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5.
reducing transaction time
## Correct answers:
using licensed software, using a personal computer instead of a public one, using antivirus programs
Question 3
Balya: 7.00
Mr. Vshokoladov earned X rubles per month throughout 2021. In addition, during this year, he won 2000000 rubles in a lottery. What is $X$ if... | Answer in rubles, without spaces and units of measurement.
Answer:
Correct answer: 600000
Question 4
Score: 3.00
Select all correct continuations of the statement.
2022 Higher Trial - qualifying stage
To file a petition to recognize a citizen as bankrupt...
## Select one or more answers:
\ulcorner | 600000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
A custodian stores securities, clients' money, and other material assets
Correct answers: A paid investment advisor consults and provides recommendations to the client on investment management, A trustee manages the client's property in their own name
Find the correspondence between the term and the statement so... | Answer in rubles, without spaces and units of measurement. Round the answer to the nearest whole number according to rounding rules.
Answer:
Correct answer: 13806
question 9
Score: 3.00
Select all possible features of an authentic ruble banknote.
Select one or more answers:
$\Gamma$ | 13806 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
watermark
Correct answers: raised relief of certain text fragments, watermark, inscriptions and ornaments
Question 10
Score: 7.00
Vladimir has saved 16,500 rubles to buy a gaming console as a birthday gift for his brother, which amounts to 220 US dollars at the current exchange rate. The birthday is not until ... | Answer in rubles, without spaces and units of measurement. Round the answer to the nearest whole number according to rounding rules.
Answer: $\qquad$
Correct answer: 76
Question 11
Score: 3.00
What services can currently be provided remotely? Select all appropriate options. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
the driver's marital status
Correct answers: bonus-malus coefficient, engine power, driver's age
Question 14
Score: 7.00
Maria Ivanovna has retired. She did not have a funded pension, only a social insurance pension, and her individual pension coefficient amount is 120. In addition, Maria Ivanovna has a bank d... | Provide the answer in rubles, without spaces and units of measurement.
Answer:
The correct answer is: 19930
Question 15
Score: 7.00
Insert the missing words from the list below (not all provided words will be needed!):
Paying
credit; preferential; higher; cash withdrawal; service; blocking; bonus; debit; freeze;... | 19930 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.
when transferring money abroad
Correct answers: when using the Fast Payment System for amounts up to 100 thousand rubles per month, when transferring funds between one's own accounts in the same bank
Question 17
Score: 7.00
Last year, a beauty salon offered a $20 \%$ discount on facial massage when purchasing ... | Answer:
The correct answer: 1
Question 18
Score: 3.00
Select all true statements regarding digital financial assets $(DFA)$.
Select one or more answers: | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
The Bank of Russia will ensure the conversion of the "digital ruble" into another form of money (rubles) at a one-to-one ratio.
Correct answers: Stablecoins, backed by cash or gold, are an example of a CBDC., The Bank of Russia will ensure the conversion of the "digital ruble" into another form of money (rubles) ... | Provide the answer in rubles, without spaces and units of measurement. Round the answer to the nearest whole number.
Answer: $\qquad$
Correct answer: 1321412
Question 20
Score: 7.00
Financial Literacy 11th Grade Day 1
Ivan and Petr, twin brothers, went on a vacation by the sea together and purchased two different ... | 1321412 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.
the higher the risk of a financial instrument, the higher its return
Correct answers: the higher the reliability of a financial instrument, the higher its return, a financial instrument can be reliable, profitable, and liquid at the same time, risk is not related to the return of a financial instrument
Question ... | Answer write in rubles as an integer without spaces and units of measurement.
Answer: $\qquad$
Correct answer: 1378
Question 16
Score: 5.00
Establish the correspondence between specific taxes and their types.
| personal income tax | federal tax; local tax; regional tax; |
| :---: | :---: |
| land tax | federal ta... | 1378 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a square of grid paper containing an integer number of cells, a hole in the shape of a square, also consisting of an integer number of cells, was cut out. How many cells did the large square contain if 209 cells remained after the cutout? | Answer: 225 cells
Solution. The side of the larger square contains $n$ sides of a cell, and the side of the smaller square contains $m$ sides of a cell. Then $n^{2}-m^{2}=209 \rightarrow(n-m)(n+m)=209=11 \cdot 19$.
Case 1. $\left\{\begin{array}{c}n+m=209 \\ n-m=1\end{array} \rightarrow\left\{\begin{array}{c}n=105 \\ ... | 225 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (mathematics) There are scales with two pans, 4 weights of 2 kg each, 3 weights of 3 kg each, and two weights of 5 kg each. In how many different ways can a 12 kg load be balanced on the scales, if the weights are allowed to be placed on both pans? | Answer: 7 ways
Solution: Let $x$ be the number of 2 kg weights used in weighing, $y$ be the number of 3 kg weights, and $z$ be the number of 5 kg weights. Then the equilibrium condition is g... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Subtract the second equation from the first:
$x^{2}-2 x+y^{4}-8 y^{3}+24 y^{2}-32 y=-17 \rightarrow(x-1)^{2}+(y-2)^{4}=0 \rightarrow\left\{\begin{array}{l}x=1 \\ y=2\end{array}\right.$
Then $z=x^{2}+y^{4}-8 y^{3}=1+16-64=-47$ | Answer: the only solution is $x=1, y=2, z=-47$. | -47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Express z from the first equation and substitute into the second:
$x^{2}-2 x+y^{2}-2 \sqrt{3} y=-4 \rightarrow(x-1)^{2}+(y-\sqrt{3})^{2}=0 \rightarrow\left\{\begin{array}{c}x=1 \\ y=\sqrt{3}\end{array} \rightarrow z=x^{2}+y^{2}+2 x=6\right.$ | Answer: $x=1, y=\sqrt{3}, z=6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 1. The administration divided the region into several districts based on the principle: the population of a large district exceeds $8 \%$ of the region's population and for any large district, there are two non-large districts with a combined population that is larger. Into what minimum number of districts was the... | Answer: 8 districts.
Solution. The number of "small" districts is no less than 2 according to the condition, and their population does not exceed $8 \%$ of the total population of the region. We will show that the number of districts in the region is no less than 8. If the number of districts in the region is no more ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Point $M$ is located on side $CD$ of a square such that $CM: MD=1: 3$. Line $AM$ intersects the circumcircle of the square at point $E$. The area of triangle $ACE$ is 14. Find the side length of the square. | Answer: 10.
Solution.
Triangles $A M D$ and $C M E$ are similar with a similarity coefficient $k=5$. Then
$$
C E=\frac{4 x}{5}, M E=\frac{3 x}{5} \rightarrow A E=5 x+\frac{3 x}{5}=\frac{28... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Seven students in the class receive one two every two days of study, and nine other students receive one two every three days each. The rest of the students in the class never receive twos. From Monday to Friday, 30 new twos appeared in the journal. How many new twos will appear in the class journal on Satur... | Answer: 9
Solution. Over the period from Monday to Saturday (six days), in the journal, there will be 3 new twos from each student of the first group (seven people) and 2 new twos from each of the 9 students of the second group. The total number of new twos for the school week is $7 \cdot 3 + 9 \cdot 2 = 39$. Then, on... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Through the vertex $B$ of an equilateral triangle $ABC$, a line $L$ is drawn, intersecting the extension of side $AC$ beyond point $C$. On line $L$, segments $BM$ and $BN$ are laid out, each equal in length to the side of triangle $ABC$. The lines $MC$ and $NA$ intersect at a common point $D$ and intersect t... | # Answer: 1.
Solution. Triangles $M B C$ and $A B N$ are isosceles, therefore
$$
\begin{gathered}
\angle B M C = \angle B C M = \alpha \rightarrow \angle N B C = 2 \alpha, \angle B A N = \angle B N A = \beta \rightarrow \angle A B M = 2 \beta \\
2 \alpha + 2 \beta + 60^{\circ} = 180^{\circ} \rightarrow \alpha + \beta... | 1 | Geometry | proof | Yes | Yes | olympiads | false |
16. The last digit of a six-digit number was moved to the beginning (for example, $456789 \rightarrow$ 945678), and the resulting six-digit number was added to the original number. Which numbers from the interval param 1 could have resulted from the addition? In the answer, write the sum of the obtained numbers.
| par... | 16. The last digit of a six-digit number was moved to the beginning (for example, $456789 \rightarrow$ 945678), and the resulting six-digit number was added to the original number. Which numbers from the interval param 1 could have resulted from the addition? In the answer, write the sum of the obtained numbers.
| par... | 1279267 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{4}+11 x+23$. | Answer: 22.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+11 x+23>0 \Leftrightarrow-\frac{1}{4}(x+2)(x-46)>0$, from which $-2<x<46$. On this interval, there are 45 natural values of $x: x=1, x=2, \ldots, x=45$. In this interval, $y$ takes integer values only for even $x$ - a tot... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can be repeated) so that the resulting 12-digit number is divisible by 45. In how many ways can this be done? | Answer: 13122.
Solution. For a number to be divisible by 45, it is necessary and sufficient that it is divisible by 5 and by 9. To ensure divisibility by 5, we can choose 0 or 5 as the last digit (2 ways).
To ensure divisibility by nine, we proceed as follows. We will choose four digits arbitrarily (this can be done ... | 13122 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{4}+3 x+\frac{253}{4}$. | Answer: 11.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+3 x+\frac{253}{4}>0 \Leftrightarrow-\frac{1}{4}(x+11)(x-23)>0$, from which $-11<x<23$. On this interval, there are 22 natural values of $x: x=1, x=2, \ldots, x=22$. During this interval, $y$ takes integer values only for ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 0 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can repeat) so that the resulting 10-digit number is divisible by 18. In how many ways can this be done? | Answer: 3645.
Solution. For a number to be divisible by 18, it is necessary and sufficient that it is divisible by 2 and by 9. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6$ or 8 (5 ways).
To ensure divisibility by nine, we proceed as follows. Choose three digits ... | 3645 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{4}+9 x+19$. | Answer: 18.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+9 x+19>0 \Leftrightarrow-\frac{1}{4}(x+2)(x-38)>0$, from which $-2<x<38$. On this interval, there are 37 natural values of $x: x=1, x=2, \ldots, x=37$. In this interval, $y$ takes integer values only for even $x$ - a tota... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 0 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,1,2,3,4,5,6,7,8$ (digits can repeat) so that the resulting 10-digit number is divisible by 45. In how many ways can this be done? | Answer: 1458.
Solution. For a number to be divisible by 45, it is necessary and sufficient that it is divisible by 5 and by 9. To ensure divisibility by 5, we can choose 0 or 5 as the last digit (2 ways).
To ensure divisibility by nine, we proceed as follows. We select three digits arbitrarily (this can be done in $9... | 1458 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{4}+5 x+39$. | Answer: 12.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{4}+5 x+39>0 \Leftrightarrow-\frac{1}{4}(x+6)(x-26)>0$, from which $-6<x<26$. On this interval, there are 25 natural values of $x: x=1, x=2, \ldots, x=25$. In this interval, $y$ takes integer values only for even $x$ - a total ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $1,2,3,4,5,6,7,8,9$ (digits can repeat) so that the resulting 12-digit number is divisible by 18. In how many ways can this be done? | Answer: 26244.
Solution. For a number to be divisible by 18, it is necessary and sufficient that it is divisible by 2 and by 9. To ensure divisibility by 2, we can choose the last digit from the available options as $2, 4, 6$ or 8 (4 ways).
To ensure divisibility by nine, we proceed as follows. Choose four digits arb... | 26244 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+13 x+42$. | Answer: 13.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+13 x+42>0 \Leftrightarrow-\frac{1}{3}(x+3)(x-42)>0$, from which $-3<x<42$. On this interval, there are 41 natural values of $x: x=1, x=2, \ldots, x=41$. In this case, $y$ takes integer values only when $x$ is divisible by... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the value of the expression $\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are the largest and smallest roots of the equation $x^{3}-7 x^{2}+7 x=1$, respectively. | Answer: 34.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}-1\right)-7\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-7 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-6 x+1\right)=0,
$$
from which $x=1$ or $x=3 \pm \sqrt{8}$. The largest root is $a=3+\sqrt{8}$, and the smallest... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2016 * * * * 02 * *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 15. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. Choose four digits arbitrarily... | 5184 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+7 x+54$. | Answer: 8.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+7 x+54>0 \Leftrightarrow-\frac{1}{3}(x+6)(x-27)>0$, from which $-6<x<27$. On this interval, there are 26 natural values of $x: x=1, x=2, \ldots, x=26$. In this interval, $y$ takes integer values only when $x$ is divisible ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the value of the expression $\frac{p}{q}+\frac{q}{p}$, where $p$ and $q$ are the largest and smallest roots of the equation $x^{3}+6 x^{2}+6 x=-1$, respectively. | Answer: 23.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}+1\right)+6\left(x^{2}+x\right)=0 \Leftrightarrow(x+1)\left(x^{2}-x+1\right)+6 x(x+1)=0 \Leftrightarrow(x+1)\left(x^{2}+5 x+1\right)=0 \text {, }
$$
from which $x=-1$ or $x=\frac{-5 \pm \sqrt{21}}{2}$. The largest root is $p=\frac{... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,6,7,8$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 2160.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 6, 8$ (5 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits ar... | 2160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+5 x+72$. | Answer: 7.
Solution. Let's find those values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+5 x+72>0 \Leftrightarrow-\frac{1}{3}(x+9)(x-24)>0$, from which $-9<x<24$. On this interval, there are 23 natural values of $x: x=1, x=2, \ldots, x=23$. During this time, $y$ takes integer values only when $x$ is divisible ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the value of the expression $\frac{a}{b}+\frac{b}{a}$, where $a$ and $b$ are the largest and smallest roots of the equation $x^{3}-9 x^{2}+9 x=1$, respectively. | Answer: 62.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}-1\right)-9\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-9 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-8 x+1\right)=0
$$
from which $x=1$ or $x=4 \pm \sqrt{15}$. The largest root is $a=4+\sqrt{15}$, the smallest is... | 62 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 15. In how many ways can this be done? | # Answer: 864.
Solution. For a number to be divisible by 15, it is necessary and sufficient that it is divisible by 5 and by 3. To ensure divisibility by 5, we can choose 0 or 5 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose three digits a... | 864 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{3}+20 x+63$. | Answer: 20.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+20 x+63>0 \Leftrightarrow-\frac{1}{3}(x+3)(x-63)>0$, from which $-3<x<63$. On this interval, there are 62 natural values of $x: x=1, x=2, \ldots, x=62$. In this case, $y$ takes integer values only when $x$ is divisible by 3... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the value of the expression $\frac{p}{q}+\frac{q}{p}$, where $p$ and $q$ are the largest and smallest roots of the equation $x^{3}-8 x^{2}+8 x=1$, respectively. | # Answer: 47.
Solution. The given equation is equivalent to the following
$$
\left(x^{3}-1\right)-8\left(x^{2}-x\right)=0 \Leftrightarrow(x-1)\left(x^{2}+x+1\right)-8 x(x-1)=0 \Leftrightarrow(x-1)\left(x^{2}-7 x+1\right)=0 \text {, }
$$
from which $x=1$ or $x=\frac{7 \pm \sqrt{45}}{2}$. The largest root is $p=\frac{... | 47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2016 * * * * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can be repeated) so that the resulting 11-digit number is divisible by 6. In how many ways can this be done? | Answer: 1728.
Solution. For a number to be divisible by 6, it is necessary and sufficient that it is divisible by 2 and by 3. To ensure divisibility by 2, we can choose the last digit from the available options as $0, 2, 4, 8$ (4 ways).
To ensure divisibility by three, we proceed as follows. Choose three digits arbit... | 1728 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{9}+50$. | Answer: 7.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{9}+50>0 \Leftrightarrow x^{2}<450$, from which $-\sqrt{450}<x<\sqrt{450}$. On this interval, there are 21 natural values of $x: x=1, x=2, \ldots, x=21$. During this time, $y$ takes integer values only when $x$ is divisible by 3... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 75. In how many ways can this be done? | Answer: 2592.
Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit from the available options (1 way).
To ensure divisibility by three, we proceed as follows. Select four digits arbitrarily (th... | 2592 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Plot on the plane $(x ; y)$ the set of points satisfying the equation $|15 x|+|8 y|+|120-15 x-8 y|=120$, and find the area of the resulting figure. | Answer: 60.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} { l... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{3}+70$. | Answer: 4.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+70>0 \Leftrightarrow x^{2}<210$, from which $-\sqrt{210}<x<\sqrt{210}$. On this interval, there are 14 natural values of $x: x=1, x=2, \ldots, x=14$. During this time, $y$ takes integer values only when $x$ is divisible by 3... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 02 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,7,8,9$ (digits can repeat) so that the resulting 11-digit number is divisible by 12. In how many ways can this be done? | Answer: 1296.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0, 4, or 8 as the last digit (3 ways).
To ensure divisibility by 3, we proceed as follows. We will choose three digits arbitrarily (this can be done... | 1296 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|3 x|+|4 y|+|48-3 x-4 y|=48$, and find the area of the resulting figure. | Answer: 96.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} { l... | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the $x O y$ plane having natural coordinates $(x, y)$ and lying on the parabola $y=-\frac{x^{2}}{9}+33$. | Answer: 5.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{9}+33>0 \Leftrightarrow x^{2}<297$, from which $-\sqrt{297}<x<\sqrt{297}$. On this interval, there are 17 natural values of $x: x=1, x=2, \ldots, x=17$. At the same time, $y$ takes integer values only when $x$ is divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 07 *$, each of the 5 asterisks needs to be replaced with any of the digits $0,2,4,5,6,7$ (digits can repeat) so that the resulting 11-digit number is divisible by 75. In how many ways can this be done? | # Answer: 432.
Solution. For a number to be divisible by 75, it is necessary and sufficient that it is divisible by 25 and by 3. To ensure divisibility by 25, we can choose 5 as the last digit (1 way) from the available options.
To ensure divisibility by three, we proceed as follows. We will choose three digits arbit... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Plot on the plane $(x ; y)$ the set of points satisfying the equation $|5 x|+|12 y|+|60-5 x-12 y|=60$, and find the area of the resulting figure. | Answer: 30.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} { l... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of points in the plane $x O y$ that have natural coordinates $(x, y)$ and lie on the parabola $y=-\frac{x^{2}}{3}+98$ | Answer: 5.
Solution. Let's find the values of $x$ for which $y$ is positive: $-\frac{x^{2}}{3}+98>0 \Leftrightarrow x^{2}<294$, from which $-\sqrt{294}<x<\sqrt{294}$. On this interval, there are 17 natural values of $x: x=1, x=2, \ldots, x=17$. At the same time, $y$ takes integer values only when $x$ is divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the number $2 * 0 * 1 * 6 * 0 * 2 *$, each of the 6 asterisks needs to be replaced with any of the digits $0,2,4,5,7,9$ (digits can be repeated) so that the resulting 12-digit number is divisible by 12. In how many ways can this be done? | Answer: 5184.
Solution. For a number to be divisible by 12, it is necessary and sufficient that it is divisible by 4 and by 3. To ensure divisibility by 4, we can choose 0 or 4 as the last digit from the available options (2 ways).
To ensure divisibility by three, we proceed as follows. We will choose four digits arb... | 5184 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the plane $(x ; y)$, plot the set of points satisfying the equation $|4 x|+|3 y|+|24-4 x-3 y|=24$, and find the area of the resulting figure.
# | # Answer: 24.
Solution. Note that the equality $|a|+|b|+|c|=a+b+c$ holds if and only if the numbers $a, b$, and $c$ are non-negative (since if at least one of them is negative, the left side is greater than the right). Therefore, the first equation is equivalent to the system of inequalities
$$
\left\{\begin{array} {... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Given a regular 20-gon $M$. Find the number of quadruples of vertices of this 20-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 765.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{20}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{20}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, ... | 765 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 291000, such that $k^{2}-1$ is divisible by 291. | Answer: 4000.
Solution. By factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 97)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 97. Let's consider two cases.
a) $(k+1): 97$, i.e., $k=97 p+96, p \in \mathrm{Z}$. Then we get $(97 p+95)(97 p+97):(3 \cdot 97) \Leftrigh... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Given a regular 16-gon $M$. Find the number of quadruples of vertices of this 16-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 364.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{16}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{16}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, ... | 364 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 445000, such that $k^{2}-1$ is divisible by 445. Answer: 4000. | Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(5 \cdot 89)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 89. Let's consider two cases.
a) $(k+1): 89$, i.e., $k=89 p+88, p \in \mathrm{Z}$. Then we get $(89 p+87)(89 p+89):(5 \cdot 89) \Leftrightarrow(89 p+87)(p+... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Given a regular 22-gon $M$. Find the number of quadruples of vertices of this 22-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 1045.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{22}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{22}$.
Consider a chord connecting two adjacent vertices of the polygon, for example,... | 1045 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 485000, such that $k^{2}-1$ is divisible by 485. | Answer: 4000.
Solution. By factoring the dividend and divisor, we get the condition $(k-1)(k+1):(5 \cdot 97)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 97. Let's consider two cases.
a) $(k+1): 97$, i.e., $k=97 p+96, p \in \mathrm{Z}$. Then we get $(97 p+95)(97 p+97):(5 \cdot 97) \Leftrigh... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Given a regular 18-gon $M$. Find the number of quadruples of vertices of this 18-gon that are the vertices of convex quadrilaterals, which have at least one pair of parallel sides. | Answer: 540.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{18}$ in a circle. Each quadrilateral with a pair of parallel sides is determined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{18}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, ... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 267000, such that $k^{2}-1$ is divisible by 267. | Answer: 4000.
Solution. By factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 89)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 89. Let's consider two cases.
a) $(k+1): 89$, i.e., $k=89 p+88, p \in \mathbb{Z}$. Then we get $(89 p+87)(89 p+89):(3 \cdot 89) \Leftrigh... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $5 x^{2}-6 x y+y^{2}=6^{100}$. | Answer: 19594.
Solution: By factoring the left and right sides of the equation, we get $(5 x-y)(x-y)=2^{100} \cdot 3^{100}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
5 x - y = 2 ^ { k } \cdot 3 ^ { l }, \\
x - y = 2 ^ { 1 0 0 - k } \cdot 3 ^ { 1 0 0 - l }
\end{array... | 19594 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $6 x^{2}-7 x y+y^{2}=10^{100}$. | Answer: 19998.
Solution: By factoring the left and right sides of the equation, we get $(6 x-y)(x-y)=2^{100} \cdot 5^{100}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
6 x - y = 2 ^ { k } \cdot 5 ^ { l } , \\
x - y = 2 ^ { 1 0 0 - k } \cdot 5 ^ { 1 0 0 - l }
\end{arra... | 19998 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $x^{2}+6 x y+5 y^{2}=10^{100}$. | Answer: 19594.
Solution: By factoring the left and right sides of the equation, we get $(x+5 y)(x+y)=2^{100} \cdot 5^{100}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
x+5 y=2^{k} \cdot 5^{l}, \\
x+y=2^{100-k} \cdot 5^{100-l}
\end{array} \text { or } \left\{\begin{arr... | 19594 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(x ; y)$ that satisfy the condition $x^{2}+7 x y+6 y^{2}=15^{50}$. | Answer: 4998.
Solution: Factoring the left and right sides of the equation, we get $(x+6 y)(x+y)=5^{50} \cdot 3^{50}$. Since each factor on the left side is an integer, it follows that
$$
\left\{\begin{array}{l}
x+6 y=5^{k} \cdot 3^{l}, \\
x+y=5^{50-k} \cdot 3^{50-l}
\end{array} \text { or } \left\{\begin{array}{l}
x... | 4998 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of natural numbers $k$, not exceeding 242400, such that $k^{2}+2 k$ is divisible by 303. Answer: 3200. | Solution. Factoring the dividend and divisor, we get the condition $k(k+2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k+2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p+2):(3 \cdot 101) \Leftrightarrow p(101 p+2): 3$. The first... | 3200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a regular 20-gon $M$. Find the number of quadruples of vertices of this 20-gon that are the vertices of trapezoids. | Answer: 720.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{20}$ in a circle. Each trapezoid is defined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{20}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 9 more chord... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given a regular 16-gon $M$. Find the number of quadruples of vertices of this 16-gon that are the vertices of trapezoids. | Answer: 336.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{16}$ in a circle. Each trapezoid is defined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{16}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 7 more chord... | 336 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of natural numbers $k$, not exceeding 333300, such that $k^{2}-2 k$ is divisible by 303. Answer: 4400. | Solution. Factoring the dividend and divisor, we get the condition $k(k-2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k-2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p-2):(3 \cdot 101) \Leftrightarrow p(101 p-2): 3$. The first... | 4400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a regular 22-gon $M$. Find the number of quadruples of vertices of this 22-gon that are the vertices of trapezoids. | Answer: 990.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{22}$ in a circle. Each trapezoid side is defined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{22}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 10 more... | 990 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505. Answer: 3600. | Solution. Factoring the dividend and divisor, we get the condition $k(k-1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k-1)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p-1):(5 \cdot 101) \Leftrightarrow p(101 p-1): 5$. The first... | 3600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given a regular 18-gon $M$. Find the number of quadruples of vertices of this 18-gon that are the vertices of trapezoids. | Answer: 504.
Solution. Let's inscribe the given polygon $K_{1} K_{2} \ldots K_{18}$ in a circle. Each trapezoid is defined by a pair of parallel chords with endpoints at points $K_{1}, \ldots, K_{18}$.
Consider a chord connecting two adjacent vertices of the polygon, for example, $K_{6} K_{7}$. There are 8 more chord... | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. Find the number of integer solutions ( $x ; y ; z$ ) of the equation param 1 , satisfying the condition param 2.
| param1 | param2 | |
| :---: | :---: | :---: |
| $60^{x} \cdot\left(\frac{500}{3}\right)^{y} \cdot 360^{z}=2160$ | $\|x+y+z\| \leq 60$ | |
| $60^{x} \cdot\left(\frac{500}{3}\right)^{y} \cdot 360^{z}=... | 12. Find the number of integer solutions ( $x ; y ; z$ ) of the equation param 1 , satisfying the condition param 2.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $60^{x} \cdot\left(\frac{500}{3}\right)^{y} \cdot 360^{z}=2160$ | $\|x+y+z\| \leq 60$ | 60 |
| $60^{x} \cdot\left(\frac{500}{3}\right)^{y} \cdot ... | 86 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
17. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+3 x^{2}+... | 17. It is known that the number $a$ satisfies the equation param 1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | 2 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19. In a football tournament held in a single round-robin format (each team must play each other exactly once), $N$ teams are participating. At some point in the tournament, the coach of team $A$ noticed that any two teams, different from $A$, have played a different number of games. It is also known that by this point... | 19. In a football tournament held in a single round-robin format (each team must play every other team exactly once), $N$ teams are participating. At a certain point in the tournament, the coach of team $A$ noticed that any two teams, different from $A$, have played a different number of games. It is also known that by... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
20. On the table, there are param 1 externally identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee getting at leas... | 20. On the table, there are param 1 externally identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee getting at leas... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. For each natural $n$, which is not a perfect square, the number of values of the variable $x$ is calculated, for which both numbers $x+\sqrt{n}$ and $x^{2}+param1 \cdot \sqrt{n}$ are natural numbers less than param2. Find the total number of such values of $x$.
| param1 | param2 | answer |
| :---: | :---: | :---: ... | 18. For each natural $n$, which is not a perfect square, the number of values of the variable $x$ is calculated, for which both numbers $x+\sqrt{n}$ and $x^{2}+param1 \cdot \sqrt{n}$ are natural numbers less than param2. Find the total number of such values of $x$.
| param1 | param2 | answer |
| :---: | :---: | :---: ... | 108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
20. Find param 1 given param 2.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| maximum $2 x+y$ | $\|4 x-3 y\|+5 \sqrt{x^{2}+y^{2}-20 y+100}=30$ | |
| maximum $x+2 y$ | $\|4 y-3 x\|+5 \sqrt{x^{2}+y^{2}+20 y+100}=40$ | |
| maximum $2 y-x$ | $\|4 y+3 x\|+5 \sqrt{x^{2}+y^{2}+10 x+25}=15$ | |
| maximum $x-5 y$... | 20. Find param 1 given param2.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| maximum $2 x+y$ | $\|4 x-3 y\|+5 \sqrt{x^{2}+y^{2}-20 y+100}=30$ | 16 |
| maximum $x+2 y$ | $\|4 y-3 x\|+5 \sqrt{x^{2}+y^{2}+20 y+100}=40$ | -12 |
| maximum $2 y-x$ | $\|4 y+3 x\|+5 \sqrt{x^{2}+y^{2}+10 x+25}=15$ | 8 |
| maximum $x... | -12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the table, there are 140 different cards with numbers $3,6,9, \ldots 417,420$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $7?$ | Answer: 1390.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a common difference of 3. Therefore, the remainders when these numbers are divided by 7 alternate. Indeed, if one of these numbers is divisible by 7, i.e., has the form $7k$, where $k \in \mathbb{N}$, then the n... | 1390 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|x-1|+|5-x| \leqslant 4 \\
\frac{x^{2}-6 x+2 y+7}{y+x-4} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 4.
Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $x<1$. Then $1-x+5-x \leqslant 4 \Leftrightarrow x \geqslant 1$, i.e., there are no solutions.
2) $1 \leqslant x \leqslant 5$. Then $x-1+5-x \leqslant 4 \Leftrightarrow 4 \leqslant 4$, which is always true, so ... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. On the table, there are 210 different cards with numbers $2,4,6, \ldots 418,420$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the sum of the numbers on the selected cards is divisible by $7?$ | Answer: 3135.
Solution. The given numbers, arranged in ascending order, form an arithmetic progression with a difference of 2. Therefore, the remainders of these numbers when divided by 7 alternate. Indeed, if one of these numbers is divisible by 7, i.e., has the form $7k$, where $k \in \mathbb{N}$, then the following... | 3135 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
|y|+|4-y| \leqslant 4 \\
\frac{y^{2}+x-4 y+1}{2 y+x-7} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. Consider the first inequality. To open the absolute values, we consider three possible cases.
1) $y < 0$. Then $-y-4+y \leqslant 4 \Leftrightarrow -4 \leqslant 4$, which is always true, so $y \in (-\infty, 0)$.
2) $0 \leqslant y \leqslant 4$. Then $y-4+y \leqslant 4 \Leftrightarrow 2y \leqslant 8... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. There are 200 different cards with numbers $2,3,2^{2}, 3^{2}, \ldots, 2^{100}, 3^{100}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a cube of an integer? | Answer: 4389.
Solution. To obtain the cube of a natural number, it is necessary and sufficient for each factor to enter the prime factorization of the number in a power that is a multiple of 3.
Suppose two cards with powers of two are chosen. We have 33 exponents divisible by 3 $(3,6,9, \ldots, 99)$, 34 exponents giv... | 4389 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y-x \geqslant|x+y| \\
\frac{x^{2}+8 x+y^{2}+6 y}{2 y-x-8} \leqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 8.
Solution. The first inequality is equivalent to the system ${ }^{1}$ $\left\{\begin{array}{l}x+y \leqslant y-x, \\ x+y \geqslant x-y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leqslant 0, \\ y \geqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x+4)... | 8 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5. There are 100 different cards with numbers $2,5,2^{2}, 5^{2}, \ldots, 2^{50}, 5^{50}$ (each card has exactly one number, and each number appears exactly once). In how many ways can 2 cards be chosen so that the product of the numbers on the chosen cards is a cube of an integer? | Answer: 1074.
Solution. To obtain the cube of a natural number, it is necessary and sufficient for each factor to enter the prime factorization of the number in a power that is a multiple of 3.
Suppose two cards with powers of two are chosen. We have 16 exponents that are divisible by 3 $(3,6,9, \ldots, 48)$, 17 expo... | 1074 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. On the coordinate plane, consider a figure $M$ consisting of all points with coordinates $(x ; y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
x+y+|x-y| \leqslant 0 \\
\frac{x^{2}+6 x+y^{2}-8 y}{x+3 y+6} \geqslant 0
\end{array}\right.
$$
Sketch the figure $M$ and find its area. | Answer: 3.
Solution. The first inequality is equivalent to the system $\left\{\begin{array}{l}x-y \leqslant-x-, \\ x-y \geqslant x+y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \leqslant 0, \\ y \leqslant 0 .\end{array}\right.\right.$
Consider the second inequality. It can be written as $\frac{(x+3)^{2}+(y-4)^... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. Two parabolas param 1 and param 2 touch at a point lying on the $O x$ axis. Through point $D$, the second intersection point of the first parabola with the $O x$ axis, a vertical line is drawn, intersecting the second parabola at point $A$ and the common tangent to the parabolas at point $B$. Find the ratio $D A: D ... | 1. Two parabolas param 1 and param 2 touch at a point lying on the $O x$ axis. Through point $D$, the second intersection point of the first parabola with the $O x$ axis, a vertical line is drawn, intersecting the second parabola at point $A$ and the common tangent to the parabolas at point $B$. Find the ratio $D A: D ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :--- |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 x^{2}+15... | 7. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the smallest possible value of the sum $a+b$.
| param1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-6 x^{2}+14 x+2=0$ | 3 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+6 ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. On the table, there are param 1 externally identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee getting at least... | 9. On the table, there are param 1 visually identical coins. It is known that among them, there are exactly param 2 counterfeit ones. You are allowed to point to any two coins and ask whether it is true that both these coins are counterfeit. What is the minimum number of questions needed to guarantee that you will get ... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
17. It is known that the number $a$ satisfies the equation param1, and the number $b$ satisfies the equation param2. Find the maximum possible value of the sum $a+b$.
| param1 | param2 | |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}+3 x^{2}+6... | 17. It is known that the number $a$ satisfies the equation param 1, and the number $b$ satisfies the equation param2. Find the greatest possible value of the sum $a+b$.
| param 1 | param2 | Answer |
| :---: | :---: | :---: |
| $x^{3}-3 x^{2}+5 x-17=0$ | $x^{3}-3 x^{2}+5 x+11=0$ | 2 |
| $x^{3}+3 x^{2}+6 x-9=0$ | $x^{3}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19. In a football tournament held in a single round-robin format (each team must play every other team exactly once), $N$ teams are participating. At a certain point in the tournament, the coach of team $A$ noticed that any two teams, different from $A$, have played a different number of games. It is also known that by... | 19. In a football tournament held in a single round-robin format (each team must play every other team exactly once), $N$ teams are participating. At a certain point in the tournament, the coach of team $A$ noticed that any two teams, different from $A$, have played a different number of games. It is also known that by... | 63 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. On the coordinate plane, squares are considered, all vertices of which have integer non-negative coordinates, and the center is located at the point $(60 ; 45)$. Find the number of such squares. | Answer: 2070.
Solution. Draw through the given point $(60 ; 45)$ vertical and horizontal lines $(x=60$ and $y=45)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 45 ways: $... | 2070 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. On the coordinate plane, consider squares all of whose vertices have natural coordinates, and the center is located at the point $(55 ; 40)$. Find the number of such squares. | Answer: 1560.
Solution. Draw through the given point $(55 ; 40)$ vertical and horizontal lines $(x=55$ and $y=40)$. There are two possible cases.
a) The vertices of the square lie on these lines (and its diagonals are parallel to the coordinate axes). Then the "lower" vertex of the square can be located in 39 ways: $... | 1560 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $x^{2}+x y=30000000$. | Answer: 256.
Solution. By factoring the left and right sides of the equation, we get $x(x+y)=$ $3 \cdot 2^{7} \cdot 5^{7}$. Then, if $x>0$, $x$ is one of the divisors of the right side. The right side has a total of $2 \cdot 8 \cdot 8=128$ divisors (since any divisor can be represented as $3^{a} \cdot 2^{b} \cdot 5^{c... | 256 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 70, 1 \leqslant b \leqslant 50$, and at the same time,
the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right... | Answer: 1260.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b \vdots 5$, so one of the numbers $a$ or $b$ mus... | 1260 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that for three consecutive natural values of the argument, the quadratic function $f(x)$ takes the values 6, 14, and 14, respectively. Find the greatest possible value of $f(x)$. | Answer: 15.
Solution. Let $n, n+1, n+2$ be the three given consecutive values of the argument. Since a quadratic function takes the same values at points symmetric with respect to the x-coordinate of the vertex of the parabola $x_{\text{v}}$, then $x_{\text{v}}=n+1.5$, and thus $f(x)$ can be represented as $f(x)=a(x-n... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two equal rectangles $P Q R S$ and $P_{1} Q_{1} R_{1} S_{1}$ are inscribed in triangle $A B C$ (with points $P$ and $P_{1}$ lying on side $A B$, points $Q$ and $Q_{1}$ lying on side $B C$, and points $R, S, R_{1}$ and $S_{1}$ lying on side $A C$). It is known that $P S=3, P_{1} S_{1}=9$. Find the area of triangle $A... | Answer: 72.
Solution. Draw the height $B F$ of triangle $A B C$. Let it intersect segments $P Q$ and $P_{1} Q_{1}$ at points $H$ and $M$ respectively. Note that $H M=6$. From the similarity of triangles $B P Q$ and $B P_{1} Q_{1}$, it follows that $\frac{B H}{P Q}=\frac{B M}{P_{1} Q_{1}}$ (the ratio of the height to t... | 72 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Find the number of pairs of integers $(x ; y)$ that satisfy the equation $y^{2}-x y=700000000$. | Answer: 324.
Solution. Factoring the left and right sides of the equation, we get $y(y-x)=$ $7 \cdot 2^{8} \cdot 5^{8}$. Then if $y>0$, $y$ is one of the divisors of the right side. The right side has a total of $2 \cdot 9 \cdot 9=162$ divisors (since any divisor can be represented as $7^{a} \cdot 2^{b} \cdot 5^{c}$, ... | 324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 80,1 \leqslant b \leqslant 30$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that ... | Answer: 864.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be di... | 864 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 267000, such that $k^{2}-1$ is divisible by 267. Answer: 4000. | Solution. Factoring the dividend and divisor, we get the condition $(k-1)(k+1):(3 \cdot 89)$. This means that one of the numbers $(k+1)$ or $(k-1)$ is divisible by 89. Let's consider two cases.
a) $(k+1): 89$, i.e., $k=89 p+88, p \in \mathrm{Z}$. Then we get $(89 p+87)(89 p+89):(3 \cdot 89) \Leftrightarrow(89 p+87)(p+... | 4000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 242400, such that $k^{2}+2 k$ is divisible by 303. | Answer: 3200.
Solution. By factoring the dividend and divisor, we get the condition $k(k+2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k+2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+2):(3 \cdot 101) \Leftrightarrow p(101 p... | 3200 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 353500, such that $k^{2}+k$ is divisible by 505. | Answer: 2800.
Solution: By factoring the dividend and divisor, we get the condition $k(k+1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k+1)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p+1):(5 \cdot 101) \Leftrightarrow p(101 p... | 2800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 333300, such that $k^{2}-2 k$ is divisible by 303. Answer: 4400. | Solution. Factoring the dividend and divisor, we get the condition $k(k-2):(3 \cdot 101)$. This means that one of the numbers $k$ or $(k-2)$ is divisible by 101. Let's consider two cases.
a) $k: 101$, i.e., $k=101 p, p \in \mathrm{Z}$. Then we get $101 p(101 p-2):(3 \cdot 101) \Leftrightarrow p(101 p-2) \vdots 3$. The... | 4400 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of natural numbers $k$, not exceeding 454500, such that $k^{2}-k$ is divisible by 505. | Answer: 3600.
Solution. By factoring the dividend and divisor, we get the condition $k(k-1):(5 \cdot 101)$. This means that one of the numbers $k$ or $(k-1)$ is divisible by 101. Let's consider two cases.
a) $k \vdots: 101$, i.e., $k=101 p, p \in \mathbb{Z}$. Then we get $101 p(101 p-1):(5 \cdot 101) \Leftrightarrow ... | 3600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 700. The answer should be presented as an integer. | Answer: 2520.
Solution. Since $700=7 \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, one seven, and three ones, or (b) one four, two fives, one seven, and four ones. We will calculate the number of variants in each case.
(a) First, we choose two places out of... | 2520 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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