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1. The robotics club accepts only those who know mathematics, physics, or programming. It is known that 8 members of the club know physics, 7 know mathematics, and 11 know programming. Additionally, it is known that at least two know both physics and mathematics, at least three know both mathematics and programming, an... | Answer 19.
2. From the sequence of natural numbers $1,2,3, \ldots$, all perfect squares (squares of integers) have been removed. What number will be in the 2018th position among the remaining numbers?
Answer: 2063. | 19 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. The sequence $a_{n}$ is defined as follows:
$$
a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}, \text { for } n \geq 1 . \text { Find } a_{200}
$$ | Answer: 20100.
## Variant 3a (Chelyabinsk) | 20100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. The sequence $a_{n}$ is defined as follows:
$a_{1}=1, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$. | Answer: 499500.
## School Olympiad "Conquer Sparrow Hills"
## Tasks for 7-8 Grades
## Variant 1b (Zheleznovodsk) | 499500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Thirteen millionaires arrived at the economic forum and settled in the "Super Luxury+" hotel. The hotel has rooms of 3 different types: 6-star, 7-star, and 8-star. The millionaires need to be accommodated in such a way that all three types of rooms are used (i.e., at least one person must be placed in a 6-star room,... | Answer: $C_{12}^{2}=66$. | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. The sequence $a_{n}$ is defined as follows:
$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{100}$. | Answer: 10100.
## Variant 2b (Saratov) | 10100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. A school coach decided to reward 12 students who ran the distance in the best time. Each of them should be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used (at least once), and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
... | Answer: $C_{11}^{2}=55$. | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Petrov lists the odd numbers: $1,3,5, \ldots, 2013$, while Vasechkin lists the even numbers $2,4, \ldots, 2012$. Each of them calculated the sum of all digits of all their numbers and told the result to the excellent student Masha. Masha subtracted the result of Vasechkin from the result of Petrov. What did she get? | Answer: 1007.
Solution: Let's break down the numbers of Petrov and Vasechkin into pairs as follows: $(2,3),(4,5), \ldots,(98,99),(100,101), \ldots$ (2012,2013), with 1 left unpaired for Petrov. Notice that in each pair, the sum of the digits of the second number is 1 greater than that of the first (since they differ o... | 1007 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The board for playing "Battleship" is $10 \times 10$. What is the maximum number of ships of size $1 \times 4$ that can be placed on it? | Answer: 24.
Solution: It is easy to show by trial that 24 ships can be placed. Let's color the board in 4 colors as shown in the figure. Note that each ship of size $1 \times 4$ contains one cell of each color, and there will be 24 yellow cells.
, 1 \leqslant x, y \leqslant 1000$, such that $x^{2}+y^{2}$ is divisible by 5. | Answer: 360000.
Solution: We have 200 numbers for each of the remainders $0,1,2,3,5$ when divided by 5. There are two cases: a) the numbers $x, y$ are both divisible by 5; b) or one number gives a remainder of 1 or 4, and the other gives a remainder of 2 or 3 when divided by 5. In the first case, we get $200 \times 20... | 360000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$
also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the s... | 999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solution. From the formulas for the sum of a geometric progression, we know
$$
\begin{aligned}
& \frac{b_{1}}{1-q}=60 \\
& \frac{b_{1}^{2}}{1-q^{2}}=1200
\end{aligned}
$$
By dividing the second equation by the first, we get $\frac{b_{1}}{1+q}=20$, which is the answer.
Remark. We could also find $b_{1}=30, q=1 / 2... | Answer: 20 (option $1-2:-30)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to restate it as a translation, here it is:
... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Thirteen millionaires arrived at the economic forum and settled in the "Super Luxury+" hotel. The hotel has rooms of 3 different types: 6-star, 7-star, and 8-star. The millionaires need to be accommodated in such a way that all three types of rooms are used (i.e., at least one person must be placed in a 6-star room,... | Answer: $C_{12}^{2}=66$. | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. A school coach decided to reward 12 students who ran the distance in the best time. Each of them must be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used, and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
In how many ways c... | Answer: $C_{11}^{2}=55$. | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Petrov and Vasechkin were repairing a fence. Each had to nail a certain number of boards (the same amount). Petrov nailed two nails into some boards and three nails into others. Vasechkin nailed three nails into some boards and five nails into the rest. Find out how many boards each of them nailed, given that Petrov... | Answer: 30.
Solution: If Petrov had nailed 2 nails into each board, he would have nailed 43 boards and had one extra nail. If he had nailed 3 nails into each board, he would have nailed 29 boards. Therefore, the desired number lies between 29 and 43 (inclusive). Similarly, if Vasechkin had nailed 3 nails into each boa... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Six natural numbers (possibly repeating) are written on the faces of a cube, such that the numbers on adjacent faces differ by more than 1. What is the smallest possible value of the sum of these six numbers? | Answer: 18.
Solution: Consider three faces that share a common vertex. The numbers on them differ pairwise by 2, so the smallest possible sum would be for $1+3+5=9$. The same can be said about the remaining three faces.
Thus, the sum cannot be less than 18. We will show that 18 can be achieved - place the number 1 on... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. In the Slytherin faculty at Hogwarts, there are 30 students. Some are friends (friendship is mutual, i.e., if A is friends with B, then B is also friends with A), but there are no 3 people who are pairwise friends with each other. On New Year's, each student sent cards to all their friends. What is the maximum numbe... | Answer: 450.
Solution: Let's find the person with the most friends. Suppose there are no fewer than 15, and denote their number as $15+a$. We will divide the students into two groups: the first group will consist of these $15+a$ students. According to the condition, they cannot be friends with each other, so each of t... | 450 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Points $A_{1}, \ldots, A_{12}$ are the vertices of a regular 12-gon. How many different 11-segment open broken lines without self-intersections with vertices at these points exist? Broken lines that can be transformed into each other by rotation are considered the same. | Answer: 1024.
Solution: The first vertex of the broken line can be chosen in 12 ways. Each subsequent vertex (except the last one) can be chosen in two ways - it must be adjacent to the already marked vertices to avoid self-intersections. The last vertex is chosen uniquely. We get $12 * 2^{10}$ ways. Considering 12 po... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest three-digit number with the property that if a number, which is 1 greater, is appended to it on the right, then the result (a six-digit number) will be a perfect square. Answer: 183 | Solution: Let the required number be \(a\), then \(1000a + a + 1 = n^2\). We can write it as: \(1001a = (n - 1)(n + 1)\). Factorize \(1001 = 7 \times 11 \times 13\), so the product \((n - 1)(n + 1)\) must be divisible by 7, 11, and 13. Moreover, for the square to be a six-digit number, \(n\) must be in the interval \([... | 183 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Pete came up with all the numbers that can be formed using the digits 2, 0, 1, 8 (each digit can be used no more than once). Find their sum. | Answer: 78331
Solution: First, consider the units place. Each of the digits 1, 2, 8 appears once in this place for single-digit numbers, twice for two-digit numbers, four times for three-digit numbers, and four times for four-digit numbers - a total of 11 times.
In the tens place, each of them appears 3 times for two... | 78331 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Given a sequence of natural numbers $a_{n}$, the terms of which satisfy
the relations $a_{n+1}=k \cdot \frac{a_{n}}{a_{n-1}}$ (for $n \geq 2$). All terms of the sequence are integers. It is known that $a_{1}=1$, and $a_{2018}=2020$. Find the smallest natural $k$ for which this is possible. | Answer: 2020
Solution: Let $a_2=x$. Then all terms of the sequence will have the form $x^{m} k^{n}$.
The powers of $k$ will repeat with a period of 6: $0,0,1,2,2,1,0,0, \ldots$
The powers of $x$ will also repeat with a period of 6: 0,1,1,0,-1,-1,0,1,...
Since 2018 gives a remainder of 2 when divided by 6, then $a_{... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality
$$
(2+\sqrt{3})^{x}+2<3(\sqrt{2-\sqrt{3}})^{2 x}
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-20 ; 53)$. | Answer: The solution to the inequality is $(-\infty ; 0)$. The desired sum $-19-18-\cdots-2-1=-190$. We write -190 as the answer. | -190 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. The area of triangle $\triangle A B C$ is 10 cm $^{2}$. What is the smallest value in centimeters that the circumference of the circle circumscribed around triangle $\triangle A B C$ can take, given that the midpoints of the heights of this triangle lie on the same line? If the answer is not an integer, round it to ... | Answer: The smallest value is $2 \pi \sqrt{10}$ cm. In the answer, we write 20, since $2 \pi \sqrt{10} \approx 19.8691 \ldots$ | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. First method: Since $x_{1}^{3}-2015 x_{1}+2016=0$ and $x_{2}^{3}-2015 x_{2}+2016=0$, then $x_{1}^{3}-x_{2}^{3}=2015 \cdot\left(x_{1}-x_{2}\right)$. Therefore, $x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}=2015$.
Second method: By Vieta's theorem (but then it needs to be justified that there are three distinct roots): $x_{1}+x_{... | Answer: 2015.
Answer to option: 4-2: 2016. | 2015 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $\alpha=\pi x$.
$$
\begin{aligned}
& \cos (8 \alpha)+2 \cos (4 \alpha)-\cos (2 \alpha)+2 \sin (\alpha)+3=0 \Longleftrightarrow \\
& 2 \cos ^{2} 4 \alpha-1+2 \cos 4 \alpha-1+2 \sin ^{2} \alpha+2 \sin \alpha+3=0 \Longleftrightarrow \\
& 2\left(\cos 4 \alpha+\frac{1}{2}\right)^{2}+2\left(\sin \alpha+\frac{1}{2}\ri... | Answer: 5050
Answer to option: 7-2: 4950.
Solutions for option 17-1 and answers to all options | 5050 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Drop perpendiculars $D D_{1}, D D_{2}, D D_{3}$ from point $D$ to the planes $S B C$, $S A C$, and $S A B$ respectively. Let $D D_{1}=x, D D_{2}=y, D D_{3}=z$. According to the condition, we form the system of equations
$$
\left\{\begin{array}{l}
y^{2}+z^{2}=5 \\
x^{2}+z^{2}=13 \\
x^{2}+y^{2}=10
\end{array}\right.
... | Answer: 27.
Answer to option 17-2: 108.
Answer to option $17-3: 27$.
Answer to option $17-4: 108$.
[^0]: ${ }^{1}$ This equality can be proven by expressing $B C^{2}$ from two triangles $B A C$ and $B D C$ using the planimetric cosine theorem. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. A natural number $N$ ends in ...70, and it has exactly 72 natural divisors (including 1 and itself). How many natural divisors does the number 80N have? ANSWER: 324. | Solution: Each divisor of the number $\mathrm{N}$ can be represented in the form $2^{a} \cdot 5^{b} \cdot q$, where q is not divisible by 2 and 5, and the numbers $a, b$ are 0 or 1. Therefore, there are only 4 different combinations for the pair a and b, which means there are $72 / 4=18$ possible values for q. These co... | 324 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality
$$
\frac{4+\sqrt{-3-x}}{3-|x+4|} \leqslant 1
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and do not exceed 12 in absolute value. | Solution. Transferring 1 to the left side and bringing to a common denominator:
$$
\frac{1+\sqrt{-3-x}+|x+4|}{3-|x+4|} \leqslant 0
$$
For $x>-3$, the expression under the square root is negative, meaning there are no solutions. For $x \leqslant-3$, the numerator is positive, and thus,
$$
3-|x+4|3 \Longleftrightarrow... | -50 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. The number 2015 can be represented as the sum of consecutive integers in various ways, for example, $2015=1007+1008$ or $2015=$ $401+402+403+404+405$. In how many ways can this be done? | Answer: 16.
Solution: The sum of $k$ numbers starting from $n$ is $S(k, n)=\frac{1}{2}(2 n+k-1) \cdot k$. That is, we need to solve the equation $(2 n+k-1) \cdot k=4030$ in integers. Obviously, $k$ can be any divisor of $4030=2 \times 5 \times 13 \times 31$. Note that each of the prime factors $(2,5,13$ and 31$)$ can ... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. Given triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$, the areas of which are 1 and 2025, respectively. It is known that rays $A B$ and $A^{\prime} B^{\prime}$ are parallel and point in opposite directions (see figure). The same is true for pairs $B C$ and $B^{\prime} C^{\prime}, C A$ and $C^{\prime} A^{\pr... | Answer: 484.
Solution: Obviously, triangles $A B C A^{\prime} B^{\prime} C^{\prime}$ are similar with a coefficient of 45.
Three trapezoids are formed, in which $A^{\prime \prime} B^{\prime \prime}, B^{\prime \prime} C^{\prime \prime}$ and $C^{\prime \prime} A^{\prime \prime}$ are segments connecting the midpoints of... | 484 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. A school coach decided to reward 12 students who ran the distance in the best time. Each of them must be awarded a "gold", "silver", or "bronze" medal. All three types of medals must be used, and the one who finished earlier cannot be awarded a less valuable medal than the one who finished later.
How many ways can ... | Answer: $C_{11}^{2}=55$.
Solution: see 9th grade. | 55 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced 1 cm apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these points lie ... | Answer: 31.
Solution: see 9th grade.
## Tasks for 7-8 grades
## Option 3b (Moscow) | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Each worker on the construction site has at least one construction specialty. 10 people have the specialty of a mason, 9 - the specialty of a painter, 8 - the specialty of a plasterer. It is known that at least four people have the specialties of a mason and a plasterer simultaneously, at least five people - the spe... | # Answer 18.
Solution: According to the principle of inclusion-exclusion, the total number of workers is K+M+S - KM-KS-MS + KMS = 10+9+8-4-5-3 + KMS = 15 + KMS. Note that the number of workers proficient in all three specialties cannot exceed 3. | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Natural numbers, for which the sum of the digits equals 5, were arranged in ascending order. What number is in the $122-$nd position? | Answer: 40001.
Solution: Let's calculate the number of such numbers for different digit counts.
Let $n$ be the number of digits. Subtract 1 from the most significant digit, we get a number (which can now start with zero), the sum of whose digits is 4. Represent this as follows - there are 4 balls, between which $n-1$... | 40001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On a line, there are 16 points $A_{1}, \ldots A_{16}$, spaced $1 \mathrm{~cm}$ apart. Misha constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{16}$.
c) None of these... | Solution: We can represent such a system of circles as a tree with 16 leaves.
In such a tree, there cannot be more than 31 nodes. It is easy to construct a binary tree in which there are exactly 31 nodes. | 31 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The sequence $a_{n}$ is defined as follows:
$a_{1}=2, a_{n+1}=a_{n}+\frac{2 a_{n}}{n}$, for $n \geq 1$. Find $a_{999}$. | Answer: 999000.
Solution: Consider the beginning of the sequence: $a_{1}=2, a_{2}=6, a_{3}=12, a_{4}=20, \ldots$
We can notice a pattern - the difference between consecutive terms forms an arithmetic progression: $4,6,8, \ldots$.
From this, we get the formula for the $\mathrm{n}$-th term $a_{n}=n \cdot(n+1)$, which ... | 999000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The robotics club accepts only those who know mathematics, physics, or programming. It is known that 8 members of the club know physics, 7 know mathematics, and 11 know programming. Additionally, it is known that at least two know both physics and mathematics, at least three know both mathematics and programming, an... | Answer 19.
3 Natural numbers, the sum of whose digits is 5, were arranged in ascending order. What number is in the $111-$th place?
Answer: 23000. | 23000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. On a line, there are 15 points $A_{1}, \ldots A_{15}$, spaced 1 cm apart. Petya constructs circles according to the following rules:
a) The circles do not intersect or touch.
b) Inside each circle, there is at least one of the specified points $\mathrm{A}_{1}, \ldots \mathrm{A}_{15}$.
c) None of these points lie o... | Answer: 29.
## Option 2b (Saratov) | 29 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Let there be $n$ voters, and $k$ votes were cast for a given candidate. Then $0.50332 \leqslant$ $\frac{k}{n} \leqslant 0.50333$, and $1.00664 \leqslant \frac{2 k}{n} \leqslant 1.00666$. If we denote $m=2 k-n$, then $0.00664 \leqslant \frac{m}{n} \leqslant$ 0.00666.
- If $m=1$, then $150.1<\frac{1}{0.00666} \leqsla... | Answer: 451. Answer to the option: $5-2: 341$.
# | 451 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the su... | 999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Igor Gorshkov has all seven books about Harry Potter. In how many ways can Igor arrange these seven volumes on three different bookshelves, so that each shelf has at least one book? (Arrangements that differ by the order of books on the shelf are considered different). | Answer: $C_{6}^{2} \times 7!=75600$.
Solution: First, we can arrange the books in any order, which gives 7! options. Place two dividers into 6 gaps between the books - this can be done in $C_{6}^{2}$ ways. The dividers will divide the books into three parts, which need to be placed on the 1st, 2nd, and 3rd shelves. In... | 75600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Find the sum of all natural numbers that have exactly four natural divisors, three of which (of the divisors) are less than 15, and the fourth is not less than 15. | Answer: $\left((2+3+5+7+11+13)^{2}-\left(2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}\right)\right) / 2-$ $6-10-14+27=649$.
Solution: The numbers $N$ specified have exactly 4 divisors either if $N=$ $p^{3}$, or if $N=p q$, where $p$ and $q$ are primes. In the first case, only 27 fits. In the second case, we need to consider ... | 649 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. A right-angled triangle is called Pythagorean if the lengths of all its sides are natural numbers. Find the greatest integer that divides the product of the lengths of the sides of any Pythagorean triangle. | # Problem 5.
Answer: 60.
Solution. Since a triangle with sides $3,4,5$ is a right triangle, the desired number cannot be greater than 60. We will show that the product of the lengths of the sides of any Pythagorean triangle is divisible by 60, i.e., by 3, by 4, and by 5.
The remainders when the square of an integer ... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Let's call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist?
# | # Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, because otherwise the product will be at least 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. Assemble a square of the smallest area from squares of size $1 \times 1$, $2 \times 2$, and $3 \times 3$, such that the number of squares of each size is the same. | Answer:
Solution. Let $n$ be the number of squares of each type. Then $n + 4n + 9n = 14n$ must be a perfect square. The smallest $n$ for which this is possible is 14. The figure shows an ex... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. The company conducted a survey among its employees - which social networks they use: VKontakte or Odnoklassniki. Some employees said they use VKontakte, some - Odnoklassniki, some said they use both social networks, and 40 employees said they do not use social networks. Among all those who use social networks, 75% ... | Answer: 540
Solution: Since 75% of social media users use VKontakte, it follows that only 25% use Odnoklassniki. Additionally, 65% use both networks, so the total percentage of users of Odnoklassniki is $65 + 25 = 90\%$ of social media users. These $90\%$ constitute $5 / 6$ of the company's employees, so $100\%$ const... | 540 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. In a regular 2017-gon, all diagonals are drawn. Petya randomly selects some $\mathrm{N}$ diagonals. What is the smallest $N$ such that among the selected diagonals, there are guaranteed to be two of the same length? | Answer: 1008.
Solution: Let's choose an arbitrary vertex and consider all the diagonals emanating from it. There are 2014 of them, and by length, they are divided into 1007 pairs. Clearly, by rotating the polygon, any of its diagonals can be aligned with one of these. Therefore, there are only 1007 different sizes of ... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A pirate schooner has taken a merchant ship by boarding. Ten pirates did not participate in the fight, while the rest lost either a hand, a leg, or both a hand and a leg in the battle. $54\%$ of the participants in the battle lost a hand, and $34\%$ lost both a hand and a leg. It is known that $2/3$ of all the pirat... | Answer: 60 pirates.
Solution - similar to option v3a. | 60 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In a regular 1000-gon, all diagonals are drawn. What is the maximum number of diagonals that can be selected such that among any three of the selected diagonals, at least two have the same length? | Answer: 2000
Solution: For the condition of the problem to be met, it is necessary that the lengths of the diagonals take no more than two different values. The diagonals connecting diametrically opposite vertices are 500. Any other diagonal can be rotated to coincide with a diagonal of the corresponding length, i.e.,... | 2000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In how many ways can the number 1024 be factored into three natural factors such that the first factor is divisible by the second, and the second is divisible by the third? | Answer: 14
Solution: Note that the factors have the form $2^{a} \times 2^{b} \times 2^{c}$, where $\mathrm{a}+\mathrm{b}+\mathrm{c}=10$ and $a \geq b \geq c$. Obviously, c is less than 4, otherwise the sum would be greater than 10. Let's consider the cases:
$c=0)$ Then $b=0, \ldots, 5, a=10-b-6$ options
c=1) Then $b... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. A rectangle $A D E C$ is described around a right triangle $A B C$ with legs $A B=5$ and $B C=6$, as shown in the figure. What is the area of $A D E C$? | # Answer 30.
Solution: The area of the rectangle is twice the area of the triangle. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. In the Empire of Westeros, there were 1000 cities and 2017 roads (each road connected some two cities). From any city, you could travel to any other. One day, an evil wizard enchanted $N$ roads, making it impossible to travel on them. As a result, 7 kingdoms formed, such that within each kingdom, you could travel fr... | Answer: 1024.
Solution: Suppose the evil wizard enchanted all 2017 roads. This would result in 1000 kingdoms (each consisting of one city). Now, imagine that the good wizard disenchants the roads so that there are 7 kingdoms. He must disenchant at least 993 roads, as each road can reduce the number of kingdoms by no m... | 1024 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Jack Sparrow needed to distribute 150 piastres into 10 purses. After placing a certain number of piastres in the first purse, he put more in each subsequent purse than in the previous one. As a result, it turned out that the number of piastres in the first purse was not less than half the number of piastres in the l... | # Answer: 16
Solution: Let there be $x$ piastres in the first purse. Then in the second there are no less than $\mathrm{x}+1$, in the third - no less than $\mathrm{x}+2$... in the 10th - no less than $\mathrm{x}+9$. Thus, on one side $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, From which $x \leq 10$. On the other side $x ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find the smallest natural number $N$, such that the decimal representation of the number $N \times 999$ consists entirely of sevens. | Answer 778556334111889667445223
Solution: $N \times 999=77 \ldots 7$, then $N$ is a multiple of 7, denote $n=N / 7$. We get $999 n=$ $1000 n-n=11 \ldots 1$, so $1000 n-111 \ldots 1=n$. Write it as a column subtraction and repeat the found digits of $\mathrm{N}$ with a shift of 3 to the left $* * * * * * * 000$
111111... | 778556334111889667445223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. How many natural numbers from 1 to 2017 have exactly three distinct natural divisors? | Answer: 14.
Solution: Only squares of prime numbers have exactly three divisors. Note that $47^{2}>2017$, so it is sufficient to consider the squares of prime numbers from 2 to 43. There are 14 of them. | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Petya is coming up with a password for his smartphone. The password consists of 4 decimal digits. Petya wants the password not to contain the digit 7, and at the same time, the password should have at least two (or more) identical digits. In how many ways can Petya do this? | # Answer 3537.
Solution: The total number of passwords not containing the digit 7 will be $9^{4}=6561$. Among these, 9x8x7x6=3024 consist of different digits. Therefore, the number of passwords containing identical digits is 6561-3024=3537 passwords. | 3537 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In the computer center, there are 200 computers, some of which (in pairs) are connected by cables, a total of 345 cables are used. We will call a "cluster" a set of computers such that a signal from any computer in this set can reach all other computers via cables (possibly through intermediate computers). Initially... | Answer: 153.
Solution: Let's try to imagine the problem this way: an evil hacker has cut all the wires. What is the minimum number of wires the admin needs to restore to end up with 8 clusters? Obviously, by adding a wire, the admin can reduce the number of clusters by one. This means that from 200 clusters, 8 can be ... | 153 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In trapezoid $A B C D$ with bases $A D / / B C$, the diagonals intersect at point $E$. It is known that the areas $S(\triangle A D E)=12$ and $S(\triangle B C E)=3$. Find the area of the trapezoid. | Answer: 27
Solution: Triangle ADE and CBE are similar, their areas are in the ratio of the square of the similarity coefficient. Therefore, this coefficient is
 | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Find the smallest natural number ending in the digit 2 that doubles when this digit is moved to the beginning. | Answer: 105263157894736842
Solution: Let's write the number in the form ***...** 2 and gradually restore the "asterisks" by multiplying by 2:
```
\(* * \ldots * * 2 \times 2=* * * . . . * * 4\)
\(* * * . . * 42 \times 2=* * * \ldots * 84\)
\(* * * . . .842 \times 2=* * * \ldots * 684\)
***... \(* 6842 \times 2=* * * ... | 105263157894736842 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. At the international StarCraft championship, 100 participants gathered. The game is played in a knockout format, meaning in each match, two players compete, the loser is eliminated from the tournament, and the winner remains. Find the maximum possible number of participants who won exactly two games? | # Answer: 49
Solution: Each participant (except the winner) lost one game to someone. There are 99 such participants, so no more than 49 participants could have won 2 games (someone must lose 2 games to them).
We can show that there could be 49 such participants. Let's say №3 won against №1 and №2, №5 won against №3 ... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. A box of sugar has the shape of a rectangular parallelepiped. It contains 280 sugar cubes, each of which is a $1 \times 1 \times 1$ cm cube. Find the surface area of the box, given that the length of each of its sides is less than 10 cm. | # Answer: 262
Solution: $\mathbf{2 8 0}=\mathbf{2}^{3} \mathbf{x} \mathbf{x} \mathbf{x}$. From the condition that the lengths of the sides are integers less than 10, it follows that its edges are equal to 5, 7, and 8. The surface area is $2 *\left(5^{*} 7+5 * 8+7 * 8\right)=262$ | 262 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. We will call a number "remarkable" if it has exactly 4 distinct natural divisors, and among them, there are two such that neither is a multiple of the other. How many "remarkable" two-digit numbers exist? | Answer 36.
Solution: Such numbers must have the form $p_{1} \cdot p_{2}$, where $p_{1}, p_{2}$ are prime numbers. Note that the smaller of these prime numbers cannot be greater than 7, otherwise the product will be no less than 121. It is sufficient to check $p_{1}=2,3,5,7$. For 2, we get the second factor: 3, 5, 7, 1... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Find all three-digit numbers $\overline{\Pi B \Gamma}$, consisting of different digits $\Pi, B$, and $\Gamma$, for which the equality $\overline{\Pi B \Gamma}=(\Pi+B+\Gamma) \times(\Pi+B+\Gamma+1)$ holds. | Answer: 156.
Solution: Note that П+В $\Gamma \geq 3$ and $\leq 24$ (since the digits are different). Moreover, the numbers
$\overline{\Pi В \Gamma}$ and $(П+B+\Gamma)$ must give the same remainder when divided by 9. This is only possible when П+В $\Gamma) \times(П+B+\Gamma+1)=90$-a two-digit number. By trying $12,15,... | 156 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. At the international table tennis championship, 200 participants gathered. The game is played in a knockout format, i.e., in each match, two players participate, the loser is eliminated from the championship, and the winner remains. Find the maximum possible number of participants who won at least three matches. | Answer: 66.
Solution: Each participant (except the winner) lost one game to someone. There are 199 such participants, so no more than 66 participants could have won 3 games (someone must lose 3 games to them).
We will show that there could be 66 such participants. Let №4 win against №1,2,3; №7 - against №4,5,6,... №1... | 66 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Kolya has 440 identical cubes with a side length of 1 cm. Kolya assembled a rectangular parallelepiped from them, with all edges having a length of at least 5 cm. Find the total length of all the edges of this parallelepiped | Answer: 96.
Solution: $440=2^{3} \times 5 \times 11$. Since all edges have a length of at least 5, their lengths are 8, 5, and 11. Each edge is included 4 times, so the total length is ( $8+5+11)^{*} 4=96$. | 96 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the inequality
$$
\frac{2^{2+\sqrt{x-1}}-24}{2^{1+\sqrt{x-1}}-8}>1 .
$$
In your answer, specify the sum of all integer values of $x$ that satisfy the given inequality and belong to the interval $(-70 ; 34)$. | Solution. Let $t=2^{1+\sqrt{x-1}}$, we get
$$
\frac{2 t-24}{t-8} \geqslant 1 \Longleftrightarrow \frac{t-16}{t-8} \geqslant 0 \Longleftrightarrow t \in(-\infty ; 8) \cup[16 ;+\infty)
$$
From this, either $2^{1+\sqrt{x-1}}<2^{3}, \sqrt{x-1}<2, x \in[1 ; 5)$, or $2^{1+\sqrt{x-1}} \geqslant 2^{4}, \sqrt{x-1} \geqslant 3... | 526 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Inside a right triangle $ABC$ with hypotenuse $AC$, a point $M$ is taken such that the areas of triangles $ABM$ and $BCM$ are one-third and one-fourth of the area of triangle $ABC$, respectively. Find $BM$, if $AM=60$ and $CM=70$. If the answer is not an integer, round it to the nearest integer. | Solution. Denoting $A B=c, B C=a$, we get
$$
\left\{\begin{array}{l}
\left(c-\frac{c}{4}\right)^{2}+\left(\frac{a}{3}\right)^{2}=60^{2} \\
\left(\frac{c}{4}\right)^{2}+\left(a-\frac{a}{3}\right)^{2}=70^{2}
\end{array}\right.
$$
We solve the system and find $B M=\left(\frac{a}{3}\right)^{2}+\left(\frac{c}{4}\right)^{2... | 38 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Find all $a$ for which the system
$$
\left\{\begin{array}{l}
x^{2}+4 y^{2}=1 \\
x+2 y=a
\end{array}\right.
$$
has a unique solution. If necessary, round it to two decimal places. If there are no solutions, put 0 in the answer. | Solution. Since $2 y=a-x$, from the first equation we get:
$$
x^{2}+(a-x)^{2}=1 \Longleftrightarrow 2 x^{2}-2 a x+a^{2}-1=0
$$
This equation has a unique solution when $\frac{D}{4}=2-a^{2}=0$. Therefore, $a= \pm \sqrt{2}$, and the smallest value is $-\sqrt{2} \approx-1.414214 \ldots$
Answer: $-1.41$.
I-A. The sum o... | 1894 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. The sum of 1265 natural numbers is $2016+33$, and their product is $-\left(2016^{2}+33\right)$. Find all possible sets of such numbers. In your answer, indicate the sum of the largest and smallest numbers from all the found sets. If such numbers do not exist, then in your answer, indicate the number 0.
Hint: $2016^... | Answer 574.
II-A. Solve the equation
$$
(1+\cos x+\cos 2 x+\cos 3 x+\cos 4 x)^{4}+(\sin x+\sin 2 x+\sin 3 x+\sin 4 x)^{4}=\frac{3}{4}+\frac{\cos 8 x}{4}
$$
Find the sum of all roots on the interval $A$, rounding to the nearest integer if necessary. If there are no roots or infinitely many roots on this interval, wri... | 574 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Equation:
$$
\cos \frac{a \pi x}{x^{2}+4}-\cos \frac{\pi\left(x^{2}-4 a x+4\right)}{4 x^{2}+16}=-\sqrt{2-\sqrt{2}}
$$
inequality $f(a) \leqslant 5$. Answer: $a \in\left[-\frac{21}{2} ; \frac{21}{2}\right]$, the length of the interval is 21.
V-A. Let $a_{n}$ be the number of permutations $\left(k_{1}, k_{2}, \ldot... | Solution. We will prove in several steps:
Step 1. Prove that there is a recurrence relation: $a_{n}=a_{n-1}+a_{n-3}+1$. The following are the possible beginnings of permutations:
- In the sequence $(1,2, \ldots, n)$. Discard the first one, and decrease the remaining numbers by 1. What remains satisfies the conditions... | 6585451 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Ivan Semenovich leaves for work at the same time every day, drives at the same speed, and arrives exactly at 9:00. One day he overslept and left 40 minutes later than usual. To avoid being late, Ivan Semenovich drove at a speed 60% faster than usual and arrived at 8:35. By what percentage should he have increased hi... | Answer: By $30 \%$.
Solution: By increasing the speed by $60 \%$, i.e., by 1.6 times, Ivan Semenovich reduced the time by 1.6 times and gained 40+25=65 minutes. Denoting the usual travel time as $T$, we get $\frac{T}{1.6}=T-65$, from which $T=\frac{520}{3}$. To arrive in $T-40=\frac{400}{3}$, the speed needed to be in... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The English club is attended by 20 gentlemen. Some of them are acquainted (acquaintances are mutual, i.e., if A knows B, then B knows A). It is known that there are no three gentlemen in the club who are pairwise acquainted.
One day, the gentlemen came to the club, and each pair of acquaintances shook hands with ea... | Answer: 100 handshakes.
Solution: Choose a gentleman with the maximum number of acquaintances (if there are several, choose any one). Suppose he has $n$ acquaintances. These acquaintances cannot be pairwise acquainted with each other. Consider the remaining $(20-n-1)$ gentlemen, each of whom has no more than $n$ acqua... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Ms. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that there are exactly 2 boys 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet, given that 5 boys are participating? | Answer: 20 girls.
Solution: Let's select and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 m away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 m. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$,... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. Given a polynomial $P(x)$, not identically zero. It is known that for all $x$ the identity $(x-2020) \cdot P(x+1)=(x+2021) \cdot P(x)$ holds. How many roots does the equation $P(x)=0$ have? Answer: 4042 | Solution:
First, we prove two auxiliary lemmas:
Lemma 1. If $x_{0} \neq 2020$ is a root of $P(x)$, then $x_{0}+1$ is also a root.
Proof: Substitute $\left(x_{0}-2020\right) \cdot P\left(x_{0}+1\right)=\left(x_{0}+2021\right) \cdot P\left(x_{0}\right)=0$ but $\left(x_{0}-2020\right) \neq 0$.
Lemma 2. If $x_{0} \neq ... | 4042 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. 2-1. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-100)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer. | Solution. Knowledge of arithmetic progression is required. It turns out to be a quadratic function
$f(x)=51 x^{2}-2(2+4+6+\ldots+100) x+(2^{2}+4^{2}+6^{2}+\ldots+100^{2})=51 x^{2}-2 \cdot 50 \cdot 51 x+4 \cdot(1^{2}+2^{2}+3^{2}+\ldots+50^{2})$.
The minimum is achieved at the point $x_{0}=50$. In this case,
$f(50)=50... | 44200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2-2. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-102)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it as the answer. | Solution. $f(x)=52 x^{2}-2(2+4+\ldots+102) x+2^{2}+4^{2}+\ldots+102^{2}=52 x^{2}-2 \cdot 51 \cdot 52 x+4\left(1^{2}+2^{2}+3^{2}+\ldots+51^{2}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=51 . f(51)=51^{2}+49^{2}+\ldots 1^{2}+1^{2}+3^{2}+\ldots+49^{2}+51^{2}=$ $2\left(1^{2}+3^{2}+\ldots 51^{... | 46852 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2-3. Find the minimum value of the function
$$
f(x)=(x-1)^{2}+(x-3)^{2}+\ldots+(x-101)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it in the answer. | Solution. $f(x)=51 x^{2}-2(1+3+\ldots+101) x+\left(1^{2}+3^{2}+\ldots+101^{2}\right)=51 x^{2}-2 \cdot 51^{2} x+\left(1^{2}+3^{2}+\ldots+101^{2}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=51$. Since $1^{2}+3^{2}+\ldots+(2 n-1)^{2}=\frac{n\left(4 n^{2}-1\right)}{3}$, then $f(51)=51^{3}-2 \c... | 44200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2-4. Find the minimum value of the function
$$
f(x)=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-104)^{2}
$$
If the result is a non-integer, round it to the nearest integer and write it in the answer. | Solution. $f(x)=53 x^{2}-2(2+4+\ldots 104) x+4\left(1^{2}+2^{2}+\ldots+52^{2}\right)=53 x^{2}-2 \cdot 52 \cdot 53 x+4 \cdot \frac{52 \cdot 53 \cdot 105}{6}$ (since $\left.1^{2}+2^{2}+3^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\right)$.
The minimum of the function $f$ is achieved at the point $x_{0}=52 . f(52)=53 \cdot ... | 49608 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Determine how many roots of the equation
$$
4 \sin 2 x + 3 \cos 2 x - 2 \sin x - 4 \cos x + 1 = 0
$$
are located on the interval $\left[10^{2014!} \pi ; 10^{2014!+2015} \pi\right]$. In your answer, write the sum of all digits of the found number. | Solution. Let $t=\sin x+2 \cos x$. Then $t^{2}=\sin ^{2} x+2 \sin 2 x+4 \cos ^{2} x=2 \sin 2 x+\frac{3}{2} \cos 2 x+\frac{5}{2}$. The original equation is equivalent to $t^{2}-t-2=0$. From which
$$
\left[\begin{array}{c}
\operatorname{sin} x + 2 \operatorname{cos} x = -1, \\
\operatorname{sin} x + 2 \operatorname{cos}... | 18135 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 4. Find the smallest natural $m$, for which there exists such a natural $n$ that the sets of the last 2014 digits in the decimal representation of the numbers $a=2015^{3 m+1}$ and $b=2015^{6 n+2}$ are the same, and $a<b$. | Solution. The coincidence of the last 2014 digits of the two specified powers means divisibility by \(10^{2014}\) of the difference
\[
2015^{6 n+2}-2015^{3 m+1}=2015^{3 m+1} \cdot\left(2015^{6 n-3 m+1}-1\right)
\]
since the first factor in the obtained decomposition is not divisible by 2, and the second is not divisi... | 671 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. In a trapezoid, the diagonals intersect at a right angle, and one of them is equal to the midline. Determine the angle this diagonal forms with the bases of the trapezoid. | Answer: $60^{\circ}$.
Solution: We perform a parallel translation of the diagonal - we will get a right triangle, in which the leg is half the hypotenuse. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. In the wallet of the merchant Hans, there are 20 silver coins worth 2 crowns each, 15 silver coins worth 3 crowns each, and 3 gold ducats (1 ducat equals 5 crowns). In how many ways can Hans pay a sum of 10 ducats? Coins of the same denomination are indistinguishable. | Answer: 26
Solution. If the merchant paid $x$ coins at 2 crowns each, $y$ coins at 3 crowns each, and $z$ ducats (i.e., $z$ times 5 crowns), we get the system: $2 x+3 y+5 z=50, x \in[0 ; 20], y \in[0 ; 15], z \in[0 ; 3]$.
a) When $z=0$, we get the equation $2 x+3 y=50$, which has suitable solutions under the conditio... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. According to the condition, these numbers are written only with the digits $0,1,2,6,8$. Then, three-digit numbers that are multiples of 4 can end with exactly 10 variants: $00,08,12,16,20,28,60,68,80,88$. In each of these 10 variants, the first place can be occupied by any of the 4 digits $1,2,6,8$. Additionally, th... | # Answer: 49.
Answer to option: 1-2: 53.
## Solutions for option 1-2.
According to the condition, these numbers are written only with the digits $0,2,3,4,5,7$. Then, three-digit numbers that are multiples of 4 can end with exactly 9 variants: $00,04,20,24,32,40,44,52,72$. In this case, the first place in each of the... | 49 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. A parallelepiped is inscribed in a sphere of radius $\sqrt{3}$, and the volume of the parallelepiped is 8. Find the surface area of the parallelepiped. | 3. $S_{\text {full }}=24$
The translation is provided while maintaining the original text's formatting and structure. | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Ms. Olga Ivanovna, the class teacher of 5B, is staging a "Mathematical Ballet." She wants to arrange the boys and girls so that exactly 2 boys are 5 meters away from each girl. What is the maximum number of girls that can participate in the ballet if it is known that 5 boys are participating? | Answer: 20 girls.
Solution: Let's select and fix two arbitrary boys - $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. Suppose they are 5 m away from some girl - G. Then $\mathrm{M}_{1}, \mathrm{M}_{2}$, and G form an isosceles triangle with the legs being 5 m. Given the fixed positions of $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$,... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1.2 Mishа noticed that the tram passed by him in 4 seconds, and a tunnel 64 meters long in 12 seconds. Find the length of the tram (in meters), assuming that its speed remains constant throughout the entire observation period. | Answer: 32. (E)
 | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1 Find $f(2013)$, if for any real $x$ and $y$ the equality holds
$$
f(x-y)=f(x)+f(y)-2xy
$$ | Solution. Substitute $x=y=0$. We get $f(0)=2 f(0)+0$, from which we obtain that $f(0)=0$. Substitute $x=y$. We get $0=f(0)=f(x)+f(x)-2 x^{2}$. Hence, $f(x)=x^{2}$.
Answer: 4052169. (C)
 | 4052169 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-1. Calculate the sum
$$
S=\frac{2014}{2 \cdot 5}+\frac{2014}{5 \cdot 8}+\frac{2014}{8 \cdot 11}+\ldots+\frac{2014}{2012 \cdot 2015}
$$
In your answer, specify the remainder from dividing by 5 the even number closest to the obtained value of $S$. | Solution. Since
$$
\begin{aligned}
& \frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{2012 \cdot 2015}= \\
& =\frac{1}{3} \cdot\left(\left(\frac{1}{2}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\ldots\left(\frac{1}{2012}-\frac{1}{2015}\righ... | 336 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4-2. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 5 more than the other. | Answer: 13. (C)
Options.
 | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-3. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 6. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is nine times smaller than the other. | Answer: 20. (E)
Answer options.
 | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-4. A circle touches the sides of an angle at points $A$ and $B$. The distance from a point $C$ lying on the circle to the line $A B$ is 8. Find the sum of the distances from point $C$ to the sides of the angle, given that one of these distances is 30 less than the other. | Answer: 34. (B)
Answer options.
 | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6-2. From three mathematicians and ten economists, a committee of seven people needs to be formed. At the same time, the committee must include at least one mathematician. In how many ways can the committee be formed? | Solution. Direct calculation gives: $C_{13}^{7}-C_{10}^{7}=1596$.
Answer: 1596. | 1596 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6-3. From twelve students and three teachers, a school committee consisting of nine people needs to be formed. At the same time, at least one teacher must be included in it. In how many ways can the committee be formed? | Solution. Direct calculation gives: $C_{15}^{9}-C_{12}^{9}=4785$.
Answer: 4785. | 4785 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6-4. From eleven students and three teachers, a school committee consisting of eight people needs to be formed. At the same time, at least one teacher must be included in it. In how many ways can the committee be formed? | Solution. Direct calculation gives: $C_{14}^{8}-C_{11}^{8}=2838$.
Answer: 2838. | 2838 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10-1. At the base of the pyramid $S A B C D$ lies a trapezoid $A B C D$ with bases $B C$ and $A D$. Points $P_{1}$, $P_{2}, P_{3}$ belong to the side $B C$, such that $B P_{1}<B P_{2}<B P_{3}<B C$. Points $Q_{1}, Q_{2}, Q_{3}$ belong to the side $A D$, such that $A Q_{1}<A Q_{2}<A Q_{3}<A D$. Denote the points of inter... | Solution. From the properties of the trapezoid, it follows that the triangles (see Fig. 2), shaded with the same color, have the same area. From this, the equality of the sums of the areas marked
$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the s... | 999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. The test consists of 5 sections, each containing the same number of questions. Pavel answered 32 questions correctly. The percentage of his correct answers turned out to be more than 70 but less than 77. How many questions were in the test?
ANSWER: 45. | Solution: from the condition $0.7<32 / x<0.77$ it follows that $41<x<46$, but $x$ is a multiple of 5, so $x=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. How many five-digit numbers of the form $\overline{a b 16 c}$ are divisible by 16? ( $a, b, c-$ are arbitrary digits, not necessarily different).
ANSWER: 90. | Solution: Note that the first digit does not affect divisibility, hence $a=1, \ldots, 9$. On the other hand, divisibility by 8 implies that $c=0$ or 8. If $c=0$, then $\mathrm{b}$ must be even, and if $\mathrm{c}=8$ - odd. In both cases, we get 5 options, from which the total number is $9 *(5+5)=90$.
Lomonosov Moscow ... | 90 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Philatelist Andrey decided to distribute all his stamps equally into 3 envelopes, but it turned out that one stamp was extra. When he distributed them equally into 5 envelopes, 3 stamps were extra; finally, when he distributed them equally into 7 envelopes, 5 stamps remained. How many stamps does Andrey have in tota... | Solution. If the desired number is $x$, then the number $x+2$ must be divisible by 3, 5, and 7, i.e., it has the form $3 \cdot 5 \cdot 7 \cdot p$. Therefore, $x=105 p-2$. Since by the condition $150<x \leq 300$, then $p=2$. Therefore, $x=208$. | 208 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Find the smallest natural number $N$ such that the number $99 N$ consists only of threes.
ANSWER: 3367. | Solution. The number 33N must consist of all ones. A number is divisible by 33 if it is divisible by 3 and by 11. A number consisting of all ones is divisible by 3 if the number of ones is a multiple of 3, and it is divisible by 11 if the number of ones is a multiple of 2. The smallest such number is 111111, so 33N = 1... | 3367 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. (9) On a plane, there are 9 points arranged in a $3 \times 3$ grid, as shown in the figure.
a) Lines are drawn through all possible pairs of points. How many different lines are obtaine... | Answer: a) 20 ; b) 76.
Solution: a) The number of pairs of points will be $C_{9}^{2}=36$, but lines coincide when three points lie on the same line. There are $8=3$ horizontals +3 verticals +2 diagonals such cases. Therefore, subtract $36-8 \times 2=20$.
b) The number of triplets of points is $C_{9}^{3}=84$, but not ... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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