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5.1. Find the smallest natural number $n$ such that the natural number $n^{2}+14 n+13$ is divisible by 68. | Answer: 21
Solution. Note that $n^{2}+14 n+13=(n+1)(n+13)$, and $68=4 \cdot 17$. Therefore, one of the numbers $n+1$ and $n+13$ must be divisible by 17. Moreover, $n$ must be odd; otherwise, $n+1$ and $n+13$ would both be odd. The smallest such natural number $n$ is $2 \cdot 17-13=21$. Note that it works because $21^{... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,028 |
6.1. On September 2nd, Robin Bobin ate 12 chickens, and starting from September 3rd, he ate every day as many as he had already eaten on average in September. On September 15th, Robin Bobin ate 32 chickens. How many chickens did he eat on September 1st? | Answer: 52
Solution. Let at the beginning of the $k$-th day he on average ate $x$ chickens. This means that by this point, he had eaten $(k-1) x$ chickens. Then on the $k$-th day, he also ate $x$ chickens, which means that over $k$ days, he ate $k x$ chickens. This means that after the $k$-th day, the average number o... | 52 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,029 |
7.1. In quadrilateral $A B C D$, $A B=B C=C D$. Let $E$ be the intersection point of $A B$ and $C D$ (with $B$ between $A$ and $E$). It turns out that $A C=C E$ and $\angle D B C=15^{\circ}$. Find $\angle A E D$. | Answer: $50^{\circ}$
Solution. We will solve the general problem. Let $\angle D B C=\alpha$, and $\angle B E C=\beta$. From the isosceles property of $\triangle D C B$, we get $\angle B D C=\alpha$. From the isosceles property of $\triangle A C E$, we get $\angle B E C=\angle B A C$, and from the isosceles property of... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,030 |
8.1. There are 15 rectangular sheets of paper. In each move, one of the sheets is chosen and divided by a straight cut, not passing through the vertices, into two sheets. After 60 moves, it turned out that all the sheets are either triangles or hexagons. How many hexagons? | Answer: 25
Solution. Let after 60 cuts, $x$ hexagons and $75-x$ triangles are formed (there were 15 sheets, and with each cut, the number of sheets increases by 1). Now let's count the number of sides. Initially, there were $15 \cdot 4=60$ sides, and with each cut, the number of sides increases by 4 (the cut divides t... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,031 |
1. The quadratic trinomial $\mathrm{x}^{2}+\mathrm{ax}+\mathrm{b}$ has integer roots whose absolute values are greater than 2. Prove that the number a + b + 1 is composite. | Let $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ be the roots of the given quadratic polynomial. Then, by Vieta's formulas, $a + \mathrm{b} + 1 = -(\mathrm{x}_{1} + \mathrm{x}_{2}) + \mathrm{x}_{1} \mathrm{x}_{2} + 1 = (\mathrm{x}_{1} - 1)(\mathrm{x}_{2} - 1)$. From the condition, it follows that each bracket is not equal to ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,032 |
2. Can the number $\left(\mathrm{x}^{2}+\mathrm{x}+1\right)^{2}+\left(\mathrm{y}^{2}+\mathrm{y}+1\right)^{2}$ be a perfect square for some integers x and y | 2. No, it cannot. If this expression is a perfect square, then it is the square of an even number. The square of any even number is divisible by 4.
Since $x^{2}+x+1=x(x+1)+1$, for any integers x and y, the value of each expression is an odd number. The square of an odd number, when divided by 4, leaves a remainder of ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,033 |
3. Vasya's parents bought him a pen, a book, a backpack, and a football. If these items had cost 12, 6, 4, and 3 times less respectively, they would have paid 1000 rubles. And if they had cost 12, 15, 20, and 30 times less respectively, they would have paid 500 rubles. Which is more expensive: the pen or the backpack? | 3. Let $x, y, z, t$ be the prices of a pen, a book, a backpack, and a ball, respectively, each increased by 60 times. For brevity, the cost of the purchase will be measured in thousands. Then, according to the first condition, $5 \mathrm{x}+10 \mathrm{y}+15 \mathrm{z}+20 \mathrm{t}=6$ or $\mathrm{x}+2 \mathrm{y}+3 \mat... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,034 |
4. In an acute triangle $\mathrm{ABC}$, the bisector $\mathrm{AD}$ and the altitude $\mathrm{BE}$ are drawn. Prove that angle $\mathrm{CED}$ is greater than 45 degrees. | 4. Let K, L, M be the feet of the perpendiculars dropped from point D to AB, AC, BE (points K, L, M lie on segments AB, AC, BE, and not on their extensions, since triangle ABC is acute-angled). Let DK intersect BE at point P. Note that DLEM is a rectangle, and point D is equidistant from sides AB and AC, as it lies on ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,035 |
5. There are 2001 coins on the table. Two players play the following game: they take turns, on each turn the first player can take any odd number of coins from 1 to 99, the second player can take any even number of coins from 2 to 100, and so on. The player who cannot make a move loses. Who wins with correct play? | 5. The first one wins. Winning strategy: the first player should take 81 coins from the table on the first move. On each subsequent move, if the second player takes x coins, then the first player should take (101 - x) coins. He can always do this because if x is an even number from 2 to 100, then (101 - x) is an odd nu... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,036 |
Problem 9.3. Given a convex quadrilateral $ABCD$, $X$ is the midpoint of diagonal $AC$. It turns out that $CD \parallel BX$. Find $AD$, if it is known that $BX=3, BC=7, CD=6$.
 | Answer: 14.
Solution. Double the median $B X$ of triangle $A B C$, to get point $M$. Quadrilateral $A B C M$ is a parallelogram (Fig. 1).
Notice that $B C D M$ is also a parallelogram, since segments $B M$ and $C D$ are equal in length (both 6) and parallel. This means that point $M$ lies on segment $A D$, since $A M... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,037 |
Problem 9.5. Point $M$ is the midpoint of side $B C$ of triangle $A B C$, where $A B=17$, $A C=30, B C=19$. A circle is constructed with side $A B$ as its diameter. An arbitrary point $X$ is chosen on this circle. What is the minimum value that the length of segment $M X$ can take?
, as shown in Fig. 2.
.
. From this similarity and the cyclic nature of the pentagon
. It is known that $AB = AN$, $BC = MC$. The circumcircles of triangles $ABM$ and $CBN$ intersect at points $B$ and $K$. How many degrees does the angle $AKC$ measure if $\angle ABC = 68^\circ$?
. | 4. Answer. $x=1938, y=77$.
Let these numbers be $x$ and $y$. Then $x+y=2015$ and $x=25 y+s$, where $s$ is the remainder. We get that $x+y=$ $26 y+s=2015.26 y=2015-s$. From the inequality $0 \leq s \leq 24$ we obtain
$$
\begin{gathered}
2015-24 \leq 26 y \leq 2015, \text { or } \\
76.5<y<77.5
\end{gathered}
$$
Thus, ... | 1938,77 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,047 |
5. The names of three classmates are Petya, Vasya, and Vitya. Masha knows this, but she doesn't know which boy is which. She can ask them questions that can only be answered with "yes" or "no." Each question is asked to one of the boys, and only he answers it. Natasha knows that Petya will always answer truthfully, Vas... | 5. Let's ask any boy twice in a row: "Is your name Vasya?" Petya will answer "No" twice, Vasya will answer "No" to the first question and "Yes" to the second, and Vitya will say "Yes" both times. This way, we can find out the name of our interlocutor in two questions. Then we will ask him, pointing to any of the other ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,048 |
# 6. Clone 1
The figure shows a hexagon composed of identical equilateral triangles, each with an area of 10. Find the area of the shaded part.
# | # Answer: 110
## Solution
We will call the equilateral triangles that make up the original hexagon unit triangles. Consider each of the highlighted triangles inside the hexagon separately. The smallest one coincides with the unit triangle and has an area of 10. The medium triangle is inscribed in a hexagon made up of... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,051 |
9.1. Let's call the main divisors of a composite number $n$ the two largest natural divisors of $n$ different from $n$. Composite natural numbers $a$ and $b$ are such that the main divisors of $a$ coincide with the main divisors of $b$. Prove that $a=b$.
(A.S. Golyanov) | Solution. Let $n>k$ be the main divisors of the number $a$; then $a / n$ and $a / k$ are the two smallest divisors of the number $a$, greater than one. Let $p$ be the smallest prime divisor of the number $a$, and $q$ be the smallest prime divisor of $a$ other than $p$ (if such exists). Then $a / n = p$. Next, $a / k$ i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,052 |
9.2. The bisectors of triangle $A B C$ intersect at point $I$, and the external bisectors of its angles $B$ and $C$ intersect at point $J$. Circle $\omega_{b}$ with center at point $O_{b}$ passes through point $B$ and is tangent to line $C I$ at point $I$. Circle $\omega_{c}$ with center at point $O_{c}$ passes through... | Answer: $1 / 3$.
First Solution. In circle $\omega_{b}$, draw the diameter $I X$, and in circle $\omega_{c}$, draw the diameter $I Y$. Note that $\angle I B J=90^{\circ} = \angle I C J$, since the internal and external bisectors of an angle are perpendicular. Therefore, point $X$ lies on $B J$, and point $Y$ lies on $... | \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,053 |
9.3. A string of 200 natural numbers is written. Among any two adjacent numbers in the string, the right one is either 9 times greater than the left one, or 2 times smaller than the left one. Can the sum of all these 200 numbers equal $24^{2022}$?
(O. Podlipsky, I. Bogdanov) | Answer: It cannot.
Solution: Let the string consist of numbers $a_{1}, a_{2}, \ldots, a_{200}$ in this order. If the number $a_{i}=2 k$ is even, then the next number can be $k$ or $18 k$; these numbers give the same remainder when divided by 17. If $a_{i}$ is odd, then $a_{i+1}=9 a_{i}$. In any case, we get that $a_{i... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,054 |
9.4. In a class, there are 18 children. The parents decided to gift the children from this class a cake. To do this, they first asked each child the area of the piece they wanted to receive. After that, they ordered a square cake, the area of which is exactly equal to the sum of the 18 named numbers. However, upon seei... | Answer. $k=12$.
Solution. We always assume that the area of the cake is 1. We will show that for some children's requests, the parents will not be able to cut out more than 12 required pieces. Choose a number $1 / 15 > x > 1 / 16$. Suppose that 15 main children ordered a piece of cake with an area of $x$ each (and the... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,055 |
Problem 4.3. Zhenya drew a square with a side of 3 cm, and then erased one of these sides. A figure in the shape of the letter "P" was obtained. The teacher asked Zhenya to place dots along this letter "P", starting from the edge, so that the next dot was 1 cm away from the previous one, as shown in the picture, and th... | Answer: 31.
Solution. Along each of the three sides of the letter "П", there will be 11 points. At the same time, the "corner" points are located on two sides, so if 11 is multiplied by 3, the "corner" points will be counted twice. Therefore, the total number of points is $11 \cdot 3-2=31$. | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,057 |
Problem 4.5. Hooligan Dima made a construction in the shape of a $3 \times 5$ rectangle using 38 wooden toothpicks. Then he simultaneously set fire to two adjacent corners of this rectangle, marked in the figure.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire construc... | Answer: 65.
Solution. In the picture below, for each "node", the point where the toothpicks connect, the time in seconds it takes for the fire to reach it is indicated. It will take the fire another 5 seconds to reach the middle of the middle toothpick in the top horizontal row (marked in the picture).
![](https://cd... | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,058 |
Problem 4.8. The figure shows a road map between the houses of five children. The shortest distance by road from Asya to Galia is 12 km, from Galia to Borya - 10 km, from Asya to Borya 8 km, from Dasha to Galia - 15 km, from Vasya to Galia - 17 km. How many kilometers is the shortest distance by road from Dasha to Vasy... | Answer: 18.
Solution. Add the distances from Dasha to Gala and from Vasya to Gala: $15+17=32$. This will include the "main" road (from Dasha to Vasya) and twice the "branch" from it to Gala.
Add the distance from Asey to Bory to the obtained sum: $32+8=40$. Now all three branches (to Gala - twice) and the main road w... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,059 |
Problem 5.3. The figure shows a plan of the road system of a certain city. In this city, there are 8 straight streets, and 11 intersections are named with Latin letters $A, B, C, \ldots, J, K$.
Three police officers need to be placed at some intersections so that at least one police officer is on each of the 8 streets... | Answer: $B, G, H$.
Solution. The option will work if police officers are placed at intersections $B, G, H$. It can be shown that this is the only possible option.
Since there are only three vertical streets, and each must have one police officer, there are definitely no police officers at intersections $C$ and $K$. T... | B,G,H | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,060 |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,061 |
Problem 5.8. Inside a large triangle with a perimeter of 120, several segments were drawn, dividing it into nine smaller triangles, as shown in the figure. It turned out that the perimeters of all nine small triangles are equal to each other. What can they be equal to? List all possible options.
The perimeter of a fig... | Answer: 40.
Solution. Let's add the perimeters of the six small triangles marked in gray in the following figure:
From the obtained value, subtract the perimeters of the other three small wh... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,062 |
Problem 6.7. Anya places pebbles on the sand. First, she placed one stone, then added pebbles to form a pentagon, then made a larger outer pentagon with pebbles, then another outer pentagon, and so on, as shown in the picture. The number of stones she had arranged on the first four pictures: 1, 5, 12, and 22. If she co... | Answer: 145.
Solution. On the second picture, there are 5 stones. To get the third picture from it, you need to add three segments with three stones on each. The corner stones will be counted twice, so the total number of stones in the third picture will be $5+3 \cdot 3-2=12$.
![](https://cdn.mathpix.com/cropped/2024_... | 145 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,065 |
Problem 7.8. On a rectangular sheet of paper, a picture in the shape of a "cross" was drawn from two rectangles $A B C D$ and $E F G H$, the sides of which are parallel to the edges of the sheet. It is known that $A B=9, B C=5, E F=3, F G=10$. Find the area of the quadrilateral $A F C H$.
![](https://cdn.mathpix.com/c... | Answer: $52.5$.
Solution. The intersection of the two original rectangles forms a "small" rectangle with sides 5 and 3. Its area is 15.
Extend the segments $D A, G H, B C, E F$ to form lines. They form a "large" rectangle with sides 9 and 10, containing the "cross" (Fig. 2). Its area is 90.
![](https://cdn.mathpix.c... | 52.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,067 |
Problem 8.2. In the chat of students from one of the schools, a poll was being held: "On which day to hold the disco: October 22 or October 29?"
The graph shows how the votes were distributed an hour after the start of the poll.
Then, 80 more people participated in the poll, voting only for October 22. After that, th... | Answer: 260.
Solution. Let $x$ be the number of people who voted an hour after the start. From the left chart, it is clear that $0.35 x$ people voted for October 22, and $-0.65 x$ people voted for October 29.
In total, $x+80$ people voted, of which $45\%$ voted for October 29. Since there are still $0.65 x$ of them, ... | 260 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,068 |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,069 |
Problem 9.4. A line $\ell$ is drawn through vertex $A$ of rectangle $ABCD$, as shown in the figure. Perpendiculars $BX$ and $DY$ are dropped from points $B$ and $D$ to line $\ell$. Find the length of segment $XY$, given that $BX=4$, $DY=10$, and $BC=2AB$.
Fig. 5: to the solution of problem 9.4
Solution. Note that since $\angle Y A D=90^{\circ}-\angle X A B$ (Fig. 5), right triangles $X A B$ and $Y D A$ are similar by the acute an... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,070 |
Problem 9.5. Leonid has a white checkered rectangle. First, he painted every other column gray, starting with the leftmost one, and then every other row, starting with the topmost one. All cells adjacent to the border of the rectangle ended up being painted.
How many painted cells could there be in the rectangle if 74... | Answer: 301 or 373.
Solution. From the condition, it follows that the rectangle has an odd number of both rows and columns. Let's number the rows from top to bottom with the numbers $1,2, \ldots, 2 k+1$, and the columns from left to right with the numbers $1,2, \ldots, 2 l+1$ (for non-negative integers $k$ and $l$). W... | 301or373 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,071 |
Problem 9.6. In triangle $A B C$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $A C$, point $K$ is marked, and on side $B C$, points $L$ and $M$ are marked such that $K L=K M$ (point $L$ lies on segment $B M$).
Find the length of segment $L M$, if it is known that $A K=4, B L=31, M C=3... | Answer: 14.
Solution. In the solution, we will use several times the fact that in a right-angled triangle with an angle of $30^{\circ}$, the leg opposite this angle is half the hypotenuse. Drop the height $K H$ from the isosceles triangle $K M L$ to the base (Fig. 7). Since this height is also a median, then $M H=H L=... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,072 |
Problem 10.2. Points $A, B, C, D, E, F, G$ are located clockwise on a circle, as shown in the figure. It is known that $A E$ is the diameter of the circle. Also, it is known that $\angle A B F=81^{\circ}, \angle E D G=76^{\circ}$. How many degrees does the angle $F C G$ measure?
=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,074 |
Problem 11.2. A square was cut into five rectangles of equal area, as shown in the figure. The width of one of the rectangles is 5. Find the area of the square.
 | Answer: 400.
Solution. The central rectangle and the rectangle below it have a common horizontal side, and their areas are equal. Therefore, the vertical sides of these rectangles are equal, let's denote them by $x$ (Fig. 13). The vertical side of the lower left rectangle is $2x$, and we will denote its horizontal sid... | 400 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,075 |
Problem 11.3. In a football tournament, 15 teams participated, each playing against each other exactly once. For a win, 3 points were awarded, for a draw - 1 point, and for a loss - 0 points.
After the tournament ended, it turned out that some 6 teams scored at least $N$ points each. What is the greatest integer value... | # Answer: 34.
Solution. Let's call these 6 teams successful, and the remaining 9 teams unsuccessful. We will call a game between two successful teams an internal game, and a game between a successful and an unsuccessful team an external game.
First, note that for each game, the participating teams collectively earn n... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,076 |
Problem 11.8. Given a parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. A point $X$ is chosen on the edge $A_{1} D_{1}$, and a point $Y$ is chosen on the edge $B C$. It is known that $A_{1} X=5, B Y=3, B_{1} C_{1}=14$. The plane $C_{1} X Y$ intersects the ray $D A$ at point $Z$. Find $D Z$.
. How can we determine t... | # Solution:
It is impossible to solve this with one weighing. Indeed, if one of the pans outweighs, it can mean both that a heavier fake coin is on it and that a lighter fake coin is on the other.
Two weighings are sufficient. For example, we can do the following. First, we place the coins worth 2 and 3 cruzeros on o... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,081 |
7.2. The year 2009 has the property that by rearranging its digits, it is impossible to form a smaller four-digit number (numbers do not start with zero). In which year will this property reoccur for the first time?
(I. V. Raskina) | Answer: in 2022.
Solution. In 2010, 2011, ..., 2019, and in 2021, the year number contains a one, and if it is moved to the first position, the number will definitely decrease. The number 2020 can be reduced to 2002. However, the number 2022 cannot be decreased by rearranging the digits.
## Comment.
3 points. For th... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,083 |
7.3. What angle do the clock hands form at 12:20? | Answer: $110^{\circ}$.
Solution. At 12:00, the clock hands coincide. After this, in 20 minutes, the minute hand travels $1 / 3$ of the circumference, i.e., it describes an angle of $120^{\circ}$. The hour hand moves 12 times slower than the minute hand (since it describes one circle in 12 hours). Therefore, in 20 minu... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,084 |
7.4. Five friends gathered. The following conversation took place between them. Andrey: "I know a secret!". Borya (to Andrey): "You don't know!". Vitya: "Boris, you are wrong!". Gosha (to Vitya): "You are wrong, Vitya!". Dima: "You are lying, Gosha!". It is known that more than half of the boys told the truth. Does And... | Answer: Andrey knows the secret.
Solution. Suppose Andrey lied. Then Borya told the truth, Vitya lied, Gosha told the truth, and Dima lied. This means that three out of five lied, which contradicts the condition. If Andrey told the truth, then Borya lied, Vitya told the truth, Gosha lied, and Dima told the truth, whic... | Andreyknowsthesecret | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,085 |
7.5. Grandfather is 31 times older than his grandson. In how many years will he be 7 times older than his grandson, given that the grandfather is more than 50 but less than 90 years old? | Answer: In 8 years.
Solution: Grandfather's age is divisible by 31. But the only such number greater than 50 and less than 90 is 62. Therefore, the grandfather is 62 years old, and the grandson is 2 years old. In x years, the grandfather will be x+62 years old, and the grandson will be x+2 years old. If at that time h... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,086 |
1. Three candles were lit simultaneously. When the first candle burned out, $\frac{2}{5}$ of the second candle remained, and $\frac{3}{7}$ of the third candle remained. What part of the third candle will remain when the second candle burns out? | # Solution
In time $t$, $\frac{3}{5}$ of the second candle and $\frac{4}{7}$ of the third candle burned.
The burning rate of the second candle is $\frac{3}{5 t}$, and the burning rate of the third candle is $\frac{4}{7 t}$.
$\frac{2}{5}$ of the second candle will burn in $\frac{2}{5} : \frac{3}{5 t} = \frac{2 t}{3}$... | \frac{1}{21} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,087 |
2. Solve the equation $\left(x^{2}-|x|-2\right)^{2}+\sqrt{|2 x+1|-3}=0$. | # Solution
Since each term of the original equation is non-negative, the equation can have solutions if and only if both terms are equal to zero. We obtain the equivalent system of equations
$$
\left\{\begin{array}{l}
x^{2}-|x|-2=0 \\
|2 x+1|-3=0
\end{array}\right.
$$
The solution to this system is $x=-2$.
## Crite... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,088 |
3. On a plane, three lines are drawn, with no more than two of them being parallel. Is it always possible to find a point in this plane that is equidistant from all three lines?
# | # Solution
There can be two cases.
1st - no two lines are parallel. In this case, a triangle is formed when the lines intersect. The point we are looking for is the center of the circle inscribed in this triangle.
2nd - exactly two lines are parallel (Fig. 1). In this case, the perpendicular AB to the parallel lines... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,089 |
4. Is it possible to divide a convex 2017-gon into black and white triangles such that any two triangles either share a side (when colored differently), a vertex, or have no common points, and each side of the 2017-gon is a side of one of the black triangles? | # Solution
Let $n$ be the number of white triangles. Then, there are $3 n$ sides of white triangles, since it follows from the condition that monochromatic triangles do not share sides.
Then, there are $3 n + 2017$ sides of black triangles, since any side of a white triangle is also a side of a black one. However, $3... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,090 |
5. Two players take turns erasing the inscription on the board: "Hello to the participants of the mathematics Olympiad!". In one move, you can erase either any letter, or an exclamation mark, or several identical letters. The player who makes the last move wins. How to play to win
# | # Solution
The second player wins by responding to the opponent's moves as follows.
All characters are divided into pairs:
R $\leftrightarrow \mathrm{B}, \quad U \leftrightarrow \mathrm{Ch}, \quad C \leftrightarrow \mathrm{N}, \quad L \leftrightarrow D, \quad Y \leftrightarrow!, \quad K K \leftrightarrow O O, \quad ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,091 |
11.1. Angles $\alpha$ and $\beta$ are such that $\operatorname{tg} \alpha+\operatorname{tg} \beta=2$, and $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=5$. Find the value of $\operatorname{tg}(\alpha+\beta)$. | Answer: $10 / 3$.
Solution: Transform the condition $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=5: \frac{1}{\operatorname{tg} \alpha}+\frac{1}{\operatorname{tg} \beta}=5$, which is $\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=5$, or $\frac{2}{... | \frac{10}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,092 |
11.2. Is the number $4^{2019}+6^{2020}+3^{4040}$ prime? | Answer: No.
Solution. Transform the expression: $\quad 4^{2019}+6^{2020}+3^{4040}=\left(2^{2}\right)^{2019}+2^{2020} \cdot 3^{2020}+3^{2 \cdot 2020}=$ $\left(2^{2019}\right)^{2}+2 \cdot 2^{2019} \cdot 3^{2020}+\left(3^{2020}\right)^{2}=\left(2^{2019}+3^{2020}\right)^{2}$. Therefore, the original number is the square o... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,093 |
11.3. Prove that in a regular pentagon, the ratio of any diagonal to the side parallel to it is $\frac{\sqrt{5}+1}{2}$. | Solution. From the condition, it follows that $E D C D^{\prime}$ and $D C B C^{\prime}$ are parallelograms (see figure).
Therefore, $E C^{\prime}=E B-C^{\prime} B=E B-D C=E B-E D^{\prime}=D^... | \frac{\sqrt{5}+1}{2} | Geometry | proof | Yes | Yes | olympiads | false | 17,094 |
11.4. Find all pairs $(m, n)$ such that any pair $(x, y)$ satisfying the equation $\frac{x}{y}=m$, satisfies the equation $(x+y)^{2}=n$. | Answer: $m=-1, n=0$.
Solution: Let the pair $(m, n)$ satisfy the condition of the problem.
Pairs $(x, y)$ equal to $(m, 1)$ and $(2m, 2)$ satisfy the first equation. Therefore,
$$
(m+1)^{2}=(2m+2)^{2}=n \Leftrightarrow (m+1)^{2}=4(m+1)^{2}=n \Leftrightarrow m=-1, n=0
$$
since $n=4n \Leftrightarrow n=0$.
Thus, the ... | =-1,n=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,095 |
11.5. A square plot of 14 by 14 cells needs to be paved with rectangular tiles of size $1 \times 4$. The tiles can only be laid along the grid (not diagonally), and the tiles cannot be broken. What is the maximum number of tiles required? Will there be any uncovered area left? | # Solution:
Evaluation. The total number of cells on the plot is $14 \times 14=196$. Dividing by the number of cells in one tile, 196:4 = 49. Therefore, the number of tiles that can be cut from the $14 \times 14$ plot is no more than 49.
We will color the cells of the plot in 4 colors, as shown in the diagram. Clearl... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,096 |
3. A $3 \times 3$ table was filled with prime numbers. It turned out that the sum of the numbers in any two adjacent cells is a prime number. What is the maximum number of different numbers that can be in the table? Provide an example and prove that there cannot be more different numbers.
Fill the $3 \times 3$ table w... | 3. Answer: 6.
Example:
| 5 | 2 | 29 | |
| :---: | :---: | :---: | :---: |
| 2 | 3 | 2 | |
| 11 | 2 | 17 | |
| | -2 | | 0 |
We will prove that it cannot be more. Divide the cells of th... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,097 |
4. Six numbers are written in a row on the board. It is known that each number, starting from the third, is equal to the product of the two preceding numbers, and the fifth number is equal to 108. Find the product of all six numbers in this row. | 4. Answer: 136048896.
Solution. Let the first number be $x$, and the second number be $y$. Then the third number is $xy$.
The fourth number is $xy^2$. The fifth number is $x^2 y^3 = 108$. The sixth number is $x^3 y^3$.
The product of all six numbers is $x^8 y^{12} = (x^2 y^3)^4 = 108^4 = 136048896$.
## Grading Crit... | 136048896 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,098 |
5. On 300 cards, all natural numbers from 1 to 300 are written. The cards are evenly distributed between two players. Each of them lays out pairs of cards such that the difference between the numbers on them is a multiple of 25, until it is no longer possible. Prove that they will lay out the same number of such pairs ... | 5. If both players have no cards left to lay out, then they have laid out pairs of cards equally. Consider a player who has cards left to lay out. Take any of these cards. Let the remainder of dividing the number on such a card by 25 be r. Since the cards have 300 consecutive numbers, there are a total of 300/25=12 num... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,099 |
1. In Grandfather Frost's bag, there are chocolate and jelly candies, a total of 2023 pieces. Chocolate candies make up 75% of the jelly candies. How many chocolate candies are in Grandfather Frost's bag? | Answer: 867. Solution. Chocolate candies make up 3/4 of the quantity of jelly candies. Then the total number of candies is 7/4 of jelly candies. Therefore, the number of jelly candies is 2023×4:7=1156, and the number of chocolate candies is 2023-1156=867.
Grading criteria. A correct and justified solution - 7 points. ... | 867 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,100 |
3. Kikimora, Leshy, and Vodyanoy met. Each of them either always tells the truth or always lies. Kikimora said: «We are all liars.» Leshy replied: «Among us, there are exactly two liars.» Vodyanoy remained silent. Who is the liar and who is not? | Answer: Kikimora and Vodyanoy are liars, Leshy is truthful.
Solution. Kikimora cannot tell the truth, because if she did, she wouldn't be able to say that everyone is a liar. Therefore, there is one or two liars. Suppose there is one liar. Then Leshy is lying. But Kikimora is also lying, so there are more than one lia... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,102 |
4. Friends Petya, Vasya, and Kolya live in the same entrance. Climbing from the 1st to the 3rd floor, Petya passes a number of steps that is a multiple of four. Vasya, climbing from the 1st to the 6th floor, passes a number of steps that is a multiple of nine. How many steps will Kolya pass, who climbs from the 1st to ... | Answer: 144. Solution. From the 1st to the 3rd floor, there are 2 flights of stairs. According to the condition, the number of steps in these two flights is divisible by four. Then the number of steps in one flight is even. From the 1st to the 6th floor, there are 5 flights of stairs. According to the condition, the nu... | 144 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,103 |
1. What is the greatest length of an arithmetic progression of natural numbers
$a_{1}, a_{2}, \ldots, a_{n}$ with a difference of 2, in which for all $k=1,2 \ldots, n$ all numbers $a_{k}^{2}+1$ are prime | Solution.
All members of the progression must be even numbers. Consider the remainders when divided by 5. Since the common difference of the arithmetic progression is 2, the remainders in the progression will appear in the order
$\ldots 2,4,1,3,0,2,4,1,3,0, \ldots$ This means that among the terms of the progression, ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,105 |
2. Given 2019 polynomials of degree 2018, no two of which have common roots. It is known that each of these polynomials has a common root with the sum of the remaining 2018. Prove that the sum of these 2019 polynomials is zero. | Solution.
Let $P_{1}(x), P_{2}(x), P_{3}(x), \ldots, P_{2019}(x)$ be the given polynomials, $H(x)=P_{1}(x)+P_{2}(x)+$ $P_{3}(x)+\ldots+P_{2019}(x)$ be their sum. Note that $H(x)$ is a polynomial of degree no higher than 2018. Let $x_{1}$ be a root of the polynomial $P_{1}(x)$, this root does not coincide with the root... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,106 |
3. On a grid paper, a convex polygon is drawn such that all its vertices are at the grid points, and none of its sides are horizontal or vertical. Prove that the sum of the lengths of the vertical segments of the grid lines enclosed within the polygon is equal to the sum of the lengths of the horizontal segments of the... | Solution.
We will prove that the area of a polygon is equal to the sum of the lengths of the horizontal segments of the grid lines enclosed within the polygon. Let the lengths of the corresponding horizontal segments be \(a_{1}, a_{2}, \ldots, a_{n-1}, a_{n}\). The polygon is divided by the horizontal segments of the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,107 |
4. Quadrilateral $ABCD$ is circumscribed around a circle $\omega \cdot P, Q, R, S$ are the points of tangency of this circle with the sides $AB, BC, CD, DA$ respectively. Incircles are inscribed in triangles $APS, BPQ, CQR, DRS$. The centers of these incircles are denoted as $X, Y, Z, V$ respectively. Prove that the di... | Solution.
Consider triangle $A P S$ and point $X_{1}$ - the midpoint of arc $P S$ of circle $\omega$ (not shown on the diagram, we will now prove that it coincides with point $X$). We will ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,108 |
5. Each spectator who bought a ticket for the first row took one of the seats in the first row. It turned out that all seats in the first row were occupied, but each spectator was sitting in the wrong seat. The usher can swap two neighbors if both are sitting in the wrong seats. Can the usher seat all the spectators in... | Solution.
Number the seats in the first row from left to right from 1 to $n$, and number the spectators by the numbers of their correct seats (written on the tickets). We will move the spectator with number $N$ to the right. If we manage to seat spectator $N$ in their place, the problem will be solved, as we can then ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,109 |
1. First, write the original equation as follows:
$\left((x-1)^{2}-2\right)^{2}+3(x-1)^{2}-16=0$.
Let $y=(x-1)^{2}$, then for $y$ we get the equation $y^{2}-y-12=0$, whose roots are 4 and -3. The real solutions are highlighted as follows: $(x-1)^{2}=$ 4. | Answer: $x_{1}=3, x_{2}=-1$. | x_{1}=3,x_{2}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,110 |
5. From the condition, we get $a^{2 n}=x^{2}+y^{2}+2 x y=a^{m}+2 x y$. Therefore, $a^{2 n}>a^{m}$, which means $a^{2 n}$ is divisible by $a^{m}$. Consequently, $2 x y$ is divisible by $a^{m}=x^{2}+y^{2}$. We obtain that $2 x y \geq x^{2}+y^{2} \geq 2 x y$. Hence, $x^{2}+y^{2}=2 x y, x=y$. Therefore, $2 x=a^{n}, 2 x^{2}... | Answer: $\left(2^{k}, 2^{k}\right)$, where $k \geq 0$. | (2^{k},2^{k}) | Algebra | proof | Yes | Yes | olympiads | false | 17,111 |
1. Vasya thought of a four-digit number and for each pair of its adjacent digits, he wrote down their product on the board. After that, he erased one product, and the numbers 20 and 21 remained on the board. What is the smallest number Vasya could have thought of? | Answer: 3745.
Solution. Since the products 20 and 21 are given, the digits of the thought-of number include 4, 5 and 3, 7, meaning we know all its digits. The digits 4 and 5, 3 and 7 must be adjacent, so the smallest number that meets the conditions is 3745, and Vasya erased the product $7 \cdot 4=28$.
Criteria. Only... | 3745 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,112 |
2. Three friends live in houses with different numbers. It turned out that each of them has a floor number that matches the house number of one of his friends. Can this situation be maintained if
$a)$ one of the friends moves up one floor in his house?
$b)$ each of the friends moves up one floor in his house? | Answer: a) can; b) cannot.
Solution. a) Let's call the three friends Anton, Boris, and Sergey. Suppose Anton lives in a house with number $a$. The number of his floor coincides with the number of the house $b$ of one of his friends, for example, Boris. By condition, $a \neq b$. Suppose Anton moved to the floor above. ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,113 |
3. The last digit of a four-digit number was moved to the beginning (for example, $1234 \rightarrow$ 4123) and the resulting number was added to the original. The sum turned out to be 3333. What is the original number, given that its representation contains no digit 0? Find all possible options. | Answer: 1212 and 2121.
Solution. Let $\overline{x y t}$ be the desired four-digit number, and $\overline{t x y z}$ be the number obtained from it. Their sum is $S=\overline{x y z t}+\overline{t x y z}=11(100 x+10 y+z)+1001 t=11(\overline{x y z}+91 t)=3333$. From this, $\overline{x y z}+91 t=303$. The digit $t$ is diff... | 12122121 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 17,114 |
4. A $10 \times 10$ square was cut into 17 rectangles, each with both side lengths greater than 1. What is the smallest number of squares that could be among these rectangles? Provide an example of such a cutting. | Answer: one square.
Solution. Suppose that among the rectangles, there is not a single square. Let $a$ and $b$ be the sides of an arbitrary rectangle, with $a > b$. Since the integer $b$ is greater than 1, then $b \geqslant 2$, and thus $a \geqslant 3$. Therefore, the area of each such rectangle $a \times b$ is at lea... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,115 |
5. The diagonals of quadrilateral $A B C D$ intersect at point $O$. It is known that $A B=B C=$ $=C D, A O=8$ and $\angle B O C=120^{\circ}$. What is $D O ?$ | Answer: $D O=8$.
Solution. Mark a point $E$ on the line $A C$ such that triangle $B O E$ is equilateral (Fig. 2). We will prove the equality of triangles $B A E$ and $B C O$. Indeed, since $A B=B C$, triangle $A B C$ is isosceles, and thus $\angle B A C=\angle B C A$. Additionally, note another pair of equal angles $\... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,116 |
1. A seagull is being fed from a moving boat. A piece of bread is thrown down, the seagull takes 3 seconds to pick up the piece from the sea surface, and then it takes 12 seconds to catch up with the boat. Upon entering the bay, the boat reduces its speed by half. How much time will it now take the seagull to catch up ... | Solution. Let the speed of the seagull relative to the boat be $x$ (m/s). Then in 12 seconds, it flies $12 x$ (m). The boat covers this distance in 3 seconds, so its speed is $4 x$ (m/s), and the speed of the seagull is $5 x$ (m/s). In the bay, the boat's speed becomes $2 x$ (m/s). In 3 seconds, it travels $6 x$ (m). T... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,117 |
4. In triangle $ABC$, a circle with center $I$ is inscribed, touching side $AB$ at point $K$. On segment $BK$, a point $P$ is chosen such that the tangent to the circle from this point intersects the extension of side $AC$ beyond point $C$ at point $Q$.
Prove that $\angle BIP = \angle CIQ$. | Solution. The center of the circle inscribed in an angle lies on the bisector of this angle. Applying the external angle theorem of a triangle to triangle $B I P$, we get: $\angle B I P=\angle A P I-\angle A B I=\frac{1}{2} \angle A P Q-\frac{1}{2} \angle A B C$ (1). Similarly, for triangle $C I Q: \angle C I Q=\angle$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,118 |
5. In the tournament, 49 chess players participated. At a certain point in the tournament, 83 games had been played, and each participant had played either three or four games (and no one played each other twice). Could it have been the case that no two players who played exactly three games played against each other?
... | Solution. Let $x$ chess players have played three games by this point, then 49-x have played four. In each game, two chess players participate, so the total number of games played is $(3x + 4(49-x))/2 = 83$. From this, $x=30$. Suppose the chess players who played three games did not play against each other. Then all th... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 17,119 |
4-1. Katya attached a square with a perimeter of 40 cm to a square with a perimeter of 100 cm as shown in the figure. What is the perimeter of the resulting figure in centimeters?
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,120 |
4-7. Along a straight alley, 100 lamps are placed at equal intervals, numbered sequentially from 1 to 100. At the same time, from different ends of the alley, Petya and Vasya started walking towards each other at different constant speeds (Petya from the first lamp, Vasya from the hundredth). When Petya was at the 22nd... | Answer. At the 64th lamppost.
Solution. There are a total of 99 intervals between the lampposts. From the condition, it follows that while Petya walks 21 intervals, Vasya walks 12 intervals. This is exactly three times less than the length of the alley. Therefore, Petya should walk three times more to the meeting poin... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,124 |
5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
 | Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$ | 500 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,125 |
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th l... | Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to ... | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 17,126 |
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right ... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,127 |
5-7. On the faces of a die, the numbers $6,7,8,9,10,11$ are written. The die was rolled twice. The first time, the sum of the numbers on the four "vertical" (that is, excluding the bottom and top) faces was 33, and the second time - 35. What number can be written on the face opposite the face with the number 7? Find al... | Answer: 9 or 11.
Solution. The total sum of the numbers on the faces is $6+7+8+9+10+11=51$. Since the sum of the numbers on four faces the first time is 33, the sum of the numbers on the two remaining faces is $51-33=18$. Similarly, the sum of the numbers on two other opposite faces is $51-35=16$. Then, the sum on the... | 9or11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,128 |
6-3. The red segments in the figure have equal length. They overlap by equal segments of length $x$ cm. What is $x$ in centimeters?
 | Answer: 2.5.
Solution. Adding up the lengths of all the red segments, we get 98 cm. Why is this more than 83 cm - the distance from edge to edge? Because all overlapping parts of the red segments have been counted twice. There are 6 overlapping parts, each with a length of $x$. Therefore, the difference $98-83=15$ equ... | 2.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,130 |
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,133 |
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment ... | Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$. | 219 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,134 |
10.5. It is known that a square grid can be cut into $n$ identical figures of $k$ cells. Prove that it can also be cut into $k$ identical figures of $n$ cells.
(S. Volchonok) | Solution. Let the side of the square be $m$. According to the condition, $m \cdot m=n \cdot k$. Let $d=$ GCD $(m, n)$; then $m=m_{1} d, n=n_{1} d$, where GCD $\left(m_{1}, n_{1}\right)=1 ;$ in this case $m_{1} m=n_{1} k$. The first multipliers on both sides of the last equation are coprime, therefore, $m$ is divisible ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 17,138 |
10.6. Does there exist an infinite sequence of natural numbers such that for any natural $k$ the sum of any $k$ consecutive terms of this sequence is divisible by $k+1$?
(S. Berlov) | Answer. No.
Solution. Suppose there exists such a sequence $a_{1}, a_{2}, a_{2}, \ldots$
Let $k$ be any natural number greater than 1. Consider the first $2 k-1$ terms of the sequence $a_{1}, a_{2}, \ldots, a_{2 k-1}$. By our assumption, the sum $a_{1}+a_{2}+\ldots+a_{2 k-1}$ is divisible by $2 k$, and each of the su... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 17,139 |
10.7. In an acute scalene triangle $ABC$, the median $AM$ and the altitude $AH$ are drawn. Points $Q$ and $P$ are marked on the lines $AB$ and $AC$ respectively such that $QM \perp AC$ and $PM \perp AB$. The circumcircle of triangle $PMQ$ intersects the line $BC$ again at point $X$. Prove that $BH = CX$.
(M. Diden) | First solution. Let $P^{\prime}$ and $Q^{\prime}$ be the points symmetric to $P$ and $Q$ with respect to $M$. Consider the triangle $M Q P^{\prime}$. In it, $Q B \perp M P^{\prime}$ by the condition; moreover, $P B P^{\prime} C$ is a parallelogram, so $P^{\prime} B \| P C \perp Q M$. Therefore, $B$ is the orthocenter o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,140 |
10.8. A numismatist has 100 coins that look identical. He knows that among them, 30 are genuine and 70 are counterfeit. Moreover, he knows that the weights of all genuine coins are the same, while the weights of all counterfeit coins are different, and any counterfeit coin is heavier than a genuine one; however, the ex... | Answer: 70.
Solution: 1. We will show that the numismatist can find the genuine coin in 70 weighings. Stack all 100 coins in a pile. With each weighing, he will select two coins from the pile and compare them. If their weights are equal, then both coins are genuine, and the required coin is found. If not, then the hea... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,141 |
11.1. At each point $A$ of the plane, there stands a real number $f(A)$. It is known that if $M$ is the point of intersection of the medians of triangle $ABC$, then $f(M)=f(A)+f(B)+f(C)$. Prove that $f(A)=0$ for all points $A$. | Solution. Let's take an arbitrary point $M$ on the plane and prove that $f(M)=0$. For this, consider an arbitrary triangle $ABC$ for which point $M$ is the centroid. Denote by $D$, $E$, and $F$ the centroids of triangles $BCM$, $CAM$, and $ABM$, respectively (see Fig. 4).
Notice that point $M$ is also
 \operatorname{tg}(a y)=1, \\
\operatorname{tg}(21 x) \operatorname{tg}(b y)=1
\end{array}\right.
$$
has at least one solution
(M. Antipov) | Answer. No.
Solution. We will show that the system will not have solutions when $a=8, b=13$. Indeed, from the equations of the system, it follows that
$$
13 x + a y = \frac{\pi}{2} + \pi k, \quad 21 x + b y = \frac{\pi}{2} + \pi \ell
$$
for integers $k$ and $\ell$. From this, we get
$(21 a - 13 b) y = 21(13 x + a y... | proof | Algebra | proof | Yes | Yes | olympiads | false | 17,143 |
11.3. Given $n$ coins of pairwise distinct masses and $n$ balance scales, $n>2$. In each weighing, it is allowed to choose some one scale, place one coin on each of its pans, look at the scale's reading, and then remove the coins. Some one of the scales (it is unknown which) is broken and may randomly give either a cor... | Answer. In $2 n-1$ weighings.
Solution. First, we prove that the heaviest coin can be found in $2 n-1$ weighings. More precisely, we will prove by induction on $n$ that the heaviest of $n \geqslant 2$ given coins can be determined in $2 n-1$ weighings, using three scales, one of which may be faulty.
If $n=2$, we weig... | 2n-1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 17,144 |
11.4. Given a tetrahedron \(ABCD\). The sphere \(\omega_A\) is tangent to the face \(BCD\) and to the planes of the other faces outside the faces themselves. Similarly, the sphere \(\omega_B\) is tangent to the face \(ACD\) and to the planes of the other faces outside the faces themselves. Let \(K\) be the point of tan... | Solution. Let us mark the points of tangency $K_{1}$ and $L_{1}$ of the inscribed sphere $\omega$ of the tetrahedron with the faces $A C D$ and $B C D$ respectively, as well as the points $K_{2}$ and $L_{2}$ of tangency of the spheres $\omega_{B}$ and $\omega_{A}$ with these faces. The spheres $\omega$ and $\omega_{A}$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,145 |
11.5. The radii of five concentric circles $\omega_{0}, \omega_{1}, \omega_{2}, \omega_{3}$, $\omega_{4}$ form a geometric progression with common ratio $q$ in the given order. For what largest $q$ can a non-closed broken line $A_{0} A_{1} A_{2} A_{3} A_{4}$, consisting of four segments of equal length, be drawn such t... | Answer. When $q=\frac{\sqrt{5}+1}{2}$.
Solution. We can assume that $q \geqslant 1$. Let the radius of $\omega_{i}$ be $R_{i}=R q^{i}$.
Choose some positive $\ell$ and try to construct the required broken line with segments of length $\ell$, starting from an arbitrary point $A_{0} \in \omega_{0}$. Suppose the point $... | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 17,146 |
11.6. On the side $AC$ of the isosceles triangle $ABC$ with base $BC$, a point $D$ is taken. On the smaller arc $CD$ of the circumcircle of triangle $BCD$, a point $K$ is chosen. The ray $CK$ intersects the line parallel to $BC$ and passing through $A$ at point $T$. Let $M$ be the midpoint of segment $DT$. Prove that $... | Solution. Extend segment $A M$ by its length beyond point $M$, obtaining point $N$ such that $A D N T$ is a parallelogram. Since $\angle A N T = \angle C A M$, to solve the problem, it is sufficient to show that $\angle A K T = \angle A N T$, or that points $A, T, N, K$ lie on the same circle (see Fig. 3).
![](https:/... | proof | Geometry | proof | Yes | Yes | olympiads | false | 17,147 |
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