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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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 $f(x+4)=f(x)+3 x+4$ holds. | Answer: 5998001
Solution: In the equation $f(x+4)-f(x)=3 x+4$, we will substitute for $x$ the numbers $0,4,8, \ldots, 3996$. We get:
$$
\begin{aligned}
& f(4)-f(0)=3 \cdot 0+4 \\
& f(8)-f(4)=3 \cdot 4+4
\end{aligned}
$$
$$
f(4000)-f(3996)=3 \cdot 3996+4
$$
Adding the equations, we get: $f(4000)-f(0)=3 \cdot(0+4+8+\... | 5998001 | Algebra | 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 12 minutes?
 The numbers $a, b, c, d$ belong to the interval $[-13.5,13.5]$. Find the maximum value of the expression $a+2 b+c+2 d-a b-b c-c d-d a$. | Answer: 756
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 $-28 \leqslant y-1 \leqslant 26$ and $-25 \leqslant 2-x \leqslant 29$. Therefore, $(y-1)(2-x)+2 \leqslant 26 \cdot 29+2=756$. ... | 756 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (8 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 | 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:
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=75 \\
y^{2}+y z+z^{2}=4 \\
z^{2}+x z+x^{2}=79
\end{array}\right.
$$
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, $A B^{2}=75$, $B C^{2}=4, A C^{2}=79$. Note that triangle $A B C$ is a right triangle with hypotenuse $A C$. The sum of t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (5 points) When dividing the numbers 312837 and 310650 by a certain three-digit natural number, the remainders were the same. Find this remainder. | Answer: 96
Solution. $312837-310650=2187=3^{7}$. The three-digit divisors of this number are also powers of three, i.e., $3^{6}=729,3^{5}=243$. By dividing these numbers by 729 and 243 with a remainder, we find that the remainder is always 96. | 96 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (13 points) On the coordinate plane, all points whose coordinates satisfy the conditions
$$
\left\{\begin{array}{l}
|2 x+3 y|+|3 x-2 y| \leq 13 \\
2 x^{2}-3 x y-2 y^{2} \leq 0
\end{array}\right.
$$
are shaded. Find the area of the resulting figure.
# | # Answer: 13
Solution. The solution to the first inequality defines the square $ABCD$, the solution to the second inequality is the shaded part of the plane between the perpendicular lines $2x + y = 0$ and $x - 2y = 0$.
 It is known that no digit of a three-digit number is zero and the sum of all possible two-digit numbers formed from the digits of this number is equal to this number. Find the largest such three-digit number. | # Answer: 396
Solution. If there are identical digits, then the sum of three two-digit numbers is less than 300. If the digits are different, then $\overline{a b c}=\overline{a b}+\overline{a c}+\overline{b a}+\overline{b c}+\overline{c a}+\overline{c b}=22(a+b+c)$. The sum of the digits gives the same remainder when... | 396 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) A square is divided into 2016 triangles, with the vertices of no triangle lying on the sides or inside another triangle. The sides of the square are sides of some triangles in the partition. How many points, which are vertices of the triangles, are located inside the square? | Answer: 1007
Solution. Let us have $k$ points inside the square. Then the sum of the angles of all triangles is $360 k+4 \cdot 90=180 \cdot 2016$ degrees. | 1007 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Anton, Boris, Vadim, Gena, Dima, and Egor gathered at the cinema. They bought 6 seats in a row. Anton and Boris want to sit next to each other, while Vadim and Gena do not want to. In how many ways can the boys sit in their seats considering these preferences?
# | # Answer: 144
Solution. The total number of seating arrangements where Anton and Boris sit next to each other is $2 \cdot 5!=240$. The number of seating arrangements where the pairs Anton-Boris and Vadim-Gena end up next to each other is $2 \cdot 2 \cdot 4!=96$. Subtracting the second set from the first gives the answ... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (12 points) On the coordinate plane, all points whose coordinates satisfy the condition
$$
|| x|-2|+|y-3| \leq 3
$$
are shaded. Find the area of the resulting figure. | Answer: 34
Solution. The polygon that results from solving the equation ||$x|-2|+$ $|y-3|=3$ consists of segments of straight lines, glued together at special points: $x=0, x=2, x=$ $-2, y=3$.
$, where $i=1,2, \ldots, n$ satisfy the following conditions:
1) $a_{i}+b_{i}+c_{i}=2017$ for all $i=1,2, \ldots, n$;
2) if $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$. What is the maximum possible value of $n$? (M. Popov) | Answer: 1343.
Solution: Note that
$$
\sum_{i=1}^{n} a_{i} \geqslant \sum_{i=1}^{n} i=\frac{n(n+1)}{2} .
$$
A similar inequality is written for the sums $b_{i}$ and $c_{i}$. Adding the three obtained inequalities, we get
$$
\begin{gathered}
3 \cdot \frac{n(n+1)}{2} \leqslant \sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}... | 1343 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Cinderella, together with her fairy godmother, released a collection of crystal slippers featuring 7 new models. The fairy tale heroines organized a presentation of the collection for some guests: the audience was supposed to say which slippers they liked. The guests wrote in a questionnaire which models they though... | Solution. Suppose that the guests were given a questionnaire where they could put a "1" next to each pair of shoes they liked and a "0" next to the pair they did not select as the best. Thus, the opinion of each guest can be recorded as a string of "1"s and "0"s. Then, the number of different sets of favorite shoes is ... | 128 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
Over two days, 50 financiers raised funds to combat a new virus. Each of them made a one-time contribution of a whole number of thousands of rubles, not exceeding 100. Each contribution on the first day did not exceed 50 thousand, while on the second day, it was more than this amount; and no pair of the 50 c... | The solution significantly depends on whether all contributions were distinct or could repeat.
Let's first consider the case where all contributions are distinct. Any natural number greater than 50 but not exceeding 100 can be represented as $50+n$, where $n \in [1, 2, 3, \ldots, 50]$. According to the condition, ther... | 2525 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. After another labor season, the electrified part of the Mediterranean Tundra doubled. At the same time, its non-electrified part decreased by $25 \%$. What fraction of the entire Tundra was not supplied with electricity at the beginning of the labor season? | # Solution
Let $x$ and $y$ be the fractions of the electrified and non-electrified parts, respectively. Clearly, $x+y=1$. According to the condition, $2x + 0.75y = 1$. We obtain the equation
$$
x+y=2x+0.75y
$$
from which we can find the ratio
$$
\frac{x}{y}=\frac{1}{4}
$$
Now we can find the required ratio
$$
\fr... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given a rectangular parallelepiped. The perimeters of each of its three mutually perpendicular faces are equal to the sides of a new rectangular parallelepiped. What can be the minimum ratio of the volume of the new parallelepiped to the volume of the original?
# | # Solution
Let $x, y, z$ be the sides of the original parallelepiped. Then the volume of the new one is
$$
V_{2}=2 \cdot(x+y) \cdot 2 \cdot(y+z) \cdot 2 \cdot(z+x)
$$
The desired ratio of volumes is
$$
\begin{gathered}
\frac{V_{2}}{V_{1}}=\frac{8(x+y)(y+z)(z+x)}{x y z}=\frac{8\left(x y+y^{2}+x z+y z\right)(z+x)}{x ... | 64 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Find the maximum value of the quantity $x^{2}+y^{2}$, given that
$$
x^{2}+y^{2}=3 x+8 y
$$ | # Solution
## Method 1
Introduce a Cartesian coordinate system and consider an arbitrary vector $\mathbf{a}$ with coordinates $(x, y)$ and a fixed vector $\mathbf{c}$ with coordinates $(3, 8)$.
Then the left side of the condition represents the square of the length of vector $\mathbf{a}$, and the right side represen... | 73 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the code of the letter. Then, the weight of a word can be defined as the sum of the codes of all the letters in that ... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have:
$$
k(C)+k(T)+k(O) \geq k(\Pi)+k(\text { Ya) }+k(T)+k(\text{ b })+k(C)+k(O)+k(T)
$$
which is equivalent to
$$
k(\Pi)+k(T)+k(\text{ b })+k(\text { Ya) }=0
$$
from which it follows that
$$
k(\Pi)=k(T)=k(\text{ b })=k(\text { Ya) }=0
$$
Th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4.
Over two days, 100 bankers collected funds to fight a new virus. Each of them made a one-time contribution of a whole number of thousands of rubles, not exceeding 200. Each contribution on the first day did not exceed 100 thousand, while on the second day it was more than this amount; and no pair of all 1... | The solution significantly depends on whether all contributions were distinct or could repeat.
Let's first consider the case where all contributions are distinct. Any natural number greater than 100 but not exceeding 200 can be represented as $100+n$, where $n \in [1,2,3, \ldots, 100]$. According to the condition, the... | 10050 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the six houses on one side of the street is connected by cable lines to each of the eight houses on the opposite side. How many pairwise intersections do the shadows of these cables form on the surface of the street, if no three of them intersect at the same point? Assume that the light causing these shadows... | # Solution
Let's take an arbitrary pair of houses on one side of the street and an arbitrary pair on the other. They are the vertices of a convex quadrilateral (since two sides of the quadrilateral, coming from each chosen pair, lie on one side of the line, i.e., the angles do not exceed $180^{\circ}$), so its diagona... | 420 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Find the maximum value of the quantity $x^{2}+y^{2}+z^{2}$, given that
$$
x^{2}+y^{2}+z^{2}=3 x+8 y+z
$$ | # Solution
## 1st Method
Introduce a Cartesian coordinate system and consider an arbitrary vector $\mathbf{a}$ with coordinates $(x, y, z)$ and a fixed vector $\mathbf{c}$ with coordinates $(3, 8, 1)$. Then, the left side of the condition represents the square of the length of vector $\mathbf{a}$, and the right side ... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4.
In modern conditions, digitalization - the conversion of all information into digital code - is considered relevant. Each letter of the alphabet can be assigned a non-negative integer, called the letter code. Then, the weight of a word can be defined as the sum of the codes of all the letters in that word. I... | # Solution.
Let $k(x)$ denote the elementary code of the letter $x$. We have:
$$
k(C)+k(T)+k(O) \geq k(\Pi)+k(\text { ( })+k(T)+k(\mathrm{~b})+k(C)+k(O)+k(T)
$$
which is equivalent to
$$
k(\Pi)+k(T)+k(\mathrm{~b})+k(\text { Я) }=0
$$
from which it follows that
$$
k(\Pi)=k(T)=k(\mathrm{~b})=k(\text { Я })=0
$$
Th... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 3.
A polynomial $P(x)$ with integer coefficients has the properties
$$
P(1)=2019, \quad P(2019)=1, \quad P(k)=k,
$$
where the number $k$ is an integer. Find this number $k$.
# | # Solution.
Since the polynomial $P(x)$ has integer coefficients, $P(a)-P(b)$ is divisible by $a-b$ for any integers $a$ and $b$.
We get that
$$
\begin{gathered}
P(k)-P(1)=(k-2019) \text { is divisible by }(k-1), \\
P(k)-P(2019)=(k-1) \text { is divisible by }(k-2019) .
\end{gathered}
$$
This can only be true if $|... | 1010 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that there are four ways to prepare magical pollen to create elixirs of kindness, joy, wisdom, luck, health, friendliness, and creativity. But elixirs of kindness, joy, and wisdom are made from fairy pollen, while elixirs of luck, health, friendliness, and creativity are made from elf pollen. Among the i... | Solution. Four methods of preparing pollen are represented by four branches of a tree. From two branches with methods of preparing fairy pollen, three branches of elixirs (of goodness, joy, and wisdom) branch off, and from two branches with methods of preparing elf pollen, four branches (of luck, health, friendliness, ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. It is known that the free term $a_{0}$ of the polynomial $P(x)$ with integer coefficients is less than 100 in modulus, and $P(20)=P(16)=2016$. Find $a_{0}$. | Solution. We can write $P(x)-2016=(x-16)(x-20) Q(x)$, where $Q(x)$ is a polynomial with integer coefficients. The constant term of the right side is $320 k$, where $k$ is an integer. Thus, $a_{0}=2016-320 k$. The condition is satisfied only by the value $k=6, a_{0}=96$.
Answer. $a_{0}=96$. | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The council of the secret pipeline village is gathering around a round table, where each arriving member can sit in any free seat. How many different seating arrangements are possible if 7 participants will attend the council? (Two seating arrangements are considered the same if the same people sit to the left and r... | # Solution
Since empty seats are not taken into account, we can consider only the ways of arranging on seven seats. The first person can sit in any of the 7 seats, the next in any of the 6 remaining seats, and so on until the last. In total, there are $7 \cdot 6 \cdot \ldots \cdot 1=7!$ ways.
However, each arrangemen... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In the summer, Ponchik eats honey cakes four times a day: instead of morning exercise, instead of a daytime walk, instead of an evening run, and instead of a nighttime swim. The quantities of cakes eaten instead of exercise and instead of a walk are in the ratio of $3: 2$; instead of a walk and instead of a run - as... | # Solution
According to the problem, we can establish the following ratios for the number of doughnuts eaten instead of engaging in a particular useful activity:
$$
\frac{\text { Morning Exercise }}{\text { Walk }}=\frac{3}{2} \quad \frac{\text { Walk }}{\text { Jog }}=\frac{5}{3} \quad \frac{\text { Jog }}{\text { S... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. In the country of "Energetika," there are 150 factories, and some of them are connected by bus routes that do not stop anywhere except at these factories. It turned out that any four factories can be divided into two pairs such that there is a bus route between the factories of each pair. Find the smallest n... | Solution. Suppose that some factory $X$ is connected by bus routes to no more than 146 factories. Then a quartet of factories, consisting of $X$ and any three with which it is not connected, does not satisfy the problem's condition, since $X$ cannot be paired with any of the three remaining factories. Therefore, each f... | 11025 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4.
Find the number of numbers $N$ in the set $\{1,2, \ldots, 2018\}$ for which there exist positive solutions $x$ to the equation
$$
x^{[x]}=N
$$
( $[x]$ is the integer part of the real number $x$, i.e., the greatest integer not exceeding $x$) | # Solution.
Notice that suitable numbers $N$ for $x$ such that $[x]=n$ are the numbers from $n^{n}$ to $(n+1)^{n-1}$, that is, exactly the numbers for which $[\sqrt[n]{N}]=n$. Such numbers (among the numbers from 1 to 2018) are the number 1, the numbers from $2^{2}$ to $3^{2-1}$ (there are exactly 5 of them), the numb... | 412 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
A rule is given by which each pair of integers $X$ and $Y$ is assigned a number $X \nabla Y$. (The symbol «»» means applying the rule to the numbers $X$ and $Y$.) It is known that for any integers $X, Y$ the following properties hold:
1) $X \nabla 0=X$
2) $X \nabla(Y-1)=(X \nabla Y)-2$
3) $X \nabla(Y+1)=... | # Solution.
Let's start writing down in order the result of applying the operation $X \nabla Y$ for $Y=0,1,2, \ldots$, using property 3.
$$
\begin{aligned}
& X \nabla 0=0, \\
& X \nabla 1=X \nabla(0+1)=(X \nabla 0)+2=X+2, \\
& X \nabla 2=X \nabla(1+1)=(X \nabla 1)+2=X+4, \\
& X \nabla 3=X \nabla(2+1)=(X \nabla 2)+2=X... | -673 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Does there exist a convex polygon with 2015 diagonals?
# | # Solution.
The number of diagonals in a convex $n$-gon is
$$
\frac{n(n-1)}{2}-n=\frac{n^{2}-3 n}{2}
$$
Solve the equation
$$
\frac{n^{2}-3 n}{2}=2015 \quad \Longleftrightarrow \quad n^{2}-3 n-4030=0
$$
Its discriminant $D=9+4 \cdot 4030=16129$. Note that $120^{2}<16129<130^{2}$ and $D$ ends with the digit 9. If $... | 65 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 2.
In a football tournament, each team is supposed to play one match against each of the others. But during the tournament, half of all the teams were disqualified and withdrew from further participation. As a result, 77 matches were played, and the teams that withdrew had managed to play all their matches a... | # Solution.
According to the condition, an even number of teams $2n$ started, of which $n$ were disqualified. The eliminated teams played $\frac{n(n-1)}{2}$ matches among themselves. The same number of matches were played by the teams that remained in the tournament.
Let each of the eliminated teams have played with ... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. In a square table with 2015 rows and columns, positive numbers are arranged. The product of the numbers in each row and each column is 1, and the product of the numbers in any 1008 x 1008 square is 2. What number is in the center of the table?
# | # Solution
Consider the first 1008 rows of the table. From the additional condition, it follows that if these rows are covered by two squares of size $1008 \times 1008$, then these squares will overlap by one column. Denote the product of the numbers in this column (1008 numbers) by $M$.
Then the product of all the n... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. Clever Dusya arranges six cheat sheets in four secret pockets so that the 1st and 2nd cheat sheets end up in the same pocket, the 4th and 5th cheat sheets also end up in the same pocket, but not in the same pocket as the 1st. The others can be placed anywhere, but only one pocket can remain empty (or all can be fill... | # Solution
Let's reason constructively and straightforwardly. Place the 1st and 2nd cheat sheets in any pocket. This can be done in 4 ways. Now place the 4th and 5th cheat sheets in any free pocket. This can be done in 3 ways. In total, there are 12 ways. There remain two cheat sheets (3rd and 6th) and two free pocket... | 144 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Four managers of responsibility shifting reported: "If they are arranged in pairs, there will be 1 left. If they are arranged in threes, there will also be 1 left. If they are arranged in fours, there will be 2 left, and if they are arranged in fives, there will also be 2 left." Should the head of the report recepti... | # Solution
Let $a$ be the desired quantity. If when dividing by 4, 2 remains, then $a$ is even. But then when dividing by two, 1 cannot remain. Therefore, these two statements are contradictory and the message as a whole is false.
Remove one of the contradictory statements.
1st var.: leave the division by 3, by 4, a... | 1042 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. (15 points) Real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ are such that
$$
\left\{\begin{array}{l}
x_{1}+x_{2} \geqslant 12 \\
x_{1}+x_{3} \geqslant 13 \\
x_{1}+x_{4} \geqslant 14 \\
x_{3}+x_{4} \geqslant 22 \\
x_{2}+x_{3} \geqslant 23 \\
x_{2}+x_{4} \geq 24
\end{array}\right.
$$
What is the smallest value t... | Answer: 37.
Solution. By adding the second equality to the last one, we get $x_{1}+x_{2}+x_{3}+x_{4} \geqslant 37$.
It is also worth noting that the value of the expression $x_{1}+x_{2}+x_{3}+x_{4}$ can be equal to 37, for example, when $x_{1}=1, x_{2}=11, x_{3}=12, x_{4}=13$. It is easy to verify that such numbers s... | 37 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
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