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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 | 15,285 |
1. Can we find five different natural numbers, each of which is not divisible by 3, 4, or 5, but the sum of any three is divisible by 3, the sum of any four is divisible by 4, and the sum of all five is divisible by 5? | Answer: for example, 1, 61, 121, 181, and 241.
Solution. The five numbers mentioned can be written in the form $60k + 1$, where $k$ takes the values 0, 1, 2, 3, 4. Therefore, the sum of any three numbers
$$
\left(60k_{1} + 1\right) + \left(60k_{2} + 1\right) + \left(60k_{3} + 1\right) = 60\left(k_{1} + k_{2} + k_{3}\... | 1,61,121,181,241 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,286 |
2. It is known that $2 x+y^{2}+z^{2} \leqslant 2$. Prove that $x+y+z \leqslant 2$. | Solution. By the condition $2 x \leqslant 2-y^{2}-z^{2}$, it follows that $2 x+2 y+2 z \leqslant 2-\left(y^{2}-2 y\right)-\left(z^{2}-2 z\right)$. Completing the squares, we get
$$
2 x+2 y+2 z \leqslant 4-(y-1)^{2}-(z-1)^{2} \leqslant 4 .
$$
Thus, $2 x+2 y+2 z \leqslant 4 \Longleftrightarrow x+y+z \leqslant 2$. Equal... | x+y+z\leqslant2 | Inequalities | proof | Yes | Yes | olympiads | false | 15,287 |
3. Find the smallest natural number $n$ such that the sum of the digits of each of the numbers $n$ and $n+1$ is divisible by 17. | Answer: 8899.
Solution. If $n$ does not end in 9, then the sums of the digits of the numbers $n$ and $n+1$ differ by 1 and cannot both be divisible by 17. Let $n=\overline{m 99 \ldots 9}$, where the end consists of $k$ nines, and the number $m$ has a sum of digits $s$ and does not end in 9. Then the sum of the digits ... | 8899 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,288 |
5. In triangle $ABC$, point $O_{1}$ is the center of the inscribed circle. On the extension of side $AB$ beyond point $B$, point $D$ is chosen. A circle with center $O_{2}$ touches segment $CD$ and the extensions of sides $AB$ and $AC$ of triangle $ABC$. Prove that if $O_{1} C=O_{2} C$, then triangle $BCD$ is isosceles... | Solution. (Fig. 2.) The angle $C O_{1} O_{2}$ is external for triangle $A C O_{1}$, therefore
$$
\angle C O_{1} O_{2}=\angle C A O_{1}+\angle A C O_{1}=\frac{1}{2} \angle C A B+\frac{1}{2} \angle A C B=\frac{1}{2}\left(180^{\circ}-\angle A B C\right)=\frac{1}{2} \angle C B D
$$
Next, let $E$ be some point on the exte... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,289 |
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 | 15,290 |
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
![](https://cdn.mathpix.com/cropped/2024_05_06_aa3023726f3a3c515454g-2.jpg?height=436&width=902&top_left_y=1606&top_left_... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,291 |
3. Excellent student Vasya invented a new chess piece called "grasshopper," which can move any number of squares diagonally, or jump over one square horizontally or vertically. Vasya claims that this piece can visit all the squares of an $8 \times 8$ board, visiting each square exactly once. Is Vasya right if the "gras... | # Solution:
If the "grasshopper" starts moving from a white cell, then according to the rules, it will move to a white cell (skipping a black cell horizontally or vertically). All cells on its diagonal are also white. Thus, the "grasshopper" can only visit white cells, without stepping on any black ones.
Answer: Vasy... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,292 |
4. When entering the shooting range, the player pays 100 rubles to the cashier. After each successful shot, the amount of money increases by $10 \%$, and after each miss - decreases by $10 \%$. Could it be that after several shots, he ends up with 80 rubles and 19 kopecks? | # Solution:
An increase of $10 \%$ means multiplying by 1.1. A decrease of $10 \%$ means multiplying by 0.9.
80 rubles 19 kopecks $=8019$ kopecks.
Let's factorize 8019: $8019=3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 11=9 \cdot 9 \cdot 9 \cdot 11$.
Therefore, after three misses and one hit, the player will ha... | 80.19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,293 |
5. Four different digits are written on two cards - one on each side of the card. Can it be that every two-digit number that can be formed from these cards is prime? (Digits cannot be flipped upside down, i.e., you cannot turn a 6 into a 9 or vice versa.)
# | # Solution:
All two-digit numbers ending in $0, 2, 4, 6$ or 8 are even, and those ending in 5 are multiples of five. Therefore, such numbers cannot be prime, and it makes no sense to write these digits on the cards. The remaining digits are 1, 3, 7, and 9. If the digits 3 and 9 are written on different cards, then the... | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,294 |
1. It is known that the quadratic trinomials $x^{2}+p x+q$ and $x^{2}+q x+p$
have different real roots. Consider all possible pairwise products of the roots of the first quadratic trinomial with the roots of the second (there are four such products in total). 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 | 15,295 |
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 | 15,296 |
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 | 15,297 |
5. On cards, Petya wrote down three natural numbers, each from 1 to 20. Moreover, there are no two cards with the same set of numbers. Prove that there will be two cards having exactly one common number, if it is known that Petya filled out 21 cards. | Proof by contradiction. Then there do not exist two cards having exactly one common number. Let's call cards "similar" if they have two common numbers.
Lemma: If cards $A, B$ are "similar" and cards $B, C$ are "similar", then cards $A$ and $C$ are also "similar".
Proof. Suppose card $A$ consists of numbers $a, b, c$,... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,298 |
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 | 15,301 |
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 | 15,302 |
3. Determine the smallest product of positive numbers $a$ and $b$ that satisfy the equation
$$
20 a b=13 a+14 b
$$ | Answer: 1.82.
## Solution.
$$
\begin{gathered}
(20 a b)^{2}=(13 a+14 b)^{2}=(13 a)^{2}+(14 b)^{2}+2 \cdot 13 a \cdot 14 b \geq \\
\geq 2 \cdot 13 a \cdot 14 b+2 \cdot 13 a \cdot 14 b=4 \cdot 13 a \cdot 14 b \Rightarrow \\
20^{2} a b \geq 4 \cdot 13 \cdot 14 \Rightarrow 20^{2} a b \geq 4 \cdot 13 \cdot 14 \Rightarrow ... | 1.82 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,303 |
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 | 15,304 |
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 | 15,305 |
# Problem 9.1 (7 points)
Each participant of the Olympiad filled out two questionnaires and put them in any two of the 4 piles of sheets on the table. Initially, the piles contained 2, 3, 1, and 1 sheets. Can it end up with an equal number of sheets in each pile? | # Solution:
The initial number of sheets is 7, each time it changes by 2, and remains odd and will never be a multiple of 4.
| Criteria | Points |
| :--- | :---: |
| Complete solution | 7 |
| Answer without justification | 0 |
## Answer: cannot | cannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,306 |
# 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 | 15,307 |
# 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 | 15,308 |
# 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 | 15,309 |
# 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 | 15,310 |
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 | 15,312 |
3.4. A natural number $n$ is written with different digits, the sum of which is 22. What can the sum of the digits of the number $n-1$ be? Find all possible options.
Answer, option 1. 15 and 24 (two answers).
Answer, option 2. 18 and 27 (two answers).
Answer, option 3. 20 and 29 (two answers).
Answer, option 4. 21 ... | Solution Option 1. If the number $n$ does not end in 0, then the sum of the digits of $n-1$ is 1 less than the sum of the digits of $n$, which means it is 15. If the number $n$ does end in 0, then it ends in only one zero (since all digits in the decimal representation of $n$ are different). Then, if the number $n$ has... | 1524 | Number Theory | MCQ | Yes | Yes | olympiads | false | 15,313 |
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 | 15,314 |
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 | 15,315 |
6.4. On an island, there live 456 aborigines, each of whom is either a knight, who always tells the truth, or a liar, who always lies. All residents have different heights. One day, each aborigine said: "All other residents are shorter than me!". What is the maximum number of aborigines who could have said a minute lat... | Solution Variant 1. Evaluation. Consider the tallest and the shortest aborigines. For each of them, one of the spoken phrases is true, the other is false. Therefore, no matter who they are - knights or liars - none of them could have spoken the second phrase a minute later. Thus, the number of spoken phrases is no more... | 454 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 15,316 |
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 $BE=6$.
![](https://cdn.mathpix.com/cropped/2024_05_06_0b1749f4383e44f39544g-7.j... | Solution Option 1. In triangle $A E D$, the angle at vertex $E$ is a right angle. In a right triangle, it is often useful to draw the median from the vertex of the right angle. If $E M$ is the median, then it is equal to half the hypotenuse $A D$, that is, $E M=M A=M D=10$.
![](https://cdn.mathpix.com/cropped/2024_05_... | 13 | Geometry | MCQ | Yes | Yes | olympiads | false | 15,317 |
8.4. Sasha is a coffee lover who drinks coffee at work, so every morning before work, he prepares a thermos of his favorite drink according to a strict ritual. He brews 300 ml of the strongest aromatic coffee in a Turkish coffee pot. Then Sasha repeats the following actions 6 times: he pours 200 ml from the pot into th... | Solution option 1. Note that by the end of the ritual, there will be 200 ml of coffee and 800 ml of water in the thermos. Therefore, the concentration of coffee is $200 / 1000 = 1 / 5 = 0.2$.
Remark. One could try to solve the problem "step by step": calculate the concentration of coffee in the thermos after the first... | 0.2 | Other | math-word-problem | Yes | Yes | olympiads | false | 15,318 |
# Task 9.1
For which natural numbers $n$ is the expression $n^{2}-4 n+11$ a square of a natural number?
## Number of points 7 | Answer:
for $n=5$
## Solution
Let $n^{2}-4 n+11=t^{2}$.
Note that $n^{2}-4 n+4=(n-2)^{2}$ 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 | 15,319 |
# Problem 9.2
Let $a, b, c$ be the lengths of the sides of a triangle. Prove that:
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+a}<\frac{5}{2}$
Number of points 7
# | # Solution
Without loss of generality (due to symmetry with respect to $a, b, c$), we can assume that: $a \leq b \leq c$
Then:
$\frac{a}{b+c} \leq \frac{a}{a+a}=\frac{1}{2}$
$\frac{b}{a+c}<1$
$\frac{c}{b+a}<1$
Here, the second and third inequalities are a consequence of the triangle inequality.
Adding these thre... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,320 |
# Task 9.3
36 circles, arranged in a circle, were sequentially connected by segments. On each segment, a certain number was written, and in each circle - the sum of the two numbers written on the segments entering it. After that, all the numbers on the segments and in one of the circles (see figure) were erased. Can t... | Answer:
It is possible
## Solution
We will color the circles, alternating between two colors, white and black. Then each segment will enter once into a circle of each color, so the sum of the numbers in the white circles will be equal to the sum of the numbers in the black circles (each of them is equal to the sum o... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,321 |
# 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 | 15,322 |
1. Solve the system $\left\{\begin{array}{l}\sqrt{2 x^{2}+2}=y+1 ; \\ \sqrt{2 y^{2}+2}=z+1 ; \\ \sqrt{2 z^{2}+2}=x+1 .\end{array}\right.$ | Answer: $(1,1,1)$.
Sketch of the solution. Squaring each of the equations and adding them, we move all the terms to the left side and factor out complete squares. We get $(x-1)^{2}+(y-1)^{2}+(z-1)^{2}=0$. From this, $x=y=z=1$.
Verification shows that this triplet works, as the left and right sides are both equal to 2... | (1,1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,324 |
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 | 15,325 |
3. Find all functions defined on the set of real numbers and taking real values such that for any $x, y$ the equality $f(x+|y|)=f(|x|)+f(y)$ holds. | Answer: $f(x)=0$ for all real $x$.
Sketch of the solution. Let $x=0, y=0$, we get $f(0)=f(0)+f(0)$, so $f(0)=0$. If $y=0$, then $f(x)=f(|x|)$. Let $x=-|y|, 0=f(0)=f(\mid-$ $|y| \mid)+f(y)=f(y)+f(y)=2 f(y)$. Therefore, $f(y)=0$. | f(x)=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,326 |
6. Vasya thought of a natural number $n \leq 2020$. Petya tries to guess it as follows: he names some natural number x and asks if his number is greater (is it true that $\mathrm{x}<\mathrm{n}$?), and Vasya answers him with "yes" or "no". Petya wins if he finds out the number, and loses if after receiving a "no" answer... | Answer: 64.
Solution. First, note that if Petya receives the first "no" answer at some point, this means that at that moment he knows that the number $n$ is in some interval [ $n_{1}, n_{2}]$. If his next attempt is a number $x > n_{1}$, and he receives a "no" answer - he loses, because he is not allowed to ask more q... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,327 |
114. a) Prove the inequality $\sqrt{n+1}+2 \sqrt{n}<\sqrt{9 n+3}$ for all natural numbers n; b) Does there exist a natural number n such that $[\sqrt{n+1}+2 \sqrt{n}]<[\sqrt{9 n+3}]$, where $[a]$ denotes the integer part of the number a?
115. Find the maximum and minimum values of the function $y=\cos x(\cos x+1)(\cos... | 114. a) Prove the inequality $\sqrt{n+1}+2 \sqrt{n}<\sqrt{9 n+3}$ for all natural numbers $n$; b) Does there exist a natural number $n$ for which $[\sqrt{n+1}+2 \sqrt{n}]<[\sqrt{9 n+3}]$, where $[a]$ denotes the integer part of the number $a$?
Answer: b) No such $n$ exists. Hint: See problem 10.5.
115. Find the maxim... | No | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 15,328 |
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 | 15,329 |
10.2. Given an acute-angled triangle $A B C$, in which $A B < A C$. Let $M$ and $N$ be the midpoints of sides $A B$ and $A C$ respectively, and $D$ be the foot of the altitude from $A$. On the segment $M N$, there is a point $K$ such that $B K = C K$. The ray $K D$ intersects the circumcircle $\Omega$ of triangle $A B ... | Solution. Let $\ell$ be the perpendicular bisector of $BC$. Note that $\ell$ passes through $K$. Let point $A'$ be symmetric to point $A$ with respect to $\ell$. Clearly, $A'$ lies on $\Omega$, and due to symmetry, arcs $AB$ and $A'C$ are equal.
Since point $D$ is symmetric to point $A$ with respect to line $MN$, $D$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,330 |
10.3. Given a natural number $k$. On a grid plane, $N$ cells are initially marked. We will call the cross of cell $A$ the set of all cells that are in the same vertical or horizontal line with $A$. If the cross of an unmarked cell $A$ contains at least $k$ other marked cells, then cell $A$ can also be marked.
, & \text { if } k=2 m ; \\ m^{2}, & \text { if } k=2 m-1\end{array}\right.$.
Solution. Let $N(k)$ be the answer to the problem; set $f(k)=\left[\frac{k+1}{2}\right] \cdot\left[\frac{k+2}{2}\right]$. First, we prove tha... | N=[\frac{k+1}{2}]\cdot[\frac{k+2}{2}]={\begin{pmatrix}(+1),&\text{if}k=2;\\^{2},&\text{if}k=2-1\end{pmatrix}.} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,331 |
10.4. Initially, a natural number is written on the board. Then every second, the product of all its non-zero digits is added to the current number. Prove that there exists a natural number $a$ such that the addition of the number $a$ will occur infinitely many times.
(D. Krachun) | Solution. Note that $9^{100}10$ by Bernoulli's inequality. Then by induction on $m$ it is easy to obtain that
$$
9^{m}200$. Let this number result from the number $A$. Since $p(A) \leqslant 9^{n}B-10^{n-1}>10^{n}$. Suppose $A$ starts with $k$ ones, then immediately after them is 0, that is, $\underbrace{11 \ldots 1}_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,332 |
10.5. In a $10 \times 10$ table, positive numbers are written such that in any row the numbers form an arithmetic progression (from left to right), and in any column they form a geometric progression (from top to bottom). Prove that the common ratios of all these geometric progressions are equal.
(P. Kozhevnikov) | Solution. It is sufficient to solve the problem for a $3 \times 3$ square. From this, it will follow that the denominators of the progressions in any two adjacent columns are equal, and therefore they are equal in all columns.
From the fact that the middle row of the square contains an arithmetic progression, and all ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,333 |
10.6. Given natural numbers $a$ and $b$. Prove that there are infinitely many natural numbers $n$ such that the number $a^{n}+1$ is not divisible by $n^{b}+1$.
(A. Golev) | Solution. Let's call a natural number $n$ bad if $a^{n}+1$ is not divisible by $n^{b}+1$. Our goal is to prove that there are infinitely many bad numbers.
First Solution. We will prove that for any even $n$, one of the numbers $n$ and $n^{3}$ is bad; from this, the required result obviously follows. Assume the opposit... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,334 |
10.7. In a convex quadrilateral $A B C D$, angles $A$ and $C$ are equal. On sides $A B$ and $B C$, there are points $M$ and $N$ respectively such that $M N \| A D$ and $M N=2 A D$. Let $K$ be the midpoint of segment $M N$, and $H$ be the orthocenter of triangle $A B C$. Prove that lines $K H$ and $C D$ are perpendicula... | The first solution. We will use the following fact.
Lemma (Monge's theorem). A convex quadrilateral XYZT is cyclic if and only if the perpendiculars dropped from the midpoints of its sides to the opposite sides of the quadrilateral have a common point.
Proof. Let $P, Q, R$ and $S$ be the midpoints of sides $XY$, $YZ$... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,335 |
10.8. The game board consists of a left and a right part. Each part has several fields; between them, several segments are drawn, each connecting two fields from different parts. From any field, it is possible to reach any other field via the segments. Initially, a lilac chip is placed on one field of the left part, an... | Answer. No.
Solution. We will present a strategy that allows Lёsha to win. Let $L$ and $P$ be the fields where the lilac ($\ell$) and purple ($p$) tokens are initially placed, respectively. According to the problem, it is possible to reach $P$ from $L$ by moving along the edges; let Lёsha choose one such path $L=L_{1}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,336 |
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 | 15,337 |
11.2. Any two of the real numbers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ differ by at least 1. It turned out that for some real $k$ the equalities
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=2 k \quad \text{and} \quad a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2}=2 k^{2}
$$
are satisfied. Prove that $k^{2} \geqslant 25 / 3$.
... | Solution. Without loss of generality, we can assume that $a_{1}<\ldots<a_{5}$. By the condition, $a_{i+1}-a_{i} \geqslant 1$ for all $i=1,2,3,4$. Therefore, $a_{j}-a_{i} \geqslant j-i$ for all $1 \leqslant i<j \leqslant 5$. Squaring each of the obtained inequalities and summing them all, we get $\sum_{1 \leqslant i<j \... | k^{2}\geqslant25/3 | Algebra | proof | Yes | Yes | olympiads | false | 15,338 |
11.3. A square grid is colored in a checkerboard pattern with black and white colors. Then the white cells are recolored in red and blue such that cells adjacent by a corner are of different colors. Let $\ell$ be a line not parallel to the sides of the cells. For each segment $I$ parallel to $\ell$, we calculate the di... | Solution. Lemma. We will color the cells of the plane in two colors: odd columns in green, and even columns in yellow. Then for any segment parallel to the line $\ell$, the difference between the sums of the lengths of its green and yellow segments does not exceed some number $D$, depending only on $\ell$.
Proof. The ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,339 |
11.4. Given a pyramid $S A_{1} A_{2} \ldots A_{n}$, the base of which is a convex polygon $A_{1} \ldots A_{n}$. For each $i=1,2, \ldots, n$, in the plane of the base, a triangle $X_{i} A_{i} A_{i+1}$ is constructed, equal to the triangle $S A_{i} A_{i+1}$ and lying on the same side of the line $A_{i} A_{i+1}$ as the ba... | First solution. Consider an arbitrary point $P$ on the base and prove that it is covered by one of the triangles. Consider a small sphere lying inside the pyramid and touching the base at point $P$ (such a sphere clearly exists). Start increasing its radius while maintaining the condition of tangency; then at some mome... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,340 |
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 | 15,343 |
8.2. One side of the rectangle (width) was increased by $10 \%$, and the other (length) - by $20 \%$. a) Could the perimeter increase by more than 20\%? b) Find the ratio of the sides of the original rectangle if it is known that the perimeter of the new rectangle is $18 \%$ larger than the perimeter of the original? | Answer. a) could not; b) 1:4. Solution. a) Let $a$ and $b$ be the sides of the original rectangle. Then the sides of the new rectangle are $1.1a$ and $1.2b$, and its perimeter is $2(1.1a + 1.2b) < 2(1.12(a + b))$. Therefore, the perimeter increased by less than $20\%$. b) See problem 7.2 b). | 1:4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,344 |
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 | 15,345 |
8.5. 20 numbers: $1,2, \ldots, 20$ were divided into two groups. It turned out that the sum of the numbers in the first group is equal to the product of the numbers in the second group. a) What is the smallest and b) what is the largest number of numbers that can be in the second group? | Answer. a) 3; b) 5. Solution. a) The sum of all numbers from 1 to 20 is 210. It is obvious that the second group contains more than one number, since otherwise it would be equal to the sum of the remaining 19 numbers, i.e., no less than \(1+2+\ldots+19=190\). We will show that in fact the second group contains more tha... | )3;b)5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,347 |
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 | 15,348 |
4-7. Along a straight alley, 100 lamps are placed at equal intervals, numbered in order 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 lam... | 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 | 15,352 |
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 | 15,353 |
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 | 15,354 |
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 | 15,355 |
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 | 15,356 |
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 | 15,358 |
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 | 15,361 |
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 | 15,362 |
# Problem 1
It is known that $\frac{1}{3 a}+\frac{2}{3 b}=\frac{\mathbf{3}}{a+2 b}$. Prove that $a=b$.
# | # Solution.
1st method. Transform the given equality by multiplying both sides by $3 \mathrm{ab}(\mathrm{a}+2 \mathrm{~b})$. We get: $\mathrm{b}(\mathrm{a}+2 \mathrm{~b})+2 \mathrm{a}(\mathrm{a}+2 \mathrm{~b})=9 \mathrm{ab}$. After expanding the brackets and combining like terms, the equation will take the form: $2 \m... | b | Algebra | proof | Yes | Yes | olympiads | false | 15,366 |
# 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
 must be a multiple ... | 34or87 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,368 |
# Problem №4
Two ferries simultaneously depart from opposite banks of a river and cross it perpendicularly to the banks. The speeds of the ferries are constant but not equal. The ferries meet at a distance of 720 meters from one bank, after which they continue their journey. On the return trip, they meet 400 meters fr... | Answer: $1760 \mathrm{M}$
## Solution.
The total distance traveled by the ferries by the time of their first meeting is equal to the width of the river, and the distance traveled by the time of their second meeting is three times the width of the river. Therefore, by the time of the second meeting, each ferry has tra... | 1760 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,369 |
# Problem №5
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 $\mathrm{DE}$ to intersect line $\mathrm{BC}$ at point $\mathrm{K}$ (see figure).
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 | 15,370 |
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 | 15,371 |
2. A dog from point $A$ chased a fox, which was 30 m away from the dog at point $B$. The dog's jump is 2 m, the fox's jump is 1 m. The dog makes 2 jumps, while the fox makes 3 jumps. At what distance from point $A$ will the dog catch the fox? | Solution. In one unit of time, the dog runs $2 \cdot 2=4$ (m), and the fox runs $3 \cdot 1=3$ (m), which means that in one unit of time, the dog catches up to the fox by 1 m. The distance of 30 m will be covered in 30 units of time.
Answer. $120 \mathrm{~m}$. | 120\mathrm{~} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,372 |
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 | 15,373 |
6. In the cells of a $3 \times 3$ square, Petya placed the numbers $1,2,3, \ldots, 9$ (each one exactly once), and then calculated the sums in each row, each column, and each of the two diagonals. Could these 8 sums be equal to $13,14, \ldots, 20$? | Solution. Suppose it were possible. Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}$ be the sums of the three rows. The sum $\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}$ is the sum of all numbers in the square, i.e., $1+2+3+\ldots+9=45$. Similarly, if $b_{1}, b_{2}, b_{3}$ are the sums of the columns, then $b_{1}+... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,374 |
8.2 In a class of 30 people, is it possible that 9 of them have 3 friends (in this class), 11 have 4 friends, and 10 have 5 friends? | Answer: cannot
Reasoning: Suppose it can. Consider the corresponding graph: 30 points on a plane (vertices of the graph) and segments connecting some vertices (edges of the graph). Let's count the total number of edges: $(3 \cdot 9 + 4 \cdot 11 + 5 \cdot 10) \cdot \frac{1}{2} = \frac{121}{2}$ - a non-integer number, w... | cannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,375 |
8.3 Consider an isosceles triangle with a vertex angle of $20^{\circ}$. Prove that the lateral side is greater than twice the base. | Solution: See fig.
We rely on the fact that in a triangle, the larger side lies opposite the larger angle. Additional construction: CD=CA.
Then angles of $50^{\circ}$ appear (see fig.). From triangle $\mathrm{ABD}$ we find $\mathrm{BD}>\mathrm{AD}$. From triangle $A D C$ we find $A D>A C$. Therefore, $\mathrm{BD}>\ma... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,376 |
Problem 3. Returning home, the hunter first walked through the swamp at a speed of 2 km/h, then - through the forest at a speed of 4 km/h, and then - along the highway at a speed of 6 km/h. In 4 hours, he walked 17 kilometers. What did he spend more time on - walking through the swamp or walking along the highway? | Answer. For walking on the highway. Solution. Let the hunter walked $a$ hours through the swamp, $-b$ hours through the forest, and $-c$ hours on the highway. In 4 hours he walked
$$
2 a+4 b+6 c=4(a+b+c)+2(c-a)=4 \cdot 4+2(c-a)=17 \text { kilometers. }
$$
It follows that $2(c-a)=1$, hence $c>a$.
- For a correct answ... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,378 | |
9.1. It is known that $a+b+c<0$ and that the equation $a x^{2}+b x+c=0$ has no real roots. Determine the sign of the coefficient $c$.
---
Translation provided as requested, maintaining the original formatting and structure. | Answer: $c<0$.
Solution. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has no real roots, which means it maintains the same sign for all values of the argument x. Since $f(1)=a+b+c<0$, then $f(0)=c<0$. | <0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,379 |
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 | 15,380 |
9.3. A circle of radius $R$ touches the base $A C$ of an isosceles triangle $A B C$ at its midpoint and intersects side $A B$ at points $P$ and $Q$, and side $C B$ at points $S$ and $T$. The circumcircles of triangles $S Q B$ and $P T B$ intersect at points $B$ and $X$. Find the distance from point $X$ to the base of t... | Answer: R.
Solution:
Obviously, the diagram shows symmetry with respect to the line $B H$, which divides the angle $A B C$ into two equal angles. Note that the angles $Q B X$ and $Q S X$ are... | R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,381 |
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 | 15,382 |
9.5. Is it possible to write ones in some cells of an 8 x 8 table and zeros in the others, so that the sums in all columns are different, and the sums in all rows are the same? | Answer: Yes, it is possible.
Solution: Let the sum of the numbers in each row be x. Then the sum of all numbers in the table is 8x, meaning the total sum is divisible by 8. Note that each column can contain between 0 and 8 ones. The sum of all numbers from 0 to 8 is 36. To get a number divisible by 8, we need to subtr... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,383 |
8.2 On a chessboard, 8 rooks were placed so that they do not attack each other. Prove that in any "square" rectangle of size $4 \times 5$ (cells) there is at least one rook. | 8.2 On a chessboard, 8 rooks were placed so that they do not attack each other. Prove that in any "square" rectangle of size $4 \times 5$ (cells) there is at least one rook.
Hint Suppose the opposite, and let, for definiteness, the rectangle be located in five rows and four columns. Then any rook is among the remainin... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,385 |
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 | 15,386 |
8.4. Prove that for all natural $n>1$ the number $n^{2016}+4$ is composite. | 84. Prove that for all natural $n>1$ the number $n^{2016}+4$ is composite.
Hint The result follows from the factorization $n^{2016}+4=n^{2016}+4 n^{1008}+4-4 n^{1008}=\left(n^{1008}+2\right)^{2}-\left(2 n^{504}\right)^{2}=$ $\left(n^{1008}+2 n^{504}+2\right)\left(n^{1008}-2 n^{504}+2\right)$, and for $n>1$ both factor... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,387 |
8.5. Given a rectangle that is not a square, where the numerical value of the area is three times the perimeter. Prove that one of the sides of the rectangle is greater than 12. | 8.5. Given a rectangle that is not a square, for which the numerical value of the area is three times the perimeter. Prove that one of the sides of the rectangle is greater than 12.
Hint See problem 7.5. | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,388 |
1. Suppose the following statements are true:
a) among people who play chess, there are those who are not interested in mathematics;
b) people who swim in a pool every day but are not interested in mathematics do not play chess.
Does it follow from these statements that not all people who play chess swim in a pool e... | # Solution
The statement "not all people who play chess bathe in a pool every day" is true. The solution is illustrated using Euler-Venn diagrams.
Set A - people who play chess, Set B - people who are not interested in mathematics, Set C - people who bathe in a pool every day.
.
# | # Solution
Let's take $x$ ml of the 70% solution, and $\mathrm{V}$ ml - the volume of the resulting 9% solution. Then we have the equation
$$
0.7 x + 0.05(V - x) = 0.09 V
$$
from which we find
$$
\frac{x}{V} = \frac{4}{65}
$$
that is, we take 4 parts of the 70% solution and 61 parts of the 5% solution.
The larges... | 406.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,392 |
5. A grid sheet $5 \times 7$ was cut into squares $2 \times 2$, three-cell corners, and strips $1 \times 3$. How many squares could have been obtained? | # Solution
Let's take $x$ as the number of squares and $y$ as the number of three-cell figures.
We get the equation $4 x + 3 y = 35$, from which we find
1) $x=8$, 2) $x=5$, 3) $x=2$.
We will show that it is impossible to cut into eight squares. For this, we introduce a coloring.
(a+b)=-\frac{a-b}{a b},(a-b \neq 0)$, $a+b=-\frac{1}{a b}, \quad a b(a+b)=-1$.
If both numbers $a$ and $b$ were $>0$, then it would be $a b(a+b)>0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,397 |
8.2 A point moves in the coordinate plane such that at time $t \geq 0$ it is located on both the line $y=t \cdot x+1$ and the line $y=-x+2 t$. Prove that it will never be to the right of the vertical line $x=2$ (i.e., in the region $x \geq 2$). | We need to prove that the abscissa of the intersection points of the lines $y=t \cdot x+1$ and $y=-x+2 t$ for any $t \geq 0$ is no more than 2. Let's find this abscissa:
$t x+1=-x+2 t, \quad(t+1) x=2 t-1, \quad x=\frac{2 t-1}{t+1}$.
We have:
$\frac{2 t-1}{t+1}=\frac{2(t+1)-3}{t+1}=2-\frac{3}{t+1}<2$, i.e., $x<2$ whi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,398 |
8.5 At the beginning of the game, the number 0 is written on the board. Two players take turns. On a turn, a player adds to the written number any natural number not exceeding 10, and writes the result on the board in place of the original number. The player who first obtains a four-digit number wins. Which of the play... | The first move: The beginner writes down the number 10 (the number 10 is the remainder of the division of the smallest four-digit number 1000 by 11). Next, if the second player adds the number $\mathrm{a}, 1 \leq \mathrm{a} \leq 10$, then the first player responds by adding the number $11-a, 1 \leq 11-\mathrm{a} \leq 1... | Thewinningstrategybelongstothebeginner | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,400 |
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 | 15,401 |
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 | 15,402 |
11.3. One side of the leaflets is painted in some color, and on the other side, a smiley face is drawn. In front of you lie 4 leaflets: the first one is yellow, the second one is black, the third one has a happy smiley face, and the fourth one has a sad smiley face. You need to verify the statement: "If one side of the... | # Solution.
The first card does not need to be flipped; if there is a happy smiley face on it, then everything is fine, if it is a sad one, then the statement does not say anything about it.
The second card needs to be flipped; if there is a happy smiley face on it, then the statement is false.
The third card needs ... | Two.Thethird | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,403 |
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