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6. (8 points) By expanding the expression $(1+\sqrt{11})^{208}$ using the binomial theorem, we obtain terms of the form $C_{208}^{k}(\sqrt{11})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 160
Solution: The ratio of two consecutive terms $\frac{C_{208}^{k+1}(\sqrt{11})^{k+1}}{C_{208}^{k}(\sqrt{11})^{k}}$ is greater than 1 when $k<$ $\frac{208 \sqrt{11}-1}{\sqrt{11}+1}$. Then the terms increase up to $\left[\frac{208 \sqrt{11}-1}{\sqrt{11}+1}\right]+1$, and then decrease. | 160 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 28 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 28 minutes? | Answer: 378.
Solution: Let's represent 28 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... | 378 | 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 $C D=8, B C=7.5 \sqrt{3}-4$. | Answer: 17
## 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... | 17 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2014
$$ | Answer: 4060224
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2014$. From which $n=(2014+1)^{2}-1=... | 4060224 | 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 30 more than the total number of students. How many valentines were given? | Answer: 64
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+30$. Then $(x-1)(y-1)=31$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 31. The number of valentines is $2 \cdot 32=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 9 blue, 7 red, and 14 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: 7779200
Solution: First, arrange all the blue and red bulbs in $C_{16}^{9}$ ways. In the gaps between them and at the ends, choose 14 positions and insert the white bulbs. There are $C_{17}^{14}$ ways to do this. In total, there are $C_{16}^{9} \cdot C_{17}^{14}$ ways to compose the garland from the available ... | 7779200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-8.5 ; 8.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 306
Solution: Notice 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 $-18 \leqslant y-1 \leqslant 16$ and $-15 \leqslant 2-x \leqslant 19$. Therefore, $(y-1)(2-x)+2 \leqslant 16 \cdot 19+2=3... | 306 | 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 $\cos \alpha = \frac{4}{5}$? | Answer: 16.
## 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_... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{5})^{209}$ using the binomial theorem, we obtain terms of the form $C_{209}^{k}(\sqrt{5})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 145
Solution: The ratio of two consecutive terms $\frac{C_{209}^{k+1}(\sqrt{5})^{k+1}}{C_{209}^{k}(\sqrt{5})^{k}}$ is greater than 1 when $k<$ $\frac{209 \sqrt{5}-1}{\sqrt{5}+1}$. Then the terms increase up to $\left[\frac{209 \sqrt{5}-1}{\sqrt{5}+1}\right]+1$, and then decrease. | 145 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 29 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 29 minutes? | Answer: 406.
Solution: Let's represent 29 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... | 406 | Number Theory | 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 $C D=15, B C=18 \sqrt{3}-7.5$. | Answer: 39
## 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... | 39 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2013
$$ | Answer: 4056195
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2013$. From which $n=(2013+1)^{2}-1=... | 4056195 | 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 36 more than the total number of students. How many valentines were given? | Answer: 76
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+36$. Then $(x-1)(y-1)=37$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 37. The number of valentines is $2 \cdot 38=76$. | 76 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 6 blue, 7 red, and 9 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: 3435432
Solution: First, arrange all the blue and red bulbs in $C_{13}^{6}$ ways. In the gaps between them and at the ends, choose 9 positions and insert the white bulbs. There are $C_{14}^{9}$ ways to do this. In total, there are $C_{13}^{6} \cdot C_{14}^{9}$ ways to compose the garland from the available bul... | 3435432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-8 ; 8]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 272
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 $-17 \leqslant y-1 \leqslant 15$ and $-14 \leqslant 2-x \leqslant 18$. Therefore, $(y-1)(2-x)+2 \leqslant 15 \cdot 18+2=272... | 272 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{13})^{210}$ using the binomial theorem, we obtain terms of the form $C_{210}^{k}(\sqrt{13})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 165
Solution: The ratio of two consecutive terms $\frac{C_{210}^{k+1}(\sqrt{13})^{k+1}}{C_{210}^{k}(\sqrt{13})^{k}}$ is greater than 1 when $k<$ $\frac{210 \sqrt{13}-1}{\sqrt{13}+1}$. Then the terms increase up to $\left[\frac{210 \sqrt{13}-1}{\sqrt{13}+1}\right]+1$, and then decrease. | 165 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) Thirty ones are written on the board. 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 30 minutes? | Answer: 435.
Solution: Let's represent 30 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... | 435 | Number Theory | 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 $C D=10, B C=12 \sqrt{3}-5$. | Answer: 26
## 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... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2012
$$ | Answer: 4052168
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2012$. From which $n=(2012+1)^{2}-1=... | 4052168 | 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 40 more than the total number of students. How many valentines were given? | Answer: 84
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+40$. Then $(x-1)(y-1)=41$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 41. The number of valentines is $2 \cdot 42=84$. | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 5 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: 468468
Solution: First, arrange all the blue and red bulbs in $C_{13}^{5}$ ways. In the gaps between them and at the ends, choose 11 positions and insert the white bulbs there. There are $C_{14}^{11}$ ways to do this. In total, there are $C_{13}^{5} \cdot C_{14}^{11}$ ways to compose the garland from the avail... | 468468 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-7 ; 7]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 210
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 $-15 \leqslant y-1 \leqslant 13$ and $-12 \leqslant 2-x \leqslant 16$. Therefore, $(y-1)(2-x)+2 \leqslant 13 \cdot 16+2=210... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{7})^{211}$ using the binomial theorem, we obtain terms of the form $C_{211}^{k}(\sqrt{7})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 153
Solution: The ratio of two consecutive terms $\frac{C_{211}^{k+1}(\sqrt{7})^{k+1}}{C_{211}^{k}(\sqrt{7})^{k}}$ is greater than 1 when $k<$ $\frac{211 \sqrt{7}-1}{\sqrt{7}+1}$. Then the terms increase up to $\left[\frac{211 \sqrt{7}-1}{\sqrt{7}+1}\right]+1$, and then decrease. | 153 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 31 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 31 minutes? | Answer: 465.
Solution: Let's represent 31 units as points on a plane. Each time we combine numbers, we will connect the points of 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 connected by $xy$ line s... | 465 | Number Theory | 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 $C D=5, B C=6 \sqrt{3}-2.5$. | Answer: 13
## 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... | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2011
$$ | Answer: 4048143
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2011$. From which $n=(2011+1)^{2}-1=... | 4048143 | 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 42 more than the total number of students. How many valentines were given? | Answer: 88
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y = x + y + 42$. Then $(x-1)(y-1) = 43$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 43. The number of valentines is $2 \cdot 44 = 88$.
 The numbers $a, b, c, d$ belong to the interval $[-6 ; 6]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 156
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 $-13 \leqslant y-1 \leqslant 11$ and $-10 \leqslant 2-x \leqslant 14$. Therefore, $(y-1)(2-x)+2 \leqslant 11 \cdot 14+2=156... | 156 | 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 $21$ and $\cos \alpha = \frac{4}{7}$? | Answer: 24.
## 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_... | 24 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{11})^{212}$ using the binomial theorem, we obtain terms of the form $C_{212}^{k}(\sqrt{11})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 163
Solution: The ratio of two consecutive terms $\frac{C_{212}^{k+1}(\sqrt{11})^{k+1}}{C_{212}^{k}(\sqrt{11})^{k}}$ is greater than 1 when $k<$ $\frac{212 \sqrt{11}-1}{\sqrt{11}+1}$. Then the terms increase up to $\left[\frac{212 \sqrt{11}-1}{\sqrt{11}+1}\right]+1$, and then decrease. | 163 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 32 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 32 minutes? | Answer: 496.
Solution: Let's represent 32 units as points on a plane. Each time we combine numbers, we will connect the points of 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 connected by $x y$ line ... | 496 | 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 $C D=21, B C=14 \sqrt{3}-10.5$. | Answer: 35
## 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... | 35 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2010
$$ | Answer: 4044120
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2010$. From which $n=(2010+1)^{2}-1=... | 4044120 | 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 46 more than the total number of students. How many valentines were given? | Answer: 96
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+46$. Then $(x-1)(y-1)=47$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 47. The number of valentines is $2 \cdot 48=96$. | 96 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 8 blue, 6 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: 1366365
Solution: First, arrange all the blue and red bulbs in $C_{14}^{8}$ ways. In the gaps between them and at the ends, choose 12 positions and insert the white bulbs. There are $C_{15}^{12}$ ways to do this. In total, there are $C_{14}^{8} \cdot C_{15}^{12}$ ways to compose the garland from the available ... | 1366365 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-5 ; 5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 110
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 $-11 \leqslant y-1 \leqslant 9$ and $-8 \leqslant 2-x \leqslant 12$. Therefore, $(y-1)(2-x)+2 \leqslant 9 \cdot 12+2=110$. ... | 110 | 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 $15$, and $\cos \alpha = \frac{3}{5}$? | Answer: 18.
## 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... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{5})^{213}$ using the binomial theorem, we obtain terms of the form $C_{213}^{k}(\sqrt{5})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 147
Solution: The ratio of two consecutive terms $\frac{C_{213}^{k+1}(\sqrt{5})^{k+1}}{C_{213}^{k}(\sqrt{5})^{k}}$ is greater than 1 when $k<$ $\frac{213 \sqrt{5}-1}{\sqrt{5}+1}$. Then the terms increase up to $\left[\frac{213 \sqrt{5}-1}{\sqrt{5}+1}\right]+1$, and then decrease. | 147 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 33 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 33 minutes? | Answer: 528.
Solution: Let's represent 33 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 $xy$ line seg... | 528 | 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 $C D=9, B C=6 \sqrt{3}-4.5$. | Answer: 15
## 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... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2019
$$ | Answer: 4080399
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\cdots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2019$. From which $n=(2019+1)^{2}-1=... | 4080399 | 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 52 more than the total number of students. How many valentines were given? | Answer: 108
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+52$. Then $(x-1)(y-1)=53$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 53. The number of valentines is $2 \cdot 54=108$. | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) There are 7 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: 1561560
Solution: First, arrange all the blue and red bulbs in $C_{14}^{7}$ ways. In the gaps between them and at the ends, choose 12 positions and insert the white bulbs. There are $C_{15}^{12}$ ways to do this. In total, there are $C_{14}^{7} \cdot C_{15}^{12}$ ways to compose the garland from the available ... | 1561560 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-4 ; 4]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 72
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 $-9 \leqslant y-1 \leqslant 7$ and $-6 \leqslant 2-x \leqslant 10$. Therefore, $(y-1)(2-x)+2 \leqslant 7 \cdot 10+2=72$. The ma... | 72 | 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 $\cos \alpha = \frac{3}{4}$? | Answer: 18.
## 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... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) By expanding the expression $(1+\sqrt{11})^{214}$ using the binomial theorem, we obtain terms of the form $C_{214}^{k}(\sqrt{11})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 165
Solution: The ratio of two consecutive terms $\frac{C_{214}^{k+1}(\sqrt{11})^{k+1}}{C_{214}^{k}(\sqrt{11})^{k}}$ is greater than 1 when $k<$ $\frac{214 \sqrt{11}-1}{\sqrt{11}+1}$. Then the terms increase up to $\left[\frac{214 \sqrt{11}-1}{\sqrt{11}+1}\right]+1$, and then decrease. | 165 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) On the board, 34 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 34 minutes? | Answer: 561.
Solution: Let's represent 34 units as points on a plane. Each time we combine numbers, we will connect the points of 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 connected by $x y$ line ... | 561 | 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 $C D=12, B C=8 \sqrt{3}-6$. | Answer: 20
## 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... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Find the smallest value of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
8(x+7)^{4}+(y-4)^{4}=c \\
(x+4)^{4}+8(y-7)^{4}=c
\end{array}\right.
$$ | Answer: $c=24$.
Solution. By the Cauchy-Bunyakovsky-Schwarz inequality, we have
(1) $\left(\frac{1}{2}+1\right)^{3}\left(8(x+\alpha)^{4}+(y-\beta)^{4}\right) \geqslant$
$$
\geqslant\left(\left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right)\right)^{2} \geqslant
$$
$$
\geqslant(|x+\alpha|+|y-\beta|)^{... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2537^{\circ}+\sin 2538^{\circ}+\cdots+\sin 6137^{\circ}\right)^{\cos } 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6120^{\circ}\right)
$$ | Answer: $73^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6119^{\circ}=0$ and in the exponent only $\cos ... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 5 zeros and have exactly 42 divisors. Find their sum. | Answer: 700000
Solution: Since the number ends with 5 zeros, it has the form $N=10^{5} k$. The smallest number of this form $10^{5}$ has 36 divisors: all divisors have the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 5. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has othe... | 700000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 40 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 40 minutes? | Answer: 780.
Solution: Let's represent 40 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... | 780 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=53$.
 | Answer: 2862
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=53(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{53 S}{k}+\frac{53 S}{l}$... | 2862 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=91
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 40
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=16, A C^{2}=91$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2018}+(x-y-z)^{2018}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1020100
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2018}+(x-t)^{2018}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2018}=x^{2018}+a_{1} x^{2017} t+\ldots+a_{2017} x t^{2017}+t^{2018} \\
& (x-t)^{2018}=x^{2018}-a_{1} x^{2017} t+\ldots-a_{2... | 1020100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2539^{\circ}+\sin 2540^{\circ}+\cdots+\sin 6139^{\circ}\right)^{\cos } 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6120^{\circ}\right)
$$ | Answer: $71^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6119^{\circ}=0$ and in the exponent only $\cos ... | 71 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 7 zeros and have exactly 72 divisors. Find their sum. | Answer: 70000000
Solution: Since the number ends with 7 zeros, it has the form $N=10^{7} k$. The smallest number of this form $10^{7}$ has 64 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 7. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has ... | 70000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 38 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 38 minutes? | Answer: 703.
Solution: Let's represent 38 units as points on a plane. Each time we combine numbers, we will connect the points of 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 connected by $xy$ line s... | 703 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=47$.
 | Answer: 2256
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=47(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{47 S}{k}+\frac{47 S}{l}$... | 2256 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=124
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 48
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=16, A C^{2}=124$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2020}+(x-y-z)^{2020}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1022121
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2020}+(x-t)^{2020}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2020}=x^{2020}+a_{1} x^{2019} t+\ldots+a_{2019} x t^{2019}+t^{2020} \\
& (x-t)^{2020}=x^{2020}-a_{1} x^{2019} t+\ldots-a_{2... | 1022121 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2541^{\circ}+\sin 2542^{\circ}+\cdots+\sin 6141^{\circ}\right)^{\cos } 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6120^{\circ}\right)
$$ | Answer: $69^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2520^{\circ}+\cos 2521^{\circ}+\cdots+\cos 6119^{\circ}=0$ and in the exponent only $\cos ... | 69 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 9 zeros and have exactly 110 divisors. Find their sum. | Answer: 7000000000
Solution: Since the number ends with 9 zeros, it has the form $N=10^{9} k$. The smallest number of this form $10^{9}$ has 100 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 9. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ h... | 7000000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 37 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 37 minutes? | Answer: 666.
Solution: Let's represent 37 units as points on a plane. Each time we combine 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 connected ... | 666 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=43$.
 | Answer: 1892
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=43(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{43 S}{k}+\frac{43 S}{l}$... | 1892 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=147 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=163
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 56
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=147$, $B C^{2}=16, A C^{2}=163$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 56 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2022}+(x-y-z)^{2022}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1024144
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2022}+(x-t)^{2022}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2022}=x^{2022}+a_{1} x^{2021} t+\ldots+a_{2021} x t^{2021}+t^{2022} \\
& (x-t)^{2022}=x^{2022}-a_{1} x^{2021} t+\ldots-a_{2... | 1024144 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2903^{\circ}+\sin 2904^{\circ}+\cdots+\sin 6503^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | # Answer: $67^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$, it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\c... | 67 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 8 zeros and have exactly 90 divisors. Find their sum. | Answer: 700000000
Solution: Since the number ends with 8 zeros, it has the form $N=10^{8} k$. The smallest number of this form $10^{8}$ has 81 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 8. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has... | 700000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 39 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 39 minutes? | Answer: 741.
Solution: Let's represent 39 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 $xy$ line seg... | 741 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=37$.
 | Answer: 1406
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=37(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{37 S}{k}+\frac{37 S}{l}$... | 1406 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=36 \\
z^{2}+x z+x^{2}=111
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 60
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=36, A C^{2}=111$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2024}+(x-y-z)^{2024}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1026169
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2024}+(x-t)^{2024}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2024}=x^{2024}+a_{1} x^{2023} t+\ldots+a_{2023} x t^{2023}+t^{2024} \\
& (x-t)^{2024}=x^{2024}-a_{1} x^{2023} t+\ldots-a_{2... | 1026169 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2905^{\circ}+\sin 2906^{\circ}+\cdots+\sin 6505^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | Answer: $65^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\cos ... | 65 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 6 zeros and have exactly 56 divisors. Find their sum. | Answer: 7000000
Solution: Since the number ends with 6 zeros, it has the form $N=10^{6} k$. The smallest number of this form $10^{6}$ has 49 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 6. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ has o... | 7000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (7 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 $\sin \alpha = \frac{\sqrt{11}}{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$. Note that points $B_{... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 45 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 45 minutes? | Answer: 990.
Solution: Let's represent 45 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... | 990 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=31$.
 | Answer: 992
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=31(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{31 S}{k}+\frac{31 S}{l}$.... | 992 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=49 \\
z^{2}+x z+x^{2}=124
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 70
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=49, A C^{2}=124$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2026}+(x-y-z)^{2026}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1028196
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2026}+(x-t)^{2026}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2026}=x^{2026}+a_{1} x^{2025} t+\ldots+a_{2025} x t^{2025}+t^{2026} \\
& (x-t)^{2026}=x^{2026}-a_{1} x^{2025} t+\ldots-a_{2... | 1028196 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2907^{\circ}+\sin 2908^{\circ}+\cdots+\sin 6507^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | Answer: $63^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\cos ... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 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 $28$, and $\sin \alpha = \frac{\sqrt{45}}{7}$? | Answer: 16.
## 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_{... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 46 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 46 minutes? | Answer: 1035.
Solution: Let's represent 46 units as points on a plane. Each time we combine numbers, we will connect the points of 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 connected by $x y$ line... | 1035 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=29$.
 | Answer: 870
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=29(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{29 S}{k}+\frac{29 S}{l}$.... | 870 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=64 \\
z^{2}+x z+x^{2}=139
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 80
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=64, A C^{2}=139$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2028}+(x-y-z)^{2028}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1030225
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2028}+(x-t)^{2028}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2028}=x^{2028}+a_{1} x^{2027} t+\ldots+a_{2027} x t^{2027}+t^{2028} \\
& (x-t)^{2028}=x^{2028}-a_{1} x^{2027} t+\ldots-a_{2... | 1030225 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 3269^{\circ}+\sin 3270^{\circ}+\cdots+\sin 6869^{\circ}\right)^{\cos } 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6840^{\circ}\right)
$$ | Answer: $61^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6839^{\circ}=0$ and in the exponent only $\cos ... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 7 zeros and have exactly 72 divisors. Find their sum.
# | # Answer: 70000000
Solution: Since the number ends with 7 zeros, it has the form $N=10^{7} k$. The smallest number of this form $10^{7}$ has 64 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 7. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ ha... | 70000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (7 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 $21$, and $\sin \alpha = \frac{\sqrt{40}}{7}$? | Answer: 18.
## 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_{... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 47 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 47 minutes? | Answer: 1081.
Solution: Let's represent 47 units as points on a plane. Each time we combine numbers, we will connect the points of 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 connected by $x y$ line... | 1081 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=23$.
 | Answer: 552
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=23(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{23 S}{k}+\frac{23 S}{l}$.... | 552 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=64 \\
z^{2}+x z+x^{2}=172
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 96
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=64, A C^{2}=172$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2030}+(x-y-z)^{2030}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1032256
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2030}+(x-t)^{2030}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2030}=x^{2030}+a_{1} x^{2029} t+\ldots+a_{2029} x t^{2029}+t^{2030} \\
& (x-t)^{2030}=x^{2030}-a_{1} x^{2029} t+\ldots-a_{2... | 1032256 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 3271^{\circ}+\sin 3272^{\circ}+\cdots+\sin 6871^{\circ}\right)^{\cos } 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6840^{\circ}\right)
$$ | Answer: $59^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6839^{\circ}=0$ and in the exponent only $\cos ... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 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 $14$, and $\sin \alpha = \frac{\sqrt{33}}{7}$? | Answer: 16
## 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_... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 48 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 48 minutes? | Answer: 1128.
Solution: Let's represent 48 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 s... | 1128 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=19$.
 | Answer: 380
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=19(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{19 S}{k}+\frac{19 S}{l}$.... | 380 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=49 \\
z^{2}+x z+x^{2}=157
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 84
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=49, A C^{2}=157$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2032}+(x-y-z)^{2032}$, the parentheses were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1034289
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2032}+(x-t)^{2032}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2032}=x^{2032}+a_{1} x^{2031} t+\ldots+a_{2031} x t^{2031}+t^{2032} \\
& (x-t)^{2032}=x^{2032}-a_{1} x^{2031} t+\ldots-a_{2... | 1034289 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2193^{\circ}+\sin 2194^{\circ}+\cdots+\sin 5793^{\circ}\right)^{\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5760^{\circ}}\right)
$$ | # Answer: $57^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$, it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5759^{\circ}=0$ and in the exponent only $\c... | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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