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class | __index_level_0__ int64 0 742k |
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# Task 5. (20 points)
Prove that if $a$ and $b$ are the legs, and $c$ is the hypotenuse of a right triangle, then the radius of the circle inscribed in this triangle can be found using the formula
$$
r=\frac{1}{2}(a+b-c)
$$
# | # Solution.
I method. Area of a triangle
$$
S=\frac{1}{2} a b=p r=\frac{1}{2}(a+b+c) r
$$
Therefore,
$$
r=\frac{a b}{a+b+c}
$$
By the Pythagorean theorem, $a^{2}+b^{2}=c^{2}$, or equivalently,
$$
(a+b)^{2}-2 a b=c^{2}
$$
Thus, $2 a b=(a+b)^{2}-c^{2}=(a+b+c)(a+b-c)$. Then (1) becomes
$$
r=\frac{2 a b}{2(a+b+c)}=... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,629 |
1. In a toy store, 125 plush bears are sold in $k$ different colors and six different sizes. For what largest $k$ can we assert that there will be at least three identical bears? (i.e., matching both in color and size) (20 points) | Solution: If $k=10$, then there are 60 varieties of teddy bears in the store. If there are no more than two of each type, then there are no more than 120 teddy bears in the store - a contradiction. Therefore, there will be at least three identical teddy bears among them.
If $k=11$, then there may not be three identica... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,630 |
3. The graphs of the functions $y=x^{2}$ and $y=a x^{2}+b x+c$ intersect at points $A$ and $B$, which lie on opposite sides of the y-axis. Point $O$ is the origin. It turns out that $\angle A O B=90^{\circ}$. Find all possible values of $c$. (20 points)
International School Olympiad URFU "Emerald" 2022, 2nd stage | Solution: Let $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ be the coordinates of points $A$ and $B$ respectively, with $x_{1} \neq 0$ and $x_{2} \neq 0$. Then $x_{1}, x_{2}$ are the roots of the equation $(a-1)x^{2}+b x+c=0$. If $a=1$, the graphs touch at point $O$, and there cannot be two points of intersecti... | \neq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,632 |
4. On a plane, two equal triangles $A B C$ and $A^{\prime} B^{\prime} C^{\prime}$ are drawn. On the extensions of the sides of triangle $A B C$, points $A_{1}, B_{1}, C_{1}$ are taken, and on the extensions of the sides of triangle $A^{\prime} B^{\prime} C^{\prime}$, points $A_{2}, B_{2}, C_{2}$ are taken, and then all... | Solution:
We will prove that the areas of triangles $A_{1} C C_{1}$ and $A_{2} B^{\prime} B_{2}$ are equal. By the Law of Sines, we have
$$
\frac{A B}{\sin \angle A_{1} C C_{1}}=\frac{A B}{\sin \angle B C A}=\frac{A C}{\sin \angle A B C}=\frac{A^{\prime} C^{\prime}}{\sin \angle A^{\prime} B^{\prime} C^{\prime}}=\frac... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,633 |
5. In the language of the "Tekimar" tribe, there are only 7 letters: A, E, I, K, M, R, T, but the order of these letters in the alphabet is unknown. A word is defined as any sequence of seven different letters from the alphabet, and no other words exist in the language. The chief of the tribe listed all existing words ... | Solution: The total number of words in the tribe's language is $7!=5040$. Note that the number of words starting with any particular letter is the same for any first letter, which is $7!\div 7=6!=720$. If a word starts with the first letter in the alphabet, the numbering of any such word starts from number 1 and ends a... | 3745 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,634 |
1. Petya painted all natural numbers in 2017 different colors. Is it true that regardless of the painting method, one can find two numbers of the same color, the ratio of which is an integer and divisible by 2016? | Solution. Consider the natural numbers $1, 2016, 2016^2, 2016^{2017}$. There are 2018 of them, and there are 2017 colors, so among these numbers, there are at least two of the same color. The ratio of these numbers is an integer and is divisible by 2016. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,635 |
3. A square is inscribed in a triangle with a base equal to a, one of its sides lying on the base of the triangle. The area of the square is $\frac{1}{6}$ of the area of the triangle. Determine the height of the triangle and the side of the square. | Solution. If $M D E L$ is a square with side $x$, inscribed in the given $\triangle A B C$ with height $h=B K, a=A C$, then we have
$$
\begin{gathered}
\triangle A E M \sim \triangle A B K \Rightarrow \frac{A E}{A B}=\frac{x}{h} \\
\triangle E B D \sim \triangle A B C \Rightarrow \frac{E B}{A B}=\frac{x}{a}
\end{gathe... | \frac{3\\sqrt{6}}{6},\quad=(5\2\sqrt{6}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,637 |
4. Given the system of equations
$$
\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=0 \\
a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=0 \\
a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=0
\end{array}\right.
$$
the coefficients of which satisfy the following conditions:
(a) $a_{11}, a_{22}, a_{33}$ are positive;
(b... | Solution. Let $\left(x_{1}, x_{2}, x_{3}\right)$ be a solution of the system and assume without loss of generality that $\left|x_{1}\right| \geqslant\left|x_{2}\right| \geqslant\left|x_{3}\right|$. If $\left|x_{1}\right|=0$, then $x_{1}=x_{2}=x_{3}=0$. Suppose that $\left|x_{1} \neq 0\right|$. Then
$$
\begin{array}{r}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,638 |
5. For all values of the parameter a, solve the inequality
$$
\log _{x}(x-a)>2
$$ | Solution. The permissible values of the variable $x$ are determined by the system
$$
\left\{\begin{array}{l}
x>0 \\
x \neq 1 \\
x-a>0
\end{array}\right.
$$
This system corresponds to a set of points on the coordinate plane $x O a$ that lie below the line $a=x$, to the right of the $a$-axis, and do not include the lin... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,639 |
2. Prove that the number
$$
1^{2017}+2^{2017}+\ldots+2016^{2017}
$$
is divisible by 2017 and not divisible by 2018. | Solution. Let's denote this number as $S$. Add to $S$ the same sum of numbers written in reverse order and group the numbers into pairs:
$$
2 S=\left(1^{2017}+2016^{2017}\right)+\ldots+\left(2016^{2017}+1^{2017}\right)
$$
Since $a^{2 n+1}+b^{2 n+1}$ is divisible by $a+b$ for any natural number $n$, we get that $2 S$ ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,640 |
3. A line parallel to the bases of a given rectangular trapezoid divides it into two trapezoids, each of which can have a circle inscribed in it. Find the bases of the original trapezoid if its lateral sides are equal to \( c \) and \( d \), where \( c < d \). | Solution. Consider the given rectangular trapezoid $ABCD$, where $c=AB, d=CD, MN \| BC$, and in $MBC L$ and $AMLD$ inscribed circles with radii $r_{1}$ and $r_{2}$. Extend the lateral sides of the trapezoid to intersect at point $O$.
Let $BC=x, MN=y, AD=z$.
Notice that $\triangle MOL \sim \triangle AOD$.
Then $\frac... | \frac{(\sqrt{+}\\sqrt{-})^{2}}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,641 |
4. Find four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, such that each, when added to the product of the others, equals two. | Solution. Let $x_{1} x_{2} x_{3} x_{4}=p$. Then our system of equations will take the form:
$$
x_{i}+\frac{p}{x_{i}}=2, \quad i=1,2,3,4
$$
The case when one of the sought numbers is zero leads to a contradiction. Indeed, substituting zero for $x_{1}$, we get
$$
x_{2} x_{3} x_{4}=2, x+2=x+3=x+4=2 \text {. }
$$
Thus,... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,642 |
5. For all values of the parameter a, solve the inequality
$$
\log _{\sqrt{2 a}}\left(a+2 x-x^{2}\right)<2
$$ | Solution. The permissible values of the variable $x$ are given by the system:
$$
\left\{\begin{array}{l}
x^{2}-2 x-a<0 \\
x^{2}-2 x+a>0 \\
a \neq \frac{1}{2}
\end{array}\right.
$$
Rewrite the original inequality in the form
$$
\log _{\sqrt{2 a}}\left(a+2 x-x^{2}\right) < \frac{1}{2} \quad \text{for} \quad a > \frac{... | notfound | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,643 |
1. On each page of the book, its number is written. The page numbering starts from one. Vasya tore out all the even-numbered sheets from the book (there are two pages on each sheet of the book). The numbers of the remaining pages in the book together contain exactly 845 digits. How many pages could there have been in t... | Solution: Among every four consecutive pages, two pages with numbers that have remainders of 1 and 2 when divided by 4 remain. The remaining pages have single-digit numbers 1, 2, 5, 6, 9, double-digit numbers $10, 13, 14, 17, \ldots, 97, 98$, and some three-digit numbers $101, 102, 105, 106, \ldots$. Among the first 10... | 598600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,644 |
2. On a piece of paper, 25 points are marked - the centers of the cells of a $5 \times 5$ square. The points are colored with several colors. It is known that no three points of the same color lie on any straight line (vertical, horizontal, or at any angle). What is the minimum number of colors that could have been use... | Solution: If no more than two colors are used, then, in particular, in the first row there will be at least three points of the same color - they lie on the same straight line. An example of using three colors (identical numbers denote points painted in the same color):
## Answer: 3 colors
| 1 | 2 | 2 | 3 | 3 |
| :--... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,645 |
3. Let's call a natural number special if one of its digits can be replaced by another digit so that all digits in the resulting number are distinct. Numbers in which all digits are already distinct are also considered special. How many special ten-digit numbers exist? (20 points) | Solution: Let's divide the numbers into groups:
1) Numbers in which all digits are different - there are a total of $9 \cdot 9 \cdot 8 \cdot 7 \cdot \ldots \cdot 1=9 \cdot 9$ !.
2) Numbers in which two non-zero digits are the same and do not stand in the highest place. There are 9 such digits, pairs of positions where... | 414\cdot9! | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,646 |
5. Does there exist a convex polygon without a center of symmetry that can be cut into two convex polygons, each of which has a center of symmetry? (20 points) | Solution: Suppose such a polygon exists. Draw a certain cut that divides it into two convex polygons. If the cut line is not a segment, then there will be three points $A, B, C$ on it, forming a triangle (see Figure 1). Both polygons, due to their convexity,
 | Solution: From the condition, it follows that $A I$ and $B I$ are the angle bisectors of triangle $A B C$. Let $\angle B A P = \angle C A P = \alpha$ and $\angle A B I = \angle C B I = \beta$. Then $\angle C B P = \alpha$, as an inscribed angle, and therefore $\angle I B P = \alpha + \beta$. Note that $\angle B I P = \... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,650 |
4. Let's call different natural numbers $m$ and $n$ related if the sum of the smallest natural divisor of $m$, different from 1, and the largest natural divisor of $m$, different from $m$, equals $n$, and the sum of the smallest natural divisor of $n$, different from 1, and the largest natural divisor of $n$, different... | Solution: Let $S(x)$ and $L(x)$ denote the smallest natural divisor of $x$ different from 1 and the largest natural divisor of $x$ different from $x$, respectively. It is clear that $x \neq$ 1, since in this case $S(x)$ and $L(x)$ do not exist. Any related pair $m$ and $n$ is a solution to the system
$$
\left\{\begin{... | 56 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,651 |
5. Positive numbers $a, b, c$ are such that $a+b+c=1$. Find the maximum value of the expression $\frac{(a+1)(b+1)(c+1)}{a b c+1} \cdot(20$ points $)$ | Solution: If $a=b=c=\frac{1}{3}$, then $\frac{(a+1)(b+1)(c+1)}{a b c+1}=\frac{16}{7}$. We will prove that $\frac{(a+1)(b+1)(c+1)}{a b c+1} \leq \frac{16}{7}$. Transform the expression to $\frac{(a+1)(b+1)(c+1)}{a b c+1}=\frac{a b c+a b+b c+a c+a+b+c+1}{a b c+1}=1+\frac{a b+b c+a c+1}{a b c+1}$. We will prove that $\fra... | \frac{16}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,652 |
1. Daria Dmitrievna is preparing a test on number theory. She promised to give each student as many problems as the number of addends they create in the numerical example
$$
a_{1}+a_{2}+\ldots+a_{n}=2021
$$
where all numbers $a_{i}$ are natural numbers, greater than 10, and are palindromes (do not change if their dig... | Solution: A student cannot receive one problem since 2021 is not a palindrome. Suppose he can receive two problems, then at least one of the numbers $a_{1}, a_{2}$ is a four-digit number. If it starts with 2, then the second digit is 0 and the number itself is 2002. In this case, the second number is 19, which is not a... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,653 |
2. Does there exist a polygon without a center of symmetry that can be cut into two convex polygons, each of which has a center of symmetry? (20 points) | Solution: It exists. Example.
The centers of symmetry of the rectangles are the points of intersection of their diagonals. The given polygon does not have a center of symmetry, because if it... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,654 |
3. Positive numbers $a, b, c, d$ are such that the numbers $a^{2}, b^{2}, c^{2}, d^{2}$ in the given order form an arithmetic progression and the numbers $\frac{1}{a+b+c}, \frac{1}{a+b+d}, \frac{1}{a+c+d}, \frac{1}{b+c+d}$ in the given order form an arithmetic progression. Prove that $a=b=c=d$. (20 points) | Solution: Let's write the characteristic property for each arithmetic progression:
$$
\begin{gathered}
a^{2}+c^{2}=2 b^{2} ;(1) \\
b^{2}+d^{2}=2 c^{2} ;(2) \\
\frac{1}{a+b+c}+\frac{1}{a+c+d}=\frac{2}{a+b+d}
\end{gathered}
$$
Transform equation (3):
$$
\begin{gathered}
(a+b+d)(a+c+d)+(a+b+c)(a+b+d)=2(a+b+c)(a+c+d) \\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,655 |
4. Find the number of triples of natural numbers $m, n, k$, which are solutions to the equation $m+$ $\sqrt{n+\sqrt{k}}=2023$. (20 points) | Solution: For the left side to be an integer, the numbers $k$ and $n+\sqrt{k}$ must be perfect squares, and since $n+\sqrt{k} \geq 2$, it follows that $\sqrt{n+\sqrt{k}} \geq 2$ and thus $m \leq 2021$. Since $1 \leq m \leq$ 2021, $\sqrt{n+\sqrt{k}}$ can take any value from 2 to 2022 - this value uniquely determines the... | 27575680773 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,656 |
4. for $a<0.5$
## Correct answer: for $a<1$
## Question 3
Correct
Points: 1.00 out of a maximum of 1.00
Question 4
Correct
Points: 1.00 out of a maximum of 1.00
## Question 5
Correct
Points: 1.00 out of a maximum of 1.00
Worker $A$ and worker $B$ can complete the work in 7.2 days, worker $A$ and worker $C$ in... | Answer: 6
Correct answer: 6
A piece of copper-tin alloy has a total mass of 12 kg and contains 45% copper. How much pure tin should be added to this piece of alloy to obtain a new alloy containing 40% copper?
Answer: 1.5
Correct answer: 1.5
Find a two-digit number that is three times the product of its digits. If ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,658 |
1. A printing house determines the cost of printing a book as follows: it adds the cost of the cover to the cost of each page, and then rounds the result up to the nearest whole number of rubles (for example, if the result is 202 rubles and 1 kopeck, it is rounded up to 203 rubles). It is known that the cost of a book ... | Solution: Let's convert all costs into kopecks. Let one page cost $x$ kopecks, and the cover cost $100 y$ kopecks. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
13300<100 y+104 x \leq 13400 \\
18000<100 y+192 x \leq 18100
\end{array}\right.
$$
Subtracting the first inequality from the second, we ... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,659 |
2. Let's call a natural number $N=\overline{a_{1} a_{2} \ldots a_{k}}$ an anti-square if $a_{k} \neq 0$ and the number $\overline{a_{k} a_{k-1} \ldots a_{1}}$ is a perfect square (for example, the numbers 94 and 441 are anti-squares, while the numbers 51 and 190 are not). Prove that the number $400 \ldots 005$ (2023 ze... | Solution: Suppose the given number can be represented as the sum of two antipowers of two. One of them (let's call it a) is not less than half of the given number, that is, not less than 200 ... 003 (2023 zeros). Squares cannot end in the digits 2 and 3, so the number $a$ starts with 4 and is not less than the number 4... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 4,660 |
3. Maxim came up with a new way to divide numbers by a two-digit number $N$. To divide any number $A$ by the number $N$, the following steps need to be performed:
1) Divide $A$ by the sum of the digits of the number $N$;
2) Divide $A$ by the product of the digits of the number $N$;
3) Subtract the second result from th... | Solution: Let $N=\overline{x y}=10 x+y$, where $x, y$ are digits. We need to find all pairs $(x, y)$ such that $\frac{A}{10 x+y}=\frac{A}{x+y}-\frac{A}{x y}$, which means $\frac{1}{10 x+y}=\frac{1}{x+y}-\frac{1}{x y}$. Note that $y \neq 0$. Transform the equation to the form
$$
\begin{gathered}
x y(x+y)=x y(10 x+y)-(x... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,661 |
4. A square of size $5 \times 5$ was divided into unit squares, and the center of each unit square was marked. Dima drew four segments with endpoints at the marked points such that no two segments intersect (even at the marked centers). Prove that after this, he can always draw a fifth segment with endpoints at the mar... | Solution: Let's draw four such segments that satisfy the condition of the problem. Now, we will extend the segments (in any order) until they intersect the boundaries of the square or already extended segments. For convenience, we will consider not the original segments, but the ones obtained after extension. The inter... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 4,662 |
5. For all natural $n$, greater than one, prove the inequality
$$
\sin \left(\frac{30^{\circ}}{n}\right)>\frac{1}{2 n} \cdot(20 \text { points })
$$ | Solution: Consider an isosceles triangle $ABC$ with an angle of $\frac{60^{\circ}}{n}$ at the vertex and a side length of 1. By drawing the height $BH$ to the base, we get $\sin \left(\frac{30^{\circ}}{n}\right) = \frac{BH}{AB} = BH$, hence $BC = 2 \sin \left(\frac{30^{\circ}}{n}\right)$.
. Consequently, the same can be said about the numbers $m$ and $n$. But in this case, $2018=m^{2}-n^{2}=(m-n)(m+n)$ must be divisible by 4 (since both factors are even), whereas 2018 is... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,664 |
2. The diagonals of a quadrilateral are equal, and the lengths of its midlines are p and q. Find the area of the quadrilateral. | Solution. Since $K L, L M, N M$ and $K N$ are the midlines of the corresponding triangles (see the figure), then $K L=N M=\frac{1}{2} A C, K N=L M=\frac{1}{2} B D$. According to the condition $A C=B D$. Therefore, $K L M N$ is a rhombus, the area of which is equal to $\frac{1}{2} p q$.
. Prove that all the delegates of the congress can b... | Solution. Let's choose three delegates of the congress. Among them, there will be two who know the same language - we will place them in one double room of the hotel. From the remaining 998 delegates, we will again select three, among whom there will again be two who can be placed in one room of the hotel - and so on, ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 4,666 |
4. Find all solutions to the equation
$$
8^{x}(3 x+1)=4
$$
and prove that there are no other solutions. | Solution. It is easy to notice that $x=\frac{1}{3}$ is a solution. For $x<-\frac{1}{3}$, the left side of the equation is negative, while the right side is positive, so there can be no roots. For $x \geq-\frac{1}{3}$, the left side monotonically increases, therefore, the equation can have only one root, so there are no... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,667 |
5. In a taxi company, there are currently 60 free cars: 10 black, 20 yellow, and 30 green. At a call, one of the cars, which happened to be the closest to the customer, drove out. Find the probability that a non-green taxi will arrive. | Solution. The probability that a car of a certain color will arrive is calculated as the ratio of the number of favorable outcomes (the arrival of a car of the desired color) to the total number of outcomes (the total number of cars). Thus, the probability of a green car arriving is $\frac{30}{60}=\frac{1}{2}$. | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,668 |
# Task 1. (20 points)
It is known that the only solution to the equation
$$
\pi / 4=\operatorname{arcctg} 2+\operatorname{arcctg} 5+\operatorname{arcctg} 13+\operatorname{arcctg} 34+\operatorname{arcctg} 89+\operatorname{arcctg}(x / 14)
$$
is a natural number. Find it. | # Solution.
Consider the equation:
$$
\operatorname{arcctg} a-\operatorname{arcctg} b=\operatorname{arcctg} y \text {. }
$$
Let $\alpha=\operatorname{arcctg} a, \beta=\operatorname{arcctg} b$. Now $y=\operatorname{ctg}(\alpha-\beta)=\frac{1+\operatorname{ctg} \alpha \operatorname{ctg} \beta}{\operatorname{ctg} \beta... | 2016 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,671 |
# Task 2. (20 points)
Let A be the point of intersection of two circles. From this point, points $X_{1}$ and $X_{2}$ start moving along each circle, clockwise, at constant speeds. After one complete revolution, both points return to $A$. Prove that there always exists a fixed point $B$ such that the equality $X_{1} B=... | # Solution.
Let two circles intersect at point $A$, and points $X_{1}, X_{2}$ move along the circles as described in the problem. We can assume that we are interested in exactly all pairs $X_{1}, X_{2}$ described in the problem for which $X_{1} \neq X_{2}$ (otherwise, there is nothing to check for any choice of $P$).
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,672 |
# Task 3. (20 points)
For the cubic polynomial $p(x)=a x^{3}+b x^{2}+c x+d$ with integer coefficients $a, b, c, d$, it is known that $p(1)=2015$ and $p(2)=2017$. Prove that the equation $p(x)=2016$ has no integer roots. | # Solution.
It is easy to see that if integers $m$ and $n$ have the same parity, then $p(m)$ and $p(n)$ also have the same parity. For all even integers $x$, the value of $p(x)$ is odd, as it is odd for $p(2)$, and for all odd integers, it is odd because $p(1)$ is odd. Therefore, $p(x)-2016$ is odd for all integers $x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 4,673 |
# Task 4. (20 points)
Prove that the largest square by area that fits inside a right-angled triangle shares a common angle with it. | # Solution.
Let the triangle have sides $a, b, \sqrt{a^{2}+b^{2}}$, and let the side of the considered square be denoted by $q$.
Consider any square in such a right triangle. Note that if the square does not touch any side, then it is not maximal, since we can always consider a square with the same center but a sligh... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,674 |
# Task 5. (20 points)
Find all values of the parameter $a$ for which the roots $x_{1}$ and $x_{2}$ of the equation
$$
2 x^{2}-2016(x-2016+a)-1=a^{2}
$$
satisfy the double inequality $x_{1}<a<x_{2}$.
# | # Solution.
Let the polynomial from the problem statement be denoted by $f(x)$. Then, due to the positivity of the coefficient of $x^{2}$, the condition of the problem is equivalent to the simple inequality $f(a)<0$. That is, $f(a)=a^{2}-2a \cdot 2016+2016^{2}-1=(a-2016)^{2}-1<0$. From this, we obtain that $a \in (201... | \in(2015,2017) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,675 |
1. Is it possible to arrange the natural numbers from 1 to 9 (each used once) in the cells of a $3 \times 3$ table so that the sum of the numbers in each row is divisible by 9 and the sum of the numbers in each column is divisible by 9? (20 points) | Solution: Yes, for example like this:
| 1 | 3 | 5 |
| :--- | :--- | :--- |
| 8 | 4 | 6 |
| 9 | 2 | 7 |
Answer: it is possible | itispossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,676 |
3. Before you is a segment-digit. For displaying time on electronic clocks, each digit uses seven segments, each of which can be lit or not; the lit segments form the digit as shown in the figure
## 8723456789
That is, to display zero, six segments are used, to display one - two segments, and so on. On electronic clo... | Solution: Let's write down the number of segments required to display each digit:
\section*{8723456789
$$
\begin{array}{llllllllll}
6 & 2 & 5 & 5 & 4 & 5 & 6 & 3 & 7 & 6
\end{array}
$$
Consider several types of displayed time:
1) The last digit of the minutes is not 9. Then the next displayed time differs by only t... | 630 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,678 |
4. Masha chose a natural number $n$ and wrote down all natural numbers from 1 to 6 n on the board. Then, Masha halved half of these numbers, reduced a third of the numbers by a factor of three, and increased all the remaining numbers by a factor of six. Could the sum of all the resulting numbers match the sum of the or... | Solution: Yes, this could have happened. Let $n=2$, then the sum of all numbers from 1 to 12 is
$$
1+2+3+\ldots+12=78
$$
Suppose Masha halved the numbers 2, 3, 5, 9, 10, 11, reduced the numbers 4, 6, 8, 12 by a factor of three, and increased the numbers 1, 7 by a factor of six. Then the sum of the resulting numbers w... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,679 |
5. Two players are playing a game. The number 2022 is written on the board. In one move, a player must replace the current number $a$ on the board with another (different from $a$), obtained as a result of one of the three operations:
1) Subtract 3;
2) Subtract the remainder of dividing $a-2$ by 7 from $a$;
3) Subtract... | Solution: A draw is impossible, as per the task requirements, the number must change after each move, which means the number on the board will decrease with each move and will eventually become negative. The player who receives the number 2018 on their turn either has a winning strategy or their opponent does. The firs... | thefirstplayer | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,680 |
1. At the vertices of a regular dodecagon, natural numbers from 1 to 12 (each exactly once) are arranged in some order. Could it happen that the sums of all pairs of adjacent numbers are prime and the sums of all pairs of numbers with exactly two numbers between them are also prime? (20 points) | Solution: Each number at the vertex participates in exactly four sums. Note that to obtain a simple sum, to the numbers 6 and 12, only $1,5,7$ and 11 can be added. This means that for the vertices where the numbers 6 and 12 are located, the sets of adjacent numbers and numbers standing two vertices away must coincide. ... | couldnot | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,681 |
2. In how many ways can the numbers from 1 to 9 be arranged in a $3 \times 3$ table (each number appearing exactly once) such that in each column from top to bottom and in each row from left to right, the numbers are in increasing order? (20 points) | Solution: Let's number the cells of the table as shown in the figure. It is clear that the number 1 is in the upper left cell, and the number 9 is in the lower right cell.
| 1 | $a_{2}$ | $a_{3}$ |
| :---: | :---: | :---: |
| $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{7}$ | $a_{8}$ | 9 |
By the condition, $a_{5}>a_{2}, a_{5... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,682 |
3. Let's call a number $x$ semi-integer if the number $2x$ is an integer. The semi-integer part of a number $x$ is defined as the greatest semi-integer number not exceeding $x$, and we will denote it as $] x[$. Solve the equation $x^{2} + 2 \cdot ] x[ = 6$. (20 points) | Solution: Consider two cases.
1) The number $x-$ is half-integer, then $] x\left[=x\right.$ and the original equation will take the form $x^{2}+2 x-6=0$. The roots of this equation are $x_{1,2}=-1 \pm \sqrt{7}$, but then the numbers $2 x_{1,2}$ are not integers, so there are no solutions.
2) There is an equality $x=\f... | \sqrt{3},-\sqrt{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,683 |
4. In a non-isosceles triangle $ABC$, point $K$ is the midpoint of side $AB$, $M$ is the centroid (point of intersection of the medians), and $I$ is the incenter (center of the inscribed circle). It is known that $\angle K I B=90^{\circ}$. Prove that $M I \perp B C$. (20 points) | Solution: Let the line $A I$ intersect the circumcircle of triangle $A B C$ at point $P$. From the condition, it follows that $A I$ and $B I$ are the angle bisectors of triangle $A B C$. Let $\angle B A P = \angle C A P = \alpha$ and $\angle A B I = \angle C B I = \beta$. Then $\angle C B P = \alpha$, as an inscribed a... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,684 |
5. Let $p_{1}, p_{2}, \ldots, p_{n}, \ldots$ be the set of all prime numbers, arranged in some order. Can it happen that for all natural $i$ the number $\frac{p_{i} p_{i+1}-p_{i+2}^{2}}{p_{i}+p_{i+1}}$ is a natural number? (20 points) | Solution: We will prove an auxiliary lemma.
Lemma: There are infinitely many prime numbers of the form $3k+2$.
Proof: Suppose there are only finitely many prime numbers of the form $3k+2$. Let these numbers be $q_{1}, q_{2}, \ldots, q_{l}$. The number $3 q_{1} q_{2} \ldots q_{l}-1$ does not divide the prime numbers $... | cannot | Number Theory | proof | Yes | Yes | olympiads | false | 4,685 |
1. How many and which digits will be needed to write down all natural numbers from 1 to $10^{2017}$ inclusive? | Solution. First, consider all natural numbers from 1 to $10^{2017}-1=\underbrace{99 \ldots 9}_{2017 \text { of them }}$. In this case, all numbers with fewer than 2017 digits will be padded with leading zeros to make them 2017-digit numbers, and we will add one more number $\underbrace{00 \ldots 0}_{2017 \text { of the... | 2017\cdot10^{2017}-\sum_{k=1}^{2016}(2017-k)\cdot10^{k-1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,686 |
2. Three circles are inscribed in a corner - a small, a medium, and a large one. The large circle passes through the center of the medium one, and the medium one through the center of the small one. Determine the radii of the medium and large circles if the radius of the smaller one is $r$ and the distance from its cen... | Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the circles mentioned in the problem, $O_{1} A=a, E, F, K$ be the points of tangency of the small, medium, and large circles with the side $A C$ of the angle, respectively. Then $O_{1} E=r, O_{1} E \perp A C, O_{2} F \perp A C, O_{3} K \perp$ $A C$ and $\triangle A ... | \frac{}{-r};\quad\frac{^{2}r}{(-r)^{2}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,687 |
3. Given $n$ positive numbers $a_{1}, a_{2}, \ldots, a_{n}$, such that
$$
a_{1} \cdot a_{2} \cdot \ldots \cdot a_{n}=1
$$
Prove that
$$
\left(1+a_{1}\right) \cdot\left(1+a_{2}\right) \cdot \ldots \cdot\left(1+a_{n}\right) \geqslant 2^{n}
$$ | Solution. Note that
$$
\left(1+a_{1}\right) \cdot\left(1+a_{2}\right) \cdot \ldots \cdot\left(1+a_{n}\right)=\left(1+\frac{1}{a_{1}}\right) \cdot\left(1+\frac{1}{a_{2}}\right) \cdot \ldots \cdot\left(1+\frac{1}{a_{n}}\right)
$$
Therefore,
$$
\left(\left(1+a_{1}\right) \cdot \ldots \cdot\left(1+a_{n}\right)\right)^{2... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 4,688 |
4. Consider the game "Battleship" on a $5 \times 5$ square grid. What is the minimum number of shots needed to guarantee hitting a ship of size $1 \times 4$ cells? | Solution. Obviously, less than five shots are not enough, since at least one shot must be made in each vertical and each horizontal. We will show that five shots are not enough. Indeed, a shot made in the first vertical leaves four consecutive cells on the given horizontal, which means at least one more shot must be ma... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,689 |
5. Let $x$ be a natural number. Solve the equation
$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x} \cdots+\frac{1}{x}=3
$$ | Solution. Multiply both sides of the equation by $x$. We get
$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$
The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1$, and $a_{n}=1$. The progression has $(x-1)$ terms.
Since $S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n$, the equation c... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,690 |
1. How many units are in the number $S=9+99+999+\cdots+\overbrace{9 \ldots 90}^{1000}$? | $$
\begin{aligned}
& S=10-1+100-1+1000-1+\cdots+10^{1000}-1=10\left(1+10+100+\cdots+10^{999}\right)- \\
& -100=10 \frac{10^{1000}-1}{9}-1000=\underbrace{111 \ldots 10}_{999 \text { ones }}-1000 \Rightarrow
\end{aligned}
$$
the number has 998 ones. Answer: $\{998\}$. | 998 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,691 |
3. Solve the equation $\sqrt{6-x}+\sqrt{x-4}=x^{2}-10 x+27$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. Given $4 \leq x \leq 6$. Find the maximum of the function $f(x)=\sqrt{6-x}+\sqrt{x-4}$
$f^{\prime}(x)=-\frac{1}{\sqrt{6-x}}+\frac{1}{\sqrt{x-4}} \Rightarrow f^{\prime}(x)=0 ; \sqrt{4-x}=\sqrt{6-x} \Rightarrow x=5$. When $x=5$, $f(x)$ has a maximum value of 2. On the other hand, $x^{2}-10 x+27=(x-5)^{2}+2 \Rightarro... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false | 4,693 |
4. Solve the inequality $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} \geq 0$. In the answer, indicate the sum of all natural numbers that are solutions to the inequality. | 4. $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} * \frac{\left|x^{2}-3 x\right|+\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|+\left|x^{2}-2 x\right|} \geq 0 \Leftrightarrow \frac{\left(x^{2}-3 x\right)^{2}-\left(x^{2}-2\right)^{2}}{\left(x^{2}-x-2\right)^{2}-\left(x^{... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,694 |
5. Find all solutions to the equation $2017^{x}-2016^{x}=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. It is obvious that $x=1$ will be a solution to the given equation. Let's show that there are no others. We have: $\left(\frac{2017}{2016}\right)^{x}=1+\frac{1}{2016^{x}} ;\left(\frac{2017}{2016}\right)^{x}-1=\frac{1}{2016^{x}}$ From this, it is clear that the function on the left side is increasing, while the functi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,695 |
6. Three circles with radii $1,2,3$ touch each other externally. Find the radius of the circle passing through the three points of tangency of these circles. | 6. Let the centers of the given circles be denoted as $O_{1}, O_{2}, O_{3}$. Let $K, M, N$ be the points of tangency of the given circles. Then $O_{1} K=O_{1} N=1, O_{2} K=O_{2} M=2, O_{3} M=O_{3} N=3$.
 set off in pursuit of the raft. The motorboat caught up with the raft and immediately turned back to point $A$. After 3.6 hours, the motorboat arrived at point $A$, while the raft reached po... | 7. Let the speed of the river current be $x$ km/h, $t$-the time elapsed from the moment the boat left until the moment of the meeting. Then, by the condition: 1) the distance traveled by the raft until the meeting, 2) the distance traveled by the motorboat until the meeting, 3) the distance traveled by the motorboat af... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,697 |
Task 5.
1000 rubles are deposited in a bank for 5 years. In which case will the depositor receive more money: if the bank accrues $3 \%$ once a year, or if it accrues $\frac{1}{4} \%$ once a month. | The answer must be justified.
Translate the text above into English, please retain the source text's line breaks and format, and output the translation result directly. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,702 |
# Task 5.
1000 rubles are deposited in a bank for 10 years. In which case will the depositor receive more money: if the bank accrues $5 \%$ once a year, or if it accrues $\frac{5}{12} \%$ once a month. | The answer must be justified.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,703 |
1. Solve the equation $n+S(n)=1964$, where $S(n)$ is the sum of the digits of the number $n$. | 1. We have $S(n) \leq 9 * 4=36 \Rightarrow n \geq 1964-36=1928 \Rightarrow n=1900+10 k+l$, where $2 \leq k \leq 9$ $0 \leq l \leq 9 ; 1900+10 k+l+10+k+l=1964,11 k+2 l=54,2 \leq 2 l \leq 18,36 \leq 11 k \leq 54$ $3 \frac{3}{11} \leq k \leq 4 \frac{10}{11} \Rightarrow k=4, l=5, n=1945$. Answer: $\{1945\}$. | 1945 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,707 |
2. Calculate
the sum: $\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\cdots+\cos 160^{\circ}+\cos 180^{\circ}$ | 2.
$\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\cos 80^{\circ}+\cos 100^{\circ}+\cos 120^{\circ}+\cos 140^{\circ}+\cos 160^{\circ}+$ $\cos 180^{\circ}=\cos 20+\cos 40+\frac{1}{2}+\sin 10^{\circ}-\sin 10^{\circ}-\frac{1}{2}-\cos 40^{\circ}-1-\cos 20=-1$ Answer: $\{-1\}$ | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,708 |
3. Two pedestrians set out simultaneously from $A$ to $B$ and from $B$ to $A$. When the first had walked half the distance, the second had 24 km left to walk, and when the second had walked half the distance, the first had 15 km left to walk. How many kilometers will the second pedestrian have left to walk after the fi... | 3. Let $v_{1}, v_{2}$ be the speeds of the first and second pedestrians, $S$ be the distance from $A$ to $B$, and $x$ be the distance the second pedestrian still needs to walk when the first pedestrian finishes the crossing. From the problem statement, we get the system $\frac{S}{2} \frac{v_{2}}{v_{1}}+24=S, \frac{S}{2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,709 |
4. The denominator of an irreducible fraction is less than the square of the numerator by one. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $\frac{1}{3}$. If 3 is subtracted from both the numerator and the denominator, the fraction will be less than $\frac{1}{1... | 4. Let $\frac{m}{n}$ be the desired irreducible fraction, where $m, n$ are single-digit numbers. According to the problem, we have $n=m^{2}-1, \frac{m+2}{n+2}>\frac{1}{3} \Rightarrow 3 m+6>m^{2}+1, m^{2}-3 m-5<0$, then $\frac{3-\sqrt{29}}{2}<m<\frac{3+\sqrt{29}}{2} \Rightarrow$ $\Rightarrow 0<m<\frac{3+5.4}{2} \Rightar... | 38 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,710 |
5. Solve the inequality: $\frac{\left(\left|x^{2}-2\right|-7\right)(|x+3|-5)}{|x-3|-|x-1|}>0$. Write the largest integer that is a solution to the inequality in your answer. | 5. $\frac{\left(\left|x^{2}-2\right|-7\right)\left(\left|x^{2}-2\right|+7\right)(|x+3|-5)(|x+3|+5)}{(|x-3|-|x-1|)(|x-3|+|x-1|)}>0 \Leftrightarrow \frac{\left(\left(x^{2}-2\right)^{2}-49\right)\left((x+3)^{2}-25\right)}{(x-3)^{2}-(x-1)^{2}}>0$ $\Leftrightarrow \frac{\left(x^{2}-2-7\right)\left(x^{2}-2+7\right)(x+3-5)(x+... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,711 |
Variant 1.
1. Let $x$ be the total amount of money that Fox Alice appropriated, and let $y$ be the amount of money each fox cub received. Then the first cub received $a+\frac{x-a}{m}=y$, the second received: $2 a+\frac{x-y-2 a}{m}=y$, and so on, the $\mathrm{n}$-th cub received: $n a+\frac{x-(n-1) y-n a}{m}=y$. Multip... | Answer: $a(m-1)^{2}$ rubles were stolen, there were $m-1$ fox cubs. | (-1)^2 | Algebra | proof | Yes | Yes | olympiads | false | 4,714 |
4. Let $A B C$ be a triangle (see Fig.1), where $A F$ and $C E$ are medians, $\angle B A F=\angle B C E=30^{\circ}$.
Prove that $A B=B C=A C$. | Solution.
$\triangle A B F$ is similar to $\triangle B C E$ by two angles ( $\angle B$ - common); from the similarity we have:
$\frac{B F}{A B}=\frac{E B}{B C} ; \frac{\frac{1}{2} B C}{A B}=\frac{\frac{1}{\frac{1}{2}} A B}{B C} \Rightarrow B C=A B$, i.e., triangle $\mathrm{ABC}$ is isosceles. Let point M be the inter... | proof | Geometry | proof | Yes | Yes | olympiads | false | 4,719 |
6. Let's highlight the solution $(0,0)$, by dividing both sides of the first equation by $x y$, and the second by $x^{2} y^{2}$, we get:
$\left\{\begin{array}{c}\left(\frac{x}{y}+\frac{y}{x}\right)(x+y)=15 \\ \left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right)\left(x^{2}+y^{2}\right)=85\end{array}\right.$
Let $x+y=a$... | Answer: $(0,0),(2,4),(4,2)$. | (0,0),(2,4),(4,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,720 |
7. By the property of absolute value, replacing $x$ with $-x$ does not change this relation. This means that the figure defined by the given inequality is symmetric with respect to the OY axis. Therefore, it is sufficient to find the area of half of the figure for $x \geq 0$. In this case, we obtain the inequality $\le... | Answer: $\{30\}$.
Fig. 2 | 30 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,721 |
8. It is clear that $b \geq 0$. Rewrite the equation as:
$\sqrt{(x+2)^{2}+4 a^{2}-4}+\sqrt{(x-2)^{2}+4 a^{2}-4}=4 b$. If $x_{0}$ is a solution to this equation, then $-x_{0}$ is also a solution to this equation. From the uniqueness, it follows that $x_{0}=0$. Then substituting $x=0$ into the original equation, we get ... | Answer: $b \in [0 ; 1) \cup (1, +\infty)$.
## Olympiad "Sails of Hope" 2016. MIIT II round.
4 variant. | b\in[0;1)\cup(1,+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,722 |
8. DOD: $a>0 ; \quad a \neq 1 ; \quad 4 y+a^{2}>0$.
Let $\left(x_{0} ; y_{0}\right)$ be a solution to the given system. Then, due to the evenness of the system with respect to $x$, $\left(-x_{0} ; y_{0}\right)$ will also be a solution. Therefore, the necessary condition for uniqueness is $x_{0}=0$. Substituting $x_{0}... | Answer: $\boldsymbol{a}=2 \sqrt{3}, x=0 ; y=-1$.
2nd variant. | 2\sqrt{3},0;-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,726 |
7. Let $\sin \boldsymbol{x}=\boldsymbol{a} ; \sin 2 \boldsymbol{x}=\boldsymbol{b} ; \sin 3 \boldsymbol{x}=\boldsymbol{c}$.
Then $\boldsymbol{a}^{3}+\boldsymbol{b}^{3}+\boldsymbol{c}^{3}=(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c})^{3} \Leftrightarrow \boldsymbol{a}^{3}+\boldsymbol{b}^{3}=(\boldsymbol{a}+\boldsymbol{... | Answer: $\boldsymbol{x}=\frac{\boldsymbol{k} \pi}{2} ; \quad \boldsymbol{x}= \pm \frac{2 \pi}{3}+2 \pi \boldsymbol{m} ; \quad \boldsymbol{x}=\frac{2 \pi \boldsymbol{n}}{5} ; \quad \boldsymbol{k}, \boldsymbol{m}, \boldsymbol{n} \in \boldsymbol{Z}$. | \frac{k\pi}{2};\quad\\frac{2\pi}{3}+2\pi;\quad\frac{2\pin}{5};\quadk,,n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 4,728 |
8. DOD: $a>0 ; a \neq 1 ; x+a>0$.
From uniqueness, it follows that $y=0$, so we get $\left\{\begin{array}{l}\log _{2}(x+a)=2 \\ |x|=2\end{array} \Rightarrow\left\{\begin{array}{l}x+a=4 \\ x= \pm 2\end{array} \Rightarrow \boldsymbol{I}\left\{\begin{array}{l}x=2 \\ \boldsymbol{a}=2\end{array} \quad \boldsymbol{I}\left\{... | Answer: $a=6, x=-2, y=0$.
## 3rd variant. | 6,-2,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,729 |
5. Let the cost of a pound of rice be x coins, a pound of brown sugar be y coins, and a pound of refined sugar be z coins. We obtain the system:
$$
\left\{\begin{array}{l}
4 x+\frac{9}{2} y+12 z=6 ; \\
12 x+6 y+6 z=8
\end{array}\right. \text { Subtract the first equation from twice the second equation and }
$$
expres... | Answer: 4 pounds sterling. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,731 |
7. Let $\operatorname{tg} x=a ; \operatorname{tg} 2 x=b ; \operatorname{tg} 3 x=c$.
Domain of definition: $\cos 3 x \neq 0 ; \quad \cos 2 x \neq 0 ; \quad \cos x \neq 0$.
Then $\boldsymbol{a}^{3}+\boldsymbol{b}^{3}+c^{3}=(\boldsymbol{a}+\boldsymbol{b}+\boldsymbol{c})^{3} \Leftrightarrow \boldsymbol{a}^{3}+\boldsymbol... | Answer: $\boldsymbol{x}=\frac{\pi \boldsymbol{m}}{3} ; \boldsymbol{x}=\frac{\pi \boldsymbol{n}}{5} ; \boldsymbol{m}, \boldsymbol{n} \in \boldsymbol{Z}$. | \frac{\pi}{3};\frac{\pin}{5};,n\inZ | Algebra | proof | Yes | Yes | olympiads | false | 4,732 |
8. $\left\{\begin{array}{l}|x-2 a| \leq 3 \\ \log _{2}(x+a) \leq 2\end{array} \Leftrightarrow\left\{\begin{array}{l}-5 \leq x-2 a \leq 5 \\ 0<x+a \leq 4\end{array} \Leftrightarrow\left\{\begin{array}{l}2 a-5 \leq x \leq 2 a+5 \\ -a<x \leq 4-a\end{array}\right.\right.\right.$
Let's represent the obtained inequalities w... | Answer: $\mathrm{a}=3, \mathrm{x}=1$.
## 4 variant. | 3,1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,733 |
3. This system is equivalent to the following system:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ | \boldsymbol { y } - 3 x | \geq | 2 \boldsymbol { y } + \boldsymbol { x } | } \\
{ - 1 \leq y - 3 \leq 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(2 \boldsymbol{y}+\boldsymbol{x})^{2}-(\boldsymbol{y}-... | Answer: $(x ; y)=(0 ; 0)$ or $(-2 ; 0)$. | (x;y)=(0;0)or(-2;0) | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,734 |
8. $\left\{\begin{array}{l}\log _{3}(a+2 x) \leq 1 \\ |a-x| \leq 2\end{array} \Leftrightarrow\left\{\begin{array}{l}0<a+2 x \leq 3 \\ -2 \leq x-a \leq 2\end{array} \Leftrightarrow\left\{\begin{array}{l}-\frac{a}{2}<x \leq \frac{3-a}{2} \\ a-2 \leq x \leq a+2\end{array}\right.\right.\right.$
To illustrate the solutions... | Answer: $a=7 / 3$ and $x=1 / 3$.
Answers.
| | 1st option | 2nd option | 3rd option | 4th option |
| :---: | :---: | :---: | :---: | :---: |
| 1 | possible | possible | possible | possible |
| 2 | 27 | 8 | 40 | 1 |
| 3 | 6 | $56 / 3$ | $32 / 3$ | 10.5 |
| 4 | $(15 ; 18)$ | $(5 ; 3 ; 19)$ | $(3 ; 3 ; 2)$ | $(0 ; 0)$ ... | 7/31/3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,736 |
1. Since you have to approach each apple and return to the basket, the number of meters walked will be equal to twice the sum of the first hundred numbers, or 101 taken a hundred times, i.e., $10100 \mathrm{M}$. | Answer: $\{10100$ meters $\}$ | 10100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,737 |
Variant 1.
1. Let $x$ be the total amount of money that Fox Alice appropriated, and let $y$ be the amount of money each fox cub received. Then the first cub received $a+\frac{x-a}{m}=y$, the second received: $2 a+\frac{x-y-2 a}{m}=y$, and so on, the $\mathrm{n}$-th cub received: $n a+\frac{x-(n-1) y-n a}{m}=y$. Multip... | Answer: $a(m-1)^{2}$ rubles were stolen, there were $m-1$ fox cubs. | (-1)^2 | Algebra | proof | Yes | Yes | olympiads | false | 4,738 |
4. Ivan, Bohdan, and Roman live in one dormitory room and smoke pipe tobacco from a box that costs 1200 rubles. If Roman did not smoke, Ivan and Bohdan would have enough tobacco for 30 days. If Bohdan did not smoke, the other two smokers would have enough tobacco for 15 days. If Ivan did not smoke, his roommates would ... | Present the answer in the form of an integer with an accuracy of 0.1.
Translate the text above into English, keep the original text's line breaks and format, and output the translation result directly. | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,740 | |
5. It is obvious that $x=y=0$ are roots of this equation. Let now $x, y>0$ be natural numbers greater than or equal to 1. Write the equation in the form: $(\sqrt{2 x}-1)(\sqrt{2 y}-1)=1$. Let for definiteness $y \geq x$, then $\sqrt{2 x}-1 \leq 1$, i.e. $2 x \leq 4, x \leq 2 \Rightarrow x=1, x=2$. For $x=1$ we get: $(\... | Answer: $x=y=0, x=y=2$. | 0,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,741 |
7. Let $\bar{a}$ be a vector with coordinates $\bar{a}(3,-4,-\sqrt{11})$, and $\bar{b}$ be a vector with coordinates $\bar{b}(\operatorname{Sin} 2 x \operatorname{Cos} y, \operatorname{Cos} 2 x \operatorname{Cos} y, \operatorname{Siny})$. The length $|\bar{a}|$ is 6, and the length of vector $\bar{b}$ is clearly 1. Sin... | Answer: $\min f(x, y)=-6, \max f(x, y)=6$. | \f(x,y)=-6,\maxf(x,y)=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,743 |
8. Let $(a, b)$ be a pair of numbers satisfying the condition of the problem. Then the equality holds for any $x$, including $x=\pi$, and $x=2 \pi$. Substituting these values of $x$, we get the system $\left\{\begin{array}{c}-2 a+b^{2}=\operatorname{Cos}\left(\pi a+b^{2}\right)-1 \\ b^{2}=\operatorname{Cos}\left(2 \pi ... | Answer: $(0 ; 0),(1 ; 0)$. | (0,0),(1,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,745 |
1. Domain of definition (ODZ) $x>-2 ; x \neq-1$
$$
\log _{2+x}\left(8+x^{3}\right) \leq \log _{2+x}(2+x)^{3} \Leftrightarrow\left(8+x^{3}-(2+x)^{3}\right)(2+x-1) \leq 0 \Leftrightarrow x(x+2)(x+1) \leq 0 \Rightarrow
$$ | Answer: $-2<x<-1 ; x \geq 0$ | -2<x<-1;x\geq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,746 |
4. Denoting $\operatorname{Sin} x=a, \operatorname{Sin} 2 x=b, \operatorname{Sin} 3 x=c$, we get $a^{3}+b^{3}+c^{3}=(a+b+c)^{3}$. This equality can easily be reduced to the form: $(a+b)(a+c)(b+c)=0$. Then, using the formula $\operatorname{Sin} \lambda+\operatorname{Sin} \beta=2 \operatorname{Sin} \frac{\lambda+\beta}{2... | Answer: $\left\{\frac{2 k \pi}{3}, \frac{2 m \pi}{5}, \frac{\pi n}{2}\right\} k, m, n \in Z$. | {\frac{2k\pi}{3},\frac{2\pi}{5},\frac{\pin}{2}}k,,n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,747 |
Task 7.
1000 rubles are deposited in a bank for 10 years. In which case will the depositor receive more money: if the bank accrues $5 \%$ once a year, or if it accrues $\frac{5}{12} \%$ once a month. | The answer must be justified.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,750 |
Task 7.
1000 rubles are deposited in a bank for 5 years. In which case will the depositor receive more money: if the bank accrues $3 \%$ once a year, or if it accrues $\frac{1}{4} \%$ once a month. | The answer must be justified.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,751 |
# Task 7.
1000 rubles are deposited in a bank for 8 years. In which case will the depositor receive more money: if the bank accrues $4 \%$ once a year, or if it accrues $\frac{1}{3} \%$ once a month. | The answer must be justified.
# | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,752 |
1. Solve the inequality: $\log _{|x-1|}\left(\frac{x-2}{x}\right)>1$. In the answer, indicate the largest negative integer that is a solution to this inequality. | 1. Find the domain: $(x-2) x>0$. From this, $x<0$ or $x>2$. If $|x-1|<1$, then $0<x<2$. But if $|x-1|>1$, then $x<0$ or $x>2$. Therefore, the inequality is equivalent to the system:
$\left\{\begin{array}{c}|x-1|>1 \\ \frac{x-2}{x}>|x-1|\end{array} \Leftrightarrow\right.$ a) $\left\{\begin{array}{c}x<0 \\ \frac{x-2}{x}... | -1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,753 |
3. Find the area of the region defined by the inequality: $|y-| x-2|+| x \mid \leq 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. Consider three intervals of change for $x$: 1) $x \leq 0$; 2) $0 \leq x \leq 2$; 3) $x \geq 2$.
1) $x \leq 0$; then $|y+x-2|-x \leq 4, |y+x-2| \leq 4+x \Rightarrow -4 \leq x \leq 0$. Squaring the inequality, we get $(y+x-2)^{2} \leq (4+x)^{2}$, from which $(y-6)(y+2x+2) \leq 0$
2) $0 \leq x \leq 2$, then $|y+x-2| \l... | 32 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,755 |
4. Find all roots of the equation
$1-\frac{x}{1}+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\frac{x(x-1)(x-2)(x-3)}{4!}-\frac{x(x-1)(x-2)(x-3)(x-4)}{5!}+$
$+\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!}=0 . \quad($ (here $n!=1 \cdot 2 \cdot 3 . . . n)$
In the Answer, indicate the sum of the found roots. | 4. Note that by substituting the numbers $1,2,3,4,5,6$ sequentially into the equation, we will get the equality: $0=0$. This means these numbers are roots of the given equation. Since the equation is of the sixth degree, there are no other roots besides the numbers mentioned above. Answer: $\{21\}$ | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,756 |
5. Solve the equation in integers: $5 x^{2}-2 x y+2 y^{2}-2 x-2 y=3$. In the answer, write the sum of all solutions $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 5. Transforming the equation, we get: $(x+y)^{2}-2(x+y)+1+y^{2}-4 x y+4 x^{2}=4 \Leftrightarrow(x+y-1)^{2}+(y-2 x)^{2}=4$. Since $x$ and $y$ are integers, this equality is possible only in four cases:
$\left\{\begin{array}{c}x+y-1=0 \\ y-2 x=2\end{array}, \quad\left\{\begin{array}{c}x+y-1=2 \\ y-2 x=0\end{array}, \quad... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,757 |
6. Two acute angles $\alpha$ and $\beta$ satisfy the condition $\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin}(\alpha+\beta)$. Find the sum of the angles $\alpha+\beta$ in degrees. | 6. Using the formula for $\operatorname{Sin}(\alpha+\beta)$, we get:
$\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin} \alpha \operatorname{Cos} \beta+\operatorname{Cos} \alpha \operatorname{Sin} \beta \Leftrightarrow \operatorname{Sin} \alpha(\operatorname{Sin} \alpha-\operatorname{Cos} \... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,758 |
7. Find $\max x^{2} y^{2} z$ subject to the condition that $x, y, z \geq 0$ and $2 x+3 x y^{2}+2 z=36$. | 7. From the inequality for the arithmetic mean and geometric mean of three numbers, we have:
$\sqrt[3]{2 x \cdot 3 x y^{2} \cdot 2 z} \leq \frac{2 x+3 x y^{2}+2 z}{3}=12 \Rightarrow 3 x y^{2} \cdot 2 z \cdot 2 x \leq 12^{3}, x^{2} y^{2} z \leq 144$
This inequality becomes an equality if the numbers are equal, i.e., $... | 144 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,759 |
8. Determine which of the numbers is greater: (1000!) $)^{2}$ or $1000^{1000}$. Write 1 in the answer if the first is greater or equal, 2 if the second is greater. | 8. Let's show that the left number is greater than the right one. For this, we will write the first number as follows: $(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \ldots \cdot 1000)(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \ldots \cdot 1000)=(1 \cdot 1000)(2 \cdot 999)(3 \cdot 998)(4 \cdot 997) \ldots(k(10... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,760 |
9. What is the maximum area that a triangle with sides $a, b, c$ can have, given that the sides are within the ranges: $0 < a \leq 1, 1 \leq b \leq 2, 2 \leq c \leq 3$? | 9. Among the triangles with two sides $a, b$, which satisfy the conditions $0<a \leq 1, 1 \leq b \leq 2$. The right triangle with legs $a=1, b=2$ has the largest area. Its area $S=1$. On the other hand, the third side $c=\sqrt{5}$ as the hypotenuse of the right triangle satisfies the condition that $2 \leq c \leq 3$ an... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,761 |
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