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7.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17. | Solution. The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\... | 17 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. First-grader Petya was laying out a contour of an equilateral triangle with the tokens he had, so that each of its sides, including the vertices, contained the same number of tokens. Then, with the same tokens, he managed to lay out the contour of a square in the same way. How many tokens does Petya have, if each si... | Answer: 24.
Solution. Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips being counted twice). From the problem statement, it follows that $y=x-2$. Therefore, we get the equation $3(x-... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov and Vasechkin were solving the same arithmetic problem. A certain number had to be divided by 2, multiplied by 7, and 1001 subtracted. Petrov performed all the operations correctly, while Vasechkin got everything mixed up: he divided by 8, squared the result, and also subtracted 1001. It is known that Petrov ... | Answer: 295.
Solution. Note that the number 1001 is divisible by 7 without a remainder. This means that the number Petrov obtained must be divisible by 7. But it is a prime number, so Petrov got 7. Let's reverse the operations Petrov performed and find the original number: $\frac{7+1001}{7} \cdot 2=288$. Now, let's re... | 295 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face? | Answer: 16.
Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).
Consider a face containing the vertex where the number 6 is placed. ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place?
Otvet: 41000. | Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number as cells, in each o... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. For the sequence $\left\{a_{n}\right\}$, it is known that $a_{1}=1.5$ and $a_{n}=\frac{1}{n^{2}-1}$ for $n \in \mathbb{N}, n>1$. Are there such values of $n$ for which the sum of the first $n$ terms of this sequence differs from 2.25 by less than 0.01? If yes, find the smallest one. | Answer: yes, $n=100$.
Solution. The general formula for the terms of the sequence (except the first) can be written as $(n \geqslant 2)$:
$$
a_{n}=\frac{1}{n^{2}-1}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right)
$$
As a result, the sum of the first $n$ terms of the sequence, except the first, takes the form:
$... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Can non-negative integers be placed on the faces of two cubes so that when they are randomly thrown, the sum of the points showing can be equal to any divisor of the number 36? If this is possible, indicate the sum of all 12 numbers on the faces; if it is impossible - indicate 0. | Answer: The sum is 111. Solution. The numbers $1,2,3,4,5$ and 6 can be placed on the faces of one die, and $0,6,12,18,24,30$ on the faces of the other. Then the sum of the points rolled can be equal to any of the numbers from 1 to 36. The sum of the numbers on all faces is 111. This sum does not depend on how the dice ... | 111 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find all three-digit numbers $\overline{L O M}$, composed of distinct digits $L, O$ and $M$, for which the following equality holds:
$$
\overline{L O M}=(L+O+M)^{2}+L+O+M
$$ | Answer: 156. Instructions. Let $x=L+O+M$. Then $\overline{L O M}=x(x+1)$. In this case, $x \geq 10$ (otherwise $x(x+1)<100)$ and $x \leq 24$ (the sum of digits does not exceed $9+8+7=24$). Therefore, $x \in[10 ; 24]$. From the relation $100 \cdot L+10 \cdot O+M=x^{2}+L+O+M$ it follows that $x^{2}=99 \cdot L+9 \cdot O$,... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Pete was given a new electric jigsaw for his birthday, with a function to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of $50 \mathrm{cm}$ and cut it into squares with sides of 10 cm and squares with sides of 20 cm. How many squares in total were obtained, i... | Answer: 16. Solution. Each protruding edge is part of the perimeter of two figures, in addition, the perimeter of the original square must be taken into account, from which we get that the total perimeter of the resulting small squares is $280 \cdot 2+200=760$. Now we can denote the number of squares through $x$ and $y... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. How many four-digit numbers exist that have the following properties: all digits of the number are even; the number is divisible by four, if the last digit is erased, the resulting three-digit number is not divisible by four? | Answer: 120. Solution. The last digit of the number must be divisible by 4 (that is, it can be 0, 4 or 8), and the second to last - not (2 or 6). In addition, the first digit is not zero. Therefore, we get $4 \cdot 5 \cdot 2 \cdot 3=120$ variants. | 120 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. [5-6.2 (20 points), 7-8.2 (15 points), 9.2 (15 points)] In how many ways can the word «РОТОР» be read from the letters in the diagram, if it is not allowed to return to the letters already passed, and readings that differ only in direction are considered the same?
| Р O Т O | дохо |
| :--- | :--- |
| O Т O P | О ... | Answer: 25.
Solution. The number of ways to read starting from the upper left letter "P" is $2^{4}=16$, as 4 steps are required to move through the letters, and each step is directed either to the right or down. If we strike out the considered letter "P", it remains to count the number of remaining ways in Fig. 1.
| ... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.1. [5-6.5 (а) - 20 points, б) - 20 points)] Vovochka adds numbers in a column in the following way: he does not remember the tens, and under each pair of digits in the same place value, he writes their sum, even if it is a two-digit number. For example, for the sum $248+208$, he would get the value 4416.
a) In how m... | Answer: a) 244620 b) 1800.
Solution. a) Vovochka will get the correct answer if and only if the sum of the last and second-to-last digits also results in a digit, i.e., there is no carry-over. Let's count how many pairs of digits have a sum that does not exceed 9. If one digit is 0, the other can be any of 10 digits (... | 244620 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.1. [7-8.7 (20 points), 9.8 (15 points), 10.8 (20 points)] There is a rotating round table with 16 sectors, on which numbers $0,1,2, \ldots, 7,8,7,6, \ldots, 2,1$ are written in a circle. 16 players are sitting around the table, numbered in order. After each rotation of the table, each player receives as many points ... | Answer: 20.
Solution. Players No. 5 and No. 9 together scored $72+84=156=12 \cdot 13$ points. In one spin, they can together score no more than 12 points. Therefore, in each of the 13 spins, they together scored 12 points. Note that the 12 points they score can be one of the sums $8+4, 7+5$, $6+6, 5+7$ or $4+8$, when ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
18.1. $[10.2$ (15 points)] A wooden parallelepiped, all sides of which are expressed in whole centimeters, was painted red, and then sawn parallel to the faces into cubes with a side of 1 cm. It turned out that one third of the resulting cubes have at least one red face, while the remaining two thirds have all faces un... | Answer: 18 cm.
Solution. Let the length of the parallelepiped be $n$ cm, then the width is $n-2$, and the height is $n-4$. This gives $n(n-2)(n-4)$ cubes. The number of uncolored cubes will be $(n-2)(n-4)(n-6)$ (a layer one cube wide is "removed" from each side). This results in the equation $(n-2)(n-4)(n-7)=\frac{2}{... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
20.1. [10.7 (15 points)] The function $f$, defined on the set of integers, satisfies the following conditions:
1) $f(1)+1>0$
2) $f(x+y)-x f(y)-y f(x)=f(x) f(y)-x-y+x y$ for any $x, y \in \mathbb{Z}$;
3) $2 f(x)=f(x+1)-x+1$ for any $x \in \mathbb{Z}$.
Find $f(10)$. | Answer: 1014.
Solution. If $h(x)=f(x)+x$, then from condition 2) we get $h(x+y)=h(x) h(y)$. Then, for $x=y=0$, this equality takes the form $h(0)^{2}=h(0)$, i.e., $h(0)=0$ or $h(0)=1$. In the first case, $h(x) \equiv 0$, which is impossible due to condition 1). If $a=h(1)$, then $h(x)=a^{x}$ for any $x \in \mathbb{Z}$... | 1014 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
20.2. The function $g$, defined on the set of integers, satisfies the conditions:
1) $g(1)>1$
2) $g(x+y)+x g(y)+y g(x)=g(x) g(y)+x+y+x y$ for any $x, y \in \mathbb{Z}$;
3) $3 g(x)=g(x+1)+2 x-1$ for any $x \in \mathbb{Z}$.
Find $g(5)$. | Answer: 248.
Solution. Let $g(1)=a$, then from condition 3) we sequentially find $g(2)=3 a-1, g(3)=$ $9 a-6, g(4)=27 a-23, g(5)=81 a-76$. From condition 2), substituting $x=4$ and $y=1$, we obtain after simplifications the equation $a^{2}-5 a+4=0$, from which $a=1$ or $a=4$. The case $a=1$ contradicts condition 1). Th... | 248 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 2.
In a school test, there are 5 sections, each containing the same number of questions. Anton answered 20 questions correctly. The percentage of his correct answers was more than 60 but less than 70. How many questions were there in the test in total? | Answer: 30
Solution. According to the condition $\frac{60}{100}<\frac{20}{x}<\frac{70}{100}$, that is, $\frac{200}{7}<x<\frac{100}{3}$, so $29 \leqslant x \leqslant 33$. Since the number of questions must be divisible by 5, then $x=30$.
## B-2
In a school test, there are 4 sections, each containing the same number o... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task 5.
Around a round table, 1001 people are sitting, each of whom is either a knight (always tells the truth) or a liar (always lies). It turned out that next to each knight sits exactly one liar, and next to each liar there is a knight. What is the minimum number of knights that can sit at the table? | Answer: 502
Solution. From the condition, it follows that a knight cannot sit between two knights or two liars, and a liar cannot sit between two liars. Thus, when moving around the table, knights will be encountered in pairs, while liars will be encountered singly or in pairs. From this, it follows that each liar can... | 502 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6.
Find the smallest natural number $n$, for which the number $n+2018$ is divisible by 2020, and the number $n+2020$ is divisible by 2018. | Answer: 2034142
Solution. By the condition $n+2018=2020 m, n+2020=2018 k$, hence $1009 k-1010 m=1$. The solution to this Diophantine equation is: $k=-1+1010 p, m=-1+1009 p$. Therefore, $n+2018=$ $2020(-1+1009 p)=-2020+2020 \cdot 1009 p$, from which $n=-2018-2020+2020 \cdot 1009 p$. The smallest natural $n$ is $n=2020 ... | 2034142 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7.
Inside a convex 13-gon, 200 points are placed such that no three of these 213 points (including the vertices of the polygon) are collinear. The polygon is divided into triangles, with the vertices of each triangle being any three of the 213 points. What is the maximum number of triangles that could result... | Answer: 411
Solution. If each of the points is a vertex of some triangle, then the total sum of the angles will be equal to $180^{\circ} \cdot(13-2)+360^{\circ} \cdot 200=180^{\circ} \cdot(11+400)=180^{\circ} \cdot 411$. This means 411 triangles are formed. If some points do not participate in the cutting, then the nu... | 411 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 8.
Let $A(n)$ denote the greatest odd divisor of the number $n$. For example, $A(21)=21$, $A(72)=9, A(64)=1$. Find the sum $A(111)+A(112)+\ldots+A(218)+A(219)$. | Answer: 12045
Solution. The largest odd divisors of any two of the given numbers cannot coincide, as numbers with the same largest odd divisors are either equal or differ by at least a factor of 2. Therefore, the largest odd divisors of the numbers $111, 112, \ldots, 218, 219$ are distinct. This means that the largest... | 12045 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.1. (13 points) A guard has detained a stranger and wants to drive him away. But the person caught said that he had made a bet with his friends for 100 coins that the guard would not drive him away (if the guard does, he has to pay his friends 100 coins, otherwise they pay him), and, deciding to bribe the guard, offer... | Answer: 199.
Solution. If the guard asks for 199 coins, then by agreeing to give him this amount, the outsider will win the dispute and receive 100 coins. In total, he will lose 99 coins. If the outsider refuses, he will lose the dispute and lose 100 coins, which is less favorable (by 1 coin) for the one caught. If th... | 199 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.1. (13 points) On the picture, the entrance and exit of the maze are marked with arrows. You can move through it such that on this picture you can only move right, down, or up (you cannot turn around). How many different ways are there to navigate this maze?
---
The translation maintains the original text's formatt... | Answer: 16.
Solution. Let's call a place in the maze a fork if from it you can move in two possible directions.
Moving up or to the right after entering, we arrive at one of the forks marked... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. (13 points) Find the smallest natural number such that after multiplying it by 9, the result is a number written with the same digits but in some different order. | Answer: 1089.
Solution. Note that the number must start with one, otherwise multiplying by 9 would increase the number of digits. After multiplying 1 by 9, we get 9, so the original number must contain the digit 9. The number 19 does not work, so two-digit numbers do not work. Let's consider three-digit numbers. The s... | 1089 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. (13 points) The surface of a round table is divided into 9 identical sectors, in which the numbers from 1 to 9 are written sequentially clockwise. Around the table sit 9 players with numbers \(1, 2, \ldots, 9\), going clockwise. The table can rotate around its axis in both directions, while the players remain in p... | Answer: 57.
Solution. If, as a result of spinning the table, one coin went to someone with numbers from 5 to 8, then player No. 8 received 5 fewer coins than player No. 9, and if one coin went to someone else, then No. 8 received 4 more coins than player No. 4. Let the number of spins where one coin went to someone wi... | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Several boys and girls gathered around a round table. It is known that exactly for 7 girls, the person sitting to their left is a girl, and for 12 - a boy. It is also known that for $75\%$ of the boys, the person sitting to their right is a girl. How many people are sitting at the table? | Answer: 35 people
Solution: From the condition, it is clear that there are exactly 19 girls.
Note that the number of girls who have boys sitting to their left is equal to the number of boys who have girls sitting to their right. Thus, $75\%$ of the boys equals 12, i.e., there are 16 boys in total sitting at the table... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$ | Answer: 18108.
Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+... | 18108 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. (4 points) Mitya is 11 years older than Shura. When Mitya was as old as Shura is now, he was twice as old as she was. How old is Mitya? | Answer: 33.
Solution. 11 years ago, Shura was half as old as she is now. So, she is 22 years old, and Mitya is 33. | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. (13 points) Find the smallest natural number that is divisible by 11 and whose representation contains 5 zeros and 7 ones. (You can use the divisibility rule for 11: a number is divisible by 11 if the difference between the sum of the digits in the even positions and the sum of the digits in the odd positions is d... | Solution. The number 7 is not divisible by 2, so the number cannot have exactly $5+7=12$ digits. Let's consider the case when the number has 13 digits. If the "new" digit is zero, we again get a contradiction. Therefore, the "new" digit must be at least 1. To find the smallest number, we should place this digit as far ... | 1000001111131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. (13 points) Yura has unusual clocks with several minute hands moving in different directions. Yura calculated that the minute hands coincided in pairs exactly 54 times in one hour. What is the maximum number of minute hands that can be on Yura's clocks? | Answer: 28.
Solution. Let some two arrows coincide, then after 30 seconds they will coincide again. Therefore, the arrows in each such pair will coincide exactly 2 times per minute. Thus, if $n$ arrows move in one direction, and $m$ arrows move in the other, then $2 m n=54, m n=27$. Therefore, $n$ can be $1,3,9$ or 27... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.1. (13 points) In how many ways can eight of the nine digits $1,2,3,4,5,6$, 7,8 and 9 be placed in a $4 \times 2$ table (4 rows, 2 columns) so that the sum of the digits in each row, starting from the second, is 1 more than in the previous one? | Answer: 64.
Solution. The sum of all nine numbers is 45. Let $x$ be the sum of the two numbers in the first row, and let $a$ be the one number out of the nine that we do not place in the figure. Then $x+x+1+x+2+x+3=45-a$, from which $4 x+a=39$. Since $a$ is an integer from 1 to 9, we get 2 possible cases: either $x=9,... | 64 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.1. Find all pairs of two-digit natural numbers for which the arithmetic mean is $25 / 24$ times greater than the geometric mean. In your answer, specify the largest of the arithmetic means for all such pairs. | Answer: 75.
Solution: Let $a, b$ be the required numbers (without loss of generality, we can assume that $a > b$) and let $\frac{a+b}{2}=25x$ and $\sqrt{ab}=24x$. Then $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=98x$ and $(\sqrt{a}-\sqrt{b})^2=$ $a+b-2\sqrt{ab}=2x$. From this, it follows that $\sqrt{a}+\sqrt{b}=7(\sqrt{a}-\... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. The vertices of a cube are labeled with numbers $\pm 1$, and on its faces are numbers equal to the product of the numbers at the vertices of that face. Find all possible values that the sum of these 14 numbers can take. In your answer, specify their product. | Answer: -20160.
Solution. It is obvious that the maximum value of the sum is 14. Note that if we change the sign of one of the vertices, the sum of the numbers in the vertices will increase or decrease by 2. On the other hand, the signs of three faces will change. If their sum was $1, -1, 3, -3$, it will become $-1, 1... | -20160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers? | Answer: 250.
Solution. Note that any odd number $2 n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4 n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4 n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of th... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6.1. On the coordinate plane, an isosceles right triangle with vertices at points with integer coordinates is depicted. It is known that there are exactly 2019 points with integer coordinates on the sides of the triangle (including the vertices). What is the smallest possible length of the hypotenuse of the triangle un... | Answer: 952.
Solution. The smallest length of the hypotenuse corresponds to the case where the distance between the points is the smallest, which is only possible when the legs follow the grid lines (for a rotated triangle, the distance between the points will be no less than $\sqrt{2}$). Then each leg has a length of... | 952 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. Natural numbers from 1 to some $n$ are written in a row. When one of the numbers was removed, it turned out that the arithmetic mean of the remaining numbers is $40 \frac{3}{4}$. Find the number that was removed. | Answer: 61.
Solution. The sum of numbers from 1 to $n$ is $S_{n}=\frac{n(n+1)}{2}$. Let the number removed be $m$, where $1 \leqslant m \leqslant n$. Then the condition of the problem can be written as
$$
\frac{S_{n}-m}{n-1}=40 \frac{3}{4}
$$
Transforming: $\frac{n+2}{2}-\frac{m-1}{n-1}=40 \frac{3}{4}$. Note that $\... | 61 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. A ballpoint pen costs 10 rubles, a gel pen costs 50 rubles, and a fountain pen costs 80 rubles. What is the maximum number of gel pens that can be bought given that exactly 20 pens must be purchased in total, and among them there must be pens of all three types, and exactly 1000 rubles must be spent on them? | Answer: 13. Solution. If $x$ ballpoint pens, $y$ gel pens, and $z$ fountain pens are bought, then we have two equations: $x+y+z=20 ; 10 x+50 y+80 z=1000$ (or $x+5 y+8 z=100$ ). Subtracting the first equation from the second, we get $4 y+7 z=80$. It follows that $z$ must be divisible by 4, i.e., $z=4 n$. Therefore, $4 y... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.1. Given the function $f(x)=|x+1|-2$. How many roots does the equation $f(f(\ldots f(f(x)) \ldots))=\frac{1}{2}$ have, where the function $f$ is applied 2013 times? | Answer: 4030. Solution. Considering the graph of the function $f$ (see Fig. 1), we conclude that $|f(x)+1|=f(x)+1$ and $|f(x)+1|-2=f(x)-1$. Therefore, the graph of the function $f(f(x))$
ha... | 4030 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. Specify the integer closest to the number $\sqrt{2012-\sqrt{2013 \cdot 2011}}+\sqrt{2010-\sqrt{2011 \cdot 2009}}+\ldots+\sqrt{2-\sqrt{3 \cdot 1}}$. | Answer: 31. Solution. The number given in the condition is equal to
$$
\frac{\sqrt{4024-2 \sqrt{2013 \cdot 2011}}+\sqrt{4010-2 \sqrt{2011 \cdot 2009}}+\ldots+\sqrt{4-2 \sqrt{3 \cdot 1}}}{\sqrt{2}}
$$
$=\frac{\sqrt{(\sqrt{2013}-\sqrt{2011})^{2}}+\sqrt{(\sqrt{2011}-\sqrt{2009})^{2}}+\ldots+\sqrt{(\sqrt{3}-\sqrt{1})^{2}... | 31 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.1. Find the area of the figure defined on the coordinate plane by the inequality $2(2-x) \geq\left|y-x^{2}\right|+\left|y+x^{2}\right|$. | Answer: 15. Solution. In the region above the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq y$. In the region below the graphs of the functions $y=x^{2}$ and $y=-x^{2}$, the original inequality takes the form $2-x \geq -y$, i.e., $y \geq x-2$. In the region lying abo... | 15 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.1. First-grader Masha, entering the school, each time climbs the school porch stairs, which have 10 steps. Being at the bottom of the stairs or on one of its steps, she can either go up to the next step or jump over one step up (Masha cannot yet jump over two or more steps). What is the minimum number of times Masha ... | Answer: 89. Solution. Note that Masha can climb a porch with one step in one way, and a porch with two steps in two ways: either stepping on each step, or, by stepping over the first step, landing directly on the second.
Let $a_{n}$ be the number of ways Masha can climb a porch with $n$ steps. Since Masha can reach th... | 89 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. Having found some polynomial of the sixth degree $x^{6}+a_{1} x^{5}+\ldots+a_{5} x+a_{6}$ with integer coefficients, one of the roots of which is the number $\sqrt{2}+\sqrt[3]{5}$, write in the answer the sum of its coefficients $a_{1}+a_{2}+\ldots+a_{6}$. | Answer: -47. Solution. $x=\sqrt{2}+\sqrt[3]{5} \Rightarrow (x-\sqrt{2})^{3}=5 \Rightarrow x^{3}-3 x^{2} \sqrt{2}+3 x \cdot 2-2 \sqrt{2}=5$ $\Rightarrow x^{3}+6 x-5=(3 x^{2}+2) \cdot \sqrt{2} \Rightarrow (x^{3}+6 x-5)^{2}=(3 x^{2}+2)^{2} \cdot 2$ $\Rightarrow P_{6}(x)=(x^{3}+6 x-5)^{2}-2(3 x^{2}+2)^{2}$. In this case, t... | -47 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Each cell of a $3 \times 3$ table is painted in one of three colors such that cells sharing a side have different colors, and not all three colors need to be used. How many such colorings exist? | Answer: 246.
Solution. The central cell can be painted in any of the three colors, let's call this color $a$. Each of the four cells that share a side with the central one can be painted in any of the two remaining colors. Let the cell located above the central one be painted in color $b$. The third color we will call... | 246 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five... | Answer: 7111765.
Solution. Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive... | 7111765 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4.1. We will call a natural number interesting if all its digits, except the first and last, are less than the arithmetic mean of the two adjacent digits. Find the largest interesting number. | Answer: 96433469.
Solution. Let $a_{n}$ be the n-th digit of the desired number. By the condition, $a_{n}9$. Similarly, there cannot be four negative differences. Therefore, the maximum number of such differences can be 7, and the desired number is an eight-digit number. To make it the largest, the first digit should ... | 96433469 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17. | Solution. The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\... | 17 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6.1. How many triangles with integer sides have a perimeter equal to 27? (Triangles that differ only in the order of the sides - for example, $7,10,10$ and $10,10,7$ - are considered the same triangle.) | Answer: 19.
Solution: Arrange the sides in ascending order: $a \leqslant b \leqslant c$. Then the smaller side $a$ does not exceed 9. If $a=1$ or $a=2$, there is one such triangle. If $a=3$ or $a=4$, there are two such triangles. If $a=5$ or $a=6$, there are three such triangles. If $a=7$, there are four such triangle... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. First-grader Petya was laying out a contour of an equilateral triangle with the chips he had, so that each of its sides, including the vertices, contained the same number of chips. Then, with the same chips, he managed to lay out the contour of a square in the same way. How many chips does Petya have, if each side o... | Answer: 24.
Solution. Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips being counted twice). From the problem statement, it follows that $y=x-2$. Therefore, we get the equation $3(x-... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. In how many different ways can integers $a, b, c \in [1,100]$ be chosen so that the points with coordinates $A(-1, a), B(0, b)$, and $C(1, c)$ form a right triangle? | Answer: 974
Solution: $A B^{2}=1+(b-a)^{2}, B C^{2}=1+(c-b)^{2}, A C^{2}=4+(c-a)^{2}$.
If triangle $A B C$ is a right triangle with hypotenuse $A C$, then by the Pythagorean theorem $A C^{2}=A B^{2}+B C^{2}, 1+(b-a)^{2}+1+(b-c)^{2}=4+(a-c)^{2}$, which simplifies to $(b-a)(b-c)=1$. Since both factors are integers, we ... | 974 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Nезнayka jumped from his house to Znayka's house. For the first three-quarters of the path, he jumped with jumps the length of two of his usual steps, and for the remaining quarter of the path, he jumped with jumps the length of three of his usual steps. It turned out that the number of jumps of two steps was 350 mo... | Answer variant l: 1200.
Answer variant 2: 1800.
Answer variant $3: 2040$.
Answer variant 4: 900.
Scoring evaluation: 15 баллов - correct solution and correct answer; 10 баллов - arithmetic error in otherwise correct solution | 1200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. What is the least number of odd numbers that can be chosen from the set of all odd numbers lying between 16 and 2016, so that no one of the chosen numbers is divisible by any other?
| Solution. We will select all odd numbers from 673 to 2015, their quantity is
$\frac{2015-671}{2}=672$. None of these numbers is divisible by another, because in division, the result should be an odd number, which is at least 3, but $673-3>2015$.
We will show that it is impossible to select more than 672 numbers satis... | 672 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.1. (4 points) Find the least common multiple of the numbers 28 and 72. | Answer: 504.
Solution. We have $28=2^{2} \cdot 7, 72=2^{3} \cdot 3^{2}$, therefore $\operatorname{LCM}(28,72)=2^{3} \cdot 3^{2} \cdot 7=504$. | 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1.4. Find the greatest common divisor of the numbers 144 and 120. | Answer: 24.
Solution. We have $144=2^{4} \cdot 3^{2}, 120=2^{3} \cdot 3 \cdot 5$, so $\text{GCD}(144,120)=2^{3} \cdot 3=24$. | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1. (16 points) A team consisting of juniors and masters from the "Vimpel" sports society went to a shooting tournament. The average number of points scored by the juniors turned out to be 22, by the masters - 47, and the average number of points for the entire team - 41. What is the proportion (in percent) of masters... | Answer: 76.
Solution. Let there be $x$ juniors and $y$ masters in the team. Then the total number of points scored by the team is $22x + 47y = 41(x + y)$, from which we find $19x = 6y$. Therefore, the proportion of masters is $\frac{y}{x+y} = \frac{19y}{19x + 19y} = \frac{19y}{25y} = 0.76$, i.e., $76\%$. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.1. (16 points) A square with a side of 36 cm was cut into three rectangles such that the areas of all three rectangles are equal and any two rectangles share a common boundary segment. What is the total length (in cm) of the cuts made? | Answer: 60.
Solution. One cut has a length of 36 cm and separates one of the rectangles (see fig.). The remaining 2 rectangles have a common side $x$, so they are both $18 \times x$ cm in size. Therefore,
 An eraser, 3 pens, and 2 markers cost 240 rubles, while 2 erasers, 4 markers, and 5 pens cost 440 rubles. What is the total cost (in rubles) of 3 erasers, 4 pens, and 6 markers? | Answer: 520.
Solution. From the condition, it follows that 3 erasers, 8 pens, and 6 markers cost $440+240=680$ rubles. Moreover, 2 erasers, 6 pens, and 4 markers will cost $2 \cdot 240=480$ rubles. Therefore, one pen costs $480-440=40$ rubles. Then, 3 erasers, 4 pens, and 6 markers cost $680-4 \cdot 40=520$ rubles. | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.1. (16 points) A squirrel jumps once a minute to a distance of 1 along the number line, starting from point 0, and in any direction. She jumped for 40 minutes, after which she fell asleep. How many points on the number line could the squirrel have fallen asleep in? | Answer: 41.
Solution. We will solve the problem in a general form. Let the squirrel have jumped for $N$ minutes. If $N$ is even $(N=2 n)$, then after $2 n$ steps, the squirrel can only reach a point with an even coordinate in the range from $-2 n$ to $2 n$, and it can reach any of these points (to a point with coordin... | 41 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then ad... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five... | Answer: 7111765.
Solution. Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive... | 7111765 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3.1. Find the largest integer solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}<1$. | Answer: -2001.
Solution. By factoring out $2^{x}$ and using the formula for the sum of the first terms of a geometric progression, we can transform the inequality to $2^{x} \cdot\left(2^{2001}-1\right)<1$. For $x \geqslant-2000$, this inequality is not satisfied. If, however, $x=-2001$, we get $1-2^{-2001}<1$, which i... | -2001 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
5.1. How many triangles with integer sides have a perimeter equal to 2017? (Triangles that differ only in the order of their sides, for example, 17, 1000, 1000 and 1000, 1000, 17, are counted as one triangle.) | Answer: 85008.
Solution. Let the sides of the triangle be $a, b, c$ ordered such that $a \leqslant b \leqslant c$. Then the smaller side $a$ does not exceed $2017 / 3=672 \frac{1}{3}$.
Consider the case of even $a$, i.e., $a=2 k$. Then for $k=1, \ldots, 252$, there will be $k$ triangles each, totaling $253 \cdot 252 ... | 85008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Three friends, weightlifters A, B, and C, came to a competition. They all competed in the same weight category, and one of them became the winner. If the weight lifted by weightlifter A is added to the weight lifted by weightlifter B, the total is 220 kg. If the weights lifted by weightlifters A and C are ad... | Answer: 135.
Solution. The sum of three weights is $(220+240+250) / 2=355$. Therefore, V lifted $355-220=135$, B lifted $355-240=115$, A lifted $355-250=105$. The winner is $\mathrm{B}=135$. | 135 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. How many solutions in integers does the equation
$$
\frac{1}{2022}=\frac{1}{x}+\frac{1}{y} ?
$$ | Answer: 53.
Solution. By eliminating the denominators, we obtain the equation
\[
(x-2022)(y-2022)=2022^{2}.
\]
Since \(2022^{2}=2^{2} \cdot 3^{2} \cdot 337^{2}\), the number \(x-2022\) can have the factors \(2^{0}, 2^{1}, 2^{2}\) - a total of three options. Similarly for the other factors. The number \(x-2022\) thus... | 53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. From the digits $a, b, c, d, e$, a five-digit number $\overline{a b c d e}$ is formed. For the two-digit numbers $\overline{a b}, \overline{b c}, \overline{c d}, \overline{d e}$, formed from the same digits, it is known that
$$
(\overline{a b}+\overline{b c})(\overline{b c}+\overline{c d})(\overline{c d}+\o... | Answer: 12345 or 21436.
Solution. Note that $157605=3 \cdot 5 \cdot 7 \cdot 19 \cdot 79$. To start, we need to understand which combinations of prime factors correspond to the sums of two-digit numbers. The sum of two-digit numbers cannot exceed $99+99=198$, and $79 \cdot 3$ is already greater than 200, so one of the ... | 12345 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Find the smallest natural number that has the following property: the remainder of its division by 20 is one less than the remainder of its division by 21, and the remainder of its division by 22 is 2. | Answer: 838.
Solution. The desired number is $20k + a = 21l + a + 1 = 22m + 2$, where $0 \leqslant a \leqslant 19$ and $l, k, m \geqslant 0$. From the first equation and the congruence modulo 20, we get that $l + 1 \equiv 0 \pmod{20}$. Since we are looking for the smallest number, let's try $l = 19$, if this $l$ does ... | 838 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Find the three last digits of the number $10^{2022}-9^{2022}$. | Answer: 119.
Solution. Since $A=10^{2022}-(10-1)^{2022}=10^{2022}-10^{2022}+2022 \cdot 10^{2021}-C_{2022}^{2} \cdot 10^{2022}+\ldots+$ $C_{2022}^{3} \cdot 10^{3}-C_{2022}^{2} \cdot 10^{2}+C_{2022}^{1} \cdot 10-1$, then $A(\bmod 1000) \equiv-C_{2022}^{2} \cdot 100+C_{2022}^{1} \cdot 10-1(\bmod 1000) \equiv$ $-\frac{202... | 119 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. There is a certain number of identical plastic bags that can be nested inside each other. If all the other bags end up inside one of the bags, we will call this situation a "bag of bags." Calculate the number of ways to form a "bag of bags" from 10 bags.
## Explanation. Denote a bag with parentheses.
If th... | Solution. If $\Pi_{n}$ denotes the number of ways for $n$ packages, then:
$$
\begin{gathered}
\Pi_{1}=1, \Pi_{2}=1, \Pi_{3}=2, \Pi_{4}=4, \Pi_{5}=9, \Pi_{6}=20, \Pi_{7}=48, \Pi_{8}=115, \Pi_{9}=286 \\
\Pi_{10}=719
\end{gathered}
$$
The problem is solved by enumerating the cases. For example, if we take $\Pi_{5}$:
$\... | 719 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits. | Answer: 27.
Solution: Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits ( $A+B$ ) does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, where $n-$
is some natural number that does not exceed 4. However, 1 and 2 do not wo... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. On a circle, 25 points are marked, painted either red or blue. Some of the points are connected by segments, with one end of each segment being blue and the other end red. It is known that there do not exist two red points that belong to the same number of segments. What is the maximum possible number of red points? | Answer: 13.
Solution: Let's take 13 red and 12 blue points. The first red point is not connected to any other, the second is connected to one blue, ..., the 13th is connected to 12 blue points. Obviously, there cannot be more red points, because if there are more than 13, the number of connection options is less than ... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Given the polynomial $P(x)=x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$, it is known that $P(2014)=1, P(2015)=2, P(2016)=3, P(2017)=4$, $P(2018)=5$. Find $P(2013)$. | Answer: $-5!=-120$. Solution. Consider the polynomial $P(x)-x+2013$, the numbers $2014, \ldots 2018$ are its roots. Therefore (since the leading coefficient is 1) it can be represented as $P(x)-x+2013=$ $(x-2014)(x-2015) \cdot \ldots \cdot(x-2018)$. Substituting $x=2013$, we get $P(2013)-2013+2013=-120$. | -120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A flea jumps along the number line, and the length of each jump cannot be less than $n$. It starts its movement from the origin and wants to visit all integer points belonging to the segment $[0,2013]$ (and only them!) exactly once. For what greatest value of $n$ will it be able to do this? | Answer: $n=1006$.
Solution: For $n=1006$, a path can be constructed as follows:
$$
0 \rightarrow 1007 \rightarrow 1 \rightarrow 1008 \rightarrow \ldots \rightarrow 1005 \rightarrow 2012 \rightarrow 1006 \rightarrow 2013
$$
We will prove that $n$ cannot be greater than 1006. Indeed, suppose $n \geqslant 1007$. Then t... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the number of 9-digit numbers in which each digit from 1 to 9 appears exactly once, the digits 1, 2, 3, 4, 5 are arranged in ascending order, and the digit 6 appears before the digit 1 (for example, 916238457). | Answer: 504
Solution: Note that after arranging the digits $7,8,9$, the remaining digits are uniquely determined. Therefore, the number of such numbers is $7 \cdot 8 \cdot 9=504$. | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.1. From the digits 1, 3, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers. | Answer: 1998 (all possible numbers: $135,153,315,351,513,531$ ). | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1.2. From the digits 1, 2, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers. | Answer: 1776 (all possible numbers: 125, 152, 215, 251, 512, 521). | 1776 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. The decreasing sequence $a, b, c$ is a geometric progression, and the sequence $577 a, \frac{2020 b}{7}, \frac{c}{7}$ is an arithmetic progression. Find the common ratio of the geometric progression. | Answer: 4039.
Solution. Let $b=a q, c=a q^{2}$. The properties of an arithmetic progression and the conditions of the problem lead to the equation $2 \cdot \frac{2020 a q}{7}=577 a+\frac{a q^{2}}{7} \Leftrightarrow q^{2}-4040 q+4039=0$, from which $q=1$ or $q=4039$. A decreasing geometric progression can only occur wh... | 4039 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.1. The numbers $p$ and $q$ are chosen such that the parabola $y=p x-x^{2}$ intersects the hyperbola $x y=q$ at three distinct points $A, B$, and $C$, and the sum of the squares of the sides of triangle $A B C$ is 324, while the centroid of the triangle is 2 units away from the origin. Find the product $p q$. | Answer: 42.
Solution. The system $y=p x-x^{2}, x y=q$ reduces to the cubic equation $x^{3}-p x^{2}+q=0$, which has, according to the problem, three distinct roots $x_{1}, x_{2}, x_{3}$, since the abscissas of any two different points on the curve $y=p x-x^{2}$ are distinct. By Vieta's theorem,
$$
x_{1}+x_{2}+x_{3}=p,... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. How many natural numbers $n \in [20182019 ; 20192018]$ are there for which the number $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right]$ is even? (Here, $[x]$ denotes the greatest integer not exceeding $x$.) | Answer: 4999.
Solution. Consider the sequences
$$
x_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad y_{n}=\left(\frac{1-\sqrt{5}}{2}\right)^{n}, \quad a_{n}=x_{n}+y_{n}
$$
Since $-1<\frac{1-\sqrt{5}}{2}<0$, we have $y_{n} \in(0 ; 1)$ for even $n$, and $y_{n} \in(-1 ; 0)$ for odd $n$. Then $\left[x_{n}\right]=\left[... | 4999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Pete was given a new electric jigsaw on his birthday, with a feature to count the length of the cuts made. To try out the gift, Pete took a square piece of plywood with a side of 50 cm and cut it into squares with sides of 10 cm and 20 cm. How many squares in total were obtained, if the electric jigsaw shows a total... | Answer: 16. Solution. The perimeter of the figures must be taken into account, in addition to the perimeter of the original square, from which we get that the total perimeter of the resulting squares is $280 \cdot 2 + 200 = 760$. Now, we can denote the number of squares through $x$ and $y$ respectively and solve the sy... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. It is known that $x+y+z=2016$ and $\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=\frac{7}{224}$. Find $\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x}$ ANSWER:60. | Solution: Add 1 to each fraction, we get $\frac{z}{x+y}+1+\frac{x}{y+z}+1+\frac{y}{z+x}+1=\frac{x+y+z}{x+y}+$ $\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}=2016 \cdot\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2016}{32}=63$. | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.1. (16 points) In a hut, several inhabitants of the island gathered, some from the Ah tribe, and the rest from the Ukh tribe. The inhabitants of the Ah tribe always tell the truth, while the inhabitants of the Ukh tribe always lie. One of the inhabitants said: "There are no more than 16 of us in the hut," and then ad... | Answer: 15.
Solution. A resident of tribe Ah cannot say "we are all from tribe Ukh," so the first one is from tribe Ukh. Therefore, there are no fewer than 17 people in the hut. So the second one spoke the truth, i.e., he is from tribe Ah. Therefore, there are no more than 17 people in the hut. Thus, there are 17 peop... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Photographs are archived in the order of their numbering in identical albums, with exactly 4 photographs per page. In this case, the 81st photograph in sequence landed on the 5th page of one of the albums, and the 171st photograph landed on the 3rd page of another. How many photographs can each album hold?
# | # Answer: 32.
Solution. Let $x, y$ be the album numbers in which the 81st and 171st photos are placed, respectively, and $n>4$ be the number of pages in the album. Then $4 n(x-1)+16<81 \leqslant 4 n(x-1)+20, 4 n(y-1)+8<171 \leqslant 4 n(y-1)+12$, i.e., $61 \leqslant 4 n(x-1)<65, 159 \leqslant 4 n(y-1)<163$. Therefore,... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. Andrei likes all numbers that are not divisible by 3, and Tanya likes all numbers that do not contain digits divisible by 3.
a) How many four-digit numbers are liked by both Andrei and Tanya?
b) Find the total sum of the digits of all such four-digit numbers. | Answer: a) 810; b) 14580.
Solution. a) The required numbers must be composed of the digits $1,2,4,5,7,8$, and according to the divisibility rule by 3, the sum of the digits in each number should not be divisible by three. The digits 1, 4, and 7 (let's call them the digits of set $A$) give a remainder of 1 when divided... | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. All natural numbers are divided into "good" and "bad" according to the following rules:
a) From any bad number, you can subtract some natural number not exceeding its half so that the resulting difference becomes "good".
b) From a "good" number, you cannot subtract no more than half of it so that it remains "good"... | Answer: 2047. Solution. Considering the first few natural numbers, we notice that good numbers have the form $2^{n}-1$ (while $2^{n}, \ldots, 2 n+1-2$ are bad).
We will prove this by mathematical induction. For $n=1$, the statement is given in the condition. Suppose the statement is proven for $n-1$. Consider a number... | 2047 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. In a convex quadrilateral $A B C D$, the areas of triangles $A B D$ and $B C D$ are equal, and the area of $A C D$ is half the area of $A B D$. Find the length of the segment $C M$, where $M$ is the midpoint of side $A B$, if it is known that $A D=12$. | Answer: $C M=18$. Solution. From the equality of the areas of $A B C$ and $C B D$, it follows that $O$ - the intersection point of the diagonals $A C$ and $B D$ bisects $A C$ (see figure). And from the fact that $S(A C D) = \frac{1}{2} S(A B D)$, it follows that $S(A O D) = \frac{1}{3} S(A O B)$, therefore, $O D = \fra... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the smallest natural number that is greater than the sum of its digits by 1755 (the year of the founding of Moscow University). | Answer: 1770.
Solution. From the condition, it follows that the desired number cannot consist of three or fewer digits. We will look for the smallest such number in the form $\overline{a b c d}$, where $a, b, c, d$ are digits, and $a \neq 0$. We will form the equation:
$$
\begin{gathered}
\overline{a b c d}=a+b+c+d+1... | 1770 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. There were 21 consecutive natural numbers written on the board. When one of the numbers was erased, the sum of the remaining numbers became 2017. Which number was erased? | Answer: 104.
Solution: Let the numbers on the board be N-10, N-9,..,N, ..., N+10. Their sum is $21 \mathrm{~N}$. When one of these numbers - x - was erased, the sum became 2017, $21 \mathrm{~N}-$ $x=2017$. Therefore, $x=21 N-2017$, since this is one of these numbers, we get $N-10 \leq 21 N-2017 \leq N+10$. Solving the... | 104 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. How many diagonals in a regular 32-sided polygon are not parallel to any of its sides | Answer: 240.
Solution: In a 32-sided polygon, there are $32*(32-3)/2=464$ diagonals in total. We can divide the sides into 16 pairs of parallel sides. It is not hard to notice that if we fix a pair, i.e., 4 vertices, the remaining vertices can be connected in pairs by diagonals parallel to this pair. There will be a t... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. For natural numbers $m$ and $n$, it is known that $3 n^{3}=5 m^{2}$. Find the smallest possible value of $m+n$. | Answer: 60.
Solution: Obviously, if $m, n$ contain prime factors other than 3 or 5, then they can be canceled out (and reduce $m+n$).
Let $n=3^{a} \cdot 5^{b}, m=3^{c} \cdot 5^{d}$. Then the condition implies that $3 a+1=2 c, 3 b=2 d+1$.
The smallest possible values are: $a=1, b=1, c=2, d=1$, from which $n=15, m=45$... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. For the function $y=f(x)$, it is known that it is defined and continuous on the entire number line, odd, and periodic with a period of 5, and that $f(-1)=f(2)=-1$. What is the minimum number of roots that the equation $f(x)=0$ can have on the interval [1755; 2017]? Answer: 210. | Solution. Since the function $f$ is odd and defined at zero, we get $f(0)=-f(0)$ $\Rightarrow f(0)=0$. Due to the 5-periodicity, we then have $f(5)=f(0)=0$. Using the oddness again: $f(1)=-f(-1)=1$, and due to the 5-periodicity $f(3)=f(-2)=1, f(4)=$ $f(-1)=-1$. Thus, at points $1, 2, 3$, and $4$, the values of the func... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Calculate $\sqrt{n}+\sqrt{n+524}$, given that this number is rational and that $n$ is a natural number. | Answer: 262.
Solution. Let the desired number be $a$. We have $\sqrt{n+524}=a-\sqrt{n}, n+524=a^{2}-2 a \sqrt{n}+n$. By the condition, $a$ is rational, so $\sqrt{n}$ is also rational. Therefore, $n=k^{2}, k \in \mathbb{N}$. Then the number $\sqrt{n+524}$ is also rational, so $n+524=m^{2}, m \in \mathbb{N}$. Thus, $m^{... | 262 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. We consider all possible sets consisting of 2017 different natural numbers, where in each set no number can be represented as the sum of two other numbers in the set. What is the smallest value that the largest number in such a set can take?
Answer: 4032. | Solution. Let for brevity $n=2017$. Consider the following set of $n$ numbers: $n-1, n, n+1, \ldots$, $2 n-3, 2 n-2$. Since $(n-1) + n = 2 n - 1 > 2 n - 2$, no number in this set can equal the sum of two others, meaning the given set satisfies the problem's condition.
Now, let there be an arbitrary set of $n$ natural ... | 4032 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In a full container, there are 150 watermelons and melons for a total of 24 thousand rubles, with all the watermelons together costing as much as all the melons. How much does one watermelon cost, given that the container can hold 120 melons (without watermelons) and 160 watermelons (without melons)? | Answer: 100 rubles. Solution. Let there be $x$ watermelons, then there will be $150-x$ melons. If a container can hold 120 melons, then one melon occupies $\frac{1}{120}$ of the container. Similarly, one watermelon occupies $\frac{1}{160}$ of the container. Therefore, if the container is fully loaded with 150 watermelo... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A $3 \times 3$ table needs to be filled with the numbers 2014, 2015, and 2016, such that the sum of the numbers in each row is the same. In how many different ways can this be done? | Answer: 831. Solution. Subtract 2015 from all numbers - the sums will remain the same, and the numbers in the table will take on values of 0 and $\pm 1$. Consider the 27 possible combinations of numbers in one row. Possible values of the sum of these numbers: $\pm 3$ - 1 combination each, $\pm 2$ - 3 combinations each,... | 831 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. At Andrei's birthday, Yana, who arrived last, gave him a ball, and Eduard, who arrived second to last, gave him a calculator. Testing the calculator, Andrei noticed that the product of the total number of his gifts and the number of gifts he had before Eduard arrived is exactly 16 more than the product of his age an... | Answer: 18. Solution. If $n$ is the number of gifts, and $a$ is Andrei's age, then $n(n-2)=a(n-1)$. From this, $a=\frac{n^{2}-2 n-16}{n-1}=n-1-\frac{17}{n-1}$. Therefore, $n-1=17, a=16$. | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In an equilateral triangle $A B C$, points $A_{1}$ and $A_{2}$ are chosen on side $B C$ such that $B A_{1}=A_{1} A_{2}=A_{2} C$. On side $A C$, a point $B_{1}$ is chosen such that $A B_{1}: B_{1} C=1: 2$. Find the sum of the angles $\angle A A_{1} B_{1}+\angle A A_{2} B_{1}$. | Answer: $30^{\circ}$. Solution. Since $A_{1} B_{1} \| A B$, then $\angle B A A_{1}=\angle A A_{1} B_{1}$. From symmetry, $\angle B A A_{1}=\angle C A A_{2}$. It remains to note that $\angle C B_{1} A_{2}=$ $\angle B_{1} A A_{2}+\angle A A_{2} B_{1}$ as an exterior angle in $\triangle A A_{2} B_{1}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Find the greatest possible value of $\gcd(x+2015 y, y+2015 x)$, given that $x$ and $y$ are coprime numbers. | Answer: $2015^{2}-1=4060224$. Solution. Note that the common divisor will also divide $(x+2015 y)-2015(y+2015 x)=\left(1-2015^{2}\right) x$. Similarly, it divides $\left(1-2015^{2}\right) y$, and since $(x, y)=1$, it divides $\left(1-2015^{2}\right)$. On the other hand, if we take $x=1, y=2015^{2}-2016$, then we get $\... | 4060224 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. A flea jumps along the number line, and the length of each jump cannot be less than $n$. It starts its movement from the origin and wants to visit all integer points belonging to the segment $[0,2013]$ (and only them!) exactly once. For what greatest value of $n$ will it be able to do this? | Answer: $n=1006$.
Solution: For $n=1006$, a path can be constructed as follows:
$$
0 \rightarrow 1007 \rightarrow 1 \rightarrow 1008 \rightarrow \ldots \rightarrow 1005 \rightarrow 2012 \rightarrow 1006 \rightarrow 2013
$$
We will prove that $n$ cannot be greater than 1006. Indeed, suppose $n \geqslant 1007$. Then t... | 1006 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In how many different ways can a chess king move from square $e 1$ to square $h 5$, if it is allowed to move only one square to the right, up, or diagonally to the right-up? | Answer: 129 ways.
Solution: Sequentially (starting from e1) find the number of ways to reach each cell. Each number (except 1) is obtained by summing the neighbors below, to the left, and diagonally left-below.
 From the digits 1, 3, and 5, different three-digit numbers are formed, each of which has all distinct digits. Find the sum of all such three-digit numbers. | Answer: 1998.
Solution. All possible numbers: $135,153,315,351,513,531$. | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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