$ on side $A C$ of triangle $A B C$, perpendiculars $K L_{1}$ and $K M_{1}$ were dropped to sides $A B$ and $B C$ respectively. From point $L_{1}$, a perpendicular $L_{1} L_{2}$ was dropped to $B C$, and from point $M_{1}$, a perpendicular $M_{1} M_{2}$ was dropped to $A B$.
It tu... | # Answer: 8
## Solution:
Notice that the quadrilateral $L_{1} M_{2} L_{2} M_{1}$ is cyclic, since $\angle L_{1} M_{2} M_{1}=\angle L_{1} L_{2} M_{1}=$ $90^{\circ}$. Therefore, $\angle B M_{2} L_{2}=180^{\circ}-\angle L_{1} M_{2} L_{2}=\angle L_{2} M_{1} L_{1}=\angle B M_{1} L_{1}$. Similarly, $\angle B L_{2} M_{2}=\a... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
The graphs of the quadratic trinomials $f(x)$ and $g(x)$ intersect at the point $(3 ; 8)$. The trinomial $f(x)+g(x)$ has a single root at 5. Find the leading coefficient of the trinomial $f(x)+g(x)$.
# | # Answer: 4
## Solution:
Since the quadratic polynomial $f(x)+g(x)$ has a unique root 5, it can be represented as $a(x-5)^{2}$, where $a$ is precisely the leading coefficient we are looking for. Additionally, the value of this quadratic polynomial at the point 3 is equal to the sum of the values of the polynomials $f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3. (3 points)
It is known that $a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c=30$ and $a^{2}+b^{2}+c^{2}=13$.
Find $a+b+c$. | Answer: 5
## Solution:
$(a+b+c)^{3}-(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)=2\left(a^{2} b+a^{2} c+b^{2} a+b^{2} c+c^{2} a+c^{2} b+3 a b c\right)$. Let $a+b+c$ be $x$, substitute the known values of the expressions from the condition, and we get the equation $x^{3}-13 x-60=0$.
It is not hard to notice that the number ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (3 points)
Circles $O_{1}$ and $O_{2}$ touch circle $O_{3}$ with radius 13 at points $A$ and $B$ respectively and pass through its center $O$. These circles intersect again at point $C$. It is known that $O C=12$. Find $A B$. | # Answer: 10
## Solution:
Since circles $O_{1}$ and $O_{2}$ touch circle $O_{3}$ at points $A$ and $B$ respectively and pass through its center $O$, $AO$ and $BO$ are their diameters. Therefore, angles $\angle OCA$ and $\angle OCB$ are right angles. These cannot be the same angle, as this would mean that circles $O_{... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8. (5 points)
32 volleyball teams participate in a tournament according to the following scheme. In each round, all remaining teams are randomly paired; if the number of teams is odd, one team skips this round. In each pair, one team wins and the other loses, as there are no draws in volleyball. After three ... | # Solution:
For all teams to be eliminated except one, they must suffer at least 93 losses, meaning at least 93 matches must be played.
We also note that after each round, the number of teams decreases by at most half, because no more than half of the teams lose. In particular, in the final round, only 2 teams could ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (2 points)
In a class, each student has either 5 or 6 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class? | Answer: 11
## Solution:
Let's look at some person. Suppose he has five friends. Then each of these five people has six friends. Similarly, there are at least another 6 people with five friends each. In total, there are 11 people.
It is quite easy to construct an example: two groups of 5 and 6 people, people from dif... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (2 points)
In a positive non-constant geometric progression, the arithmetic mean of the second, seventh, and ninth terms is equal to some term of this progression. What is the minimum possible number of this term? | Answer: 3
Solution:
The second element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first element is even less suitable.
To prove that the answer "3" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we need to solv... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (2 points)
In a class, each student has either 5 or 7 friends (friendship is mutual), and any two friends have a different number of friends in the class. What is the smallest number of students, greater than 0, that can be in the class?
Answer: 12 | Solution:
Let's look at some person. Suppose he has five friends. Then each of these five people has seven friends. Similarly, there are at least seven more people with five friends each. In total, there are 12 people.
It is quite easy to construct an example: two groups of 5 and 7 people, people from different group... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2. (2 points)
In a positive non-constant geometric progression, the arithmetic mean of the third, fourth, and eighth terms is equal to some term of this progression. What is the minimum possible number of this term? | Answer: 4
Solution:
The third element is either the larger or the smaller of the three specified, so it cannot be equal to the arithmetic mean of all three. The first and second elements are even less suitable.
To prove that the answer "4" is possible, let's introduce the notation: let $b_{n}=$ $b q^{n-1}$. Then we ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. $\mathrm{ABCD}$ is a trapezoid with bases $\mathrm{AD}=6$ and $\mathrm{BC}=10$. It turns out that the midpoints of all four sides of the trapezoid lie on the same circle. Find its radius.
If there are multiple correct answers, list them in any order separated by a semicolon. | Answer: 4.
## Examples of answer notation:
45
$4 ; 5$
# | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Three consecutive terms of a geometric progression with a common ratio $q$ were used as coefficients of a quadratic trinomial, with the middle term being the leading coefficient. For what largest integer $q$ will the resulting trinomial have two distinct roots regardless of how the other two coefficients are arrange... | Answer: -1
## Examples of answer notation:
2
5;9
# | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) On the Island of Misfortune, there live knights, who always tell the truth, and liars, who always lie. One day, $n$ islanders gathered in a room. $\quad 30$
The first one said: "Exactly every first person present in this room is a liar."
The second one said: "Exactly every second person present in this ... | # Answer: 2
## Solution:
Let's note that among these people, there is definitely one knight, otherwise it would mean that the first islander is a liar telling the truth, which is impossible. Moreover, the first person is definitely a liar.
Furthermore, there is exactly one knight, since all the speakers contradict e... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, $n$ islanders gathered in a room.
The first one said: "Exactly every second person in this room is a liar."
The second one said: "Exactly every third person in this room is a liar."
and so on
... | # Answer: 2
Solution:
According to the condition, there is at least one knight among these people. Therefore, there must be at least two people, otherwise the first person would be a knight who is lying.
Moreover, there is exactly one knight, as all the speakers contradict each other. Thus, there cannot be more than... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. (4 points) At a knight's tournament, each knight gave each of his acquainted ladies as many flowers as she had acquainted knights, except for him. After this, every two knights arranged as many duels as they had common acquainted ladies. What was more: the given flowers or the arranged duels, and by how many times?
... | Solution: For each trio consisting of a lady and two of her acquainted knights, one duel will take place. Regarding the flowers, the first knight will give the lady one flower for introducing him to the second, and the second will give one for being introduced to the first, making a total of two flowers.
## Grade 9 2... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. (4 points) With the number written on the board, one of the following operations is allowed:
1) If there is a digit in the original number that is not 9 and has two neighboring digits greater than 0, you can increase this digit by 1, and decrease the neighboring digits by 1.
2) Subtract 1 from any non-zero digit exc... | Answer: 3
Solution:
The first operation corresponds to subtracting a number of the form $910 \cdot 0$, the second and third - subtracting $70 \cdot 0$. Both these operations do not change the remainder of the original number when divided by 7, since 91 is divisible by 7.
$1001=c d o t 91$, and $888888=888 \cdot 1001... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. (4 points) With the number written on the board, one of the following operations is allowed:
1) If there is a digit in the original number that is not equal to 9 and has two neighboring digits greater than 0, you can increase this digit by 1, and decrease the neighboring digits by 1.
2) Subtract 1 from any non-zero ... | # Answer: 3
Solution:
The first operation corresponds to subtracting a number of the form $910 \cdot 0$, the second and third - subtracting $70 \cdot 0$. Both these operations do not change the remainder of the original number when divided by 7, since 91 is divisible by 7.
$1001 = 91 \cdot 11$, and $999999 = 999 \cd... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. (4 points) Find what $x+y$ can be equal to, given that $x^{3}-6 x^{2}+15 x=12$ and $y^{3}-6 y^{2}+15 y=16$. | Answer: 4.
Solution:
Let $u=x-2$ and $v=y-2$. Then the original equations transform into $u^{3}+3 u=-2$ and $v^{3}+3 v=2$. Adding these equations, we get $(u+v)\left(u^{2}-u v+v^{2}+3\right)=0$. The second bracket is always positive, so the first one must be zero, from which we have $x+y=u+v+4=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (4 points) Find what $x+y$ can be equal to, given that $x^{3}+6 x^{2}+16 x=-15$ and $y^{3}+6 y^{2}+16 y=-17$. | Answer: -4
Solution: Let $u=x+2$ and $v=y+2$. Then the original equations transform into $u^{3}+4 u=1$ and $v^{3}+4 v=-1$. Adding these equations, we get $(u+v)\left(u^{2}-u v+v^{2}+4\right)=0$. The second bracket is always positive, so the first one must be zero, from which we have $x+y=u+v-4=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Point $O$ is the center of a certain circle, $A$ is a point outside the circle, $B$ is a point on the circle such that $AB$ is a tangent. $AO=6$. Find the greatest possible value of the area of triangle $AOB$. | Answer: 9.
## Examples of answer notations:
17
$1 / 7$
1.7 | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Circle $S_{1}$ intersects circle $S_{2}$ at points $A$ and $B$ and is tangent to circle $S_{3}$ at point $Z$. The common tangent of circles $S_{1}$ and $S_{3}$ intersects line $A B$ at point $C$. Also, through $C$ passes line $X Y$, which is tangent to circle $S_{2}$ at point $X$ and to circle $S_{3}$ at point $Y$.
... | Answer: 12
Allowed for input are digits, a period or comma, a division sign, a semicolon
# | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangle $\mathrm{ABCD}$. The length of side $\mathrm{BC}$ is one and a half times less than the length of side $\mathrm{AB}$. Point $\mathrm{K}$ is the midpoint of side AD. Point $\mathrm{L}$ on side $\mathrm{CD}$ is such that $\mathrm{CL}=\mathrm{AK}$. Point $\mathrm{M}$ is the intersection of line BL and... | Answer: 8
## Examples of answer recording:
# | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The function $f(x)$ is such that $f(x+1)+f(x-1)=5.2 f(x)$. It is known that $f(0)=-98$, and $\mathrm{f}(3)=249.2$. Find $f(1)$. | Answer: -10
## Examples of how to write answers:
17
$-1.7$
$1 / 7$
# | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (3 points) Prove that for $n=6002$ the sum of binomial coefficients with a step of 6, i.e., $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$, gives a remainder of 1 when divided by 3.
$C_{n}^{k}-$ the number of ways to choose $k$ items from $n$, which is $\frac{n!}{k!(n-k)!}$ if $0 \leqslant k \leqslant n$ and 0 in all oth... | # Solution:
It is easy to verify that $C_{n}^{k}=C_{n-3}^{k-3}+3 C_{n-3}^{k-2}+3 C_{n-3}^{k-1}+C_{n-3}^{k}$, therefore, $C_{n}^{k}$ gives the same remainder when divided by 3 as $C_{n-3}^{k-3}+C_{n-3}^{k}$.
Thus, $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$ gives the same remainder when divided by 3 as $C_{n-3}^{1}+C_{n-... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. (3 points) Prove that for $n=9002$ the sum of binomial coefficients with a step of 6, i.e., $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$, gives a remainder of 1 when divided by 3.
$C_{n}^{k}-$ the number of ways to choose $k$ items from $n$, which is $\frac{n!}{k!(n-k)!}$ if $0 \leqslant k \leqslant n$ and 0 in all oth... | Solution:
It is easy to verify that $C_{n}^{k}=C_{n-3}^{k-3}+3 C_{n-3}^{k-2}+3 C_{n-3}^{k-1}+C_{n-3}^{k}$, therefore, $C_{n}^{k}$ gives the same remainder when divided by 3 as $C_{n-3}^{k-3}+C_{n-3}^{k}$.
Thus, $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$ gives the same remainder when divided by 3 as $C_{n-3}^{1}+C_{n-3}... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
4. (3 points) Prove that for $n=12002$ the sum of binomial coefficients with a step of 6, i.e., $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$, gives a remainder of 1 when divided by 3.
$C_{n}^{k}-$ the number of ways to choose $k$ items from $n$, which is $\frac{n!}{k!(n-k)!}$ if $0 \leqslant k \leqslant n$ and 0 in all ot... | # Solution:
It is easy to verify that $C_{n}^{k}=C_{n-3}^{k-3}+3 C_{n-3}^{k-2}+3 C_{n-3}^{k-1}+C_{n-3}^{k}$, therefore, $C_{n}^{k}$ gives the same remainder when divided by 3 as $C_{n-3}^{k-3}+C_{n-3}^{k}$.
Thus, $C_{n}^{1}+C_{n}^{7}+\ldots+C_{n}^{n-1}$ gives the same remainder when divided by 3 as $C_{n-3}^{1}+C_{n-... | 1 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. A grid rectangle $4 \times 11$ is divided into $2 \times 2$ squares and strips of three cells. What is the maximum number of strips that can participate in this division? | Answer: 12
## Examples of answer notation:
100
90
# | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Point $O$ is the center of a certain circle, $A$ is a point outside the circle, $B$ is a point on the circle such that $AB$ is a tangent. $AO=6$. Find the greatest possible value of the area of triangle $AOB$.
| Answer: 9.
## Examples of answer notations:
17
$1 / 7$
1.7
# | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the number $\sqrt{3}+\sqrt{5}$ is a root of a polynomial of the fourth degree with integer coefficients, the leading coefficient of which is 1. What is the sum of the coefficients of this polynomial? | Answer: -11
2. It is known that the number $\sqrt{3}+\sqrt{7}$ is a root of a polynomial of the fourth degree with integer coefficients, the leading coefficient of which is 1. What is the sum of the coefficients of this polynomial?
Answer: -3 | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that the function $f(x)$ satisfies the equality for any $x$:
$3 \sin x+f(x)=f(x+2)$
Find the value of $\mathrm{f}(2017) \sin 1$, given that $f(1)=\frac{3 \cos ^{2} 1008}{\sin 1}$. | Answer: 3
## Examples of answer notations:
17
$1 / 7$
1.7 | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) In space, there is a cube $1000 \times 1000 \times 1000$ with a vertex at the origin and faces parallel to the coordinate planes. Vectors are drawn from the origin to all integer points inside and on the boundary of this cube. Find the remainder when the sum of the squares of the lengths of these vectors ... | # Answer: 0
Solution:
The sum of the squares of the lengths of these vectors is the sum of the squares of all their coordinates, that is, $3 \cdot 1001^{2} \cdot\left(0^{2}+1^{2}+\right.$ $2^{2}+\ldots+1000^{2}$ ), which is divisible by 13, since 1001 is divisible by 13. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) In space, there is a cube $1000 \times 1000 \times 1000$ with a vertex at the origin and faces parallel to the coordinate planes. Vectors are drawn from the origin to all integer points inside and on the boundary of this cube. Find the remainder when the sum of the squares of the lengths of these vectors ... | # Answer: 0
## Solution:
The sum of the squares of the lengths of these vectors is the sum of the squares of all their coordinates, that is, $3 \cdot 1001^{2} \cdot\left(0^{2}+1^{2}+\right.$ $\left.2^{2}+\ldots+1000^{2}\right)$, which is divisible by 11, since 1001 is divisible by 11. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) Let $x, y, z$ and $t$ be non-negative numbers such that $x+y+z+t=5$. Prove the inequality
$$
\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}+1}+\sqrt{z^{2}+y^{2}}+\sqrt{z^{2}+t^{2}}+\sqrt{t^{2}+9} \geqslant 10
$$ | # Solution:
Consider the following points on the plane: $A(0,0) ; B(x, 1) ; C(x+y, 1+x) ; D(x+y+z, 1+$ $x+y) ; E(x+y+z+t, 1+x+y+z) ; F(x+y+z+t+3,1+x+y+z+t)$. Then the length of the broken line $A B C D E F$ coincides with the expression that needs to be evaluated. By the triangle inequality, the length of the broken l... | 10 | Inequalities | proof | Yes | Yes | olympiads | false |
6. (3 points) Let $x, y, z$ and $t$ be non-negative numbers such that $x+y+z+t=2$. Prove the inequality
$$
\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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
40.
$$
\frac{x^{2}}{x^{2}+2 y z}+\frac{y^{2}}{y^{2}+2 z x}+\frac{z^{2}}{z^{2}+2 x y} \geqslant 1
$$ | 40. Since $2 x y \leqslant x^{2}+y^{2}$, we get
$$
\begin{aligned}
\frac{x^{2}}{x^{2}+2 y z}+\frac{y^{2}}{y^{2}+2 z x} & +\frac{z^{2}}{z^{2}+2 x y} \geqslant \\
& \geqslant \frac{x^{2}}{x^{2}+y^{2}+z^{2}}+\frac{y^{2}}{x^{2}+y^{2}+z^{2}}+\frac{z^{2}}{x^{2}+y^{2}+z^{2}}=1
\end{aligned}
$$ | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
41.
$$
\frac{x^{2}+2 y^{2}+2 z^{2}}{x^{2}+y z}+\frac{y^{2}+2 z^{2}+2 x^{2}}{y^{2}+z x}+\frac{z^{2}+2 x^{2}+2 y^{2}}{z^{2}+x y}>6
$$ | 41. Transform the left side using $2 x y \leqslant x^{2}+y^{2}$, multiplying the numerator and denominator by 2:
$$
\begin{aligned}
& \frac{x^{2}+2 y^{2}+2 z^{2}}{x^{2}+y z}+\frac{y^{2}+2 z^{2}+2 x^{2}}{y^{2}+z x}+\frac{z^{2}+2 x^{2}+2 y^{2}}{z^{2}+x y} \geqslant \\
& \geqslant 2\left(\frac{x^{2}+2 y^{2}+2 z^{2}}{2 x^... | 6 | Inequalities | proof | Yes | Yes | olympiads | false |
65. What is the smallest degree that a polynomial $P$ can have if it is known that there exists an integer $b$ such that $A_{P}$ contains elements both greater and less than $b$, but does not contain $b$? | 65. Answer: 4.
For example, consider the polynomial $P(x)=(x-3)(x-1)(x+1)(x+3)$. It is even. Therefore, if $c \neq P(0)$, then $l_{P}(c)$ is even. Therefore, in $A$ there can only be one odd number, namely, $l_{P}(P(0))$. Thus, $b=1$ or $b=3$ is the desired number, since $0,4 \in A_{P}$. Why can't the degree be less? ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
65. What is the smallest degree that a polynomial $P$ can have if it is known that there exists an integer $b$ such that $A_{P}$ contains elements both greater and smaller than $b$, but does not contain $b$? | 65. Answer: 4.
For example, consider the polynomial $P(x)=(x-3)(x-1)(x+1)(x+3)$. It is even. Therefore, if $c \neq P(0)$, then $l_{P}(c)$ is even. Therefore, in $A$ there can only be one odd number, namely, $l_{P}(P(0))$. Thus, $b=1$ or $b=3$ is the desired number, since $0,4 \in A_{P}$. Why can't the degree be less? ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The number 2015 was divided into 12 addends, after which all numbers that can be obtained by adding some of these addends (from one to nine) were written down. What is the minimum number of numbers that could have been written down? | 5.
Answer: 10
Example: Let's break down 2015 into 11 numbers of 155 and one number of 310. In this case, the sum of the addends can range from $155 \times 1$ to $155 \times 10$.
We will prove that it is impossible to get fewer than 10 different numbers: first, note that 2015 is not divisible by 12, so we will have ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Seven natives from several tribes are sitting in a circle by the fire. Each one says to the neighbor on their left: “Among the other five, there are no members of my tribe.” It is known that the natives lie to foreigners and tell the truth to their own. How many tribes are represented around the fire? | Solution. If there are at least 4 natives from one tribe, then two of them sit next to each other, and one of them will lie to the other, although they should tell the truth. If there is only one native from a certain tribe, then he tells the truth to his left neighbor, although he should lie. Therefore, each tribe has... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. There are 20 chocolate candies on the table. Masha and Bear are playing a game according to the following rules. Players take turns. In one move, a player can take one or several candies from the table and eat them. Masha goes first, but on this move, she cannot take all the candies. In all other moves, players cann... | Solution. Masha will win if she eats 4 candies on her first move.
Let's arrange the candies in a row, number them, and assume that the players take candies in a row from left to right.
Suppose Masha eats one candy on her first move. In this case, she will lose, as until the end of the game, the players will take one ... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Petya thought of a 9-digit number obtained by rearranging the digits of the number 123456789. Vitya is trying to guess it. For this, he chooses any 9-digit number (possibly with repeated digits and zeros) and tells it to Petya, who then responds with how many digits of this number match the ones in his thought numbe... | Solution. Vitya's first move is to name the number 122222 222. Then Petya can only respond with 0, 1, or 2.
If Petya answers 0, then neither the one nor the two hit their places. This means the two must be in the first position. There's no need to ask further.
If Petya answers 2, then both the one and the two hit the... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Given a string of 2021 letters A and B. Consider the longest palindromic substring. What is its minimum possible length? A palindrome is a string that reads the same from right to left and from left to right. | Solution. The minimum possible length of the maximum palindrome is 4.
We will prove that it cannot be less than 4. Consider the 5 letters in the center of the string. If these are alternating letters, then it is a palindrome of length 5. Suppose among these five letters there are two identical letters standing next to... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. On the bus route, there are only four stops - "Initial", "First", "Final", and "Last". At the first two stops, passengers only got on, and at the remaining stops, they only got off. It turned out that 30 passengers got on at the "Initial" stop, and 14 passengers got off at the "Last" stop. At the "First" stop, three... | Answer: Those traveling from "First" to "Final" are six more.
Let $x$ be the number of people who got on at "First". Then, $3x$ people got off at "Final". Since the number of people getting on equals the number of people getting off, we have $30 + x = 3x + 14$, from which $x = 8$. Let $y$ be the number of people trave... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Baron Munchausen placed a horse in some cells of an $N \times N$ board. He claims that no one will find two different $4 \times 4$ squares on this board (with sides along the grid lines) with the same number of horses. For what largest $N$ can his words be true?
# | # Solution:
Answer: $N=7$.
The number of knights in a $4 \times 4$ square can range from 0 to 16, i.e., there are 17 possible variants. The number of $4 \times 4$ squares on an $N \times N$ board is $(N-3)^{2}$ (since the top-left cell of the square can occupy positions from the far left to the fourth from the right ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In a $4 \times 4$ square, cells are colored in several colors such that in any $1 \times 3$ rectangle, there are two cells of the same color. What is the maximum number of colors that can be used? | Solution. Maximum 9 colors. See example in the picture. We will prove that more is not possible. Any row (row or column) gives a maximum of three colors, so the first row + first column will give a maximum of $3+3-1=5$ colors.
We will prove that the remaining $3 \times 3$ square will give a maximum of 4 colors. Indeed... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In the company, there are elves, fairies, and gnomes. Each elf is friends with all fairies except for three, and each fairy is friends with twice as many elves. Each elf is friends with exactly three gnomes, and each fairy is friends with all gnomes. Each gnome is friends with exactly half of the elves and fairies c... | Solution. Answer: 12.
Let $n$ be the number of elves, $m$ be the number of fairies, and $k$ be the number of gnomes. Then the number of friendly pairs "elf-fairy" is $n(m-3)$, and "fairy-elf" is $-m \cdot 2(m-3)$. But these are the same pairs, therefore,
$$
n(m-3)=2 m(m-3)
$$
From which, $n=2 m$. Counting the friend... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Today's date is written as: 22.11.2015. How many other days in this year can be written with the same set of digits? | Solution. The month number cannot start with a two, so it is either 11 or 12. In the first case, it is 22.11, in the second case 12.12 and 21.12. Answer: two. | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.4. Let initially each island is inhabited by one colony, and let one of the islands have $d$ neighboring islands. What can the maximum possible number of colonies that can settle on this island be equal to? | Solution. Answer: $d+1$.
Example. We will prove that in a vertex of degree $d$, $d+1$ colonies can gather. Suspend the tree from this vertex as the root and prove that in each vertex from which $e$ edges go down, $e+1$ colonies can gather, conducting only migrations within its subtree. We will prove this by "induction... | +1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The gnomes went to work, and Snow White is feeling lonely. She laid out a pile of 36 stones on the table. Every minute, Snow White splits one of the existing piles into two, and then adds a new stone to one of them. After some time, she had seven piles, each with an equal number of stones. How many stones ended up i... | Solution. There will be seven piles after six moves. After six moves, there will be $36+6=42$ stones - meaning, 6 stones in each pile. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Seven people stood in a circle, each of whom is either a knight, who always tells the truth, or a liar, who always lies, or a traveler, who alternates between truth and lies.
The first and second said in unison: "Among us there is exactly 1 liar," the second and third: "Among us there are exactly 2 knights," the th... | Solution. If all 7 statements are lies, then everyone lied, meaning everyone is a liar, and then 7 and 1 told the truth. This means that at least one statement is true. On the other hand, there are no more than 2 true statements (one about the liars, the other about the knights). That is, there are either 0 or 1 knight... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. When the child was born, their parents were not yet 40 years old, but they were already adults. When the child turned 2 years old, the age of exactly one of the parents was divisible by 2; when the child turned 3 years old, the age of exactly one of the parents was divisible by 3, and so on. How long could such a pa... | Solution. The condition means that at the moment of the child's birth, the age of exactly one of the parents was divisible by 2, the age of exactly one of the parents - by 3, and so on (as long as this pattern continued).
Let one of the parents be 24 years old, and the other 35 years old. Then this pattern could conti... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$, angle $A$ is equal to $50^{\circ}, B H$ is the altitude. Point $M$ on $B C$ is such that $B M=B H$. The perpendicular bisector of segment $M C$ intersects $A C$ at point $K$. It turns out that $A C=2 \cdot H K$. Find the angles of triangle $A B C$. | Solution. Draw a perpendicular to $BC$ from point $M$. Let $K^{\prime}$ be the point of its intersection with $AC$. Then the segment $MK$ is the midline in $\triangle K^{\prime}MC$, that is, $K^{\prime}K = KC$. From the fact that $HK = KC + AH$, we get $AH = HK^{\prime}$. Right triangles $BHA$ and $BHK^{\prime}$ are eq... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Konstantin pronounced the names of all natural numbers from 180 to 220 inclusive, while Mikhail - from 191 to 231 inclusive. Who pronounced more words and by how many? | Solution. Let's remove the numbers that both have: 191-220. Then each will have 11 numbers left: Konstantin has 180-190, and Mikhail has 221-231. Note that in the names of the numbers 181-189, 221-229, and 231, there are three words each, while in 180, 190, and 230, there are two words each. This means that Mikhail sai... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. At the festival "Hobbits - for cultural diversity!", more than 20 participants arrived. A correspondent found out that among any 15 participants of the festival, there are at least 4 humans and at least 5 elves. How many hobbits participated in the festival? Provide all possible answers and prove that there are no o... | Solution. Suppose there is at least one hobbit. If there are 10 people among the participants, then in their company with the hobbit and 4 other participants, there will not be 5 elves, which contradicts the condition. Therefore, there are no more than 9 people. Since the total number of participants is more than 20, t... | 0 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Yesterday at the market, with one hundred tugriks you could buy 9 gingerbreads and 7 pastries (and even get some change), but today this amount is no longer enough. However, with the same one hundred tugriks today, you can buy two gingerbreads and 11 pastries (also with some change), but yesterday this amount would ... | Solution. Note that something has become more expensive (otherwise, the situation where we could buy a set yesterday but not today could not have arisen), and another sweet has become cheaper for similar reasons. Nine cookies and seven cakes together have become more expensive, so the cookie has become more expensive, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.