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Problem 4.7. Denis threw darts at four identical dartboards: he threw exactly three darts at each board, where they landed is shown in the figure. On the first board, he scored 30 points, on the second - 38 points, on the third - 41 points. How many points did he score on the fourth board? (For hitting each specific zo... | Answer: 34.
Solution. "Add" the first two dart fields: we get 2 hits in the central field, 2 hits in the inner ring, 2 hits in the outer ring. Thus, the sum of points on the first and second fields is twice the number of points obtained for the fourth field.
From this, it is not difficult to get the answer
$$
(30+38... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,863 |
Problem 4.8. In a grove, there are trees of four species: birches, firs, pines, and aspens. There are a total of 100 trees. It is known that among any 85 trees, there will be trees of all four species. Among what minimum number of any trees in this grove will there definitely be trees of at least three species? | Answer: 69.
Solution. Suppose there are no more than 15 birches in the grove. Then there are at least 85 other trees, and according to the problem's condition, among them, there must be trees of all four types. This is a contradiction. Therefore, there must be at least 16 birches in the grove. Similarly, we can conclu... | 69 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,864 |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,865 |
Problem 5.2. Alina, Bohdan, Vika, and Grisha came to the physical education lesson in shorts and T-shirts, and each of these items of clothing was either blue or red. Alina and Bohdan had red T-shirts, and their shorts were of different colors. Vika and Grisha had T-shirts of different colors, and their shorts were blu... | Answer: Alina - red T-shirt and red shorts, Bogdan - red T-shirt and blue shorts, Vika - blue T-shirt and blue shorts, Grisha - red T-shirt and blue shorts.
Solution. Alina and Vika have different T-shirts according to the condition, so Vika has a blue T-shirt. Then Grisha has a red T-shirt. Therefore, the blue T-shir... | Alina-redT-shirtredshorts,Bogdan-redT-shirtblueshorts,Vika-blueT-shirtblueshorts,Grisha-redT-shirtblueshorts | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,866 |
Problem 5.3. By September 1st, Vlad bought himself several ballpoint and gel pens. He noticed that if all the pens he bought were gel pens, he would have paid 4 times more than he actually did. And if all the pens were ballpoint pens, the purchase would have cost him 2 times less than the actual price. How many times m... | Answer: 8.
Solution. If all the pens were gel pens, their price would be 4 times the actual price, which in turn is 2 times more than if all the pens were ballpoint pens. Therefore, gel pens cost $4 \cdot 2=8$ times more than ballpoint pens. Consequently, one gel pen is 8 times more expensive than one ballpoint pen. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,867 |
Problem 5.4. Arrange the digits from 1 to 6 (each must be used exactly once) so that the sum of the three numbers located on each of the 7 lines is equal to 15. In your answer, indicate which digits should be placed at positions $A-F$.
![](https://cdn.mathpix.com/cropped/2024_05_06_6da73bfd3e09e8b55e3fg-09.jpg?height=... | Answer: $A=4, B=1, C=2, D=5, E=6, F=3$.
Solution. According to the condition, $A, D, E$ are different digits, not exceeding 6, the sum of which is 15. If these digits are taken to be the maximum possible, then their sum is $4+5+6=15$. Therefore, $A, D, E$ are $4,5,6$ in some order (if at least one of the digits is no ... | A=4,B=1,C=2,D=5,E=6,F=3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,868 |
Problem 5.5. The houses of Andrey, Boris, Vova, and Gleb are located in some order on a straight street. The distance between Andrey's and Boris's houses, as well as the distance between Vova's and Gleb's houses, is 600 m. What can the distance in meters between Andrey's and Gleb's houses be, if it is known that it is ... | Answer: $900, 1800$.
Solution. For brevity, we will denote the houses of the residents by the first letter of their names. We will denote the distance between the houses of the corresponding people by two capital letters in a row.
Without loss of generality, A is to the left of B (otherwise, we will look at everythin... | 900,1800 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,869 |
Problem 5.6. Vanya received three sets of candies as a New Year's gift. In the sets, there are three types of candies: lollipops, chocolate, and jelly. The total number of lollipops in all three sets is equal to the total number of chocolate candies in all three sets, as well as the total number of jelly candies in all... | Answer: 29.
Solution. There are more lollipops than jelly candies in the first set by 7, and in the second set by 15. Since the total number of each type of candy is the same in all sets, and there are 0 lollipops in the third set, there must be $7+15=22$ jelly candies in the third set.
Similarly, there are more loll... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,870 |
Problem 5.7. Jerry the mouse decided to give Tom the cat a square cake $8 \times 8$ for his birthday. In three pieces marked with the letter "P," he put fish, in two pieces marked with the letter "K," he put sausage, and in one piece, he added both, but did not mark it (all other pieces are without filling). Jerry also... | # Answer: 5.
Solution. Let's call a piece with fish and sausage a coveted piece.
According to the problem, in any $6 \times 6$ square, there are at least 2 pieces with fish. In any such square, at least one known piece with fish is already included; consider the squares that contain only 1 known piece of fish (all of... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,871 |
Problem 5.8. There are 7 completely identical cubes, on which 3 points are marked on one face, 2 points on two faces, and 1 point on the rest. These cubes were glued together to form a figure in the shape of the letter "П", as shown in the image, with the number of points on any two adjacent faces being the same.
 is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the leng... | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,877 |
Problem 6.6. Four girls and eight boys came to take a class photo. The children approach the photographer in pairs and take a joint photo. Among what minimum number of
photos will there definitely be either a photo of two boys, or a photo of two girls, or two photos of the same children? | Answer: 33.
Solution. Suppose at some point there is neither a photo of two boys, nor a photo of two girls, nor two photos of the same children. Then, on each photo, there is a boy and a girl, and on different photos, there are different pairs. However, the total number of possible pairs consisting of a boy and a girl... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,878 |
Problem 6.8. A natural number $n$ is called good if 2020 gives a remainder of 22 when divided by $n$. How many good numbers exist? | Answer: 10.
Solution. In the subsequent solution, the expression of the form $a^{k}$ - the number $a$ raised to the power of $k$ - is the number $a$ multiplied by itself $k$ times. We will also consider $a^{0}=1$. For example, $3^{2}=3 \cdot 3=9$, and $2^{0}=1$.
Since 2020 gives a remainder of 22 when divided by $n$,... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,880 |
Problem 7.1. Petya wrote 20 natural numbers 1, 2, .., 20 on the board. Vasya first erased all even numbers, and then erased all numbers that give a remainder of 4 when divided by 5. How many numbers are left on the board | Answer: 8.
Solution. After erasing all even numbers, the following remain on the board:
$$
\begin{array}{llllllllll}
1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 & 17 & 19 .
\end{array}
$$
Among them, exactly two give a remainder of 4 when divided by 5 - these are 9 and 19. After erasing these as well, 8 numbers remain on the b... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,881 |
Problem 7.2. Denis divided a triangle into nine smaller triangles, as shown in the figure, and placed numbers in them, with the numbers in the white triangles being equal to the sums of the numbers in the adjacent (by sides) gray triangles. After that, Lesha erased the numbers 1, 2, 3, 4, 5, and 6 and wrote the letters... | Answer: a1 b3 c2 d5 e6 f4.
Solution. Note that the number 6 can be uniquely represented as the sum of three numbers from the set of numbers from 1 to 6, which is $6=1+2+3$ (or the same numbers in a different order).
Now let's look at the numbers $B, D$, and $E$. The maximum value of the sum $D+E$ is the sum $5+6=11$,... | a1b3c2d5e6f4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,882 |
Problem 7.3. The pages in a book are numbered as follows: the first sheet is two pages (with numbers 1 and 2), the second sheet is the next two pages (with numbers 3 and 4), and so on. The hooligan Petya tore out several consecutive sheets from the book: the first torn page has the number 185, and the number of the las... | Answer: 167.
Solution. Since any leaf ends with a page number that is an even number, the number of the last torn-out page is either 158 or 518. But 158 does not work, since 158 < 185. Therefore, the last page ends with the number 518. Now let's calculate the number of torn-out pages. Among the pages from 1 to 518, th... | 167 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,883 |
Problem 7.5. A rectangle was cut into nine squares, as shown in the figure. The lengths of the sides of the rectangle and all the squares are integers. What is the smallest value that the perimeter of the rectangle can take?
 = d$, then $(x + y) \vdots d$, i.e., the distance between the cities is divisible by all the calculated GCDs.
Now suppose the distance between the cities (let's call it $S$) is divisible by some n... | 39 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,886 |
Problem 7.7. In the election for class president, Petya and Vasya competed. Over three hours, 27 students in the class voted for one of the two candidates. In the first two hours, Petya received 9 more votes than Vasya. In the last two hours, Vasya received 9 more votes than Petya. In the end, Petya won. By what maximu... | Answer: 9.
Solution. In the last two hours, at least 9 people voted for Vasya. This means that in the end, he received at least 9 votes. Then Petya received no more than $27-9=18$ votes. Therefore, his lead in votes does not exceed $18-9=9$.
Now let's provide an example of how a lead of 9 votes could be achieved. Sup... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,887 |
Problem 7.8. Carlson and Little Man have several jars of jam, each weighing an integer number of pounds.
The total weight of all the jars of jam that Carlson has is 13 times the total weight of all the jars that Little Man has. Carlson gave Little Man the jar with the smallest weight (from those he had), after which t... | Answer: 23.
Solution. All variables in the solution will be natural numbers, since the weights of all jars are integers by condition.
Let Little have a total of $n$ pounds of jam initially, then Karlson had $13 n$ pounds of jam. Let Karlson give away his smallest jar with $a$ pounds of jam. Then, by condition, $13 n-... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,888 |
Problem 8.1. A square was cut into four equal rectangles, and from them, a large letter P was formed, as shown in the figure, with a perimeter of 56.
What is the perimeter of the original squ... | Answer: 32.
Solution. Let the width of the rectangle be $x$. From the first drawing, we understand that the length of the rectangle is four times its width, that is, it is equal to $4 x$. Now we can calculate the dimensions of the letter P.
. Therefore, it must contain the num... | 25 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,890 |
Problem 8.3. Four children were walking along an alley and decided to count the number of firs planted along it.
- Anya said: "There are a total of 15 firs along the alley."
- Borya said: "The number of firs is divisible by 11."
- Vера said: "There are definitely fewer than 25 firs."
- Gena said: "I am sure that their... | Answer: 11.
Solution. Let $N-$ be the number of elms along the alley.
Suppose Genya told the truth, and $N$ is divisible by 22. But then $N$ is also divisible by 11, i.e., Borya also told the truth. But according to the problem, one of the boys was wrong. Therefore, Genya must have been wrong, but Borya was right. So... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,891 |
Problem 8.4. In a class, there are 20 students. Thinking about which girls to send a Valentine's card to on February 14, each boy made a list of all the girls in the class he finds attractive (possibly an empty list). It is known that there do not exist three boys whose lists have the same number of girls. What is the ... | Answer: 6.
Solution. Let the number of girls in the class be $d$. According to the condition, there are no three boys whose lists match in the number of girls, so there can be a maximum of 2 empty lists, 2 lists with one girl, 2 lists with two girls, ..., 2 lists with $d$ girls. This means that the number of lists, an... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,892 |
Problem 8.5. At the ball, ladies and gentlemen arrived - in total less than 50 people. During the first dance, only a quarter of the ladies were not invited to dance, and 2/7 of the total number of gentlemen did not invite anyone. How many people came to the ball? (For the dance, a certain gentleman invites a certain l... | Answer: 41.
Solution. Let the number of ladies be $n$, and the number of gentlemen be $m$. We will count the number of pairs who danced. On one hand, it is $\frac{3}{4} n$, and on the other hand, $\frac{5}{7} m$. By equating, we find the ratio
$$
\frac{n}{m}=\frac{20}{21}
$$
Since the fraction on the right side is i... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,893 |
Problem 8.6. For quadrilateral $ABCD$, it is known that $AB=BD, \angle ABD=\angle DBC, \angle BCD=90^{\circ}$. A point $E$ is marked on segment $BC$ such that $AD=DE$. What is the length of segment $BD$, if it is known that $BE=7, EC=5$?
Fig. 3: to the solution of problem 8.6
Solution. In the isosceles triangle $ABD$, drop a perpendicular from point $D$, let $H$ be its foot (Fig. 3). Since this triangle is acute... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,894 |
Problem 8.7. For three real numbers $p, q$, and $r$, it is known that
$$
p+q+r=5, \quad \frac{1}{p+q}+\frac{1}{q+r}+\frac{1}{p+r}=9
$$
What is the value of the expression
$$
\frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r} ?
$$ | Answer: 42.
Solution. Multiply the two given equalities, we get
$$
5 \cdot 9=\frac{p+q+r}{p+q}+\frac{p+q+r}{q+r}+\frac{p+q+r}{p+r}=\left(1+\frac{r}{p+q}\right)+\left(1+\frac{p}{q+r}\right)+\left(1+\frac{q}{p+r}\right)
$$
Subtract 3 from both sides of the equation, we get
$$
\frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r}... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,895 |
Problem 8.8. Masha wrote on the board in ascending order all natural divisors of some number $N$ (the very first divisor written is 1, the largest divisor written is the number $N$ itself). It turned out that the third from the end divisor is 21 times greater than the second from the beginning. What is the largest valu... | Answer: 441.
Solution. The second divisor from the beginning is the smallest prime divisor of the number $N$, let's denote it as $p$. The third divisor from the beginning is either $p^{2}$ or the second largest prime divisor of the number $N$, let's denote it as $q$.
Case 1. The third divisor from the beginning is $p... | 441 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,896 |
Problem 9.2. Anton, Vasya, Sasha, and Dima were driving from city A to city B, each taking turns at the wheel. The car traveled the entire distance at a constant speed.
Anton drove for half the time Vasya did, and Sasha drove as much as Anton and Dima combined. Dima was behind the wheel for only one-tenth of the journ... | Answer: 0.4.
Solution. Let Anton drive $a$ kilometers, Vasya - $b$ kilometers, Sasha - $c$ kilometers, and Dima - $d$ kilometers.
Anton drove the car twice as little as Vasya. Therefore, $2a = b$.
Sasha drove the car as much as Anton and Dima combined. Therefore, $c = a + d$. Writing the chain of equalities $2a + 2c... | 0.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,898 |
Problem 9.3. To 30 palm trees in different parts of an uninhabited island, a sign is nailed.
- On 15 of them, it is written: "Exactly under 15 signs, treasure is buried."
- On 8 of them, it is written: "Exactly under 8 signs, treasure is buried."
- On 4 of them, it is written: "Exactly under 4 signs, treasure is burie... | Answer: 15.
Solution. Suppose the treasure is not buried under at least 16 plaques. Then there are two plaques with different inscriptions under which there is no treasure. According to the condition, the inscriptions on both should be true, but they contradict each other. Contradiction.
Therefore, the treasure is no... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,899 |
Problem 9.4. From left to right, intersecting squares with sides $12, 9, 7, 3$ are depicted respectively. By how much is the sum of the black areas greater than the sum of the gray areas?
 | Answer: 103.
Solution. Let's denote the areas by $A, B, C, D, E, F, G$.
We will compute the desired difference in areas:
$$
\begin{aligned}
A+E-(C+G) & =A-C+E-G=A+B-B-C-D+D+E+F-F-G= \\
& =... | 103 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,900 |
Problem 9.5. Buratino has many coins of 5 and 6 soldi, more than 10 of each type. Coming to the store and buying a book for $N$ soldi, he realized that he could not pay for it without receiving change. What is the greatest value that the natural number $N$ can take if it is no more than 50? | Answer: 19.
Solution. It is easy to check that with $N=19$ change is necessary.
Notice that numbers from 20 to 24 do not meet the condition, as they can be paid for without change: $N=20=4 \cdot 5, 21=3 \cdot 5+6, 22=2 \cdot 5+2 \cdot 6, 23=5+3 \cdot 6, 24=4 \cdot 6$.
It is clear that then numbers from 25 to 50 also... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,901 |
Problem 9.6. At a ball, 29 boys and 15 girls arrived. Some boys danced with some girls (no more than once in each pair). After the ball, each person told their parents how many times they danced. What is the maximum number of different numbers the children could have mentioned? | Answer: 29.
Solution. The largest possible number that could have been named is 29 (in the case where there is a girl who danced with all the boys), and the smallest is 0. Thus, we have that the number of different numbers named is no more than 30.
We will prove that it cannot be exactly 30. Suppose the opposite, tha... | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,902 |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,903 |
Problem 9.8. How many pairs of natural numbers $a$ and $b$ exist such that $a \geqslant b$ and
$$
\frac{1}{a}+\frac{1}{b}=\frac{1}{6} ?
$$ | Answer: 5.
Solution. From the condition, it follows that $\frac{1}{6}=\frac{1}{a}+\frac{1}{b} \leqslant \frac{2}{b}$, from which $b \leqslant 12$. Also, $\frac{1}{6}=\frac{1}{a}+\frac{1}{b}>\frac{1}{b}$, so $b>6$. Therefore, $b$ can take values from 7 to 12 inclusive.
Using $\frac{1}{a}=\frac{1}{6}-\frac{1}{b}=\frac{... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,904 |
Problem 10.1. In each cell of a $5 \times 5$ table, a natural number is written in invisible ink. It is known that the sum of all the numbers is 200, and the sum of three numbers inside any $1 \times 3$ rectangle is 23. What is the central number in the table?
We get 8 rectangles $1 \... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,905 |
Problem 10.2. It is known that $\frac{a+b}{a-b}=3$. Find the value of the expression $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$. | Answer: 0.6.
Solution. Multiplying the equality $\frac{a+b}{a-b}=3$ by the denominator, we get $a+b=3a-3b$. Moving $a$ to the right and $3b$ to the left, we get $4b=2a$, from which $a=2b$. Substituting $a=2b$ into the second expression, we get
$$
\frac{4b^2-b^2}{4b^2+b^2}=0.6
$$ | 0.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,906 |
Problem 10.3. Yura has $n$ cards, on which numbers from 1 to $n$ are written. After losing one of them, the sum of the numbers on the remaining cards turned out to be 101. What number is written on the lost card? | Answer: 4.
Solution. Suppose that $n \leqslant 13$. Then $1+2+\ldots+n=\frac{n(n+1)}{2} \leqslant 91101$, a contradiction.
Therefore, $n=14$, and the missing number is $1+2+\ldots+14-101=105-101=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,907 |
Task 10.4. In the central cell of a $21 \times 21$ board, there is a chip. In one move, the chip can be moved to an adjacent cell by side. Alina made 10 moves. How many cells can the chip end up in? | Answer: 121.
Solution. We will paint the entire board in a checkerboard pattern so that the central cell of the board is black. With each move of the chip to an adjacent cell, the color of the cell on which the chip is standing will change. After an odd number of moves, the chip will always be on a white cell, and aft... | 121 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,908 |
Problem 10.5. Hooligan Vasya loves to run on the escalator in the subway, and he runs down twice as fast as he runs up. If the escalator is not working, it takes Vasya 6 minutes to run up and down. If the escalator is moving down, it takes Vasya 13.5 minutes to run up and down. How many seconds will it take Vasya to ru... | Answer: 324.
Fig. 6: to the solution of problem 10.4
Solution. Let the length of the escalator be 1, the speed of Vasya going down the escalator be $x$, and the speed of the escalator be $y$... | 324 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,909 |
Problem 10.6. The graph of the quadratic trinomial $y=\frac{2}{\sqrt{3}} x^{2}+b x+c$ intersects the coordinate axes at three points $K, L$, and $M$, as shown in the figure below. It turns out that $K L=K M$ and $\angle L K M=120^{\circ}$. Find the roots of the given trinomial.
. According to the problem, triangle $O M K$ is a right triangle with a $30^{\circ}$ angle at vertex $M$, so $O M=\sqrt{3} K O=\sqrt{3} p$ and $K M=2 K O=2 p$. Al... | 0.51.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,910 |
Problem 10.7. Oleg has four cards, on each of which natural numbers are written on both sides (a total of 8 numbers are written). He considers all possible quadruples of numbers, where the first number is written on the first card, the second on the second, the third on the third, and the fourth on the fourth. Then, fo... | Answer: 21.
Solution. Let the numbers on one card be $a$ and $b$, on another card - $c$ and $d$, on the third card - $e$ and $f$, and on the fourth card - $g$ and $h$. According to the problem, the sum of 16 terms of the form $a c e g + a c e h + \ldots + b d f h$ equals 330. Note that this sum is also obtained by exp... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,911 |
Problem 10.8. Rectangle $ABCD$ is such that $AD = 2AB$. Point $M$ is the midpoint of side $AD$. Inside the rectangle, there is a point $K$ such that $\angle AMK = 80^{\circ}$ and ray $KD$ is the bisector of angle $MKC$. How many degrees does angle $KDA$ measure?
.
Using the fact that in the inscribed quadrilateral $K M D C$ the sum of opposite angles is $180^{\circ}$, we get $\angle M K D=\frac{\angle M K C}{2}=\frac{180^{\circ}-\... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,912 |
Problem 11.1. Inside a circle, 16 radii of the circle and 10 concentric circles, whose centers coincide with the center of the circle, are drawn. Into how many regions do the radii and circles divide the circle? | Answer: 176.
Solution. 10 circles divide the circle into 10 rings and one smaller circle, a total of 11 parts. 16 radii divide each of the 11 parts into 16 more. In total, there are $11 \cdot 16=176$ regions. | 176 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,913 |
Problem 11.2. Along a road, 25 poles are standing in a row. Sometimes a goldfinch lands on one of the poles, and immediately a goldfinch flies away from one of the adjacent poles (if there was at least one goldfinch on the adjacent poles at that moment). Also, no more than one goldfinch can sit on each pole.
Initially... | Answer: 24.
Solution. First, we will show that it is impossible to occupy all the poles. Suppose this happened. Consider the last tit that sat down. Since it occupied the last unoccupied pole, there must have been an occupied pole next to it. Consequently, the tit sitting on that pole must have flown away. Contradicti... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,914 |
Problem 11.3. A natural number $n$ is called interesting if $2 n$ is a perfect square, and $15 n$ is a perfect cube. Find the smallest interesting number.
Answer: 1800. | Solution. Let's factorize the number $n$ into prime factors. For a number to be a square, it is necessary that in this factorization all prime numbers appear in even powers, and for a number to be a cube, it is necessary that all prime numbers appear in powers divisible by 3.
Let's see what power of two divides $n$. F... | 1800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,915 |
Problem 11.4. Senya has three straight sticks, each 24 centimeters long. Senya broke one of them into two pieces so that he could form the outline of a right-angled triangle with the two pieces of this stick and the two whole sticks. How many square centimeters is the area of this triangle? | Answer: 216.
Solution. Let Senya break one of the sticks into parts of lengths $a$ and $24-a$, which we will call small, while sticks of length 24 will be called large. From four sticks, Senya formed a triangle, so one of the sides consists of two sticks, and the other two sides consist of one stick each.
Let's see w... | 216 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,916 |
Problem 11.5. At the call of the voivode, 55 soldiers arrived: archers and swordsmen. All of them were dressed either in golden or black armor. It is known that swordsmen tell the truth when wearing black armor and lie when wearing golden armor, while archers do the opposite.
- In response to the question "Are you wea... | Answer: 22.
Solution. To the first question, affirmative answers will be given by archers in gold armor and archers in black armor, that is, all archers.
To the second question, affirmative answers will be given by archers in gold armor and swordsmen in gold armor, that is, all soldiers in gold armor.
To the third q... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,917 |
Problem 11.6. Inside the cube $A B C D A_{1} B_{1} C_{1} D_{1}$, there is the center $O$ of a sphere with radius 10. The sphere intersects the face $A A_{1} D_{1} D$ along a circle with radius 1, the face $A_{1} B_{1} C_{1} D_{1}$ along a circle with radius 1, and the face $C D D_{1} C_{1}$ along a circle with radius 3... | Answer: 17.
Solution. Let $\omega$ be the circle that the sphere cuts out on the face $C D D_{1} C_{1}$. From point $O$
Fig. 10: to the solution of problem 11.6
drop a perpendicular $O X$ ... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,918 |
Task 11.7. For what least natural $x$ is the expression
$$
\sqrt{29+\sqrt{x}}+\sqrt{29-\sqrt{x}}
$$
an integer? | Answer: 400.
Solution. If we square the given expression, we get $58+2 \sqrt{29^{2}-x}$. If the initial expression was an integer, then this number is a perfect square. We need to find the smallest $x$, i.e., we need to make this expression the largest possible perfect square. Notice that $58+2 \sqrt{29^{2}-x}<58+2 \s... | 400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,919 |
Problem 11.8. Given an increasing sequence of 8 real numbers. Diana wrote down all possible sequences of 4 consecutive numbers from it. It turned out that two of the five new sequences are arithmetic progressions with differences of 4 and 36, respectively, and one of the sequences is a geometric progression. Find the l... | Answer: 126, 6.
Solution. Suppose the arithmetic and geometric progressions in the condition can intersect in at least three consecutive elements. Let there be three numbers $a, b$, $c$ that form both an arithmetic and a geometric progression. On one hand, this is an arithmetic progression, so $b-a=c-b$, i.e., $2b=a+c... | 126,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,920 |
1. There were 2013 empty boxes. In one of them, 13 new boxes (not nested inside each other) were placed. Thus, there became 2026 boxes. In one of them, 13 new boxes (not nested inside each other) were placed again, and so on. After several such operations, there became 2013 non-empty boxes. How many boxes are there in ... | # Solution.
After each operation, the number of non-empty boxes increases by 1. Initially, there were no non-empty boxes. Therefore, a total of 2013 operations were performed. With each operation, 13 new boxes are added. Thus, we have
$2013 + 13 \times 2013 = 2013 \times 14 = 28182$.
Recommendations for checking.
A... | 28182 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,922 |
2. Prove that the inequality $x^{2}-2 x \sqrt{y-5}+y^{2}+y-30 \geq 0$ holds for any valid values of $x$ and $y$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | Solution. Domain of definition: $x$ - any number, $y \geq 5$.
$x^{2}-2 x \sqrt{y-5}+y^{2}+y-30=(x-\sqrt{y-5})^{2}+y^{2}-25 \geq 0$,
since $(x-\sqrt{y-5})^{2} \geq 0$ and $y^{2}-25 \geq 0$ for $y \geq 5$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,923 |
3. Plot on the coordinate plane the set of points whose coordinates satisfy the equation: $x^{2}\left(y+y^{2}\right)=y^{3}+x^{4}$. | # Solution.
$x^{2} y+x^{2} y^{2}-y^{3}-x^{4}=0$, | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,924 |
3. An electronic clock displays time from 00.00.00 to 23.59.59. How much time during a day does the clock display exactly four digit 3s? | 3. If the display shows ab.cd.mn, then $a \neq 3$, so there are 5 cases where one of the digits $b, c, d, m, n$ is not 3, while the others are 3. a) On the display ab.33.33, where $b \neq 3$. There are 21 such sets.
b) On the display a3.c3.33, where c is not 3. Here a can take any of the three values 0, 1, or 2, and c... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,925 |
4. The numbers from 1 to 10 are written on a board. It is allowed to erase any two numbers $x$ and $y$, and instead write the numbers $x-1$ and $y+3$ on the board. Could it be that after some time, the numbers on the board are $2,3, \ldots, 9,10,2012$? | 4. Suppose we managed to get the numbers $2,3, \ldots$, 9,10,2012 on the board. Notice that after each operation, the sum of the numbers on the board increases by 2. Initially, it was 55. That is, after each operation, the sum of the numbers on the board will be odd. However, the sum $2+3+\ldots+10+2012=2066$ is even. ... | couldnot | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,926 |
5. Around a round table, $2 n(n>5)$ people are sitting - knights and liars. Liars give a false answer to any question, while knights give a true answer. Each of them knows who is a knight and who is a liar. Each of them answered two questions: "Who is your left neighbor?", "Who is your right neighbor?". A wise man, who... | 5. Two adjacent knights (R), like two adjacent liars (L), give answers "R" and "R", while adjacent R and L give answers "L" and "L". If the number of L is less than \( n-1 \), then in a group of consecutive R, the outermost R can be replaced by L, and we get the same set of answers with a different number of R. Also, L... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,927 |
9.5. Find all numbers $a$ such that for any natural $n$ the number $\operatorname{an}(n+2)(n+4)$ is an integer.
(O. Podlipsky) | Answer. $a=\frac{k}{3}$, where $k$ is any integer.
Solution. Substituting $n=1$ and $n=2$, we get that the numbers $15a$ and $48a$ are integers. Therefore, the number $48a - 3 \cdot 15a = 3a$ is also an integer. Thus, $a = \frac{k}{3}$ for some integer $k$.
It remains to show that all numbers of this form work. Indee... | \frac{k}{3} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,928 |
9.6. Initially, three distinct points were marked on a plane. Every minute, some three of the marked points were chosen - let's denote them as $A, B$, and $C$, after which a point $D$ symmetric to $A$ with respect to the perpendicular bisector of $B C$ was marked on the plane.
After a day, it turned out that among the... | Solution. Suppose the opposite; then the initial three points lie on some circle $\omega$. We will prove by induction on the number of minutes that all marked points also lie on $\omega$. Indeed, initially, this is true. Let at some moment points $A, B, C$ be used to construct point $D$. Then the perpendicular bisector... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,929 |
9.7. Find all triples of prime numbers $p, q, r$ such that the fourth power of any of them, decreased by 1, is divisible by the product of the other two.
(V. Senderov) | # Answer. $2,3,5$.
Solution. It is clear that any two numbers in the triplet are distinct (if $p=q$, then $p^{4}-1$ does not divide $q$). Let $p$ be the smallest of the numbers in the triplet. We know that the number $p^{4}-1=(p-1)(p+1)\left(p^{2}+1\right)$ is divisible by $q r$. Note that $p-1$ is less than any of th... | 2,3,5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,930 |
9.8. A straight rod 2 meters long was sawn into $N$ sticks, the length of each of which is expressed as an integer number of centimeters. For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle?
(A. Magazinov) | Answer. $N=102$.
Solution. First solution. Let $N \leqslant 101$. Cut the stick into $N-1$ sticks of 1 cm each and one stick of $(201-N)$ cm. It is impossible to form a rectangle from this set, as each side of the rectangle is less than half the perimeter, and thus the stick of length $201-N \geqslant 100$ cm cannot b... | 102 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,931 |
9.1 A stock market player doubles his capital every working day (from Monday to Friday inclusive). But on Saturday and Sunday, he parties, and for each party day, he spends $75\%$ of his wealth. By the evening of December 31, he became a millionaire for the first time in his life. On which day of the week did September... | Solution. The trader's capital decreases by a factor of 4 each weekend day, meaning it decreases by a factor of 16 over Saturday and Sunday. Over the next 4 days, by Thursday evening, it can only recover. Therefore, an increase in capital (compared to the previous week) is only possible on Friday. This means that Decem... | Wednesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,932 |
9.2 About the coefficients $a, b$, $c$, and $d$ of the two quadratic trinomials $x^{2}+b x+c$ and $x^{2}+a x+d$, it is known that $0<a<b<c<d$. Can these trinomials have a common root? Justify your answer. | Solution. Let $x_{0}$ be the common root of these trinomials. Then the system of equations $\left\{\begin{array}{l}x_{0}^{2}+b x_{0}+c=0 \\ x_{0}^{2}+a x_{0}+d=0\end{array}\right.$ is satisfied. By equating the left parts of the equations, we get the equality $b x_{0}+c=a x_{0}+d$, from which $x_{0}=\frac{d-c}{b-a}$. T... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,933 |
9.3 Vasya is training on the rink. He placed three pucks at the vertices of a triangle, and then hits one of the pucks so that it (moving in a straight line) passes through the goal formed by the other two pucks.
a) Can all three pucks return to their original positions after 7 shots?
b) Can they return to the vertic... | Solution. a) Suppose it is possible. Let the initial (and final) positions of the pucks be denoted by the letters $A, B, C$. We will traverse the triangle along the path $A \rightarrow$ $B \rightarrow C \rightarrow A$; without loss of generality, we can assume that the traversal is clockwise. Consider what will happen ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,934 |
9.4 In the thirtieth kingdom, there are three types of coins in circulation: bronze rubles, silver coins worth 9 rubles, and gold coins worth 81 rubles. From the treasury, which contains an unlimited supply of each type of coin, a certain amount was issued with 23 coins, which is less than 700 rubles. Find this amount,... | Solution. Since the issued amount is less than 700 rubles, the number of gold coins is less than $700: 81$, which means no more than 7. Then the number of bronze and silver coins together is at least $23-7=16$. Note that the number of bronze coins cannot exceed 8: otherwise, we can replace 9 bronze coins with 1 silver ... | 647 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,935 |
9.5 In some cells of a $10 \times 10$ table, crosses are placed such that each of them is the only one either in its row or in its column. What is the maximum number of crosses that can be in such a table? Justify your answer. | Solution. We will place crosses in all cells of the first row and the first column, excluding the top-left cell - a total of 18 crosses. Then each cross in the first row is unique in its column, and each cross in the first column is unique in its row. Therefore, it is possible to place 18 crosses.
We will show that if... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,936 |
9.6 Two equal segments $A B$ and $C D$ are perpendicular, with point $C$ lying on segment $A B$. Point $X$ is such that $B X = X C$ and $A X = X D$.
a) Prove that triangle $B X C$ is a right triangle.
b) Prove that triangle $A X D$ is a right triangle. | Solution. Method 1. Triangles $A X B$ and $C X D$ are equal by three sides. Hence, angle $X C D$ is equal to angle $X B A$. Since triangle $B X C$ is isosceles, $\angle X B C = \angle X C B$, therefore ray $C X$ is the bisector of the right angle $B C D$. Triangle $B X C$ is then isosceles with base angles of $45^{\cir... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,937 |
1. Solve the equation: $\sin \left(\frac{\pi}{2} \cos x\right)=\cos \left(\frac{\pi}{2} \sin x\right)$. | Answer: $\frac{\pi}{2}+\pi k, \quad k \in Z ; \quad 2 \pi k, \quad k \in Z$.
## Solution:
Using the reduction formula, we transform the right side of the equation:
$$
\sin \left(\frac{\pi}{2} \cos x\right)=\sin \left(\frac{\pi}{2}-\frac{\pi}{2} \sin x\right)
$$
Equality of sines is possible in one of the following ... | \frac{\pi}{2}+\pik,\quadk\inZ;\quad2\pik,\quadk\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,938 |
2. Dima claims that he came up with a quadratic equation that has two negative roots. The next day, Dima rearranged the coefficients of this equation and claims that now the equation has two positive roots. Could Dima be right? | Answer: No.
## Solution:
If both roots of the quadratic equation $a x^{2}+b x+c=0$ are positive, then the sum of the roots is positive, but by Vieta's theorem it is equal to $-b / a$, so $a$ and $b$ have different signs.
If both roots of the equation $a_{1} x^{2}+b_{1} x+c_{1}=0$ are negative, then the sum of the ro... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,939 |
3. The teacher wrote on the board the fraction $\frac{a n+b}{c n+d}$ (numbers $a, b, c, d, n$ are natural). Vova was able to reduce the fraction (numerator and denominator) by 2017. Prove that then $a d - b c$ is divisible by 2017. | # Solution:
The numerator and denominator are divisible by 2017, so the value expressed through them, $a d - b c = a \cdot (c n + d) - c \cdot (a n + b)$, is also divisible by 2017.
## Criteria:
Considering particular cases of values $a, b, c, d, n$ in points is not evaluated. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,940 |
4. On the sides $B C$ and $A D$ of the convex quadrilateral $A B C D$, their midpoints - points $M$ and $N$ respectively - are marked. Segments $M N$ and $A C$ intersect at point $O$, and $M O=O N$. It is known that the area of triangle $A B C$ is 2017. Find the area of quadrilateral $A B C D$. | Answer: 4034.
## Solution:
Let $A C$ and $M N$ intersect at point $O$ (see the first figure on the right), $S_{\triangle A B C}=2017=S$.
We will prove that $S_{\triangle A D C}=S_{\triangle A B C}$, then $S_{A B C D}=2 S$.
There are various ways to reason.
$. It turns out that $\angle A B D=\angle B C D$. Find the length of segment $B D$, if $B C=36$ and $A D=64$.
 | Answer: 48.
Solution. Since $A D \| B C$, we have $\angle C B D=\angle B D A$. Then triangles $A B D$ and $D C B$ are similar by the first criterion. Therefore, $\frac{64}{B D}=\frac{A D}{B D}=\frac{B D}{B C}=\frac{B D}{36}$, from which we find $B D=\sqrt{64 \cdot 36}=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,953 |
Problem 8.3.1. As a homework exercise, Tanya was asked to come up with 20 examples of the form $*+*=*$, where $*$ should be replaced with different natural numbers (i.e., a total of 60 different numbers should be used). Tanya loves prime numbers very much, so she decided to use as many of them as possible, while still ... | # Answer: 41.
Solution. Note that in each example, instead of asterisks, three odd numbers cannot be used, i.e., at least one even number must be used. There is exactly one even prime number, which is 2. Therefore, among the 60 different numbers in the examples, at least 19 even composite numbers will be used, and the... | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,954 |
Problem 8.4.1. A rectangle was cut into six smaller rectangles, the areas of five of them are marked on the diagram. Find the area of the remaining rectangle.
 | Answer: 101.
Solution. Let's introduce the notation as shown in the figure.
Notice that
$$
2=\frac{40}{20}=\frac{S_{H I L K}}{S_{D E I H}}=\frac{H K \cdot H I}{H D \cdot H I}=\frac{H K}{H ... | 101 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,955 |
Problem 8.5.1. In the cells of a $12 \times 12$ table, natural numbers are arranged such that the following condition is met: for any number in a non-corner cell, there is an adjacent cell (by side) that contains a smaller number. What is the smallest number of different numbers that can be in the table?
(Non-corner c... | Answer: 11.
Solution. First, let's provide an example with the arrangement of 11 different numbers.
| 1 | 2 | 3 | 4 | 5 | 6 | 6 | 5 | 4 | 3 | 2 | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 3 | 4 | 5 | 6 | 7 | 7 | 6 | 5 | 4 | 3 | 2 |
| 3 | 4 | 5 | 6 | 7 ... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,956 |
Problem 8.6.1. Even natural numbers $a$ and $b$ are such that $\operatorname{GCD}(a, b) + \operatorname{LCM}(a, b) = 2^{23}$. How many different values can $\operatorname{LCM}(a, b)$ take? | Answer: 22.
Solution. Note that LCM $(a, b) \vdots$ GCD $(a, b)$, therefore
$$
2^{23}=\text { GCD }(a, b)+\text { LCM }(a, b) \vdots \text { GCD }(a, b) .
$$
From this, it follows that GCD $(a, b)$ is a natural divisor of the number $2^{23}$.
At the same time, GCD $(a, b) \neq 1$ (since $a$ and $b-$ are even number... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,957 |
Problem 8.7.1. Given trapezoid $A B C D (B C \| A D)$. Point $H$ on side $A B$ is such that $\angle D H A=$ $90^{\circ}$. It is known that $C H=C D=13$ and $A D=19$. Find the length of segment $B C$.
.
Figure for the solution to problem 8.7.1
In the isosceles triangle $HCD$, we have $\angle C... | 9.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,958 |
Problem 8.8.1. Different positive numbers $a, b, c$ are such that
$$
\left\{\begin{array}{l}
a^{2}+b c=115 \\
b^{2}+a c=127 \\
c^{2}+a b=115
\end{array}\right.
$$
Find $a+b+c$. | Answer: 22.
Solution. Subtract the third equation from the first and transform:
$$
\begin{gathered}
\left(a^{2}+b c\right)-\left(c^{2}+a b\right)=0 \\
a^{2}-c^{2}+b c-a b=0 \\
(a-c)(a+c)+b(c-a)=0 \\
(a-c)(a+c-b)=0
\end{gathered}
$$
By the condition $a \neq c$, therefore $b=a+c$. Now add the first two equations:
$$
... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,959 |
9.1. Find the ratio $\frac{b^{2}}{a c}$ if it is known that one of the roots of the equation $a x^{2}+b x+c=0$ is four times the other $(a c \neq 0)$. | Answer: $\frac{25}{4}$.
Solution: Let the equation have roots $x_{1}, x_{2}\left(x_{1}=4 x_{2}\right)$. Then, from Vieta's formulas, we get: $\quad x_{1}+x_{2}=5 x_{2}=-\frac{b}{a}, \quad x_{1} x_{2}=4 x_{2}^{2}=\frac{c}{a} . \quad$ Hence, $\quad x_{2}=-\frac{b}{5 a}, \quad$ and $4 x_{2}^{2}=\frac{c}{a}=4\left(-\frac{... | \frac{25}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,962 |
9.2. Let $x, y, z$ be non-zero numbers. Prove that among the inequalities: $x+y>0$, $y+z>0, z+x>0, x+2 y<0, y+2 z<0, z+2 x<0$ at least two are false. | First solution. Assume the opposite. Among three non-zero numbers, there will be two of the same sign - let these be $x$ and $y$. Then one of the inequalities $x+y>0$ and $x+2y>0$ holds. Similarly, if $a$ and $b$ are of the same sign, one of the inequalities $a+b>0$ and $a+2b>0$ holds. Contradiction.
In the second cas... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,963 |
9.4. In triangle $A B C$, where $A B>A C$, the bisector $A L$ is drawn. A point $K$ is chosen on side $A B$ such that $A K=A C$. Let $O$ be the center of the circumcircle of triangle $A L B$. Prove that angles $K C B$ and $A B O$ are equal. | First solution. The inscribed angle $B A L$ is half the central angle $B O L$, so $\angle B O L = 2 \angle B A L = \angle K A C$ (see Fig. 2). Therefore, the angles $\angle K A C$ and $\angle B O L$ are equal angles at the vertices of isosceles triangles $K A C$ and $B O L$, so $\angle A K C = \angle O B C$. But the an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,965 |
9.5. The chess piece "centaur" moves alternately as a knight and as a white pawn (i.e., strictly one square upwards). Can it, starting from some square on an $8 \times 8$ chessboard, visit every square exactly once, if the first move is made as a pawn? The starting square is considered visited. | Answer: It cannot.
Solution: Consider the chessboard coloring of our board. Note that both the pawn and the knight change the color of the square with each move. Suppose the centaur starts its tour of the board with a pawn move from a white square. Then it will land on a black square, and the next move (as a knight) w... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,966 |
11.6. Given four consecutive natural numbers greater than 100. Prove that from these numbers, one can choose three numbers whose sum can be represented as the product of three different natural numbers greater than 1.
(N. Agakhanov) | Solution. Let $n, n+1, n+2, n+3$ be the given numbers. The sum of the three smallest of them is $3n+3=3(n+1)$, and the sum of the three largest numbers is $3(n+2)$. But at least one of the numbers $n+1$ and $n+2$ is even, that is, equal to the product of the numbers 2 and $k$, where $k>3$. Therefore, the given sum can ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,967 |
11.7. Given a positive number $a \neq 1$. Prove that the sequence $x_{1}, x_{2}, \ldots$, where $x_{n}=2^{n}(\sqrt[2^{n}]{a}-1)$, is decreasing. (A. Khryabrov) | Solution. Let $t=\sqrt[2^{n+1}]{a}$. Note that $t \neq 1$. Then $x_{n+1}=2^{n+1}(t-1)$ and $x_{n}=2^{n}\left(t^{2}-1\right)$. Therefore,
\[
\begin{aligned}
x_{n}-x_{n+1}= & 2^{n}\left(t^{2}-1\right)-2^{n+1}(t-1)= \\
& =2^{n}\left(t^{2}-2 t+1\right)=2^{n}(t-1)^{2}>0
\end{aligned}
\]
which is what we needed to prove.
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,968 |
11.8. On the sides $AB$ and $AC$ of triangle $ABC$, points $D$ and $E$ are found such that $DB = BC = CE$. Segments $BE$ and $CD$ intersect at point $P$. Prove that the circumcircles of triangles $BDP$ and $CEP$ intersect at the center of the inscribed circle of triangle $ABC$.
(R. Zhenodarov) | Solution. Let $I$ be the center of the inscribed circle of triangle $ABC$, point $I$ is the intersection of the angle bisectors. We will prove that points $B, D, P, I$ lie on the same circle. Similarly, we will show that points $C, E, P, I$ lie on the same circle, and the problem will be solved.
It is sufficient to es... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,969 |
11.9. In a class, there are $m$ students. During September, each of them went to the swimming pool several times; no one went twice on the same day. On October 1st, it turned out that all the numbers of swimming pool visits by the students were different. Moreover, for any two of them, there was definitely a day when t... | Answer. $m=28$.
Solution. For each natural $n$, let $X_{n}=$ $=\{1,2, \ldots, n\}$. To each student, we associate the set of all days when he went to the pool (this will be a subset of $X_{30}$). Thus, we have obtained a set of $m$ (according to the condition, non-empty) subsets of $X_{30}$. The condition is equivalen... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,970 |
11.10. Given a natural number $n \geqslant 2$. Petya and Vasya play the following game. Petya chooses $2 n$ (not necessarily distinct) non-negative numbers $x_{1}, x_{2}, \ldots, x_{2 n}$, the sum of which is 1. Vasya arranges these numbers in a circle in some order of his choosing. After that, he calculates the produc... | Answer: $\frac{1}{8(n-1)}$.
Solution: If Petya chooses the numbers $0, \frac{1}{2}, \frac{1}{4(n-1)}, \frac{1}{4(n-1)}, \ldots, \frac{1}{4(n-1)}$, then no matter how Vasya arranges these numbers, the number $\frac{1}{2}$ will be paired with the number $\frac{1}{4(n-1)}$. Therefore, one of the products will be $\frac{1... | \frac{1}{8(n-1)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,971 |
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