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2. Six numbers are given, the arithmetic mean of which is equal to some number A. Petrov calculated the arithmetic mean of the first four numbers - it turned out to be A+10. Vasechkin calculated the arithmetic mean of the last four numbers - it is A - 7. In which direction and by how much does the arithmetic mean of th... | Answer: 3 less.
Solution - similar to option v3a. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,581 |
5. In a regular $1000-$gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length? | Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e.,... | 2000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,583 |
5. Vasya is coming up with a 4-digit password for a code lock. He doesn't like the digit 2, so he doesn't use it. Additionally, he doesn't like when two identical digits are next to each other. He also wants the first digit to match the last digit. How many options need to be tried to guarantee guessing Vasya's passwor... | Answer: 504
Solution: The password must have the form ABCA, where A, B, C are different digits (not equal to 2). They can be chosen in $9 * 8 * 7=504$ ways.
Comment for graders: Half a point can be given to those who consider that the first digit cannot be zero. | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,584 |
1. If you open the cold water tap, the bathtub will fill up in 10 minutes, if you open the hot water tap, it will take 15 minutes. If you pull the plug, the bathtub will completely drain in 12 minutes. How long will it take to fill the bathtub if you open both taps and pull the plug?
# | # Answer: 12 min.
Solution: Let's take one bathtub as a unit of volume. The cold water tap flows at a rate of $1 / 10$ of a bathtub per minute, and the hot water tap flows at a rate of $1 / 15$ of a bathtub per minute. Water drains from the drain hole at a rate of $1 / 12$ of a bathtub per minute. Therefore, the total... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,585 |
2. How many natural numbers from 1 to 2017 have exactly three distinct natural divisors? | Answer: 14.
Solution: Only squares of prime numbers have exactly three divisors. Note that $47^{2}>2017$, so it is sufficient to consider the squares of prime numbers from 2 to 43. There are 14 of them. | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,586 |
3. Vovochka approached a slot machine, on the screen of which the number 0 was glowing. The game rules stated: «The screen shows the number of points. If you throw a 1-ruble coin, the number of points will increase by 1. If you throw a 2-ruble coin, the number of points will double. If you score 50 points, the machine ... | Solution: Let's try to solve it from the end - how to get the number 1 from 50 with the least amount of rubles, if you can only divide by 2 and subtract 1. We get: 50>25->24->12->6->3->2->1. That is, it will require 4 two-ruble and 3 one-ruble coins. It is obvious that if you use 3 two-ruble coins and fewer than 5 one-... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,587 |
4. Petya is coming up with a password for his smartphone. The password consists of 4 decimal digits. Petya wants the password not to contain the digit 7, and at the same time, the password should have at least two (or more) identical digits. In how many ways can Petya do this? | # Answer 3537.
Solution: The total number of passwords not containing the digit 7 will be $9^{4}=6561$. Among these, 9x8x7x6=3024 consist of different digits. Therefore, the number of passwords containing identical digits is 6561-3024=3537 passwords. | 3537 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,588 |
5. In the computer center, there are 200 computers, some of which (in pairs) are connected by cables, a total of 345 cables are used. We will call a "cluster" a set of computers such that a signal from any computer in this set can reach all the others via the cables. Initially, all computers formed one cluster. But one... | Answer: 153.
Solution: Let's try to imagine the problem this way: an evil hacker has cut all the wires. What is the minimum number of wires the admin needs to restore to end up with 8 clusters? Obviously, by adding a wire, the admin can reduce the number of clusters by one. This means that from 200 clusters, 8 can be ... | 153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,589 |
3. At the international StarCraft championship, 100 participants gathered. The game is played in a knockout format, meaning in each match, two players compete, the loser is eliminated from the tournament, and the winner remains. Find the maximum possible number of participants who won exactly two games? | Answer: 49
Solution: Each participant (except the winner) lost one game to someone. There are 99 such participants, so no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We will show that there could be 49. Let's say №3 won against №1 and №2, №5 - against №3 and №4, ... №99 - aga... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,590 |
4. The robot moves along straight segments, making turns of 90 degrees to the right or left every minute (neglect the time for turning). The robot travels 10 meters per minute. What is the minimum distance from the starting position the robot can be after 9 minutes, if it did not turn during the first minute? | Answer: $10 \mathrm{~m}$
Solution: We can consider a coordinate grid with nodes spaced 10m apart. Clearly, the robot moves along the nodes of this grid. It cannot return to the initial position in 9 minutes. However, it can be at a distance of 10 m - it is easy to construct an example. | 10\mathrm{~} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,591 |
5. On graph paper, a right-angled triangle with legs equal to 7 cells was drawn (see fig.). Then all the grid lines inside the triangle were outlined. What is the maximum number of triangles that can be found in this drawing? | Answer: 28 triangles
Solution: One of the sides of the triangle must go at an
angle, i.e., lie on the segment BC. If we fix some diagonal segment, the remaining vertex is uniquely determine... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,592 |
6. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of distinct digits $\Pi, B$, and $\Gamma$, for which the equality $\overline{\Pi B \Gamma}=(\Pi+B+\Gamma) \times(\Pi+B+\Gamma+1)$ holds. | Answer: 156.
Solution: Note that П+В $\overline{\Pi В \Gamma}$ and (П + В + Г) should give the same remainder when divided by 9. This is only possible when $П+В+\Gamma$ is a multiple of 3. Note that $П+В+\Gamma=9$ - does not work, because (П + В + $\Gamma) \times(П+B+\Gamma+1)=90$-two-digit. By trying $12,15,18,21,24$... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,593 |
2. How many three-digit natural numbers have an even number of distinct natural divisors? | Answer: 878.
Solution: Note that only perfect squares have an odd number of divisors (for non-squares, divisors can be paired with their complements). There are 900 three-digit numbers in total. Among them, the perfect squares are $10^{2}, 11^{2}, \ldots, 31^{2}=961$ ( $32^{2}=1024-$ is a four-digit number). There are... | 878 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,594 |
3. At the international table tennis championship, 200 participants gathered. The game is played in a knockout format, i.e., in each match, two players participate, the loser is eliminated from the championship, and the winner remains. Find the maximum possible number of participants who won at least three matches. | Answer: 66.
Solution: Each participant (except the winner) lost one game to someone. There are 199 such participants, so no more than 66 participants could have won 3 games (someone must lose 3 games to them).
We will show that there could be 66 such participants. Let №4 win against №1,2,3; №7 - against №4,5,6,... №1... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,595 |
4. The turtle crawled out of its house and moved in a straight line at a constant speed of 5 m/hour. After an hour, it turned $90^{\circ}$ (right or left) and continued moving, then crawled for another hour, then turned $90^{\circ}$ (right or left) again... and so on. It crawled for 11 hours, turning $90^{\circ}$ at th... | Answer 5 m.
Solution: We can consider a coordinate grid with nodes spaced 5 m apart. It is clear that the turtle crawls along
the nodes of this grid. The turtle cannot return to the initial... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,596 |
1. Can you use the four arithmetic operations (and also parentheses) to write the number 2016 using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in sequence? | Answer: $1 \cdot 2 \cdot 3 \cdot(4+5) \cdot 6 \cdot 7 \cdot 8: 9=2016$. | 2016 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,598 |
2. Anya did not tell Misha how old she is, but she informed him that on each of her birthdays, her mother puts as many coins into the piggy bank as Anya is turning years old. Misha estimated that there are no fewer than 110 but no more than 130 coins in the piggy bank. How old is Anya? | Answer: 15. Solution. Either use the formula for the sum of an arithmetic progression: $110 \leq \frac{1+n}{2} n \leq 130$, or simply calculate the sum "brute force". | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,599 |
4. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{2016}$, and the other leg and the hypotenuse are expressed as natural numbers
ANSWER: 12. | Solution. According to the condition $c^{2}-b^{2}=a^{2}=2016$, that is, $(c-b)(c+b)=2^{5} \cdot 3^{2} \cdot 7$. The system $\left\{\begin{array}{l}c-b=n, \\ c+b=k\end{array}\right.$ (here $n-$ is one of the divisors of the number 2016, and $k=\frac{2016}{n}$) has natural solutions $c=\frac{n+k}{2}, b=\frac{k-n}{2}$, if... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,601 |
5. Find all 4-digit numbers that are 7182 less than the number written with the same digits in reverse order.
ANSWER: 1909 | Solution. Let's write the desired number as $\overline{a b c d}$
Then $\overline{a b c d}=\overline{d c b a}-7182$, from which $111(d-a)+10(c-b)=798$.
Obviously, $(d-a)$ can only be 7 or 8.
In the case $d-a=7$, we get $10(c-b)=21-$ which does not fit.
In the case $d-a=8$, we get $10(c-b)=-90$, hence $b-c=9$, from w... | 1909 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,602 |
7. The number $n+2015$ is divisible by 2016, and the number $n+2016$ is divisible by 2015. Find the smallest natural $n$ for which this is possible.
ANS: 4058209. | Solution. According to the condition $\left\{\begin{array}{l}n+2015=2016 m, \\ n+2016=2015 k .\end{array}\right.$ From this, $2016 m-2015 k=-1$. The solution of this equation in integers: $m=-1+2015 p, k=-1+2016 p$. Therefore, $n+2015=2016(-1+2015 p)=-2016+2016 \cdot 2015 p$, which means $n=-2015-2016+2016 \cdot 2015 p... | 4058209 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,604 |
1. Insert arithmetic operation signs and parentheses into the expression consisting of three numbers, $\frac{1}{8} \ldots \frac{1}{9} \cdots \frac{1}{28^{\prime}}$ so that the result of the calculations is equal to $\frac{1}{2016}$.
ANSWER: $\frac{1}{8} \times \frac{1}{9} \times \frac{1}{28}$ or $\left(\frac{1}{8}-\fr... | Solution: factorize 2016 into prime factors and notice that 2016=8x9x28. | \frac{1}{8}\times\frac{1}{9}\times\frac{1}{28} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,605 |
2. Kolya is twice as old as Olya was when Kolya was as old as Olya is now. And when Olya is as old as Kolya is now, their combined age will be 36 years. How old is Kolya now?
ANS: 16 years. | Solution: Let $x$ be Kolya's current age, $y$ be Olya's age. We can set up the system $\mathrm{x}=2(y-(x-y)) ; x+(x-y)+y+(x-y)=36$. Solving it: $x=16, y=12$. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,606 |
3. We will call the oscillation of a function the difference between its maximum and minimum values. What can be the maximum oscillation of the function $f(x) \times g(x)$, if it is known that the interval $[-8,4]$ is the range of the function $f(x)$, and the interval $[-2,6]$ is the range of the function $g(x)$. ANSWE... | Solution: The maximum value of $f(x) \times g(x)$ is 24=4x6, minimum $-48=(-8)$ x 6. | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,607 |
4. How many different right-angled triangles exist, one of the legs of which is equal to $\sqrt{1001}$, and the other leg and the hypotenuse are expressed as natural numbers
ANSWER: 4. | Solution: Let's write down the Pythagorean theorem: $a^{2}+1001=b^{2}$. From this, we get $(b-a)(b+a)=1001=7 \times 11 \times 13$. We can represent 1001 as the product of two factors $1 \times 1001=7 \times 143=11 \times 91=13 \times 77$ - the first factor must be smaller - there are 4 options in total. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,608 |
5. Find all 4-digit numbers that are 8802 more than the number written with the same digits in reverse order.
ANSWER: 1099 | Solution: Note that $x+8802<10000$, therefore, $x<1198$, so the first digit is 1, and the second digit is 0 or 1. The sum of x+8802 ends in 1, so the last digit is 9. We get a number of the form 10a9 or 11a9, we get an equation of the form $10 \mathrm{a} 9+8802=9 \mathrm{a} 01$ or $11 \mathrm{a} 9+8802=9 \mathrm{a} 11$... | 1099 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,609 |
6. Given a right triangle $K L M$ with legs $L M=60$ and $K M=80$. On the sides $K L, L M$, and $M K$, points $M_{1}, K_{1}, L_{1}$ are chosen, respectively, such that $K M_{1}=L K_{1}=\mathrm{ML}_{1}=2$. Find the area of triangle $K_{1} L_{1} M_{1}$.
ANSWER: 2216.8 | Solution: Let $S(K L M)=2400=S$. Then $S\left(K L_{1} M_{1}\right)=2 / 100 \times 78 / 80 \times S$; $S\left(K_{1} L M_{1}\right)=98 / 100 \times 2 / 60 \times S ; S\left(K_{1} L_{1} M\right)=58 / 60 \times 2 / 80 \times S$. We get $S\left(K_{1} L_{1} M_{1}\right)=S(1-$ $468 / 24000-580 / 24000-784 / 24000)=22168 / 240... | 2216.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,610 |
7. Find all pairs of integers ( $x, y$ ) for which the equation $x^{2}+y^{2}=x+y+2$ holds.
ANSWER: $(-1,0) ;(-1,1) ;(0,-1) ;(0,2) ;(1,-1),(1,2),(2,0) ;(2,1)$ | Solution: The function $2-\mathrm{x}$ takes values from -0.25 to infinity, $-\mathrm{y} 2+\mathrm{y}+2$ from minus infinity to 2.25. Common values (integers) are $0,1,2$, further by trial.
Lomonosov Moscow State University
## School Students' Olympiad "Conquer Sparrow Hills" in Mathematics
Final stage tasks for the ... | (-1,0);(-1,1);(0,-1);(0,2);(1,-1),(1,2),(2,0);(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,611 |
1. Know-it-all told Don't-know-it that to convert kilolunes (a unit of mass used by the little people on the Moon) to kilograms, one needs to divide the mass in kilolunes by 4 and then decrease the obtained number by $4 \%$. Don't-know-it decided that to convert from kilograms to kilolunes, one should multiply the mass... | Solution: One kilolun constitutes $0.25 * 0.96=0.24$ kg. Therefore, in one kilogram there are 25/6 kiloluns. If Nезнайка converts 1 kg, he will get $4 * 1.04=4.16$, which is $99,84 \%$ of $25 / 6$. | 0.16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,612 |
2. Find the largest natural number that cannot be represented as the sum of two composite numbers.
OTBET: 11 | Solution: Even numbers greater than 8 can be represented as the sum of two even numbers greater than 2. And odd numbers greater than 12 can be represented as the sum of 9 and an even composite number. By direct verification, we find that 11 cannot be represented in this way. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,613 |
3. Solve the equation $\frac{\sqrt{(-x)^{2}}+(\sqrt{-x})^{2}}{x^{2}+(-x)^{2}}=\frac{1}{2016}$
ANSWER: $x=-2016$. | Solution: Note that $\mathrm{x}<0$ from the domain, we can represent the equation as $\frac{-2 x}{2 x^{2}}=\frac{1}{2016}$. | -2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,614 |
4. Let $f(x)=x^{2}+p x+q$ where $p, q$ are some coefficients. By how much can the maximum value of the function $g(x)=|f(x)|$ differ from the minimum value of this function on the interval $[2 ; 6]$?
ANSWER: by 2. | Solution: For $f(x)=x^{2}+p x+q$ the difference between the maximum and the minimum value is at least 4 (this can be shown graphically). By choosing $q$,
we find that the maximum value of the modulus of the minimum differs by no more than 2. Example: $f(x)=(x-4)^{2}-2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,615 |
5. How many five-digit numbers of the form $\overline{a b 16 c}$ are divisible by 16? $(a, b, c-$ arbitrary digits, not necessarily different).
ANSWER: 90. | Solution: Note that the first digit does not affect divisibility, hence, a=1,..,9. On the other hand, divisibility by 8 implies that c=0 or 8. If c=0, then $b$ must be even, and if $c=8$ - odd. In both cases, we get 5 options, from which the total number is $9 *(5+5)=90$. | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,616 |
7. Solve the equation $x^{6}=y^{3}+217$ in integers.
ANSWER: $(-1,-6)(1,-6)(-3,8),(3,8)$. | Solution: Note that $y^{3}+217$ >= 0, hence, $y >= -6$.
We check that $Y=-1,-2,-3,-4,-5$ do not provide solutions, and for $y=-6$ we get $x=+/-1$.
Note that $\mathrm{y}^{3}+217=\mathrm{x}^{6}>=(\mathrm{y}+1)^{3}$, from which $\mathrm{y}^{2}+\mathrm{y}$ in mathematics
Final stage tasks for the 2015/2016 academic year... | (-1,-6),(1,-6),(-3,8),(3,8) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,618 |
2. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in tota... | Solution. If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105 p-2$. Since by the condition $150<x \leq 300$, then $p=2$. Therefore, $x=208$. | 208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,619 |
5. Find all natural numbers that are 36 times the sum of their digits. ANSWER: $324 ; 648$. | Solution. Let $S(x)$ denote the sum of the digits of the number $x$. Then the equation $x=36 \cdot S(x)$ has no solutions for $n \geq 5$, since $x \geq 10^{n-1}$, while $36 \cdot S(x) \leq 36 \cdot 9 \cdot n=324 n36(a+b+c+d) \Leftrightarrow 964 a+64 b>26 \cdot 9+35 \cdot 9 \geq 26 c+35 d$.
For $n=3: a \cdot 100+b \cdo... | 324;648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,620 |
7. Solve the equation in natural numbers $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}$
ANSWER: $n=1$. | Solution: $2 n=3-\frac{2}{n}+\frac{1}{n^{5}} \leq 3$, only $\mathrm{n}=1$ fits
Lomonosov Moscow State University
## School Olympiad "Conquer Sparrow Hills" in Mathematics
Final Stage Tasks for the 2015/2016 Academic Year for 9th Grade | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,621 |
1. Find the smallest natural number $N$ such that $N+2$ is divisible (without remainder) by 2, $N+3$ by 3, ..., $N+10$ by 10.
ANSWER: 2520. | Solution: Note that $N$ must be divisible by $2,3,4, \ldots, 10$, therefore, N= LCM $(2,3,4, . ., 10)=2^{3} \times 3^{2} \times 5 \times 7=2520$. | 2520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,622 |
6. A circle with diameter $A B$ intersects segments $A C$ and $B C$ at points $M$ and $N$, respectively, and the length of segment $M N$ is equal to the radius of the circle. Find the area of quadrilateral $A B N M$, given that $A C=12$ and $B C=8$.
 for which the equation $x^{2}+x y=y+92$ holds.
OTBET: $(2,88) ;(8,4)$. | Solution: Transform $x^{2}-1+x y-y=91$. Factorize: $(x-1)(x+y+1)=7 \times 13$. Both factors are positive, and the first factor must be smaller, so the possible cases are $x-1=1, x+y+1=91$ or $x-1=7$, $x+y+1=13$. We get the solutions $x=2, y=88$ or $x=8, y=4$.
Lomonosov Moscow State University
## School Olympiad "Conq... | (2,88);(8,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,626 |
1. Captain Jack Sparrow found a cave with a pirate treasure. In it, there are 6 chests, and the treasure is only in one of them, while the other chests are home to venomous snakes, ready to attack anyone who disturbs their peace.
On the first chest, it says “The treasure is in the third chest.”
On the second, “The tr... | Solution: Let's create a $6 \times 6$ table. In the і-th row and ј-th column, we will place a cross if the і-th statement is true when the treasure is in the ј-th chest:
| | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | | | X | | | |
| 2 | X | X | | | | |
| 3 | X | X... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,627 |
3. $A B C D E F$ - a regular hexagon, point O - its center. How many different isosceles triangles with vertices at the specified seven points can be constructed? Triangles that differ only in the order of vertices are considered as
, place the numbers $2016, 2017, \ldots$, 2024, in such a way that for any three vertices forming an equilateral triangle, one of the numbers is equal to the arithmetic mean of the other two. ANSWER: place the numbers in sequence (other options are possible).
.
,(6,-12),(24,96)$.
Let Vasya perform operations with the number $x$, and Petya with the number $y$. Then $\left\{\begin{array}{l}2 x^{3}=3 y^{2}, \\ |x-y| \leqslant 100 .\end{array}\right.$ From the equation of the system, it follows that $x=3 a, y=2 b, a, b \in \mathbb{Z}$. Substituting, we get $54 ... | (6,12),(6,-12),(24,96) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,639 |
5. Solve the system
$$
\left\{\begin{array}{c}
4 \cos ^{2} x+\cos ^{2} 5 y=4 \cos x \cdot \cos ^{6} 5 y \\
\lg (x-y)^{2}<2 \lg (2 \pi)-\lg 5-\lg 45
\end{array}\right.
$$ | 5. Answer: $\left(-\frac{1}{2}, 3\right),\left(3,-\frac{1}{2}\right)$.
If $x y>0$, then $\frac{|x y|}{x y}=1$ and the left side of the inequality is not less than 1, while the right side of the inequality is negative. If $x y<0$, then $\frac{|x y|}{x y}=-1$ and the equation is equivalent to the system $\left\{\begin{a... | (-\frac{1}{2},3),(3,-\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,641 |
16. Given 2024 sets, each consisting of 44 elements. The union of any two of these sets contains 87 elements. How many elements does the union of all 2024 sets contain? | Answer: 87033.
II. Find the number of natural numbers $n$, not exceeding 500, for which the equation $x^{[x]}=n$ has a solution. Here $[x]$ is the greatest integer not exceeding $x$.
Solution. If $[x]=0$, then the solution is: $0 \leqslant x < 1$, and $n=1$.
If $[x]=1$, then the solution is: $1 \leqslant x < 2$, and... | 288 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,643 |
30. Find the number of natural numbers $n$, not exceeding 3134, for which the equation $x^{[x]}=n$ has a solution. Here $[x]$ is the greatest integer not exceeding $x$. | Answer: 422.
III. Solve the inequality
$$
\frac{4^{-|x-2|}}{\sqrt{x^{2}-x-2}+2} \leqslant \frac{2^{1-|x|}}{\sqrt{x^{2}+6 x}+4}.
$$
In the answer, write the smallest root by absolute value, rounding it to two decimal places if necessary. If there are no solutions, then write the number 0.
Solution. Rewrite the inequ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,644 |
25. Solve the inequality
$$
\frac{4^{-|x-6|}}{\sqrt{x^{2}-6 x}+3} \leqslant \frac{2^{1-|x-3|}}{\sqrt{x^{2}+6 x-27}+6} .
$$
In the answer, write the sum of the lengths of the intervals of the solution that belong to the set $[-2020,2018]$. If the inequality has no solutions, write -1. | Answer: 4020.
IV. In triangle $ABC$, the height $BH$ is drawn, point $O$ is the center of the circumscribed circle around it, and the length of its radius is $R$. Find the largest of the angles $\angle BAC, \angle ACB$, expressed in radians, if it is known that $R = (6/5) \cdot BH = 4 \cdot OH$. Round the found value ... | 0.81 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,645 |
25. In triangle $ABC$, the altitude $BH$ is drawn, point $O$ is the center of the circumscribed circle around it, and the length of its radius is $R$. Find the measure in radians of the largest of the angles $ACB$ and $BAC$, given that $R=(5 / 8) \cdot B H=(5 / 6) \cdot O H$. If necessary, round the found value to two ... | Answer: 1.82.
V. For their summer vacation, the residents of Flower City, Znayka and Neznayka, stayed in a large 15-story hotel by the sea. Znayka noticed that the sum of all room numbers from the first to his own, inclusive, is twice the sum of all room numbers from the first to the one where Neznayka stayed, inclusi... | 1.82 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,647 |
4. Between the pairs of solutions $(x, y)$ and $f(x, y)$ in this chain, there are no other solutions to the equation.
Suppose such a solution $(\alpha, \beta)$ does exist, then $(x, y)<(\alpha, \beta)<f(x, y)$. Apply the mapping $g$, which is the inverse of $f: g(x, y)=(3 x-2 y,-4 x+3 y)$, to all parts of this inequal... | Answer: 696.
V-2. The summer vacation residents of Flower City, Know-it-all and Don't-know-it, spent in a large 15-story hotel by the sea. Know-it-all noticed that the sum of all room numbers from the first to his own, inclusive, is twice the sum of all room numbers from the first to the one where Don't-know-it stayed... | 539 | Number Theory | proof | Yes | Yes | olympiads | false | 5,648 |
1. Place the numbers $1,2,3,4,5,6,7,8$ and 9 in the nine cells of the figure shown in the diagram, so that the sum of the numbers in each column, starting from the second, is 1 more than in the previous one. It is sufficient to find at least one such arrangement. In your answer, indicate the number in the first column.... | Answer: 7.
Solution: For now, we will not pay attention to the order of numbers in one column.
The sum of the given numbers is 45. Let $x$ be the number in the bottom-left cell. Then $5x + 10 = 45$, from which $x = 7$. Therefore, the sum of the numbers in the second column is $8 = 5 + 3 = 6 + 2$. If the second column... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,650 |
1. A certain three-digit number was added to the number written with the same digits but in reverse order, and the result was 1777. What numbers were added
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Answer: 839 and 938.
Solution: Let the number be $\overline{a b c}=100 a+10 b+c$, then $\overline{a b c}+\overline{c b a}=$ $101 a+20 b+101 c=1777$. The sum $a+c$ must end in 7, i.e., $a+c=7$ or $a+c=17$. The first case is impossible, since then $101 a+20 b+101 c \leqslant 7 \cdot 101+9 \cdot 20=777+180<1777$. Therefo... | 839938 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,654 |
3. Points $A, B, C$ are located sequentially, with the distance $A B$ being 3 km and the distance $B C$ being 4 km. A cyclist departed from point $A$ and headed to point $C$. At the same time, a pedestrian left point $B$ and headed to point $A$. It is known that the pedestrian and the cyclist arrived at points $A$ and ... | Answer: 2.1 km.
Solution: The speeds of the cyclist and the pedestrian are in the ratio $7: 3$, the point of their meeting divides $A B$ in the same ratio. | 2.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,656 |
4. A certain 4-digit number is a perfect square. If you remove the first digit from the left, it becomes a perfect cube, and if you remove the first 2 digits, it becomes a fourth power of an integer. Find this number. | Answer: 9216.
Solution: Only 16 and 81 are two-digit fourth powers. But 81 does not work, since no three-digit cube ends in $81\left(5^{3}=125,7^{3}=343,9^{3}=729\right)$. But 16 is the ending of $6^{3}=216$. Next, we look for a perfect square that ends in 216. | 9216 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,657 |
5. How many natural numbers from 1 to 2015 inclusive have a sum of digits that is a multiple of 5? | Answer: 402.
Solution: Note that among ten numbers of the form $\overline{a 0}, \ldots, \overline{a 9}$, exactly two numbers have a sum of digits that is a multiple of five. Thus, among the numbers from 10 to 2009, there are exactly 200 such tens, and therefore, 400 such numbers. Considering also the numbers 5 and 201... | 402 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,658 |
7. The numbers $1,2, \ldots, 2016$ are paired in such a way that the product of the numbers in each pair does not exceed some natural number $N$. What is the smallest $N$ for which this is possible? | Answer: $1017072=1008 \times 1009$.
Solution: If we consider the numbers $1008,1009, \ldots, 2016$, then some two must fall into one pair. Therefore, $N$ cannot be less than $1008 \cdot 1009$. We will show that such a partition is possible when $N=1008 \cdot 1009$. We can partition into pairs: $(1,2016),(2,2016), \ldo... | 1017072=1008\times1009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,660 |
4. On the base plane of a cone with height $h$ and base radius $R$, a point (outside the cone) is given, which is at a distance of $aR$ from the circumference of the base. Find the angle between the tangent planes to the cone passing through the given point.
First method:
Let $L$ be the given point, $B$ and $C$ be th... | Answer: $2 \operatorname{arctg}(\sqrt{5} / 2)$.
Answer to the variant: $5-2: 2 \operatorname{arctg}(\sqrt{2} / \sqrt{3})$, (here $h=2 R, a=1)$). | 2\operatorname{arctg}(\sqrt{5}/2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,665 |
5. Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
... | Answer: 27.
Answer to option 17-2: 108.
Answer to option $17-3: 27$.
Answer to option $17-4: 108$.
[^0]: ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,668 |
1. Let $a=\operatorname{tg} x, b=\operatorname{tg} \frac{y}{2}$. Then $a+b=\frac{4}{\sqrt{3}}$ and $\frac{1}{a}+\frac{1}{b}=\frac{4}{\sqrt{3}}$. From the second equation, it follows that $a b=1$. By Vieta's theorem, $a$ and $b$ satisfy the equation $t^{2}-\frac{4}{\sqrt{3}} t+1=0$, whose roots are $\sqrt{3}$ and $\frac... | Answer: $x=\frac{\pi}{3}+\pi n, y=\frac{\pi}{3}+2 \pi k$ or $x=\frac{\pi}{6}+\pi n, y=\frac{2 \pi}{3}+2 \pi k$. Answer to option 1-2: $x=\frac{\pi}{3}+2 \pi k, y=\frac{\pi}{3}+\pi n$ or $x=\frac{2 \pi}{3}+2 \pi k, y=\frac{\pi}{6}+\pi n$. | \frac{\pi}{3}+\pin,\frac{\pi}{3}+2\pikor\frac{\pi}{6}+\pin,\frac{2\pi}{3}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,672 |
5. Note that $x=0$ is not a root of the equation for any $a$. Transform the equation by dividing it by $x^{3}$:
$$
\left(a^{2}+1\right)\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)+2=12 a
$$
Thus, if $x_{0}$ is a root of the equation, then $\frac{1}{x_{0}}$ is also a root. Therefore, the only possible ro... | Answer: $a=-\frac{3}{2}, 0, \frac{1}{2}, 1$.
Answer to option: $5-2: a=-\frac{2}{3}, 1, 2$.
## Lomonosov Moscow State University
Olympiad "Conquer Sparrow Hills"
Variant 2-1 (Chelyabinsk) | -\frac{3}{2},0,\frac{1}{2},1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,676 |
1. $3 y^{2}+3 x y+x+y=2019 \Leftrightarrow(x+y)(3 y+1)=2019$. Since $2019=1 \cdot 2019=3 \cdot 673$ and $3 y+1 \equiv 1(\bmod 3)$, the possible cases are $3 y+1=1$, or $3 y+1=673$. From which $(x, y)=(-221,224)$, or $(x, y)=(2019,0)$. | Answer: $(x, y)=(-221,224),(2019,0)$. Answer to variant: 2-2: $(x, y)=(-224,221)$, $(0,-2019)$. | (x,y)=(-221,224),(2019,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,677 |
3. Using the method of auxiliary argument, we arrive at the equation
$$
\sqrt{\sin ^{4} b+\cos ^{4} b} \cos (x-\varphi(b))=a
$$
The answer to the problem will be $a$ satisfying the relation
$$
|a|>\min _{b} \sqrt{\sin ^{4} b+\cos ^{4} b}=\min _{b} \sqrt{1-\frac{1}{2} \sin ^{2} 2 b}=\frac{1}{\sqrt{2}}
$$ | Answer: $a \in(-\infty,-1 / \sqrt{2}) \cup(1 / \sqrt{2},+\infty)$.
Remark: It is implied that the equation may have no solutions at all for any $b$. Answer to the variant: $2-2: a \in(-\infty, 1-\sqrt{2}) \cup(\sqrt{2}-1,+\infty)$. | \in(-\infty,-1/\sqrt{2})\cup(1/\sqrt{2},+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,679 |
Let $x, y, z$ be the number of students in the categories of biology, physics, and chemistry, respectively. Then, according to the problem, we get the system of equations:
$$
\left\{\begin{array} { l }
{ 5 x = 2 ( y + z ) , } \\
{ 7 z = 3 ( x + y ) . }
\end{array} \Rightarrow \left\{\begin{array} { l }
{ 5 x - 2 y =... | Answer: 29. Answer to option: 4-2: 11.
# | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,682 |
4. Let's introduce the notations: $B H=2 a, H C=a, B F=y, F C=x$. Since angle $B F H$ is a right angle, by the theorem of relations in a right-angled triangle for the two legs $B H, H C$, we have:
$$
\left\{\begin{array}{l}
a^{2}=x(y+x) \\
4 a^{2}=y(x+y)
\end{array} \quad \Rightarrow \frac{y}{x}=4 \Rightarrow y=4 x\ri... | Answer: $\frac{1}{10} ; \frac{2\left(S_{1}+S_{2}\right)}{S_{A B C}}=\frac{1}{2}\left(\arcsin \frac{4}{5}+\frac{4}{5}\right)$. Answer to the variant: $4-2: \frac{1}{20}$; $\frac{3}{4}\left(\arcsin \frac{3}{5}+\frac{3}{5}\right)$ | \frac{1}{10};\frac{1}{2}(\arcsin\frac{4}{5}+\frac{4}{5}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,685 |
Let $b>0$ and $q>0$ be the first term and the common ratio of the progression, respectively. Note that the case $q=1$ does not work. If $b>0$ and $q>0$ (with $q \neq 1$), then from the problem statement we have
$$
b \frac{q^{6}-1}{q-1}=344 \cdot b \frac{q^{3}-1}{q-1} \Leftrightarrow\left\{\begin{array}{l}
q^{6}-344 q^... | Answer: 7. Answer to the option: $5-2: 3$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,686 | |
2. Transform each equation by bringing the fractions to a common denominator. We obtain
$$
\left\{\begin{array}{l}
(\sqrt{2}+1)(x-y)=\sqrt{2}\left(x^{2}+y^{2}\right)-3 x y \\
(\sqrt{2}-1)(x-y)=\sqrt{2}\left(x^{2}+y^{2}\right)+3 x y
\end{array}\right.
$$
if $x \neq \pm \sqrt{2} y$ and $y \neq \pm \sqrt{2} x$. Adding a... | Answer: $(x, y)=((5+\sqrt{5}) / 6,(-5+\sqrt{5}) / 6) ;((5-\sqrt{5}) / 6,(-5-\sqrt{5}) / 6)$. Answer to the variant: $5-2:(x, y)=((3+\sqrt{3}) / 8,(-3+\sqrt{3}) / 8) ;(x, y)=$ $(3-\sqrt{3}) / 8,(-3-\sqrt{3}) / 8)$. | (x,y)=(\frac{5+\sqrt{5}}{6},\frac{-5+\sqrt{5}}{6});(\frac{5-\sqrt{5}}{6},\frac{-5-\sqrt{5}}{6}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,687 |
4. Since the height $S H$ of the pyramid is seen from vertices $A$ and $C$ at the same angle, point $H$ lies on the median (which is also the bisector and altitude) $B M$ of triangle $A B C$ or its extension. If $S H=h, A B=B C=a$, $A C=b$ and $\angle S A H=\angle S C H=\alpha$, and the desired angle $\angle S B H=\bet... | Answer: $\frac{\pi}{8}$ or $\frac{3 \pi}{8}$. Answer to the variant: $5-2: \frac{\pi}{12}$ or $\frac{5 \pi}{12}$. | \frac{\pi}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,689 |
5. The solution can only exist if $a \in\left\{-\frac{\pi}{12}\right\} \cup\left(0 ; \frac{\pi}{12}\right]$, since otherwise the left side of the equation is either undefined or strictly positive. When $a=-\frac{\pi}{12}$, the equation becomes $2|x-16|=0$. Therefore, when $a=-\frac{\pi}{12}$, $x=16$. If $a \in\left(0 ;... | Answer: $x=16$ when $a=-\frac{\pi}{12}$. For other $a$, there are no solutions. Answer to option $5-2: x=-16$ when $a=\frac{\pi}{12}$. For other $a$, there are no solutions.
Lomonosov Moscow State University
## Olympiad "Conquer Sparrow Hills"
Option $6-1$ (Nizhny Novgorod) | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,690 |
3. For $x<0$ there are no solutions, since the function on the right is negative, while the one on the left is non-negative. For $x \geqslant 0$ we have $\arccos \left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \operatorname{arctg} x$, therefore the original inequality is equivalent to
$$
(x-a)^{2} \leqslant 0
$$
From which $x... | Answer: When $a<0$, there are no solutions; when $a \geqslant 0$, there is one solution $x=a$. | for\geqslant0 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,692 |
5. Let's make the substitution of variables:
$$
3^{2 x^{2}-3 x+1}=u, \alpha=\pi x+\frac{7 \pi n}{6}
$$
The equation can be transformed into:
$$
\log _{2}\left(u^{2}-6 u+17\right)=\frac{6\left(2 \log _{2} \sin \alpha+1\right)}{\left(2 \log _{2} \sin \alpha+1\right)^{2}+1}
$$
Now let's introduce the variable $t: t=2 ... | Answer: $\{6,9\}$.
## Answers to option $6-2$ | {6,9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,694 |
5. Let's introduce the function $g(x)=f(x)+(x-2021)^{2}-4$. For this function, the conditions $g(2019)=g(2020)=g(2021)=g(2022)=g(2023)=0$ are satisfied, meaning that the function $g(x)$ has 5 roots. Since it is a polynomial of the 5th degree, it has no other roots. Therefore,
$$
g(x)=(x-2019)(x-2020)(x-2021)(x-2022)(x... | Answer: -125. Answer to option: 7-2: -115. 7-3: 115. 7-4: 125. | -125 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,698 |
2. Find the smallest possible value of $\left|2015 m^{5}-2014 n^{4}\right|$, given that $m, n$ are natural numbers. | Answer: 0.
Solution: Let's find $N=2014^{x} \cdot 2015^{y}$ such that $m^{5}=2014^{x-1} \cdot 2015^{y}$ and $n^{4}=2014^{x} \cdot 2015^{y-1}$. For this, $x$ and $y-1$ must be multiples of 4, and $x-1$ and $y-5$ must be multiples of 5. For example, $x=16$ and $y=5$ work. Then, if we take $m=2014^{3} \cdot 2015$ and $n=... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,706 |
4. Before leaving for work, Mom entrusted Misha, Petya, and Vasya with the following tasks: a) sweep the floor in the hallway; b) wash the dishes; c) buy bread; d) pay for electricity; e) take out the trash; f) vacuum the carpet in the living room. In how many different ways can they distribute the tasks so that each t... | Answer: 540.
Solution: There are a total of $3^{6}=729$ ways to distribute the tasks. However, in $2^{6}=64$ ways, all tasks will be performed by Misha and Petya. There are also 64 ways when all tasks will be performed by Petya and Vasya, and 64 ways when Misha and Vasya. If we subtract $3 \times 64$, we get that the ... | 540 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,708 |
5. Find the largest three-digit number that is divisible by the sum of its digits and in which the first digit matches the third, but does not match the second. | Answer: 828.
Solution: Let this number be $\overline{a b a}=100 a+10 b+a$, where $a \neq b$. It must be divisible by $2 a+b$, so $101 a+10 b-10(2 a+b)=81 a$ is also divisible by $2 a+b$.
Since we need to find the largest such number, consider $a=9$. Then $81 a=729=3^{6}$, i.e., all divisors are powers of three, so $1... | 828 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,709 |
6. Solve the equation in natural numbers
$$
a b c + a b + b c + a c + a + b + c = 164
$$
In your answer, specify the product $a b c$. | Answer: 80.
Solution: $(a+1) \times(b+1) \times(c+1)=a b c+a b+b c+a c+a+b+c+1=$ $165=3 \times 5 \times 11$, therefore, $a=2, b=4$ and $c=10$. Note that the solution is unique up to the permutation of $a, b$ and $c$, since $3,5,11$ are prime numbers.
## 2013/2014 Academic Year CRITERIA FOR DETERMINING WINNERS AND PRI... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,710 |
9.1. Prove that the sum of the pairwise products of three consecutive natural numbers cannot equal 3000000. | Solution: The sum of the pairwise products of three consecutive natural numbers $n-1, n$, and $n+1$ is given by $n(n-1)+n(n+1)+(n-1)(n+1)=3n^2-1$, i.e., it is not divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 5,711 |
9.2. On the board in the laboratory, two numbers are written. Every day, the senior researcher Pyotr Ivanovich erases both numbers from the board and writes down their arithmetic mean and harmonic mean ${ }^{2}$. In the morning of the first day, the numbers 1 and 2 were written on the board.
Find the product of the nu... | # Solution:
The product of the numbers on the board does not change.
Indeed, $\frac{a+b}{2} \times \frac{2}{\frac{1}{a}+\frac{1}{b}}=a b$. Therefore, the desired product is 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,712 |
9.3. The numbers $a$ and $\mathrm{b}$ satisfy the equation $\frac{2 a}{a+b}+\frac{b}{a-b}=2$. Find all possible values of the expression $\frac{3 a-b}{a+5 b}$. | Solution. From the given inequality, it follows that
$$
2 a(a-b)+b(a+b)=2\left(a^{2}-b^{2}\right) \quad b(3 b-a)=0
$$
From which $\mathrm{b}=0$ or $\mathrm{a}=3 \mathrm{~b}$. If $\mathrm{b}=0$, then the given equality holds for all $a \neq 0$, which means the expression $\frac{3 a-b}{a+5 b}=3$.
If $a=3 b$ and $\math... | 3or1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,713 |
9.4. Let x and y be positive numbers for which
$$
x+y=1
$$
Prove that
$$
\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right) \geq 9
$$ | Solution. The inequality is easily transformed into the form
$$
x^{2}+y^{2}+8 x^{2} y^{2} \leq 1
$$
## Since
${ }^{2}$ The arithmetic mean of two numbers $a$ and $b$ is the number $\frac{a+b}{2}$, and the harmonic mean is the number $\frac{2}{\frac{1}{a}+\frac{1}{b}}$
$$
x y \leq\left(\frac{x+y}{2}\right)^{2}=\frac... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 5,714 |
9.5. The sides of a rectangle are expressed in integers. What should their lengths be so that the perimeter of the rectangle is numerically equal to its area? | Solution: Denoting the sides of the rectangle by $x$ and $y$, we form the equation
$$
2 x+2 y=x y, \text{ from which } x=\frac{2 y}{y-2}
$$
Since $x$ and $y$ must be positive, the number $y-2$ must also be positive, i.e., $y$ must be greater than 2.
Notice now that
$$
x=\frac{2 y}{y-2}=\frac{2(y-2)+4}{y-2}=2+\frac{... | rectangle\with\sides\3\\6;\\with\side\4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,715 |
9.6. In a certain company, there are 100 shareholders, and any 66 of them own no less than $50 \%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? | Solution. Let M be the shareholder owning the largest percentage of shares - x percent of shares. Divide the other 99 shareholders into three groups A, B, and
C, each with 33 shareholders. Let them own a, b, c percent of shares, respectively. Then
$$
2(100-x)=2(a+b+c)=(a+b)+(b+c)+(c+a) \geq 50+50+50
$$
That is, $x \... | 25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,716 |
6.1. A ticket to the Historical Museum costs 300 rubles for an adult and 60 rubles for a schoolchild, while pensioners can visit the museum for free. There is also a
"family ticket" for two adults and two children, which costs 650 rubles. What is the minimum amount in rubles that a family, including a father, a mother... | Solution. All possible options are presented in the table.
| Grandmother | Father | Mother | 1st child | 1st child | 1st child | 1st child | Total cost, RUB |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| free | "adult" ticket 300 | "adult" ticket 300 | "child" 60 | "child" 60 | "child" 60 | "chi... | 770 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,717 |
6.2. The lake is home to carp and perch. Two fishermen caught 70 fish, with 5/9 of the first fisherman's catch being carp, and 7/17 of the second fisherman's catch being perch. How many fish did each fisherman catch? | Solution. The number of fish caught by the second fisherman is a multiple of 17, so it can be: 17, 34, 51, or 68. The number of fish caught by the first can be (respectively) 53, 36, 19, or 2. But the number of fish caught by the first must be a multiple of 9, from which we get the answer: the first fisherman caught 36... | 3634 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,718 |
6.3. Oleg, Igor, and Anya are in the 6th grade. Among them, there is the best mathematician, the best chess player, and the best artist. It is known that:
a) the best artist did not paint their own portrait, but painted Igor's portrait
b) Anya has never lost to boys in chess.
Who in the class is the best mathematicia... | Solution: Since Anya did not lose to the boys in chess, she is the best chess player. Since the artist did not paint his own portrait but painted Igor's portrait, Igor is the best mathematician, and Oleg is the best artist.
Answer: Oleg is the best artist, Anya is the best chess player, Igor is the best mathematician. | Oleg | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,719 |
6.4. Is it possible to measure out exactly 4 liters of water into the larger of the two containers, using only a 3-liter and a 5-liter container, from a tap? If so, provide an example of how to do it. | Solution. You can, fill up 5 liters, pour 3 liters into the second container, then pour out this water, leaving 2 liters in the 5-liter container. Transfer these 2 liters to the 3-liter container. Fill the 5-liter container to the top and top up the second container (exactly 1 liter), then you will have 4 liters left i... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,720 |
6.5. In the record $52 * 2 *$, replace the asterisks with digits so that the resulting number is divisible by 36. List all possible solutions. | Solution: A number is divisible by 36 if it is divisible by both 4 and 9. Since the sum of the digits $5,2,2$ is 9, the sum of the two missing digits must be 0, 9, or 18. Considering that the number must be divisible by 4 and the second-to-last digit is 2, the last digit can only be 0, 4, or 8.
Answer: $52524,52128,52... | 52524,52128,52020,52920 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,721 |
6.6. $a, b, c$ are three different digits. If you add all six two-digit numbers that can be written using them without repeating the same digit twice in one number, the result is 528. Find these digits. | Solution. Let's assume for definiteness $a<b<c$. From these digits, we can form the numbers
$\overline{a b}=10 a+b, \overline{a c}=10 a+c, \overline{b c}=10 b+c, \overline{b a}=10 b+a, \overline{c a}=10 c+a, \overline{c b}=10 c+b$.
Their sum, obviously, gives $22 a+22 b+22 c$, which, according to the condition, equals... | 7,8,9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,722 |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be | Solution. Let the first number be x, and the second y. Then the equation $1996 x+\frac{y}{96}=x+y$ must hold, from which we find that 2016x=y. Therefore, their quotient is 2016 or $\frac{1}{2016}$.
Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 7 points; correct answer without solution 1 point. | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,723 |
2. According to analysts' forecasts, next year the number of the economically active population (employed and unemployed) in a certain city will increase by $4 \%$, while the number of unemployed will decrease by $9 \%$. What percentage of the economically active population next year will be unemployed, if this year th... | Solution. Let x people be the initial population of the city. Then $0.08x$ people are the number of unemployed initially. Currently, the population is $-1.04x$ people, and the number of unemployed is $(0.91 \cdot 0.056x)$ people. $(0.91 \cdot 0.056x \cdot 100) \div 1.04x = 4.9\%$ - the percentage of unemployed from the... | 4.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,724 |
3. The function $f(x)$ is such that for all natural $n>1$ there exists a prime divisor $p$ of the number $n$ such that
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)
$$
It is known that $f(1001)=1$. What is $f(1002) ?$ | Solution. Note that for any prime number p, the value $f(p)=f(1)-$ $f(p)$. Therefore, $f(p)=\frac{f(1)}{2}$ for any prime number. For prime numbers p and q, we get that either $\mathrm{f}(\mathrm{pq})=\mathrm{f}(\mathrm{p})-\mathrm{f}(\mathrm{q})=0$, or $\mathrm{f}(\mathrm{pq})=\mathrm{f}(\mathrm{q})-\mathrm{f}(\mathrm... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,725 |
4. On a line, 3025 points are marked. The midpoints of every two of the marked points are painted green, blue, or red. Prove that the number of points painted in one of the colors on the line is at least 2016. | Solution. Let's start by considering three points. Obviously, for three points on a line, the midpoints of each pair of them are three different points. Consider the extreme point on the line. The midpoints between it and the two nearest points to it are two points that are not midpoints of any other points. If we remo... | 2016 | Combinatorics | proof | Yes | Yes | olympiads | false | 5,726 |
5. Solve the equation:
$$
x+\frac{7}{x}=[x]+\frac{7}{[x]}
$$
where $x=[x]+\{x\}$. | Solution. After equivalent transformations, we arrive at the equation
$$
(x-[x])\left(1-\frac{7}{x[x]}\right)=0
$$
which is equivalent to the system of equations $x-[x]=0$ and $[x]=\frac{7}{x}$. The solutions to the first equation are all non-zero integers. In the second equation, express the integer part of the numb... | x\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,727 |
6. In the inscribed quadrilateral $\mathrm{ABCD}$, the circumcircle of triangle COD (O - the intersection point of the diagonals) passes through the center of the circumcircle of quadrilateral ABCD. Prove that quadrilateral $\mathrm{ABCD}$ is a trapezoid. | Solution. Let $\mathrm{O}_{1}$ be the center of the circumscribed circle of quadrilateral $\mathrm{ABCD}$. From the condition, it follows that $\angle \mathrm{COD}=\angle \mathrm{CO}_{1} \mathrm{D}=\cup \mathrm{CD}$. On the other hand, $\angle \mathrm{COD}=(\cup \mathrm{AB}+\cup C D) / 2$. From the two equalities, we g... | proof | Geometry | proof | Yes | Yes | olympiads | false | 5,728 |
7. Do there exist 2016 consecutive natural numbers among which there are exactly 16 prime numbers | Solution. Let's introduce the function $\mathrm{S}(\mathrm{n})$, equal to the number of prime numbers from $\mathrm{n}$ to n+2015. Note that $\mathrm{S}(\mathrm{n})$ differs from $\mathrm{S}(\mathrm{n}+1)$ by no more than 1, $\mathrm{S}(2017!+2)=0, \quad S(1)>16 \quad(2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53$,
prime... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,729 |
1. When one of two integers was increased 1996 times, and the other was reduced 96 times, their sum did not change. What can their quotient be? | Solution. Let the first number be x, and the second y. Then the equation $1996 x + \frac{y}{96} = x + y$ must hold, from which we find that $2016 x = y$. Therefore, their quotient is 2016 or $\frac{1}{2016}$.
Answer: 2016 or $\frac{1}{2016}$.
Criteria: Full solution - 7 points; correct answer without solution 1 point... | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,730 |
2. According to analysts' forecasts, next year the number of the economically active population (employed and unemployed) in a certain city will increase by $4 \%$, while the number of unemployed will decrease by $9 \%$. What percentage of the economically active population next year will be unemployed, if this year th... | Solution. Let x people be the initial population of the city. Then $0.08 x$ people are the number of unemployed initially. Currently, the population is $-1.04 x$ people, and the number of unemployed is $(0.91 \cdot 0.056 x)$ people. $(0.91 \cdot 0.056 x \cdot 100) \div 1.04 x = 4.9\%$ - the percentage of unemployed fro... | 4.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,731 |
4. In a regular quadrilateral pyramid $S A B C D$, the height is equal to the side of the base. Points $M$ and $N$ are marked on the lateral edges $SD$ and $SB$ of the pyramid, respectively, such that the lines $AM$ and $CN$ are perpendicular to each other. Prove that
$$
2 \mathrm{SA}(\mathrm{SM}+\mathrm{SN})=\mathrm{... | Solution. We will place the coordinate system such that the origin coincides with vertex A, and the axes are as shown in the figure.
We will assume that the length of the base side of the py... | proof | Geometry | proof | Yes | Yes | olympiads | false | 5,733 |
6. In the inscribed quadrilateral $\mathrm{ABCD}$, the circumcircle of triangle COD (O - the intersection point of the diagonals) passes through the center of the circumcircle of quadrilateral ABCD. Prove that quadrilateral ABCD is a trapezoid. | Solution. Let $\mathrm{O}_{1}$ be the center of the circumscribed circle of quadrilateral $\mathrm{ABCD}$. From the condition, it follows that $\angle \mathrm{COD}=\angle \mathrm{CO}_{1} \mathrm{D}=\cup \mathrm{CD}$. On the other hand, $\angle \mathrm{COD}=(\cup \mathrm{AB}+\cup \mathrm{CD}) / 2$. From the two equaliti... | proof | Geometry | proof | Yes | Yes | olympiads | false | 5,735 |
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