problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|
30. What is the minimum number of colors needed to color the cells of a $5 \times 5$ square so that among any three consecutive cells in a row, column, or diagonal, there are no cells of the same color?
(M. Antipov) | 30. Answer: in five colors.
Consider a cross of five cells, the central cell of which coincides with the central cell of the square. By the condition, three cells in its column have different colors. Similarly, the colors of the three cells in its row are also different. Finally, any two cells at the "ends" of the cro... | 5 | Combinatorics | 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 |
36. Many city residents engage in dancing, many in mathematics, and at least one in both. Those who engage only in dancing are exactly $p+1$ times more than those who engage only in mathematics, where $p-$ is some prime number. If you square the number of all mathematicians, you get the number of all dancers. How many ... | 36. Answer: 1 person is engaged in both dancing and mathematics.
Let $a$ people be engaged only in mathematics, and $b \geqslant 1$ people be engaged in both dancing and mathematics. Then, according to the condition, $(a+b)^{2}=(p+1) a+b$. Subtract $a+b$ from both sides: $(a+b)^{2}-(a+b)=p a$. Factor out the common te... | 1 | Number Theory | 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 |
66. The sum
$$
\begin{gathered}
2 \\
3 \cdot 6
\end{gathered}+\begin{gathered}
2 \cdot 5 \\
3 \cdot 6 \cdot 9
\end{gathered}+\ldots+\begin{gathered}
2 \cdot 5 \cdot \ldots \cdot 2015 \\
3 \cdot 6 \cdot \ldots \cdot 2019
\end{gathered}
$$
was written as a decimal fraction. Find the first digit after the decimal point. | 66. Answer: the first digit after the decimal point is 5.
To start, let's simplify the given sum. Each term can be written as a difference
$$
\begin{aligned}
\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}=\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1) \cdot(3 k+3)}{3 \cdot 6 \cdot ... | 5 | Algebra | 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 |
5. From the edge of a large square sheet, a small square was cut off, as shown in the figure, and as a result, the perimeter of the sheet increased by $10 \%$. By what percent did the area of the sheet decrease? | 5. Answer: by $4 \%$.
Let the side of the larger square be denoted by $a$, and the side of the smaller square by $b$. As a result of cutting out the smaller square from the perimeter of the sheet, one segment of length $b$ disappears, and three such segments appear instead, meaning the perimeter increases by $2 b$. Th... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Sasha went to bed at 10 PM and set the alarm clock (with hands and a 12-hour dial) for 7 AM. During the night, at some point, the alarm clock, which had been working properly, broke, and its hands started moving in the opposite direction (at the same speed). Nevertheless, the alarm rang exactly at the scheduled time... | 1. Answer: the alarm clock broke at 1 o'clock at night.
Let's imagine that the minute hand on the alarm clock is missing, and the hour hand, at the moment when the alarm clock broke, split into two halves, one of which (as stated in the condition) started moving in the opposite direction, while the other continued its... | 1 | Logic and Puzzles | 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. Solve the inequality $2021 \cdot \sqrt[202]{x^{2020}}-1 \geq 2020 x$ for $x \geq 0$. (10 points) | Solution. Transform the inequality into the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021} \leq \sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1
\end{aligned}
$$
But by the r... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation $2021 x=2022 \cdot \sqrt[2022]{x^{2021}}-1 .(10$ points $)$ | Solution. $x \geq 0$. Transform the equation to the form:
But ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Solve the equation $2021 \cdot \sqrt[202]{x^{2020}}-1=2020 x$ for $x \geq 0 \cdot$ (10 points) | Solution. Transform the equation to the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021}=\sqrt[202]{x^{2020}}, \text { from which } \\
& x+x+\ldots+x+1 \\
& \frac{(2020 \text { instances) }}{2021}=\sqrt[202]{x^{2020}}
\end{aligned}
$$
But by the relation for the arithmetic mean and the geometric mean
 | 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 |
4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing
 | 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 |
3. Solve the equation $2021 \cdot \sqrt[202]{x^{2020}}-1=2020 x$ for $x \geq 0$. (10 points) | Solution. Transform the equation to the form:
$$
\begin{aligned}
& \frac{2020 x+1}{2021}=\sqrt[202]{x^{2020}}, \text { from which } \\
& \frac{\begin{array}{l}
x+x+\ldots+x+1 \\
(2020 \text { terms) }
\end{array}}{2021}=\sqrt[202]{x^{2020}} .
\end{aligned}
$$
But by the inequality between the arithmetic mean and the ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. An athlete with a mass of 78.75 kg is testing a net used by firefighters to save people. The net sags by 100 cm when the athlete jumps from a height of 15 m. Assuming the net behaves elastically like a spring, calculate how much it will sag when a person with a mass of 45 kg jumps from a height of 29 m.
Given:
$m_... | Solution. The mechanical system "Earth-athlete-net" $x_{2}-?$ can be considered closed. According to the law of conservation of energy, when the athlete jumps, his potential energy should completely transform into the energy of the elastic deformation of the net: $m_{2} g\left(h_{2}+x_{2}\right)=\frac{k x_{2}^{2}}{2} ;... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $9^{x}+4 \cdot 3^{x+1}=13$. | Solution: $3^{2 x}+4 \cdot 3^{x+1}=13, 3^{2 x}+12 \cdot 3^{x}-13=0$,
$\left(3^{x}\right)_{1,2}=\frac{-12 \pm \sqrt{144+52}}{2}=\frac{-12 \pm 14}{2}=\left[\begin{array}{l}1, \\ -13 .- \text { n.s. }\end{array} 3^{x}=1 \Rightarrow x=0\right.$. Answer: $x=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=3 \sin x+4 \cos x$. | Solution: $f(x)=3 \sin x+4 \cos x=\sqrt{3^{2}+4^{2}} \sin \left(x+\operatorname{arctg} \frac{4}{3}\right)=5 \sin \left(x+\operatorname{arctg} \frac{4}{3}\right)$.
Answer: The maximum value is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $9^{x-1}+3^{x+2}=90$. | Solution: $9^{x-1}+3^{x+2}=90,\left(3^{x-1}\right)^{2}+27 \cdot 3^{x-1}=90$, $3^{x-1}=\frac{-27 \pm \sqrt{3^{6}+4 \cdot 3^{2} \cdot 10}}{2}=\frac{-27 \pm 3 \sqrt{121}}{2}=\frac{-27 \pm 33}{2}=\left[\begin{array}{l}3, \\ -30 .\end{array} \quad x-1=1, x=2\right.$ Answer: $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{x-3}{2 x+1}}+2=3 \sqrt{\frac{2 x+1}{x-3}}$. | Solution: $t=\frac{x-3}{2 x+1}>0 ; \sqrt{t}+2=\frac{3}{\sqrt{t}}, \sqrt{t}+2 \sqrt{t}-3=0, \sqrt{t}=\left[\begin{array}{l}1 \\ -3 \text {, but this is not valid }\end{array}\right.$
$\frac{x-3}{2 x+1}=1 ; x-3=2 x+1, x=-4, \sqrt{\frac{-7}{-7}}+2=3 \sqrt{\frac{-7}{-7}}$. Answer: $x=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{2 x+2}{x+2}}-\sqrt{\frac{x+2}{2 x+2}}=\frac{7}{12}$. | Solution: $\sqrt{\frac{2 x+2}{x+2}}-\sqrt{\frac{x+2}{2 x+2}}=\frac{7}{12} ; t=\frac{2 x+2}{x+2}>0, \sqrt{t}-\sqrt{\frac{1}{t}}=\frac{7}{12}$,
$\sqrt{t}=\frac{7 \pm \sqrt{49+4 \cdot 144}}{24}=\frac{7 \pm 25}{24}=\left[\begin{array}{l}\frac{32}{24}=\frac{4}{3} \\ -\frac{18}{24}=-\frac{3}{4}\end{array}, \frac{2 x+2}{x+2}... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find the maximum value of the function $f(x)=6 \sin x+8 \cos x$. | Solution: $f(x)=6 \sin x+8 \cos x=\sqrt{6^{2}+8^{2}} \sin \left(x+\operatorname{arctg} \frac{8}{6}\right)$.
Answer: The maximum value is 10. | 10 | Algebra | 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 |
3. Find the greatest and least values of the function $y=x^{3}-3 x^{2}+5$ on the interval $[1 ; 3]$. | Solution: $y=x^{3}-3 x^{2}+5,[1 ; 3] ; y^{\prime}=3 x^{2}-6 x=3 x(x-2) \Rightarrow x=0 ; 2$
$y(1)=3 ; y(3)=5 ; y(2)=1$. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate $\sqrt{4+\sqrt{12}}-\sqrt{4-\sqrt{12}}$. | Solution: $\sqrt{4+\sqrt{12}}-\sqrt{4-\sqrt{12}}=A ; A^{2}=8-2 \sqrt{16-12}=4 ; \quad A= \pm 2$ Answer: $A=2$. | 2 | Algebra | 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 |
4. Calculate $\sqrt{3+\sqrt{8}}-\sqrt{3-\sqrt{8}}$. | Solution: $\sqrt{3+\sqrt{8}}-\sqrt{3-\sqrt{8}}=A ; A^{2}=6-2 \sqrt{9-8}=4, \quad A= \pm 2, \quad A=2$. | 2 | Algebra | 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 |
1. Solve the equation $3 \cdot 9^{x}+2 \cdot 3^{x}=1$. | Solution: $3 \cdot 9^{x}+2 \cdot 3^{x}=1,3 \cdot\left(3^{x}\right)^{2}+2 \cdot 3^{x}-1=0,3^{x}=\frac{-2 \pm \sqrt{4+12}}{6}=\left[\begin{array}{l}\frac{1}{3}, \\ -1-\text { not valid }\end{array} \quad x=-1\right.$.
Answer: $x=-1$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Two vertices of a square with an area of $256 \mathrm{~cm}^{2}$ lie on a circle, while the other two vertices lie on a tangent to this circle. Find the radius of the circle. | Solution: $S_{A B C D}=256 \, \text{cm}^{2} \Rightarrow a=16 \, \text{cm}=x$,
$E F=2 R-x, F O=R-E F=R-(2 R-x)=x-R$,
$\triangle F C O: F O^{2}=R^{2}-F C^{2} \Rightarrow(x-R)^{2}=R^{2}-8^{2}$,
$x^{2}-2 R x+R^{2}=R^{2}-8^{2}, 2 R x=16^{2}+8^{2} \Rightarrow R=\frac{16^{2}+8^{2}}{2 \cdot 16}=10$. Answer:
$R=10$.
-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 |
6. Solve the equation $\sqrt{\frac{x+3}{3 x-5}}+1=2 \sqrt{\frac{3 x-5}{x+3}}$. | Solution:
$$
\sqrt{\frac{x+3}{3 x-5}}+1=2 \sqrt{\frac{3 x-5}{x+3}}
$$
$\frac{x+3}{3 x-5}>0$
$t=\frac{x+3}{3 x-5}, \sqrt{t}+1=\frac{2}{\sqrt{t}}, \frac{t+\sqrt{t}-2}{\sqrt{t}}=0, \sqrt{t}=\frac{-1 \pm \sqrt{1+8}}{2}=\frac{-1 \pm 3}{2}=\left[\begin{array}{l}-2 \\ 1\end{array}\right.$
$t=1, \frac{x+3}{3 x-5}=1,3 x-5=x... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The distance from the point of intersection of the diameter of a circle with a chord of length 18 cm to the center of the circle is $7 \mathrm{~cm}$. This point divides the chord in the ratio $2: 1$. Find the radius.
$$
A B=18, E O=7, A E=2 B E, R=?
$$ | Solution: $2 B E \cdot B E=(R-7)(7+R)$
$$
A E \cdot B E=D E \cdot E C, \quad A E+B E=18, \quad B E=6
$$
$$
2 \cdot 6 \cdot 6=\left(R^{2}-7^{2}\right), R^{2}=72+49=121=11^{2}
$$
Answer: $R... | 11 | Geometry | 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 |
1. Solve the equation $7^{-x}-3 \cdot 7^{1+x}=4$. | Solution: $7^{-x}-3 \cdot 7^{1+x}=4, \frac{1}{7^{x}}-21 \cdot 7^{x}=4,-21\left(7^{x}\right)^{2}-4 \cdot 7^{x}+1=0$,
$\left(7^{x}\right)=\frac{4 \pm \sqrt{16+84}}{-42}=\frac{4 \pm 10}{-42}=\left[\begin{array}{l}-\frac{14}{42}=-\frac{1}{3} \text {, not valid } \\ \frac{1}{7}\end{array}, x=-1\right.$. Answer: $x=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Calculate $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}$. | Solution: $\sqrt{6+\sqrt{20}}-\sqrt{6-\sqrt{20}}=A, A^{2}=12-2 \sqrt{36-20}=4, A= \pm \sqrt{4}=2$. Answer: 2 . | 2 | 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 |
1. Solve the equation $9 \cdot 3^{2 x-1}+3^{x}-30=0$. | Solution: $9 \cdot 3^{2 x-1}+3^{x}-30=0,3 \cdot 3^{2 x}+3^{x}-30=0$, $3^{x}=\frac{-1 \pm \sqrt{1+360}}{6}=\frac{-1 \pm 19}{6}=\left[\begin{array}{l}3 \\ -\frac{10}{3} \text {, not valid }\end{array}, x=1\right.$.
Answer: $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Solve the equation $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$. | Solution: $\sqrt{\frac{x^{2}-16}{x-3}}+\sqrt{x+3}=\frac{7}{\sqrt{x-3}}$, Domain of Definition $\frac{x^{2}-16}{x-3} \geq 0, x>3 \Rightarrow x \geq 4$,
$\sqrt{x^{2}-16}+\sqrt{x^{2}-9}=7, x^{2}-16=t, \sqrt{t}+\sqrt{t+7}=7, t+2 \sqrt{t(t+7)}+t+7=49$,
$2 \sqrt{t(t+7)}=42-2 t, \sqrt{t^{2}+7 t}=21-t, t \leq 21, t^{2}+7 t=21... | 5 | Algebra | 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 |
4. Two adjacent faces of a tetrahedron, which are isosceles right triangles with a hypotenuse of 2, form a dihedral angle of 60 degrees. The tetrahedron is rotated around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron onto the plane containing
-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 |
2. Given 50 numbers. It is known that among their pairwise products, exactly 500 are negative. Determine the number of zeros among these numbers. | Solution. Let $m, n$ and $p-$ be the number of negative, zero, and positive numbers among the given 50 numbers. Then from the condition of the problem we have: $m+n+p=50$ and $m \cdot p=500$. It follows that $m$ and $p$ are divisors of 500, the sum of which does not exceed 50. Among all divisors of the number 500, only... | 5 | Number Theory | 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 |
3. Solve the equation
$$
2(x-6)=\frac{x^{2}}{(1+\sqrt{x+1})^{2}}
$$ | Solution 1. Rewrite the equation as $2(x-6)(1+\sqrt{x+1})^{2}=x^{2}$ and make the substitution $\sqrt{x+1}=y$. Then $x=y^{2}-1$ and $y \geq 0$. After the substitution, we get $2\left(y^{2}-7\right)(y+1)^{2}=$ $\left(y^{2}-1\right)^{2}$. Factor the right side as $(y+1)^{2}(y-1)^{2}$, move it to the left, and factor out ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. How many positive numbers are there among the 2014 terms of the sequence: $\sin 1^{\circ}, \sin 10^{\circ}$, $\sin 100^{\circ}, \sin 1000^{\circ}, \ldots ?$ | Solution. For $n>3$ we have
$10^{n}-1000=10^{3}\left(10^{n-3}-1\right)=25 \cdot 40 \cdot\left(10^{n-3}-1\right)$.
Since $10^{n-3}-1$ is divisible by 9, it follows that $10^{n}-1000$ is divisible by 360. Therefore, all terms of the sequence, starting from the fourth, coincide with $\sin 1000^{\circ}=\sin \left(3 \cdot... | 3 | Number Theory | 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.
. The result was 1440 g. When Mom decorated the cake with identical candle... | Answer: 4 years.
Solution: From the condition, it follows that the cake weighed from 1436 to 1445 grams before decoration, and from 1606 to 1615 grams after decoration. Therefore, the total weight of the candles is between 161 and 179 grams. Since the weight of each candle is given as 40 grams, the weight of one candl... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 5. In class 5A, a survey was conducted on what fruits the students like. It turned out that 13 students like apples, 11 like plums, 15 like peaches, and 6 like melons. A student can like more than one fruit. Every student who likes plums also likes either apples or peaches (but not both at the same time). 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 |
Problem 5. Timofey placed 10 grid rectangles on a grid field, with areas of $1, 2, 3, \ldots, 10$ respectively. Some of the rectangles overlapped each other (possibly completely, or only partially). After this, he noticed that there is exactly one cell covered exactly once; there are exactly two cells covered exactly t... | Answer: 5 cells.
Solution. Estimation. There are a total of $1+2+\ldots+10=55$ cell coverings. The ten cells described in the condition are covered in total exactly $1 \cdot 1+2 \cdot 2+3 \cdot 3+4 \cdot 4=30$ times. There remain $55-30=25$ cell coverings. Therefore, the number of cells covered at least five times is ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) How many zeros does the number $4^{5^{6}}+6^{5^{4}}$ end with in its decimal representation?
# | # Answer: 5
Solution. Consider the number $4^{25}+6$. Let's check that it is divisible by 5 and not divisible by 25. We have
$$
\begin{gathered}
4^{25}+6 \equiv(-1)^{25}+6 \equiv 0 \quad(\bmod 5) \\
4^{25} \equiv 1024^{5}+6 \equiv(-1)^{5}+6 \equiv 5 \quad(\bmod 25)
\end{gathered}
$$
We will write $5^{t} \| c$, if $5... | 5 | Number Theory | 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 |
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 $10$, and $\cos \alpha = \frac{2}{5}$? | Answer: 8.
## 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_{... | 8 | Geometry | 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 |
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 $8$ and $\cos \alpha = \frac{3}{4}$? | Answer: 12.
## 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 $\angle A M B=\alpha$. Note that points $B_{1... | 12 | Geometry | 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 |
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