problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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Problem 4. The median $\mathrm{AA}_{0}$ of triangle $\mathrm{ABC}$ is laid off from point $\mathrm{A}_{0}$ perpendicular to side $\mathrm{BC}$ to the outside of the triangle. Denote the second endpoint of the constructed segment as $\mathrm{A}_{1}$. Similarly, points $\mathrm{B}_{1}$ and $\mathrm{C}_{1}$ are constructe... | Answer._ $\triangle \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$ is equilateral, all angles are $60^{\circ}$.
## Solution.
Since $\triangle \mathrm{ABC}$ is isosceles, $\mathrm{BB}_{0}$ is the perpendicular bisector of the base AC. Therefore, $\mathrm{B}_{1}$ lies on this perpendicular and $\mathrm{CB}_{0} \perp \m... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.5. A circle is inscribed with 103 numbers. It is known that among any five consecutive numbers, there are at least two positive numbers. What is the minimum number of positive numbers that can be among these 103 written numbers? | Answer: 42.
Solution. We will show that there will be 3 consecutive numbers, among which there are at least 2 positive ones. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them, there are at least 6 positive numbers. But th... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.2. In the bottom left corner of a $7 \times 7$ chessboard, there is a king. In one move, he can move one square to the right, or one square up, or one square diagonally to the right and up. In how many different ways can the king travel to the top right corner of the board, if he is forbidden to visit the central ce... | Answer: 5020.
Solution. We will construct a $7 \times 7$ table, in each cell of which we will write the number equal to the number of allowed paths by which the king can reach this cell from the bottom left corner. First, we fill the left column and the bottom row with ones and write 0 in the central cell (as per the ... | 5020 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the maximum value of the expression $x^{2}+y^{2}+z^{2}$, if $x, y, z$ are integers satisfying the system
$$
\left\{\begin{array}{l}
x y+x+y=20 \\
y z+z+y=6 \\
x z+x+z=2
\end{array}\right.
$$ | Solution. Adding 1 to both sides of each equation in the system, we get:
$$
\left\{\begin{array}{l}
x y + x + y + 1 = 21 \\
y z + z + y + 1 = 7 \\
x z + x + z + 1 = 3
\end{array}\right.
$$
Factoring the left-hand sides, we obtain:
$$
\left\{\begin{array}{l}
(x+1)(y+1) = 21 \\
(y+1)(z+1) = 7 \\
(z+1)(x+1) = 3
\end{ar... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.3. While Cheburashka eats two portions of ice cream, Winnie-the-Pooh manages to eat five such portions, and while Winnie-the-Pooh eats three portions, Karlson eats seven. Working together, Cheburashka and Karlson ate 82 portions. How many portions did Winnie-the-Pooh eat during this time? Justify your answer. | Solution: Suppose Winnie-the-Pooh ate exactly 15 portions of ice cream. Since $15=3 \cdot 5$, Cheburashka ate $3 \cdot 2=6$ portions, and Karlson ate $-5 \cdot 7=35$ portions. Together, Karlson and Cheburashka ate $6+35=41$ portions, which is half of the 82 portions given in the problem. Therefore, the feast actually l... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Points $O$ and $I$ are the centers of the circumcircle and incircle of triangle $ABC$, and $M$ is the midpoint of the arc $AC$ of the circumcircle (not containing $B$). It is known that $AB=15, BC=7$, and $MI=MO$. Find $AC$. | Answer: $A C=13$.
Solution. (Fig. 3). First, we prove that $M I=M A$ (trident lemma).
Indeed, the external angle $A I M$ of triangle $A I B$ is equal to the sum of angles $B A I$ and $A B I$, and since $A I$ and $B I$ are angle bisectors, $\angle A I M=\frac{1}{2} \angle A+\frac{1}{2} \angle B$. Angle $I A M$ is equa... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.3. There are 55 boxes standing in a row, numbered in order from 1 to 55. Each box contains no more than 10 balls, and the number of balls in any two adjacent boxes differs by exactly 1. It is known that the boxes numbered $1,4,7,10, \ldots, 55$ contain a total of 181 balls. What is the minimum total number of balls t... | Solution. Since the number of balls in two adjacent boxes differs by 1, the parity of this number always differs. Therefore, throughout the entire row of boxes, the parity alternates - the number of balls in boxes with different parity numbers also has different parity. We will divide the boxes into pairs: 1 with 4, 7 ... | 487 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find the smallest six-digit number that is a multiple of 11, where the sum of the first and fourth digits is equal to the sum of the second and fifth digits and is equal to the sum of the third and sixth digits. | Solution. Let the desired number have the form $\overline{1000 x y}$, where $x, y$ are some digits. Then, by the condition, the sum of the first and fourth digits is 1, from which $x=y=1$. But the number 100011 is not divisible by 11. Therefore, we will look for a number of the form $\overline{1001 x y}$. Then $x=y=2$,... | 100122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. Evgeny is tiling the floor of his living room, which measures 12 by 16 meters. He plans to place 1 m $\times$ 1 m square tiles along the perimeter of the room, and the rest of the floor will be covered with 2 m $\times$ 2 m square tiles. How many tiles will he need in total? | Solution. Let's step back 1 meter from the living room's boundaries. We will be left with a rectangle of size $10 \times 14$, which needs to be covered with $2 \times 2$ tiles. The area of the rectangle is 140, and the area of one tile is 4, so we will need $140: 4=35$ large tiles. The area of the boundary strip is $12... | 87 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.4. Daniil and Pasha were driving in a car along a straight road at a constant speed. While Daniil was driving, Pasha had nothing to do and was looking at the kilometer markers. Pasha noticed that exactly at noon they passed a marker with the number $X Y$ (where $X, Y$ are some digits), at $12:42$ - a marker with the ... | Solution. Method 1. In 42 minutes from 12:00 to 12:42, the car traveled no more than 100 km, and in the remaining 18 minutes, it traveled even less (in $42 / 18=7 / 3$) times. Therefore, by 13:00, the car had not reached the sign with the number 200. Then $X=1$. In this case, regardless of the number $Y$, the car trave... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
40. At the vertices of a convex 2020-gon, numbers are placed such that among any three consecutive vertices, there is both a vertex with the number 7 and a vertex with the number 6. On each segment connecting two vertices, the product of the numbers at these two vertices is written. Andrey calculated the sum of the num... | 40. Answer: 1010. This answer is achieved if the numbers 7 and 6 alternate.
We can assume that all numbers are equal to 6 or 7. Indeed, if a number is different from 6 and 7, then among its neighbors, one is 6 and one is 7, and the same is true for the neighbors two steps away. Then, replacing this number with 7 does ... | 1010 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. On a circle, 2021 points are marked. Kostya marks a point, then marks the next point to the right, then he marks the point to the right of the last marked point, skipping one, then the point to the right of the last marked point, skipping two, and so on. On which move will a point be marked for the second time? | Answer: on the 67th move. The question is about the smallest natural $b$ for which $(a+1)+(a+2)+\cdots+(b-1)+b$ is divisible by $2021=43 \cdot 47$. This sum is equal to $b(b+1) / 2-a(a+1) / 2=$ $(b-a)(b+a+1) / 2$ and it is not hard to see that the smallest $b=(b-a) / 2+(b+a) / 2$ is achieved with the factorization $(66... | 67 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. On the table, there are 100 weights of different masses. A weight is called successful if its mass is equal to the sum of the masses of some two other weights on the table. What is the smallest number of successful weights that would allow us to assert with certainty that the masses of some two weights differ by at ... | Answer: 87. Let there be 87 lucky numbers. If one of the two addends of a lucky number $a=b+c$ is itself lucky, for example, $b=d+e$, then $a=c+d+e \geqslant 3 \min (c, d, e)$, which means there will be two numbers differing by at least a factor of three. If, however, all the addends of the eighty-seven lucky numbers a... | 87 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
25. The sequence $\left(x_{n}\right)$ is defined by the conditions: $x_{1}=1$, and for each natural number $n$, the number $x_{n+1}$ is the largest number that can be obtained by rearranging the digits of the number $x_{n}+1$. Find the smallest $n$ for which the decimal representation of the number $x_{n}$ has exactly ... | 25. Answer: the smallest such number $n$ is $18298225=9(1+2+3+\ldots+2016)+1$.
To understand how this sequence is constructed, let's write out the first few terms. Initially, the sequence contains single-digit numbers:
$$
1, \quad 2, \quad 3, \quad \ldots, \quad 9 .
$$
Next come the two-digit numbers:
$. Clearly, this difference will be a composite number. | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
34. In the gym, 200 schoolchildren gathered. Every pair of acquaintances shook hands. It turned out that any two strangers made at least 200 handshakes in total. Prove that there were at least 10000 handshakes in total. | 34. Let's write the solution in the language of graphs. If the degree of a vertex is not less than 100, we call it rich; if it is less than 100, we call it poor. From the condition, it follows that all poor vertices are pairwise adjacent to each other.
We will prove that each poor vertex can be paired with a non-adjac... | 10000 | Combinatorics | proof | Yes | Yes | olympiads | false |
6. Find the minimum value of the expression $\left[\frac{7(a+b)}{c}\right]+\left[\frac{7(a+c)}{b}\right]+\left[\frac{7(b+c)}{a}\right]$, where $a, b$ and $c$ are arbitrary natural numbers. | Answer: 40. Evaluation: the sum of the fractions under the integer parts is no less than $14 \cdot 3=42$ (we add three inequalities of the form $7(a / b+b / a) \geqslant 14$, and each integer part is greater than the fraction reduced by 1. Therefore, the desired value is greater than $42-3=39$, i.e., not less than 40. ... | 40 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. The cells of a $100 \times 100$ table are painted white. In one move, it is allowed to select any 99 cells from one row or one column and repaint each of them to the opposite color - from white to black, and from black to white. What is the minimum number of moves required to obtain a table with a checkerboard patte... | Answer: in 100 moves. Evaluation: to repaint each of the cells on the black diagonal, a move is required. Example: repaint all rows and columns with odd numbers. In this case, in the $k$-th row and $u$-th column, we will not repaint their common cell. It is easy to see that as a result, we will get a chessboard colorin... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. A $10 \times 10$ table is filled with numbers from 1 to 100: in the first row, the numbers from 1 to 10 are written in ascending order from left to right; in the second row, the numbers from 11 to 20 are written in the same way, and so on; in the last row, the numbers from 91 to 100 are written from left to right. C... | 1. Answer: Yes. If $x$ is in the center of the fragment, then the sum of the numbers in it is $7x$ and then with 646566 $x=65$ the sum will be exactly 455 (see figure). | 65 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
34. A company gathered for a meeting. Let's call a person sociable if in this company they have at least 20 acquaintances, and at least two of them are acquainted with each other. Let's call a person shy if in this company they have at least 20 strangers, and at least two of them are strangers to each other. It turned ... | 34. Answer: 40.
Evaluation: Suppose that a certain person (let's call him Kostya) is acquainted with at least 20 people. If some of Kostya's acquaintances know each other, then Kostya is sociable. Let's consider the case where all of Kostya's acquaintances do not know each other. In this case, if Kostya is acquainted ... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
50. In the city, 2019 metro stations have been built. Some pairs of stations are connected by tunnels, and from any station, you can reach any other station via the tunnels. The mayor ordered the organization of several metro lines: each line must include several different stations, sequentially connected by tunnels (t... | 50. Answer: $k=1008$. Let's provide an example of a connected graph with 2019 vertices that cannot be covered by 1008 simple paths: one vertex is connected to 2018 vertices of degree 1. Any simple path contains no more than two pendant vertices, so at least 1009 paths are required for coverage.
We will prove that 1009... | 1008 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
55. In a $10 \times 10$ grid (the sides of the cells have a unit length), $n$ cells were chosen, and in each of them, one of the diagonals was drawn and an arrow was placed on this diagonal in one of two directions. It turned out that for any two arrows, either the end of one coincides with the beginning of the other, ... | 55. Answer: when $n=48$.
For each arrow, consider the three-cell corner obtained by removing from a $2 \times 2$ square, centered at the end of the arrow, a $1 \times 1$ square, the diagonal of which is this arrow. Note that such corners do not intersect and are contained within a $12 \times 12$ square. Therefore, the... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
46. In a convex quadrilateral $A B C D$, point $M$ is the midpoint of side $A D$, $C M \| A B$, $A D=B D$ and $3 \angle B A C=\angle A C D$. Find the angle $A C B$.
(S. Berlov) | 46. Answer: $\angle A C B=90^{\circ}$.
Notice that $\angle B A C=\angle A C M$, hence $\angle D C M=\angle A C D-\angle A C M=2 \angle A C M$. Let $N$ be the midpoint of segment $A B$. Then $D N$ is the median, bisector, and altitude of the isosceles triangle $A B D$. Next, point $C$ lies on the midline of this triang... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
31. Around a round table, 300 people are sitting: some of them are knights, and the rest are liars. Anton asked each of them: "How many liars are among your neighbors?" and added up the numbers he received. Then Anya did the same. When answering the question, knights always tell the truth, while liars always lie, but t... | 31. Answer: there are 200 liars at the table.
It is clear that the knights gave two identical answers, so the difference between the sums of Anton and Anya could have arisen because some liars gave Anton and Anya different answers. At the same time, the answers of any liar differ by only 1 or 2. The total difference o... | 200 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Inside an isosceles triangle $A B C (A B=A C)$, a point $K$ is marked. Point $L$ is the midpoint of segment $B K$. It turns out that $\angle A K B=\angle A L C=90^{\circ}, A K=C L$. Find the angles of triangle $A B C$. | Answer: the triangle is equilateral. Note that $\triangle A L C=\triangle B K A$ by the leg and hypotenuse. Therefore, $A L=B K=2 K L$, so $\angle K A L=30$. In addition, the sum of angles $B A K$ and $C A L$ in these triangles is 90, so $\angle B A C=90-30-60$. | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. The figure shows a city plan. Nodes are intersections, and the 35 lines connecting them are streets. There are $N$ buses operating on these streets. All buses start simultaneously at intersections and move to adjacent intersections along the streets every minute. Each bus follows a closed, non-self-intersecting rout... | Answer: 35 (by the number of streets), i.e., it is possible to launch the minibuses so that at any moment in time, exactly one minibus is moving along each street. Obviously, $N$ cannot be greater than 35, because otherwise, in the first minute, some two minibuses will inevitably end up on the same street. Let's provid... | 35 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
35. In the cells of a $9 \times 9$ square, non-negative numbers are placed. The sum of the numbers in any two adjacent rows is at least 20, and the sum of the numbers in any two adjacent columns does not exceed 16. What can the sum of the numbers in the entire table be?
(A. Chukhnov) | 35. The total sum is no less than $4 \cdot 20=80$ (since the entire table is divided into 4 pairs of columns) and does not exceed $5 \cdot 16=80$ (since the table is covered by five pairs of rows with an overlap at the 8th row), therefore it is equal to 80. | 80 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
64. A nearsighted rook attacks all the cells in its row and column that can be reached in no more than 60 steps, moving from cell to adjacent cell by side. What is the maximum number of non-attacking nearsighted rooks that can be placed on a $100 \times 100$ square? | 64. Answer: 178 myopic rooks.
Evaluation. Divide the $100 \times 100$ square into a central $22 \times 22$ square and $4 \cdot 39=156$ rectangles of $1 \times 61$. In each rectangle of the partition and in each row of the $22 \times 22$ square, no more than one rook can be placed, so there are no more than 178 rooks.
... | 178 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
13. Leshа wrote on the board in ascending order all natural divisors of a natural number $n$, and Dima erased several first and several last numbers of the resulting sequence, leaving 151 numbers. What is the maximum number of these 151 divisors that could be fifth powers of natural numbers?
(M. Achtipov) | 13. Answer: 31.
Lemma. If $n$ is divisible by $a^{5}$ and $b^{5}$, then $n$ is also divisible by $a^{4} b, a^{3} b^{2}, a^{2} b^{3}, a b^{4}$.
Proof of the lemma. Note that $n^{5}=n^{4} \cdot n$ is divisible by $a^{20} \cdot b^{5}$; taking the fifth root, we get that $n$ is divisible by $a^{4} b$. Similarly, $n^{5}=n... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
60. In a class, there are 25 students. The teacher wants to stock $N$ candies, conduct an olympiad, and distribute all $N$ candies for success in it (students who solve the same number of problems should receive the same number of candies, those who solve fewer should receive fewer, including possibly zero candies). Wh... | 60. Answer: $600=25 \cdot 24$ candies.
Let's show that a smaller number of candies might not be enough. If all participants solved the same number of problems, the number of candies must be a multiple of 25. Let $N=25 k$. Imagine that 24 people solved the same number of problems, while the 25th solved fewer. If each o... | 600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
47. Inside an equilateral triangle $A B C$, points $P$ and $Q$ are chosen such that $P$ is inside triangle $A Q B$, $P Q = Q C$, and $\angle P A Q = \angle P B Q = 30$. Find $\angle A Q B$. | 47. We construct regular triangles $A P X$ and $B P Y$ on the sides $A P$ and $P B$ of triangle $A P B$. Triangle $A X C$ is obtained by rotating triangle $A P B$ by $60^{\circ}$ around point $A$, so these triangles are equal. Therefore, $X C = P B = P Y$. Similarly, $C Y = A P = X P$. Consequently, quadrilateral $P X ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of all natural numbers in which each subsequent digit is less than the previous one. (8 points) | Solution. The largest possible number that satisfies the condition of the problem is 9876543210. In addition, the number must be at least two digits. All other such numbers can be obtained from 9876543210 by deleting one, two, three, four, five, six, seven, or eight digits out of ten. Then the total number
$$
\begin{g... | 1013 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,14, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points $)$ | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equali... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the sum of all four-digit numbers in which only the digits $1,2,3,4,5$ appear, and each digit appears no more than once. (8 points) | Solution. Any of these digits appears in any place as many times as the remaining four digits can be distributed among the remaining three places. This number is $4 \cdot 3 \cdot 2=24$. Therefore, the sum of the digits in each of the four places, taken over all four-digit numbers satisfying the conditions of the proble... | 399960 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points $)$ | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equality ... | 3495250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the number of all natural numbers in which each subsequent digit is greater than the previous one. (8 points) | Solution. The largest possible number satisfying the condition of the problem is 123456789. Moreover, the number must be at least two digits. All other such numbers can be obtained from 123456789 by erasing one, two, three, four, five, six, or seven digits out of nine. Then the total number
$1+C_{9}^{1}+C_{9}^{2}+C_{9... | 502 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both among the members of the arithmetic progression $\{5,8,11,13, \ldots\}$, and among the members of the geometric progression $\{20,40,80,160, \ldots\} \cdot(10$ points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80,160, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For common elements, the equality ... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\} \cdot(10$ points $)$ | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For common elements, the equation $4+3 n=... | 13981000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,14, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equati... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{10,20,40,80, \ldots\} \cdot(10$ points) | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{10,20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=10 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equality ... | 3495250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are common to both the arithmetic progression $\{5,8,11,13, \ldots\}$ and the geometric progression $\{20,40,80,160, \ldots\}$. (10 points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80,160, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equat... | 6990500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{4,7,10,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\} .(10$ points $)$ | Solution. The members of the arithmetic progression $\{4,7,10,13,16,19, \ldots\}$ are given by the formula
$$
a_{n}=4+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $(20,40,80, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For common elements, the equality $4+3 n=2... | 13981000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. For a sports parade, the coach decided to line up the children in rows of 8, but 5 children were left over. Then he lined them up in rows of 10, but 3 places were left empty. It is known that there were no fewer than 100 and no more than 150 children. How many children were there? | Solution: $a=8 x+5=8(x+1)-3, a+3 \vdots 8$,
$$
\begin{array}{lc}
a=8 x+5=8(x+1)-3, & a+3 \vdots 8 \\
a=10 y-3, & a+3 \vdots 10,
\end{array} \Rightarrow a+3 \vdots 40, a+3=\underline{120,160}
$$
Answer: 117.
TICKET № 4 | 117 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. In the packaging workshop, there are 2 types of boxes: one for 20 parts and one for 27 parts. A batch of parts between 500 and 600 pieces has arrived for packaging. When the parts are packed in the first type of box, 13 parts are left unpacked, and when packed in the second type of box, 7 spaces are left unfilled. ... | Solution: ${ }^{a=20 x+13=20(x+1)-7,} a+7 \vdots 20 ; a+7 \vdots 27, \Rightarrow a+7 \vdots 540$.
$$
a=27 x-7
$$
Answer: $a=533$.
## TICKET № 5 | 533 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=5 \sin x+12 \cos x$. | Solution: $f(x)=5 \sin x+12 \cos x=\sqrt{5^{2}+12^{2}}\left(\sin \left(x+\operatorname{arctg} \frac{12}{5}\right)\right)=\sqrt{13^{2} \sin x}$.
Answer: 13 . | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. A florist received between 300 and 400 roses for a celebration. When he arranged them in vases with 21 roses in each, 13 roses were left. But when arranging them in vases with 15 roses in each, 8 roses were missing. How many roses were there in total? | Solution: $\left\{\begin{array}{ll}a=21 x+13=21(x+1)-8, & a+8 \vdots 21, \\ a=15 y-8, & a+8 \vdots 15,\end{array}\right\} \Rightarrow a+8 \vdots 105$. Answer: $a=307$.
$$
a+8=105,210, \underline{315,} 420
$$
## TICKET № 6 | 307 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Koschei is counting his gold coins. When he counts them by tens, there are 7 coins left, and he is 3 coins short of a whole number of dozens. Koschei's wealth is estimated at $300-400$ coins. How many coins does Koschei have? | Solution: $\left\{\begin{array}{ll}a=10 x+7=10(x+1)-3, & a+3 \mid 10, \\ a=12 y-3, & a+3 \mid 12,\end{array}\right\} \Rightarrow a+3 \mid 60$. Answer: $a=357$.
$$
a+3=300,360,420
$$
## TICKET № 7 | 357 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the eleventh term of the arithmetic progression if the sum of the first seven terms $S_{7}=77$, and the first term $a_{1}=5$. | Solution: $S_{7}=77 ; a_{1}=5, a_{11}=$ ?
$S_{7}=\frac{2 a_{1}+d(7-1)}{2} \cdot 7=\frac{10+6 d}{2} \cdot 7=(5+3 d) 7=77 \Rightarrow d=2, a_{11}=a_{1}+10 d=25$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=9 \sin x+12 \cos x$. | Solution: $f(x)=9 \sin x+12 \cos x=\sqrt{9^{2}+12^{2}} \sin \left(x-\operatorname{arctg} \frac{12}{9}\right)$. Answer: 15 . | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=8 \sin x+15 \cos x$. | Solution: $f(x)=8 \sin x+15 \cos x=\sqrt{8^{2}+15^{2}} \sin \left(x+\operatorname{arctg} \frac{15}{8}\right)=17 \sin \left(x+\operatorname{arctg} \frac{15}{8}\right)$
Answer: The maximum value of $f(x)=17$. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Between 200 and 300 children enrolled in the first grade of school. It was decided to form classes of 25 students, but it turned out that ten would not have a place. Then they formed classes of 30 students, but in one of the classes, there were 15 fewer students. How many children enrolled in the first grade? | Solution: $\left\{\begin{array}{ll}a=25 R+10=25(R+1)-15, & a+15: 25, \\ a=30 l-15, & a+15: 30,\end{array}\right\} \Rightarrow a+15 \vdots 150$. Answer: $a=285$.
$$
a+15=150, \underline{300,} 450
$$ | 285 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. Two vertices of a square lie on a circle with a radius of $5 \mathrm{~cm}$, while the other two lie on a tangent to this circle. Find the area of the square. | Solution: $R=5, A B C D$-square, $S_{A B C D}=$ ?
$A B=x, E F=2 R-x, E O=R-E F=x-R$
$\Delta E O C: E C^{2}=O C^{2}-E O^{2}=R^{2}-(x-R)^{2} \Rightarrow\left(\frac{x}{2}\right)^{2}=R^{2}-x^{2}+2 R x-R^{2}$,
-5, & a+5 \vdots: 12, \\ a=25 l-5, & a+5 \vdots 25,\end{array}\right\} a+5: 300$. Answer: $a=595$.
$$
a+5=300, \underline{600,} 900
$$ | 595 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. To pack books when moving a school library, you can buy small boxes that hold 12 books or large ones that hold 25 books. If all the books are packed in small boxes, 7 books will remain, and if all the books are packed in large boxes, there will be room for 5 more books. The library's collection contains between 500... | Solution: $\left\{\begin{array}{ll}a=12 R+7=12(R+1)-5, & a+5 \vdots: 12, \\ a=25 l-5, & a+5 \vdots 25,\end{array}\right\} a+5: 300$. Answer: $a=595$.
$$
a+5=300, \underline{600,} 900
$$ | 595 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of the first 10 elements that are found both in the arithmetic progression $\{5,8,11,13, \ldots\}$ and in the geometric progression $\{20,40,80,160, \ldots\}$. (10 points) | Solution. The members of the arithmetic progression $\{5,8,11,14,17,20,23, \ldots\}$ are given by the formula
$$
a_{n}=5+3 n, n=0,1,2, \ldots
$$
The members of the geometric progression $\{20,40,80,160, \ldots\}$ are given by the formula
$$
b_{n}=20 \cdot 2^{k}, k=0,1,2, \ldots
$$
For the common elements, the equat... | 6990500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest possible value of the function
$$
f(x)=|x+1|+|x+2|+\ldots+|x+100|
$$
$(25$ points. $)$ | Answer: 2500.
Using the known inequality $|a|+|b| \geqslant|a-b|$, then $|x+k|+|x+m| \geqslant|k-m|$. Grouping in the original sum the terms equally distant from the ends, we get
$$
\begin{aligned}
f(x) & =(|x+1|+|x+100|)+(|x+2|+|x+99|)+\ldots+(|x+50|+|x+51|) \geqslant \\
& \geqslant(100-1)+(99-2)+\ldots+(51-50)
\end... | 2500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.
The answer to the task should be an integer. Enter all characters without spaces. Do not indicate units of measurement.
Doughnut eats a cake in 5 minutes, and Nosy eats it in 7 minutes. How many seconds will it take for the boys to eat the cake together if they do not conflict? | # 1. /2 points/
The answer to the task should be an integer. Enter all characters without spaces. Do not indicate units of measurement.
Doughnut eats a cake in 5 minutes, and Nosy eats it in 7 minutes. How many seconds will it take for the boys to eat the cake together if they do not conflict
Answer: 175
# | 175 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 3.
The answer to the task should be an integer. Enter all characters without spaces.
Find $n$, if $9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}=3^{2012}$. | # 3. /2 points/
The answer to the task should be an integer. Enter all characters without spaces.
Find $n$, if $9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}+9^{n}=3^{2012}$.
Answer: 1005
# | 1005 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 7.
Among the statements given below, select the correct ones. In your answer, write down the numbers of these statements (in ascending order, without using spaces, commas, or other separators).
Example of answer format: 12345
1) Among the numbers $123,365,293,18$, exactly three numbers are divisible by 3.
2) If th... | # 7. /3 points/
Among the statements given below, select the correct ones. In your answer, write down the numbers of these statements (in ascending order, without using spaces, commas, or other separators). Example of answer format: $\quad 12345$
1) Among the numbers $123,365,293,18$, exactly three numbers are divisi... | 2456 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 8.
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (comma, digits) without spaces. There is no need to indicate units of measurement.
The base and the side of the triangle are 30 and ... | # 8. /3 points/
The answer to the problem should be a certain integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (comma, digits) without spaces. Do not specify units of measurement.
The base and the side of the triangle are 30... | 168 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 9.
The answer to the task should be some integer or a number written as a finite decimal fraction. If the answer contains a fractional number, use a comma when writing it. Enter all characters (minus sign, comma, digits) without spaces.
Find all distinct values of the parameter $p$, for each of which the equation g... | # 9. /3 points/
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (minus sign, comma, digits) without spaces.
Find all distinct values of the parameter $p$ for which the equation given be... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 2. What is the smallest natural number that is divisible by 2022 and whose notation starts with 2023? | Answer: 20230110.
Solution. Let the number we are looking for have $n+4$ digits, then it has the form $2023 \cdot 10^{n}+a, a<10^{n}$. Subtract from it $2022 \cdot 10^{n}$, we get that $b=10^{n}+a$ is also divisible by 2022. That is, we need to find a number that starts with 1 and is divisible by 2022. Numbers divisib... | 20230110 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. (3 points)
Find the smallest natural number that starts with the digit 3 and which, after erasing this digit, decreases by 25 times. | Answer: 3125
## Interregional Subject Olympiad KFU 2013-2014, MATHEMATICS 9th grade, 2nd variant, Internet round
The answer in each task should be an integer or a number written as a finite decimal. If the answer contains a fractional number, then when writing it, use a comma. All symbols (minus sign, comma, digits) ... | 3125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. In the club, 36 schoolchildren are attending. If any 33 of them come to the lesson, then girls will always be more than half. But if 31 students come to the lesson, it may turn out that more than half of them are boys. How many girls are attending the club? (20 points) | Answer: 20.
Solution: Since among any 33 children, there are more girls than half, then boys among them are less than half, that is, no more than 16. Therefore, there are no more than 16 boys in total, because if there were at least 17, it could happen that 17 boys and another 16 students came, and the boys would alre... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 3. Sasha chose five numbers from the numbers 1, 2, 3, 4, 5, 6, and 7 and told Anna their product. Based on this information, Anna realized that she could not uniquely determine the parity of the sum of the numbers chosen by Sasha. What number did Sasha tell Anna? (20 points) | Answer: 420.
Solution: Let's look at the two remaining numbers. Since Anya knows the sum of all numbers from 1 to 7 (which is 28), these two remaining numbers are such that their product cannot determine the parity of their sum. Therefore, their product can be represented in two ways: \( ab = xy \), where numbers \( a... | 420 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 10. $/ 3$ points/
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all characters (comma, digits) without spaces. There is no need to specify units of measurement.
The bisector of angle $N$ of triang... | Answer: 70
The answer to the task should be some integer or a number written as a finite decimal. If the answer contains a fractional number, use a comma when writing it. Enter all symbols (minus sign, comma, digits) without spaces.
Find all different values of the parameter $p$ for which the equation given below has... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $ABC$, two altitudes $AK$ and $CL$ are drawn. Find the measure of angle $B$, given that $AC = 2 \cdot LK$. | Solution. Construct a circle on side $AC$ as its diameter, which will pass through points $L$ and $K$, since $\angle ALC = \angle AKC = 90^{\circ}$. According to the condition, $AC = 2 \cdot LK$, and thus, the segment $LK$ is equal to the radius of the constructed circle, so the arc subtended by the chord $LK$ is $60^{... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can 8 identical rooks be placed on an $8 \times 8$ board symmetrically with respect to the diagonal passing through the bottom-left corner? | Solution. On an $8 \times 8$ board, there are 8 diagonal and 56 non-diagonal cells, the latter of which can be divided into 28 pairs of cells symmetric relative to the diagonal. All placements of rooks will be divided into 5 non-overlapping classes - in the $m$-th class, we will include placements where $m$ pairs of ro... | 139448 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Let's call a year interesting if a person turns as many years old as the sum of the digits of their birth year in that year. A certain year turned out to be interesting for Ivan, who was born in the 20th century, and for Vovochka, who was born in the 21st century. What is the difference in their ages?
Note. Fo... | Answer: 18 years.
Solution. Let Ivan's year of birth be $\overline{19 x y}$, and Vovochka's $-\overline{20 z t}$. An interesting year for Ivan will be $1900+10 x+y+10+x+y$, and for Vovochka $-2000+10 z+t+2+z+t$, according to the condition these values are equal, that is, $2002+11 z+2 t=1910+11 x+2 y$, from which $11(x... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Given a triangle $ABC$ with sides $AB=5$; $BC=8$. The bisector of angle $B$ intersects the circumcircle of the triangle at point $D$. a) It is known that the area of triangle $ABD$ is 10. Find the area of triangle $BCD$. b) Can it be that $S_{ABD}$ equals 100? | Solution. a) The fact that point $D$ lies on the circle is irrelevant. The areas of triangles $A B O$ and $O B C$ are in the ratio $A O: O C$. The same applies to triangles $A O D$ and $D O C$. Therefore,
$$
S_{A B D}: S_{B C D}=A O: O C=A B: B C=5: 8
$$
by the property of the angle bisector.
. And ever... | Answer: 22 people.
Solution. Estimate. Since 15 people like peaches and 6 people like melons, there are at least 9 people who like peaches but do not like melons. Then all these people like plums but do not like apples. Therefore, in addition to these people, there are at least 13 people who like apples. Thus, the tot... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 1. Masha has a piggy bank where she puts a 50 or 100 ruble bill every week. At the end of every 4 weeks, she picks the bill of the smallest denomination from the piggy bank and gives it to her little sister. After a year, it turned out that she had given her sister 1250 rubles. What is the minimum amount of money ... | Answer: 3750 rubles.
Solution. Let's call a 4-week period a "month", there are 13 such "months" in a year. If all the bills given by Masha were hundred-ruble bills, her sister would have received 1300 rubles. This means Masha gave 100 rubles twelve times and 50 rubles once. If in any "month" Masha gave 100 rubles, it ... | 3750 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 2. Amir is 8 kg heavier than Ilnur, and Daniyar is 4 kg heavier than Bulat. The sum of the weights of the heaviest and lightest boys is 2 kg less than the sum of the weights of the other two. The total weight of all four is 250 kg. How many kilograms does Amir weigh? (20 points) | Answer: 66 kg.
Solution. Let the weight of Bulat in kilograms be denoted by $x$, and the weight of Ilnur by $y$. Then Amir weighs $y+8$, and Daniyar weighs $-x+4$. Note that the heaviest is either Amir or Daniyar, and the lightest is either Ilnur or Bulat.
If the heaviest is Daniyar, then $x+4>y+8$, so $x>y$ and the ... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2018
$$ | Answer: 4076360
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2018$. From which $n=(2018+1)^{2}-1=... | 4076360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) On Valentine's Day, every student in the school gave a valentine to every female student. It turned out that the number of valentines was 16 more than the total number of students. How many valentines were given? | Answer: 36
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the condition, $x y=x+y+16$. Then $(x-1)(y-1)=17$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 17. The number of valentines is $2 \cdot 18=36$. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 5 blue, 6 red, and 7 white bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 365904
Solution: First, arrange all the blue and red bulbs in $C_{11}^{5}$ ways. In the gaps between them and at the ends, choose 7 positions and insert the white bulbs. There are $C_{12}^{7}$ ways to do this. In total, there are $C_{11}^{5} \cdot C_{12}^{7}$ ways to compose the garland from the available bulb... | 365904 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-4.5 ; 4.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 90
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-10 \leqslant y-1 \leqslant 8$ and $-7 \leqslant 2-x \leqslant 11$. Therefore, $(y-1)(2-x)+2 \leqslant 8 \cdot 11+2=90$. The m... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{7})^{205}$ using the binomial theorem, we obtain terms of the form $C_{205}^{k}(\sqrt{7})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 149
Solution: The ratio of two consecutive terms $\frac{C_{205}^{k+1}(\sqrt{7})^{k+1}}{C_{205}^{k}(\sqrt{7})^{k}}$ is greater than 1 when $k<$ $\frac{205 \sqrt{7}-1}{\sqrt{7}+1}$. Then the terms increase up to $\left[\frac{205 \sqrt{7}-1}{\sqrt{7}+1}\right]+1$, and then decrease. | 149 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 25 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 25 minutes? | Answer: 300.
Solution: Let's represent 25 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=18, B C=12 \sqrt{3}-9$. | Answer: 30
## Solution:
The quadrilateral is inscribed, hence $\angle A+\angle C=180^{\circ}$. From the given ratio, $\angle A=2 x, \angle C=4 x$. Therefore, $x=30^{\circ}$ and $\angle A=60... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2017
$$ | Answer: 4072323
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2017$. From which $n=(2017+1)^{2}-1=... | 4072323 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) On Valentine's Day, every student in the school gave each female student a valentine. It turned out that the number of valentines was 18 more than the total number of students. How many valentines were given? | Answer: 40
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the condition, $x y=x+y+18$. Then $(x-1)(y-1)=19$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 19. The number of valentines is $2 \cdot 20=40$. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 7 blue, 6 red, and 10 white light bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 1717716
Solution: First, arrange all the blue and red bulbs in $C_{13}^{7}$ ways. In the gaps between them and at the ends, choose 10 positions and insert the white bulbs. There are $C_{14}^{10}$ ways to do this. In total, there are $C_{13}^{7} \cdot C_{14}^{10}$ ways to compose the garland from the available ... | 1717716 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-5.5 ; 5.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 132
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-12 \leqslant y-1 \leqslant 10$ and $-9 \leqslant 2-x \leqslant 13$. Therefore, $(y-1)(2-x)+2 \leqslant 10 \cdot 13+2=132$... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{5})^{206}$ using the binomial theorem, we obtain terms of the form $C_{206}^{k}(\sqrt{5})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 143
Solution: The ratio of two consecutive terms $\frac{C_{206}^{k+1}(\sqrt{5})^{k+1}}{C_{206}^{k}(\sqrt{5})^{k}}$ is greater than 1 when $k<$ $\frac{206 \sqrt{5}-1}{\sqrt{5}+1}$. Then the terms increase up to $\left[\frac{206 \sqrt{5}-1}{\sqrt{5}+1}\right]+1$, and then decrease. | 143 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, there are 26 ones. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 26 minutes? | Answer: 325.
Solution: Let's represent 26 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 325 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=20, B C=24 \sqrt{3}-10$. | Answer: 52
## Solution:
The quadrilateral is inscribed, so $\angle A + \angle C = 180^{\circ}$. From the given ratio, $\angle A = 2x, \angle C = 4x$. Therefore, $x = 30^{\circ}$ and $\angle... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2016
$$ | Answer: 4068288
Solution: Note that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2016$. From which $n=(2016+1)^{2}-1=40... | 4068288 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) On Valentine's Day, every student in the school gave a valentine to every female student. It turned out that the number of valentines was 22 more than the total number of students. How many valentines were given? | Answer: 48
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+22$. Then $(x-1)(y-1)=23$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 23. The number of valentines is $2 \cdot 24=48$. | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 8 blue, 8 red, and 11 white bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 159279120
Solution: First, arrange all the blue and red bulbs in $C_{16}^{8}$ ways. In the gaps between them and at the ends, choose 11 positions and insert the white bulbs. There are $C_{17}^{11}$ ways to do this. In total, there are $C_{16}^{8} \cdot C_{17}^{11}$ ways to compose the garland from the availabl... | 159279120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-6.5 ; 6.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 182
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-14 \leqslant y-1 \leqslant 12$ and $-11 \leqslant 2-x \leqslant 15$. Therefore, $(y-1)(2-x)+2 \leqslant 12 \cdot 15+2=182... | 182 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{7})^{207}$ using the binomial theorem, we obtain terms of the form $C_{207}^{k}(\sqrt{7})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 150
Solution: The ratio of two consecutive terms $\frac{C_{207}^{k+1}(\sqrt{7})^{k+1}}{C_{207}^{k}(\sqrt{7})^{k}}$ is greater than 1 when $k<$ $\frac{207 \sqrt{7}-1}{\sqrt{7}+1}$. Then the terms increase up to $\left[\frac{207 \sqrt{7}-1}{\sqrt{7}+1}\right]+1$, and then decrease. | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 27 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 27 minutes? | Answer: 351.
Solution: Let's represent 27 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line se... | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=16, B C=15 \sqrt{3}-8$. | Answer: 34
## Solution:
The quadrilateral is inscribed, hence $\angle A+\angle C=180^{\circ}$. From the given ratio, $\angle A=2 x, \angle C=4 x$. Therefore, $x=30^{\circ}$ and $\angle A=60... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2015
$$ | Answer: 4064255
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2015$. From which $n=(2015+1)^{2}-1=... | 4064255 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) On Valentine's Day, every student in the school gave each female student a valentine. It turned out that the number of valentines was 28 more than the total number of students. How many valentines were given? | Answer: 60
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the condition, $x y=x+y+28$. Then $(x-1)(y-1)=29$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 29. The number of valentines is $2 \cdot 30=60$. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 8 blue, 7 red, and 12 white light bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 11711700
Solution: First, arrange all the blue and red bulbs in $C_{15}^{8}$ ways. In the gaps between them and at the ends, choose 12 positions and insert the white bulbs. There are $C_{16}^{12}$ ways to do this. In total, there are $C_{15}^{8} \cdot C_{16}^{12}$ ways to compose the garland from the available... | 11711700 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-7.5 ; 7.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 240
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-16 \leqslant y-1 \leqslant 14$ and $-13 \leqslant 2-x \leqslant 17$. Therefore, $(y-1)(2-x)+2 \leqslant 14 \cdot 17+2=240... | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\cos \alpha = \frac{5}{6}$? | Answer: 20.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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