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8. We will call a natural number beautiful if in its decimal representation there are an equal number of digits $0,1,2$, and no other digits. Can the product of two beautiful numbers be beautiful? (K. Sukhov) | Answer. It cannot. Solution. Obviously, the number of digits in a beautiful number is divisible by 3. If a beautiful number $x$ has $3 n$ digits in its decimal representation, then it satisfies the inequality $10^{3 n-1}<x<3 \cdot 10^{3 n-1}(*)$. Therefore, the product of two beautiful numbers, written with $3 k$ and $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,982 |
9. Petya and Vasya wrote 100 different natural numbers each on the board. Petya divided all his numbers by Vasya's with a remainder and wrote down all 10000 resulting remainders in his notebook. Vasya divided all his numbers by Petya's with a remainder and wrote down all 10000 resulting remainders in his notebook. It t... | Solution. Let Petya write down the numbers $a_{1}>a_{2}>\ldots>a_{100}$, and Vasya $-b_{1}>b_{2}>\ldots>b_{100}$. If $a_{1}>b_{1}$, then one of Vasya's remainders will be $b_{1}$, while all of Petya's remainders will be less than $b_{1}$ - a contradiction. Similarly, the assumption that $a_{1}<b_{1}$ leads to a contrad... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,983 |
10. At the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip is ... | Answer: 50. Solution: Example: The chip 50 is sequentially exchanged 99 times with the next one counterclockwise.
Evaluation. We reason by contradiction. Let $k<50$. First proof. We will consider the shifts of the chips relative to their initial positions, with shifts clockwise counted as positive and counterclockwise... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 6,984 |
1. The operation of doubling a digit of a natural number consists in multiplying this digit by 2 (if the product turns out to be a two-digit number, the digit in the next place of the number is increased by 1, as in column addition). For example, from the number 9817, by doubling the digits 7, 1, 8, and 9, one can obta... | Answer. Possible. Solution. Triple the first digit of the number 22... 22 (20 twos). We get the number 1622... 22 (21 digits). Now double the digit 6 and we get the desired number $22 . .22$ (21 twos). Note. There are other ways as well. | 22..22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,985 |
2. Each of the 10 people is either a knight, who always tells the truth, or a liar, who always lies. Each of them thought of some natural number. Then the first said: “My number is greater than 1”, the second said: “My number is greater than 2”, ..., the tenth said: “My number is greater than 10”. After that, they, in ... | Answer: 8. Solution: Those who in the first series of answers said that their numbers are greater than 9 and 10 are definitely liars, because these answers are incompatible with any of the answers in the second series. Therefore, there are no more than eight knights. An example when there are exactly 8 knights: the fir... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,986 |
3. One hundred natural numbers are arranged in a circle. Each of them is divided with a remainder by the next one in the clockwise direction. Could it happen that 100 identical non-zero remainders are obtained? (N. Agakhanov) | Answer. It could not. Solution. Let $r$ be the remainder specified in the condition. Then each of the numbers standing in a circle is greater than $r$. Therefore, the quotient in each division with remainder is greater than 0, and hence each number is greater than the one following it clockwise. But this is impossible,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,987 |
4. There is a cube, each face of which is divided into 4 identical square cells. Oleg wants to mark 8 cells with invisible ink so that no two marked cells share a side. Rustem has detectors. If a detector is placed in a cell, the ink on it becomes visible. What is the minimum number of detectors Rustem can place in the... | Answer: 16. Solution: Example. Let's divide all 24 cells into eight triples, where each triple consists of three cells adjacent to one vertex of the cube. Any two cells in the same triple share a common side. Since the number of marked cells is the same as the number of triples, there must be exactly one marked cell in... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,988 |
5. The perimeter of triangle $ABC$ is 2. On side $AC$, point $P$ is marked, and on segment $CP$, point $Q$ is marked such that $2AP = AB$ and $2QC = BC$. Prove that the perimeter of triangle $BPQ$ is greater than 1. (A. Kuznetsov) | Solution. Let $A B=c, A C=b, B C=a$. We need to prove that $B P+B Q+P Q>1=(a+b+c) / 2$. Since $P Q=A C-A P-C Q=b-(a+c) / 2$, we need to prove that $B P+B Q>(a+b+c) / 2-b+(a+c) / 2=a+c-b / 2$.
Let $M$ and $N$ be the midpoints of sides $A B$ and $B C$ respectively, and let $R$ and $S$ be points on rays $A C$ and $C A$ r... | proof | Geometry | proof | Yes | Yes | olympiads | false | 6,989 |
2. In the country of Euleria, there are 101 cities. Every two cities are connected by a two-way non-stop flight of one of 99 airlines. It is known that from each city, flights of all 99 companies depart. We will call a triangle three cities that are pairwise connected by flights of the same company. Prove that in Euler... | Solution. Let's call a tick two flights of the same airline departing from one city. From each city, exactly 100 flights depart, representing all 99 airlines. Therefore, each city serves as the center of exactly one tick, meaning there are a total of 101 ticks.
Suppose there are at least two triangles in Euleria. Each... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 6,991 |
3. Given an equilateral triangle ABC. Point $D$ is chosen on the extension of side $A B$ beyond point $A$, point $E$ is on the extension of $B C$ beyond point $C$, and point $F$ is on the extension of $A C$ beyond point $C$ such that $C F=A D$ and $A C+E F=D E$. Find the angle BDE. (A. Kuznetsov) | Answer: 60 - . Solution: Complete triangle $A C E$ to parallelogram $A C E G$. Since $C F=A D$, $C E=A G$ and $\cdot F C E=\cdot D A G=60 \cdot$, triangles $D A G$ and $F C E$ are equal, from which $G D=E F$. Therefore, $D E=A C+E F=G E+G D$. This means that point $G$ lies on segment $D E$, and thus $D E \| A C$, from ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,992 |
4. Given a $2 n$-digit natural number $a$ and a natural number $k$. The numbers $a$ and $k a$ were written on a tape, and each of the two records was cut into two-digit numbers, starting from the last digits (here, the numbers 00, 01, ..., 09 are also considered two-digit; if the number $k a$ ended up with an odd numbe... | The first solution. Let's write the numbers $a$ and $k$ in base 100. The two-digit numbers, into which the records of numbers $a$ and $k a$ are cut according to the condition, will be digits in this system. Hereinafter, we will understand by "digits" the digits in the base-100 system. Let in this system $a=\overline{a_... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,993 |
1. Once Baron Munchausen, returning from a walk, told that half of the way he walked at a speed of 5 km/h, and half of the time spent on the walk - at a speed of 6 km/h. Was the baron mistaken? (I. Rubanov) | Answer. Mistake. Solution. Solution. If the baron walked half of the time spent at a speed of 6 km/h, then he walked at a speed of 5 km/h, at most, the second half of this time, that is, no longer than at a speed of 6 km/h. But this means that the baron, walking at a speed of 5 km/h, covered a shorter distance than at ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 6,994 |
2. Find any seven consecutive natural numbers, each of which can be changed (increased or decreased) by 1 in such a way that the product of the seven resulting numbers equals the product of the seven original numbers. (Method Commission of the All-Russian Olympiad) | Solution. Solution. For example, the numbers from 3 to 9 fit: replace 3 with 2, 4 with 5, 5 with 6, and the numbers in each of the pairs $(6,7)$ and $(8,9)$ with each other. In the end, we get $3 \cdot 4 \cdot 5 \cdot 6$ $\cdot 7 \cdot 8 \cdot 9=2 \cdot 5 \cdot 6 \cdot 7 \cdot 6 \cdot 9 \cdot 8$
Remark. If it was poss... | 3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9=2\cdot5\cdot6\cdot7\cdot6\cdot9\cdot8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,995 |
3. On the hypotenuse $BC$ of the right triangle $ABC$, a point $K$ is chosen such that $AB = AK$. Segment $AK$ intersects the bisector $CL$ at its midpoint. Find the acute angles of triangle $ABC$. (I. Bogdanov) | Answer. $\angle B=54^{\circ}, \angle C=36^{\circ}$. Solution. Let the midpoint of the bisector $C L$ be $P$, and the angle $A B C$ be $\beta$; then $\angle A C L=\left(90^{\circ}-\beta\right) / 2$. In the right triangle $A C L$, the segment $A P$ is a median, so $A P=C P=L P$. Now from the isosceles triangles $A P L$ a... | \angleB=54,\angleC=36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 6,996 |
4. Given natural numbers a and $b$, and a < 1000. Prove that if $a^{21}$ is divisible by $b^{10}$, then $a^{2}$ is divisible by $b$. (P. Kozhevnikov) | Solution. Suppose the statement of the problem is false; then there exists a prime number $p$ that enters the prime factorization of the number $a^{2}$ with an exponent less than in the prime factorization of the number $b$. That is, if $a$ is divisible by $p^{k}$ but not by $p^{k+1}$, and $b$ is divisible by $p^{m}$ b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 6,997 |
1. Does the number 1... (1000 ones) have a ten-digit divisor, all digits of which are different? | Answer: No. Solution: If all the digits of a ten-digit number are different, then their sum is 45. This means that such a number is divisible by 9. The number $1 ... 1$ (1000 ones) with a digit sum of 1000 is not even divisible by 3. | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 6,998 |
2. Positive numbers $a, b$, and $c$ are such that $a^{2}<b$ and $b^{2}<c$ and $c^{2}<a$. Prove that all three numbers $a$, $b$, and $c$ are less than 1. | Solution. Let $a \geq 1$. Then $b>a^{2} \geq 1, c>b^{2} \geq 1$ and $a>c^{2}>1$. But then $a^{2}<b<b^{2}<c<c^{2}<a$, from which $a<1$. Contradiction. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 6,999 |
4. Prove that the number
$12345678987654321^{2} 987654321012345679^{2}+\left(12345678987654321^{2}+987654321012345679^{2}\right) 10^{36}$ is a square of an integer. | Solution. Let $x=12345678987654321, y=987654321012345679$. Then $x+y=10^{18}$, and therefore the given expression equals $x^{2} y^{2}+\left(x^{2}+y^{2}\right)(x+y)^{2}=\left(x y+x^{2}+y^{2}\right)^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,001 |
5. It is known that among 100 balls, exactly 51 are radioactive. There is a device into which two balls can be placed, and if both are radioactive, a light will turn on (if at least one of the two balls is not radioactive, the light will not turn on). Can all the radioactive balls be found using the device no more than... | Answer. Yes. Solution. Let's divide the balls into 50 pairs and test them. Consider two possible cases.
1) Exactly one of these tests revealed two radioactive balls. Then in each of the remaining 49 pairs, there is exactly one radioactive ball. Testing one of the found radioactive balls with one ball from each of the ... | 145 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,002 |
3. Two angle bisectors of a triangle intersect at an angle of 60 degrees. Prove that one of the angles of this triangle is 60 degrees. | Solution. Let the angle bisectors $A A_{1}$ and $C C_{1}$ of triangle
Fig. 4 $A B C$ intersect at point $I$. Suppose that $\angle A I C_{1}=60^{\circ}$. By the exterior angle theorem of a t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,004 |
4. When Winnie-the-Pooh came to visit Rabbit, he ate 3 bowls of honey, 4 bowls of condensed milk, and 2 bowls of jam, and after that, he couldn't get out of Rabbit's burrow because he had become too fat from all the food. But it is known that if he had eaten 2 bowls of honey, 3 bowls of condensed milk, and 4 bowls of j... | Answer: From condensed milk. Solution. According to the condition $3 m+4 c+2 v>2 m+3 c+4 v$, from which $\mathbf{~}^{+}+>2 v (*)$. According to the condition $3 m+4 c+2 v>4 m+2 c+3 v$, from which
$2 c>$ m+v. Adding the last inequality to the inequality (*), we get m +3 c $>$ m +3 v, from which c $>$ v. | v | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,005 |
2. The arithmetic mean of several consecutive natural numbers is 5 times greater than the smallest of them. How many times is the arithmetic mean less than the largest of these numbers? | Answer: $9 / 5 = 1.8$ times. Solution. Let the total number of numbers be $d$, and $n-$ the smallest of them. Then their arithmetic mean is
$$
(n+(n+1)+\ldots+(n+d-1)) / d=(n d+(1+2+\ldots+d-1)) / d=n+d(d-1) / 2 d=n+(d-1) / 2
$$
On the other hand, it equals $5 n$, from which $d-1=8 n$. Therefore, the largest number i... | 1.8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,008 |
3. In a box, there are balls of 10 colors. It is known that you can take out 100 balls from the box so that the remaining balls of all 10 colors are equal. Prove that you can add 900 balls to the box so that the number of balls of all colors becomes equal. | Solution. Let's say that in order to have $k$ balls of each of the 10 colors, 100 balls need to be removed, among which $a_{1}$ balls are of the first color, $a_{2}$ - of the second color, ..., $a_{10}$ - of the tenth color. Then, if 100 - $a_{1}$ balls of the first color, 100 - $a_{2}$ - of the second color, ..., 100 ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 7,009 |
4. Inside angle BAC, equal to $45^{\circ}$, a point $D$ is taken such that each of the angles ADB and ADC is $45^{\circ}$. Points $D_{1}$ and $D_{2}$ are symmetric to point $D$ with respect to lines AB and AC, respectively. Prove that points $D_{1}, D_{2}, B$ and C lie on the same line. | Solution. Since triangles $ABD$ and $ABD_{1}$ are symmetric by condition, $AD_{1}=AD, \angle BAD_{1}=\angle BAD, \quad \angle AD_{1}B=\angle ADB=45^{\circ}$. Similarly, $AD_{2}=AD, \angle CAD_{2}=\angle CAD, \angle AD_{2}C=\angle ADC=45^{\circ}$. From the equalities $\angle BAD_{1}=\angle BAD$, and $\angle CAD_{2}=\ang... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,010 |
5. In the country of Dumulandia, exactly 10 roads led out of each city, and each road connected exactly two cities. The road network was connected, meaning that from any city, one could travel to any other city, possibly through other cities. However, during a flood, two cities connected by a road were flooded, after w... | Solution. In total, 18 roads led out of the flooded cities to other cities (let's call these roads flooded). After the flood, all cities were divided into at least two parts, and it was possible to travel from one part to others only through the flooded cities. At least one of these parts had no more than 9 flooded roa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 7,011 |
1. Along the road from Yolkiino to Palkino, there is an oak tree, from which it is twice as close to Yolkiino as to Palkino. Fyodor, traveling at a constant (and greater than 0) speed from Yolkiino to Palkino, was twice as close to the oak tree as to Yolkiino at 12:00. At 12:40, it was again found that Fyodor was twice... | Answer: 13.10. Solution: On the road from Yolkiino to Palkino, there are two points that are twice as close to the oak tree as to Yolkiino. One of them $\left(\Phi_{1}\right)$ is between Yolkiino and the oak tree, and the other ( $\Phi_{2}$ ) is between the oak tree and Palkino (see figure). For
 These are six consecutive natural numbers.
2) The sum of these six numbers is even.
Vasya: 1) These are the numbers 1, 2, 3, 4, 5, 6.
2) Kolya is a liar.
K... | Answer: $3,6,9,12,15,18$. Solution. Among six consecutive natural numbers, there are exactly three odd numbers. Therefore, their sum is odd. This means that Petya lied either the first time or the second time, and thus he is a liar, that is, he lied both times. But then Vasya is also a liar because in his first stateme... | 3,6,9,12,15,18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,013 |
3. There are six visually identical coins. Four of them are genuine, two are counterfeit. All genuine coins weigh the same, each counterfeit coin weighs less than a genuine one, and the weights of the counterfeit coins are different. There are scales with two pans: if coins are placed on them, the pan with the heavier ... | Solution. Divide the coins into three pairs and compare two coins in each pair. If the fake coins end up in two different pairs, we will find them as the lighter ones in their respective pairs (while the third pair will balance). If, however, they end up in the same pair, then there will be balance in two pairs, and we... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,014 |
4. Inside the acute angle BAC, a point $D$ is taken such that angle CAD is twice the angle BAD. Could point D be twice as far from line $A C$ as from line $A B$? | Answer: It could not. Solution: Drop a perpendicular $D E$ from point $D$ to line $A B$, and on ray $A C$, lay off segment $A F = A D$. In
the isosceles triangle $A D F$, draw the median $A ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,015 |
5. The numerator of each of the 48 fractions is one of the numbers 2, 3, ..., 49, and the denominator is also one of these numbers, with each of these 48 numbers appearing both among the numerators and among the denominators. Prove that either one of these fractions is an integer, or from them, no more than 25 fraction... | Solution. Consider the fraction $a_{1} / 2$ with the denominator 2. If $a_{1}$ is even, we have already obtained an integer. Otherwise, multiply $a_{1} / 2$ by the fraction $a_{2} / a_{1}$, the result - by the fraction $a_{3} / a_{2}$, and so on until some numerator $a_{n}$ becomes even (this will eventually happen bec... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,016 |
1. On the side $B C$ of triangle $A B C$, a point $D$ is taken in such a way that the perpendicular bisector of segment $A D$ passes through the center of the inscribed circle of triangle $A B C$. Prove that this perpendicular bisector passes through the vertex of triangle $A B C$. (L. Emelyanov) | Solution. Drop a perpendicular $J N$ from the center $J$ of the inscribed circle of the triangle to the side $C B$. Suppose the point $N$ lies on the segment $D C$ (the case where the point $N$ lies on the segment $D B$ is considered similarly.). Drop a perpendicular $J M$ from the center $J$ to the side $C A$. Since $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,017 |
2. Oleg and Sergey take turns writing down one digit from left to right until a nine-digit number is formed. At the same time, they cannot write down digits that have already been written. Oleg starts (and finishes). Oleg wins if the resulting number is divisible by 4; otherwise, Sergey wins. Who will win with correct ... | Answer. Sergey. First solution. In his first three moves, Sergey can ensure that the digits 2 and 6, and another even digit, for example, 0, are definitely used. With his last move, Sergey places any unused odd digit. If Oleg places an odd digit last, the number will be odd. If he places one of the remaining even digit... | Sergey | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,018 |
3. In each cell of a $2012 \times 2012$ table, either a zero or a one is written, with the condition that in each column and each row there are both zeros and ones. Prove that in this table, there exist two rows and two columns such that at the ends of one of the diagonals of the rectangle they form, there are zeros, a... | Solution. Consider the row $C_{1}$, which has the most ones. In it, there is a cell $X$ containing a zero, and in the column where cell $X$ is located, there is a cell $Y$ containing a one. If in the row $C_{2}$, where cell $Y$ is located, there is a zero under some one in row $C_{1}$, then the proof is complete: the d... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 7,019 |
4. Do there exist two polygons (not necessarily convex) with the following property: by attaching them to each other (without overlapping), one can obtain polygons with any number of sides from 3 to 100 inclusive? (S. Volchonkov, S. Berlov, I. Bogdanov)
 | Answer: 3. Solution: If the numbers are written, for example, in the order 2, 7, 5, 6, 1, 4, 3, then the numbers 7, 6, and 4 will be good. It remains to show that there cannot be more than three good numbers.
Note that a good number is greater than both of its neighbors, so two good numbers cannot stand next to each o... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,021 |
2. In each cell of a 100 x 100 table, one of the numbers 1 or -1 is written. Could it be that the sums of the numbers are negative in exactly 99 rows and positive in exactly 99 columns? (D. Nenashov) | Answer. No, it could not. Solution. Suppose the desired arrangement exists. Since in each row and each column of the table there is an even number of odd numbers, all the sums of the numbers in the rows and columns are even. Therefore, in each row with a negative sum, this sum is no more than -2. Consequently, the sum ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 7,022 |
3. In trapezoid $ABCD$, point $M$ is the midpoint of base $AD$. It is known that $\cdot ABD=90^{\circ}$ and $BC=CD$. On segment $BD$, a point $F$ is chosen such that $\cdot BCF=90^{\circ}$. Prove that $MF \perp CD$. (N. Chernega) | Solution. Consider the right triangle $ABD$. In it, $M$ is the midpoint of the hypotenuse, so $AM = MD = BM$. Therefore, $M$ lies on the perpendicular bisector of $BD$. On the other hand, since $BC = CD$, point $C$ also lies on the perpendicular bisector of $BD$. Thus, $MC \perp BD$. Next, since $AD \parallel BC$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,023 |
4. Petya chose 10 consecutive natural numbers and wrote each one either in red or blue pencil (both colors are present). Can the sum of the least common multiple of all red numbers and the least common multiple of all blue numbers end in 2016? (O. Dmitriev, R. Zhenodarov) | Answer: No, it cannot. Solution: Assume the opposite. Note that a number ending in 2016 is necessarily divisible by 16. Among the ten numbers Petya has, there is either one or two numbers divisible by 8. In the first case, one of the obtained least common multiples (LCM) is divisible by 8, and the other is not, so thei... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,024 |
6. On an infinite tape, all natural numbers with the sum of digits equal to 2018 are written in ascending order. What number is written in the 225th position? (Method Commission) | Answer: 39... 998 (223 nines). Solution. Since $2018=9 \cdot 224+2$, the smallest number with a digit sum of 2018 will be 29... (224 nines), and the second smallest - the number 389...9 (223 nines). In the next 223 largest numbers with a digit sum of 2018, the eight "travels" from the beginning to the end of the row of... | 39..998(223nines) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,025 |
7. At noon, Vasya put 10 paper-cut convex decagons on the table. Then, from time to time, he took scissors, cut one of the polygons lying on the table along a straight line into two, and put both resulting pieces back on the table. By midnight, Vasya had performed this operation 51 times. Prove that by midnight, among ... | Solution. Since after each cut the number of polygons increased by 1, there were 61 polygons on the table at midnight. Suppose each of these polygons had at least five vertices. Then the total number of vertices of the polygons on the table at midnight was at least $5 \times 61 = 305$. However, with each cut, the total... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,026 |
8. On the bisector $A L$ of triangle $A B C$, a point $D$ is chosen. It is known that $\angle B A C=2 \alpha, \angle A D C=3 \alpha$, $\angle A C B=4 \alpha$. Prove that $B C+C D=A B$. (A. Kuznetsov) | Solution. On the extension of segment $B C$ beyond point $C$, choose point $E$ such that $C D = C E$. Then $\angle A C D = 180^{\circ} - \angle D A C - \angle A D C = 180^{\circ} - 4 \alpha = \angle A C E$. Therefore, triangles $A C D$ and $A C E$ are congruent by two sides and the included angle, so $\angle A E C = \a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 7,027 |
9. On a white checkered board of size $25 \times 25$ cells, several cells are painted black, with exactly 9 cells painted black in each row and each column. What is the smallest $k$ such that it is always possible to repaint $k$ cells to white in such a way that it is impossible to cut out a black $2 \times 2$ square? ... | Solution. Evaluation. Note that if 9 cells are shaded in a row, then four of them can be repainted so that no two shaded cells are adjacent: it is enough to renumber the shaded cells from left to right and repaint the cells with even numbers. If such repainting is done with all even rows, then 48 cells will be repainte... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,028 |
10. Prove that there exists a natural number $n$, greater than $10^{100}$, such that the sum of all prime numbers less than $n$ is coprime with $n$. (R. Salimov) | Solution. Let $S(n)$ denote the sum of all prime numbers less than $n$. Notice that
$$
S(n)p>10^{100}$. Suppose $S(p)$ is not coprime with $p$, and $S(q)$ is not coprime with $q$. Then $S(p)$ is divisible by $p$, and $S(q)$ is divisible by $q$. Let $S(p)=k p$. From inequality (*) it follows that $k<p-1$. Then, since $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,029 |
1. Find three irreducible fractions with numerators and denominators not equal to 1, such that their sum is an integer, and the sum of the fractions reciprocal to them is also an integer. | Solution. For example, 11/2, 11/6, 11/3. Remark. The problem setters, of course, meant that the fractions should be different. But since this was not explicitly stated in the problem, answers with equal fractions, such as $-1 / 3,-1 / 3,-1 / 3 ; 5 / 2,5 / 4,5 / 4$ and so on, are also accepted. | 11/2,11/6,11/3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,030 |
2. From the natural numbers from 1 to 25, Dasha chose six such that the difference between any two chosen numbers is divisible by 4. What is the maximum number of prime numbers that Dasha could have chosen? | Answer: Five. First solution. The difference of two numbers is divisible by 4 if and only if these numbers have the same remainder when divided by 4. Let's list all prime numbers less than 25 and their remainders when divided by 4: 2-2, 3-3, 5-1, 7-3, 11-3, 13-1, 17-1, 19-3, 23-3. The most common remainder is 3, which ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,031 |
3. In triangle ABC, the median BM is drawn. It is known that $\angle A B M=40^{\circ}$, and $\angle C B M=70^{\circ}$. Find the ratio $A B: B M$. | Answer: 2. First solution. Complete the triangle $A B C$ to a parallelogram $A B C D$. Since the diagonals of a parallelogram bisect each other, point $M$ is the point of their intersection and $B D=2 B M$. On the other hand, $\angle B D A=\angle C B D=70^{\circ}$, and $\angle B A D=180^{\circ}-\angle B D A-\angle A B ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,032 |
5. The cells of an $n \times n$ square are colored black and white with the condition that no four cells, located at the intersection of two rows and two columns, can all be the same color. What is the largest possible value of $n$? | Solution. An example for $n=4$ is shown in the figure on the right. We will prove by contradiction that it is impossible to color a $5 \times 5$ square in this way. Let's call a row predominantly black if it has more black squares than white, and predominantly white otherwise. Among the five rows, there will be either ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,034 |
4. A two-digit number $N$ was multiplied by 2, the digits of the result were swapped, and then the number was divided by 2. The result was the same number $N$. How many such numbers $N$ exist?
Answers:
A) none (-) B) exactly 4 (-) C) at least 10 (+) D) at least 14 (+) E) at least 15 (-) | Solution. Since the result turned out to be the same number, two identical digits were swapped. This means that $2 \mathrm{~N}$ should have two such digits. Let's consider several cases:
1) When multiplying by 2, there was no carry-over to the next place value. Obviously, as $N$, the numbers 11, 22, 33, 44 fit and onl... | 14 | Number Theory | MCQ | Yes | Yes | olympiads | false | 7,037 |
5. How many solutions can the equation ||$|x-a|-1|-1|=|b|$ have?
Answers:
A) 2 (+) B) 3 (+) C) $4(+)$ D) 5 (-) E) $6(+)$ | Solution. The graph of the function $y=|||x|-1|-1|$ is shown in the figure. The graph of $y=|||x-a|-1|-1|$ is obtained by shifting along the $X$-axis and does not affect the intersection with any line parallel to the $X$-axis. The graph shown in the figure can intersect the horizontal line $\mathrm{y}=|\mathrm{b}|$ at ... | notfound | Algebra | MCQ | Yes | Yes | olympiads | false | 7,038 |
6. How many pairs of prime (not necessarily distinct) numbers ( $p ; q$) exist such that $p^{q}-p q$ is also prime? Answers:
A) at least one (+) B) at least two (+) C) at least three (-) D) none (-) E) infinitely many (-) | # Solution.
pq-pq is divisible by p, so for this expression to be a prime number, it is necessary that pq-pq=p, from which $\mathrm{p}^{\mathrm{q}-1}-\mathrm{q}=1$. Therefore, $\mathrm{p}$ and $\mathrm{q}$ must be of different parity, so one of them must be 2. If $\mathrm{q}=2$, then $\mathrm{p}^{1}-2=1, \mathrm{p}=3$... | p=2,q=3;p=3,q=2 | Number Theory | MCQ | Yes | Yes | olympiads | false | 7,039 |
7. On a $6 \times 6$ chessboard, there are 9 rooks. How many "unattacked" cells can be on the board?
Answers:
A) 9 (+) B) 8 (-) C) $6(+)$ D) 4 (+) E) 1 (+) | Solution. If the rooks are in $k$ columns and $m$ rows, then $(6-k)(6-m)$ cells are free. 9 rooks cannot fit in fewer than 2 rows and fewer than 2 columns. Moreover, it is impossible to fit them in a $4 \times 2$ rectangle. Therefore, each of the factors does not exceed 3. All other options are possible, that is, there... | 9,6,4,1,0 | Combinatorics | MCQ | Yes | Yes | olympiads | false | 7,040 |
8. On the side $A B$ of rectangle $A B C D$, a point $L$ is taken, and on the sides $A D$ and $B C$, points $M$ and $N$ are taken such that the rays $L M$ and $L N$ divide the straight angle $A L B$ into three equal angles. The circle with diameter $M N$ intersects these rays at points $E$ and $F$. Which is greater: $E... | Solution. Angle LNF = LME $=30^{\circ}$, so LE = 1/2 LM, LF = 1/2 LN.
In turn, in triangles ALN and $B L M A L=1 / 2 L N$, and $B L=1 / 2 L M$ | Geometry | MCQ | Yes | Yes | olympiads | false | 7,041 | |
9. For what $n$ can $2 n$ distinct points be marked on a plane so that for any natural $k$ from 1 to $n$ there exists a line containing exactly $k$ marked points?
Answers:
A) for $\mathrm{n}$ less than 6 (+) B) for n= 6 (+) C) for $\mathrm{n}$ less than 7 (+) D) for any $\mathrm{n}$ (-) E) for no $\mathrm{n}$ (-)
VA... | # Solution.
If $\mathrm{k}>4$, then there must be different lines with k, $k-1$ and k-2 points. These lines can only intersect at three points, that is, be the sides of some triangle. Thus, $\mathrm{k}+(\mathrm{k}-2)+(\mathrm{k}-4)$ points have already been used. To be able to draw the next line, it is necessary to ma... | notfound | Combinatorics | MCQ | Yes | Yes | olympiads | false | 7,042 |
10. Vanya, Kostya, and Lesha are playing a game: there are 2008 matches on the table. In one move, Vanya and Lesha can take 1 or 2 matches, Kostya - 1, 2, or 3. Vanya moves first, Kostya second, and Lesha third. The player who takes the last match wins. Which two players can combine their efforts against the third to p... | Solution. We will show that any team with Kostya always wins. Note that any two players together can take any number of matches from 2 to 4 in one round (each moves once). If the third can only take 1 or 2, then the two who have teamed up can ensure taking 5 matches in one round. Then, if Vanya and Kostya team up, Vany... | proof | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 7,043 |
1. Anthracene - aromatic, Fluoranthene - antiaromatic, Aceanthrylene - antiaromatic. Total: (10 points)
## TASK № 5
The compound $\mathrm{C}_{9} \mathrm{H}_{10}$ decolorizes bromine water and a solution of $\mathrm{KMnO}_{4}$ at room temperature. Upon heating with an aqueous solution of $\mathrm{KMnO}_{4}$, para-phth... | # Solution:
para-Phthalic acid is an aromatic compound with side chains. The aromatic ring does not decolorize bromine water and does not react with potassium permanganate solution at room temperature. These properties are characteristic of an unsaturated compound. An unsaturated fragment built from two carbon atoms i... | 1-vinyl-4-methylbenzene,para-methylstyrene | Other | math-word-problem | Yes | Yes | olympiads | false | 7,057 |
# Problem 3. Cryptoarithmetic
Problem text. Nurlan has submitted an application to the Spring School of the University for Information Security and needs to solve a test assignment on decrypting the proposed message. As such, a mathematical operation over words is used, where each character hides a certain numerical v... | Solution to the problem. From the structure of the addition operation, it can be immediately noticed that $\mathrm{A}=9, \mathrm{P}=0, Д=\mathrm{M}+1$. S cannot be zero, and since the sum $\mathrm{C}+\mathrm{C}+\mathrm{C}$ returns the same digit, the value of $\mathrm{C}=5$.
For the remaining unknown letters, a system... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,074 | |
# Task 4. Synthesis of Discrete Devices
Task text. When comparing drilling data received at the Central Processing Center with actual production values, a significant discrepancy was identified. A service inspection conducted at the field revealed a malfunction of the sensors, which were decided to be completely repla... | To solve the problem, we first need to obtain a formula for the function \( y(a, b, c) \) that contains only inversions, conjunctions, and modulo-two sums. Notice that when \( a = 0 \) (the upper half of the table), the column of values matches the column of values for the modulo-two sum \( b \oplus c \), and when \( a... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,075 |
# Task 5. Business Process Automation
Problem statement. The Sustainability Committee of an oil and gas company has announced a competition to fill three intern positions. The position is an entry-level role within this structure, but it already requires considerable erudition, backed by real achievements. To send out... | ```
def \(\operatorname{concurs}(n):\)
commands \(=[]\)
for i in range( \(n\) ):
name_command = ' '.join(list(sorted(input().split())))
commands.append (name_command)
n_commands = commands.copy()
commands = set(commands)
winer = max(commands, key = lambda x: n_commands.count(x))
... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,076 |
# Problem 6. Game Theory
Problem text. A subsidiary of an oil and gas holding company is in charge of building gas transportation infrastructure oriented towards new markets. The first of its divisions has expertise only in pipeline, while the second has expertise only in maritime transportation and have difficulties ... | Solution to the problem. First, we determine the upper and lower game values and conclude whether a solution in pure strategies is possible. For Pipeline Route Schemes (PRS), the maximin, i.e., the lower game value, can be calculated by selecting the maximum number from the two minimum numbers in each row: \(\max \min ... | \mathrm{PRS}^{*}=(\frac{3}{8},\frac{5}{8})\mathrm{MRS}^{*}=(\frac{19}{24},\frac{5}{24},0,0)v=\frac{19}{8} | Other | math-word-problem | Yes | Yes | olympiads | false | 7,077 |
# Task 8. Number Systems in System Administration
Task text. In a workers' settlement of shift workers, a network failure occurred as a result of which one of the devices in the gas extraction complex began to return not the standard display of the device's IP address and subnet mask, but its binary representation. Th... | To convert an IP address and subnet mask from binary to decimal, each of them needs to be divided into octets — combinations of 8 numbers separated by dots. Thus, we get:
IP address: 10110010.10110000.11100110.10101010
Subnet mask: 11111111.11111111.11111111.10000000
Next, each bit of each octet is converted by summ... | 178.176.230.128 | Other | math-word-problem | Yes | Yes | olympiads | false | 7,079 |
Task 10. Logic
Text of the task. Sigismund, Asema, Marina, and Sasha received an invitation to Career Days at their university. Due to conference week, they had little time, so on that day each could have detailed talks with only one company. Two students went to meet with Eastern Gas, one with Transgas Space Systems,... | Solution to the problem. This problem can be solved by comparing the facts in the table. Let's mark the number of students who attended the meeting with each company. Then exclude for Sizigmund the meeting with Transgas Space Systems. By comparing combinations, we establish that Asema and Marina attended the meeting wi... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,081 |
Task 2. (10 points) A numerical sequence is given:
$x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1$.
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2021$. | # Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,083 |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2021)$, if $f(1)=5, f(4)=2$. | # Solution.
Substituting the pairs of numbers \(a=4, b=1\) and \(a=1, b=4\) into the given equation, respectively, we get
If \(a=4, b=1\), then \(f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{2+2 \cdot 5}{3}=4, f(2)=4\).
If \(a=1, b=4\), then \(f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3... | -2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,084 |
Task 4. (20 points) A two-meter gas pipe has rusted in two places. Determine the probability that all three resulting sections can be used as offsets for gas stoves, if according to regulations, the stove should not be located closer than 50 cm to the main gas pipe.
# | # Solution.
Let's denote the sizes of the parts into which the pipe was cut as $x, y$, and (200-x-y).
Obviously, the values of $x$ and $y$ can take any values from the interval [0;200]. Then, the set of all possible combinations ( $x ; y$ ) can be represented on the coordinate plane XOY as a right-angled triangle wit... | \frac{1}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,085 |
Task 5. (20 points) Three compressor stations are located not on the same straight line, but are connected by straight roads. The distance from the first station to the third via the second is three times longer than the direct route between them; the distance from the first station to the second via the third is $a$ k... | # Solution.
Let $x$ be the distance between the first and second compressor stations, $y$ the distance between the second and third, and $z$ the distance between the first and third (Fig. 2).
 A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere of radius 3. Segment $C D$ is a diameter of this sphere. Find the volume of the prism if $A D=2 \sqrt{6}$. | Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (Fig.... | 6\sqrt{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,087 |
Task 1. (5 points) Solve the equation $x^{9}-2021 x^{3}+\sqrt{2022}=0$. | Solution.
Rewrite the equation as $x^{9}-2022 x^{3}+x^{3}+\sqrt{2022}=0$. Then
$$
\begin{aligned}
& x^{3}\left(x^{6}-2022\right)+x^{3}+\sqrt{2022}=0 \\
& x^{3}\left(x^{3}-\sqrt{2022}\right)\left(x^{3}+\sqrt{2022}\right)+\left(x^{3}+\sqrt{2022}\right)=0 \\
& \left(x^{3}+\sqrt{2022}\right)\left(x^{6}-x^{3} \sqrt{2022}+... | {-\sqrt[6]{2022};\sqrt[3]{\frac{\sqrt{2022}\\sqrt{2018}}{2}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,088 |
Task 2. (10 points) A numerical sequence is given:
$$
x_{0}=\frac{1}{n} ; x_{k}=\frac{1}{n-k}\left(x_{0}+x_{1}+\ldots+x_{k-1}\right) ; k=1,2, \ldots, n-1
$$
Find $S_{n}=x_{0}+x_{1}+\ldots+x_{n-1}$, if $n=2022$. | # Solution.
We will prove by mathematical induction that $x_{0}+x_{1}+\ldots+x_{k-1}+x_{k}=\frac{1}{n-k}$. For $k=0$, this equality holds. For $k=1$, $x_{0}+x_{1}=\frac{1}{n}+\frac{1}{n-1} \cdot \frac{1}{n}=\frac{1}{n-1}$. Suppose this equality holds for all $m \leq k$, then it should also hold for $k+1$. Let's check ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,089 |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2021)$, if $f(1)=1, f(4)=7$.
# | # Solution.
Substituting the pairs of numbers $a=4, b=1$ and $a=1, b=4$ into the given equation, respectively, we get
If $a=4, b=1$, then $f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{7+2 \cdot 1}{3}=3, f(2)=3$.
If $a=1, b=4$, then $f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\fr... | 4041 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,090 |
Task 4. (20 points) A three-meter gas pipe has rusted in two places. Determine the probability that all three resulting sections can be used as connections to gas stoves, if according to regulations, the stove should not be located closer than 75 cm to the main gas pipe.
# | # Solution.
Let the sizes of the parts into which the pipe was cut be denoted as $x, y$, and $(300-x-y)$.
Obviously, the values of $x$ and $y$ can take any values from the interval (0; 300). Then, the set of all possible combinations $(x; y)$ can be represented on the coordinate plane $OXY$ as a right-angled triangle... | \frac{1}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,091 |
Task 5. (20 points) Three compressor stations are located not on the same straight line, but are connected by straight roads. The distance from the first station to the third via the second is three times longer than the direct route between them; the distance from the first station to the second via the third is $a$ k... | # Solution.
Let $x$ be the distance between the first and second compressor stations, $y$ the distance between the second and third, and $z$ the distance between the first and third (Fig. 2).
 A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere. Segment $C D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $A A_{1}$. Find the volume of the prism if $C K=2 \sqrt{6}, D K=4$. | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (Fi... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,093 |
Task 1. (5 points) Solve the equation $x^{9}-2022 x^{3}+\sqrt{2021}=0$.
# | # Solution.
Rewrite the equation as $x^{9}-2021 x^{3}-x^{3}+\sqrt{2021}=0$.
## then
$x^{3}\left(x^{6}-2021\right)-\left(x^{3}-\sqrt{2021}\right)=0$,
$x^{3}\left(x^{3}-\sqrt{2021}\right)\left(x^{3}+\sqrt{2021}\right)-\left(x^{3}-\sqrt{2021}\right)=0$,
$\left(x^{3}-\sqrt{2021}\right)\left(x^{6}+x^{3} \sqrt{2021}-1\r... | {\sqrt[6]{2021};\sqrt[3]{\frac{-\sqrt{2021}\45}{2}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,094 |
Task 2. (10 points) It is known that the function $f(x)$ for each value of $x \in(-\infty ;+\infty)$ satisfies the equation $f(x)+(0.5+x) f(1-x)=1$. Find all such functions $f(x)$. | # Solution.
Substitute the argument $(1-x)$ into the equation and write the system of equations:
$\left\{\begin{array}{l}f(x)+(0.5+x) f(1-x)=1, \\ f(1-x)+(0.5+1-x) f(1-1+x)=1 ;\end{array}\left\{\begin{array}{l}f(x)+(0.5+x) f(1-x)=1, \\ f(1-x)+(1.5-x) f(x)=1 .\end{array}\right.\right.$
Solve the system by the method ... | f(x)={\begin{pmatrix}\frac{1}{0.5-x},x\neq0.5,\\0.5,0.50\end{pmatrix}.} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,095 |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 55 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | # Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similar... | 1705 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,096 |
Task 4. (20 points) A meter-long gas pipe has rusted in two places. Determine the probability that all three resulting sections can be used as offsets for gas stoves, if according to regulations, the stove should not be located closer than 25 cm to the main gas pipe.
# | # Solution.
Let's denote the sizes of the parts into which the pipe was cut as $x, y$, and $(100-x-y)$.
Obviously, the values of $x$ and $y$ can take any values from the interval (0; 100). Then, the set of all possible combinations $(x; y)$ can be represented on the coordinate plane $OXY$ as a right-angled triangle w... | \frac{1}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,097 |
Task 5. (20 points) A lot consisting of three packages of shares from oil extraction companies, Razneft, Dvaneft, and Trineft, is up for auction. The total number of shares in the packages of Razneft and Dvaneft is equal to the number of shares in the Trineft package. The package of Dvaneft shares is 4 times cheaper th... | # Solution.
Let's introduce the following notations:
$x$ - the price of one share of Dvanefte,
$y$ - the price of one share of Raznefte,
$z$ - the price of one share of Trinefte,
$n$ - the number of shares in a Dvanefte package,
$m$ - the number of shares in a Raznefte package.
The other conditions of the proble... | 12.5to15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,098 |
Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere of radius 6. Segment $C D$ is a diameter of this sphere. Find the volume of the prism if $A D=4 \sqrt{6}$. | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (Fi... | 48\sqrt{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,099 |
Task 1. (5 points) Solve the equation $x^{9}-2022 x^{3}-\sqrt{2021}=0$.
# | # Solution.
Rewrite the equation as $x^{9}-2021 x^{3}-x^{3}-\sqrt{2021}=0$.
$$
\begin{aligned}
& \text { then } \\
& x^{3}\left(x^{6}-2021\right)-\left(x^{3}+\sqrt{2021}\right)=0, \\
& x^{3}\left(x^{3}-\sqrt{2021}\right)\left(x^{3}+\sqrt{2021}\right)-\left(x^{3}+\sqrt{2021}\right)=0, \\
& \left(x^{3}+\sqrt{2021}\righ... | {-\sqrt[6]{2021};\sqrt[3]{\frac{\sqrt{2021}\45}{2}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,100 |
Task 2. (10 points) It is known that the function $f(x)$ for each value of $x \in(-\infty ;+\infty)$ satisfies the equation $f(x)-(x-0.5) f(-x-1)=1$. Find all such functions $f(x)$. | # Solution.
Substitute the argument ($-x-1$) into the equation and write the system of equations:
$\left\{\begin{array}{l}f(x)-(x-0.5) f(-x-1)=1, \\ f(-x-1)-(-x-1-0.5) f(x+1-1)=1 ;\end{array}\left\{\begin{array}{l}f(x)-(x-0.5) f(-x-1)=1, \\ f(-x-1)+(x+1.5) f(x)=1 .\end{array}\right.\right.$
Solve the system by the m... | f(x)={\begin{pmatrix}\frac{1}{0.5+x},x\neq-0.5,\\0.5,-0.50\end{pmatrix}.} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,101 |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 41 and 24, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | Solution.
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of tria... | 984 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,102 |
Task 4. (20 points) A four-meter gas pipe has rusted in two places. Determine the probability that all three resulting sections can be used as offsets for gas stoves, if according to regulations, the stove should not be located closer than 1 m to the main gas pipe.
# | # Solution.
Let's denote the sizes of the parts into which the pipe was cut as $x, y$, and $(400 - x - y)$.
Obviously, the values of $x$ and $y$ can take any values from the interval $(0; 400)$. Then, the set of all possible combinations $(x; y)$ can be represented on the coordinate plane $OXY$ as a right-angled tria... | \frac{1}{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,103 |
Task 5. (20 points) A lot consisting of three packages of shares from oil extraction companies, Razneft, Dvaneft, and Trineft, is up for auction. The total number of shares in the packages of Razneft and Dvaneft is equal to the number of shares in the Trineft package. The package of Dvaneft shares is 3 times cheaper th... | # Solution.
Let's introduce the following notations:
$x$ - the price of one share of Dvanefte,
$y$ - the price of one share of Raznefte,
$z$ - the price of one share of Trinefte,
$n-$ the number of shares in the Dvanefte package,
$m-$ the number of shares in the Raznefte package.
The other conditions of the prob... | 15to25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,104 |
Task 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere. Segment $C D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $A A_{1}$. Find the volume of the prism if $C K=2 \sqrt{3}, D K=2 \sqrt{2... | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along circles circumscribed around the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sp... | 9\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,105 |
Task 1. (5 points) Solve the equation $x^{6}-2020 x^{2}-\sqrt{2021}=0$.
---
The above text has been translated into English, maintaining the original format and line breaks. | Solution.
Rewrite the equation as $x^{6}-2021 x^{2}+x^{2}-\sqrt{2021}=0$. Then
$$
\begin{aligned}
& x^{2}\left(x^{4}-2021\right)+\left(x^{2}-\sqrt{2021}\right)=0, \\
& x^{2}\left(x^{2}-\sqrt{2021}\right)\left(x^{2}+\sqrt{2021}\right)+\left(x^{2}-\sqrt{2021}\right)=0, \\
& \left(x^{2}-\sqrt{2021}\right)\left(x^{4}+x^{... | {\\sqrt[4]{2021}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,106 |
Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds. The sequence must contain a term $a_{k}=2021$. Determine the maximum number of three-digit numbers, divisible by 25, that ... | # Solution.
The final sequence can contain all three-digit numbers, as it can consist of a given number of natural numbers starting from the chosen number $a_{i}$.
We will prove that for any term of the arithmetic progression $1,2,3, \ldots$ defined by the formula for the $n$-th term $a_{n}=n$, the equality $a_{k+2}=... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,107 |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2022)$, if $f(1)=1, f(4)=7$.
# | # Solution.
Substituting the pairs of numbers \(a=4, b=1\) and \(a=1, b=4\) into the given equation, respectively, we get
If \(a=4, b=1\), then \(f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{7+2 \cdot 1}{3}=3, f(2)=3\).
If \(a=1, b=4\), then \(f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3... | 4043 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,108 |
Task 4. (20 points) In the center of a circular field stands a geologists' cabin. From it, 8 straight roads extend, dividing the field into 8 equal sectors. Two geologists set out on a journey from their cabin at a speed of 5 km/h along a road each arbitrarily chooses. Determine the probability that the distance betwee... | # Solution.
Let's find the distance between the geologists after 1 hour if they are walking on adjacent roads (Fig. 1).
Fig. 1
By the Law of Cosines: \( x^{2}=5^{2}+5^{2}-2 \cdot 5 \cdot 5... | 0.375 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,109 |
Task 5. (20 points) At the first deposit, equipment of the highest class was used, and at the second deposit, equipment of the first class was used, with the highest class being less than the first. Initially, $40 \%$ of the equipment from the first deposit was transferred to the second. Then, $20 \%$ of the equipment ... | # Solution.
Let there initially be $x$ units of top-class equipment at the first deposit and $y$ units of first-class equipment at the second deposit $(x1.05 y$, from which $y48 \frac{34}{67} .\end{array}\right.\right.$
This double inequality and the condition "x is divisible by 5" is satisfied by the unique value $x... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,110 |
Task 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1} \mathrm{c}$ is inscribed in a sphere with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$. Segment $C D$ is the diameter of this sphere, point $K$ and $L$ are the midpoints of edge $A A_{1}$ and $A B$ respectively. Find the volume of the... | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$, with their centers at points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (F... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,111 |
Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this se... | # Solution.
Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that an arithmetic progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equation $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.
In... | 225 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,112 |
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2022)$, if $f(1)=5, f(4)=2$. | # Solution.
Substituting the pairs of numbers $a=4, b=1$ and $a=1, b=4$ into the given equation, respectively, we get
If $a=4, b=1$, then $f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{2+2 \cdot 5}{3}=4, f(2)=4$.
If $a=1, b=4$, then $f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\fr... | -2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,113 |
Task 4. (20 points) In the center of a circular field stands a geologists' cabin. From it, 8 straight roads extend, dividing the field into 8 equal sectors. Two geologists set out on a journey from their cabin at a speed of 4 km/h along a road each arbitrarily chooses. Determine the probability that the distance betwee... | # Solution.
Let's find the distance between the geologists after 1 hour if they are walking on adjacent roads (Fig. 1).
Fig. 1
By the cosine theorem: \( x^{2}=4^{2}+4^{2}-2 \cdot 4 \cdot 4... | 0.375 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,114 |
Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ is inscribed in a sphere with the base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$. The segment $C_{1} D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $C C_{1}$. Find the volume of the prism if $D K=2, D A=\sq... | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$, with their centers at points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (F... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,116 |
Task 2. (10 points) Find the greatest value of the parameter $b$ for which the inequality $b \sqrt{b}\left(x^{2}-10 x+25\right)+\frac{\sqrt{b}}{\left(x^{2}-10 x+25\right)} \leq \frac{1}{5} \cdot \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right|$ has at least one solution. | Solution.
$$
\begin{aligned}
& b \sqrt{b}\left(x^{2}-10 x+25\right)+\frac{\sqrt{b}}{\left(x^{2}-10 x+25\right)} \leq \frac{1}{5} \cdot \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right| \\
& b \sqrt{b}(x-5)^{2}+\frac{\sqrt{b}}{(x-5)^{2}} \leq \frac{1}{5} \sqrt[4]{b^{3}} \cdot\left|\sin \frac{\pi x}{10}\right| \\
... | 0.0001 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,117 |
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 367 and 6, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$. | # Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the simila... | 2202 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,118 |
Task 4. (20 points) In the center of a circular field stands a geologists' cabin. From it, 6 straight roads extend, dividing the field into 6 equal sectors. Two geologists set out on a journey from their cabin at a speed of 5 km/h along a road each arbitrarily chooses. Determine the probability that the distance betwee... | # Solution.
Let's find the distance between the geologists after 1 hour if they are walking on adjacent roads (Fig. 2).
$60^{\circ}$
Fig. 2
Since the triangle is equilateral, $x=5$, whic... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,119 |
Task 5. (20 points) Three compressor stations are located not on the same straight line, but are connected by straight roads. The distance from the first station to the third via the second is four times longer than the direct route between them; the distance from the first station to the second via the third is longer... | Solution.
Let $x$ be the distance between the first and second compressor stations, $y$ the distance between the second and third, and $z$ the distance between the first and third (Fig. 4).
... | 0<<68;60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,120 |
Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ is inscribed in a sphere with the base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$. Segment $C D$ is the diameter of this sphere, point $K$ and $L$ are the midpoints of edge $A A_{1}$ and $A B$ respectively. Find the volume of the pri... | # Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$, with their centers at points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (F... | 12\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,121 |
Task 2. (10 points) Find the greatest value of the parameter $m$ for which the inequality $m \sqrt{m}\left(x^{2}-6 x+9\right)+\frac{\sqrt{m}}{\left(x^{2}-6 x+9\right)} \leq \sqrt[4]{m^{3}} \cdot\left|\cos \frac{\pi x}{5}\right|$ has at least one solution.
# | # Solution.
1st method.
$m \sqrt{m}\left(x^{2}-6 x+9\right)+\frac{\sqrt{m}}{\left(x^{2}-6 x+9\right)} \leq \sqrt[4]{m^{3}} \cdot\left|\cos \frac{\pi x}{5}\right|$,
$m \sqrt{m}(x-3)^{2}+\frac{\sqrt{m}}{(x-3)^{2}} \leq \sqrt[4]{m^{3}} \cdot\left|\cos \frac{\pi x}{5}\right|$,
$\sqrt[4]{m^{3}}(x-3)^{2}+\frac{1}{\sqrt[4... | 0.0625 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,122 |
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