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1. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the abscissa of the point of intersection.
# | # Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint of exa... | Answer. 1.
First solution. We will prove by induction on $N$ that there is one more even matching than odd. For $N=1$, the statement is obvious: there is only one matching, and it is even. Now we will prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. Let the marked points be $A_{1}, A_{2}, ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.6. Petya chose a natural number $a>1$ and wrote down fifteen numbers $1+a, 1+a^{2}, 1+a^{3}, \ldots, 1+a^{15}$ on the board. Then he erased several numbers so that any two remaining numbers are coprime. What is the maximum number of numbers that could remain on the board?
(O. Podlipsky) | # Answer. 4 numbers.
Solution. First, we will show that there cannot be more than four such numbers. Note that if $k$ is odd, then the number $1+a^{n k}=1^{k}+\left(a^{n}\right)^{k}$ is divisible by $1+a^{n}$. Next, each of the numbers $1,2, \ldots, 15$ has one of the forms $k, 2 k, 4 k, 8 k$, where $k$ is odd. Thus, ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$, the median $A M$ is perpendicular to the bisector $B D$. Find the perimeter of the triangle, given that $A B=1$, and the lengths of all sides are integers. | Answer: 5.
Solution. Let $O$ be the point of intersection of the median $A M$ and the bisector $B D$. Triangles $A B O$ and $M B O$ are congruent (by the common side $B O$ and the two adjacent angles). Therefore, $A B = B M = 1$ and $C M = 1$, since $A M$ is a median. Thus, $A B = 1, B C = 2$. By the triangle inequali... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. Given a $5 \times 5$ square grid of cells. In one move, you can write a number in any cell, equal to the number of cells adjacent to it by side that already contain numbers. After 25 moves, each cell will contain a number. Prove that the value of the sum of all the resulting numbers does not depend on the order in... | Solution. Consider all unit segments that are common sides for two cells. There are exactly forty such segments - 20 vertical and 20 horizontal. If a segment separates two filled cells, we will say that it is "painted." Note that when we write a number in a cell, it indicates the number of segments that were not painte... | 40 | Combinatorics | proof | Yes | Yes | olympiads | false |
3. Find the smallest natural number $n$ such that the sum of the digits of each of the numbers $n$ and $n+1$ is divisible by 17. | Answer: 8899.
Solution. If $n$ does not end in 9, then the sums of the digits of the numbers $n$ and $n+1$ differ by 1 and cannot both be divisible by 17. Let $n=\overline{m 99 \ldots 9}$, where the end consists of $k$ nines, and the number $m$ has a sum of digits $s$ and does not end in 9. Then the sum of the digits ... | 8899 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. In the Rhind Papyrus (Ancient Egypt), among other information, there are decompositions of fractions into the sum of fractions with a numerator of 1, for example,
$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$
One of the denominators here is replaced by the letter $x$. Find this denominator. | # Solution:
First, find $\stackrel{1}{-.}$ from the equation:
$\frac{2}{73}=\frac{1}{60}+\frac{1}{219}+\frac{1}{292}+\frac{1}{x}$
$\frac{2}{73}-\frac{1}{219}-\frac{1}{292}-\frac{1}{60}=\frac{1}{x}$
$\frac{2}{73}-\frac{1}{73 \cdot 3}-\frac{1}{73 \cdot 4}-\frac{1}{60}=\frac{1}{x}$
$\frac{2 \cdot 3 \cdot 4-4-3}{73 \c... | 365 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. At a certain moment, Anna measured the angle between the hour and minute hands of her clock. Exactly one hour later, she measured the angle between the hands again. The angle turned out to be the same. What could this angle be? (Consider all cases).
# | # Solution:
After 1 hour, the minute hand remains in its place. During this time, the hour hand has turned $30^{\circ}$. Since the angle has not changed, the minute hand must be dividing one of the
. Prove that the sum of the reciprocals of these products doe... | Let $x_{1}, x_{2}$ be the roots of the first quadratic polynomial and $x_{3}, x_{4}$, then we need to prove that $\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}$ does not depend on $p$ and $q$. Transform the given expression:
$\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3. Solve the equation in integers $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=0$.
## Answer: $x=y=z=t=0$ | Note that $x$ is even. Let $x=2x_{1}$, then we get
$4x_{1}^{4}-8y_{1}^{4}-z^{4}-2t^{4}=0$. Therefore, $z=2z_{1}$ and $2x_{1}^{4}-4y_{1}^{4}-8z_{1}^{4}-t^{4}=0$. Similarly, $t=2t_{1}$ and $x_{1... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On the table lie three balls, touching each other pairwise. The radii of the balls form a geometric progression with a common ratio $q \neq 1$. The radius of the middle one is 2012. Find the ratio of the sum of the squares of the sides of the triangle formed by the points of contact of the balls with the table to th... | Let the radius of the smaller of the balls be $r$, then the radii of the others are $r q=2012$ and $r^{2}$, and the points of contact of the balls with the table are denoted as $A, B, C$ respectively. Consider the section of two balls by a plane perpendicular to the table and passing through the centers of these balls ... | 4024 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A natural number has a digit sum of 2013. The next number has a smaller digit sum and is not divisible by 4. What is the digit sum of the next natural number. | Answer: 2005.
Solution. The sum of the digits of the next number decreases only if it ends in nine.
It cannot end in two or more nines, as the next number would end in two zeros and be divisible by 4.
Therefore, the 9 is replaced by 0, but the tens digit increases by 1. The sum of the digits of the next number is 20... | 2005 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of all seven-digit natural numbers for which the digits in the decimal representation are in strictly increasing order up to the middle, and then in strictly decreasing order. For example, the number 1358620 would fit.
# | # Solution.
If the middle digit of a seven-digit number is 4 (it cannot be less), the number of desired numbers can be obtained by erasing one digit to the right of the " " "middle digit of the final number" in the number 12343210. There are 4 such numbers.
If the middle digit of a seven-digit number is 5, the number... | 7608 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the following equation for positive $x$.
$$
x^{2014} + 2014^{2013} = x^{2013} + 2014^{2014}
$$ | Answer: 2014.
Solution. Rewrite the equation in the following form $x^{2013}(x-1)=2014^{2013}(2014-1)$. The solution is $x=2014$. Since the function on the left, for $x>1$ this function is increasing, as the product of two positive increasing functions, the intersection with a constant function can be no more than one... | 2014 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that in the pyramid $A B C D$ with vertex $D$, the sum $\angle A B D+\angle D B C=\pi$. Find the length of the segment $D L$, where $L$ is the base of the bisector $B L$ of triangle $A B C$, if it is known that
$$
A B=9, B C=6, A C=5, D B=1 .
$$ | Answer: 7.
Solution. Let point $M$ lie on the line, outside the segment $B C$, beyond point $B$. From the condition, we have the equality of angles $\angle A B D=\angle D B M$. Then the projection of line $B D$ onto the plane $A B C$ is the bisector $B K$ of angle $A B M$. The bisectors $B K$ and $B L$ are perpendicul... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.2 (7 points)
In a bag, there are 70 balls that differ only in color: 20 red, 20 blue, 20 yellow, and the rest are black and white.
What is the minimum number of balls that need to be drawn from the bag, without seeing them, to ensure that among them there are at least 10 balls of the same color? | # Solution:
By drawing 37 balls, we risk getting 9 red, 9 blue, and 9 yellow balls, and we won't have ten balls of one color. If we draw 38 balls, the total number of red, blue, and yellow balls among them will be at least 28, and the number of balls of one of these colors will be at least ten (since $28 > 3 \cdot 9$)... | 38 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.3 (7 points)
$H$ is the point of intersection of the altitudes of an acute-angled triangle $ABC$. It is known that $HC = BA$. Find the angle $ACB$.
# | # Solution:
Let $\mathrm{D}$ be the foot of the altitude dropped from vertex A to side BC. Angles НСВ and DAB are equal as acute angles with mutually perpendicular sides. Therefore, $\triangle \mathrm{CHD}=\Delta \mathrm{ABD}$ (by hypotenuse and acute angle). Hence, $=\mathrm{AD}$, i.e., triangle $\mathrm{ACD}$ is iso... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.4 (7 points)
A palindrome is a number, letter combination, word, or text that reads the same in both directions. How much time in a day do palindromes appear on the clock display, if the clock shows time from 00.00.00 to 23.59.59? | # Solution:
If the digits on the display are ab.cd.mn, then $\mathrm{a}=0,1,2,0 \leq \mathrm{b} \leq 9,0 \leq \mathrm{c} \leq 5,0 \leq \mathrm{d} \leq 9,0 \leq \mathrm{m} \leq 5$, $0 \leq \mathrm{n} \leq 9$. Therefore, if $\mathrm{a}=\mathrm{n}, \mathrm{b}=\mathrm{m}, \mathrm{c}=\mathrm{d}$, then the symmetric number ... | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 9.5 (7 points)
In a store, there are buttons of six colors. What is the smallest number of buttons that need to be bought so that they can be sewn in a row, such that for any two different colors in the row, there are two adjacent buttons of these colors? | # Solution:
From the condition, it follows that for each fixed color A, a button of this color must appear in a pair with a button of each of the other 5 colors. In a row, a button has no more than two neighbors, so a button of color A must appear at least 3 times. The same applies to each other color. Thus, there sho... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.4. From sticks of the same length, a row of 800 hexagons was laid out as shown in the figure. How many sticks were used in total?
Answer, option 1. 5001.
Answer, option 2. 2501.
Answer, option 3. 3001.
Answer, option 4. 4001. | Solution option 1. In the first hexagon, there are 6 sticks. Building each subsequent hexagon adds 5 sticks. In total, it will be $6+5 \cdot 999=5001$ sticks. | 5001 | Geometry | MCQ | Yes | Yes | olympiads | false |
4.4. In a $3 \times 4$ rectangle, natural numbers $1,2,3, \ldots, 12$ were written, each exactly once. The table had the property that in each column, the sum of the top two numbers was twice the bottom number. Over time, some numbers were erased. Find all possible numbers that could have been written in place of $\sta... | Solution 1. According to the condition, the number 2 is written in the bottom-left cell. The sum of the two unknown numbers in the second column is 10, and in the third column, it is 16. Therefore, the sum of the numbers in the first three columns is $6+15+24$, and the sum of all numbers in the table is $1+2+\ldots+12=... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.4. 10 children stood in a circle. Each thought of an integer and told it to their clockwise neighbor. Then each loudly announced the sum of their number and the number of the counterclockwise neighbor. The first said "10", the next clockwise - "9", the next clockwise - "8", and so on, the ninth said "2". What number ... | Solution option 1. The sum of all ten thought-of numbers will be obtained if we add what the first child said to what the third, fifth, seventh, and ninth children said. Similarly, we add what the second, fourth, sixth, eighth, and tenth children said. Again, we get the sum of all ten thought-of numbers, so the equatio... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
7.4. In isosceles triangle $ABC$, the base $AC$ is equal to $x$, and the lateral side is 12. On the ray $AC$, point $D$ is marked such that $AD=24$. A perpendicular $DE$ is dropped from point $D$ to line $AB$. Find $x$ if it is known that $BE=6$.
^{2}$ is also the square of some integer $r=n-2$, which is less than $t$.
We get that $t^{2}-r^{2}=(t+r)(t-r)=7$.
The numbers $(t+r)$ and $(t-r)$ are natural and the first is greater than the second.
Thus, $(t+r)=7$, and $(t-r)=1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 9.4
On the sides $B C$ and $C D$ of the square $A B C D$, points $M$ and $K$ are marked such that $\angle B A M = \angle C K M = 30^{\circ}$. Find the angle $A K D$.
## Number of points 7
# | # Answer:
$75^{\circ}$
## Solution
## First Solution
If we find angle $A K D$, it is only present in triangle $A K D$, and we do not know another angle in this triangle.
All the information in the problem is above line $A K$. Therefore, we need to move upwards.
And we need to find angle $A K M$. Triangle $A K M$ ... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In a beach soccer tournament, 17 teams participate, and each team plays against each other exactly once. Teams are awarded 3 points for a win in regular time, 2 points for a win in extra time, and 1 point for a win on penalties. The losing team does not receive any points. What is the maximum number of teams that ca... | Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 5 points each. The total points include all points scored by these teams in matches against each other (at least 1) and possibly in matches against other teams: $5 n \geq(n-1) n / 2$. Hence, $n \leq 11$.
Example: Place eleven teams in a... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the number of roots of the equation
$$
|x|+|x+1|+\ldots+|x+2018|=x^{2}+2018 x-2019
$$
(V. Dubinskaya) | Answer: 2.
Solution. For $x \in(-2019,1)$ there are no roots, since on the given interval the left side is non-negative, while the right side is negative.
For $x \in[1, \infty)$ all absolute values are resolved with a positive sign, so the equation will take the form $g(x)=0$, where $g(x)=x^{2}-x-2009+(1+2+\ldots+201... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Initially, there are 111 pieces of clay of the same mass on the table. In one operation, you can choose several groups with the same number of pieces and in each group, combine all the clay into one piece. What is the minimum number of operations required to get exactly 11 pieces, any two of which have different ... | Answer. In two operations.
Solution. Let the mass of one original piece be 1. If in the first operation there are $k$ pieces in each group, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to obtain eleven pieces of different masses in one operation.
We will show that the required re... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. The younger brother takes 25 minutes to reach school, while the older brother takes 15 minutes to walk the same route. How many minutes after the younger brother leaves home will the older brother catch up to him if he leaves 8 minutes later? | Answer. in 17 minutes. Solution. See problem 7.1 | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$, side $A C$ is the largest. Points $M$ and $N$ on side $A C$ are such that $A M=A B$ and $C N=C B$. It is known that angle $N B M$ is three times smaller than angle $A B C$. Find $\angle A B C$. | Answer: $108^{\circ}$. Solution: Let $\angle N B M=x$. Then $\angle A B M+\angle N B C=\angle A B C+\angle N B M=4 x . \quad$ On the other hand, $\angle A B M+\angle N B C=\angle B M N+\angle B N M=180^{\circ}-\angle N B M=180^{\circ}-x$. Therefore, we have the equation $4 x=180^{\circ}-x$. Thus, $5 x=180^{\circ}, x=36... | 108 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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.
We will call suc... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
# Problem №2
In an $8 \times 8$ frame that is 2 cells wide (see figure), there are a total of 48 cells.
How many cells are in a $254 \times 254$ frame that is 2 cells wide? | Answer: 2016.
## Solution
First method. Cut the frame into four identical rectangles as shown in the figure. The width of the rectangles is equal to the width of the frame, i.e., 2 cells. The length of each
.
Since $\mathrm{BK} \| \mathrm{AD}$, then $\angle \mathrm{KBE}=\angle \mathrm{DAE}$.
Moreover, $\angle \mathrm{KEB}=\angle \mathrm{DEA}$ and $\mathrm{AE}=\mathrm{BE}$, thus triangles $\mathrm{BKE}$ ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Four princesses each thought of a two-digit number, and Ivan thought of a four-digit number. After they wrote their numbers in a row in some order, the result was 132040530321. Find Ivan's number. | Solution. Let's go through the options. Option 1320 is not suitable because the remaining part of the long number is divided into fragments of two adjacent digits: 40, 53, 03, 21, and the fragment 03 is impossible, as it is not a two-digit number. Option 3204 is impossible due to the invalid fragment 05 (or the fragmen... | 5303 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. There are more than 30 people in the class, but less than 40. Any boy is friends with three girls, and any girl is friends with five boys. How many people are in the class | Solution. Let $\mathrm{m}$ be the number of boys, $\mathrm{d}$ be the number of girls, and $\mathrm{r}$ be the number of friendly pairs "boy-girl". According to the problem, $\mathrm{r}=3 \mathrm{~m}$ and $\mathrm{r}=5 \mathrm{~d}$. Therefore, $\mathrm{r}$ is divisible by 3 and 5, and thus by 15: $\mathrm{r}=15 \mathrm... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.2. Little kids were eating candies. Each one ate 11 candies less than all the others together, but still more than one candy. How many candies were eaten in total? | Answer: 33 candies.
Solution: Let $S$ be the total number of candies eaten by the children. If one of the children ate $a$ candies, then according to the condition, all the others ate $a+11$ candies, and thus all together ate $S=a+(a+11)=2a+11$ candies. This reasoning is valid for each child, so all the children ate t... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. The number 49 is written on the board. In one move, it is allowed to either double the number or erase its last digit. Is it possible to get the number 50 in several moves? | Answer: Yes.
Solution: The number 50 can be obtained by doubling 25, and 25 can be obtained by erasing the last digit of the number 256, which is a power of two. Thus, the necessary chain of transformations can look like this: $49 \rightarrow 4 \rightarrow 8 \rightarrow 16 \rightarrow 32 \rightarrow 64 \rightarrow 128... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Given an acute-angled triangle $A B C$. Point $M$ is the intersection point of its altitudes. Find the angle $A$, if it is known that $A M=B C$.
---
The text has been translated while preserving the original formatting and line breaks. | 83. Given an acute-angled triangle $A B C$. Point $M$ is the intersection point of its altitudes. Find the angle $A$, if it is known that $A M=B C$.
Answer: $45^{\circ}$. Hint Let $\mathrm{K}$ be the foot of the altitude from point В. We will prove that triangles АМ К and BKC are equal. Indeed, we have right triangles... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. The numbers $2^{2021}$ and $5^{2021}$ are written one after another. How many digits are written in total?
# | # Solution
Let the number $2^{2021}$ have $\mathrm{k}$ digits, and the number $5^{2021}$ have $\mathrm{m}$ digits, then the number of digits in the desired number is $\mathrm{k}+\mathrm{m}$. $10^{k-1}<2^{2021}<10^{k}, 10^{m-1}<5^{2021}<10^{m}$, therefore, $10^{k+m-2}<$ $10^{2021}<10^{m+k}$ and $\mathrm{k}+\mathrm{m}=2... | 2022 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the inequality $\sqrt{2 x^{2}-8 x+6}+\sqrt{4 x-x^{2}-3}<x-1$ | Note that all solutions to the original inequality exist if the expressions under the square roots are non-negative. These inequalities are simultaneously satisfied only under the condition $x^{2}-4 x+3=0$. This equation has two roots, 1 and 3. Checking shows that the original inequality has a unique solution 3.
Answe... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11.1. At first, there were natural numbers, from 1 to 2021. And they were all white. Then the underachiever Borya painted every third number blue. Then the top student Vova came and painted every fifth number red. How many numbers remained white? (7 points)
# | # Solution
Borya repainted the numbers divisible by 3, a total of [2021:3]=673 numbers. Vova repainted [2021:5]=404 numbers. 673+404=1077. However, numbers divisible by 15 were counted twice. [2021:15]=134. Therefore, the number of white numbers remaining is 2021-1077+134=1078.
Answer: 1078 white numbers
| criteria ... | 1078 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. Doctor Vaccinov and Doctor Injectionov vaccinated all the residents of the village of Covido. Vaccinov thought: if I had given 40% more vaccinations, then Injectionov's share would have decreased by 60%. And how would Injectionov's share change if Vaccinov had given 50% more vaccinations? (7 points)
# | # Solution
$40 \%$ of the injections given by Privevkin equals $60 \%$ of the number of injections given by Ukolkin, so Privevkin gave 1.5 times more injections. Therefore, an increase in Privevkin's share by $n \%$ would decrease Ukolkin's share by $1.5 n \%$.
Answer: It would decrease by $75 \%$.
| criteria | poin... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. Solve the equation $2021 x^{2021}-2021+x=\sqrt[2021]{2022-2021 x}$. (7 points)
# | # Solution
The function $f(x)=2021x^{2021}-2021+x$ is increasing, while the function $g(x)=\sqrt[2021]{2022-2021x}$ is decreasing. Therefore, the equation $f(x)=g(x)$ has no more than one root. However, it is obvious that $f(1)=g(1)$.
Answer: $x=1$.
| criteria | points |
| :--- | :---: |
| correct solution | 7 |
| P... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Four non-zero numbers are written on the board, and the sum of any three of them is less than the fourth number. What is the smallest number of negative numbers that can be written on the board? Justify your answer. | Solution: Let the numbers on the board be $a \geqslant b \geqslant c \geqslant d$. The condition of the problem is equivalent to the inequality $a+b+c < d$ for optimality | 7 points |
| There is a proof that there are no fewer than three negative numbers (in the absence of an example with three numbers) | 3 points |
| ... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.5. At the exchange office, only the following operations can be performed:
1) exchange 2 gold coins for three silver coins and one copper coin;
2) exchange 5 silver coins for three gold coins and one copper coin.
Nikolai had only silver coins. After several visits to the exchange office, he had fewer silver coins, ... | Solution: As a result of each operation, Nikolai acquires exactly 1 copper coin, which means there were exactly 50 operations in total. Of these, some (let's say \(a\)) were of the first type, and the rest \(50-a\) were of the second type. On operations of the first type, Nikolai spent \(2a\) gold coins, and on operati... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. On an $8 \times 8$ chessboard, tokens are placed according to the following rule. Initially, the board is empty. A move consists of placing a token on any free square. With this move, exactly one of the tokens that ends up adjacent to it is removed from the board (if there is such an adjacent token). What is the max... | Answer: No more than 61 chips can be placed. | 61 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In a certain country, there are 47 cities. Each city has a bus station from which buses run to other cities in the country and possibly abroad. A traveler studied the schedule and determined the number of internal bus routes departing from each city. It turned out that if the city of Lake is not considered, then for... | Solution.
Note that external lines are not considered in this problem.
There are a total of 47 variants of the number of internal lines - from 0 to 46. Note that the existence of a city with 46 lines excludes the existence of a city with 0 lines and vice versa.
Suppose there is a city with 46 lines. Then the smalles... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Find all natural numbers $n \geq 2$ such that $20^{n}+19^{n}$ is divisible by $20^{n-2}+19^{n-2}$. | Solution.
Consider the expression
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)
$$
By the condition, it is divisible by $20^{n-2}+19^{n-2}$. On the other hand,
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)=20^{n-2}\left(20^{2}-19^{2}\right)=20^{n-2} \cdot 39
$$
Note that $20^{n-2}$ and ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. The teacher suggested that his students - Kolya and Seryozha - solve the same number of problems during the lesson. After some time from the start of the lesson, it turned out that Kolya had solved a third of what Seryozha had left to solve, and Seryozha had left to solve half of what he had already completed. Ser... | # 8.1. Answer: 16.
Solution. Let Tanya have solved x problems, then she has $\frac{x}{2}$ problems left to solve.
Let $\mathrm{t}_{1}$ be the time interval after which Tanya and Kolya evaluated the shares of solved and remaining problems, and $\mathrm{t}_{2}$ be the remaining time. Since Tanya's problem-solving speed... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. At a round table, 10 people are sitting, some of them are knights, and the rest are liars (knights always tell the truth, while liars always lie). It is known that among them, there is at least one knight and at least one liar. What is the maximum number of people sitting at the table who can say: "Both of my neig... | # Answer. 9.
Solution. Note that all 10 could not have said such a phrase. Since at the table there is both a knight and a liar, there will be a liar and a knight sitting next to each other. But then this knight does not have both neighbors as knights. If, however, at the table there are 9 liars and 1 knight, then eac... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.3. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $5 \times 5$ square so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.) | Answer: 4.
Solution: Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked in Fig. 7. Since none of these corners can be painted, at least one cell in each of these corners must be painted. Note that one corner cannot paint cells of two marke... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime | Answer: 5.
Solution: Note that an odd semiprime number can only be the sum of two and an odd prime number.
We will show that three consecutive odd numbers $2n+1$, $2n+3$, and $2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers $2n-1$, $2n+1$, and $2n+3$ are... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.8. In the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip i... | # Answer. 50.
Solution. Example. The chip 50 is sequentially exchanged 99 times with the next one counterclockwise. We get the required arrangement.
There are several ways to prove the estimate, below we provide two of them.
The first way. Suppose that for some $k<50$ the required arrangement is obtained.
At any mo... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.4. Point $E$ is the midpoint of side $A B$ of parallelogram $A B C D$. On segment $D E$, there is a point $F$ such that $A D = B F$. Find the measure of angle $C F D$. | Answer: $90^{\circ}$.
Solution. Extend $D E$ to intersect line $B C$ at point $K$ (see Fig. 8.4). Since $B K \| A D$, then $\angle K B E = \angle D A E$. Moreover, $\angle K E B = \angle D E A$ and $A E = B E$, therefore, triangles $B K E$ and $A D E$ are equal. Thus, $B K = A D = B C$.
Therefore, in triangle $CFK$, ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.5. Kuzya cut a convex paper 67-gon along a straight line into two polygons, then similarly cut one of the two resulting polygons, then one of the three resulting ones, and so on. In the end, he got eight $n$-gons. Find all possible values of $n$. | Answer: $n=11$.
Solution. A straight-line cut can be of three types: from side to side, from a vertex to a side, and from a vertex to a vertex. Therefore, after one cut, the total number of sides of the polygons increases by 4, 3, or 2, respectively. Kuzya made 7 cuts, so the number of sides added is at least 14 but n... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.6. A triangle is divided into triangular cells as shown in the figure. A natural number is written in each cell. For each side of the triangle, there are four layers parallel to this side, containing seven, five, three, and one cell, respectively. It turns out that the sum of the numbers in each of these twelve layer... | Answer: 22.
Solution. Example. In each of the three corner cells, we write the number 3, and in each of the others, we write the number 1. Then the sum of the written numbers is $3 \cdot 3 + 13 \cdot 1 = 22$, and the sums of the numbers in the layers are: 11, 5, 3, and 3, respectively.
Estimation. Any corner cell is ... | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 4.2. Find the smallest number in which all digits are different, and the sum of all digits is 32. | Answer: 26789.
Solution. For a four-digit number with different digits, the maximum possible sum of digits is $9+8+7+6=30<32$, so the number we need must be at least five digits.
We will try to make the first digit as small as possible. It is clear that it is no less than 2. We will place 2 in the first position. The... | 26789 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
Problem 4.4. Lev Alex decided to count the stripes on Marty the zebra (black and white stripes alternate). It turned out that there is one more black stripe than white ones. Alex also noticed that all white stripes are of the same width, while black stripes can be wide or narrow, and there are 7 more white stripes than... | # Answer: 8.
Solution. First, let's look only at the wide black stripes. There are 7 fewer of them than white ones. If we add the narrow black stripes to the wide black ones, we get all the black stripes, which are 1 more than the white ones. Therefore, to find the number of narrow black stripes, we need to first "com... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.5. Hooligan Dima laid out a structure 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 diagram.
It is known that one toothpick burns for 10 seconds. How many seconds will it take for the entire struct... | 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).
, 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 |
Problem 5.6. Several stones are arranged in 5 piles. It is known that
- in the fifth pile, there are six times more stones than in the third;
- in the second pile, there are twice as many stones as in the third and fifth combined;
- in the first pile, there are three times fewer stones than in the fifth, and ten fewer... | Answer: 60.
Solution. Let's denote the number of stones in the third pile as a rectangle. Then, since there are 6 times more stones in the fifth pile, this can be represented as 6 rectangles. In the second pile, the number of stones is twice the sum of the third and fifth piles, which is $(1+6) \cdot 2=14$ rectangles.... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
Problem 6.1. Five consecutive natural numbers are written in a row. The sum of the three smallest of them is 60. What is the sum of the three largest? | Answer: 66.
Solution. The fifth number is 4 more than the first, and the fourth is 2 more than the second. Then the sum of the three largest numbers is $2+4=6$ more than the sum of the three smallest, and it is equal to $60+6=66$. | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.6. On an island, there live knights who always tell the truth, and liars who always lie. One day, 65 inhabitants of the island gathered for a meeting. Each of them, in turn, made the statement: "Among the previously made statements, there are exactly 20 fewer true statements than false ones." How many knights... | Answer: 23.
Solution. The first 20 people are liars, because before them, there were no more than 19 statements made, which means there cannot be exactly 20 fewer true statements than false ones.
Next, the 21st person tells the truth (since before him, there were exactly 20 false statements), so he is a knight. Then ... | 23 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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 $5+3 \cdot 3-2=12$.
$ candies. Since the total number of candies for both boys did not change after the game, we get $7 x=3(50-x)$. Solvin... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.3. Numbers were placed in the cells of a square such that the sums of the numbers in each vertical, horizontal, and each diagonal of three cells are equal. Then some numbers were hidden. What is the sum of the numbers in the two shaded cells?
| 16 | | |
| :--- | :--- | :--- |
| | | 10 |
| 8 | | 12 | | Answer: 34.
Solution. Let $a$ be the number in the top right cell, and $b$ be the number in the central cell. The sum of the diagonal containing the number 8 and the sum of the numbers in the right column have a common term $a$, so $8+b=10+12$, from which $b=14$. The sums of the three numbers along both diagonals have... | 34 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.4. Masha and Olya bought many identical pens for the new school year at the store. It is known that one pen costs an integer number of rubles, more than 10. Masha bought pens for exactly 357 rubles, and Olya - for exactly 441 rubles. How many pens did they buy in total? | Answer: 38.
Solution. Let the pen cost $r$ rubles. Then the numbers 357 and 441 are divisible by $d$. Since the greatest common divisor of the numbers $357=3 \cdot 7 \cdot 17$ and $441=3^{2} \cdot 7^{2}$ is $3 \cdot 7$, then 21 is also divisible by $r$. Since $r>10$, then $r=21$. Therefore, the total number of pens bo... | 38 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. In each room of the hotel, no more than 3 people can be accommodated. The hotel manager knows that a group of 100 football fans, who support three different teams, will soon arrive. In one room, only men or only women can be accommodated; also, fans of different teams cannot be accommodated together. How m... | Answer: 37.
Solution. All fans are divided into six groups: men/women, supporting the first team; men/women, supporting the second team; men/women, supporting the third team. Only people from the same group can be housed in one room. We divide the number of people in each of the six groups by 3 with a remainder. If th... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. The numbers from 1 to 200 were arranged in a random order on a circle such that the distances between adjacent numbers on the circle are the same.
For any number, the following is true: if you consider 99 numbers standing clockwise from it and 99 numbers standing counterclockwise from it, there will be an... | Answer: 114.
Solution. Consider the number 2. Less than it is only the number 1. Since it is unique, it cannot be in any of the groups relative to the number 2. Therefore, 1 must be opposite 2.
Consider the number 4. The numbers less than it are three in number - an odd count. This means that one of them should not b... | 114 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. In the chat of students from one of the schools, a vote was 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 voting.
Then, 80 more people participated in the voting, voting only for October 22. After that, the ... | 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 |
Problem 8.3. In the class, there are 29 students: some are excellent students and some are troublemakers. Excellent students always tell the truth, while troublemakers always lie.
All students in this class sat around a round table.
- Several students said: “There is exactly one troublemaker next to me.”
- All other ... | Answer: 10.
Solution. If there were three straight-A students in a row, the middle one would definitely lie. Therefore, among any three consecutive people, there must be at least one troublemaker.
Choose an arbitrary troublemaker. Assign this person the number 29, and number all the people following them clockwise fr... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively, such that $A D=E C$. It turns out that $B D=E D, \angle B D C=\angle D E B$. Find the length of segment $A C$, given that $A B=7$ and $B E=2$. | Answer: 12.
Solution. Note that triangles $D E C$ and $B D A$ are equal. Indeed, $D E=B D, E C=$ $D A$ and $\angle D E C=180^{\circ}-\angle B E D=180^{\circ}-\angle B D C=\angle B D A$. From this, it follows that $D C=A B=7$ and $\angle D C E=\angle B A D$ (Fig. 3). From the equality of these angles, it follows that t... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. The numbers $1, 2, 3, \ldots, 235$ were written on a board. Petya erased some of them. It turned out that among the remaining numbers, no number is divisible by the difference of any two others. What is the maximum number of numbers that could remain on the board? | Answer: 118.
Solution. 118 odd numbers could remain on the board: none of them is divisible by the difference of any two others, because this difference is even.
Suppose at least 119 numbers could remain. Consider 118 sets: 117 pairs $(1,2)$, $(3,4),(5,6), \ldots,(233,234)$ and one number 235. By the Pigeonhole Princ... | 118 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 |
Problem 8.8. In how many ways can all natural numbers from 1 to 200 be painted in red and blue so that the sum of any two different numbers of the same color is never equal to a power of two? | Answer: 256.
## Solution.
Lemma. If the sum of two natural numbers is a power of two, then the powers of 2 in the prime factorizations of these two numbers are the same.
Proof of the lemma. Let the sum of the numbers $a=2^{\alpha} \cdot a_{1}$ and $b=2^{\beta} \cdot b_{1}$, where $a_{1}$ and $b_{1}$ are odd, be $2^{... | 256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. On an island, there live red, yellow, green, and blue chameleons.
- On a cloudy day, either one red chameleon changes its color to yellow, or one green chameleon changes its color to blue.
- On a sunny day, either one red chameleon changes its color to green, or one yellow chameleon changes its color to b... | Answer: 11.
Solution. Let $A$ be the number of green chameleons on the island, and $B$ be the number of yellow chameleons. Consider the quantity $A-B$. Note that each cloudy day it decreases by 1, and each sunny day it increases by 1. Since there were $18-12=6$ more sunny days than cloudy days in September, the quanti... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. Denis has cards with numbers from 1 to 50. How many ways are there to choose two cards such that the difference between the numbers on the cards is 11, and the product is divisible by 5?
The order of the selected cards does not matter: for example, the way to choose cards with numbers 5 and 16, and the wa... | Answer: 15.
Solution. For the product to be divisible by 5, it is necessary and sufficient that one of the factors is divisible by 5. Let $n$ be the chosen number divisible by 5, then its pair should be $n-11$ or $n+11$, and both chosen numbers must be natural numbers. Clearly, for $n=5,10,40,45,50$ there is only one ... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. Traders Andrey and Boris bought 60 bags of potatoes each from the same farmer. All the bags cost the same.
Andrey sold all his bags, increasing their price by $100 \%$. Boris first increased the price by $60 \%$, and after selling 15 bags, increased the price by another $40 \%$ and sold the remaining 45 b... | Answer: 250.
Solution. Let the bag cost the farmer $x$ rubles. Andrei and Boris spent the same amount on buying 60 bags.
From the condition, it follows that Andrei sold 60 bags for $2 x$ rubles each, i.e., he received $60 \cdot 2 x$ rubles. Boris, on the other hand, sold 15 bags at a price of $1.6 x$ rubles and then ... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
Problem 9.6. In triangle $ABC$, the angles $\angle B=30^{\circ}$ and $\angle A=90^{\circ}$ are known. On side $AC$, point $K$ is marked, and on side $BC$, points $L$ and $M$ are marked such that $KL=KM$ (point $L$ lies on segment $BM$).
Find the length of segment $LM$, if it is known that $AK=4$, $BL=31$, and $MC=3$.
... | 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 |
Problem 9.8. Integers $n$ and $m$ satisfy the inequalities $3 n - m \leq 26$, $3 m - 2 n < 46$. What can $2 n + m$ be? List all possible options. | Answer: 36.
Solution. Since the numbers $m$ and $n$ are integers, the values of the expressions in the condition $3 n-m$, $n+m$, $3 m-2 n$ are also integers. Then
$$
\left\{\begin{array}{l}
3 n-m \leqslant 4 \\
n+m \geqslant 27 \\
3 m-2 n \leqslant 45
\end{array}\right.
$$
By adding three times the first inequality ... | 36 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Task 10.1. Find the largest 12-digit number $N$ that satisfies the following two conditions:
- In the decimal representation of the number $N$, there are six digits «4» and six digits «7»;
- In the decimal representation of the number $N$, no four consecutive digits form the number «7444». | Answer: 777744744744.
Solution. It is clear that the number 777744744744 meets the condition of the problem.
Suppose there exists a larger number, then it must start with four sevens. In this number, we select the largest number of consecutive fours, let there be $k$ of them. If $k \geqslant 3$, then by selecting amo... | 777744744744 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 |
Problem 10.8. For what least natural $a$ does the numerical interval ( $a, 3 a$ ) contain exactly 50 perfect squares? | Answer: 4486.
Solution. Let's choose the largest natural $n$ such that $n^{2} \leqslant a$. Then
$$
n^{2} \leqslant a\frac{49}{\sqrt{3}-1}=\frac{49(\sqrt{3}+1)}{2}>\frac{49(1.7+1)}{2}>\frac{132}{2}=66
$$
From the fact that $n+1>66$, it follows that $n \geqslant 66$ and $a \geqslant 66^{2}$.
If on the corresponding ... | 4486 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 11.1. The product of nine consecutive natural numbers is divisible by 1111. What is the smallest possible value that the arithmetic mean of these nine numbers can take? | Answer: 97.
Solution. Let these nine numbers be $-n, n+1, \ldots, n+8$ for some natural number $n$. It is clear that their arithmetic mean is $n+4$.
For the product to be divisible by $1111=11 \cdot 101$, it is necessary and sufficient that at least one of the factors is divisible by 11, and at least one of the facto... | 97 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 as $x$ (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 |
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