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5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=31$.
 | Answer: 992
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=31(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{31 S}{k}+\frac{31 S}{l}$.... | 992 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=49 \\
z^{2}+x z+x^{2}=124
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 70
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=49, A C^{2}=124$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2026}+(x-y-z)^{2026}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1028196
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2026}+(x-t)^{2026}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2026}=x^{2026}+a_{1} x^{2025} t+\ldots+a_{2025} x t^{2025}+t^{2026} \\
& (x-t)^{2026}=x^{2026}-a_{1} x^{2025} t+\ldots-a_{2... | 1028196 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2907^{\circ}+\sin 2908^{\circ}+\cdots+\sin 6507^{\circ}\right)^{\cos } 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6480^{\circ}\right)
$$ | Answer: $63^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2880^{\circ}+\cos 2881^{\circ}+\cdots+\cos 6479^{\circ}=0$ and in the exponent only $\cos ... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $28$, and $\sin \alpha = \frac{\sqrt{45}}{7}$? | Answer: 16.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 46 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 46 minutes? | Answer: 1035.
Solution: Let's represent 46 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line... | 1035 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=29$.
 | Answer: 870
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=29(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{29 S}{k}+\frac{29 S}{l}$.... | 870 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=64 \\
z^{2}+x z+x^{2}=139
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 80
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=75$, $B C^{2}=64, A C^{2}=139$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2028}+(x-y-z)^{2028}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1030225
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2028}+(x-t)^{2028}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2028}=x^{2028}+a_{1} x^{2027} t+\ldots+a_{2027} x t^{2027}+t^{2028} \\
& (x-t)^{2028}=x^{2028}-a_{1} x^{2027} t+\ldots-a_{2... | 1030225 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 3269^{\circ}+\sin 3270^{\circ}+\cdots+\sin 6869^{\circ}\right)^{\cos } 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6840^{\circ}\right)
$$ | Answer: $61^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6839^{\circ}=0$ and in the exponent only $\cos ... | 61 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Two different natural numbers end with 7 zeros and have exactly 72 divisors. Find their sum.
# | # Answer: 70000000
Solution: Since the number ends with 7 zeros, it has the form $N=10^{7} k$. The smallest number of this form $10^{7}$ has 64 divisors: all divisors are of the form $2^{a} 5^{b}$, where $a$ and $b$ range from 0 to 7. We will show that $k$ does not have any prime divisors other than 2 and 5. If $k$ ha... | 70000000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $21$, and $\sin \alpha = \frac{\sqrt{40}}{7}$? | Answer: 18.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 47 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 47 minutes? | Answer: 1081.
Solution: Let's represent 47 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line... | 1081 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=23$.
 | Answer: 552
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=23(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{23 S}{k}+\frac{23 S}{l}$.... | 552 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=64 \\
z^{2}+x z+x^{2}=172
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 96
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=64, A C^{2}=172$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2030}+(x-y-z)^{2030}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1032256
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2030}+(x-t)^{2030}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2030}=x^{2030}+a_{1} x^{2029} t+\ldots+a_{2029} x t^{2029}+t^{2030} \\
& (x-t)^{2030}=x^{2030}-a_{1} x^{2029} t+\ldots-a_{2... | 1032256 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 3271^{\circ}+\sin 3272^{\circ}+\cdots+\sin 6871^{\circ}\right)^{\cos } 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6840^{\circ}\right)
$$ | Answer: $59^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 3240^{\circ}+\cos 3241^{\circ}+\cdots+\cos 6839^{\circ}=0$ and in the exponent only $\cos ... | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $14$, and $\sin \alpha = \frac{\sqrt{33}}{7}$? | Answer: 16
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Notice that points $B_... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 48 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 48 minutes? | Answer: 1128.
Solution: Let's represent 48 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line s... | 1128 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Rectangles $A B C D, D E F G, C E I H$ have equal areas and integer sides. Find $D G$, if $B C=19$.
 | Answer: 380
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=19(a+b)$. According to the condition, $S$ is divisible by $a$ and $b$, that is, $S=a k$ and $S=b l$, where $k=D G$ and $l=C H-$ are natural numbers. Then $a=\frac{S}{k}$ and $b=\frac{S}{l}$. We get that $S=\frac{19 S}{k}+\frac{19 S}{l}$.... | 380 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=49 \\
z^{2}+x z+x^{2}=157
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 84
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=108$, $B C^{2}=49, A C^{2}=157$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum o... | 84 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2032}+(x-y-z)^{2032}$, the parentheses were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1034289
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2032}+(x-t)^{2032}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2032}=x^{2032}+a_{1} x^{2031} t+\ldots+a_{2031} x t^{2031}+t^{2032} \\
& (x-t)^{2032}=x^{2032}-a_{1} x^{2031} t+\ldots-a_{2... | 1034289 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2193^{\circ}+\sin 2194^{\circ}+\cdots+\sin 5793^{\circ}\right)^{\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5760^{\circ}}\right)
$$ | # Answer: $57^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$, it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5759^{\circ}=0$ and in the exponent only $\c... | 57 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 $16$, and $\sin \alpha = \frac{\sqrt{55}}{8}$? | 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_{... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 49 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 49 minutes? | Answer: 1176.
Solution: Let's represent 49 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... | 1176 | 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 $B C=17$.
 | Answer: 306
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=17(a+b)$. 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 $S=\frac{17 S}{k}+\frac{17 S}{l}$.... | 306 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=108 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=117
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 36
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, $A B^{2}=108$, $B C^{2}=9, A C^{2}=117$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2034}+(x-y-z)^{2034}$, 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: 1036324
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2034}+(x-t)^{2034}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2034}=x^{2034}+a_{1} x^{2033} t+\ldots+a_{2033} x t^{2033}+t^{2034} \\
& (x-t)^{2034}=x^{2034}-a_{1} x^{2033} t+\ldots-a_{2... | 1036324 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the degree measure of the angle
$$
\delta=\arccos \left(\left(\sin 2195^{\circ}+\sin 2196^{\circ}+\cdots+\sin 5795^{\circ}\right)^{\cos } 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5760^{\circ}\right)
$$ | Answer: $55^{\circ}$
Solution: From the statement $\cos \alpha+\cos \left(\alpha+180^{\circ}\right)=0$ it follows that $\cos \alpha+\cos \left(\alpha+1^{\circ}\right)+$ $\cdots+\cos \left(\alpha+179^{\circ}\right)=0$. Then $\cos 2160^{\circ}+\cos 2161^{\circ}+\cdots+\cos 5759^{\circ}=0$ and in the exponent only $\cos ... | 55 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) On the radius $A O$ of a circle with center $O$, a point $M$ is chosen. On one side of $A O$ on the circle, points $B$ and $C$ are chosen such that $\angle A M B = \angle O M C = \alpha$. Find the length of $B C$ if the radius of the circle is $16$, and $\sin \alpha = \frac{\sqrt{39}}{8}$? | Answer: 20.
## Solution:
Consider point $B_{1}$, which is symmetric to point $B$ with respect to the line $O A$. It also lies on the circle and $\angle A M B=\alpha$. Note that points $B_{... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. (7 points) On the board, 50 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 50 minutes? | Answer: 1225.
Solution: Let's represent 50 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 conne... | 1225 | 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=13$.
 | Answer: 182
Solution: Let $D E=a$ and $E C=b$. Then the area of the rectangles $S=13(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{13 S}{k}+\frac{13 S}{l}$.... | 182 | 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}=147 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=156
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 42
Solution: Let there be three rays with vertex $O$, forming angles of $120^{\circ}$ with each other. On these rays, we lay off segments $O A=x, O B=y, O C=z$. Then, by the cosine theorem, $A B^{2}=147$, $B C^{2}=9, A C^{2}=156$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) In the expression $(x+y+z)^{2036}+(x-y-z)^{2036}$, the brackets were expanded and like terms were combined. How many monomials $x^{a} y^{b} z^{c}$ with a non-zero coefficient were obtained? | Answer: 1038361
Solution: Let $t=y+z$, then the polynomial can be rewritten as $(x+t)^{2036}+(x-t)^{2036}$. We expand both brackets using the binomial theorem and get
$$
\begin{aligned}
& (x+t)^{2036}=x^{2036}+a_{1} x^{2035} t+\ldots+a_{2035} x t^{2035}+t^{2036} \\
& (x-t)^{2036}=x^{2036}-a_{1} x^{2035} t+\ldots-a_{2... | 1038361 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In a $4 \times 4$ table, 16 different natural numbers are arranged. For each row and each column of the table, the greatest common divisor (GCD) of the numbers located in it was found. It turned out that all eight found numbers are different. For what largest $n$ can we assert that there is a number in such a table ... | # Answer: 32.
Solution. If the greatest common divisor (GCD) in some row is $n$, then there are four numbers in that row that are divisible by $n$, meaning there is a number no less than $4n$. Since the greatest common divisors in all rows are distinct, one of them is certainly no less than 8. Then, in the correspondi... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) Given an isosceles right triangle with a leg of 10. An infinite number of equilateral triangles are inscribed in it as shown in the figure: the vertices lie on the hypotenuse, and the bases are sequentially laid out on one of the legs starting from the right angle vertex. Find the sum of the areas of the ... | Answer: 25 .
 | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. (20 points) Inside an acute triangle $A B C$, a point $M$ is marked. The lines $A M, B M$, $C M$ intersect the sides of the triangle at points $A_{1}, B_{1}$ and $C_{1}$ respectively. It is known that $M A_{1}=M B_{1}=M C_{1}=3$ and $A M+B M+C M=43$. Find $A M \cdot B M \cdot C M$. | Answer: 441.
Solution. Let $A M=x, B M=y, C M=z$. Note that $\frac{M C_{1}}{C C_{1}}=\frac{S_{A M B}}{S_{A B C}}$ and similarly for the other two segments. From the equality $\frac{S_{A M B}}{S_{A B C}}+\frac{S_{B M C}}{S_{A B C}}+\frac{S_{A M C}}{S_{A B C}}=1$ it follows that $\frac{3}{x+3}+\frac{3}{y+3}+\frac{3}{z+3... | 441 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Find all values of the parameter $c$ such that the system of equations has a unique solution
$$
\left\{\begin{array}{l}
2|x+7|+|y-4|=c \\
|x+4|+2|y-7|=c
\end{array}\right.
$$ | Answer: $c=3$.
Solution. Let $\left(x_{0} ; y_{0}\right)$ be the unique solution of the system. Then $\left(-y_{0} ;-x_{0}\right)$ also satisfies the conditions of the system. This solution coincides with the first, so $y_{0}=-x_{0}$. The equation $2|x+7|+|x+4|=$ $c$ has a unique solution $x_{0}=-7$ when $c=|7-4|$, si... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the minimum number of cells on a $3 \times 2016$ board that can be painted so that each cell has a side-adjacent painted cell?
(A. Khryabrov) | Answer: 2016.
Solution: Let's divide our board as follows:
We have obtained two three-cell corners and 1007 D-hexomino figures. In our three-cell corners, at least one cell must be shaded ... | 2016 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) Find the remainder when $20^{16}+201^{6}$ is divided by 9. | Answer: 7
Solution. $201^{16}$ is divisible by 9. 20 gives a remainder of $2.2^{6}$ gives a remainder of $1,2^{16}=2^{6} \cdot 2^{6} \cdot 2^{4}$ gives the same remainder as 16. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) A hare jumps in one direction along a strip divided into cells. In one jump, it can move either one or two cells. In how many ways can the hare reach the 12th cell from the 1st cell?
 On the coordinate plane, all points whose coordinates satisfy the condition
$$
|2 x-2|+|3 y-3| \leq 30
$$
are shaded. Find the area of the resulting figure. | Answer: 300
 | 300 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For a natural number $n$, the smallest divisor $a$, different from 1, and the next largest divisor $b$ were taken. It turned out that $n=a^{a}+b^{b}$. Find $n$. | Answer: $n=2^{2}+4^{4}=260$.
Solution. If $n$ is odd, then all its divisors are also odd. Then $a$ and $b$ are odd and $a^{a}+b^{b}$ is even. Therefore, $n$ is even. Then its smallest divisor, different from 1, is 2, and thus $a=2$. Therefore, $n=2^{2}+b^{b}$. Consequently, $4=n-b^{b}$ is divisible by $b$. Then $b=4$,... | 260 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. (20 points) A four-digit number $\overline{a b c d}$ is called perfect if $a+b=c+d$. How many perfect numbers can be represented as the sum of two four-digit palindromes? | Solution: Let the number $\overline{a b c d}=\overline{n m m n}+\overline{x y y x}$, then
$$
\overline{a b c d}=1001(n+x)+110(m+y) \vdots 11
$$
From the divisibility rule by 11, it follows that $b+d=a+c$. Since the number $\overline{a b c d}$ is perfect, we get $a=d$ and $b=c$, hence the original number is a palindro... | 80 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=1000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+4 x+2$ holds. | Answer: 999001
Solution: In the equation $f(x+2)-f(x)=4 x+2$, we will substitute for $x$ the numbers $0,2,4, \ldots, 998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=4 \cdot 0+2 \\
& f(4)-f(2)=4 \cdot 2+2
\end{aligned}
$$
$$
f(1000)-f(998)=4 \cdot 998+2
$$
Adding the equations, we get: $f(1000)-f(0)=4 \cdot(0+2+4+\cdot... | 999001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 70 negative numbers were recorded. What is the ... | Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=70+62$, which means $x=12$.
2) Let there be $y$ people with "positive tempe... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 30 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-4.5,4.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 90
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-10 \leqslant y-1 \leqslant 8$ and $-7 \leqslant 2-x \leqslant 11$. Therefore, $(y-1)(2-x)+2 \leqslant 8 \cdot 11+2=90$. The m... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) In $\triangle A B C, A B=86$, and $A C=97$. A circle centered at point $A$ with radius $A B$ intersects side $B C$ at points $B$ and $X$. Moreover, $B X$ and $C X$ have integer lengths. What is the length of $B C ?$ | Answer: 61
Solution: Let $x=B X$ and $y=C X$. We will calculate the power of point $C$ in two ways
$$
y(y+x)=97^{2}-86^{2}=2013
$$
Considering all divisors of the number 2013 and taking into account the triangle inequality $\triangle A C X$, we obtain the unique solution 61. | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=25 \\
z^{2}+x z+x^{2}=37
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 20
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, $A B^{2}=12$, $B C^{2}=25$, and $A C^{2}=37$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The s... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=2000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+4 x+2$ holds. | Answer: 3998001
Solution: In the equation $f(x+2)-f(x)=4 x+2$, we will substitute for $x$ the numbers $0, 2, 4, \ldots, 1998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=4 \cdot 0+2 \\
& f(4)-f(2)=4 \cdot 2+2
\end{aligned}
$$
$$
f(2000)-f(1998)=4 \cdot 1998+2
$$
Adding the equations, we get: $f(2000)-f(0)=4 \cdot(0+2+4... | 3998001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 68 positive and 64 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=64+68$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 28 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-5.5,5.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 132
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-12 \leqslant y-1 \leqslant 10$ and $-9 \leqslant 2-x \leqslant 13$. Therefore, $(y-1)(2-x)+2 \leqslant 10 \cdot 13+2=132$... | 132 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (8 points) In $\triangle A B C, A B=86$, and $A C=97$. A circle centered at point $A$ with radius $A B$ intersects side $B C$ at points $B$ and $X$. Additionally, $B X$ and $C X$ have integer lengths. What is the length of $B C ?$ | Answer: 61
Solution: Let $x=B X$ and $y=C X$. We will calculate the power of point $C$ in two ways
$$
y(y+x)=97^{2}-86^{2}=2013
$$
Considering all divisors of the number 2013 and taking into account the triangle inequality $\triangle A C X$, we obtain the unique solution 61. | 61 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) On the board, 33 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 33 minutes? | Answer: 528.
Solution: Let's represent 33 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $xy$ line seg... | 528 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=28
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 16
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, $A B^{2}=12$, $B C^{2}=16, A C^{2}=28$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=3000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+3 x+2$ holds. | Answer: 6748501
Solution: In the equation $f(x+2)-f(x)=3 x+2$, we will substitute for $x$ the numbers $0, 2, 4, \ldots, 2998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=3 \cdot 0+2 \\
& f(4)-f(2)=3 \cdot 2+2
\end{aligned}
$$
$$
f(3000)-f(2998)=3 \cdot 2998+2
$$
Adding the equations, we get: $f(3000)-f(0)=3 \cdot(0+2+4... | 6748501 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 78 positive and 54 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=54+78$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 26 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-6.5,6.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 182
Solution: Note that $a+2 b+c+2 d-a b-b c-c d-d a=(a+c)+2(b+d)-(a+$ $c)(b+d)$. Let $x=a+c, y=b+d$, then we will find the maximum value of the expression $x+2 y-x y=(y-1)(2-x)+2$, where $-14 \leqslant y-1 \leqslant 12$ and $-11 \leqslant 2-x \leqslant 15$. Therefore, $(y-1)(2-x)+2 \leqslant 12 \cdot 15+2=182... | 182 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) On the board, 32 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 32 minutes? | Answer: 496.
Solution: Let's represent 32 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line ... | 496 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=12 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=21
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 12
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, $A B^{2}=12$, $B C^{2}=9, A C^{2}=21$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=4000$, if $f(0)=1$ and for any $x$ the equality $f(x+2)=f(x)+3 x+2$ holds. | Answer: 11998001
Solution: In the equation $f(x+2)-f(x)=3 x+2$, we will substitute for $x$ the numbers $0,2,4, \ldots, 3998$. We get:
$$
\begin{aligned}
& f(2)-f(0)=3 \cdot 0+2 \\
& f(4)-f(2)=3 \cdot 2+2
\end{aligned}
$$
$$
f(4000)-f(3998)=3 \cdot 3998+2
$$
Adding the equations, we get: $f(4000)-f(0)=3 \cdot(0+2+4+... | 11998001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 92 positive and 40 negative numbers were recorded. What is the ... | Answer: 2
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=40+92$, which means $x=12$.
2) Let there be $y$ people with "positive t... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 24 minutes?
 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 |
6. (8 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 eat in 31 minutes? | Answer: 465.
Solution: Let's represent 31 units as points on a plane. Each time we combine numbers, we will connect the points of one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $xy$ line s... | 465 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=27 \\
y^{2}+y z+z^{2}=25 \\
z^{2}+x z+x^{2}=52
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 30
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, $A B^{2}=27$, $B C^{2}=25, A C^{2}=52$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=1500$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 750001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute the numbers $0,3,6, \ldots, 1497$ for $x$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(1500)-f(1497)=2 \cdot 1497+3
$$
Adding the equations, we get: $f(1500)-f(0)=2 \cdot(0+3+6+\c... | 750001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 50 positive and 60 negative numbers were recorded. What is the ... | Answer: 5
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers arise from their interactions.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=60+50$, which means $x=11$.
2) Let there be $y$ people with "positive tempe... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 22 minutes?
 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 |
6. (8 points) On the board, 30 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 30 minutes? | Answer: 435.
Solution: Let's represent 30 units as points on a plane. Each time we combine numbers, we will connect the points of one group to all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connected by $x y$ line se... | 435 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=27 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=43
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 24
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, $A B^{2}=27$, $B C^{2}=16, A C^{2}=43$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=3000$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 3000001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 2997$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(3000)-f(2997)=2 \cdot 2997+3
$$
Adding the equations, we get: $f(3000)-f(0)=2 \cdot(0+3+6+\... | 3000001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 54 positive and 56 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=56+54$, which means $x=11$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 20 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-9.5,9.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 380
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 $-20 \leqslant y-1 \leqslant 18$ and $-17 \leqslant 2-x \leqslant 21$. Therefore, $(y-1)(2-x)+2 \leqslant 18 \cdot 21+2=380... | 380 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) On the board, 29 ones are written. Every minute, Karlson erases two arbitrary numbers and writes their sum on the board, and then eats a number of candies equal to the product of the two erased numbers. What is the maximum number of candies he could have eaten in 29 minutes? | Answer: 406.
Solution: Let's represent 29 units as points on a plane. Each time we combine two numbers, we will connect the points corresponding to one group with all the points of the second group with line segments. Note that if we replace numbers $x$ and $y$ with $x+y$, the groups " $x$ " and " $y$ " will be connec... | 406 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=27 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=36
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 18
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, $A B^{2}=27$, $B C^{2}=9, A C^{2}=36$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=4500$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 6750001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 4497$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(4500)-f(4497)=2 \cdot 4497+3
$$
Adding the equations, we get: $f(4500)-f(0)=2 \cdot(0+3+6+\... | 6750001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 62 positive and 48 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+62$, which means $x=11$.
2) Let there be $y$ people with "positive t... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 18 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-10.5,10.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 462
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 $-22 \leqslant y-1 \leqslant 20$ and $-19 \leqslant 2-x \leqslant 23$. Therefore, $(y-1)(2-x)+2 \leqslant 20 \cdot 23+2=462... | 462 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=48 \\
y^{2}+y z+z^{2}=25 \\
z^{2}+x z+x^{2}=73
\end{array}\right.
$$
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, $A B^{2}=48$, $B C^{2}=25, A C^{2}=73$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=6000$, if $f(0)=1$ and for any $x$ the equality $f(x+3)=f(x)+2 x+3$ holds. | Answer: 12000001
Solution: In the equation $f(x+3)-f(x)=2 x+3$, we will substitute for $x$ the numbers $0,3,6, \ldots, 5997$. We get:
$$
\begin{aligned}
& f(3)-f(0)=2 \cdot 0+3 \\
& f(6)-f(3)=2 \cdot 3+3
\end{aligned}
$$
$$
f(6000)-f(5997)=2 \cdot 5997+3
$$
Adding the equations, we get: $f(6000)-f(0)=2 \cdot(0+3+6+... | 12000001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 42 positive and 48 negative numbers were recorded. What is the ... | Answer: 4
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each giving $x-1$ answers, so $x(x-1)=48+42$, which means $x=10$.
2) Let there be $y$ people with "positive t... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 16 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-11.5,11.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 552
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 $-24 \leqslant y-1 \leqslant 22$ and $-21 \leqslant 2-x \leqslant 25$. Therefore, $(y-1)(2-x)+2 \leqslant 22 \cdot 25+2=552$... | 552 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=48 \\
y^{2}+y z+z^{2}=16 \\
z^{2}+x z+x^{2}=64
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 32
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, $A B^{2}=48$, $B C^{2}=16, A C^{2}=64$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of ... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Find the value of the function $f(x)$ at the point $x_{0}=2000$, if $f(0)=1$ and for any $x$ the equality $f(x+4)=f(x)+3 x+4$ holds. | Answer: 1499001
Solution: In the equation $f(x+4)-f(x)=3 x+4$, we will substitute for $x$ the numbers $0, 4, 8, \ldots, 1996$. We get:
$$
\begin{aligned}
& f(4)-f(0)=3 \cdot 0+4 \\
& f(8)-f(4)=3 \cdot 4+4
\end{aligned}
$$
$$
f(2000)-f(1996)=3 \cdot 1996+4
$$
Adding the equations, we get: $f(2000)-f(0)=3 \cdot(0+4+8... | 1499001 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) At the World Meteorologists Conference, each participant in turn announced the average monthly temperature in their hometown. At this moment, all the others recorded the product of the temperatures in their and the speaker's cities. In total, 48 positive and 42 negative numbers were recorded. What is the ... | Answer: 3
Solution: When solving, we will not consider cities with a zero temperature, as no positive or negative numbers appear when interacting with them.
1) Let there be $x$ participants at the conference, each of them gave $x-1$ answers, so $x(x-1)=42+48$, which means $x=10$.
2) Let there be $y$ people with "posi... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) Four mole burrows $A, B, C, D$ are connected sequentially by three tunnels. Each minute, the mole runs through a tunnel to one of the adjacent burrows. In how many ways can the mole get from burrow $A$ to $C$ in 14 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-12.5,12.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 650
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 $-26 \leqslant y-1 \leqslant 24$ and $-23 \leqslant 2-x \leqslant 27$. Therefore, $(y-1)(2-x)+2 \leqslant 24 \cdot 27+2=650$. ... | 650 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 points) On the board, 26 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 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Let for positive numbers $x, y, z$ the following system of equations holds:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=48 \\
y^{2}+y z+z^{2}=9 \\
z^{2}+x z+x^{2}=57
\end{array}\right.
$$
Find the value of the expression $x y+y z+x z$. | Answer: 24
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, $A B^{2}=48$, $B C^{2}=9, A C^{2}=57$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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