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3. The line $l$ intersects the graph of the function $y=\frac{k}{x}(k \neq 0)$ at points with abscissas $x_{1}$ and $x_{2}$, and the x-axis at a point with abscissa $x_{3}$. Prove that $x_{1}+x_{2}=x_{3}$. | Solution. The line is obviously not vertical, otherwise it would intersect the hyperbola at only one point. Let the equation of the line be $y=t x+b$. Then $0=t x_{3}+b$, from which $b=-t x_{3}$ and the equation of the line is $y=t x-t x_{3}$. To find the coordinates of the points of intersection of the line and the gi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,634 |
4. Scrooge McDuck placed a gold coin in one of the vertices of a hexagon, and there is nothing in the other vertices. Every day, he removes an arbitrary number of coins from one of the vertices of the hexagon and immediately places six times as many coins in an adjacent vertex. If on some day Scrooge McDuck can achieve... | Answer: He cannot.
Solution. Let's number the vertices of the hexagon, starting from the one where the coin lies, with consecutive natural numbers from 1 to 6 (moving, for example, counterclockwise). Denote by $n_{1}, n_{2}, \ldots, n_{6}$ the numbers of coins lying in vertices $1,2, \ldots, 6$ respectively. Let $N_{1... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,635 |
5. A semicircle with diameter $A B$ and center at point $O$ is divided into three parts by points $C$ and $D$ such that point $C$ lies on the arc $A D$. Perpendiculars $D E$ and $D F$ are dropped from point $D$ to segments $O C$ and $A B$ respectively. It turns out that $D E$ is the angle bisector of triangle $A D C$, ... | Answer: $20^{\circ}$.
Solution. Triangle $A O D$ is isosceles ($O D=O A$, as radii), hence, $\angle O A D=\angle O D A$. Since $D O$ is the bisector of angle $A D F$, then $\angle O A D=$ $\angle O D F$. Calculation of angles in the right triangle $A F D$ shows that $\angle O A D=30^{\circ}$. Let $G$ be the point of i... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,636 |
Problem 7.1. Solve the riddle
$$
C, B A+A, A A=B, A
$$
(Different letters represent different digits, the same letters represent the same digits.) | Answer: $A=5, B=9, C=3$.
Solution. It is obvious that $A \neq 0$ (otherwise, for example, $C=B$).
The hundredths place can disappear only if the digits in the hundredths place sum to a number ending in 0. This is only possible if $A=5$. Then the puzzle can be rewritten as follows:
$$
\begin{gathered}
C, B 5+5,55=B, ... | A=5,B=9,C=3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,637 |
Problem 7.2. Vlad and Dima decided to earn some money. Each of them decided to deposit 3000 rubles in the bank and withdraw all the money after a year.
Vlad chose the deposit "Confidence": the amount increases by $20\%$ over the year, but the bank charges a $10\%$ fee upon withdrawal.
Dima chose the deposit "Reliabil... | Answer: Dima will earn 120 rubles more.
Solution. Vlad's deposit amount will increase to $3000 \cdot 1.2$ rubles in a year, and after withdrawal, it will decrease to $3000 \cdot 1.2 \cdot 0.9=3240$ rubles.
Dima's deposit amount will increase to $3000 \cdot 1.4$ rubles in a year, and after withdrawal, it will decrease... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,638 |
Problem 7.3. The Smeshariki Kros, Yozhik, Nyusha, and Barash ate a total of 86 candies, and each of them ate no fewer than 5 candies. It is known that:
- Nyusha ate more candies than each of the other Smeshariki;
- Kros and Yozhik together ate 53 candies.
How many candies did Nyusha eat | Answer: 28.
Solution. Krosh or Yozhik ate at least 27 candies (otherwise, they would have eaten no more than $26+26=52$ candies in total), so Nusha ate at least 28 candies. Considering that Barash ate at least 5 candies, we get that all of them together ate at least $53+28+5=86$ candies. Therefore, this is only possib... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,639 |
Problem 7.5. A store sells four types of nuts: hazelnuts, almonds, cashews, and pistachios. Stepan wants to buy 1 kilogram of nuts of one type and another 1 kilogram of nuts of a different type. He calculated how much such a purchase would cost him depending on which two types of nuts he chooses. Five out of six possib... | Answer: 2290.
Solution. Let $a, b, c, d$ be the cost of 1 kilogram of hazelnuts, almonds, cashews, and pistachios, respectively. From the condition, it follows that the set $A=\{1900,2070,2110,2330,2500\}$ is contained in the set $B=\{a+b, b+c, c+d, d+a, a+c, b+d\}$.
Note that the 6 elements of set $B$ can be divided... | 2290 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,641 |
Problem 7.6. A magic square is a $3 \times 3$ table in which numbers are arranged so that the sums of all rows, columns, and the two main diagonals are the same. The figure shows a magic square in which all numbers except three have been erased. Find what the number in the upper left corner of the square is.
| $?$ | 3... | Answer: 14.
Solution. Let the unknown number be $x$, then the sums in all rows, columns, and on the main diagonals are $9+31+x=40+x$.
1) Considering the left column, we get that the number in the lower left corner of the square is 27.
2) Considering the main diagonal going up to the right, we get that the number in t... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,642 |
Problem 7.7. All 25 students in class 7A participated in a quiz consisting of three rounds. In each round, each participant scored a certain number of points. It is known that in each round, as well as in the total of all three rounds, all participants scored a different number of points.
Student Kolya from 7A was thi... | Answer: 10.
Solution. In the first round, 2 classmates overtook Kolya, in the second - 3, in the third - 4. Then, in the sum of all three rounds, he could be overtaken by no more than $2+3+4=9$ classmates, i.e., in the sum of the three rounds, he could not end up lower than 10th place.
Now let's provide an example of... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,643 |
Problem 7.8. A set of 28 different dominoes looks like this:
All 28 dominoes were arranged such that the number of dots on the touching halves of the dominoes is the same. On some halves, the ... | Answer: $A=2, B=5$.
Solution. First, note that any number must appear exactly 8 times on the halves of the dominoes (on 6 dominoes paired with other numbers and on one domino twice). This means each number appears an even number of times.
Consider the highlighted red areas consisting of 1 or 3 cells. According to the... | A=2,B=5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,644 |
8.3. Does there exist a triangle with heights equal to a) 2; 3; 6? b) 2; 3; 5? | Answer a) does not exist, b) exists. Hint a) Suppose such a triangle exists. Denote the sides to which the heights are drawn as a, b, c. Then from the formula for the area of a triangle, it follows that $S=\frac{2 a}{2}=\frac{3 b}{2}=\frac{6 c}{2}$, i.e., the ratios of the sides $\mathrm{a}: \mathrm{b}: \mathrm{c}=\fra... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,646 |
84. a) Prove that there exists a pair of two-digit numbers such that if 20 is added to the first number and 15 is subtracted from the second, the resulting numbers will remain two-digit, and their product will be equal to the product of the original numbers? b) How many such pairs are there? | Answer: b) 16 pairs. Hint For part a), it is sufficient to provide a specific example (see problem 7.4). b) Let $a, b$ be the desired pair of numbers. Then $(a+20)(b-15)=a b$. From this, $20 b-15 a=20 \cdot 15 \Leftrightarrow 4 b-3 a=60$.
Since 60 and $3a$ are divisible by 3, $b$ must also be divisible by 3, i.e., $b=... | 16 | Algebra | proof | Yes | Yes | olympiads | false | 18,647 |
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 | 18,648 |
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$?
. This means that currently, Masha and Petya have the same a... | 7.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,657 |
10.1. On the side $B C$ of parallelogram $A B C D$, points $E$ and $F$ are marked, with $E$ lying between $B$ and $F$. Diagonals $A C$ and $B D$ intersect at point $O$. Lines $A E$ and $D F$ are tangent to the circumcircle of triangle $A O D$. Prove that they are also tangent to the circumcircle of triangle $E O F$.
(... | The first solution. We will denote $(X Y Z)$ as the circumcircle of triangle $X Y Z$.
From the tangency of the circle $(A O D)$ and the line $A E$, we have $\angle E A O = \angle A D O$. From the parallelism $B C \| A D$, we have $\angle E B O = \angle A D O$ (see Fig. 2). Thus, $\angle E A O = \angle E B O$, which me... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,659 |
10.2. Find all sets of natural numbers $x_{1}, x_{2}, \ldots, x_{20}$ such that
$$
x_{i+2}^{2}=\operatorname{LCM}\left(x_{i+1}, x_{i}\right)+\operatorname{LCM}\left(x_{i}, x_{i-1}\right)
$$
for $i=1,2, \ldots, 20$, where $x_{0}=x_{20}, x_{21}=x_{1}, x_{22}=x_{2} . \quad$ (P. Kozlov) | Answer. $x_{1}=x_{2}=\ldots=x_{20}=2$.
Solution. From the condition, it follows that all $x_{i}$ are greater than 1, and $x_{i+2}^{2}$ is divisible by $x_{i}$ for $i=1,2, \ldots, 20$ (here and below $x_{j+20}=$ $=x_{j}=x_{j-20}$ for $\left.j=1, \ldots, 20\right)$.
Let $x_{k}$ be the largest of the numbers $x_{1}, \ld... | x_{1}=x_{2}=\ldots=x_{20}=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,660 |
10.3. In a country with $N$ cities, some pairs of cities are connected by two-way air routes, with no more than one route between any pair. Each air route belongs to one of $k$ companies. It turned out that from any city, one can reach any other city (possibly with layovers), but if all the air routes of any company ar... | Answer. The construction is possible only for $k<N$, and then the maximum number of edges is $C_{N}^{2}-C_{k}^{2}$.
First Solution. Consider a graph where the vertices are cities, and the edges are air routes, with edges corresponding to air routes of the $i$-th company colored in the $i$-th color.
Example. Suppose i... | C_{N}^{2}-C_{k}^{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,661 |
11.1. In the product of seven natural numbers, each factor was decreased by 3. Could the product have increased exactly 13 times as a result?
( | Answer. Yes, it could.
Solution. As an example, the product $1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 16=32$ fits. After the specified operation, it becomes $(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot(-2) \cdot(-1) \cdot 13=13 \cdot 32$.
Remark 1. Let's explain how to come up with this example. Suppose five of t... | 1,1,1,1,1,2,16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,663 |
11.2. Vasya has thought of 8 cells on a chessboard, none of which lie in the same row or column. On his turn, Petya places 8 rooks on the board, none of which attack each other, and then Vasya indicates all the rooks standing on the thought-of cells. If the number of rooks indicated by Vasya on this turn is even (i.e.,... | Answer. In 2 moves.
Solution. First, we will show how Pete can win in 2 moves. On his first move, he will place 8 rooks along the diagonal of the board. If he hasn't won yet, there is an odd number of cells on the diagonal that Vasya has chosen. In particular, there is both a cell $A$ chosen by Vasya and a cell $B$ no... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,664 |
11.3. The teacher is going to give the children a problem of the following type. He will inform them that he has thought of a polynomial $P(x)$ of degree 2017 with integer coefficients, the leading coefficient of which is 1. Then he will tell them $k$ integers $n_{1}, n_{2}, \ldots, n_{k}$, and separately inform them o... | Answer. For $k=2017$.
Solution. First, we prove that $k>2016$. Suppose the teacher used some $k \leqslant 2016$, thinking of the polynomial $P(x)$. Consider the polynomial $Q(x)=P(x)+(x-n_{1})(x-n_{2}) \ldots(x-n_{k})$. Note that the degree of the polynomial $Q(x)$ is also 2017, and its leading coefficient is also 1. ... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,665 |
11.4. An equilateral triangle $ABC$ is inscribed in a circle $\Omega$ and circumscribed around a circle $\omega$. Points $P$ and $Q$ are chosen on sides $AC$ and $AB$ respectively such that the segment $PQ$ passes through the center of triangle $ABC$. Circles $\Gamma_{b}$ and $\Gamma_{c}$ are constructed on segments $B... | Solution. Let $O$ be the center of triangle $ABC$. Denote by $B_2$ and $C_2$ the points of tangency of $\omega$ with $AC$ and $AB$ respectively, and by $B_1$ and $C_1$ the points of $\Omega$ diametrically opposite to points $B$ and $C$ respectively (see Fig. 6). Then points $B_1$ and $C_1$ are symmetric to $O$ with res... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,666 |
9.2. Given a triangle $A B C$, where $\frac{B C}{A C}<\frac{1}{2}$. Prove that $\angle A<30^{\circ}$. | Solution. Let $a=BC, b=AC$. The first method of solving. Construct a circle of radius $a$ passing through points $B$ and $C$ such that its center $O$ lies on the same side of the line $BC$ as point $A$. Then $A$ cannot lie inside this circle or on the circle itself, since $AC$ is greater than its diameter. Therefore, a... | \angleA<30 | Geometry | proof | Yes | Yes | olympiads | false | 18,668 |
9.4. What is the minimum number of unit-radius circles required to completely cover a triangle with sides $2 ; 3 ; 4$? | Answer: three circles. Solution. Let $A C=4, A B=2, B C=3$ and let $C_{1}, A_{1}$ and $B_{1}$ be the midpoints of sides $A B, B C$ and $A C$ respectively. Note that angle $B$ is obtuse, since $A C^{2}>A B^{2}+B C^{2}$. Therefore, points $B$ and $B_{1}$ lie inside the circle of radius 1 with center at point $O$ - the mi... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,670 |
9.5. In 9a and $9b$ classes, there are 25 people each. In 9a, each student has at least 13 friends in the class, and in $9b$, each has at least 12 friends in the class. Is it necessary that there will be three friends (where each in the trio is friends with the other two) a) in 9a; b) in $9b$? | Answer: a) yes; b) no. Solution. a). Let's take any two friends $A$ and $B$. From the remaining 23 people, $A$ has at least 12 friends, and $B$ has at least 12 friends. Therefore, among the friends of $A$ and $B$, there is at least one common friend (otherwise, it would be $12+12 \leq 23$). Together with $A$ and $B$, t... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,671 |
1. At what moment after noon do the hour and minute hands form a right angle for the first time? | Answer: $180 / 11$ minutes $=164 / 11$ minutes.
## Solution:
The hour hand makes 1 revolution in 12 hours, while the minute hand makes 12 revolutions in the same time. Therefore, the minute hand moves 12 times faster. We will measure everything in minutes, so a right angle corresponds to 15 minutes. Let the hour and ... | \frac{180}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,674 |
3. The quadratic trinomial $f(x)$ is such that each of the equations $f(x)=x-1$ and $f(x)=2-2x$ has exactly one solution. Prove that the trinomial $f(x)$ has no roots.
# | # Solution:
Let $f(x)=a x^{2}+b x+c$.
The equations from the problem can be transformed (by moving to the left side of the equation, grouping) to the following form: $a x^{2}+(b-1) x+(c+1)=0$ and $a x^{2}+(b+2) x+(c-2)=0$.
From the condition, it follows that their discriminants are equal to $0: D_{1}=(b-1)^{2}-4 a(c... | proof | Algebra | proof | Yes | Yes | olympiads | false | 18,676 |
4. In square $A B C D$, a point $E$ is taken on side $B C$, points $F$ and $G$ are taken on side $C D$ (point $F$ is between $C$ and $G$), and a point $H$ is taken on side $A D$. Given that $C E=C F$ and $D G=D H$, prove that a circle can be circumscribed around the quadrilateral formed by the intersection of angles $H... | # Solution:
Let the vertices of the quadrilateral formed by the intersection of angles $H B G$ and $E A F$ be denoted as $K, L, M, N$ (see the figure on the right). Right triangles $A B E$ and $A D F$, as well as $B A H$ and $B C G$, are pairwise equal by two legs. In the first pair of triangles, the angles at vertex ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,677 |
5. In a state, some cities are connected by roads, and some three cities are connected to each other. One-way traffic has been introduced on all roads, but in such a way that it is possible to travel from any city to any other. Prove that in this state, there is a closed route consisting of an odd number of roads.
# | # Solution:
Consider three roads connecting three cities as described in the problem. If the introduction of one-way traffic already forms a cycle of 3 cities, the proof is complete. If not,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,678 |
10.5. Eleven integers (not necessarily distinct) are written on a board. Can it happen that the product of any five of them is greater than the product of the remaining six?
(I. Bogdanov) | Answer: It can.
Solution: Let one of the numbers be 10, and each of the others be -1. Then the product of any five of them is greater than the product of the remaining six. Indeed, if the number 10 is included in the product of five numbers, then this product is 10, and the product of the remaining six numbers is 1, a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,679 |
10.6. Given a natural number $n>5$. On a ring-shaped strip of paper, a sequence of zeros and ones is written. For each sequence $w$ of $n$ zeros and ones, the number of ways to cut a fragment from the strip on which $w$ is written was calculated. It turned out that the maximum number $M$ is achieved on the sequence $11... | Solution. Let $N$ be the number of ways to cut out from a strip the sequence $1 \underbrace{00 \ldots 0}_{\geqslant n-2} 1$ (i.e., the number of sequences with at least $n-2$ zeros, before and after which there are ones). Before each of them, there can be either 1 or 0; denote the number of those with 1 before them by ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,680 |
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 | 18,684 |
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 | 18,688 |
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 | 18,689 |
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 | 18,690 |
5-6. A rectangular table of size $x$ cm $\times 80$ cm is covered with identical sheets of paper of size 5 cm $\times 8$ cm. 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 | 18,691 |
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 | 18,692 |
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 | 18,694 |
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 | 18,697 |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,700 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,701 |
1. Solve the equation $3 \cdot 2^{x}+1=y^{2}$ in integers. | Answer: $(0 ;-2),(0 ; 2),(3 ;-5),(3 ; 5),(4 ;-7),(4 ; 7)$.
## Solution
Obviously, $x \geq 0$, otherwise the left side would be a fraction. It is also clear that $y=0$ is not a solution.
Since $y$ is in an even power, the solutions will come in pairs, for example, when $x=0$, we get two solutions $y= \pm 2$.
Now let... | (0,-2),(0,2),(3,-5),(3,5),(4,-7),(4,7) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,702 |
3. Find all pairs of real numbers $a$ and $b$, for which there exists a polynomial $f(x)$ such that $f(f(x))=\left(x^{2}+a x+b\right)^{2}$. | Answer: $a$ - arbitrary, $b=\frac{a}{2}+\frac{a^{2}}{4}$.
## Solution (outline)
The polynomial has the form (this needs to be explained) $f(x)=x^{2}+p x+q$. By expanding both sides of the equation, we arrive at four equations involving $p$ and $q$ with parameters $a$ and $b$. This system must be solvable, from which ... | a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,704 |
4. Circles with radii $\mathrm{a}>\mathrm{b}>0$ touch each other externally. Three common tangents are drawn to them. Find the perimeter of the triangle formed by them. | Answer: $P=\frac{4 a \sqrt{a b}}{a-b}$.
According to the theorem of tangent segments, the given triangle ABC is isosceles, and all common tangent segments are equal, i.e., $\mathrm{CE}=\math... | \frac{4\sqrt{}}{-b} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,705 |
5. Given an arbitrary 2019-digit natural number, in the decimal representation of which the digits $0,1,2,3$ are absent. If the number contains two consecutive identical digits or two consecutive identical two-digit numbers, they can be deleted. It is also allowed to insert two identical digits or two consecutive ident... | # Solution
In the record of the number, 6 digits are used: 4,5,6,7,8,9. Consider a pair of adjacent digits ab. Add two digits bb to the left and two digits aa to the right, resulting in the block bbabaa, which contains two identical two-digit numbers baba that can be removed, i.e., we arrive at the combination ba. Thu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,706 |
11.1. Can the number 2023 be represented as the sum of three natural numbers $a, b, c$ such that $a$ is divisible by $b+c$ and $b+c$ is divisible by $b-c+1?$
(A. Kuznetsov) | Answer: No.
Solution: Suppose such three numbers exist. Since $a$ is divisible by $b+c$, the sum $a+(b+c)=2023$ is also divisible by $b+c$, which implies that $b+c$ is odd. Therefore, $b-c+1$ is an even number, and the odd number $b+c$ cannot divide it.
Comment: Correct answer without justification - 0 points. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,707 |
11.2. Given different real numbers $a_{1}, a_{2}, a_{3}$ and $b$. It turned out that the equation $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)=b$ has three different real roots $c_{1}, c_{2}, c_{3}$. Find the roots of the equation $(x+$ $\left.+c_{1}\right)\left(x+c_{2}\right)\left(x+c_{3}\right)=b$.
... | Answer: $-a_{1}, -a_{2}$, and $-a_{3}$.
First Solution. Since the polynomial $\left(x-a_{1}\right)\left(x-a_{2}\right)(x- \left.-a_{3}\right)-b$ has a leading coefficient of 1 and roots $c_{1}, c_{2}, c_{3}$, we have $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)-b=\left(x-c_{1}\right)\left(x-c_{2}\righ... | -a_{1},-a_{2},-a_{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,708 |
11.3. In city $\mathrm{N}$, 50 city olympiads in different subjects were held, with exactly 30 schoolchildren participating in each of these olympiads, but there were no two olympiads with the same set of participants. It is known that for any 30 olympiads, there is a schoolchild who participated in all these 30 olympi... | Solution. Suppose the opposite, and let there be 30-element subsets $A_{1}, A_{2}, \ldots$, $A_{50}$ (sets of participants of each Olympiad) in the set of all schoolchildren such that the intersection of any 30 of them is non-empty, while the intersection of all of them is empty.
Suppose among the sets $A_{1}, A_{2}, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,709 |
11.4. Pairs of numbers are written on a board. Initially, the pair of numbers $(1,2)$ is written on the board. If the pair of numbers $(a, b)$ is written on the board, then the pair $(-a,-b)$ and the pair $(-b, a+b)$ can also be written on the board. Additionally, if the pairs of numbers $(a, b)$ and $(c, d)$ are writt... | Answer. Could not.
First solution. We will prove that for any pair $(x, y)$ written on the board, the number $2 x-y$ is divisible by 7.
Indeed, for the pair $(1,2)$, the number $2 \cdot 1-1=0$ is divisible by 7.
Suppose for the pair $(a, b)$ the number $2 a-b$ is divisible by 7. Then for the pair $(-a,-b)$ the numbe... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,710 |
11.5. In an acute scalene triangle $ABC$, the altitude $AH$, the median $AM$, and the center $O$ of its circumcircle $\omega$ are drawn. The segments $OH$ and $AM$ intersect at point $D$, the lines $AB$ and $CD$ intersect at point $E$, and the lines $BD$ and $AC$ intersect at point $F$. The rays $EH$ and $FH$ intersect... | Solution. Let $P$ be a point on the ray $H E$ such that $P B \perp B C$ (see Fig. 7). We will prove that points $C, O$, and $P$ lie on the same line.
Indeed, by Menelaus' theorem for triangle $A D E$ and line $C M B$, we have $\frac{E C}{C D} \cdot \frac{D M}{M A} \cdot \frac{A B}{B E}=1$. Since lines $P B, A H$, and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,711 |
1. The numbers a, b, c are such that a>b and (a-b)(b-c)(c-a)>0. Which is greater: a or c? | 1. Obviously, among the three given numbers there are no equal ones. Since a > b, a - b > 0, from which (b - c)(c - a) > 0. Suppose c > a. Then c - a > 0 and $\mathrm{b}-\mathrm{c}>0$, i.e., $\mathrm{b}>\mathrm{c}>\mathrm{a}$ - a contradiction to the condition.
Answer: a ass | a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,712 |
2. The truant Vasya skipped one lesson every Monday in September of a certain year, two lessons every Tuesday, three lessons every Wednesday, four lessons every Thursday, and five lessons every Friday. Could it have turned out that he skipped exactly 64 lessons for the entire September? (All Saturdays and Sundays in Se... | 2. Suppose Vasya skipped exactly 64 lessons. Note that September contains four full weeks and two consecutive days. In one full week, Vasya skipped \(1+2+3+4+5=15\) lessons, i.e., in four weeks Vasya skipped exactly 60 lessons, and in the two consecutive days - 4 lessons. But if one of these two days is a weekend, then... | couldnot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,713 |
3. After teacher Mary Ivanovna moved Vovochka from the first row to the second, Vanechka from the second row to the third, and Mashenka from the third row to the first, the average age of students sitting in the first row increased by one week, those sitting in the second row increased by two weeks, and those sitting i... | 3. Let there be x people in the third row. Since the average age is the sum of the ages divided by the number of people, after the rearrangement, the total age of the children in the first row increased by 12 weeks, in the second row by 24 weeks, and in the third row by -4x weeks. Since the total sum of the ages of all... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,714 |
4. In triangle $ABC$, angle $C$ is right. On side $AC$ there is a point $D$, and on segment $\mathrm{BD}$ there is a point $\mathrm{K}$ such that $\angle \mathrm{ABC}=\angle \mathrm{KAD}=\angle \mathrm{AKD}$. Prove that $\mathrm{BK}=2 \mathrm{DC}$. | 4. Extend $\mathrm{DC}$ beyond point $\mathrm{C}$ and mark a segment $\mathrm{CM}=\mathrm{CD}$ (see the figure). Then $\mathrm{BD}=\mathrm{BM}$ (in triangle $\mathrm{BDM}$, the median coincides with the altitude). We have $\angle B A K=\angle A K D-\angle A B K=\angle A B C-\angle A B K=\angle K B C=\angle C B M, \quad... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,715 |
5. Prove that the numbers from 1 to 16 can be written in a line, but cannot be written in a circle such that the sum of any two adjacent numbers is a perfect square. | 5. If next to 16 stands the number $\mathrm{x}$, then $16+1 \leq 16+\mathrm{x}=\mathrm{a}^{2} \leq 16+15$, from which $\mathrm{a}^{2}=25$ and $\mathrm{x}=9$. Therefore, 16 cannot have more than one neighbor, and the arrangement of numbers in a circle that satisfies the condition is impossible. An example of an arrangem... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,716 |
1. On some cells of a $4 \times 4$ square board, golden coins are placed in a stack, and on the remaining cells, silver coins are placed. Is it possible to arrange the coins so that in each $3 \times 3$ square, there are more silver coins than golden ones, and on the entire board, there are more golden coins than silve... | Answer: Yes.
Solution. For example, place a stack of nine silver coins on one of the cells of the central $2 \times 2$ square, and place one gold coin on each of the other cells of the board. Then, in each $3 \times 3$ square, there will be 9 silver coins and 8 gold coins, and on the entire board, there will be 15 gol... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,717 |
2. Can the products of all non-zero digits of two consecutive natural numbers differ by exactly 54 times
Can the products of all non-zero digits of two consecutive natural numbers differ by exactly 54 times | Answer: They can, for example, the numbers 299 and 300. | 299300 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,718 |
3. A circle with center $O$ is inscribed in triangle $A B C$. A point $P$ is chosen on side $A B$, and a point $Q$ is chosen on the extension of side $A C$ beyond point $C$ such that segment $P Q$ is tangent to the circle. Prove that $\square B O P = \square C O Q$. | Solution. The center of the circle inscribed in an angle lies on its bisector. Applying the external angle theorem to triangle $B O P$ (see figure), we get
$\square B O P=\square A P O-\square A B O=1 / 2 \square A P Q-1 / 2$
$\square B$. Similarly, $\square C O Q=\square A C O-\square A Q O=1 / 2 \square C-1 / 2$
$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,719 |
4. From Zlatoust to Miass, "GAZ", "MAZ", and "KamAZ" set off simultaneously. "KamAZ", having reached Miass, immediately turned back and met "MAZ" 18 km from Miass, and "GAZ" - 25 km from Miass. "MAZ", having reached Miass, also immediately turned back and met "GAZ" 8 km from Miass. What is the distance from Zlatoust to... | Answer: 60 km.
Solution. Let the distance between the cities be $x$ km, and the speeds of the trucks: "GAZ" $-g$ km/h, "MAZ" - $m$ km/h, "KAMAZ" - $k$ km/h. For each pair of vehicles, we equate their travel time until they meet.
We get $\frac{x+18}{k}=\frac{x-18}{m}, \frac{x+25}{k}=\frac{x-25}{g}$ and $\frac{x+8}{m}=... | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,720 |
5. While waiting for customers, the watermelon seller sequentially weighed 20 watermelons (weighing 1 kg, 2 kg, 3 kg, ..., 20 kg), balancing the watermelon on one scale pan with one or two weights on the other pan (possibly identical). The seller recorded on a piece of paper the weights of the weights he used. What is ... | Answer: 6.
Solution. With one or two weights of 1 kg, 3 kg, 5 kg, 7 kg, 9 kg, and 10 kg, any of the given watermelons can be weighed. Indeed, $2=1+1, 4=3+1$, $6=5+1, 8=7+1, 11=10+1, 12=9+3, 13=10+3, 14=9+5, 15=10+5, 16=9+7, 17$ $=10+7, 18=9+9, 19=10+9, 20=10+10$. Thus, six different numbers could have been recorded.
... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,721 |
Task 7.2
Can a two-digit and a three-digit number be formed from the digits $1,2,3,4,5$ such that the second number is divisible by the first? Each digit must be used exactly once.
Number of points 7 | Answer:
It is possible.
## Solution
It is easy to check that 532 is divisible by 14, and 215 is divisible by 43.
To find these pairs, we can apply a brute-force approach. To reduce the number of options, we use divisibility properties, which in particular imply that: if a two-digit number ends in 5, the second numb... | 532isdivisible14,215isdivisible43 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,723 |
# Task 7.3
Vasya and Petya each have 55 weights weighing $1, 2, \ldots, 55$ kg. They take turns placing their weights on their own pan of a two-pan balance. Vasya goes first. Petya wins if the difference in the mass of the weights on the pans is 50 kg. Can he achieve this?
## Number of points 7 | Answer:
Yes.
## Solution
## First method
Petya can simply repeat Vasya's moves. At some point, Vasya will be forced to move the 50 kg weight and immediately lose.
## Second method
Petya sets aside his 50 kg weight and moves the other weights as he pleases. At the end of the game, Vasya will lay out all the weight... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,724 |
# Task 7.5
Along a straight section of the border, 15 poles are installed. Near each pole, several nearsighted spies were caught. Each of them honestly said how many other spies they saw. However, any spy could only see those who were near their pole and near the nearest neighboring poles. Can the data be used to dete... | Answer:
It is possible.
## Solution
Number the poles from 1 to 15 from left to right. From the interrogation of all spies caught at the second pole, we learn the total number of spies at the first three poles, and from the interrogation of spies caught at the first pole, we learn the number of spies caught at the fi... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,726 |
1. Two types of tiles were laid on the wall in a checkerboard pattern. Several tiles fell off the wall. The remaining tiles are shown in the picture. How many striped tiles fell off? Be sure to explain your answer.
Method 2. Look at the number of fallen cells by rows: in the second row from the top, 2 tiles fell out, of which 1 ... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,727 |
3. Mowgli asked five monkeys to bring him some nuts. The monkeys collected an equal number of nuts and carried them to Mowgli. On the way, they quarreled, and each monkey threw one nut at each of the others. As a result, they brought Mowgli half as many nuts as they had collected. How many nuts did Mowgli receive? Be s... | Answer: 20 nuts.
## Solution.
Each monkey threw 4 nuts, so the monkeys threw a total of $5 \cdot 4=20$ nuts together.
If half of the nuts remained, it means that Mowgli brought as many nuts as were thrown, which is 20 nuts.
## Grading Criteria.
- Correct solution - 7 points.
- Very brief solution (such as "5 - $4=... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,729 |
4. In the picture, we see four children: Kolya, Vasya, Senya, and Yan. It is known that we see Senya to the right of Kolya, and Kolya gave Vasya his left hand. Find out who is who, and explain why you think so.
Kolya cannot be the farthest to the right in the picture, as we see Sanya to his right. But Kolya cannot be the farthest to the left either, as the boy on the far left... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,730 |
5. The seller has 3 packs of stickers, with 100 stickers in each pack. Three customers approached him. The first customer needs 70 stickers, and the second and third customers need 60 stickers each. How can the seller count out the required number of stickers for each customer in 70 seconds, if he counts exactly one st... | # Solution.
To count out 70 stickers from a pack of 100, it is enough to count out 30 stickers and give the rest to the customer.
Let's write down the seller's actions. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,731 |
1. The distance from the home of Vintik and Shpuntik to school is 6 kilometers. Vintik and Shpuntik left for school at the same time, with Vintik spending half of the time riding a scooter at a speed of 10 km/h and then walking, while Shpuntik traveled half the distance by bicycle and then walked. They arrived at schoo... | Answer: 15 km/h.
## Solution:
First method.
Since Shtyubik rides twice as fast as he walks, he also covers twice the distance he walks (since he spends the same amount of time on both), which is 4 km. Since Shtyubik and Shpuntyk walk at the same speed, the last 2 km they walked together. Therefore, while Shtyubik wa... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,732 |
2. In the notebook, all irreducible fractions with the numerator 15 are written down, but which are greater than $\frac{1}{16}$ and less than $\frac{1}{15}$. How many such fractions are written in the notebook? | Answer: 9 fractions.
## Solution:
We are looking for all suitable irreducible fractions of the form $\frac{15}{n}$. Since $\frac{1}{16}<\frac{15}{n}<\frac{1}{15}$, then $15 \cdot 15<n<15 \cdot 16$ or $225<n<240$ (with the fraction $\frac{15}{n}$ being irreducible, meaning $n$ is not divisible by 3 or 5). It is not di... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,733 |
3. In each cell of a $2 \times 2$ table, a number was written, and all the numbers are different. It turned out that the sum of the numbers in the first row is equal to the sum of the numbers in the second row, and the product of the numbers in the first column is equal to the product of the numbers in the second colum... | # Answer: 0.
## Solution:
| $a$ | $b$ |
| :--- | :--- |
| $c$ | $d$ |
Let's denote the numbers in the table as shown on the left. According to the condition, $a+b=c+d$, $ac=bd$. Then $a=c+d-b$, substitute this into the product equality: $(c+d-b)c=bd$. Expand the brackets and move everything to the left: $c^2 + cd - ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,734 |
4. In triangle $A B C$, the median $B M$ was drawn. It turned out that $A B=2 B M$ and $\angle M B A=40^{\circ}$. Find $\angle C B A$.
# | # Answer: $110^{\circ}$.
## Solution:
Extend the median $B M$ beyond point $M$ by its length and obtain point $D$. Since $A B=2 B M$, then $A B=B D$, which means triangle $A B D$ is isosceles. Therefore, angles $B A D$ and $B D A$ are each equal to $\left(180^{\circ}-40^{\circ}\right): 2=70^{\circ}$. $A B C D$ is a p... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,735 |
5. The organizers of a mathematics olympiad decided to photograph 60 participants. It is known that no more than 30 participants can fit in a single photograph, however, any two students must appear together in at least one photograph. What is the minimum number of photographs needed to achieve this? | # Answer: 6.
## Solution:
Example with 6 photos: divide 60 participants into 4 groups of 15 people (groups $A, B, B$, Г). Take 6 photos of all possible pairs of groups: $A+D, A+B, A+\Gamma, B+B, B+\Gamma, B+\Gamma$ - in each photo, there will be 30 people, and it is easy to see that in this way any two people will be... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,736 |
10.1. Can a natural number that has exactly 1234 divisors be a perfect square? | Answer: No.
Solution. Let $n=m^{2}$, then $n$ has $k$ divisors less than $m$, and the same number greater than $m$. Taking $m$ into account, we get an odd number of divisors.
Comment. Answer without justification - 0 points. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,737 |
10.2. Prove the inequality $\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{k^{2}}<\frac{k-1}{k}$. | Solution. Since $\frac{1}{n^{2}}<\frac{1}{(n-1) n}=\frac{1}{n-1}-\frac{1}{n}$ for any $\mathrm{n} \geq 2$, the sum $\frac{1}{2^{2}}+\frac{1}{3^{2}}+\cdots+\frac{1}{k^{2}}$ does not exceed the value
$$
\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\ldots+\left(\frac{1}{k-1}-\frac{1}{k}\right)=1-\frac{... | \frac{k-1}{k} | Inequalities | proof | Yes | Yes | olympiads | false | 18,738 |
10.3. Prove that for integers $a, b, c, d$ the product $A=(b-a)(c-a)(d-a)(b-c)(d-c)(d-b)$ is divisible by 12. | Solution. Let's divide the set of integers into four classes $\{4 t\}$, $\{4 t+1\},\{4 t+2\},\{4 t+3\}$.
If among the numbers $a, b, c, d$ there are two that belong to the same class, then their difference, and therefore the number $A$, is divisible by 4.
If no two of the numbers $a, b, c, d$ belong to the same class... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,739 |
10.4. Given an acute triangle $A B C$. From point $D$ on side $A B$, perpendiculars $D E$ and $D F$ are drawn to sides $A C$ and $B C$ respectively. At what position of point $D$ will the distance between points $E$ and $F$ be the smallest | Answer: $D$ is the foot of the altitude dropped from vertex $C$.
Solution. Points $E$ and $F$ lie on the circle constructed with segment $C D$ as its diameter. In this circle, the constant angle $C$ subtends the chord $E F$, so the length of chord $E F$ will be minimal if the diameter $C D$ of the circle is minimal, i... | DisthefootofthealtitudedroppedfromvertexC | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,740 |
10.5. Five athletes came to training with their balls, and when leaving, each took someone else's. In how many ways is this possible. | Answer: 44.
Solution. First, let's assume that no two athletes have exchanged balls. Imagine them sitting around a round table. Then, depending on their different relative positions, they would choose the ball of the next (for example, clockwise) athlete. However, this is only possible if 5 are seated at one table, wh... | 44 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 18,741 |
10.1 Let \( f(x, y) = kx + \frac{1}{y} \). Prove that if \( f(a, b) = f(b, a) \) for \( a \neq b \), then \( f(ab, 1) = 0 \). | Solution. According to the condition $\mathrm{ka}+\frac{1}{\mathrm{~b}}=\mathrm{kb}+\frac{1}{\mathrm{a}}$. Transform: $(\mathrm{a}-\mathrm{b})\left(\mathrm{k}+\frac{1}{\mathrm{ab}}\right)=0$. Since $\mathrm{a}-\mathrm{b} \neq 0$, then $\mathrm{k}+\frac{1}{\mathrm{ab}}=0, \mathrm{kab}+1=0$, which means that $\mathrm{f}(... | f(,1)=0 | Algebra | proof | Yes | Yes | olympiads | false | 18,744 |
10.2 The sequence of numbers $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \ldots$ is formed according to the rule: $\mathrm{x}_{1}=1, \mathrm{x}_{\mathrm{n}+1}=1+\frac{\mathrm{x}_{\mathrm{n}}^{2}}{\mathrm{n}}$ for $\mathrm{n}=1,2,3, \ldots$ Find $\mathrm{x}_{2019}$. | Answer: 2019
Reasoning. $x_{2}=1+\frac{1^{2}}{1}=2, x_{3}=1+\frac{2^{2}}{2}=3$. Hypothesis: $x_{n}=n$. We proceed by induction. If $x_{n}=n$, then $x_{n+1}=1+\frac{n^{2}}{n}=1+n$. The hypothesis is confirmed. | 2019 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,745 |
10.3 All integers from 1 to $2 \mathrm{n}$ were written in a row in a random order. Then, to each number, the number of the position it occupies was added. Prove that among the resulting sums, there will be at least two that give the same remainder when divided by $2 \mathrm{n}$. | Solution. Let $\mathrm{S}_{1}, \mathrm{~S}_{2}, \ldots, \mathrm{S}_{2 \mathrm{n}}$ be the obtained sums. Then $\mathrm{S}_{1}+\mathrm{S}_{2}+\ldots+\mathrm{S}_{2 \mathrm{n}}=$ $=2(1+2+\ldots+2 n)=2 \cdot \frac{1+2 n}{2} \cdot 2 n=(1+2 n) \cdot 2 n$ is divisible by $2 n$. Suppose that the remainders of the numbers $\mat... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 18,746 |
10.4 On one of the days of the year, it turned out that each resident of the city made no more than one phone call. Prove that the population of the city can be divided into no more than three groups so that residents belonging to the same group did not talk to each other on the phone that day. | Solution: We will use induction on the number of city residents, $\mathrm{n}$. If $\mathrm{n} \leq 2$, there is nothing to prove. Let $\mathrm{n} \geq 3$. Let $\mathrm{m}$ be the total number of calls made on that day. By the condition, $\mathrm{m} \leq \mathrm{n}$. Therefore, there is a resident $\mathrm{A}$ who spoke... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,747 |
10.5 In a convex quadrilateral $\mathrm{ABCD} \quad \angle \mathrm{CBD}=\angle \mathrm{CAB}, \angle \mathrm{ACD}=\angle \mathrm{ADB}$. Prove that a right triangle can be formed from the segments $\mathrm{BC}$, $\mathrm{AD}, \mathrm{AC}$. | Solution. See fig. Triangles AOD and ADC are similar: the angle at vertex A is common to both, and by condition, angles ADO and ACD are equal. We write the equality of the ratios of corresponding sides in these triangles:
$\frac{A O}{A D}=\frac{A D}{A C}$, from which $A O \cdot A C=A D^{2}$.
Similarly, from the simil... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,748 |
10.6. Inside triangle $ABC$, a point $K$ is taken, lying on the bisector of angle $BAC$. Line $CK$ intersects the circumcircle $\omega$ of triangle $ABC$ again at point $M$. Circle $\Omega$ passes through point $A$, is tangent to line $CM$ at point $K$, and intersects segment $AB$ again at point $P$, and circle $\omega... | 10.6. Let $R$ be the second point of intersection of the circle $\Omega$ and the segment $A C$ (see Fig. 4). From the tangency and the equality of inscribed angles subtending the same arc, we have $\angle M K P = \angle K A P = \angle K A R = \angle K P R$, hence $P R \parallel C M$. Further, $\angle C M Q = \angle C A... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,750 |
10.7. Given $n \geqslant 3$ pairwise coprime numbers. It is known that when the product of any $n-1$ of them is divided by the remaining number, the same remainder $r$ is obtained. Prove that $r \leqslant n-2$.
(V. Senderov) | 10.7. If $r=0$, then the statement of the problem is obviously true. Let $r>0$. Let $a_{1}, \ldots, a_{n}$ be the given numbers; set $P=$ $V$ stage, 2009-2010 academic year. Second day
$=a_{1} a_{2} \ldots a_{n}, P_{i}=P / a_{i}$ for $i=1,2, \ldots, n$. Note that $a_{i}>r$, because the number $P_{i}$ gives a remainder... | r\leqslantn-2 | Number Theory | proof | Yes | Yes | olympiads | false | 18,751 |
10.8. In a country, some pairs of cities are connected by two-way non-stop flights. Moreover, from any city, it is possible to fly to any other city (possibly with layovers). It is known that if any closed route consisting of an odd number of flights is chosen and all these flights are closed, then it will no longer be... | 10.8. Consider a graph $G$, whose vertices are cities, and two vertices are connected by an edge if there is an airline between the cities. We know that the graph is connected, but removing all edges of any odd cycle violates this condition; we need to prove that the vertices of the graph can be properly colored with 4... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,752 |
8.1. (7 points)
Compare $127^{23}$ and $513^{18}$. | Answer: $513^{18}>127^{23}$.
Solution: $127^{23}127^{23}$.
# | 513^{18}>127^{23} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,753 |
# 8.2. (7 points)
Twelve people are carrying 12 loaves of bread. Each man carries 2 loaves, each woman carries half a loaf, and each child carries a quarter of a loaf. How many men, women, and children were there? | Answer: 5 men, one woman, and 6 children.
Solution: Let $x$ be the number of men, $y$ be the number of women, and $z$ be the number of children; $x, y, z$ are natural numbers.
Then $x+y+z=12$ and $2 x+\frac{y}{2}+\frac{z}{4}=12$. From the last equation, it follows that $8 x+2 y+z=48$.
Transform the last equation: $7... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,754 |
# 8.3. (7 points)
On the median $B M$ of triangle $A B C$, a point $K$ is chosen such that $A K = B C$. Ray $A K$ intersects side $B C$ at point $P$. Prove that triangle $B K P$ is isosceles.
# | # Solution:
Method 1. Extend the median $B M$ by its length: $M D=B M$, then the quadrilateral $A B C D$ is a parallelogram (see figure).
Since $A K=B C=A D$, triangle $D A K$ is isosceles, ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 18,755 |
# 8.4. (7 points)
We will call a bus ticket with a six-digit number lucky if the sum of the digits of its number is divisible by 7. Can two consecutive tickets be lucky? If such tickets exist, provide examples of the numbers of such tickets. | Answer: for example, 429999 and 430000.
Solution: If two consecutive numbers differ only in the last digit, then their sums of digits differ by 1, and therefore cannot both be divisible by 7 simultaneously. Another case is when adding 1 to a number turns several nines at the end into zeros, and the next digit increase... | 429999430000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 18,756 |
# 8.5. (7 points)
A large square was cut out from graph paper along the grid lines. From it, a smaller square was also cut out along the grid lines. After this, exactly 79 cells remained from the large square. Was it necessary for the cut-out square to contain one of the corner cells of the large square? | Answer: Yes, definitely.
Solution: Let the large square contain $N^{2}$ cells, and the cut-out square - $M^{2}$ cells, then $N^{2}-M^{2}=79$, that is, $(N-M)(N+M)=79$.
Since 79 is a prime number, then $\left\{\begin{array}{c}N+M=79 \\ N-M=1\end{array}\right.$. The solution to this system is $N=40, M=39$. Therefore, t... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 18,757 |
Problem 1. A student started working on an exam between 9 and 10 o'clock and finished between 13 and 14 o'clock. Find the exact time spent by the applicant on completing the task, given that at the beginning and the end of the work, the hour and minute hands, having swapped places, occupied the same positions on the cl... | # Solution.
From the initial and final positions of the hour hand, it is clear that the student started writing before 9:30 and finished after 1:30 PM. Therefore, the exact time \( x \), spent on the work, satisfies the inequalities \( 4 < x < 5 \). If after completing the work, the minute hand travels the distance th... | \frac{60}{13} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 18,758 |
Problem 2. Solve the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+2}\left(1+\frac{1}{x+1}\right)\right)=x$.
Answer: 2 | Solution.
$1+\frac{1}{x+1}=\frac{x+2}{x+1}$, therefore the given equation is equivalent to the equation $1+\frac{3}{x+3}\left(1+\frac{2}{x+1}\right)=x$ under the condition that $\mathrm{x} \neq-2$. Acting similarly, we get that $1+\frac{3}{x+3}=x$, where $\mathrm{x} \neq-2$ and $\mathrm{x} \neq-3$. The roots of this e... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 18,759 |
Problem 3. At the olympiad, $\mathrm{m}>1$ schoolchildren solved $\mathrm{n}>1$ problems. All schoolchildren solved a different number of problems. All problems were solved by a different number of schoolchildren. Prove that one of the schoolchildren solved exactly one problem. | # Solution.
If a student who did not solve any problems is found, we will not consider him. Then, if there is a problem that was not solved by any of the students, we will not consider it. Still, all students solved a different number of problems, and all problems were solved by a different number of students. Let $m^... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 18,760 |
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