problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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2. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$ and $2 \times 2$ in a checkered square of size $10 \times 10$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initi... | 2. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$ and $2 \times 2$ in a checkered square of size $10 \times 10$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initi... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,375 |
3. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$, and $2 \times 2$ in a checkered square of size $9 \times 9$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initia... | 3. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$, and $2 \times 2$ in a checkered square of size $9 \times 9$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initia... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,376 |
4. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$, and $2 \times 2$ in a checkered square of size $8 \times 8$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initia... | 4. Peter and Victor are playing a game, taking turns in painting out rectangles of size $1 \times 1, 1 \times 2$, and $2 \times 2$ in a checkered square of size $8 \times 8$. Each of the players paints in their own color (Peter's color is red, and Victor's green). Recoloring already colored cells is not allowed, initia... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 19,377 |
9. (20 points) On the sides of triangle $A B C$, as bases, similar isosceles triangles $A P B, A Q C$, and $B R C$ are constructed. Triangles $A P B$ and $A Q C$ are constructed outward, while triangle $B R C$ is constructed on the same side of $B C$ as triangle $A B C$. Prove that quadrilateral $P A Q R$ is a parallel... | Solution. We will show that $A Q \| P R$. Consider triangles $B P R$ and $B A C$. Note that $\angle P B R=\angle A B C$, and also $B P / B R=A B / B C$, from the similarity of triangles $P A B$ and $R C B$. Therefore, triangles $B P R$ and $B A C$ are similar and, moreover, similarly oriented. Then $\angle(A C, P R)=\a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,378 |
10. (20 points) How many zeros does the number $4^{5^{6}}+6^{5^{4}}$ end with in its decimal representation?
# | # Answer: 5
Solution. Consider the number $4^{25}+6$. Let's check that it is divisible by 5 and not divisible by 25. We have
$$
\begin{gathered}
4^{25}+6 \equiv(-1)^{25}+6 \equiv 0 \quad(\bmod 5) \\
4^{25} \equiv 1024^{5}+6 \equiv(-1)^{5}+6 \equiv 5 \quad(\bmod 25)
\end{gathered}
$$
We will write $5^{t} \| c$, if $5... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,379 |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2018
$$ | Answer: 4076360
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2018$. From which $n=(2018+1)^{2}-1=... | 4076360 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,380 |
2. (5 points) On Valentine's Day, every student in the school gave a valentine to every female student. It turned out that the number of valentines was 16 more than the total number of students. How many valentines were given? | Answer: 36
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the condition, $x y=x+y+16$. Then $(x-1)(y-1)=17$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 17. The number of valentines is $2 \cdot 18=36$. | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,381 |
3. (7 points) There are 5 blue, 6 red, and 7 white bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 365904
Solution: First, arrange all the blue and red bulbs in $C_{11}^{5}$ ways. In the gaps between them and at the ends, choose 7 positions and insert the white bulbs. There are $C_{12}^{7}$ ways to do this. In total, there are $C_{11}^{5} \cdot C_{12}^{7}$ ways to compose the garland from the available bulb... | 365904 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,382 |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-4.5 ; 4.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 90
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-10 \leqslant y-1 \leqslant 8$ and $-7 \leqslant 2-x \leqslant 11$. Therefore, $(y-1)(2-x)+2 \leqslant 8 \cdot 11+2=90$. The m... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,383 |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $10$, and $\cos \alpha = \frac{2}{5}$? | Answer: 8.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_{... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,384 |
6. (8 points) By expanding the expression $(1+\sqrt{7})^{205}$ using the binomial theorem, we obtain terms of the form $C_{205}^{k}(\sqrt{7})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 149
Solution: The ratio of two consecutive terms $\frac{C_{205}^{k+1}(\sqrt{7})^{k+1}}{C_{205}^{k}(\sqrt{7})^{k}}$ is greater than 1 when $k<$ $\frac{205 \sqrt{7}-1}{\sqrt{7}+1}$. Then the terms increase up to $\left[\frac{205 \sqrt{7}-1}{\sqrt{7}+1}\right]+1$, and then decrease. | 149 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,385 |
7. (10 points) On the board, 25 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 25 minutes? | Answer: 300.
Solution: Let's represent 25 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 300 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,386 |
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=18, B C=12 \sqrt{3}-9$. | Answer: 30
## Solution:
The quadrilateral is inscribed, hence $\angle A+\angle C=180^{\circ}$. From the given ratio, $\angle A=2 x, \angle C=4 x$. Therefore, $x=30^{\circ}$ and $\angle A=60... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,387 |
9. (20 points) From point $D$, the midpoint of the base $B C$ of isosceles triangle $A B C$, a perpendicular $D E$ is dropped to the lateral side $A C$. The circumcircle of triangle $A B D$ intersects line $B E$ at points $B$ and $F$. Prove that line $A F$ passes through the midpoint of segment $D E$. | Solution: In the right triangle $A C D: \angle A D E=90^{\circ}-\angle C D E=\angle A C D=\angle A B D$. It follows that $D E$ is a tangent to the circumcircle of triangle $A B D$. Then $G D^{2}=G F \cdot G A$, where $G$ is the intersection point of lines $D E$ and $A F$. $\angle A F B=\angle A D B=90^{\circ}$ - as ins... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,388 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 101 to 200. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 200: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,389 |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2017
$$ | Answer: 4072323
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2017$. From which $n=(2017+1)^{2}-1=... | 4072323 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,390 |
2. (5 points) On Valentine's Day, every student in the school gave each female student a valentine. It turned out that the number of valentines was 18 more than the total number of students. How many valentines were given? | Answer: 40
Solution: Let $x, y$ be the number of boys and girls in the school, respectively. According to the condition, $x y=x+y+18$. Then $(x-1)(y-1)=19$. Therefore, the numbers $x-1$ and $y-1$ are 1 and 19. The number of valentines is $2 \cdot 20=40$. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,391 |
3. (7 points) There are 7 blue, 6 red, and 10 white light bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 1717716
Solution: First, arrange all the blue and red bulbs in $C_{13}^{7}$ ways. In the gaps between them and at the ends, choose 10 positions and insert the white bulbs. There are $C_{14}^{10}$ ways to do this. In total, there are $C_{13}^{7} \cdot C_{14}^{10}$ ways to compose the garland from the available ... | 1717716 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,392 |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-5.5 ; 5.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 132
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-12 \leqslant y-1 \leqslant 10$ and $-9 \leqslant 2-x \leqslant 13$. Therefore, $(y-1)(2-x)+2 \leqslant 10 \cdot 13+2=132$... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,393 |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $8$ and $\cos \alpha = \frac{3}{4}$? | Answer: 12.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{1... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,394 |
6. (8 points) By expanding the expression $(1+\sqrt{5})^{206}$ using the binomial theorem, we obtain terms of the form $C_{206}^{k}(\sqrt{5})^{k}$. Find the value of $k$ for which such a term attains its maximum value. | Answer: 143
Solution: The ratio of two consecutive terms $\frac{C_{206}^{k+1}(\sqrt{5})^{k+1}}{C_{206}^{k}(\sqrt{5})^{k}}$ is greater than 1 when $k<$ $\frac{206 \sqrt{5}-1}{\sqrt{5}+1}$. Then the terms increase up to $\left[\frac{206 \sqrt{5}-1}{\sqrt{5}+1}\right]+1$, and then decrease. | 143 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,395 |
7. (10 points) On the board, there are 26 ones. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 26 minutes? | Answer: 325.
Solution: Let's represent 26 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 325 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,396 |
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=20, B C=24 \sqrt{3}-10$. | Answer: 52
## Solution:
The quadrilateral is inscribed, so $\angle A + \angle C = 180^{\circ}$. From the given ratio, $\angle A = 2x, \angle C = 4x$. Therefore, $x = 30^{\circ}$ and $\angle... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,397 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 102 to 201. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 201: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,398 |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2016
$$ | Answer: 4068288
Solution: Note that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2016$. From which $n=(2016+1)^{2}-1=40... | 4068288 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,399 |
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 | 19,400 |
3. (7 points) There are 8 blue, 8 red, and 11 white bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 159279120
Solution: First, arrange all the blue and red bulbs in $C_{16}^{8}$ ways. In the gaps between them and at the ends, choose 11 positions and insert the white bulbs. There are $C_{17}^{11}$ ways to do this. In total, there are $C_{16}^{8} \cdot C_{17}^{11}$ ways to compose the garland from the availabl... | 159279120 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,401 |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-6.5 ; 6.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 182
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-14 \leqslant y-1 \leqslant 12$ and $-11 \leqslant 2-x \leqslant 15$. Therefore, $(y-1)(2-x)+2 \leqslant 12 \cdot 15+2=182... | 182 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,402 |
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_{... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,403 |
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 | 19,404 |
7. (10 points) On the board, 27 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 27 minutes? | Answer: 351.
Solution: Let's represent 27 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line se... | 351 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,405 |
8. (10 points) In the inscribed quadrilateral $A B C D$, the degree measures of the angles are in the ratio $\angle A: \angle B: \angle C=2: 3: 4$. Find the length of $A C$, if $C D=16, B C=15 \sqrt{3}-8$. | Answer: 34
## Solution:
The quadrilateral is inscribed, hence $\angle A+\angle C=180^{\circ}$. From the given ratio, $\angle A=2 x, \angle C=4 x$. Therefore, $x=30^{\circ}$ and $\angle A=60... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,406 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 103 to 202. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 202: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,407 |
1. (5 points) Find the value of $n$ for which the following equality holds:
$$
\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=2015
$$ | Answer: 4064255
Solution: Notice that $\frac{1}{\sqrt{k}+\sqrt{k+1}}=\sqrt{k+1}-\sqrt{k}$. Then $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+$ $\ldots+\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\ldots+\sqrt{n+1}-\sqrt{n}=\sqrt{n+1}-1=2015$. From which $n=(2015+1)^{2}-1=... | 4064255 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,408 |
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 | 19,409 |
3. (7 points) There are 8 blue, 7 red, and 12 white light bulbs. In how many ways can they be arranged into a garland (using all the bulbs) so that no two white bulbs are adjacent? | Answer: 11711700
Solution: First, arrange all the blue and red bulbs in $C_{15}^{8}$ ways. In the gaps between them and at the ends, choose 12 positions and insert the white bulbs. There are $C_{16}^{12}$ ways to do this. In total, there are $C_{15}^{8} \cdot C_{16}^{12}$ ways to compose the garland from the available... | 11711700 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 19,410 |
4. (7 points) The numbers $a, b, c, d$ belong to the interval $[-7.5 ; 7.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 240
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-16 \leqslant y-1 \leqslant 14$ and $-13 \leqslant 2-x \leqslant 17$. Therefore, $(y-1)(2-x)+2 \leqslant 14 \cdot 17+2=240... | 240 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,411 |
5. (8 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $12$, and $\cos \alpha = \frac{5}{6}$? | Answer: 20.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,412 |
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 | 19,413 |
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 | 19,414 |
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 | 19,415 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 104 to 203. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 203: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,416 |
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 | 19,417 |
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 | 19,418 |
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 | 19,419 |
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 | 19,420 |
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 | 19,421 |
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 | 19,422 |
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 | 19,423 |
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 | 19,424 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 105 to 204. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 204: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,425 |
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 | 19,426 |
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 | 19,427 |
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 | 19,428 |
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 | 19,429 |
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}... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,430 |
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 | 19,431 |
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 | 19,432 |
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 | 19,433 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 106 to 205. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 205: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,434 |
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 | 19,435 |
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 | 19,436 |
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 | 19,437 |
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 | 19,438 |
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_{... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 19,439 |
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 | 19,440 |
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 | 19,441 |
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 | 19,442 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 107 to 206. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 206: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,443 |
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 | 19,444 |
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 | 19,446 |
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 | 19,447 |
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 | 19,448 |
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 | 19,449 |
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 | 19,450 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 108 to 207. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 207: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,451 |
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 | 19,452 |
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 | 19,453 |
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 | 19,454 |
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 | 19,455 |
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 | 19,456 |
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 | 19,457 |
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 | 19,458 |
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 | 19,459 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 109 to 208. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 208: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,460 |
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 | 19,461 |
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 | 19,462 |
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 | 19,463 |
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 | 19,464 |
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 | 19,465 |
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 | 19,466 |
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 | 19,467 |
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 | 19,468 |
10. (20 points) Vasya filled the cells of a $10 \times 10$ table with all natural numbers from 110 to 209. He calculated the products of the numbers in each row of the table and obtained a set of ten numbers. Then he calculated the products of the numbers in each column of the table and also obtained a set of ten numbe... | Answer: No, they could not.
Solution: Each of the products of the numbers in the ten rows of the table can be represented as a product of prime factors. Let's list the last eleven prime numbers not exceeding 209: $151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199$. Note that each of these eleven numbers can only a... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,469 |
1. In a bag, there are 20 balls of three colors: white, black, and red. There is at least one ball of each color. It turned out that if the number of white balls in the bag is doubled, the probability of drawing a white ball will be $\frac{1}{5}$ less than the initial probability of drawing a red ball (before doubling ... | Answer: 5 white, 3 black, and 12 red.
Solution: Let there be $x$ white balls, $y$ black balls, and $z$ red balls. Since there were a total of 20 balls, then $x+y+z=20$. The initial probability of drawing a red ball is
$$
p_{1}=\frac{z}{x+y+z}=\frac{z}{20}
$$
and the probability of drawing a white ball after doubling... | (5,3,12) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,470 |
2. A total of 40 girls arrived at the winter school. Each of them has an even number of friends among the other arrivals. It turned out that all the girls could be paired up so that in each room, friends lived together. After this, two friends who lived in the same room had a quarrel. Is it true that it is still possib... | Solution: No, it is not correct. Let there be 4 special girls A, B, V, and G. Moreover, A is friends with B and V, B is friends with A and B, B is friends with A, B, and G, G is friends only with V. The other 36 girls are friends, for example, in a circle. Obviously, if girls A and B have a quarrel, it is impossible to... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 19,471 |
3. Find all natural numbers $a$ and $b$ such that
$$
a^{3}-b^{3}=633 \cdot p
$$
where $p$ is some prime number. | Answer: $a=16, b=13$.
Solution: Factorize the left side of the equation
$$
\begin{gathered}
(a-b)\left(a^{2}+a b+b^{2}\right)=3 \cdot 211 \cdot p \Longrightarrow \\
\Longrightarrow(a-b)\left((a-b)^{2}+3 a b\right)=3 \cdot 211 \cdot p
\end{gathered}
$$
Notice that the right side is divisible by 3, so at least one of ... | =16,b=13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 19,472 |
4. Find all values of the parameter $a$ for which the equation
$$
\left(x^{2}-6 x+8-a\right)\left(x-a^{2}+6 a-8\right)=0
$$
has exactly two distinct real roots.
(P. Alishev) | Answer: $a \in\{-1\} \cup\left\{\frac{5}{2} \pm \frac{\sqrt{13}}{2}\right\} \cup\left\{\frac{7}{2} \pm \frac{\sqrt{17}}{2}\right\}$
Solution: We will solve the problem graphically. Replace $a$ with $y$ and draw the set of solutions of the equation in the $Oxy$ plane.
The original equation is equivalent to the followi... | \in{-1}\cup{\frac{5}{2}\\frac{\sqrt{13}}{2}}\cup{\frac{7}{2}\\frac{\sqrt{17}}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 19,473 |
5. The tangent at point $A$ to the circumcircle of triangle $A B C$ intersects line $B C$ at point $K$. On the perpendicular to segment $B C$ at point $B$, a point $L$ is taken such that $A L = B L$. On the perpendicular to segment $B C$ at point $C$, a point $M$ is taken such that $A M = C M$. Prove that $K, L$, and $... | # Solution:
Since $A L = L B$ and $\angle L B C = 90^{\circ}$, $L$ is the intersection point of the perpendicular bisector of side $A B$ and the perpendicular to side $B C$ at point $B$. Si... | proof | Geometry | proof | Yes | Yes | olympiads | false | 19,474 |
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