problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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10.2. Which is greater, $2021^{2021}$ or $2022^{2020}$? | Answer. $2021^{2021}>2022^{2020}$
Solution. $\quad$ Consider $\quad$ the ratio $\quad \frac{2022^{2020}}{2021^{2021}}=\frac{2022^{2021} \cdot 2022^{-1}}{2021^{2021}}=\left(\frac{2022}{2021}\right)^{2021} \cdot \frac{1}{2022}=$ $\left(\frac{2021+1}{2021}\right)^{2021} \cdot \frac{1}{2022}=\left(1+\frac{1}{2021}\right)^... | 2021^{2021}>2022^{2020} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 12,289 |
10.3. Prove the inequality: $\frac{(a+1)^{2}}{b}+\frac{(b+1)^{2}}{a} \geq 8$, where $a>0$ and $b>0$. | Solution. To prove this, we will use the inequalities: $x^{2}+1 \geq 2 x$ and $x+\frac{1}{x} \geq 2$. $\frac{(a+1)^{2}}{b}+\frac{(b+1)^{2}}{a}=\frac{a^{2}+1+2 a}{b}+\frac{b^{2}+1+2 b}{a} \geq \frac{4 a}{b}+\frac{4 b}{a}=4 \cdot\left(\frac{a}{b}+\frac{b}{a}\right) \geq 8$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,290 |
10.4. Given a convex quadrilateral $A B C D$. It is known that the radii of the circles inscribed in triangles $A B C, B C D, C D A$ and $D A B$ are equal. Prove that $A C=B D$. | Solution. Let $r$ be the radii of the circles specified in the condition. Then $S_{A B C D}=S_{A B C}+S_{C D A}=\frac{1}{2} r(A B+B C+C A)+\frac{1}{2} r(C D+D A+A C)=$ $=\frac{1}{2} r \cdot P+r \cdot A C$, where $S$ and $P$ are the area and perimeter of quadrilateral $A B C D$. Similarly, $S_{A B C D}=S_{B D C}+S_{D A ... | AC=BD | Geometry | proof | Yes | Yes | olympiads | false | 12,291 |
10.5. At the ceremonial assembly, each student was given their individual festive photographs taken the day before. It turned out that each student received a photograph of another student. They decided to exchange the photographs so that everyone could get their own. However, to do this discreetly, photographs could o... | Answer: Yes, it will always be able to.
## Solution. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,292 |
1. Sasha, Zhenya, and Valya are sitting at a triangular table, which is divided into small triangles, as shown in the figure.
They fill all the small triangles with the numbers -1 and 1. Afte... | Solution. One of the possible arrangements is shown in the figure. It is necessary
to place -1 so that in each row in one direction they appear an even number of times, in each row in the se... | Itispossible | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,293 |
2. Let $a, b, c$ be positive numbers. Prove the inequality
$$
\frac{1}{2 a}+\frac{1}{2 b}+\frac{1}{2 c} \geq \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}
$$ | Solution. We will use the inequality between the arithmetic mean and the harmonic mean. It is known that for positive $x, y$ the following holds
$$
\frac{x+y}{2} \geq \frac{2}{\frac{1}{x}+\frac{1}{y}}
$$
In our case, we apply this inequality in turn to the pairs $\left(x=\frac{1}{a}, y=\frac{1}{b}\right)$; $\left(x=\... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,294 |
3. A circle with a radius of 26 touches two adjacent sides of a rectangle, the lengths of which are 36 and 60. Find the segments into which the circle divides the sides of the rectangle. | Solution.
Note that the radius of the circle is greater than half the length of the shorter side of the rectangle, while the diameter of the circle is less than its longer side. This means that the circle intersects the longer side of the rectangle, opposite to the touching side, at two points, and does not intersect ... | 2634;2610;2,4810 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,295 |
5. For three different integers $x, y, z$, it is known that $xy$ is divisible by $576$, $yz$ is divisible by $324$, and $xz$ is divisible by $5184$. Is $(x-y)(y-z)(z-x)$ divisible by 48? Solution.
$576=2^{6} \cdot 3^{2}, 324=2^{2} \cdot 3^{4}, 5184=2^{6} \cdot 3^{4}$. Let's represent the numbers $x, y, z$ in terms of ... | Answer: Divides.
## Municipal Stage of the All-Russian Mathematics Olympiad 2018-2019 Academic Year
9th Grade
Grading Criteria
| Problem 1 | Score | For what it is given |
| :---: | :---: | :---: |
| | 7 | Complete solution, correct arrangement is provided. |
| | 2 | An arrangement is provided that contains an ob... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,297 |
1. The bathtub fills with cold water in 6 minutes 40 seconds, with hot water - in 8 minutes. In addition, if the plug is removed from a full bathtub, the water will drain in 13 minutes 20 seconds. How long will it take to fill the bathtub completely, provided that both taps are open, but the bathtub is not plugged? | Solution: First, we will convert the time in seconds to minutes: 6 minutes 40 seconds will be replaced by $6+2 / 3$, or $20 / 3$, and 13 minutes 20 seconds will be replaced by $13+1 / 3$, or $40 / 3$. Then, in one minute, the cold water will fill $3 / 20$ of the bathtub, the hot water will fill $1 / 8$ of the bathtub, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,298 |
2. The sides of a rectangle were reduced: the length - by $10 \%$, the width - by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$? | Answer: by $18 \%$.
Solution. Let $a$ and $b$ be the length and width of the rectangle. After decreasing the length by $10 \%$ and the width by $20 \%$, we get a rectangle with sides $0.9 a$ and $0.8 b$, the perimeter of which is $0.88$ of the perimeter of the original rectangle. Therefore, $2(0.9 a + 0.8 b) = 0.88 \c... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,299 |
3. In a six-digit number, one digit was crossed out to obtain a five-digit number. The five-digit number was subtracted from the original number, and the result was 654321. Find the original number. | Answer: 727023.
Solution. Note that the last digit was crossed out, as otherwise the last digit of the number after subtraction would have been zero. Let $y$ be the last digit of the original number, and $x$ be the five-digit number after crossing out. Then the resulting number is $10 x + y - x = 9 x + y = 654321$. Di... | 727023 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,300 |
4. On the sides $A B, B C$ and $A C$ of triangle $A B C$, points $C_{1}, A_{1}$ and $B_{1}$ are taken respectively such that $C_{1} A_{1} \| A C$. Prove that $S_{A_{1} B_{1} C_{1}} \leq \frac{1}{4} S_{A B C}$. | Solution. From the condition $C_{1} A_{1} \| A C$, it follows that triangles $B C_{1} A_{1}$ and $B A C$ are similar. Let $x = C_{1} A_{1} / A C$ be the similarity coefficient. Then the height in triangle $B C_{1} A_{1}$, dropped from point $B$ to $C_{1} A_{1}$, is $x h$, where $h$ is the height of triangle $B C A$ fro... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,301 |
5. Each point of the plane is painted in one of three colors, and all three colors are used. Is it true that for any such coloring, one can choose a circle on which there are points of all three colors? | Answer: Correct.
Suppose that it is impossible to choose a circle on which there are points of all three colors. Let's choose a point $A$ of the first color and a point $B$ of the second color and draw a line $l$ through them. If there is a point $C$ of the third color outside the line $l$, then on the circle circumsc... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,302 |
1. Each of the three friends either always tells the truth or always lies. They were asked the question: "Is there at least one liar among the other two?" The first answered: "No," the second answered: "Yes." What did the third answer? | Solution Since the first and second friends gave different answers, one of them is a liar, and the other is a knight. Moreover, the knight could not have answered “No” to the question posed, as in that case, he would have lied (there is definitely a liar among the two remaining). Therefore, the first one is a liar. He ... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,303 |
# 3. It is known that a, b, c are such numbers that $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$. Prove that in this case the relation $a b+a c+b c \leq 0$ holds. | Solution Clearly, $0=(\mathrm{a}+\mathrm{b}+\mathrm{c})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}+2(\mathrm{ab}+\mathrm{ac}+\mathrm{bc})$. Therefore, $2(\mathrm{ab}+\mathrm{ac}+\mathrm{bc})$ $=-a^{2}-b^{2}-c^{2} \leq 0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,304 |
4. In a convex quadrilateral $A B C D$, sides $AB$ and $CD$ are parallel, and diagonals $A C$ and $B D$ are perpendicular. Prove that $AD + BC = AB + CD$. | Inscribe quadrilateral $ABCD$ in rectangle $EFGH$ with sides parallel to the diagonals ($\mathrm{EF} \| \mathrm{AC}$ and $\mathrm{EH} \| \mathrm{BD}$) - see the figure. Let $\mathrm{L}$ be the intersection point of lines $DC$ and $EF$, and $M$ be a point on line $HG$ such that $LM \| FG$. Then $ABLC$ is a parallelogram... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,305 |
9.1 The parabola $y=4 a x^{2}+4(a+1) x+a^{2}+a+3+\frac{1}{a} \quad(a \neq 0)$ intersects the x-axis at two points. In which direction do the branches of this parabola point - upwards or downwards? | From the condition, it follows that the discriminant $\mathrm{D}$ of the quadratic trinomial $4 a x^{2}+4(a+1) x+a^{2}+a+3+\frac{1}{a}$ is positive:
$$
\begin{aligned}
& \frac{1}{16} D=(a+1)^{2}-a\left(a^{2}+a+3+\frac{1}{a}\right)=a^{2}+2 a+1-a^{3}-a^{2}-3 a-1= \\
& =-a^{3}-a=-a\left(a^{2}+1\right)>0
\end{aligned}
$$
... | downwards | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,306 |
9.4 On the hypotenuse $AB$ of a right triangle $ABC$, a point $D$ is taken. Segments $DE$ and $DF$ are the angle bisectors of triangles $ADC$ and $BDC$. It turns out that $CD = EF$. Prove that point $D$ is the midpoint of the hypotenuse $AB$. | Solution See Fig.
Angle EDF is a right angle, as it is half of the straight angle ADB. Consider the circle with diameter $EF$. Points $C$ and $D$ lie on this circle (as points from which the diameter $EF$ is seen at a right angle).
(m+n)=-\frac{(m-n)}{m n}$.
If $m-n=0$, then $n=m$, and everything is proven.
Let $n \neq m$. Then $m+n=-\frac{1}{m n}, \quad m n(m+n)=-1$.
Suppose $n$ is rational, i.e., $n=\frac{p}{q}$, where $p$ is an integer, $q$ is a natural number, and $p$ and $q$ ar... | proof | Algebra | proof | Yes | Yes | olympiads | false | 12,308 |
1. Tourists arrived at the campsite. For lunch, each of them ate half a can of soup, a third of a can of stew, and a quarter of a can of beans. In total, they ate 39 cans of food. How many tourists were there? Answer: 36. | Solution. Note that 12 tourists ate 6 cans of soup, 4 cans of stew, and 3 cans of beans - that is, a total of 13 cans of food.
39 cans is 3 sets of 13 cans. One set is eaten by 12 people. Therefore, the total number of tourists is 36. | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,309 |
2. Six people are standing in a circle, each of whom is either a knight - who always tells the truth, or a liar - who always lies. Each of them said one of two phrases: "There is a liar next to me" or "There is a liar opposite me." What is the minimum number of liars among them? Provide an example and prove that there ... | Answer: 2.
Solution. Let's number all the people standing clockwise (this way, people with numbers 1 and 4, 2 and 5, 3 and 6 will stand opposite each other).
Zero liars is obviously impossible (then there would be only knights and no one could say any of the phrases).
If there is one liar, let's say his number is 1,... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,310 |
3. Four people with a chest want to cross a river. The people weigh 45, 50, 60, and 65 kg, and the chest weighs 100 kg. The boat can carry no more than 200 kg. The chest can be loaded into or unloaded from the boat only by all four people together. How can they all cross the river with the chest? | Solution.
Load the chest into the boat.
45 and 50 cross to the other side. 45 stays, 50 returns.
60 crosses to the other side. 45 returns.
45 and 50 cross to the other side. 45 stays, 50 returns.
65 crosses to the other side. 45 returns.
45 and 50 cross to the other side.
Unload the chest from the boat. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,311 |
4. Cut a square $8 \times 8$ along the cell boundaries into 7 parts with equal perimeters. (The parts can be different in shape) | Solution. See the example of cutting in the figure.
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 7 |
| 1 | 2 | 3 | 4 | 5 | 6 | 6 | 7 |
| 1 | 2 | 3 | 4 | 5 | 5 | 6 | 7 |
| 1 | 2 | 3 | 4 | 4 | 5 | 6 | 7 |
| 1 | 2 | 3 | 3 | 4 | 5 | 6 | 7 |
| 1 |... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,312 |
5. Children in the class were treating each other with candies. Each boy gave a candy to everyone who is taller than him, and each girl - to everyone who is shorter than her (all children are of different heights). It turned out that Sasha, Zhenya, and Varya received the same number of candies, while all others receive... | Solution.
Let's line up all the children by height.
Notice that a girl and a boy standing next to each other in this line received the same number of candies. This is because they received candies from all boys who are shorter than them and from all girls who are taller than them. If the girl is taller than the boy, ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,313 |
9.1 a) Given a triangle with heights equal to 4, 5, and 6. What type of triangle is it: acute, right, or obtuse? b) Does a triangle exist with heights equal to 2, 3, and 6? | Answer: a) acute-angled; b) does not exist. Hint: From the formula for the area of a triangle $S=a h / 2$, it follows that the sides of the given triangle are $\frac{2 S}{4} ; \frac{2 S}{5} ; \frac{2 S}{6}$ (such a triangle exists, since $\frac{1}{4}<\frac{1}{5}+\frac{1}{6}$). To determine what kind of triangle it is, ... | ) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,314 |
7.1. The numbers $1,2,3,4,5,6$ were written in some order and denoted by the letters $a, b, c, d, e, f$. Can the equality
$$
(a-1)(b-2)(c-3)(d-4)(e-5)(f-6)=75 ?
$$
# | # Answer. It can.
Solution. Arrange the numbers as follows:
$$
(6-1)(3-2)(4-3)(5-4)(2-5)(1-6)=75 .
$$
Comment. A correct answer without justification (without providing an example) - 0 points. | (6-1)(3-2)(4-3)(5-4)(2-5)(1-6)=75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,315 |
7.2. In Solar City, 6 dwarfs eat donuts daily, 8 dwarfs eat donuts every other day, and the rest do not eat donuts at all. Yesterday, 11 dwarfs ate donuts. How many dwarfs will eat donuts today? | Answer: 9.
Solution: Of the 11 dwarfs who ate donuts yesterday, 6 dwarfs eat them daily, so the remaining $11-6=5$ eat them every other day. Therefore, these five will not eat donuts today, while the other $8-5=3$ from those who eat every other day will. So today, these three will eat donuts, as well as the six who al... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,316 |
7.3. Vasya and Misha's phones show $15 \%$ charge. And after an hour, Vasya's is at $11 \%$, and Misha's is at $-12 \%$. Could Misha's phone discharge before Vasya's, if the phones discharge uniformly, and the displayed percentage of charge is the charge value rounded to the nearest whole number? | Answer. It can.
Solution. We will show that Misha's phone can run out of charge earlier. Let's assume that initially, Misha's phone had a charge of $15.4 \%$, Vasya's - $14.6 \%$, and after an hour, Misha's - $-11.6 \%$, and Vasya's - $11.4 \%$. Thus, in one hour, Misha's phone discharges by $15.4-11.6=3.8 \%$, and Va... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,317 |
7.4. Given nine cards with the numbers $5,5,6,6,6,7,8,8,9$ written on them. From these cards, three three-digit numbers $A, B, C$ were formed, each with all three digits being different. What is the smallest value that the expression $A+B-C$ can have? | # Answer: 149.
Solution. By forming the smallest sum of numbers $A$ and $B$, and the largest number $C$, we get the smallest value of the expression $A+B-C$. This is $566+567-988=145$. However, this partition is not valid: two numbers have the same digits. By swapping the digits 6 and 8 in the units place, we get the ... | 149 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,318 |
7.5. A round table was sat by 10 people - liars and knights. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of those sitting passed their coin to one of their two neighbors. After that, 5 people said: “I have one coin,” while the other 5 said: “I have no coins.” What is ... | # Answer: 7.
Solution. After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 10. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs fro... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,319 |
8.1. Around a circle, 2019 natural numbers are written. Prove that there will be two adjacent numbers whose sum is even. | Solution. The sum of two numbers will be even if they are both even or both odd. The sum of two numbers will be odd if one is even and the other is odd. Suppose the sum of any two adjacent numbers is odd, then even and odd numbers must alternate. This means the total number of numbers will be even, but according to the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,320 |
8.2. Little kids were eating candies. Each one ate 7 candies less than all the others together, but still more than one candy. How many candies were eaten in total? | Answer: 21 candies.
Solution: Let $S$ denote 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 $a+7$ candies, and thus together they ate $S=a+(a+7)=2a+7$ candies. This reasoning is valid for each child, so all the children ate... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,321 |
8.3. Through point $B$, four lines are drawn such that $A B \perp B D, B E \perp B C$, and line $A C$ intersects these lines so that $A B = B C$. Line $A C$ intersects $B D$ at point $D$, and $A C$ intersects $B E$ at point $E$. Prove that $\triangle A B E = \triangle B C D$. | Solution. Since $A B=B C$, then $\angle B A C=\angle B C A$ (see Fig.1). Next, $\angle A B E=90^{\circ}-\angle E B D$, $\angle C B D=90^{\circ}-\angle E B D$. Therefore, $\angle A B E=\angle C B D$. Thus, we have: $A B=B C, \angle B A C=\angle B C A, \angle A B E$ $=\angle C B D$. Hence, $\triangle A B E=\triangle B C ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,322 |
8.4. a) Write a non-zero digit in each circle (see Fig. 2) so that the sum of the digits in the two top circles is 7 times less than the sum of the other digits, and the sum of the digits in the two left circles is 5 times less than the sum of the other digits.
b) Prove that the problem has a unique solution.
 See Fig. 3:
Fig. 3
Solution. b) If the sum of the digits in the two upper circles is 7 times less than the sum of the other digits, then it is 8 times less than the sum of all fiv... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,323 |
8.5. The placement of chess kings on a board is called "correct" if no king attacks another, and every square of the board is either under attack or occupied by one of the kings. What is the minimum and maximum number of kings that can be correctly placed on an $8 \times 8$ chessboard? | Answer. The minimum and maximum number of kings is 9 and 16, respectively.
Solution. Divide the board into 16 squares of size $2 \times 2$. In each of them, there can be no more than one king, otherwise they will attack each other, so their number in any correct arrangement does not exceed 16. For example, if there is... | 916 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,324 |
9.1 If $x$ and $y$ are positive numbers, then at least one of the numbers $x$, $y$, and $\frac{1}{x+y}$ is greater than 0.7. Prove this. | Solution: If $x \leq 0.7$ and $y \leq 0.7$, then $x+y \leq 1.4$ and $\frac{1}{x+y} \geq \frac{1}{1.4}>0.7$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,325 |
9.3 In triangle $A B C$, the median $A M$ is drawn (point $M$ lies on side $\mathrm{BC}$). It is known that angle $C A M$ is $30^{\circ}$, and side $A C$ is 2. Find the distance from point $B$ to the line $A C$.
Omвem: 1. | Solution: See fig.
Triangle СKM is equal to triangle ВHM (these are right triangles, the hypotenuses СM and ВM of which are equal, and the angles are the same).
Therefore, $\mathrm{BH}=$ СK. But in triangle СKA, the leg СK lies opposite the angle $30^{\circ}$ and is equal to half the hypotenuse АC: СK=1.
 | Solution. We will show that $\angle A_{1} C_{1} B_{1}=45^{\circ}$. Specifically, from the isosceles triangles $A B_{1} C_{1}$ and $B A_{1} C_{1}$, we have $\angle A C_{1} B_{1} = 90^{\circ} - \frac{1}{2} \angle B A C$ and $\angle B C_{1} A_{1} = 90^{\circ} - \frac{1}{2} \angle A B C$. Therefore, $\angle A_{1} C_{1} B_{... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,328 |
9.3. Can a $12 \times 12$ checkerboard be divided into corners made of three adjacent cells $\square$ such that each horizontal and each vertical row of cells on the board intersects the same number of corners? (A row intersects a corner if it contains at least one of its cells.) | # Answer: No.
Solution. Suppose such a partition exists. Consider the first and second rows from the bottom of the board; denote them as $H_{1}$ and $H_{2}$. Each L-tromino on the board intersects with two adjacent rows. Therefore, if an L-tromino intersects with $H_{1}$, it must also intersect with $H_{2}$. Now, if r... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 12,329 |
9.4. One thousand numbers are written in a circle. Petya calculated the absolute differences of adjacent numbers, Vasya - the absolute differences of numbers standing one apart, and Tolya - the absolute differences of numbers standing two apart. It is known that any of Petya's numbers is at least twice as large as any ... | Solution. Let $v$ be the largest of Vasya's numbers, and $t$ be one of Tolya's (say, $t=|a-d|$, where $a, b, c, d$ are four consecutive numbers). It is sufficient to prove that $t \geqslant v$.
Among Petya's numbers, there is the number $|a-b|$; hence, $|a-b| \geqslant 2 v$. On the other hand, $|b-d|$ is one of Vasya'... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 12,330 |
1. The sum of two integers was multiplied by their product. Could the result be 20182017? | 1. Let $a \cdot b \cdot(a+b)=20182017$. Since the number 20182017 is odd, each of the factors on the left must also be odd. However, if $a$ and $b$ are odd numbers, then their sum $a+b$ is even, and the entire product $a \cdot b \cdot(a+b)$ will be even. This leads to a contradiction. Therefore, the number 20182017 cou... | couldnothave | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,331 |
2. It is known that the numbers $a, b, c$ are integers and their sum is divisible by six. Prove that $a^{5}+b^{3}+c$ is also divisible by six. | 2. $a^{5}+b^{3}+c=\left(a^{5}-a\right)+\left(b^{3}-b\right)+(a+b+c)=$
$=(a-1) a(a+1)\left(a^{2}+1\right)+(b-1) b(b+1)+(a+b+c)$
Each term of the sum is divisible by six. Therefore, the sum is also divisible by six. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,332 |
3. The seller has 10 melons and scales that can determine the total weight of any three melons in one weighing. How can you find the total weight of all the melons in six weighings? | 3. Let's number all the melons: $D_{1}, D_{2}, \ldots, D_{10}$.
We conduct six weighings:
1 - melons $D_{1}, D_{2}, D_{3} ; \quad 4$ - melons $D_{1}, D_{2}, D_{4} ;$
2 - melons $D_{2}, D_{3}, D_{4} ; \quad 5$ - melons $D_{5}, D_{6}, D_{7} ;$
3 - melons $D_{1}, D_{3}, D_{4} ; \quad 6$ - melons $D_{8}, D_{9}, D_{10}$... | \frac{S_{1}}{3}+S_{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,333 |
4. The angle at the vertex B of the isosceles triangle ABC is $108^{\circ}$. The perpendicular to the bisector $\mathrm{AD}$ of this triangle, passing through the point $\mathrm{D}$, intersects the side $\mathrm{AC}$ at the point $\mathrm{E}$. Prove that $\mathrm{DE}=\mathrm{BD}$. | 4.
Let $F$ be the intersection point of $DE$ and $AB$. $\angle FBD = 72^\circ$.
Consider $\triangle ADF$:
$\angle AFD = \angle ADF - \angle FAD = 90^\circ - 18^\circ = 72^\circ$.
Since $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,334 |
5. Ali-Baba is playing a game with one of the bandits: on the table lies a pile of 2017 diamonds; in one move, it is allowed to divide one pile into any two; the one who cannot make a move loses; the winner takes all the diamonds. How should Ali-Baba play to get 2017 diamonds? | 5. Initially, there is one pile of 2017 diamonds on the table. After each move, the number of piles increases by one. In the end, there will be 2017 piles, each with 1 diamond. Therefore, regardless of the players' moves, 2016 moves will be made and the player who started second will win. Thus, to take all the diamonds... | Ali-Babashouldyieldtherighttomakethefirstmovetothebit | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,335 |
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 | 12,336 |
4-2. There are several packages with apples. Each contains either 12 or 6 apples. How many apples can there be in total if it is known that there are no fewer than 70 and no more than 80? List all the options. | Answer: 72 or 78.
Solution. The number of apples in each bag is divisible by 6, so the total number of apples is also divisible by 6. In this range, there are only two numbers divisible by $6:$ these are 72 and 78. | 72or78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,337 |
4-4. In the box, there are colored pencils.
Vasya said: "There are at least four blue pencils in the box."
Kolya said: "There are at least five green pencils in the box."
Petya said: "There are at least three blue and at least four green pencils in the box."
Misha said: "There are at least four blue and at least fo... | Answer: Kolya.
Solution. There should be at least four blue pencils, since otherwise at least two boys - Vasya and Misha - will be wrong. Therefore, all statements about blue pencils are correct.
There should be at least four green pencils, since otherwise three boys - Kolya, Petya, and Misha - will be wrong. There c... | Kolya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,339 |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
. When Petya was at the 22nd... | Answer. At the 64th lamppost.
Solution. There are a total of 99 intervals between the lampposts. From the condition, it follows that while Petya walks 21 intervals, Vasya walks 12 intervals. This is exactly three times less than the length of the alley. Therefore, Petya should walk three times more to the meeting poin... | 64 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,342 |
4-8. In a chess club, 90 children attend. During the session, they divided into 30 groups of 3 people, and in each group, everyone played one game with each other. No other games were played. In total, there were 30 games of "boy+boy" and 14 games of "girl+girl". How many "mixed" groups were there, that is, groups wher... | Answer: 23.
Solution: There were a total of 90 games, so the number of games "boy+girl" was $90-30-14=46$. In each mixed group, two "boy+girl" games are played, while in non-mixed groups, there are no such games. In total, there were exactly $46 / 2=23$ mixed groups.
## Grade 5 | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,343 |
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 | 12,344 |
5-2. In a sports tournament, a team of 10 people participates. The regulations stipulate that 8 players from the team are always on the field, changing from time to time. The duration of the match is 45 minutes, and all 10 participants on the team must play an equal number of minutes. How many minutes will each player ... | Answer: 36.
Solution. In total, the players will spend $8 \cdot 45=360$ minutes on the field. This time needs to be divided equally among 10 players, so each will be on the field $360 / 10=36$ minutes. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,345 |
5-3. How many two-digit numbers exist where at least one of the digits is smaller than the corresponding digit in the number $35?$
For example, the numbers 17 and 21 are valid, while the numbers 36 and 48 are not. | Answer: 55.
Solution. First, let's find the number of two-digit numbers that do not meet the condition. In the units place, any digit from 5 to 9 can stand, and in the tens place, from 3 to 9. The total number of numbers that do not suit us will be exactly $7 \cdot 5=35$. Now we can count the number of two-digit numbe... | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,346 |
5-4. Andrei, Boris, Vladimir, and Dmitry each made two statements. For each boy, one of his statements turned out to be true, and the other false.
Andrei: "Boris is not the tallest among us four." "Vladimir is the shortest among us four."
Boris: "Andrei is the oldest in the room." "Andrei is the shortest in the room.... | Answer: Vladimir.
Solution. The first statement of Dmitry is incorrect, as it contradicts the condition of the problem. Therefore, Dmitry is the oldest. Consequently, the first statement of Boris is incorrect, so Andrey is the shortest. From this, it follows that the second statement of Andrey is incorrect, so Boris i... | Vladimir | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,347 |
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 | 12,348 |
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right ... | Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such ... | 77 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,349 |
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 | 12,350 |
5-8. In the "Young Photographer" club, 300 children attend. During the class, they divided into 100 groups of 3 people, and in each group, each person took one photo of the other two in their group. No one took any other photos. In total, there were 100 photos of "boy+boy" and 56 photos of "girl+girl". How many "mixed"... | Answer: 72.
Solution: There were a total of 300 photos, so the number of photos of "boy+girl" was $300-100-56=144$. Each mixed group provides two photos of "boy+girl", while non-mixed groups do not provide such photos. Therefore, there were exactly $144 / 2=72$ mixed groups.
## 6th grade | 72 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,351 |
6-1. The physical education teacher lined up the class so that everyone was facing him. To the right of Kolya, there are 12 people, to the left of Sasha - 20 people, and to the right of him - 8 people. How many people are standing to the left of Kolya? | Answer: 16.
Solution: Since there are 20 people to the left of Sasha and 8 people to the right of him, there are a total of 28 people in the row, not counting Sasha. Therefore, including Sasha, there are 29 people in the class. Then, to the left of Kolya, there are $29-12-1=16$ people (first subtracting the 12 people ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,352 |
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 | 12,354 |
6-4. In the store "Kopeechka", any item is sold for some number of rubles and 99 kopecks (possibly for 0 rubles and 99 kopecks). Different items may have different prices. Kolya made a purchase for 200 rub 83 kopecks. How many items could he have bought? Find all possible options. | Answer: 17 or 117.
Solution: We will buy items one by one. Then the number of kopecks will decrease by 1 each time, and after 0 kopecks comes 99 kopecks. The first time we get a whole number of rubles and 83 kopecks, by buying $100-83=17$ items. This option works: for example, 16 items could be bought for 0 rubles 99 ... | 17or117 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,355 |
6-5. Grandma baked 21 batches of dumplings, with $N$ dumplings in each batch, $N>70$. Then she laid out all the dumplings on several trays, with 70 dumplings on each tray. What is the smallest possible value of $N$? | Answer: 80.
Solution: The total number of baked buns is $21 \cdot N$. This number must be divisible by 70 to be able to distribute them into several trays of 70 each. $70=2 \cdot 5 \cdot 7$, and 21 is already divisible by 7. Therefore, $N$ must be divisible by 10, and the smallest such $N$ is 80. | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,356 |
6-7. In a confectionery store, the saleswoman laid out 91 candies of several varieties in a row on the counter. It turned out that between any two candies of the same variety, there was an even number of candies. What is the smallest number of varieties there could have been? | Answer: 46.
Solution. We will prove that there could not have been three or more candies of the same type. Indeed, let candies of the same type $A, B$, and $C$ lie in that exact order. Suppose there are $2x$ candies between $A$ and $B$, and $2y$ candies between $B$ and $C$, then there are $2x + 2y + 1$ candies between... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,358 |
6-8. Five girls played several table tennis matches at one table. At any given time, two of the girls played, while the other three rested. The girl who lost a match went to rest, and her place at the table was taken by the girl who had rested the most; if there were several such girls, any one of them could take the p... | Answer: $4,8,12,16$.
Solution. Note that the girls participated in a total of $4+6+7+10+11=38$ games, with two girls participating in each game, so there were 19 games in total. The key insight in the solution is that a girl cannot skip four games in a row. Therefore, Anya could only play 4 games in the following way:... | 4,8,12,16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,359 |
7-1. Petya has stickers. If he gives 5 stickers to each of his friends, he will have 8 stickers left. If he wants to give 6 stickers to each of his friends, he will be short of 11 stickers. How many friends does Petya have? | Answer: 19.
Solution I. Suppose Petya gave 5 stickers to each of his friends. Next, he wants to give each friend one more sticker. For this, he needs to spend $8+11=19$ stickers, and he gives one sticker to each friend, so he has 19 friends.
Solution II. Let Petya have $x$ friends. Then $5 x+8=6 x-11$, from which $x=... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,360 |
7-3. A secret object is a rectangle measuring $200 \times 300$ meters. Outside the object, there is a guard at each of the four corners. An intruder approached the perimeter of the secret object from the outside, and all the guards ran to him by the shortest paths along the external perimeter (the intruder remained in ... | Answer: 150.
Solution. Note that no matter where the violator is, two guards in opposite corners will run a distance equal to half the perimeter in total.
Therefore, all four guards will ru... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,362 |
7-4. In a giraffe beauty contest, two giraffes, Tall and Spotted, made it to the final. 135 voters are divided into 5 districts, each district is divided into 9 precincts, and each precinct has 3 voters. The majority of voters on each precinct choose the winner in their precinct; in the district, the giraffe that wins ... | Answer: 30.
Solution: For High to win the final, he must win in 3 districts. To win in a district, High must win in 5 precincts of that district. In total, he needs to win in at least $3 \cdot 5=15$ precincts. To win in a precinct, at least 2 voters must vote for him. Therefore, at least 30 voters are needed.
Comment... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,363 |
7-5. In a row, there are 1000 toy bears. The bears can be of three colors: white, brown, and black. Among any three consecutive bears, there is a toy of each color. Iskander is trying to guess the colors of the bears. He made five guesses:
- The 2nd bear from the left is white;
- The 20th bear from the left is brown;
... | Answer: 20.
Solution: Since among any three consecutive bears there is a bear of each color, the numbers of all bears of a certain color have the same remainder when divided by 3. Indeed, let's look at the bears with numbers $n, n+1$, and $n+2$, as well as with numbers $n+1, n+2$, and $n+3$. In both cases, there will ... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,364 |
7-6. In an ornithological park, there are several species of birds, a total of 2021 individuals. The birds sat in a row, and it turned out that between any two birds of the same species, there was an even number of birds. What is the smallest number of bird species that could have been there? | Answer: 1011.
Solution. Estimation. We will prove that there could not have been three or more birds of the same species. Indeed, suppose birds of the same species $A, B$, and $C$ sit in that exact order. Let there be $2x$ birds between $A$ and $B$, and $2y$ birds between $B$ and $C$, then there are $2x + 2y + 1$ bird... | 1011 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,365 |
7-7. The recruits stood in a row, facing the same direction. Among them were three brothers: Peter, Nikolai, and Denis. In front of Peter were 50 people, in front of Nikolai 100, and in front of Denis 170. On the command "About face!" everyone turned to face the opposite direction. As a result, it turned out that in fr... | Answer: 211.
Solution. Let there be $x$ people in front of Peter, $y$ people in front of Nicholas, and $z$ people in front of Denis. There are three possible cases.
- $x=4 y$. Then $4 y+5... | 211 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,366 |
7-8. Grandma and her favorite grandson Vasyutka agreed to show a trick to Mom. Grandma had 10 fillings for pies, and she baked one pie with each pair of these fillings. In total, she baked 45 pies. Only Grandma can determine the filling of a pie by looking at it.
Grandma lays out $n$ of the 45 pies on a plate, and Vas... | Solution. Let's assume Vasyutka broke less than 36 pies, i.e., there are still 10 or more pies left. Suppose Vasyutka told his mother that the pie she took has the first type of filling. Why could he be wrong? Because among the 10 or more remaining pies, there is a pie without the first type of filling (since there are... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,367 |
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 | 12,369 |
8-3. What will be the result if we expand and combine like terms in the expression $(a+1)(a+3)(a+4)(a+5)(a+6)$?
a) $a^{5}+360$.
b) $a^{5}+19 a^{4}+137 a^{3}+461 a^{2}+702 a+360$
c) $a^{5}+19 a^{4}+138 a^{3}+476 a^{2}+776 a+480$.
d) $a^{4}+18 a^{3}+119 a^{2}+342 a+360$.
e) $a^{5}+18 a^{4}+123 a^{3}+402 a^{2}+656 a+... | Answer: b).
Solution. We will discard all options except b):
- option a) is incorrect because, in addition to the two terms mentioned, there should be some others;
- option d) is incorrect, as it resulted in a polynomial of the 4th degree;
- options c) and e) are incorrect, as they have the wrong constant term. | ^{5}+19^{4}+137^{3}+461^{2}+702+360 | Algebra | MCQ | Yes | Yes | olympiads | false | 12,370 |
8-4. In a giraffe beauty contest, two giraffes, Tall and Spotted, made it to the final. 105 voters are divided into 5 districts, each district is divided into 7 precincts, and each precinct has 3 voters. The majority of voters on each precinct choose the winner of their precinct; in a district, the giraffe that wins th... | Answer: 24.
Solution. For High to win the final, he must win in 3 districts. To win a district, High must win in 4 precincts of that district. In total, he needs to win in at least $3 \cdot 4=12$ precincts. To win in a precinct, at least 2 voters must vote for him. Therefore, at least 24 voters are needed.
Comment. I... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,371 |
8-5. In a tournament, 6 teams $P, Q, R, S, T$ and $U$ participate, and each team must play against every other team exactly once. Each day, they are divided into 3 pairs, and all three matches are played simultaneously. The "Sports" channel has chosen which match it will broadcast each day:
$$
\begin{array}{c|c|c|c|c}... | Answer. Only in the 1st.
Solution. Let's look at team $P$: on the 1st, 3rd, and 5th days, it will play against teams $Q, T$, and $R$. Therefore, in the remaining two days, it must play against teams $S$ and $U$. Since on the 2nd day $S$ plays against $R$, $P$ has no choice but to play against $U$ on the 2nd day, and a... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,372 |
8-6. Inside triangle $A B C$, a point $D$ is chosen such that $\angle B A D=60^{\circ}$ and $\angle A B C=\angle B C D=30^{\circ}$. It is known that $A B=15$ and $C D=8$. Find the length of segment $A D$. If necessary, round the answer to 0.01 or write the answer as a common fraction | Answer: 3.5.
Solution I. Let the line $A D$ intersect the segment $B C$ at point $X$. Since in triangle $A B X$ the angles $A$ and $B$ are $30^{\circ}$ and $60^{\circ}$, angle $X$ is a right angle. Therefore, $A X = A B / 2 = 7.5$, since $A X$ is the leg of the right triangle $A B X$ opposite the $30^{\circ}$ angle. S... | 3.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,373 |
8-7. Petya thought of four different digits, not equal to 0. Then he formed all possible four-digit numbers from these digits without repeating any digits. The sum of all these numbers turned out to be 73326. What 4 digits did Petya think of? | Answer: $1,2,3,5$.
Solution. Let Petya have thought of the digits $a, b, c$ and $d$. In total, there will be 24 numbers composed of these digits. If we add all these numbers, each of the digits $a, b, c$ and $d$ will appear in each place value 6 times. Therefore, the sum of all 24 numbers will be equal to $6 \cdot(a+b... | 1,2,3,5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,374 |
8-8. A chess tournament is held according to the following system: each of the 15 students from the "White Rook" school must play one game with each of the 20 students from the "Black Bishop" school, i.e., a total of 300 games should be played. At any given time, no more than one game is played.
After $n$ games, a spe... | Answer: 280.
Solution: Estimation. Suppose fewer than 280 games have passed, i.e., more than 20 games remain. Then, among the participants of the "White Rook" school, there are at least two students who have not yet played all their games. In this case, Sasha cannot accurately name the participant of the next game fro... | 280 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,375 |
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 | 12,376 |
9-3. Arina wrote down all the numbers from 71 to 81 in a row without spaces, forming a large number 717273...81. Sofia started appending the next numbers to it (i.e., she first appended 82, then 83, ...). She stopped when the large number became divisible by 12. The last number she appended was $N$. What is $N$? | Answer: 88.
Solution. A number is divisible by 12 if and only if it is divisible by 3 and by 4. For a number to be divisible by 4, the number formed by its last two digits must also be divisible by 4. Therefore, the last number that Sofia writes must be divisible by 4.
The nearest number that is divisible by 4 is 84,... | 88 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,378 |
9-5. A circle is divided into 100 equal arcs by 100 points. Next to the points, numbers from 1 to 100 are written, each exactly once. It turns out that for any number $k$, if a diameter is drawn through the point with the number $k$, then the number of numbers less than $k$ on either side of this diameter will be equal... | Answer: Only 84.
Solution: Consider the odd number $2 m+1$. Let's mentally discard it and the number diametrically opposite to it. According to the condition, among the remaining numbers, all numbers less than $2 m+1$ are divided into two groups of equal size. Therefore, among the remaining numbers, there is an even n... | 84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,380 |
9-6. Petya wants to place 99 coins in the cells of a $2 \times 100$ board so that no two coins are in cells that share a side, and no more than one coin is in any cell. How many ways are there to place the coins? | Answer: 396.
Solution. Note that there will be exactly 1 empty column. Then, to the left of it, there are exactly two ways to arrange the tiles, and to the right of it, there are also exactly two ways to arrange the tiles.
, lines $M N$ and $A D$ are parallel; but this is imposs... | \frac{7}{12}\approx0.58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,382 |
9-8. All digits in the notation of 6-digit natural numbers $a$ and $b$ are even, and in the notation of any number between them, there is an odd digit. Find the largest possible value of the difference $b-a$.
---
Translation:
9-8. All digits in the notation of 6-digit natural numbers $a$ and $b$ are even, and in the ... | Answer: 111112.
Solution. Estimation. We will prove that to a 9-digit number $a$, less than 888 888, all digits of which are even, one can add a number not exceeding 111112 so that all its digits will again be even. If among the digits of the number $a$, except for the first one, there is a digit less than 8, then it ... | 111112 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,383 |
10-1. There are several bowls on the table, each containing several grapes. Different bowls may contain different numbers of grapes. If 12 bowls are each added 8 more grapes, the average number of grapes in all the bowls will increase by 6. How many bowls are on the table | Answer: 16.
Solution: Let the number of bowls be $n$. The total number of berries increased by $12 \cdot 8=96$. Since the average number of berries increased by 6, the total number of berries should have increased by $6n$. Therefore, $6n=96$, from which we find $n=16$. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,384 |
10-2. The large rectangle in the figure consists of 20 identical smaller ones. The perimeter of figure $A$ is 56 cm, the perimeter of figure $B$ is 56 cm. What is the perimeter of figure $C$? Provide your answer in cm.
| | $A$ | | |
| ---: | :--- | :--- | :--- |
| | | | |
| | | | |
| $B$ | | | $C$ |
| | ... | Answer: 40.
Solution: Let the horizontal size of the rectangle be $x$, and the vertical size be $y$. From the condition, we get the system of equations
$$
\left\{\begin{array}{l}
6 x+2 y=56 \\
4 x+6 y=56
\end{array}\right.
$$
We need to find what $2 x+6 y$ equals. Solving the system, we find $x=8, y=4$, from which $... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,385 |
10-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,386 |
10-4. Initially, a natural number was displayed on the calculator screen. Each time, Olya added a natural number to the current number \( n \) on the calculator screen, which \( n \) did not divide. For example, if the screen showed the number 10, Olya could add 7 to get 17.
Olya repeated this operation five times, an... | Answer: 189.
Solution. Estimation. Note that Olya increased the number on the screen by at least 2 each time, because any number is divisible by 1. If Olya added two five times, the initial number would have been 190, and it would not have been possible to add two to it. Therefore, Olya must have added a number greate... | 189 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,387 |
10-5. For each natural number from 1 to 999, Damir subtracted the last digit from the first digit and wrote all 1000 differences on the board. For example, for the number 7, Damir wrote the number 0 on the board, for the number 105 he wrote $(-4)$, and for the number 61 he wrote 5.
What is the sum of all the numbers o... | Answer: 495.
Solution: Note that for single-digit numbers, zeros are recorded on the board, which do not affect the sum. For numbers where the first and last digit are the same, zeros are also recorded on the board, which do not affect the sum.
Almost all other numbers can be paired: a number and the number obtained ... | 495 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,388 |
10-6. The places of cyclists in the race are determined by the sum of the times on all stages: the first place goes to the rider with the smallest total time,..., the last place goes to the rider with the largest total time. There were 500 cyclists, the race took place over 15 stages, and there were no cyclists with th... | Answer: 91.
Solution. Estimation. Note that if driver A has a higher position than driver B, then A must have overtaken B in at least one race. Vasya was overtaken by no more than $6 \cdot 15=90$ other drivers in 15 races (no more than because some drivers could have overtaken Vasya in several races). Therefore, there... | 91 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,389 |
10-7. Parabola $\Pi_{1}$ with branches directed upwards passes through points with coordinates $(10,0)$ and $(13,0)$. Parabola $\Pi_{2}$ with branches directed upwards also passes through the point with coordinates $(13,0)$. It is also known that the vertex of parabola $\Pi_{1}$ bisects the segment connecting the origi... | Answer: 33.
Solution. We will use the following fact twice: if $x_{1}$ and $x_{2}$ are the x-coordinates of the points where the parabola intersects the x-axis, then the x-coordinate of the vertex is $\frac{x_{1}+x_{2}}{2}$ (the x-coordinate of the vertex is the midpoint of the segment with endpoints $x_{1}$ and $x_{2... | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,390 |
10-8. In trapezoid $A B C D$, the bases $A D$ and $B C$ are 8 and 18, respectively. It is known that the circumcircle of triangle $A B D$ is tangent to the lines $B C$ and $C D$. Find the perimeter of the trapezoid. | Answer: 56.
Solution. Let's make the following observation. Through point $B$ on the circle, a line passes parallel to the chord $AD$. It is clear that then $B$ is the midpoint of the arc $AD$, that is, $BA = BD$ (indeed, $\angle 1 = \angle 2$ as alternate interior angles, $\angle 1 = \angle 3$ by the theorem on the a... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,391 |
11-1. Twins Paolo and Sevilla are celebrating their birthday at a cafe with friends. If the final bill amount is divided equally among everyone, then each person should pay 12 euros. But if the bill is divided equally among everyone except Paolo and Sevilla, then each person should pay 16 euros. How many friends came t... | Answer: 6.
Solution. Let $n$ be the number of friends who arrived. Then we get the equation $12(n+2)=16n$, from which $n=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,392 |
11-2. At a round table, 14 conference participants were seated. During the break, some of them (but not all) went to have coffee. It turned out that each participant who remained at the table had exactly one neighbor who left. How many participants could have gone for coffee? Provide all possible answers. | Answer: 6, 8, 10, or 12 participants.
Solution. Let participant A remain at the table. Then, one of his neighbors, let's say B, also remained. But one of B's neighbors also remained, and it is participant A. Therefore, all the participants who remained at the table can be paired.
Thus, the remaining participants sit ... | 6,8,10,12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,393 |
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
 | Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they... | 58 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,394 |
11-4. At first, a natural number was displayed on the calculator screen. Each time, Tanya added to the current number \( n \) on the screen a natural number that \( n \) did not divide. For example, if the screen showed the number 10, Tanya could add 7 to get 17.
Tanya repeated this operation five times, and the numbe... | Answer: 89.
Solution. Estimation. Note that Tanya increased the number on the screen by at least 2 each time, because any number is divisible by 1. If Tanya added two five times, the initial number would have been 90, and it would not have been possible to add two to it. Therefore, Tanya must have added a number great... | 89 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,395 |
11-5. The function $f(x)$ is defined for all positive values of $x$. It turns out that $f\left(\frac{4 y+1}{y+1}\right)=\frac{1}{y}$ for any $y>0$. Find $f(3)$. | Answer: 0.5.
Solution: We need to substitute for $y$ such a number that $\frac{4 y+1}{y+1}=3$, which means $y=2$. We get $f(3)=\frac{1}{2}$.
Comment: Note that such a function exists. The proof of this is almost identical to the solution: $\frac{4 y+1}{y+1}=a, y=\frac{a-1}{4-a}$, from which for $a \neq 4$ we get $f(a... | 0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,396 |
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