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2. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part. | Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In five 15-liter buckets, there are 1, 2, 3, 4, and 5 liters of water respectively. It is allowed to triple the amount of water in any container by pouring water from one other container (if there is not enough water to triple the amount, then it is not allowed to pour from this bucket). What is the maximum amount o... | # Answer
The answer depends on the interpretation of the condition -- whether it is allowed to pour NOT all the contents of the bucket (essentially -- whether it is possible to measure OUT ONE LITER of water)
A) If it is not allowed, then the answer is 9 liters
B) If it is allowed, then the answer is 12 liters
## S... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter Г on the board, at least one colored cell is found? | Solution: Consider a $2 \times 3$ rectangle. In it, obviously, a minimum number of cells need to be colored. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown in the figure.
... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. Two painters are painting a 15-meter corridor. Each of them moves from the beginning of the corridor to its end and starts painting at some point until the paint runs out. The first painter has red paint, which is enough to paint 9 meters of the corridor; the second has yellow paint, which is enough for 10 m... | Answer: 5 meters.
Solution. The first painter starts painting 2 meters from the beginning of the corridor and finishes at $2+9=11$ meters from the beginning of the corridor.
The second painter finishes painting 1 meter from the end of the corridor, which is 14 meters from the beginning of the corridor, and starts at ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. As is known, balance scales come to equilibrium when the weight on both pans is the same. On one pan, there are 9 identical diamonds, and on the other, 4 identical emeralds. If one more such emerald is added to the diamonds, the scales will be balanced. How many diamonds will balance one emerald? The answer ... | Answer: 3 diamonds.
Solution. From the condition of the problem, it follows that 9 diamonds and 1 emerald weigh as much as 4 emeralds. Thus, if we remove one emerald from each side of the scales, the equality will be preserved, that is, 9 diamonds weigh as much as 3 emeralds. This means that 3 diamonds weigh as much a... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. On the faces $BCD, ACD, ABD$, and $ABC$ of the tetrahedron $ABCD$, points $A_{1}, B_{1}, C_{1}$, and $D_{1}$ are marked, respectively. It is known that the lines $AA_{1}, BB_{1}, CC_{1}$, and $DD_{1}$ intersect at point $P$, and $\frac{AP}{A_{1}P}=\frac{BP}{B_{1}P}=\frac{CP}{C_{1}P}=\frac{DP}{D_{1}P}=r$. Find all po... | Solution. Let V be the volume* of tetrahedron ABCD. We introduce the consideration of the partition of the original tetrahedron ABCD into tetrahedra PBCD, PACD, PABD, and PABC. Then for the volumes of the specified tetrahedra, the following is true:
$$
V=V_{\mathrm{PBCD}}+V_{\mathrm{PACD}}+V_{\mathrm{PABD}}+V_{\mathrm... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 7.1. (7 points)
Find the value of the expression
$$
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \ldots\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)
$$ | Answer: 1.
Solution: Notice that
$$
\begin{array}{r}
\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=\frac{3}{2} \cdot \frac{2}{3}=1,\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{5}{4} \cdot \frac{4}{5}=1, \ldots \\
\left(1+\frac{1}{2 m}\right)\left(1-\frac{1}{2 m+1}\right)=\frac{2 m+1}{2 m} \cdot \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7.2. (7 points)
Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon and, without stopping, arrived: one - in $B$ at 4 PM, and the other - in $A$ at 9 PM. At what hour was dawn that day? | Answer: at 6 o'clock.
Solution: Let $x$ be the number of hours from dawn to noon. The first pedestrian walked $x$ hours before noon and 4 after, the second - $x$ before noon and 9 after. Note that the ratio of times is equal to the ratio of the lengths of the paths before and after the meeting point, so $\frac{x}{4}=\... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7.5. (7 points)
In a family, there are six children. Five of them are older than the youngest by 2, 6, 8, 12, and 14 years, respectively. How old is the youngest if the ages of all the children are prime numbers? | Answer: 5 years.
Solution. First, let's check the prime numbers less than six. It is obvious that the number we are looking for is odd. The number 3 does not satisfy the condition because $3+6=9-$ is not a prime number. The number 5 satisfies the condition because $5+2=7, 5+6=11, 5+8=13, 5+12=17, 5+14=19$, which means... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The second term of an infinite decreasing geometric progression is 3. Find the smallest possible value of the sum $A$ of this progression, given that $A>0$. | Answer: 12.
Solution. Let the first term of the progression be $a$, and the common ratio be $q$. The sum of the progression $A$ is $\frac{a}{1-q}$. From the condition, we have $3=a q$, from which $a=3 / q$. Therefore, we need to find the minimum value of $A=\frac{3}{q(1-q)}$. Note that from the condition it follows: $... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.6. The sheriff believes that if the number of bandits he catches on a certain day is a prime number, then he is lucky. On Monday and Tuesday, the sheriff was lucky, and starting from Wednesday, the number of bandits he caught was equal to the sum of the number from the day before yesterday and twice the number from ... | Solution: Let's say the sheriff caught 7 bandits on Monday and 3 on Tuesday. Then on Wednesday, Thursday, and Friday, he caught 13, 29, and 71 bandits, respectively. All these numbers are prime, and the sheriff has been lucky for five days in a row.
We will show that the sheriff cannot be lucky for six days in a row. ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. 9 knights and liars stood in a row. Each said that there is exactly one liar next to him. How many liars are there among them, if knights always tell the truth, and liars always lie? | Answer: 3 liars.
Sketch of the solution. Consider the partition of all into groups of people of the same type standing in a row, with people of different types in adjacent groups. In such a group, there can only be one liar. If there are at least two, the extreme liars tell the truth. There are no more than two knight... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the smallest integer $x$ that satisfies the inequality $\frac{100}{|x|}>x^{2}+1$. | Answer: -4.
If $|x| \geq 5$, then $\frac{100}{|x|} \leq \frac{100}{5}=20 \leq 17 = (-4)^{2} + 1$ we verify the correctness of the answer.
Comment. The correct answer without a rigorous justification up to 2 points. The inequality is solved, but the wrong answer is chosen 2 points. | -4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.3. At a drama club rehearsal, 8 people gathered. Some of them (honest people) always tell the truth, while the others always lie. One of those present said: "There is not a single honest person here." The second said: "There is no more than one honest person here." The third said: "There is no more than two honest pe... | 3. The standard evaluation methodology for solutions is provided below.
| Points | Correctness (Incorrectness) of the Solution |
| :---: | :--- |
| 7 | Fully correct solution. |
| $6-7$ | Correct solution, but with minor flaws that do not significantly affect the solution. |
| $5-6$ | The solution is generally correct... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Two-headed and seven-headed dragons came to a meeting. At the very beginning of the meeting, one of the heads of one of the seven-headed dragons counted all the other heads. There were 25 of them. How many dragons in total came to the meeting? | Answer: 8 dragons. Solution. Subtract the 6 heads belonging to the seven-headed dragon from the 25 heads counted by the seven-headed dragon. 19 heads remain. The remaining dragons cannot all be two-headed (19 is an odd number). There can only be one more seven-headed dragon (if there were two, an odd number of heads wo... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Vovodya is running on a circular track at a constant speed. There are two photographers standing at two points on the track. After the start, Vovodya was closer to the first photographer for 2 minutes, then closer to the second photographer for 3 minutes, and then closer to the first photographer again. How long did... | Solution. Let's divide both arcs of the circle between the photographers in half. The halves of the arcs adjacent to the second photographer make up half the distance. Vasya ran this half in 3 minutes, so he will run the entire circle in 6 minutes.
Answer: in 6 minutes.
Grading criteria: Full solution - 7 points. Onl... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are pan scales without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings? | Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the count... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
Problem 5.3. Irina did poorly in math at the beginning of the school year, so she had 3 threes and 2 twos in her journal. But in mid-October, she pulled herself together and started getting only fives. What is the minimum number of fives Irina needs to get so that her average grade is exactly 4? | Answer: 7.
Solution. Let Irina need to get $x$ fives. Then the sum of her grades will be $3 \cdot 3+2 \cdot 2+5 x=5 x+13$, and the number of grades $-3+2+x=x+5$.
We get the equation
$$
\begin{aligned}
(5 x+13) & :(x+5)=4 \\
5 x+13 & =4(x+5) \\
5 x+13 & =4 x+20 \\
x & =7
\end{aligned}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
# 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 |
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?
Fig. 1: to the solution of problem 8.4
Notice that $\angle A B K=\angle C B L$, since they both complement $\angle... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given the functions $f(x)=x^{2}+4 x+3$ and $g(x)=x^{2}+2 x-1$. Find all integer solutions to the equation $f(g(f(x)))=g(f(g(x)))$. | 4. Answer. $x=-2$. Solution. Let's represent the functions as $f(x)=x^{2}+4 x+3=(x+2)^{2}-1$ and $g(x)=x^{2}+2 x-1=(x+1)^{2}-2 . \quad$ Then $\quad f(g(x))=\left((x+1)^{2}-2+2\right)^{2}-1=(x+1)^{4}-1$, $g(f(x))=\left((x+2)^{2}-1+1\right)^{2}-2=(x+2)^{4}-2$. Performing similar operations, we get $g(f(g(x)))=(x+1)^{8}-2... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. Sasha chose a natural number $N>1$ and wrote down in ascending order all its natural divisors: $d_{1}<\ldots<d_{s}$ (so that $d_{1}=1$ and $d_{s}=N$). Then, for each pair of adjacent numbers, he calculated their greatest common divisor; the sum of the resulting $s-1$ numbers turned out to be $N-2$. What values cou... | Answer: $N=3$.
Solution. Note immediately that $d_{s+1-i}=N / d_{i}$ for all $i=$ $=1,2, \ldots, s$.
The number $d_{i+1}-d_{i}$ is divisible by the GCD $\left(d_{i}, d_{i+1}\right)$, so the GCD $\left(d_{i}, d_{i+1}\right) \leqslant d_{i+1}-d_{i}$. For $i=1, \ldots, s-1$, let $r_{i}=\left(d_{i+1}-d_{i}\right)-$ GCD $... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Two-digit numbers are written on the board. Each number is composite, but any two numbers are coprime. What is the maximum number of numbers that can be written? | Answer: four numbers.
Solution. Evaluation. Since any two of the written numbers are coprime, each of the prime numbers 2, 3, 5, and 7 can appear in the factorization of no more than one of them. If there are five or more numbers on the board, then all prime factors in the factorization of some of them must be at leas... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In volleyball competitions, where there are no ties, 5 teams participate. All teams played against each other. The team that took 1st place won all their matches, and the teams that took 2nd and 3rd place each won exactly two matches. In the case of equal points, the position is determined by the result of the match... | Solution. Let's denote a team by a point. If team A won against team B, we draw an arrow from A to B.
The number of outgoing arrows equals the number of wins. In total, 10 arrows can be draw... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.5. The following numbers are written on the wall: $1,3,4,6,8,9,11,12,16$. Four of these numbers were written by Vova, four numbers were written by Dima, and one number was simply the house number of the local police officer. The police officer found out that the sum of the numbers written by Vova is three times the s... | Solution. The sum of all numbers on the wall is $1+3+4+6+8+9+11+12+16=70$. Let $n$ be the sum of the numbers written by Dima, then $3n$ is the sum of the numbers written by Vova, $4n$ is the sum of the numbers written by both, $70-4n-$ is the house number of the district police officer, which when divided by 4 leaves a... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Masha left the house for school. A few minutes later, Vanya ran out of the same house to school. He overtook Masha at one-third of the way, and when he arrived at school, Masha still had half of the way left to go. How many times faster does Vanya run compared to how Masha walks? | Solution. At one third of the way, Masha and Vanya were at the same time. After that, Vanya ran $2 / 3$ of the way, while Masha walked $1 / 2 - 1 / 3 = 1 / 6$ of the way in the same time. This means that he runs $2 / 3$ : $1 / 6 = 4$ times more in the same amount of time than Masha. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a football tournament, 17 teams participate, and each team plays against each other exactly once. A team earns 3 points for a win. For a draw, 1 point is awarded. The losing team gets no points. What is the maximum number of teams that can accumulate exactly 10 points? | Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 10 points each. The total points include all points scored by these teams in matches against each other (at least 2) and possibly in matches against other teams: $10 n \geq 2(n-1) n / 2$. Hence, $n \leq 11$.
Example: All eleven teams pl... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Given three non-zero real numbers $a, b, c$ such that the equations: $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ each have two roots. How many of the roots of these equations can be negative? | Answer: 2.
Sketch of the solution.
If the numbers $a, b, c$ are replaced by their opposites, then the "new" equations will have the same set of roots as the original ones. There are two possible cases: the numbers $a, b, c$ are of the same sign; among them, there are both positive and negative numbers.
First case. D... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. Two cyclists, Andrey and Boris, are riding at a constant and identical speed along a straight highway in the same direction, so that the distance between them remains constant. There is a turnoff to a village ahead. At some point in time, the distance from Andrey to the turnoff was equal to the square of the dista... | Answer: 2 or 0 km.
Solution: Let the first mentioned distances be $a$ and $b$, then $a=b^{2}$. When each of them had traveled another kilometer, the remaining distance to the turn was $a-1$ and $b-1$ km, respectively, so $a-1=3(b-1)$, which means $b^{2}-1=3(b-1),(b-1)(b+1)=3(b-1)$, from which $b=1$ or $b=2$. When $b=1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. It is known that for real numbers $a$ and $b$ the following equalities hold:
$$
a^{3}-3 a b^{2}=11, \quad b^{3}-3 a^{2} b=2
$$
What values can the expression $a^{2}+b^{2}$ take? | Answer: 5.
Solution. We have
$$
\left(a^{2}+b^{2}\right)^{3}=a^{6}+3 a^{4} b^{2}+3 a^{2} b^{4}+b^{6}=\left(a^{3}-3 a b^{2}\right)^{2}+\left(b^{3}-3 a^{2} b\right)^{2}=11^{2}+2^{2}=125
$$
From this, $a^{2}+b^{2}=5$.
Comment. A correct and justified solution - 7 points. The correct answer obtained based on a simple e... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (7 points) The graphs of the functions $y=k x+b$ and $y=b x+k$ intersect. Find the abscissa of the point of intersection.
# | # Solution.
Method 1. The desired abscissa is the solution to the equation $k x + b = b x + k$. This equation can be reduced to $(k - b) x = k - b$. Since the given graphs intersect (do not coincide), $k \neq b$, so $x = 1$.
Method 2. Notice that $x = 1$ is a solution to the problem, because when $x = 1$, both given ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. On a circle, $2 N$ points are marked ($N$ is a natural number). It is known that through any point inside the circle, no more than two chords with endpoints at the marked points pass. We will call a matching a set of $N$ chords with endpoints at the marked points such that each marked point is the endpoint of exa... | Answer. 1.
First solution. We will prove by induction on $N$ that there is one more even matching than odd. For $N=1$, the statement is obvious: there is only one matching, and it is even. Now we will prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. Let the marked points be $A_{1}, A_{2}, ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.6. Petya chose a natural number $a>1$ and wrote down fifteen numbers $1+a, 1+a^{2}, 1+a^{3}, \ldots, 1+a^{15}$ on the board. Then he erased several numbers so that any two remaining numbers are coprime. What is the maximum number of numbers that could remain on the board?
(O. Podlipsky) | # Answer. 4 numbers.
Solution. First, we will show that there cannot be more than four such numbers. Note that if $k$ is odd, then the number $1+a^{n k}=1^{k}+\left(a^{n}\right)^{k}$ is divisible by $1+a^{n}$. Next, each of the numbers $1,2, \ldots, 15$ has one of the forms $k, 2 k, 4 k, 8 k$, where $k$ is odd. Thus, ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. In triangle $A B C$, the median $A M$ is perpendicular to the bisector $B D$. Find the perimeter of the triangle, given that $A B=1$, and the lengths of all sides are integers. | Answer: 5.
Solution. Let $O$ be the point of intersection of the median $A M$ and the bisector $B D$. Triangles $A B O$ and $M B O$ are congruent (by the common side $B O$ and the two adjacent angles). Therefore, $A B = B M = 1$ and $C M = 1$, since $A M$ is a median. Thus, $A B = 1, B C = 2$. By the triangle inequali... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the quadratic trinomials $x^{2}+p x+q$ and $x^{2}+q x+p$
have different real roots. Consider all possible pairwise products of the roots of the first quadratic trinomial with the roots of the second (there are four such products in total). Prove that the sum of the reciprocals of these products doe... | Let $x_{1}, x_{2}$ be the roots of the first quadratic polynomial and $x_{3}, x_{4}$, then we need to prove that $\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{2} x_{4}}$ does not depend on $p$ and $q$. Transform the given expression:
$\frac{1}{x_{1} x_{3}}+\frac{1}{x_{1} x_{4}}+\frac{1... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3. Solve the equation in integers $x^{4}-2 y^{4}-4 z^{4}-8 t^{4}=0$.
## Answer: $x=y=z=t=0$ | Note that $x$ is even. Let $x=2x_{1}$, then we get
$4x_{1}^{4}-8y_{1}^{4}-z^{4}-2t^{4}=0$. Therefore, $z=2z_{1}$ and $2x_{1}^{4}-4y_{1}^{4}-8z_{1}^{4}-t^{4}=0$. Similarly, $t=2t_{1}$ and $x_{1... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. It is known that in the pyramid $A B C D$ with vertex $D$, the sum $\angle A B D+\angle D B C=\pi$. Find the length of the segment $D L$, where $L$ is the base of the bisector $B L$ of triangle $A B C$, if it is known that
$$
A B=9, B C=6, A C=5, D B=1 .
$$ | Answer: 7.
Solution. Let point $M$ lie on the line, outside the segment $B C$, beyond point $B$. From the condition, we have the equality of angles $\angle A B D=\angle D B M$. Then the projection of line $B D$ onto the plane $A B C$ is the bisector $B K$ of angle $A B M$. The bisectors $B K$ and $B L$ are perpendicul... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.4. In a $3 \times 4$ rectangle, natural numbers $1,2,3, \ldots, 12$ were written, each exactly once. The table had the property that in each column, the sum of the top two numbers was twice the bottom number. Over time, some numbers were erased. Find all possible numbers that could have been written in place of $\sta... | Solution 1. According to the condition, the number 2 is written in the bottom-left cell. The sum of the two unknown numbers in the second column is 10, and in the third column, it is 16. Therefore, the sum of the numbers in the first three columns is $6+15+24$, and the sum of all numbers in the table is $1+2+\ldots+12=... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5.4. 10 children stood in a circle. Each thought of an integer and told it to their clockwise neighbor. Then each loudly announced the sum of their number and the number of the counterclockwise neighbor. The first said "10", the next clockwise - "9", the next clockwise - "8", and so on, the ninth said "2". What number ... | Solution option 1. The sum of all ten thought-of numbers will be obtained if we add what the first child said to what the third, fifth, seventh, and ninth children said. Similarly, we add what the second, fourth, sixth, eighth, and tenth children said. Again, we get the sum of all ten thought-of numbers, so the equatio... | 5 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
# Task 9.1
For which natural numbers $n$ is the expression $n^{2}-4 n+11$ a square of a natural number?
## Number of points 7 | Answer:
for $n=5$
## Solution
Let $n^{2}-4 n+11=t^{2}$.
Note that $n^{2}-4 n+4=(n-2)^{2}$ is also the square of some integer $r=n-2$, which is less than $t$.
We get that $t^{2}-r^{2}=(t+r)(t-r)=7$.
The numbers $(t+r)$ and $(t-r)$ are natural and the first is greater than the second.
Thus, $(t+r)=7$, and $(t-r)=1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In a beach soccer tournament, 17 teams participate, and each team plays against each other exactly once. Teams are awarded 3 points for a win in regular time, 2 points for a win in extra time, and 1 point for a win on penalties. The losing team does not receive any points. What is the maximum number of teams that ca... | Answer: 11.
Sketch of the solution. Estimation. Let $n$ teams have scored exactly 5 points each. The total points include all points scored by these teams in matches against each other (at least 1) and possibly in matches against other teams: $5 n \geq(n-1) n / 2$. Hence, $n \leq 11$.
Example: Place eleven teams in a... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.1. Find the number of roots of the equation
$$
|x|+|x+1|+\ldots+|x+2018|=x^{2}+2018 x-2019
$$
(V. Dubinskaya) | Answer: 2.
Solution. For $x \in(-2019,1)$ there are no roots, since on the given interval the left side is non-negative, while the right side is negative.
For $x \in[1, \infty)$ all absolute values are resolved with a positive sign, so the equation will take the form $g(x)=0$, where $g(x)=x^{2}-x-2009+(1+2+\ldots+201... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Initially, there are 111 pieces of clay of the same mass on the table. In one operation, you can choose several groups with the same number of pieces and in each group, combine all the clay into one piece. What is the minimum number of operations required to get exactly 11 pieces, any two of which have different ... | Answer. In two operations.
Solution. Let the mass of one original piece be 1. If in the first operation there are $k$ pieces in each group, then after it each piece will have a mass of 1 or $k$; therefore, it is impossible to obtain eleven pieces of different masses in one operation.
We will show that the required re... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8-2. In square $A B C D$, a segment $C E$ is drawn such that the angles shown in the diagram are $7 \alpha$ and $8 \alpha$. Find the value of angle $\alpha$ in degrees. 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 |
3. Solve the inequality $\sqrt{2 x^{2}-8 x+6}+\sqrt{4 x-x^{2}-3}<x-1$ | Note that all solutions to the original inequality exist if the expressions under the square roots are non-negative. These inequalities are simultaneously satisfied only under the condition $x^{2}-4 x+3=0$. This equation has two roots, 1 and 3. Checking shows that the original inequality has a unique solution 3.
Answe... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
11.5. Solve the equation $2021 x^{2021}-2021+x=\sqrt[2021]{2022-2021 x}$. (7 points)
# | # Solution
The function $f(x)=2021x^{2021}-2021+x$ is increasing, while the function $g(x)=\sqrt[2021]{2022-2021x}$ is decreasing. Therefore, the equation $f(x)=g(x)$ has no more than one root. However, it is obvious that $f(1)=g(1)$.
Answer: $x=1$.
| criteria | points |
| :--- | :---: |
| correct solution | 7 |
| P... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. Four non-zero numbers are written on the board, and the sum of any three of them is less than the fourth number. What is the smallest number of negative numbers that can be written on the board? Justify your answer. | Solution: Let the numbers on the board be $a \geqslant b \geqslant c \geqslant d$. The condition of the problem is equivalent to the inequality $a+b+c < d$ for optimality | 7 points |
| There is a proof that there are no fewer than three negative numbers (in the absence of an example with three numbers) | 3 points |
| ... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
9.5. At the exchange office, only the following operations can be performed:
1) exchange 2 gold coins for three silver coins and one copper coin;
2) exchange 5 silver coins for three gold coins and one copper coin.
Nikolai had only silver coins. After several visits to the exchange office, he had fewer silver coins, ... | Solution: As a result of each operation, Nikolai acquires exactly 1 copper coin, which means there were exactly 50 operations in total. Of these, some (let's say \(a\)) were of the first type, and the rest \(50-a\) were of the second type. On operations of the first type, Nikolai spent \(2a\) gold coins, and on operati... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Find all natural numbers $n \geq 2$ such that $20^{n}+19^{n}$ is divisible by $20^{n-2}+19^{n-2}$. | Solution.
Consider the expression
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)
$$
By the condition, it is divisible by $20^{n-2}+19^{n-2}$. On the other hand,
$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)=20^{n-2}\left(20^{2}-19^{2}\right)=20^{n-2} \cdot 39
$$
Note that $20^{n-2}$ and ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. At a round table, 10 people are sitting, some of them are knights, and the rest are liars (knights always tell the truth, while liars always lie). It is known that among them, there is at least one knight and at least one liar. What is the maximum number of people sitting at the table who can say: "Both of my neig... | # Answer. 9.
Solution. Note that all 10 could not have said such a phrase. Since at the table there is both a knight and a liar, there will be a liar and a knight sitting next to each other. But then this knight does not have both neighbors as knights. If, however, at the table there are 9 liars and 1 knight, then eac... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.3. What is the minimum number of L-shaped corners consisting of 3 cells that need to be painted in a $5 \times 5$ square so that no more L-shaped corners can be painted? (Painted L-shaped corners should not overlap.) | Answer: 4.
Solution: Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked in Fig. 7. Since none of these corners can be painted, at least one cell in each of these corners must be painted. Note that one corner cannot paint cells of two marke... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.4. We will call a number greater than 25 semi-prime if it is the sum of some two different prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime | Answer: 5.
Solution: Note that an odd semiprime number can only be the sum of two and an odd prime number.
We will show that three consecutive odd numbers $2n+1$, $2n+3$, and $2n+5$, greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers $2n-1$, $2n+1$, and $2n+3$ are... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.5. Kuzya cut a convex paper 67-gon along a straight line into two polygons, then similarly cut one of the two resulting polygons, then one of the three resulting ones, and so on. In the end, he got eight $n$-gons. Find all possible values of $n$. | Answer: $n=11$.
Solution. A straight-line cut can be of three types: from side to side, from a vertex to a side, and from a vertex to a vertex. Therefore, after one cut, the total number of sides of the polygons increases by 4, 3, or 2, respectively. Kuzya made 7 cuts, so the number of sides added is at least 14 but n... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.4. Lev Alex decided to count the stripes on Marty the zebra (black and white stripes alternate). It turned out that there is one more black stripe than white ones. Alex also noticed that all white stripes are of the same width, while black stripes can be wide or narrow, and there are 7 more white stripes than... | # Answer: 8.
Solution. First, let's look only at the wide black stripes. There are 7 fewer of them than white ones. If we add the narrow black stripes to the wide black ones, we get all the black stripes, which are 1 more than the white ones. Therefore, to find the number of narrow black stripes, we need to first "com... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In the class, there are 29 students: some are excellent students and some are troublemakers. Excellent students always tell the truth, while troublemakers always lie.
All students in this class sat around a round table.
- Several students said: “There is exactly one troublemaker next to me.”
- All other ... | Answer: 10.
Solution. If there were three straight-A students in a row, the middle one would definitely lie. Therefore, among any three consecutive people, there must be at least one troublemaker.
Choose an arbitrary troublemaker. Assign this person the number 29, and number all the people following them clockwise fr... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Points $D$ and $E$ are marked on sides $A C$ and $B C$ of triangle $A B C$ respectively, such that $A D=E C$. It turns out that $B D=E D, \angle B D C=\angle D E B$. Find the length of segment $A C$, given that $A B=7$ and $B E=2$. | Answer: 12.
Solution. Note that triangles $D E C$ and $B D A$ are equal. Indeed, $D E=B D, E C=$ $D A$ and $\angle D E C=180^{\circ}-\angle B E D=180^{\circ}-\angle B D C=\angle B D A$. From this, it follows that $D C=A B=7$ and $\angle D C E=\angle B A D$ (Fig. 3). From the equality of these angles, it follows that t... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. On an island, there live red, yellow, green, and blue chameleons.
- On a cloudy day, either one red chameleon changes its color to yellow, or one green chameleon changes its color to blue.
- On a sunny day, either one red chameleon changes its color to green, or one yellow chameleon changes its color to b... | Answer: 11.
Solution. Let $A$ be the number of green chameleons on the island, and $B$ be the number of yellow chameleons. Consider the quantity $A-B$. Note that each cloudy day it decreases by 1, and each sunny day it increases by 1. Since there were $18-12=6$ more sunny days than cloudy days in September, the quanti... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. Lёsha cut a cube $n \times n \times n$ into 153 smaller cubes. Moreover, all the cubes, except one, have an edge length of 1. Find $n$. | Answer: 6.
Solution. Let the remaining cube have an edge of $s$. Obviously, the numbers $n$ and $s$ are natural, and $n>s$.
The difference in the volumes of the cubes with edges $n$ and $s$ is equal to the total volume of the unit cubes. Therefore, $n^{3}-s^{3}=152$.
Since $n^{3}>152>5^{3}$, then $n>5$. Note that $n... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 10.5. Vika has 60 cards with numbers from 1 to 60. She wants to divide all the cards into pairs so that the modulus of the difference of the numbers in all pairs is the same. How many ways are there to do this? | # Answer: 8.
Solution. Let $d$ be the absolute difference between the numbers. It is clear that the number $d$ is a natural number.
It is clear that the number 1 must be paired with $d+1$, the number 2 must be paired with $d+2, \ldots$, the number $d$ must be paired with $2d$. Therefore, the first $2d$ natural number... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6.2. On the plate, there were 15 doughnuts. Karlson took three times more doughnuts than Little Man, and Little Man's dog Bibbo took three times fewer than Little Man. How many doughnuts are left on the plate? Explain your answer. | Answer: 2 doughnuts are left.
From the condition of the problem, it follows that Little One took three times as many doughnuts as Bimbo, and Karlson took three times as many doughnuts as Little One. We can reason in different ways from here.
First method. If Bimbo took one doughnut, then Little One took three doughnu... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6.5. Ladybugs gathered on a clearing. If a ladybug has 6 spots on its back, it always tells the truth, and if it has 4 spots, it always lies, and there were no other ladybugs on the clearing. The first ladybug said: "Each of us has the same number of spots on our backs." The second said: "Together, we have 30 spots on ... | Answer: 5 ladybugs.
If the first ladybug is telling the truth, then the second and third should also be telling the truth, as they should have the same number of spots as the first. However, the second and third ladybugs contradict each other, so at least one of them is lying, which means the first ladybug is also lyi... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) Cheburashka and Gena ate a cake. Cheburashka ate twice as slowly as Gena, but started eating a minute earlier. In the end, they each got an equal amount of cake. How long would it take Cheburashka to eat the cake alone? | Answer. In 4 minutes.
Solution.
First method. If Cheburashka eats twice as slowly as Gena, then to eat the same amount of cake as Gena, he needs twice as much time. This means that the time Cheburashka ate alone (1 minute) is half of the total time it took him to eat half the cake. Thus, he ate half the cake in 2 min... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. There are seven cards on the table. In one move, it is allowed to flip any five cards. What is the minimum number of moves required to flip all the cards? | 9.1. Answer. 3 moves.
Obviously, one move is not enough. After two moves, there will be at least three cards that have been flipped twice, which means these cards will be in their original position.
Let's provide an example of flipping all the cards in three moves. Number the cards from 1 to 7 and flip cards numbered... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the country, there are 20 cities. An airline wants to organize two-way flights between them so that from any city, it is possible to reach any other city with no more than $k$ transfers. At the same time, the number of air routes from any city should not exceed four. What is the smallest $k$ for which this is p... | 9.4. Answer. $k=2$.
Note that at least two transfers will be required. Indeed, from an arbitrary city $A$ without a transfer, one can reach no more than 4 cities, and with exactly one transfer - no more than $4 \cdot 3=12$ cities (since one of the flights from each of these cities leads back to $A$). Therefore, if usi... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the 5 brothers owns a plot of land. One day, they pooled their money, bought a neighbor's garden, and divided the new land equally among themselves. As a result, Andrey's plot increased by $10 \%$, Boris's plot - by $\frac{1}{15}$, Vladimir's plot - by $5 \%$, Grigory's plot - by $4 \%$, and Dmitry's plot - ... | Answer: By $5 \%$.
Solution. Let A, B, V, G, D be the areas of the plots of each brother, respectively. Then, according to the problem, $\frac{1}{10} \mathrm{~A}=\frac{1}{15} \mathrm{~B}=\frac{1}{20} \mathrm{~V}=\frac{1}{25} \mathrm{~G}=\frac{1}{30} \mathrm{~D}$ (*). Denoting $\mathrm{A}=x$, we find: $\mathrm{B}=1.5 x... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The participants of the Olympiad left 9 pens in the office. Among any four pens, at least two belong to the same owner. And among any five pens, no more than three belong to the same owner. How many students forgot their pens, and how many pens does each student have? | Answer. There are three students, each owning three pens.
Solution. No student owned more than three pens, as otherwise the condition "among any five pens, no more than three belonged to one owner" would not be met. There are a total of 9 pens, so there are no fewer than 3 students. On the other hand, among any four p... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the minimum number of cells that need to be colored on a $6 \times 6$ board so that, for any placement (including rotations and flips) of a figure consisting of 4 cells in the shape of the letter $\Gamma$, there is at least one colored cell? | Answer: 12.
Solution: Consider a $2 \times 3$ rectangle. It is obvious that at least 2 cells need to be colored in it. Divide the $6 \times 6$ board into 6 rectangles of $2 \times 3$. In each, at least 2 cells need to be colored, so in total, at least 12 cells need to be colored. An example with 12 cells is shown on t... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
In the village, 7 people live. Some of them are liars who always lie, while the rest are knights (who always tell the truth). Each resident said whether each of the others is a knight or a liar. A total of 42 answers were received, and in 24 cases, a resident called a fellow villager a liar. What is the... | Answer: 3.
Solution: The phrase "He is a knight" would be said by a knight about a knight and by a liar about a liar, while the phrase "He is a liar" would be said by a knight about a liar and by a liar about a knight. Therefore, in each pair of "knight-liar," the phrase "He is a liar" will be said twice. Since this p... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1.
A line parallel to the leg $A C$ of the right triangle $A B C$ intersects the leg $B C$ at point $K$, and the hypotenuse $A B$ at point $N$. On the leg $A C$, a point $M$ is chosen such that $M K=M N$. Find the ratio $\frac{A M}{M C}$, if $\frac{B K}{B C}=14$. | Answer: 7.
Solution.
Drop the altitude $M H$ from point $M$ in the isosceles triangle $M N K$. Then $M H K C$ is a rectangle and $M C=K H=H N$. Let $M C=K H=H N=y$. Let $K B=x$, then $C K=C ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task No. 1.1
## Condition:
Five friends - Masha, Nastya, Irina, Olya, and Anya - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Irina: I was the first in line!
Olya: No one was after me.
Anya: Only one person was after me.
Masha: There wer... | Answer: 3
Exact match of the answer - 1 point
## Solution.
From the statements of Irina and Olya, it is clear that they were first and last, respectively. Since there was only one person after Anya, it was Olya. Nastya stood next to Irina, but she could not have stood in front of her, so Nastya was second. This mean... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 1.2
## Condition:
Five friends - Katya, Polina, Alyona, Lena, and Svetlana - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Polina: I stood next to Alyona.
Alyona: I was the first in line!
Lena: No one was after me.
Katya: There... | # Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 1.3
## Condition:
Five friends - Sasha, Yulia, Rita, Alina, and Natasha - meet in the park every day after buying ice cream from the little shop around the corner. One day, the girls had a conversation.
Sasha: There were five people in front of me.
Alina: There was no one after me.
Rita: I was the first i... | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 1.4
## Condition:
Five friends - Kristina, Nadya, Marina, Liza, and Galia - meet in the park every day after buying ice cream from the shop around the corner. One day, the girls had a conversation.
Kristina: There were five people in front of me.
Marina: I was the first in line!
Liza: No one was behind ... | # Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task №1.1
# | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.2
## Condition:
Artyom makes watches for a jewelry store on order. Each watch consists of a bracelet and a dial. The bracelet can be leather, metal, or nylon. Artyom has round, square, and oval dials. Watches can be mechanical, quartz, or electronic.
Artyom is only satisfied when the watches are arranged ... | Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №3.1
# | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Given a sequence $x_{n}$ such that $x_{1}=1, x_{2}=2, x_{n+2}=\left|x_{n+1}\right|-x_{n}$. Find $x_{2015}$. | 5. Answer. ${ }^{x_{2015}}=0$.
We will show that the given sequence is periodic with a period of 9. Let's find
$x_{1}=1, x_{2}=2, x_{3}=1, x_{4}=-1, x_{5}=0, x_{6}=1, x_{7}=1, x_{8}=0, x_{9}=-1, x_{10}=1, x_{11}=2, \ldots$.
Since the sequence is completely determined by any two consecutive terms, we have obtained th... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. Several different real numbers are written on the board. It is known that the sum of any three of them is rational, while the sum of any two of them is irrational. What is the largest number of numbers that can be written on the board? Justify your answer. | Solution: The set of three numbers $\sqrt{2}, \sqrt{3}, -\sqrt{2}-\sqrt{3}$, as is easily verified, satisfies the condition of the problem. We will show that no more than three numbers can be written on the board.
Assume the contrary, and let $a_{i} (i=\overline{1,4})$ be some four numbers from this set. Then the numb... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Plot the graph of the function $y=(4 \sin 4 x-2 \cos 2 x+3)^{0.5}+(4 \cos 4 x+2 \cos 2 x+3)^{0.5}$. | Solution:
$\mathbf{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x+1}$
$\mathrm{y}=2 \sin ^{2} x+1+2 \cos ^{2} x+1, \m... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Two very small fleas are jumping on a large sheet of paper. The first jump of the fleas is along a straight line towards each other (their jumps may have different lengths). The first flea first jumps to the right, then up, then to the left, then down, then to the right again, and so on. Each jump is 1 cm longer tha... | 4. Answer: 2 meters. Solution: Let's observe the first four jumps of the first flea: 2 cm more to the left than to the right, and 2 cm more down than up. That is, after four jumps, the first flea moves 2 cm to the left and 2 cm down. Therefore, after 100 jumps, it will move 50 cm to the left and 50 cm down. Similarly, ... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10.4. Find the maximum value of the expression $a+b+c+d-ab-bc-cd-da$, if each of the numbers $a, b, c$ and $d$ belongs to the interval $[0 ; 1]$. | Answer: 2.
Solution. The first method. The value 2 is achieved, for example, if $a=c=1, b=d=0$. We will prove that with the given values of the variables $a+b+c+d-ab-bc-cd-da \leqslant 2$.
Notice that $a+b+c+d-ab-bc-cd-da=(a+c)+(b+d)-(a+c)(b+d)$. Let $a+c=x, b+d=y$, then we need to prove that $x+y-xy \leqslant 2$, if... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. There was a whole number of cheese heads on the kitchen. At night, rats came and ate 10 heads, and everyone ate equally. Several rats got stomachaches from overeating. The remaining seven rats the next night finished off the remaining cheese, but each rat could eat only half as much cheese as the night before. How m... | Solution. Let there be $k$ rats in total $(k>7)$, then each rat ate $\frac{10}{k}$ pieces of cheese on the first night. On the second night, each rat ate half as much, that is, $\frac{5}{k}$ pieces. Then seven rats ate $\frac{35}{k}$ pieces. This is an integer. The only divisor of the number 35 that exceeds 7 is the nu... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2. On a line, there are blue and red points, with no fewer than 5 red points. It is known that on any segment with endpoints at red points, containing a red point inside, there are at least 4 blue points. And on any segment with endpoints at blue points, containing 3 blue points inside, there are at least 2 red point... | Answer: 4.
Solution: Note that on a segment with endpoints at red points, not containing other red points, there cannot be 5 blue points. Indeed, in this case, between the outermost blue points, there would be 3 blue points, which means there would be at least 2 more red points. Therefore, on such a segment, there are... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.4. In the castle, there are 9 identical square rooms, forming a $3 \times 3$ square. Nine people, consisting of liars and knights (liars always lie, knights always tell the truth), each occupied one of these rooms. Each of these 9 people said: "At least one of the neighboring rooms to mine is occupied by a liar." Wha... | Answer: 6 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 3 liars (thus showing that there are no more than 6 knights). Suppose there are no more than 2 liars, then there will be a "vertical row" of rooms where only knights live. But then... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.1. For what largest $k$ can we assert that in any coloring of $k$ cells in black in a white $7 \times 7$ square, there will necessarily remain a completely white $3 \times 3$ square with sides along the grid lines? | Answer: 3
Solution. Let's highlight four $3 \times 3$ squares that are adjacent to the corners of the $7 \times 7$ square. These squares do not overlap, so if no more than three cells are sha... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3.1. A pedestrian left point $A$ for point $B$. At the same time, a cyclist left point $B$ for point $A$. After one hour, the pedestrian was three times farther from the cyclist than from point $A$. Another 30 minutes later, they met, after which both continued their journey. How many hours did it take the pedestrian t... | Answer: 9
Solution. Let the distance from $A$ to $B$ be 1 km, the pedestrian moves at a speed of $x$ km/h, and the cyclist at $y$ km/h. Then in one hour, the pedestrian has walked $x$ km, the cyclist has traveled $y$ km, and the distance between them is $1-x-y$, which should be three times $x$. Therefore, $3 x=1-x-y, ... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. We will call a natural number interesting if it is the product of exactly two (distinct or equal) prime numbers. What is the greatest number of consecutive numbers, all of which are interesting | Solution. One of four consecutive numbers is divisible by 4. However, among the numbers divisible by 4, only the number 4 itself is interesting. But the numbers 3 and 5 are not interesting, so four consecutive interesting numbers do not exist. An example of three consecutive interesting numbers: $33,34,35$. Answer: thr... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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