problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
7. (10 points) On the board, there are 26 ones. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 26 minutes?
|
Answer: 325.
Solution: Let's represent 26 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line segments. This is the same number of candies that Karlson eats. After 26 minutes, all points will be connected. In total, $\frac{26 \cdot(26-1)}{2}=325$ line segments will be drawn. Therefore, Karlson will eat 325 candies.
|
325
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=20, B C=24 \sqrt{3}-10$.
|
Answer: 52
## Solution:
The quadrilateral is inscribed, so $\angle A + \angle C = 180^{\circ}$. From the given ratio, $\angle A = 2x, \angle C = 4x$. Therefore, $x = 30^{\circ}$ and $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 48\sqrt{3} - 20, KD = KC + CD = 48\sqrt{3}$, and $AD = 48$. Using the Pythagorean theorem, we get $AC = 52$.
## Solve the following problems with justification for your answer
|
52
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2016
$$
|
Answer: 4068288
Solution: Note that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2016$. From which $n=(2016+1)^{2}-1=4068288$.
|
4068288
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) On Valentine's Day, every student in the school gave a valentine to every female student. It turned out that the number of valentines was 22 more than the total number of students. How many valentines were given?
|
Answer: 48
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the problem, $x y=x+y+22$. Then $(x-1)(y-1)=23$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 23. The number of valentines is $2 \cdot 24=48$.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) There are 8 blue, 8 red, and 11 white bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent?
|
Answer: 159279120
Solution: First, arrange all the blue and red bulbs in $C_{16}^{8}$ ways. In the gaps between them and at the ends, choose 11 positions and insert the white bulbs. There are $C_{17}^{11}$ ways to do this. In total, there are $C_{16}^{8} \cdot C_{17}^{11}$ ways to compose the garland from the available bulbs.
|
159279120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-6.5 ; 6.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$.
|
Answer: 182
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-14 \leqslant y-1 \leqslant 12$ and $-11 \leqslant 2-x \leqslant 15$. Therefore, $(y-1)(2-x)+2 \leqslant 12 \cdot 15+2=182$. The maximum value of the expression is achieved when $a=c=-6.5, b=d=6.5$.
|
182
|
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 $6$ and $\cos \alpha = \frac{2}{3}$?
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{2}{3} \cdot 6=8$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (8 points) By expanding the expression $(1+\sqrt{7})^{207}$ using the binomial theorem, we obtain terms of the form $C_{207}^{k}(\sqrt{7})^{k}$. Find the value of $k$ for which such a term attains its maximum value.
|
Answer: 150
Solution: The ratio of two consecutive terms $\frac{C_{207}^{k+1}(\sqrt{7})^{k+1}}{C_{207}^{k}(\sqrt{7})^{k}}$ is greater than 1 when $k<$ $\frac{207 \sqrt{7}-1}{\sqrt{7}+1}$. Then the terms increase up to $\left[\frac{207 \sqrt{7}-1}{\sqrt{7}+1}\right]+1$, and then decrease.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. (10 points) On the board, 27 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 27 minutes?
|
Answer: 351.
Solution: Let's represent 27 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line segments. Carlson eats as many candies as the number of line segments. After 27 minutes, all points will be connected. In total, $\frac{27 \cdot(27-1)}{2}=351$ line segments will be drawn. Therefore, Carlson will eat 351 candies.
|
351
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=16, B C=15 \sqrt{3}-8$.
|
Answer: 34
## Solution:
The quadrilateral is inscribed, hence $\angle A+\angle C=180^{\circ}$. From the given ratio, $\angle A=2 x, \angle C=4 x$. Therefore, $x=30^{\circ}$ and $\angle A=60^{\circ}, \angle B=3 x=90^{\circ}$. Extend $A B$ and $C D$ until they intersect at point $K$. Then $\angle K=30^{\circ}, K C=2 B C=30 \sqrt{3}-16, K D=K C+C D=30 \sqrt{3}$ and $A D=30$. Using the Pythagorean theorem, we get $A C=34$.
## Solve the following problems with justification for your answer
|
34
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2015
$$
|
Answer: 4064255
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2015$. From which $n=(2015+1)^{2}-1=4064255$.
|
4064255
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. (5 points) On Valentine's Day, every student in the school gave each female student a valentine. It turned out that the number of valentines was 28 more than the total number of students. How many valentines were given?
|
Answer: 60
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the condition, $x y=x+y+28$. Then $(x-1)(y-1)=29$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 29. The number of valentines is $2 \cdot 30=60$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) There are 8 blue, 7 red, and 12 white light bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent?
|
Answer: 11711700
Solution: First, arrange all the blue and red bulbs in $C_{15}^{8}$ ways. In the gaps between them and at the ends, choose 12 positions and insert the white bulbs. There are $C_{16}^{12}$ ways to do this. In total, there are $C_{15}^{8} \cdot C_{16}^{12}$ ways to compose the garland from the available bulbs.
|
11711700
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-7.5 ; 7.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$.
|
Answer: 240
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-16 \leqslant y-1 \leqslant 14$ and $-13 \leqslant 2-x \leqslant 17$. Therefore, $(y-1)(2-x)+2 \leqslant 14 \cdot 17+2=240$. The maximum value of the expression is achieved when $a=c=-7.5, b=d=7.5$.
|
240
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\cos \alpha = \frac{5}{6}$?
|
Answer: 20.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{5}{6} \cdot 12=20$
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 segments. This is the same number of candies that Karlson eats. After 28 minutes, all points will be connected. In total, $\frac{28 \cdot(28-1)}{2}=378$ line segments will be drawn. Therefore, Karlson will eat 378 candies.
|
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 $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 15\sqrt{3} - 8, KD = KC + CD = 15\sqrt{3}$, and $AD = 15$. Using the Pythagorean theorem, we get $AC = 17$.
## Solve the following problems with justification for your answer
|
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 $n=(2014+1)^{2}-1=4060224$.
|
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 bulbs.
|
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, $(y-1)(2-x)+2 \leqslant 16 \cdot 19+2=306$. The maximum value of the expression is achieved when $a=c=-8.5, b=d=8.5$.
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{4}{5} \cdot 10=16$
|
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 connected by $x y$ line segments. This is the same number of candies that Karlson eats. After 29 minutes, all points will be connected. In total, $\frac{29 \cdot(29-1)}{2}=406$ line segments will be drawn. Therefore, Karlson will eat 406 candies.
|
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 $\angle A=60^{\circ}, \angle B=3 x=90^{\circ}$. Extend $A B$ and $C D$ until they intersect at point $K$. Then $\angle K=30^{\circ}, K C=2 B C=36 \sqrt{3}-15, K D=K C+C D=36 \sqrt{3}$ and $A D=36$. Using the Pythagorean theorem, we get $A C=39$.
## Solve the following problems with justification for your answer
|
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 $n=(2013+1)^{2}-1=4056195$.
|
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 bulbs.
|
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, $(y-1)(2-x)+2 \leqslant 15 \cdot 18+2=272$. The maximum value of the expression is achieved when $a=c=-8, b=d=8$.
|
272
|
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 $\cos \alpha = \frac{1}{4}$?
|
Answer: 6.
## 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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{1}{4} \cdot 12=6$
|
6
|
Geometry
|
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 segments. This is the same number of candies that Karlson eats. After 30 minutes, all points will be connected. In total, $\frac{30 \cdot(30-1)}{2}=435$ line segments will be drawn. Therefore, Karlson will eat 435 candies.
|
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 $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 24\sqrt{3} - 10, KD = KC + CD = 24\sqrt{3}$, and $AD = 24$. Using the Pythagorean theorem, we get $AC = 26$.
## Solve the following problems with justification for your answer
|
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 $n=(2012+1)^{2}-1=4052168$.
|
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 available bulbs.
|
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, $(y-1)(2-x)+2 \leqslant 13 \cdot 16+2=210$. The maximum value of the expression is achieved when $a=c=-7, b=d=7$.
|
210
|
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 $9$ and $\cos \alpha = \frac{1}{3}$?
|
Answer: 6.
## 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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{1}{3} \cdot 9=6$
|
6
|
Geometry
|
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 segments. This is the same number of candies that Karlson eats. After 31 minutes, all points will be connected. In total, $\frac{31 \cdot (31-1)}{2}=465$ line segments will be drawn. Therefore, Karlson will eat 465 candies.
|
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 $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 12\sqrt{3} - 5, KD = KC + CD = 12\sqrt{3}$, and $AD = 12$. Using the Pythagorean theorem, we get $AC = 13$.
## Solve the following problems with justification for your answer
|
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 $n=(2011+1)^{2}-1=4048143$.
|
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 $2 \cdot 44 = 88$.
(using all the bulbs) a garland can be formed so that no two white bulbs are adjacent.
Answer: 910910
Solution: First, arrange all the blue and red bulbs in $C_{14}^{5}$ 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 $C_{14}^{5} \cdot C_{15}^{12}$ ways to form a garland from the available bulbs.
|
88
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (7 points) 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, $(y-1)(2-x)+2 \leqslant 11 \cdot 14+2=156$. The maximum value of the expression is achieved when $a=c=-6, b=d=6$.
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{4}{7} \cdot 21=24$
|
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 segments. This is the same number of candies that Karlson eats. After 32 minutes, all points will be connected. In total, $\frac{32 \cdot(32-1)}{2}=496$ line segments will be drawn. Therefore, Karlson will eat 496 candies.
|
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 $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 28\sqrt{3} - 21, KD = KC + CD = 28\sqrt{3}$, and $AD = 28$. Using the Pythagorean theorem, we get $AC = 35$.
## Solve the following problems with justification for your answer
|
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 $n=(2010+1)^{2}-1=4044120$.
|
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 bulbs.
|
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, $(y-1)(2-x)+2 \leqslant 9 \cdot 12+2=110$. The maximum value of the expression is achieved when $a=c=-5, b=d=5$.
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{3}{5} \cdot 15=18$
|
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 segments. This is the same number of candies that Karlson eats. After 33 minutes, all points will be connected. In total, $\frac{33 \cdot (33-1)}{2}=528$ line segments will be drawn. Therefore, Karlson will eat 528 candies.
|
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 $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 12\sqrt{3} - 9, KD = KC + CD = 12\sqrt{3}$, and $AD = 12$. Using the Pythagorean theorem, we get $AC = 15$.
## Solve the following problems with justification for your answer
|
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 $n=(2019+1)^{2}-1=4080399$.
|
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 bulbs.
|
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 maximum value of the expression is achieved when $a=c=-4, b=d=4$.
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha$. We find the base of the isosceles triangle using the formula $B C=2 \cdot \cos \alpha \cdot B O=2 \cdot \frac{3}{4} \cdot 12=18$
|
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 segments. This is the same number of candies that Karlson eats. After 34 minutes, all points will be connected. In total, $\frac{34 \cdot(34-1)}{2}=561$ line segments will be drawn. Therefore, Karlson will eat 561 candies.
|
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 $\angle A = 60^{\circ}, \angle B = 3x = 90^{\circ}$. Extend $AB$ and $CD$ to intersect at point $K$. Then $\angle K = 30^{\circ}, KC = 2BC = 16\sqrt{3} - 12, KD = KC + CD = 16\sqrt{3}$, and $AD = 16$. Using the Pythagorean theorem, we get $AC = 20$.
## Solve the following problems with justification for your answer
|
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) $\left(\frac{1}{2}+1\right)^{3}\left(8(x+\alpha)^{4}+(y-\beta)^{4}\right) \geqslant$
$$
\geqslant\left(\left(\frac{1}{2}+1\right)\left(2(x+\alpha)^{2}+(y-\beta)^{2}\right)\right)^{2} \geqslant
$$
$$
\geqslant(|x+\alpha|+|y-\beta|)^{4}
$$
(2) $\left(1+\frac{1}{2}\right)^{3}\left(8(x+\beta)^{4}+(y-\alpha)^{4}\right) \geqslant$
$$
\geqslant\left(\left(1+\frac{1}{2}\right)\left(2(x+\beta)^{2}+(y-\alpha)^{2}\right)\right)^{2} \geqslant
$$
$$
\geqslant(|x+\beta|+|y-\alpha|)^{4}
$$
Therefore, for any solution $(x, y)$ of the system, we have
(3) $\quad \frac{27}{4} c \geqslant(|x+\alpha|+|y-\beta|)^{4}+(|x+\beta|+|y-\alpha|)^{4} \geqslant$
$$
\begin{aligned}
& \geqslant(\alpha-\beta+x+y)^{4}+(\alpha-\beta-(x+y))^{4}= \\
& \quad=2(\alpha-\beta)^{4}+2(x+y)^{4}+12(\alpha-\beta)^{2}(x+y)^{2} \geqslant 2(\alpha-\beta)^{4}
\end{aligned}
$$
Thus, if $c<\frac{8}{27}(\alpha-\beta)^{4}$, the system has no solutions.
If $c=\frac{8}{27}(\alpha-\beta)^{4}$, then equality must be achieved in all inequalities (1)-(3). Therefore, $x+y=0$, the numbers $x+\alpha$ and $x+\beta$ must have different signs, and
$$
\frac{|x+\alpha|}{|x+\beta|}=\frac{1}{2}
$$
From these conditions, it follows that when $c=\frac{8}{27}(\alpha-\beta)^{4}$, there is only one solution $\left(-\frac{2 \alpha+\beta}{3}, \frac{2 \alpha+\beta}{3}\right)$. Remark. One can immediately notice that the solution of the system is symmetric with respect to the line $y=-x$ (two closed curves with centers at points $(-\alpha ; \beta)$ and $(-\beta ; \alpha)$). The solution will be unique if one of these curves touches this line. This is satisfied if $c=\min f(x)$, where $f(x)=8(x+\alpha)^{4}+(x+\beta)^{4} \cdot f^{\prime}(x)=32(x+\alpha)^{3}+4(x+\beta)^{3}=0$ at the point $x_{0}=-\frac{2 \alpha+\beta}{3}$. Then $c=f\left(x_{0}\right)$. https://ggbm.at/FVmUGS4Y
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (7 points) Find all pairs of natural numbers $x$ and $y$ such that
$$
\log _{2} a x+\log _{2} b y=\log _{2}\left(b x+a y+p_{1} p_{2}-1\right), \text { where } p_{1}, p_{2} \in \mathbb{P} \quad a, b>2
$$
In your answer, write the smallest possible value of $x+y$.
|
Solution: Let's get rid of the logarithms in the equation
$$
a x b y=b x+a y+p_{1} p_{2}-1
$$
Transform and factorize
$$
(a y-1)(b x-1)=p_{1} p_{2}
$$
Since $p_{1}, p_{2} \in \mathbb{P}$, we get four possible solutions
$$
\left\{\begin{array} { l }
{ x = \frac { p _ { 1 } p _ { 2 } + 1 } { b } } \\
{ y = \frac { 2 } { a } }
\end{array} \quad \left\{\begin{array} { l }
{ x = \frac { 2 } { b } } \\
{ y = \frac { p _ { 1 } p _ { 2 } + 1 } { a } }
\end{array} \left\{\begin{array} { l }
{ x = \frac { p _ { 1 } + 1 } { b } } \\
{ y = \frac { p _ { 2 } + 1 } { a } }
\end{array} \quad \left\{\begin{array}{l}
x=\frac{p_{2}+1}{b} \\
y=\frac{p_{1}+1}{a}
\end{array}\right.\right.\right.\right.
$$
Since $a, b>2$ and $x, y \in \mathbb{N}$, the first two solutions are definitely not suitable. From the remaining two, we choose the one where $x, y$ are natural numbers and the sum $x+y$ is minimal.
Answers:
| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 5 | 4 | 5 | 6 | 7 | 6 | 4 | 6 | 10 | 4 | 8 |
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. (10 points) Find the sum of the real roots of the equation
$$
2 \cdot 3^{3 x}-a \cdot 3^{2 x}-3(a+4) \cdot 3^{x}+18=0
$$
|
Solution: Let's make the substitution $t=3^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation
$$
2 t^{3}-a t^{2}-3(a+4) t+18=0
$$
It is obvious that the number $t_{1}=-3$ is a root of the equation and the corresponding $x_{1}$ is not real. We get
$$
(t+3)\left(2 t^{2}-(a+6) t+6\right)=0
$$
The parameter $a$ was chosen such that the remaining two roots $t_{2}$ and $t_{3}$ are real and strictly greater than zero. Then, using Vieta's theorem, we get
$$
3^{x_{2}+x_{3}}=t_{2} \cdot t_{3}=\frac{6}{2}=3
$$
Therefore, $x_{2}+x_{3}=1$.
(10 points) Find the sum of the real roots of the equation
$$
2 \cdot 4^{3 x}-a \cdot 4^{2 x}-4(a+6) \cdot 4^{x}+32=0
$$
Solution: Let's make the substitution $t=4^{x}$, and since $x \in \mathbb{R}$, then $t>0$. We obtain the following equation
$$
2 t^{3}-a t^{2}-4(a+6) t+32=0
$$
It is obvious that the number $t_{1}=-4$ is a root of the equation and the corresponding $x_{1}$ is not real. We get
$$
(t+4)\left(2 t^{2}-(a+8) t+8\right)=0
$$
The parameter $a$ was chosen such that the remaining two roots $t_{2}$ and $t_{3}$ are real and strictly greater than zero. Then, using Vieta's theorem, we get
$$
4^{x_{2}+x_{3}}=t_{2} \cdot t_{3}=\frac{8}{2}=4
$$
Therefore, $x_{2}+x_{3}=1$.
Answers:
| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 0.5 |
|
1
|
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 $\cos 6120^{\circ}=1$ remains. Similarly, $\sin \alpha+\sin \left(\alpha+1^{\circ}\right)+\cdots+\sin \left(\alpha+179^{\circ}\right)=0$ and $\sin 2537^{\circ}+\sin 2538^{\circ}+\cdots+\sin 6136^{\circ}=0 \cdot \sin 6137^{\circ}=\sin 17^{\circ}$. Then $\delta=\arccos \left(\sin 17^{\circ}\right)=73^{\circ}$.
|
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 other prime divisors, the number of divisors of $N$ increases by at least a factor of two: in addition to the original $1,2,5,10,20 \ldots$, there will also be $k, 2 k, 5 k, 10 k, 20 k$ and so on.
Therefore, $10^{5} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 42 can be represented as a product of two factors greater than 5 in only one way: $42=6 \cdot 7$. Then, $N=2^{5} 5^{6}$ or $N=2^{6} 5^{5}$. Their sum is 700000.
|
700000
|
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 $15$, and $\sin \alpha = \frac{\sqrt{21}}{5}$?
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{2}{5} * 15=12$
|
12
|
Geometry
|
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 connected by $x y$ line segments. Carlson will eat as many candies as there are line segments. After 40 minutes, all points will be connected. In total, $\frac{40 \cdot(40-1)}{2}=780$ line segments will be drawn. Therefore, Carlson will eat 780 candies.
|
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 $S=\frac{53 S}{k}+\frac{53 S}{l}$. Therefore, $k l=53(k+l)=>$ $k l-53 k-53 l+2809=2809=>(k-53)(l-53)=2809$. Since $l<k$, that is, $l-53<k-53$. Therefore, $l-53=1$ and $k-53=2809$. Thus, $k=D G=2862$.
|
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 the areas of triangles $A O B, B O C, A O C$ is equal to the area of triangle $A B C$. This gives us the relation $\frac{1}{2}(x y+y z+x z) \sin 120^{\circ}=\frac{1}{2} \cdot 5 \sqrt{3} \cdot 4$. From this, we obtain the answer.
|
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
$$
\begin{aligned}
& (x+t)^{2018}=x^{2018}+a_{1} x^{2017} t+\ldots+a_{2017} x t^{2017}+t^{2018} \\
& (x-t)^{2018}=x^{2018}-a_{1} x^{2017} t+\ldots-a_{2017} x t^{2017}+t^{2018}
\end{aligned}
$$
Adding them up, we get
$$
(x+t)^{2018}+(x-t)^{2018}=2\left(x^{2018}+a_{2} x^{2016} t^{2}+\ldots+a_{2016} x^{2} t^{2016}+t^{2018}\right)
$$
Notice that when expanding different $t_{1}^{n}$ and $t_{2}^{n}$, we will get different monomials (since the degrees of $x$ will be different). Also, when expanding $t^{n}$, we will get $n+1$ different monomials, so the final answer will be
$$
1+3+\ldots+2017+2019=1010^{2}=1020100
$$
## Solve the following problems with justification of the answer
|
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 $\cos 6120^{\circ}=1$ remains. Similarly, $\sin \alpha+\sin \left(\alpha+1^{\circ}\right)+\cdots+\sin \left(\alpha+179^{\circ}\right)=0$ and $\sin 2539^{\circ}+\sin 2540^{\circ}+\cdots+\sin 6138^{\circ}=0 \cdot \sin 6139^{\circ}=\sin 19^{\circ}$. Then $\delta=\arccos \left(\sin 19^{\circ}\right)=71^{\circ}$.
|
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 other prime divisors, the number of divisors of $N$ increases by at least a factor of two: in addition to the original $1,2,5,10,20 \ldots$, there will also be $k, 2 k, 5 k, 10 k, 20 k$ and so on.
Therefore, $10^{7} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 72 can be represented as a product of two factors greater than 7 in only one way: $72=8 \cdot 9$. Then, $N=2^{7} 5^{8}$ or $N=2^{8} 5^{7}$. Their sum is 70000000.
|
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 $10$, and $\sin \alpha = \frac{\sqrt{24}}{5}$?
|
Answer: 4.
## Solution:
Consider the point $B_{1}$ 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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{1}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{1}{5} *$ $10=4$
|
4
|
Geometry
|
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 segments. This is the same number of candies that Karlson eats. After 38 minutes, all points will be connected. In total, $\frac{38 \cdot (38-1)}{2}=703$ line segments will be drawn. Therefore, Karlson will eat 703 candies.
|
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 $S=\frac{47 S}{k}+\frac{47 S}{l}$. Therefore, $k l=47(k+l)=>$ $k l-47 k-47 l+2209=2209=>(k-47)(l-47)=2209$. Since $l<k$, that is, $l-47<k-47$. Therefore, $l-47=1$ and $k-47=2209$. Thus, $k=D G=2256$.
|
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 of the areas of triangles $A O B, B O C, A O C$ is equal to the area of triangle $A B C$. This gives us the relation $\frac{1}{2}(x y+y z+x z) \sin 120^{\circ}=\frac{1}{2} \cdot 6 \sqrt{3} \cdot 4$. From this, we obtain the answer.
|
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
$$
\begin{aligned}
& (x+t)^{2020}=x^{2020}+a_{1} x^{2019} t+\ldots+a_{2019} x t^{2019}+t^{2020} \\
& (x-t)^{2020}=x^{2020}-a_{1} x^{2019} t+\ldots-a_{2019} x t^{2019}+t^{2020}
\end{aligned}
$$
Adding these, we get
$$
(x+t)^{2020}+(x-t)^{2020}=2\left(x^{2020}+a_{2} x^{2018} t^{2}+\ldots+a_{2018} x^{2} t^{2018}+t^{2020}\right)
$$
Notice that when expanding different $t_{1}^{n}$ and $t_{2}^{n}$, we will get different monomials (since the degrees of $x$ will be different). Also, when expanding $t^{n}$, we will get $n+1$ different monomials, so the final answer will be
$$
1+3+\ldots+2019+2021=1011^{2}=1022121
$$
## Solve the following problems with justification of the answer
|
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 $\cos 6120^{\circ}=1$ remains. Similarly, $\sin \alpha+\sin \left(\alpha+1^{\circ}\right)+\cdots+\sin \left(\alpha+179^{\circ}\right)=0$ and $\sin 2541^{\circ}+\sin 2542^{\circ}+\cdots+\sin 6140^{\circ}=0 \cdot \sin 6141^{\circ}=\sin 21^{\circ}$. Then $\delta=\arccos \left(\sin 21^{\circ}\right)=69^{\circ}$.
|
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 $k$ has other prime divisors, the number of divisors of $N$ increases by at least a factor of two: in addition to the original $1,2,5,10,20 \ldots$, there will also be $k, 2 k, 5 k, 10 k, 20 k$ and so on.
Therefore, $10^{9} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 110 can be represented as a product of two factors greater than 9 in only one way: $110=10 \cdot 11$. Thus, $N=2^{9} 5^{1} 0$ or $N=2^{1} 05^{9}$. Their sum is 7000000000.
|
7000000000
|
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 $10$, and $\sin \alpha = \frac{\sqrt{21}}{5}$?
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{2}{5}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{2}{5} * 10=8$
|
8
|
Geometry
|
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 by $x y$ line segments. This is the same number of candies that Karlson eats. After 37 minutes, all points will be connected. In total, $\frac{37 \cdot(37-1)}{2}=666$ line segments will be drawn. Therefore, Karlson will eat 666 candies.
|
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 $S=\frac{43 S}{k}+\frac{43 S}{l}$. Therefore, $k l=43(k+l)=>$ $k l-43 k-43 l+1849=1849=>(k-43)(l-43)=1849$. Since $l<k$, that is, $l-43<k-43$. Therefore, $l-43=1$ and $k-43=1849$. Thus, $k=D G=1892$.
|
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 of the areas of triangles $A O B, B O C, A O C$ is equal to the area of triangle $A B C$. This gives us the relation $\frac{1}{2}(x y+y z+x z) \sin 120^{\circ}=\frac{1}{2} \cdot 7 \sqrt{3} \cdot 4$. From this, we obtain the answer.
|
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
$$
\begin{aligned}
& (x+t)^{2022}=x^{2022}+a_{1} x^{2021} t+\ldots+a_{2021} x t^{2021}+t^{2022} \\
& (x-t)^{2022}=x^{2022}-a_{1} x^{2021} t+\ldots-a_{2021} x t^{2021}+t^{2022}
\end{aligned}
$$
Adding them up, we get
$$
(x+t)^{2022}+(x-t)^{2022}=2\left(x^{2022}+a_{2} x^{2020} t^{2}+\ldots+a_{2020} x^{2} t^{2020}+t^{2022}\right)
$$
Notice that when expanding different $t_{1}^{n}$ and $t_{2}^{n}$, we will get different monomials (since the degrees of $x$ will be different). Also, when expanding $t^{n}$, we will get $n+1$ different monomials, so the final answer will be
$$
1+3+\ldots+2021+2023=1012^{2}=1024144
$$
## Solve the following problems with justification of the answer
|
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 $\cos 6480^{\circ}=1$ remains. Similarly, $\sin \alpha+\sin \left(\alpha+1^{\circ}\right)+\cdots+\sin \left(\alpha+179^{\circ}\right)=0$ and $\sin 2903^{\circ}+\sin 2904^{\circ}+\cdots+\sin 6502^{\circ}=0 \cdot \sin 6503^{\circ}=\sin 23^{\circ}$. Then $\delta=\arccos \left(\sin 23^{\circ}\right)=67^{\circ}$.
|
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 other prime divisors, the number of divisors of $N$ increases by at least a factor of two: in addition to the original $1,2,5,10,20 \ldots$, there will also be $k, 2 k, 5 k, 10 k, 20 k$ and so on. Therefore, $10^{8} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 90 can be represented as a product of two factors greater than 8 in only one way: $90=9 \cdot 10$. Then, $N=2^{8} 5^{9}$ or $N=2^{9} 5^{8}$. Their sum is 700000000.
|
700000000
|
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 $\sin \alpha = \frac{\sqrt{35}}{6}$?
|
Answer: 4.
## 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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{1}{6}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{1}{6} *$ $12=4$
|
4
|
Geometry
|
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 segments. This is the same number of candies that Karlson eats. After 39 minutes, all points will be connected. In total, $\frac{39 \cdot (39-1)}{2}=741$ line segments will be drawn. Therefore, Karlson will eat 741 candies.
|
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 $S=\frac{37 S}{k}+\frac{37 S}{l}$. Therefore, $k l=37(k+l)=>$ $k l-37 k-37 l+1369=1369=>(k-37)(l-37)=1369$. Since $l<k$, that is, $l-37<k-37$. Therefore, $l-37=1$ and $k-37=1369$. Thus, $k=D G=1406$.
|
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 the areas of triangles $A O B, B O C, A O C$ is equal to the area of triangle $A B C$. This gives us the relation $\frac{1}{2}(x y+y z+x z) \sin 120^{\circ}=\frac{1}{2} \cdot 5 \sqrt{3} \cdot 6$. From this, we obtain the answer.
|
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
$$
\begin{aligned}
& (x+t)^{2024}=x^{2024}+a_{1} x^{2023} t+\ldots+a_{2023} x t^{2023}+t^{2024} \\
& (x-t)^{2024}=x^{2024}-a_{1} x^{2023} t+\ldots-a_{2023} x t^{2023}+t^{2024}
\end{aligned}
$$
Adding them up, we get
$$
(x+t)^{2024}+(x-t)^{2024}=2\left(x^{2024}+a_{2} x^{2022} t^{2}+\ldots+a_{2022} x^{2} t^{2022}+t^{2024}\right)
$$
Notice that when expanding different $t_{1}^{n}$ and $t_{2}^{n}$, we will get different monomials (since the degrees of $x$ will be different). Also, when expanding $t^{n}$, we will get $n+1$ different monomials, so the final answer will be
$$
1+3+\ldots+2023+2025=1013^{2}=1026169
$$
## Solve the following problems with justification of the answer
|
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 $\cos 6480^{\circ}=1$ remains. Similarly, $\sin \alpha+\sin \left(\alpha+1^{\circ}\right)+\cdots+\sin \left(\alpha+179^{\circ}\right)=0$ and $\sin 2905^{\circ}+\sin 2906^{\circ}+\cdots+\sin 6504^{\circ}=0 \cdot \sin 6505^{\circ}=\sin 25^{\circ}$. Then $\delta=\arccos \left(\sin 25^{\circ}\right)=65^{\circ}$.
|
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 other prime divisors, the number of divisors of $N$ increases by at least a factor of two: in addition to the original $1,2,5,10,20 \ldots$, there will also be $k, 2 k, 5 k, 10 k, 20 k$ and so on.
Therefore, $10^{6} k=2^{a} 5^{b}$, and the number of its divisors is $(a+1)(b+1)$. The number 56 can be represented as a product of two factors greater than 6 in only one way: $56=7 \cdot 8$. Then, $N=2^{6} 5^{7}$ or $N=2^{7} 5^{6}$. Their sum is 7000000.
|
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 $B_{1}, M, C$ lie on the same straight line, and $\Delta B B_{1} M$ is isosceles. Therefore, the inscribed angle $\angle B B_{1} M=90^{\circ}-\alpha$, and the central angle $\angle B O C=180^{\circ}-2 \alpha$. $\triangle B O C$ is isosceles and $\angle O B C=\alpha \cdot \cos \alpha=\sqrt{1-\sin ^{2} \alpha}=\frac{5}{6}$. We find the base of the isosceles triangle using the formula $B C=2 * \cos \alpha * B O=2 * \frac{5}{6} * 12=20$
|
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 segments. Carlson eats as many candies as there are line segments. After 45 minutes, all points will be connected. In total, $\frac{45 \cdot(45-1)}{2}=990$ line segments will be drawn. Therefore, Carlson will eat 990 candies.
|
990
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.