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8.2. Petya came home from school today at $16:45$, looked at the clock and wondered: after what time will the hands of the clock be in the same position for the seventh time since he came home from school? | Answer: 435 minutes.
Solution. The speed of the minute hand is 12 divisions/hour (one division here refers to the distance between adjacent numbers on the clock face), and the hour hand is 1 division/hour. Before the seventh meeting of the minute and hour hands, the minute hand must first "lap" the hour hand 6 times, ... | 435 | Other | math-word-problem | Yes | Yes | olympiads | false | 15,060 |
8.3. Find all non-negative integer solutions of the equation
$$
2 x^{2}+2 x y-x+y=2020
$$ | Answer: $x=0, y=2020 ; x=1, y=673$.
Solution. Rewrite the equation as
$$
y(2 x+1)=2020-2 x^{2}+x
$$
For integer values of $x$, the expression $(2 x+1) \neq 0$, so by dividing it by $2 x+1$ and isolating the integer part, we get:
$$
y=\frac{2019}{2 x+1}-x+1
$$
For $y$ to be an integer, $2 x+1$ must be a divisor of ... | (0,2020),(1,673) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,061 |
8.4. On a plane, there are 20 points, the distances between any two of which are distinct. Connect each point with the nearest one by a segment. Prove that no point will be connected to more than Five neighbors. | Solution. Suppose that point $M$ is connected by segments $M A, M B$, $M C, M D, \ldots$ to points $A, B, C, D, \ldots$
Connect points $A$ and $B$ with a segment. We have $A M\alpha$ and $\g... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,062 |
8.5. Miron and Varya are playing the following game. There is a pile of 32 stones on the table. The players take turns, with Miron starting first. On their turn, a player divides any pile that has more than one stone into several equal piles. The player who cannot make a move (when there is exactly one stone in each pi... | Answer: Varya.
Solution. Varya wins regardless of the moves the players make. Since the number 32 has no odd divisor, each player will divide one pile into an even number of piles on their turn, and the parity of the number of piles will change. Before Miron's move, there will always be an odd number of piles, and bef... | Varya | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,063 |
1. Answer: $\frac{99^{2019}+1}{99^{2020}+1}>\frac{99^{2020}+1}{99^{2021}+1}$. | Solution. Consider the difference of these fractions:
$\frac{99^{2019}+1}{99^{2020}+1}-\frac{99^{2020}+1}{99^{2021}+1}=\frac{\left(99^{2019}+1\right)\left(99^{2021}+1\right)-\left(99^{2020}+1\right)\left(99^{2020}+1\right)}{\left(99^{2020}+1\right)\left(99^{2021}+1\right)}$.
Simplifying, we get the fraction $\frac{99... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,064 |
1. (7 points) Place the parentheses so that the equation is correct:
$$
90-72: 6+3=82
$$ | Solution: $90-72:(6+3)=82$.
Criteria. Correct placement of parentheses: 7 points.
Incorrect placement of parentheses: 0 points. | 90-72:(6+3)=82 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,066 |
2. (7 points) There are apples in five boxes, with an equal number of apples in each. When 60 apples were taken out of each box, after that, the total number of apples left was the same as the number of apples that were originally in two boxes. How many apples were in each box?
Answer: 100. | Solution. In total, $60 \cdot 5=300$ apples were taken out, and this is equal to the number of apples that were in three boxes. Therefore, there were 100 apples in each box.
Criteria. Any correct solution: 7 points.
If it is not justified that there were 100 apples in each box, but it is verified that the condition i... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,067 |
4. (7 points) A gnome in shoes weighs 2 kg more than a gnome without shoes. If you put five identical gnomes in shoes and five such gnomes without shoes on the scales, the scales will show 330 kg. How much does a gnome in shoes weigh? | Answer: 34 kg.
Solution. Let's put boots on five gnomes, then the weight will increase by 10 kg. It turns out that ten gnomes in boots weigh 340 kg. Therefore, one gnome in boots weighs 34 kg.
Criteria. Correctly found the weight of a gnome in boots by any method: 7 points.
Correctly found the weight of a gnome with... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,069 |
Task 1. Two runners, starting simultaneously at constant speeds, run on a circular track in opposite directions. One of them runs the loop in 5 minutes, while the other takes 8 minutes. Find the number of different meeting points of the runners on the track, if they ran for at least an hour. | Answer: 13 points.
Solution.
Let the length of the track be $\mathrm{S}$ meters. Then the speeds of the runners are $\mathrm{S} / 5$ and S/8 meters per minute, respectively. | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,071 |
Problem 2. Three families of parallel lines have been drawn, with 10 lines in each. What is the maximum number of triangles they can cut out from the plane?
Answer: 150 triangles
# | # Solution.
Consider 100 nodes - the intersection points of lines in the first and second directions. Divide them into 10 sectors: the first sector - nodes lying on the first lines of the first and second
 x^{2}+(c-a) x+(a+b)$ have a common root (not necessarily an integer). Prove that $a+b+2c$ is divisible by 3. | # Solution.
Subtract the second quadratic from the first. We get that they both have a common root with the quadratic
$a x^{2}+b x+c-\left((c-b) x^{2}+(c-a) x+(a+b)\right)=(a+b-c)\left(x^{2}+x-1\right)$. Therefore, either $a+b-c=0$, or their common root coincides with one of the roots of the quadratic $x^{2}+x-1$.
I... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,073 |
Problem 4. Given an acute triangle $ABC$. On the extensions of $BB_{1}$ and $CC_{1}$ of its altitudes beyond points $B_{1}$ and $C_{1}$, points $P$ and $Q$ are chosen respectively such that angle $PAQ$ is a right angle. Let $AF$ be the altitude of triangle $APQ$. Prove that angle $BFC$ is a right angle. | # Solution
Points $B_{1}$ and $C_{1}$ lie on the circle constructed with $B C$ as its diameter (see figure). To solve the problem, it is sufficient to prove that $F$ also lies on this circle.
Points $B_{1}$ and $F$ lie on the circle with diameter $AP$. Therefore, $\angle P F B_{1} = \angle P A B_{1}$. Similarly, $\an... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,074 |
Task 1. In a numerical expression, some digits were replaced with letters (different digits with different letters, the same digits with the same letters). The result is as follows:
$$
2018 A: B C D=A A \text {. }
$$
What numerical expression was originally written? (It is sufficient to provide an example. $2018 A$ w... | Answer: $20185: 367=55$.
Remark. No other examples exist.
## Criteria
## 4 p. A correct example is provided. | 20185:367=55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,076 |
Problem 2. In Grandfather Frost's bag, there are fewer than a hundred gifts for Petya, Vasya, Borya, and Lёsha. Grandfather Frost gave half of the gifts to Petya, a fifth to Vasya, and a seventh to Borya. How many gifts did Lёsha receive? | Answer: 11.
Solution. For Santa Claus to be able to give half of the gifts to Pete, the total number of gifts in his bag must be divisible by 2. Also, since he gave a fifth to Vasya and a seventh to Borya, the total number of gifts must be divisible by 5 and 7. Therefore, the number of gifts must be divisible by $\ope... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,077 |
Problem 4. All 25 seats in the first row at the school play are occupied by schoolchildren. It is known that
- no two girls sit next to each other in this row;
- next to each boy, there is at least one more boy;
- there are 9 girls in total in the first row.
Could it be that a boy is sitting in the central seat of th... | Answer: No, it could not.
Solution. Since no two girls sit next to each other, each girl sits between two boys. Thus, the entire row consists of "groups" of boys sitting consecutively, with exactly one girl sitting between adjacent groups of boys.
According to the condition, next to each boy sits another boy, so ther... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,079 |
Problem 5. By definition $n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$. Prove that the expression $1008! \cdot 1009! \cdot 2017! \cdot 2018!$ is not a square of a natural number. | Solution. Note that 1008! $\cdot$ 1009! = $(1008!)^{2} \cdot 1009$. Similarly, we get that $2017!\cdot 2018!=(2017!)^{2} \cdot 2018$. Thus, 1008! $\cdot$ 1009! $\cdot$ 2017! $\cdot$ 2018! = $(1008!)^{2} \cdot (2017!)^{2} \cdot 1009 \cdot 2018 = (1008! \cdot 2017! \cdot 1009)^{2} \cdot 2$.
The last expression is not a ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,080 |
Problem 6. A convex quadrilateral $ABCD$ is such that $\angle BAC = \angle BDA$ and $\angle BAD = \angle ADC = 60^{\circ}$. Find the length of $AD$, given that $AB = 14$ and $CD = 6$.
Answer: 20. | Solution. Extend $A B$ and $C D$ to intersect at point $P$. Since $\angle P A D = \angle A D P = 60^{\circ}$, triangle $A D P$ is equilateral. Next, we note that triangle $A P C$ is congruent to triangle $D A B$, because $A P = A B$, $\angle A P C = 60^{\circ} = \angle D A B$, and $\angle P A C = \angle A D B$ (Fig. 2)... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,081 |
8.1. What is the sum of the digits of the number $A=10^{50}-10^{40}+10^{30}-10^{20}+10^{10}-1$? | # Answer: 270.
Solution. The number is the sum of three numbers: a number composed of 10 nines followed by 40 zeros, a number composed of 10 nines followed by 20 zeros, and finally, a number composed of 10 nines. All the nines fall on the zeros in the other addends, so there is no carry-over, and the answer is $90+90+... | 270 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,082 |
8.2. Find the largest natural number with all distinct digits such that the sum of any two of its digits is a prime number. | Answer: 520.
Solution: If the desired number is at least a four-digit number, then it either has three digits of the same parity or two pairs of digits of the same parity. In each of these cases, we get that two of the sums of the digits are even. Therefore, they must equal 2. The number 2 can be represented as the su... | 520 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,083 |
8.3. Vasya and Misha's phones show $15 \%$ charge. An hour later, Vasya's phone shows $11 \%$, and Misha's $-12 \%$. Could Misha's phone discharge before Vasya's, if the phones discharge uniformly, and the displayed percentage of charge is the integer part of the charge value? The integer part of a number $A$ is the gr... | 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 $15.9\%$, Vasya's phone had a charge of $15.0\%$, and after an hour, Misha's phone had a charge of $-12.0\%$, and Vasya's phone had a charge of $-11.9\%$. Therefore, in on... | proof | Other | math-word-problem | Yes | Yes | olympiads | false | 15,084 |
8.4. Given a right triangle $A B C$ ( $A B$ - hypotenuse). On the larger leg $A C$ of triangle $A B C$, a point $K$ is chosen such that $A K=B K$. Let $C H$ be the altitude of triangle $A B C$, and point $M$ is symmetric to point $B$ with respect to point $H$. Prove that segments $B K$ and $C M$ are perpendicular. | Solution. Let $P$ be the intersection point of segments $B K$ and $C M$. We need to prove that angle $B P M$ is a right angle, that is, that the sum of angles $\angle P M B = \angle C M B$ and $\angle P B M = \angle K B M$ is $90^{\circ}$. But angle $C M B$ is equal to angle $C B M (= \angle C B A)$, since triangle $B ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,085 |
8.5. A round table was sat at by 10 people - liars and knights. Liars always lie, and knights always tell the truth. Each of them was given a coin. Then each of them passed their coin to one of their two neighbors. After that, each one said: “I have more coins than my right neighbor.” What is the maximum number of knig... | Answer: 6.
Solution: After the coins are passed, each person sitting at the table can have 0, 1, or 2 coins. Note that 3 knights cannot sit in a row. Indeed, let knights $A, B, C$ sit next to each other, with $B$ sitting to the right of $A$, $C$ to the right of $B$, and $D$ to the right of $C$. If $A$ has $x$ coins, $... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,086 |
10.2 Prove that if $a, b, c$ are pairwise distinct numbers, then the system of equations
$$
\left\{\begin{array}{l}
x^{3} - a x^{2} + b^{3} = 0 \\
x^{3} - b x^{2} + c^{3} = 0 \\
x^{3} - c x^{2} + a^{3} = 0
\end{array}\right.
$$
has no solutions. | Solution. Suppose the system has a solution $\mathrm{x}_{0}$. At that, $\mathrm{x}_{0} \neq 0$ is obvious (since among $\mathrm{a}, \mathrm{b}$, c there are numbers different from zero). Subtracting the second equation from the first, and the third from the second, we get for $\mathrm{x}=\mathrm{x}_{0}$
$$
(b-a) x_{0}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,087 |
10.4 Given ten different integers. For each pair of numbers, their difference (the larger minus the smaller) was calculated. Among these differences, there turned out to be exactly 44 different ones. Prove that one of the original ten numbers is equal to the half-sum of two others. | Solution. A total of 45 differences were calculated. This means that some two differences coincided. If the differences $\mathrm{a}-\mathrm{b}$ and $\mathrm{b}-\mathrm{c}$ coincided, then $\mathrm{b}=\frac{\mathrm{a}+\mathrm{c}}{2}$, and everything is proven. But if the differences $\mathrm{a}-\mathrm{b}$ and $\mathrm{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,089 |
10.5 For non-negative numbers x, y, prove the inequality
$$
x^{2}+x y+y^{2} \leq 3(x-\sqrt{x y}+y)^{2}
$$ | Solution. The case $x=y=0$ is obvious. Let at least one of the numbers $x, y$ be positive. Note that, firstly, $x+y \geq 2 \sqrt{x y}$, and secondly, $x+y-\sqrt{x y}>0$. We have:
$$
x^{2}+x y+y^{2}=(x+y)^{2}-x y=(x+y)^{2}-(\sqrt{x y})^{2}=(x+y+\sqrt{x y})(x+y-\sqrt{x y})
$$
so it is sufficient to establish the inequa... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,090 |
2. Various numbers $y_{1}, y_{2}, \ldots, y_{2021}$ are arranged in a circle. Can it happen that
$$
y_{1}+\left|y_{2}\right|=3, \quad y_{2}+\left|y_{3}\right|=3, \quad \ldots, \quad y_{2020}+\left|y_{2021}\right|=3 \quad \text { and } \quad y_{2021}+\left|y_{1}\right|=3 ?
$$ | Answer: No. Suppose one of the numbers, for example, $y_{1}$, is negative. Then $\left|y_{2}\right|>3$. From the second equation, it follows that $y_{2} \leqslant 3$. This is only possible if $y$ is also negative, which means all numbers are negative. But this implies that $y_{1}-y_{2}=3, y_{2}-y_{3}=3$, and so on, mea... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,091 |
8.1. In the class, there are more than 20 but fewer than 30 students. In the class, those who attend the chess club are half the number of those who do not attend. And those who attend the checkers club are one-third the number of those who do not attend. How many students are in the class? Provide all possible answers... | Solution: Let $n$ be the number of students in the class who attend the chess club, then $2n$ students do not attend, and the total number of students in the class is $3n$, meaning the total number of students in the class is divisible by 3. Similarly, from the fact that the number of people attending the checkers club... | 24 | Other | math-word-problem | Yes | Yes | olympiads | false | 15,092 |
8.2. Are there real numbers $x, y, z$ such that
$$
\frac{1}{x^{2}-y^{2}}+\frac{1}{y^{2}-z^{2}}+\frac{1}{z^{2}-x^{2}}=0 ?
$$ | Justify the answer.
## Solution:
Method 1. Let $a=x^{2}-y^{2}, b=y^{2}-z^{2}$. Then $z^{2}-x^{2}=-(a+b)$ and the equation becomes $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, from which $(a+b)^{2}=a b$, and $\left(a+\frac{1}{2} b\right)^{2}+\frac{3 b^{2}}{4}=0$, which is impossible since $b \neq 0$.
Method 2. The absolu... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,093 |
8.3. On the island of knights and liars, each resident was asked about each of the others: is he a knight or a liar. In total, 42 answers of "knight" and 48 answers of "liar" were received. What is the maximum number of knights that could have been on the island? Justify your answer. (It is known that knights always te... | Solution: Each of the $n$ residents gave $n-1$ answers; in total, there were $n(n-1)$ answers, which, according to the problem, equals $42+48=90$. Hence, $n=10$, meaning there are 10 residents on the island. Let the number of knights be $x$, then the number of liars is $10-x$. Answers of "liar" arise in two cases: when... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,094 |
8.4. Does there exist a nine-digit natural number without zero digits, the remainders of the division of which by each of its digits (first, second, ..., ninth) will be pairwise distinct? Justify your answer. | Solution: According to the condition, in such a number (if it exists), all digits except zero occur only once. This means the sum of all digits is 45. By the divisibility rules for 3 and 9, the number is divisible by both 3 and 9. This means the remainders from dividing the number by these digits are 0, and therefore, ... | Doesnotexist | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,095 |
8.5. Let $A B C$ be an equilateral triangle with a side length of 1 cm. Find all points in the plane for each of which the greatest of the distances to the vertices of this triangle is 1 cm. Justify your answer. | # Solution:
Method 1. The set of points on the plane, the distance from each of which to vertex $A$ is greater than the distance to vertex $B$, is a half-plane with a boundary passing through vertex $C$ of triangle $ABC$ and its median - see the left diagram. Similarly, we mark the second half-plane - the set of point... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,096 |
8.6. Can a rectangular block of size $3 \times 4 \times 5$
a) be cut into 10 rectangular blocks of pairwise different volumes and with integer edges;
b) be cut into 11 rectangular blocks of pairwise different volumes and with integer edges? | Justify the answer.
Solution: The minimum 11 different volumes of blocks are the natural numbers from 1 to 11. Their sum is 66, which is greater than the volume of the block to be cut. Therefore, it is impossible to cut the original block into 11 blocks of different integer volumes, and the answer to part b) is negati... | )can,b)cannot | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,097 |
9.1 How many five-digit natural numbers are there that are divisible by 9, and for which the last digit is 2 more than the second last digit? | Answer: 800. Hint The second last digit (the tens digit) can be any from 0 to 7 (so that after adding two, the last digit makes sense). The third and second digits can be any (from 0 to 9). After choosing the specified three digits (the second, third, and fourth), the last digit is uniquely determined by the second las... | 800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,098 |
9.2 a) Prove that if natural numbers $x$, $y$ satisfy the equation $x^{3}+5 y=y^{3}+5 x$, then $x=y$. b) Do there exist different positive real numbers $x, y$ that satisfy this equation? | Answer. b) exist. Hint a) We have $x^{3}-y^{3}=5(x-y)$. If $\mathbf{x} \neq \mathrm{y}$, then after dividing by ( $\mathbf{x}-\mathbf{y}$ ) we get $x^{2}+x y+y^{2}=5$. The smallest (and different) natural numbers 1 and 2, when substituted into the left side, give $7>5$, so for larger values, the left side will also be ... | \sqrt{\frac{5}{7}},2\sqrt{\frac{5}{7}} | Algebra | proof | Yes | Yes | olympiads | false | 15,099 |
1. A cake was divided into three parts. If two-thirds are cut from the first part and added to the second, then the third part will be twice as large as the first, but half the size of the second. What fraction of the cake do the cut pieces represent? | Answer: $3 / 7, 2 / 7, 2 / 7$.
## Solution:
Let the cake be divided into parts $x, y$, and $z$.
According to the problem: $x+y+z=1, \frac{1}{3} x \cdot 2=z, \frac{2}{3} x+y=2 z$.
Sequentially express everything in terms of $x: z=\frac{2}{3} x, \frac{2}{3} x+y=2 \cdot \frac{2}{3} x \Rightarrow y=\frac{2}{3} x$.
Sub... | \frac{3}{7},\frac{2}{7},\frac{2}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,100 |
2. Prove that for all positive numbers $a, b, c, d$ the inequality holds:
$$
\frac{a^{2}}{b}+\frac{c^{2}}{d} \geq \frac{(a+c)^{2}}{b+d}
$$ | # Solution:
Since the numbers $a, b, c, d$ are positive, multiplying both sides of the inequality by the product of the denominators yields an equivalent inequality:
$$
(b+d)\left(d a^{2}+b c^{2}\right) \geq b d(a+c)^{2}
$$
Expanding the brackets: $a^{2} b d+a^{2} d^{2}+b^{2} c^{2}+b c^{2} d \geq a^{2} b d+2 a b c d... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 15,101 |
3. Find all triples of consecutive natural numbers not exceeding 100, the product of which is divisible by 1001. | Answer: $76,77,78$ and $77,78,79$.
## Solution:
Since $1001=7 \cdot 11 \cdot 13$, among this triplet of numbers, there must be a number divisible by 7, a number divisible by 11, and a number divisible by 13. At the same time, a number cannot be divisible by both 11 and 13, so the divisibility by these numbers must be... | 76,77,7877,78,79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,102 |
4. On the extension of side $DC$ beyond point $C$ of rectangle $ABCD$, point $K$ is marked such that $BD = DK$. Prove that the bisector of angle $BAC$ passes through the midpoint of segment $BK$.
# | # Solution:
Let's describe the circle around $A B C D$. $M$ is the intersection point of $B K$ and this circle. $D M$ is the height in the isosceles triangle $B D K$.
Thus, $M$ is the midpoint of $B K$. Additionally, $D M$ is the angle bisector, so angles $B D M$ and $M D K$ are equal, and therefore, arcs $B M$ and $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,103 |
5. In the equation $* X^{2}+* X+*=0$, two players take turns replacing any asterisk with an arbitrary number (zero cannot be placed before $X^{2}$). The first player wins if the resulting equation has no roots, and the second player wins otherwise. Can either of them win, regardless of the opponent's play? | Answer: The second player wins.
## Solution:
Yes, the second player wins regardless of the first player's moves. If the first player does not place a number $C$, different from zero, in the position of the free term, then the second player will place the number 0 in this position, and regardless of the further game, ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,104 |
1. In a certain month, three Sundays fell on even dates. What day of the week was the 20th of this month? | Solution. Sundays in one month alternate between falling on even and odd dates. Since 3 of them fell on even dates, there were a total of 5 Sundays in this month, so the first one could only have been on the 2nd, from which it follows that the 20th was a Thursday.
Answer: Thursday.
## CONDITION | Thursday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,105 |
5. Given an angle $A O B$ and two points $M$ and $N$ inside it. How should a light ray be directed from point $M$ so that, after reflecting first off side $A O$ and then off side $B O$, it hits point $N$? | Solution. Construct point $M_{1}$, symmetric to point $M$ with respect to $A O$, then $N_{1}$ - symmetric to $N$ with respect to $B O$, connect them with segment $M_{1} N_{1}$, and denote by $K$ the point of intersection of it with ray $A O$; determine the required direction $М K$.
## CONDITION | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,106 |
6. A tourist goes on a hike from $A$ to $B$ and back, and completes the entire journey in 3 hours and 41 minutes. The route from $A$ to $B$ first goes uphill, then on flat ground, and finally downhill. Over what distance does the road run on flat ground, if the tourist's speed is 4 km/h when climbing uphill, 5 km/h on ... | Solution. Let $x$ km of the path be on flat ground, then $9-x$ km of the path (uphill and downhill) the tourist travels twice, once (each of the ascent or descent) at a speed of 4 km/h, the other at a speed of 6 km/h, and spends $(9-x) / 4 + (9-x) / 6$ hours on this part. Since the tourist walks $2x / 5$ hours on flat ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,107 |
1-0. The number 111123445678 is written on the board, and several digits (not all) need to be erased to get a number that is a multiple of 5. In how many ways can this be done? | Answer: 60
Solution. The digits 6, 7, and 8 must be crossed out, and 5 must be kept (otherwise, the number will not be divisible by 5). Each digit before the five can be crossed out or not. There are two options for each of the digits 2 and 3 (cross out or not), three options for the digit 4 (do not cross out, cross o... | 60 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,108 |
2-0. The number $n$ is such that $8n$ is a 100-digit number, and $81n$ is a 102-digit number. What can the second digit from the beginning of $n$ be? | Answer: 2
Solution. Since $8 n10^{101}$ (equality here is obviously impossible), it means $n>123 \cdot 10^{97}$. Therefore, the second digit from the beginning of the number $n$ is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,109 |
3-0. In an isosceles triangle, the base is twice the diameter of the inscribed circle. Find the sine of the largest angle of the triangle. | Answer: 0.96
Solution. Let $AC$ be the base of the given triangle, $B$ its vertex, $I$ the center of the inscribed circle, $H$ the midpoint of side $AC$, $r$ the radius of the inscribed circle, $\angle A=\alpha, \angle B=\beta$. From the condition $AH=2r$. In triangle $AIH$, $AI=\sqrt{r^{2}+(2r)^{2}}=r\sqrt{5}$. There... | \frac{24}{25} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,110 |
4-0. Kisa and Busya came to the cafeteria during the break, where only muffins and doughnuts were sold, costing a whole number of rubles. Kisa bought 8 muffins and 3 doughnuts, spending less than 200 rubles, while Busya bought 4 muffins and 5 doughnuts, spending more than 150 rubles. Name the highest possible price of ... | Answer: 19
Solution. Let the price of a cake be $k$, and the price of a bun be $p$ rubles. Then $8 k+3 p<150$. Multiplying the first inequality by 5, and the second by 3, we get $40 k+15 p<1000, -12 k-15 p<-450$. Adding these inequalities: $28 k<550$, from which, taking into account the integer nature, $k \leqslant 19... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,111 |
5-0. A certain quadratic trinomial $x^{2}-p x+q$ has integer roots $x_{1}$ and $x_{2}$. It turns out that the numbers $x_{1}$, $x_{2}$, and $q$ form a decreasing arithmetic progression. Find the sum of all possible values of $x_{2}$. | Answer: -5
Solution. By Vieta's theorem $q=x_{1} x_{2}$. Then $2 x_{2}=x_{1}+x_{1} x_{2}$, from which $x_{1}=\frac{2 x_{2}}{1+x_{2}}$. Since the roots are integers, $2 x_{2}$ is divisible by $1+x_{2}$. But $2+2 x_{2}$ is divisible by $1+x_{2}$, which means 2 is also divisible by $1+x_{2}$. Therefore, $x_{2}=-3,-2,0$ o... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,112 |
6-0. Quadrilateral $A B C D$ is inscribed in a circle, and the areas of triangles $A B C$ and $A C D$ are equal. Three sides of the quadrilateral are 5, 8, and 10. Find all possible values of the length of the fourth side. | Answer: $4 ; 6.25 ; 16$
Solution. Since the quadrilateral is inscribed in a circle, its opposite angles $B$ and $D$ sum up to $180^{\circ}$. Therefore, their sines are equal. Then, from the equality of the areas of triangles $A B C$ and $A C D$, it follows that $A B \cdot B C=A D \cdot C D$, which gives three possible... | 4;6.25;16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,113 |
6-2. Quadrilateral $A B C D$ is inscribed in a circle, and the areas of triangles $A B C$ and $A C D$ are equal. Three sides of the quadrilateral are 4, 6, and 8. Find all possible values of the length of the fourth side. | Answer: $3 ; 16 / 3 ; 12$ | 3;16/3;12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,114 |
7-0. The number $n$ has exactly six divisors (including 1 and itself). They were arranged in ascending order. It turned out that the third divisor is seven times greater than the second, and the fourth is 10 more than the third. What is $n$? | Answer: 2891
Solution. If $n$ has six divisors, then either $n=p^{5}$ or $n=p \cdot q^{2}$ (where $p$ and $q$ are prime numbers). In the first case, $p=7$ (since the third divisor is seven times the second), but then the second condition is not satisfied.
Therefore, $n$ has two prime divisors (one of which is squared... | 2891 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,116 |
8-0. Once, King Shahryar said to Scheherazade: "Here is a paper circle with 1001 points on its boundary. Each night, you must cut the figure you have along a straight line containing any two of the marked points, keeping only one fragment and discarding the other. And make sure that the figure you keep is not a polygon... | # Answer: 1999
Solution. First, let's describe the strategy that Scheherazade will use to meet Shahryar's conditions for 1998 nights. Initially, she will cut along the lines connecting adjacent vertices and discard the smaller part (let's call such a part a segment). She will do this for 1000 nights. After this, a 100... | 1999 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,117 |
1. Prove that for any natural number $n$ the number $n^{3}+3 n^{2}+6 n+8$ is composite | 1. The statement of the problem follows from the factorization of the given expression into factors, each of which is greater than one for all natural $\mathrm{n}:$
$$
\left(n^{3}+8\right)+\left(3 n^{2}+6 n\right)=(n+2)\left(n^{2}-2 n+4\right)+3 n(n+2)=(n+2)\left(n^{2}+n+4\right)
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,118 |
2. Each of the three boys either always tells the truth or always lies. They were told six natural numbers. After that, each boy made two statements.
Petya: 1) These are six consecutive natural numbers. 2) The sum of these numbers is even.
Vasya: 1) These numbers are $1,2,3,5,7,8$. 2) Kolya is a liar.
Kolya: 1) All ... | 2. Among six consecutive natural numbers, exactly three are odd. Therefore, their sum is odd. This means that Petya lied either the first time or the second time, and therefore he is a liar, i.e., he lied both times. But then Vasya is also a liar because in his first statement, he mentioned six natural numbers, and the... | 1,2,4,5,6,7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,119 |
3. A right triangle ABC is inscribed in a circle with hypotenuse AB. On the larger leg BC, a point D is taken such that AC = BD, and point E is the midpoint of the arc AB containing point C. Find the angle DEC. | 3. Point $\mathrm{E}$ is the midpoint of arc $\mathrm{AB}$, so $\mathrm{AE}=\mathrm{BE}$. Moreover, the inscribed angles $\mathrm{CAE}$ and $\mathrm{EBC}$, which subtend the same arc, are equal. Also, by the given condition, $\mathrm{AC}=\mathrm{BD}$. Therefore, triangles $\mathrm{ACE}$ and $\mathrm{BDE}$ are congruent... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,120 |
4. For numbers $a, b, c, d$, it is known that $a^{2}+b^{2}=1, c^{2}+d^{2}=1, a c+b d=0$. Calculate $a b+c d$.
| 4. Consider the equality: $(a c+b d)(a d+b c)=0$, since $a c+b d=0$.
We get $a^{2} c d+b^{2} c d+c^{2} a b+d^{2} a b=0$ or $\left(a^{2}+b^{2}\right) c d+\left(c^{2}+d^{2}\right) a b=0$.
Since $a^{2}+b^{2}=1$ and $c^{2}+d^{2}=1$ we obtain $a b+c d=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,121 |
5. There are 40 visually identical coins, among which 3 are counterfeit - they weigh the same and are lighter than the genuine ones (the genuine coins also weigh the same). How can you use three weighings on a balance scale without weights to select 16 genuine coins? | 5. First solution: Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, the... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,122 |
1. Find the value of the expression $a^{3}+b^{3}+12 a b$, given that $a+b=4$. | 1. $a^{3}+b^{3}+12 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+12 a b=4\left(a^{2}-\right.$ $\left.a b+b^{2}\right)+12 a b=4 a^{2}+4 b^{2}+8 a b=4(a+b)^{2}=4 \cdot 16=64$ | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,124 |
2. The boy went to the shooting range with his father. The father bought him 10 bullets. Later, the father took away one bullet for each miss and gave one additional bullet for each hit. The son fired 55 times, after which he ran out of bullets. How many times did he hit the target? | 2. Each time the boy hit the target, the number of bullets he had remained the same (he used one and received one from his father). Each time the boy missed, the number of bullets he had decreased by 2 (he used one and his father took one). This means that the son missed $10: 2=5$ times out of 55 shots, so he hit 55 - ... | 50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,125 |
3. On the side $B C$ of triangle $A B C$, a point $E$ is marked, and on the bisector $B D$ - a point $F$ in such a way that $E F \| A C$ and $A F=A D$. Prove that $A B=B E$. | 3. From the condition of the problem, it follows that $\angle \mathrm{EFD}=\angle \mathrm{ADF}=\angle \mathrm{AFD}$ (the first equality is true because EF is parallel to AC and the alternate interior angles formed by the intersection of two parallel lines with a third line are equal, the second equality is true because... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,126 |
4. Prove that for any natural number $n$ the number $4^{\text {n }}+5$ is divisible by 3. | 4. The number $4^{\text {n }}+5$ is divisible by 3 if the number $4^{\text {n }}-1=\left(4^{\mathrm{n}}+5\right)-6$ is divisible by 3. But $4^{\mathrm{n}}-$ $1=\left(2^{\mathrm{n}}-1\right)\left(2^{\mathrm{n}}+1\right)$. Since $2^{\mathrm{n}}$ is not divisible by 3, one of the numbers $2^{\mathrm{n}}-1$ or $2^{\mathrm{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,127 |
5. In the company, 11 people gathered. It turned out that each person is friends with at least six of those present. Prove that in this company, there will be three friends (each is friends with the other two). | 5. Let's take any two friends $A$ and $B$. Among the remaining nine people, $A$ has at least 5 friends, and $B$ has at least 5 friends. Therefore, among the friends of $A$ and $B$, there is at least one common friend (otherwise, it would be $5+5 \leq 9$). Together with $A$ and $B$, this common friend forms the required... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 15,128 |
10.2. Given a right triangle $A B C$ with legs $A C=a$ and $C B=b$. Find a) the side of the square with vertex $C$ of the largest area, entirely lying within the triangle $A B C$; b) the dimensions of the rectangle with vertex $C$ of the largest area, entirely lying within the triangle $A B C$. | Answer. a) $\frac{a b}{a+b}$; b) $\frac{a}{2}$ and $\frac{b}{2}$. Solution. a) Introduce a coordinate system: the origin - at vertex $C$, the $x$-axis - along $C A$, the $y$-axis - along $C B$. Then the hypotenuse lies on the line $y=b-\frac{b}{a} x$. The diagonal of the square can be written as the equation of the lin... | )\frac{}{+b};b)\frac{}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,129 |
10.4. Given a triangle with sides $a, b, c$. On its sides as diameters, semicircles are constructed outward, resulting in a figure $F$ composed of the triangle and three semicircles. Find the diameter of $F$ (the diameter of a set in the plane is the greatest distance between its points). | Answer: $\frac{a+b+c}{2}$. Solution. See problem 9.4 | \frac{+b+}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,131 |
10.5. How many solutions in integers \(x, y\) does the equation \(6 x^{2}+2 x y+y+x=2019\) have? | Answer. 4 solutions. Solution. Express $y$ from the given equation: $y=\frac{2019-6 x^{2}-x}{2 x+1}$. Dividing $6 x^{2}+x$ by $2 x+1$ with a remainder, we get $6 x^{2}+x=(3 x-1)(2 x+1)+1$. Thus, the expression for $y$ will take the form: $y=\frac{2019-(3 x-1)(2 x+1)-1}{2 x+1}=\frac{2018}{2 x+1}-3 x+1$. Therefore, for $... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,132 |
Problem 5.1. Denis placed the numbers from 1 to 9 in the cells of a $3 \times 3$ square such that the sum of the numbers in all rows and all columns is 15. And then Lesha erased the numbers from 1 to 5 and wrote the letters $A, B, C, D$, and $E$ instead. The resulting square is shown in the figure.
| $\mathbf{9}$ | $A... | Answer: a1 b5 c2 d4 e3.
Solution. Let's look at the second row. The sum of the numbers in it is $C+6+7=15$, from which $C=2$. Similarly, in the second column, we have the sum of the numbers $A+6+8=15$, from which $A=1$.
Now we can write down the sums of the numbers in the first row and in the first column - these are... | a1\b5\c2\d4\e3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,133 |
Problem 5.2. From a square with a side of 10, a green square with a side of 2, a blue square, and a yellow rectangle were cut out (see figure). What is the perimeter of the remaining figure?
The perimeter of a figure is the sum of the lengths of all its sides.
 & :(x+5)=4 \\
5 x+13 & =4(x+5) \\
5 x+13 & =4 x+20 \\
x & =7
\end{aligned}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,135 |
Problem 5.4. Alyona, Borya, Vera, and Polina were collecting apples in the garden. Someone collected 11 apples, another - 17, a third - 19, and the fourth - 24. It is known that
- one of the girls collected 11 apples;
- Alyona collected more apples than Borya;
- the total number of apples collected by Alyona and Vera ... | Answer: a3 b2 c1 d4.
Solution. The total number of apples collected by Alyona and Vера is divisible by 3. Thus, if Alyona collected 24 apples, the number of apples collected by Vера must also be divisible by 3; but among our numbers, there are no others that are multiples of 3, so this is impossible.
Since Alyona col... | a3b2c1d4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,136 |
Problem 5.5. On some trees in the magical forest, coins grow. The number of trees that do not grow any coins at all is twice as many as the trees that grow three coins. On three trees, two coins grow, on four trees - four coins, and no tree grows more than four coins. By how much is the total number of coins in the mag... | Answer: 15.
Solution. Let $x$ be the number of trees in the forest on which three coins grow, and on which one coin grows. Then in the forest, $2 x$ trees do not grow any coins at all.
Thus, the total number of coins is
$$
2 x \cdot 0+y \cdot 1+3 \cdot 2+x \cdot 3+4 \cdot 4=3 x+y+22
$$
and the total number of trees... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,137 |
Problem 5.6. In the zoo, there are red, yellow, and green parrots (there is at least one parrot of each of the listed colors; there are no parrots of other colors in the zoo). It is known that among any 10 parrots, there is definitely a red one, and among any 12 parrots, there is definitely a yellow one. What is the ma... | Answer: 19.
Solution. Let there be $x$ red, $y$ yellow, and $z$ green parrots in the zoo.
Since among any 10 parrots there is a red one, the number of non-red parrots does not exceed 9, that is, $y+z \leqslant 9$. By similar reasoning, we get that the number of non-yellow parrots does not exceed 11, that is, $x+z \le... | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,138 |
Problem 5.8. A boastful fisherman says the same phrase every day: "Today I caught more perch than I did the day before yesterday (2 days ago), but less than I did 9 days ago." What is the maximum number of days in a row that he can tell the truth? | Answer: 8.
Solution. First, let's provide an example where he tells the truth for 8 days in a row. The numbers of perch he caught on consecutive days are indicated:
$$
2020202020202061 \underbrace{728394105}_{\text {truth }} \text {. }
$$
Now let's prove that he could not have told the truth for 9 days in a row. Sup... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,140 |
8.1. The graphs of the functions $y=a x+b$ and $y=c x+d$ intersect at the point with coordinates $(1 ; 0)$. Compare the values of the expressions $a^{3}+c^{2}$ and $d^{2}-b^{3}$? | Answer: they are equal.
Solution: Since the graphs pass through the point with coordinates $(1 ; 0)$, then $0=a+b$ and $0=c+d$. Therefore, $a=-b$, and $c=-d ; a^{3}+c^{2}=(-b)^{3}+(-d)^{2}=-b^{3}+d^{2}=d^{2}-b^{3}$. Thus, the values of the expressions $a^{3}+c^{2}$ and $d^{2}-b^{3}$ are equal.
Criteria: Only the answ... | ^{3}+^{2}=^{2}-b^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,141 |
8.2. Borya found the smallest prime number \( p \) such that \( 5 p^{2} + p^{3} \) is a square of some natural number. What number did Borya find? | Answer: 11.
Solution: Since $5 p^{2}+p^{3}=p^{2}(5+p)$, the original number is a perfect square if and only if the number $5+p$ is a perfect square. Since $p$ is a prime number, $p+5 \geq 7$. It is sufficient to verify that if $p+5=9$, then $p=4$ is not a prime; if $p+5=16$, then $p=11$ satisfies the condition, and th... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,142 |
8.3. Given a rhombus $A B C D$, and $\angle B C D=60$ . Points $P$ and $Q$ are taken on sides $A B$ and $B C$ respectively, such that $A P=B Q$. Prove that triangle $D P Q$ is equilateral. | Solution: The diagonal BD of the rhombus divides it into two equilateral triangles ABD and CBD.
Consider triangles BPD and CQD. They are equal by the first criterion of triangle congruence. Therefore, $\angle \mathrm{PDB}=\angle \mathrm{QDC}$ and $\mathrm{DB}=\mathrm{DQ}$. Then triangle DPQ is isosceles.
Consider tri... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,143 |
8.4. After the Olympiad, the children dispersed in pairs, sharing impressions. In each pair, there was a boy and a girl. It turned out that in each pair, the boys solved either twice as many problems as the girls or half as many. Could the total number of problems solved by all the children be equal to 2000? | Answer: It could not.
Solution: Note that the number of problems solved by each pair of children is divisible by 3. This means that the total number of problems solved should be divisible by 3. However, 2000 is not divisible by 3.
Criteria: Only the answer - 0 points. 7 points for correct reasoning. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,144 |
8.5. Little Egor has eight colored cubes, on the faces of each of which the numbers from 1 to 6 are written. Egor assembled a large cube (a cube twice the size of the original cubes) from the available cubes, as shown in the figure. It turned out that the numbers written on the adjacent faces of the cubes are the same.... | Answer: It cannot.
Solution: The sum of all numbers written on the faces of these eight cubes is an even number $\left(8^{*}(1+2+3+4+5+6)=8^{*} 21=168\right)$. Since the numbers on the adjacent faces of the cubes are the same, all numbers inside the large cube are divided into pairs of identical numbers. That is, the ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,145 |
8.1. Provide an example of six different natural numbers such that the product of any two of them is not divisible by the sum of all the numbers, while the product of any three of them is divisible. | # Solution.
Let $\mathrm{p}$ be a sufficiently large odd prime number. Represent the number $\mathrm{p}^{2}$ as the sum $\mathrm{a}_{1}+\ldots+\mathrm{a}_{6}$ of distinct natural numbers, none of which are divisible by $\mathrm{p}$. The numbers $\mathrm{pa}_{1}, \ldots, \mathrm{pa}_{6}$ will be the desired ones: the p... | 5,10,15,20,30,45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,146 |
8.2. In a convex quadrilateral $\mathrm{ABCD}$, the bisector of angle $\mathrm{B}$ passes through the midpoint of side $\mathrm{AD}$, and $\angle \mathrm{C}=\angle \mathrm{A}+\angle \mathrm{D}$. Find the angle $\mathrm{ACD}$.
$, from which $\angle \mathrm{AEB}=180-\angle \mathrm{A}-\angle \mathrm{B} / 2=\angle \mathrm{D}$. Therefor... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,147 |
8.3. There are 11 empty boxes. In one move, you can place one coin in any 10 of them. Two players take turns. The winner is the one who, after their move, first has 21 coins in one of the boxes. Who wins with correct play? | # Solution.
Number the boxes: $1, \ldots, 11$ and denote the move by the number of the box where we did not put a coin. We can assume that the first player started the game with move 1. To win, the second player needs to, regardless of the first player's moves, make moves $2, \ldots, 11$. With these ten moves, togethe... | Theplayerwins | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,148 |
8.4. In an isosceles trapezoid, one of the bases is three times larger than the other. The angle at the larger base is $45^{\circ}$. Show how to cut this trapezoid into three pieces and form a square from them. Justify your solution. | # Solution. Method 1.
Let ABCD be the given trapezoid (Fig. 1). Draw the heights BE and CH. Since the trapezoid is isosceles, \( AE = DH = \frac{AD - BC}{2} = \frac{3BC - BC}{2} = BC \). In ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,149 |
8.5. The robbers have 13 gold bars. There are scales that can be used to determine the total weight of any two bars. Come up with a way to find out the total weight of all the bars in 8 weighings.
# | # Solution.
We will take the first three ingots and weigh them in pairs: $\mathrm{C} 1+\mathrm{C} 2, \mathrm{C} 1+\mathrm{C} 3, \mathrm{C} 2+\mathrm{C} 3$, using three weighings. By adding the results of these weighings and dividing by two, we will find the total weight of these three ingots: $((\mathrm{C} 1+\mathrm{C... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,150 |
9.1. Given various real numbers $a, b, c$. Prove that at least two of the equations $(x-a)(x-b)=x-c,(x-b)(x-c)=$ $=x-a,(x-c)(x-a)=x-b$ have a solution. (I. Bogdanov) | The first solution. Let $f_{1}(x)=(x-b)(x-c)-(x-a)$, $f_{2}(x)=(x-c)(x-a)-(x-b)$, and $f_{3}(x)=(x-a)(x-b)-(x-c)$. Suppose the statement of the problem is false, that is, the maximum of one of these functions is a root. Then, two of them, say $f_{1}$ and $f_{2}$, do not have roots. Since the leading coefficients of the... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,151 |
9.2. An acute triangle $ABC$ is inscribed in a circle $\Omega$. The tangents to $\Omega$ at points $B$ and $C$ intersect at point $P$. Points $D$ and $E$ are the feet of the perpendiculars dropped from point $P$ to the lines $AB$ and $AC$. Prove that the orthocenter of triangle $ADE$ is the midpoint of segment $BC$.
(... | Solution. Let $M$ be the midpoint of $BC$. Triangle $BPC$ is isosceles ($BP = PC$ as segments of tangents); hence, its median $PM$ is also an altitude. Since $\angle PMC = \angle PEC = 90^\circ$, quadrilateral $MCPE$ is cyclic; thus, $\angle MEP = \angle MCP$. Furthermore, $CP$ is a tangent to $\Omega$, so $\angle MCP ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,152 |
9.3. On the board, 100 pairwise distinct natural numbers $a_{1}, a_{2}, \ldots, a_{100}$ were written. Then, under each number $a_{i}$, a number $b_{i}$ was written, obtained by adding to $a_{i}$ the greatest common divisor of the remaining 99 original numbers. What is the smallest number of pairwise distinct numbers t... | # Answer. 99.
First solution. If we set $a_{100}=1$ and $a_{i}=2 i$ for $i=1,2, \ldots, 99$, then $b_{1}=b_{100}=3$, so there will be no more than 99 different numbers among the numbers $b_{i}$. It remains to prove that there will always be 99 different numbers among the numbers $b_{i}$.
Without loss of generality, w... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 15,153 |
9.4. On a plane, $n$ lines are drawn, none of which are parallel. No three of them intersect at the same point. Prove that there exists an $n$-segment non-self-intersecting broken line $A_{0} A_{1} A_{2} \ldots A_{n}$, such that each of the $n$ lines contains exactly one segment of this broken line. | Solution. We will prove a stronger fact. Let $A_{0}$ be an arbitrary point on one of the given lines, through which no other lines pass. We will prove that there exists the required broken line starting from $A_{0}$.
Induction on $n$. If $n=1$, then the broken line (consisting of one segment) is constructed trivially.... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,154 |
# 9.1. (7 points)
Two different natural numbers are subjected to four operations:
a) their sum is found;
b) the smaller is subtracted from the larger;
c) their product is found;
d) the larger number is divided by the smaller.
The sum of the results of all four operations is equal to $3^{5}$. What are these number... | Answer: 54 and $2 ; 24$ and 8.
Solution: Let $x$ and $y$ be the required natural numbers, $x>y$ :
$$
\begin{gathered}
x+y+x-y+x y+\frac{x}{y}=3^{5} \\
2 x+x y+\frac{x}{y}=3^{5} \\
2 x y+x y^{2}+x=3^{5} y \\
x\left(y^{2}+2 y+1\right)=3^{5} y \\
x(y+1)^{2}=3^{5} y \\
x=\frac{3^{5} y}{(y+1)^{2}}
\end{gathered}
$$
Since... | 542;248 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,155 |
# 9.2. (7 points)
Prove that if $a, b, c$ are the lengths of the sides of a triangle, then the equation $b^{2} x^{2}+\left(b^{2}+c^{2}-a^{2}\right) \cdot x+c^{2}=0$ has no solutions. | Solution: Let's find the discriminant of the given equation
$$
\begin{gathered}
D=\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 b^{2} c^{2}=\left(b^{2}+c^{2}-a^{2}-2 b c\right)\left(b^{2}+c^{2}-a^{2}+2 b c\right)= \\
=\left((b-c)^{2}-a^{2}\right)\left((b+c)^{2}-a^{2}\right)= \\
=(b-c-a)(b-c+a)(b+c-a)(b+c+a)
\end{gathered}
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 15,156 |
# 9.3. (7 points)
On a circular track 60 cm long, ants are moving in both directions at a speed of 1 cm/s. When two ants collide, they instantly turn around and continue moving at the same speed in opposite directions. It turned out that 48 pairwise collisions occurred in one minute. How many ants could have been on t... | Answer: 10 or 11 or 14 or 25.
Solution: Note that the situation will not change if we assume that after colliding, the ants continue their movement without changing direction or speed. Suppose two ants collide, then 30 seconds after this, each of them will have crawled half the circle and they will collide again.
The... | 10or11or14or25 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 15,157 |
# 9.4. (7 points)
At a joint conference of the party of liars and the party of truth-tellers, 32 people were elected to the presidium and seated in four rows of eight. During the break, each member of the presidium claimed that among their neighbors there are representatives of both parties. It is known that liars alw... | Answer: with eight liars.
Solution: Divide all the seats in the presidium into eight groups as shown in the figure. If there are fewer than eight liars, then in one of these groups, only truth-tellers will be sitting, which is impossible. The contradiction obtained shows that there are no fewer than eight liars. The f... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,158 |
# 9.5. (7 points)
On the shore of a circular lake, there are four piers $K, L, P, Q$. A motorboat departs from pier $K$, and a rowboat from pier $L$. If the motorboat heads straight to $P$, and the rowboat heads straight to $Q$, they will collide at some point $X$ on the lake. Prove that if the motorboat heads to $Q$,... | Solution: Let $u$ and $v$ be the speeds of the motorboat and the rowboat. Triangles $K X Q$ and $L X P$ are similar. Let $t$ be the time of movement of the rowboat and the motorboat until the moment they meet at point $X$, then $K X = u t, L X = v t$.
Then, from the similarity of triangles $K X Q$ and $L X P$ it follo... | proof | Geometry | proof | Yes | Yes | olympiads | false | 15,159 |
Task 2. Replace the asterisk (*) in the expression $\left(x^{4}-3\right)^{2}+\left(x^{3}+*\right)^{2}$ with a monomial so that after squaring and combining like terms, there are four terms. | # Solution.
Let's replace $*$ with $3 x:$
$\left(x^{4}-3\right)^{2}+\left(x^{3}+3 x\right)^{2}=x^{8}-6 x^{4}+9+x^{6}+6 x^{4}+9 x^{2}=x^{8}+x^{6}+9 x^{2}+9$
Another answer is possible, for example, $\sqrt{6} x^{2}$
Analysis of the solution: Let's analyze the situation assuming that the coefficient of a monomial can ... | 3x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,160 |
Problem 3. Prove that the number $9^{8 n+4}-7^{8 n+4}$ is divisible by 20 for all natural n. | # Solution.
$9^{8 n+4}-7^{8 n+4}=\left(9^{4 n+2}-7^{4 n+2}\right)\left(9^{4 n+2}+7^{4 n+2}\right)=\left(9^{4 n+2}-7^{4 n+2}\right)\left(81^{2 n+1}+49^{2 n+1}\right)$.
In the first parenthesis, there is the difference of two odd numbers, which is an even number. In the second parenthesis, the first power ends in a one... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 15,161 |
Problem 4. In triangle $\mathrm{ABC}$, $C D$ is the bisector of angle $\mathrm{ACB}$, $\mathrm{AB} = \mathrm{BC}$, $\mathrm{BD} = \mathrm{BK}$, $\mathrm{BL} = \mathrm{CL}$. Prove that $\mathrm{BF}$ is the bisector of angle $\mathrm{CBE}$. | # Solution.
Let $\angle \mathrm{BCD}=\angle \mathrm{DAC}=\mathrm{x}$. Then $\angle \mathrm{BAC}=2 \mathrm{x}$.
$\angle \mathrm{BDC}=\angle \mathrm{DAC}+\angle \mathrm{DCA}=3 \mathrm{x} \Rightarrow \angle \mathrm{BKD}=3 \mathrm{x}$ (triangle
: 2=2$. Then the entire large rectangle has dimensions $8 \times 18$, and its perimete... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 15,167 |
Problem 5.6. There are 4 absolutely identical cubes, each of which has 6 dots marked on one face, 5 dots on another, ..., and 1 dot on the remaining face. These cubes were glued together to form the figure shown in the image.
How many dots are on the four left faces?
![](https://cdn.mathpix.com/cropped/2024_05_06_d55... | Answer: On face $A$ there are 3 points, on face $B-5$, on face $C-6$, on face $D-5$. Solution. Let's consider the arrangement of the faces on one die. We will denote the faces by numbers corresponding to the number of dots on them.
From the picture, it is clear that face 1 borders with faces $2,3,4$ and 5. Therefore, ... | 3,5,6,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,168 |
Problem 6.1. The set includes 8 weights: 5 identical round, 2 identical triangular, and one rectangular weight weighing 90 grams.
It is known that 1 round and 1 triangular weight balance 3 round weights. Additionally, 4 round weights and 1 triangular weight balance 1 triangular, 1 round, and 1 rectangular weight.
How... | Answer: 60.
Solution. From the first weighing, it follows that 1 triangular weight balances 2 round weights.
From the second weighing, it follows that 3 round weights balance 1 rectangular weight, which weighs 90 grams. Therefore, a round weight weighs $90: 3=30$ grams, and a triangular weight weighs $30 \cdot 2=60$ ... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 15,169 |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. How many emeralds are there in total in the boxes?
![](htt... | # Answer: 12.
Solution. The total number of rubies is no more than $13+8=21$, and the number of diamonds is no less than $2+4=6$. According to the condition, their quantities differ by 15. This is only possible if the rubies are in the boxes with 13 and 8 stones, and the diamonds are in the boxes with 2 and 4 stones. ... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,170 |
Problem 7.1. In the picture, nine small squares are drawn, with arrows on eight of them. The numbers 1 and 9 are already placed. Replace the letters in the remaining squares with numbers from 2 to 8 so that the arrows from the square with the number 1 point in the direction of the square with the number 2 (the number 2... | Answer: In square $A$ there is the number 6, in $B-2$, in $C-4$, in $D-5$, in $E-3$, in $F-8$, in $G-7$.
Solution. Let's order all the squares by the numbers in them. This "increasing chain" contains all nine squares.
Notice that in this chain, immediately before $C$ can only be $E$ (only the arrows from $E$ point to... | In\\A\there\is\the\\6,\in\B-2,\in\C-4,\in\D-5,\in\E-3,\in\F-8,\in\G-7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 15,172 |
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