problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
6. The set contained weights of 43, 70, and 57 grams, with an equal number of each type. Maly lost several weights (less than half), weighed the remaining on the scales, and got 20172 grams. How many and which weights were lost? | Solution (1). If the weights had not been lost, the total weight would end in 0. Therefore, the weight of the lost weights ends in 8. This can only be the case if 4 weights of 57 g are lost (losing 70 g or the pair $43+57$ does not affect the last digit of the total weight).
Solution (2). Let's divide the initial set ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Each boy is friends with 5 girls, while all girls are friends with a different number of boys. What is the minimum number of children that can be in this group? | Solution. If there is a girl who does not have any friends, we can exclude her (the number of children will decrease, but the condition of the problem will remain).
Let us have $m$ boys and $d$ girls. Draw a segment between a boy and a girl if they are friends. Then $5m$ segments are drawn from the boys, and at least ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a volleyball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game against each other team. The ratio of the number of wins by teams from city $B$ to the number of wins by teams from city $A$ is $3: 4$. Find $n$, given that there were no draws in the tou... | (12 points) Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ games. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural numbers $n \geq 2$, for which the equality $4 x_{n}+2 y_{n}=20 n^{2}+13 n-33$ holds, where $x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n, y_{n}=1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2}$.
(20 points) | Solution: Let $z_{n}=1+2+\cdots+n=\frac{(n+1) n}{2}=\frac{n^{2}+n}{2}$. We have
$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\c... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Shooting at a target, with each shot the athlete scored only eight, nine, or ten points (all these points were scored at least once). Making more than 11 shots, in total he scored 100 points. How many 8-point shots did the athlete make? | Solution. $8 x+9 y+10 z=100, x, y, z \in \mathbb{N} \Rightarrow 8(x+y+z)<100$
$$
\begin{gathered}
\Rightarrow 11<x+y+z<\frac{100}{8}=12.5 \Rightarrow x+y+z=12 \\
\Rightarrow\left\{\begin{array} { c }
{ x + y + z = 1 2 } \\
{ 8 x + 9 y + 1 0 z = 1 0 0 }
\end{array} \Rightarrow \left\{\begin{array} { c }
{ x + y + z =... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. The Fibonacci sequence is defined recursively: $a_{1}=a_{2}=1$, $a_{n+2}=a_{n+1}+a_{n}$ for all natural $n$. What is the last digit of $a_{2020}$? | Solution. By induction, it is proved that
1) the parity of the numbers alternates as: OOEEOOEEO...
2) the number with an index divisible by 5 is divisible by 5.
Since 2020 is divisible by 5 and leaves a remainder of 1 when divided by 3, the number $a_{2020}$ is odd and divisible by 5.
Answer: 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1 ; 0) \), and its legs lying on the lines \( y = -2 \) and \( x = 0 \)?
(12 points) | Solution. $A B: \quad y=k x+b, \quad M \in A B \Rightarrow b=-k$
$$
\begin{aligned}
& A(a ;-2) \in A B \Rightarrow-2=k a+b \Rightarrow a=1-\frac{2}{k} \\
& S_{A B C}=\frac{1}{2} B C \cdot C A=\frac{1}{2}(b+2) a=\frac{1}{2}(2-k)(1-2 / k) \\
& S^{\prime}=\frac{(2-k)(2+k)}{2 k^{2}}=0, k_{\min }=-2, \quad S_{\min }=4
\end... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Specify the greatest value of the parameter $a$ for which there is a unique solution to the system
$$
\left\{\begin{array}{l}
y=1-\sqrt{x} \\
a-2(a-y)^{2}=\sqrt{x}
\end{array}\right.
$$ | # Solution:
Equation: $\left\{\begin{array}{l}\sqrt{x}=1-y \\ a-2(a-y)^{2}=1-y\end{array}\right.$, considering that $1-y \geq 0$, we solve the resulting quadratic equation:
$$
\begin{align... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, with its hypotenuse containing the point \( M(1; 3) \), and its legs lying on the lines \( y = x \) and \( y = -x \)?
(12 points) | Solution.
$A B: \quad y=k x+d, \quad M \in A B \Rightarrow d=3-k$
$A(a ;-a) \in A B \Rightarrow-a=k a+3-k \Rightarrow a=\frac{3-k}{k+1}$,
$B(b ; b) \in A B \Rightarrow b=k b+3-k \Rightarrow b=\frac{k-3}{k-1}$,
 ^ { 2 } } \\
{ ( y + 1 ) ^ { 2 } + ( x - 0.5 a ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x-1=(y+1)^{2} \\
x-2+(x-0.5 a)^{2}=0
\end{array}\right.\right.
$$
Consider the second equation of the system
$x^{2}+x(1-a)+\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. What is the minimum value that the function $F(x ; y)=6 y+8 x-9$ can take, given that $x^{2}+y^{2}+25=10(x+y)$.
# | # Solution:
$x^{2}+y^{2}+25=10(x+y) \Leftrightarrow (x-5)^{2}+(y-5)^{2}=5^{2}$ - this is a circle with center $(5 ; 5)$ and radius 5. Let $F(x ; y)=\mathrm{M}$, then $\mathrm{M}=6 y+8 x-9$ - this is a straight line.
The condition for the minimum of the function is equivalent to the condition for the minimum of $\math... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Ivan Ivanovich approached a source with two empty cans; one held 10 liters, and the other held 8 liters. Water from the source flowed in two streams - one stronger, the other weaker. Ivan Ivanovich simultaneously placed the cans under the streams and, when half of the smaller can was filled, he switched the cans. To... | Solution: Let x liters of water fill the larger can while 4 liters fill the smaller can. After the switch, (10-x) liters fill the larger can while 4 liters again fill the smaller can. Since the flow rates are constant, the ratio of the volumes of water filled in the same time is also constant. We can set up the equatio... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. In the first alloy of copper and lead, the mass ratio is $1: 3$; in the second alloy, it is $1: 2$. How many grams of the first alloy should be taken to obtain 10 g of a new alloy with a mass ratio of copper to lead of $3: 7$? | Solution. Let there be x g of copper and 3x g of lead in the first alloy. In the second alloy, there are y g of copper and 2y g of lead. Then $k \cdot 4x + n \cdot 3y = 10 ; \frac{kx + n \cdot y}{k \cdot 3x + n \cdot 2y} = \frac{3}{7}$; we need to find $k \cdot 4x$ and 3ny. Let $ny = b ; kx = a$
$\cdot \frac{a + b}{3a ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. During the shooting practice, each soldier fired 10 times. One of them successfully completed the task and scored 90 points. How many times did he score 9 points, if there were 4 tens, and the results of the other hits were sevens, eights, and nines. There were no misses at all. | # Solution:
Since the soldier scored 90 points and 40 of them were scored in 4 shots, he scored 50 points with the remaining 6 shots. As the soldier only hit the 7, 8, and 9, let's assume he scored 24 points with three shots (one each in 7, 8, and 9). Then, for the remaining three shots, he scored 26 points, which is ... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Players divide the chips. The first player takes $m$ chips and a sixth of the remainder; the second $-2 m$ chips and a sixth of the new remainder; the third $-3 m$ chips and a sixth of the new remainder, and so on. It turned out that the chips were divided equally in this way. How many players were there? | # Solution:
Let $x$ be the number of players, and $y$ be the number of chips each has. The last player took $y=m x$ chips, with no remainder, otherwise the condition of equal distribution would not be met. The second-to-last player took $y=(x-$ 1)m plus one-sixth of the remainder, and $\frac{5}{6}$ of the remainder eq... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In rectangle $A B C D$, point $E$ is located on diagonal $A C$ such that $B C=E C$, point $M$ is on side $B C$ such that $E M=M C$. Find the length of segment $M C$, if $B M=6, A E=3$. If the answer is not an integer, round the result to tenths according to rounding rules. | # Solution:
Draw $A F$ parallel to $B E$ (point $F$ lies on line $B C$), then $\angle C B E=\angle C F A, \angle C E B=$ $\angle C A F$. Considering that $B C=C E$, we get that triangle $F C ... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. For what values of the parameter a does the equation $f(x)=p(x)$ have one solution, if $f(x)=$ $\left|\frac{2 x^{3}-5 x^{2}-2 x+5}{(1.5 x-3)^{2}-(0.5 x-2)^{2}}\right|, \quad p(x)=|2 x+5|+a$. If there are more than one value of the parameter, then in the answer, specify their sum.
# | # Solution:
Simplify $f(x)=\left|\frac{2 x^{3}-5 x^{2}-2 x+5}{(1.5 x-3)^{2}-(0.5 x-2)^{2}}\right|$, we get
$$
f(x)=|x+1|, \text { where } x \neq 1, x \neq 2.5 \text {. }
$$
Solve the equation $|x+1|=|5+2 x|+a$, where $x \neq 1, x \neq 2.5$ graphically in the system $x O a$.
1) $\left\{\begin{array}{c}x<-2.5 \\ -x-1... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find all values of $n, n \in N$, for which the sum of the first terms of the sequence $a_{k}=3 k^{2}-3 k+1, \quad k \in N$, is equal to the sum of the first $n$ terms of the sequence $b_{k}=2 k+89, k \in N . \quad(12$ points) | Solution. Note that $a_{k}=3 k^{2}-3 k+1=k^{3}-(k-1)^{3}$, and the sum is $S_{n}=n^{3}$. For the second sequence, $b_{k}=2 k+89=(k+45)^{2}-(k+44)^{2}$, the sum is $S_{n}=(n+45)^{2}-45^{2}=n(n+90)$.
We get the equation $n^{3}=n(n+90) \Rightarrow n^{2}-n-90=0 \Rightarrow n=10$.
Answer: 10. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) For what values of the parameter $a$ is the sum of the squares of the roots of the equation $x^{2}+a x+2 a=0$ equal to $21?$ | Solution: By Vieta's theorem: $x_{1}+x_{2}=-a, x_{1} \cdot x_{2}=2 a, \quad$ therefore, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=a^{2}-4 a$. From the condition, we get that $a^{2}-4 a=21$ or $a \in\{-3 ; 7\}$.
The discriminant of the equation is $D=a^{2}-8 a$, when $a=7$ it is negative, meaning ... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) Dasha added 158 numbers and got 1580. Then Seryozha tripled the largest of these numbers and decreased another number by 20. The resulting sum did not change. Find the smallest of the original numbers.
# | # Solution:
Let x be the largest of the original numbers, and y be the number that Seryozha decreased. Then: $x+y=3 x+y-20$, i.e., $x=10$.
Since the arithmetic mean of the original numbers is 10, and the largest of these numbers is also 10, each of the given numbers is 10.
Answer: 10.
## Criteria:
| 20 points | An... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}$, but according to Vieta's theorem $\left\{\begin{array}{l}D \geq 0 \\ x_{1}+x_{2}=-(m+1) . \text { Then, } \mathrm{c} \\ x_{1} x_{2}=2 m-2\end{array}\right.$ considering that $D=(m+3)^{2} \geq 0$, we have $x_{1}^{2}+x_{2}^{2}=$
$(-(m+1))^{2}-2(2 m-2)=... | Answer: For the equation $x^{2}+(m+1) x+2 m-2=0$, the smallest sum of the squares of its roots is 4 when $m=1$.
## Grading Criteria.
| 15 points | Correct and justified solution. |
| :--- | :--- |
| 10 points | Using Vieta's theorem, the expression for the sum of the squares of the roots is correctly written, but the... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Solve the equation $8 \sin ^{4}(\pi x)-\sin ^{2} x=\cos ^{2} x-\cos (4 \pi x)$. In your answer, specify the sum of the roots that belong to the interval $[-1 ; 2]$.
points) | Solution. Considering the basic trigonometric identity, we get
$8 \sin ^{4}(\pi x)-1+\cos (4 \pi x)=0 \quad \Rightarrow \quad 8 \sin ^{4}(\pi x)-2 \sin ^{2}(2 \pi x)=0 \quad \Rightarrow$
$\left(2 \sin ^{2}(\pi x)-2 \sin (\pi x) \cos (\pi x)\right)\left(2 \sin ^{2}(\pi x)+2 \sin (\pi x) \cos (\pi x)\right)=0 \Rightarr... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. What is the smallest area that a right triangle can have, if its hypotenuse lies on the tangent to the graph of the function $y=\sqrt{x-3}$, one of its legs lies on the $y$-axis, and one of its vertices coincides with the point of tangency
points) | Solution. $\quad f(x)=\sqrt{x-3}, \quad f^{\prime}\left(x_{0}\right)=\frac{1}{2 \sqrt{x-3}}$
$S_{A B C}=\frac{1}{2} A B \cdot B C, x_{0}-$ abscissa of the point of tangency $A$,
$A\left(x_{0}, f\left(x_{0}\right)\right), \quad B\left(0, f\left(x_{0}\right)\right), \quad C \quad$ - the point of intersection of the tan... | 4 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $1+\sqrt{3}$. The distances from the center of the inscribed circle in triangle $D E F$ to points $A$ and $C$ are $\sqrt{2}$ and 2, respectively. Find the length of side $A B$.
points) | Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=\sqrt{2}, C O=2$. Le... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A vessel with a capacity of 10 liters is filled with air containing $24\%$ oxygen. A certain volume of air was pumped out of the vessel and an equal volume of argon was added. Then, the same volume of the mixture as the first time was pumped out and again topped up with the same volume of argon. In the new mixture, ... | Solution. Let $x$ be the volume of the mixture released each time from the vessel. Then, the first time, the amount of oxygen left in the vessel is $2.4 - 0.24x$. The percentage of oxygen in the mixture after adding argon is $(2.4 - 0.24x) \times 10$. The second time, the amount of oxygen left in the vessel is $2.4 - 0... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given 2019 indistinguishable coins. All coins have the same weight, except for one, which is lighter. What is the minimum number of weighings required to guarantee finding the lighter coin using a balance scale without weights? (12 points) | Solution. We will prove the following statement by induction on $k$: if there are $N$ visually identical coins, with $3^{k-1}<N \leq 3^{k}$, and one of them is lighter, then it can be found in $k$ weighings. Base case: $k=0, N=1$, no weighing is needed for a single coin. Inductive step: suppose the statement is true fo... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Solve the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$ | Solution. Transform the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$
$10 x-6+x^{2}=\sqrt{x^{2}+x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}} \Rightarrow 6 x-6+x^{2}+4 x=\sqrt{x^{2}+(x-1)^{2}} \cdot \sqrt{6^{2}+(x+4)^{2}}$,
This equation can be interpreted as the scalar product of vectors $\vec{a}=(... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) The area of an isosceles trapezoid is 100, and its diagonals are perpendicular to each other. Find the height of this trapezoid. | Solution. Let $ABCD$ be an isosceles trapezoid, the diagonals $AC$ and $BD$ of which are perpendicular to each other. Denote $O = AC \cap BD$, point $M$ is the midpoint of $BC$, and point $H$ is the midpoint of $AD$. Draw the segment $MH: O \in MH, MH \perp AD$. Since $MH$ is the axis of symmetry of the given trapezoid... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Find all values of the parameter $a$ for which the equation
$$
\left(\left|\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}\right|-a\right) \cdot|4 a-3 x-19|=0
$$
has one solution. In your answer, write the largest value of the parameter $a$ | Solution:
Simplify $\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}=\frac{a x^{2}-x^{2}+(-a x+x)+(-12 a+12)}{a(x+3)-(x+3)}=$
$=\frac{x^{2}(a-1)-x(a-1)-12(a-1)}{(x+3)(a-1)}=\frac{(a-1)\left(x^{2}-x-12\right)}{(x+3)(a-1)}=$
$=\frac{x^{2}+3 x-4 x-12}{x+3}=\frac{(x+3)(x-4)}{x+3}=x-4$
We get $\left\{\begin{array}{c}(|x-... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. When adding two natural numbers, Alyosha accidentally added an extra digit, so instead of the number 23456, he got the number 55555.
Determine which digit Alyosha accidentally added.
## solution.
Let Alyosha be adding numbers $x$ and $y$. When he appended an extra digit $a$ to one of them (let's say to $x$), the ... | Answer: 36 liters
## №6: Plane Geometry.
№2. From points B and C of the acute angle BAC, perpendiculars CM and BK are drawn, intersecting at point E, such that point M lies on AB, and K lies on AC. A circle is drawn through points A, M, and C, intersecting BK at point O, and $BE=17, OK=72$. AK: $KC=1:4$. Find the tan... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Determine for which $a$ the system $x+y-144=0, xy-5184-0.1a^2=0$ has one solution. Answer: 0.
Solution. $\left\{\begin{array}{c}x+y=144 \\ xy=5184+0.1a^2\end{array}\right.$
This means that $x$ and $y$ are equal to the roots of the quadratic equation
$p^2 - 144p + (5184 + 0.1a^2) = 0$. Therefore, $D = 72^2 - (5184 ... | Answer. 2023.
## Solution.
We will prove that $2021!! + 2022!!$ is divisible by the number 2023. Indeed,
$$
2022!! = (2023-1)(2023-3)(2023-5) \ldots (2023-2021)
$$
the remainder of this number when divided by 2023 is the same as the remainder of the product
$$
(-1)(-3)(-5) \ldots (-2021) = (-1)^{\frac{2022}{2}} \c... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The sequence of real numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$ satisfies the inequalities $a_{n}-2022 a_{n+1}+2021 a_{n+2} \geq 0 \quad$ for $\quad n=1,2,3, \ldots, 98$, and $\quad a_{99}-2022 a_{100}+2021 a_{1} \geq 0$, $a_{100}-2022 a_{1}+2021 a_{2} \geq 0$. Find $a_{22}$, if $a_{10}=10$. (10 points) | # Solution.
Let $b=a_{1}+a_{2}+a_{3}+\cdots+a_{100}, b_{n}=a_{n}-2022 a_{n+1}+2021 a_{n+2}, n=1,2,3, \ldots, 98$, $b_{99}=a_{99}-2022 a_{100}+2021 a_{1}, b_{100}=a_{100}-2022 a_{1}+2021 a_{2}$. By the condition $b_{n} \geq 0$, for $n=$ $1,2,3, \ldots, 100$. We have $b_{1}+b_{2}+b_{3}+\cdots+b_{100}=b-2022 b+2021 b=0 .... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. During the shooting practice, each soldier fired 10 times. One of them completed the task successfully and scored 90 points. How many times did he score 8 points, if there were 4 tens, and the results of the hits were sevens, eights, and nines. There were no misses at all. | Solution. Since the soldier scored 90 points and 40 of them were scored in 4 attempts, he scored 50 points with the remaining 6 shots. Since the soldier only hit the seven, eight, and nine, let's assume that for three shots (once in the seven, once in the eight, and once in the nine), he scored 24 points. Then, for the... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The grandmother and granddaughter were picking strawberries. The granddaughter was using a children's bucket with a capacity of 2.5 liters, while the grandmother was using a two-liter jar. The grandmother had poor eyesight and found it hard to bend down, so the granddaughter was always picking berries faster than he... | Solution. Let the granddaughter collect $X$ liters of berries in the time it takes the grandmother to collect 1 liter, $x>1$ according to the problem. After swapping containers, the granddaughter collected 1 liter of berries, while the grandmother collected $(2.5-x)$ liters. Since their work efficiency does not change,... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. A chemistry student conducted an experiment: from a bottle filled with syrup solution, he poured out one liter of liquid, refilled the bottle with water, then poured out one liter of liquid again and refilled the bottle with water. As a result, the percentage of syrup decreased from 9 to 4 percent. Determine the vol... | # Solution.
1) Let $X$ be the volume of the bottle in liters.
2) Then after pouring out one liter of liquid, there are $(x-1)$ liters of solution left, and in it $(x-1) \cdot \frac{9}{100}$ liters of syrup and $(x-1) \cdot \frac{91}{100}$ liters of water.
3) After adding one liter of water to the bottle, there are $X$... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In triangle $\mathrm{ABC}$, angle $\mathrm{C}$ is a right angle, and $\mathrm{CD}$ is the altitude. Find the length of the radius of the circle inscribed in triangle $\mathrm{ABC}$, if the lengths of the radii of the circles inscribed in triangles $\mathrm{ACD}$ and $\mathrm{BCD}$ are 6 and 8, respectively. | Answer: 10
## №4: Progression.
The numbers $5 \mathrm{x}-\mathrm{y} ; 2 \mathrm{x}+3 \mathrm{y} ; \mathrm{x}+2 \mathrm{y}$ are consecutive terms of an arithmetic progression. The numbers $(\mathrm{y}+1)^{2} ; \mathrm{xy}+1 ;(\mathrm{x}-1)^{2}$ are consecutive terms of a geometric progression. Find the numbers x and y... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Find the value of the expression $\left(\left(\frac{3}{a-b}+\frac{3 a}{a^{3}-b^{3}} \cdot \frac{a^{2}+a b+b^{2}}{a+b}\right)\right.$ ? $\left.\frac{2 a+b}{a^{2}+2 a b+b^{2}}\right) \cdot \frac{3}{a+b}$ when $a=2023, b=2020$ | Solution:
$$
\begin{aligned}
& \left(\left(\frac{3}{a-b}+\frac{3 a}{(a-b)(a+b)}\right) \text { back } \frac{2 a+b}{(a+b)^{2}}\right) \cdot \frac{3}{a+b}=\left(\left(\frac{3(a+b)+3 a}{(a-b)(a+b)}\right) \cdot \frac{(a+b)^{2}}{2 a+b}\right) \cdot \frac{3}{a+b}=\left(\frac{3(2 a+b)}{(a-b)} \cdot\right. \\
& \left.\frac{(... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. . Find all values of the parameter $a$ for which the equation
$$
\left(\left|\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}\right|-a\right) \cdot|4 a-3 x-19|=0
$$
has one solution. In your answer, write the largest value of the parameter $a$ | Solution:
Simplify $\frac{a x^{2}-a x-12 a+x^{2}+x+12}{a x+3 a-x-3}=\frac{a x^{2}-x^{2}+(-a x+x)+(-12 a+12)}{a(x+3)-(x+3)}=$
$=\frac{x^{2}(a-1)-x(a-1)-12(a-1)}{(x+3)(a-1)}=\frac{(a-1)\left(x^{2}-x-12\right)}{(x+3)(a-1)}=$
Preliminary (correspondence) online stage of the "Step into the Future" School Students' Olympi... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In the right-angled triangle $ABC$ with the right angle at vertex $B$, on the longer leg $BC$, a segment $BD$ equal to the shorter leg is laid off from point $B$. Points $K$ and $N$ are the feet of the perpendiculars dropped from points $B$ and $D$ to the hypotenuse, respectively. Find the length of segment $BK$, if... | Solution:
Drop a perpendicular from point $D$ to segment $B K$, let the foot of this perpendicular be point $M$. Then the right triangles $A K B$ and $B M D$ are equal by hypotenuse and acute angle, therefore, $B M=A K=2$. Quadrilateral $M K N D$ is a rectangle and $M K=N D=2$, then $B K=B M+M K=2+2=4$.
Answer: 4. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. The production of x thousand units of a product costs $q=0.5 x^{2}-2 x-10$ million rubles per year. At a price of p thousand rubles per unit, the annual profit from selling this
Preliminary (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics
product (in million rubl... | # Solution:
Let the annual profit $f(x)=p x-q=p x-0.5 x^{2}+2 x+10=$
$-0.5 x^{2}+x(p+2)+10=-0.5\left(x^{2}-2 x(p+2)-20\right)=$
$=-0.5\left(x^{2}-2 x(p+2)+(p+2)^{2}-(p+2)^{2}-20\right)=$
$=-0.5(x-(p+2))^{2}+(p+2)^{2} / 2+10$
The quadratic trinomial $\mathrm{f}(\mathrm{x})$ reaches its maximum value at $\mathrm{x}=... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. In triangle $A B C$, sides $A B, A C, B C$ are equal to 5, 6, and 7 respectively. On the median $A M$, segment $A K$ equal to 2 is laid off. Find the ratio of the areas of triangles $A B K$ and $A C K$. Write the answer as $\frac{S_{A B K}}{S_{A C K}}$. | # Solution:
The median of a triangle divides the triangle into two equal-area (equal in area) triangles. In triangle $ABC$, the areas of triangles $ABM$ and $ACM$ are equal because $AM$ is its median. In triangle $KBC$, the segment $KM$ is the median, so the areas of triangles $KBM$ and $KCM$ are equal. $S_{ABK} = S_{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. A truck and a car are moving in the same direction on adjacent lanes at speeds of 65 km/h and 85 km/h, respectively. At what distance from each other will they be 3 minutes after they are side by side? | Answer: 1 km. Solution. The difference in the speeds of the cars is 20 km/h, so in 3 minutes they will be separated by a distance equal to 20 (km/h) $\cdot \frac{1}{20}$ (hour) $=1$ km. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. On a line, several points were marked, after which two points were placed between each pair of adjacent points, and then the same procedure (with the entire set of points) was repeated again. Could there be 82 points on the line as a result? | Answer: Yes. Solution. Let there be x points marked on the line initially. Then after the first procedure, $2(x-1)$ points are added to them, and there are a total of $3x-2$ points. After the second procedure, $2(3x-3)$ points are added to these points. Thus, there are a total of $3x-2+2(3x-3)=9x-8$ points on the line.... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. In triangle $A B C$, the angles $A$ and $C$ at the base are $20^{\circ}$ and $40^{\circ}$ respectively. It is known that $A C - A B = 5$ (cm). Find the length of the angle bisector of angle $B$. | Answer: 5 cm. Solution. Let $B M$ be the bisector of angle $B$. Mark a point $N$ on the base $A C$ such that $A N=A B$. Then triangle $A B N$ is isosceles and $\angle A B N=\angle A N B=80^{\circ}$. Since $\angle A B M=\frac{180^{\circ}-20^{\circ}-40^{\circ}}{2}=60^{\circ}$, then $\angle B M N=\angle A+\angle A B M=20^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3. What is the smallest number of digits that need to be appended to the right of the number 2014 so that the resulting number is divisible by all natural numbers less than $10?$ | Answer: 4 digits. Solution. If three digits are appended to 2014, the resulting number will be $\leq 2014$ 999. Dividing 2014999 by $2520=$ LCM $(1,2, \ldots, 9)$, we get a quotient of 799 and a remainder of 1519. Since $1519>1000$, there is no integer multiple of 2520 between the numbers 2014000 and 2014999. Therefore... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Solve the equation $\sin ^{2} x+1=\cos (\sqrt{2} x)$. | The answer is $x=0$. Solution. The left side of the equation $\geq 1$, and the right side $\leq 1$. Therefore, the equation is equivalent to the system: $\sin x=0, \cos \sqrt{2} x=1$. We have $x=\pi n, \sqrt{2} x=2 \pi k$ ( $n, k$ - integers). From this, $n=k \cdot \sqrt{2}$. Since $\sqrt{2}$ is an irrational number, t... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. Andrey and Seva are going to visit Borya. Andrey is at point $A$, and Borya is at point $B$, 30 km away from point $A$ along a straight highway. Seva is at point $C$, exactly halfway between $A$ and $B$. The friends decided to leave simultaneously: Andrey on a bicycle, and Seva on foot, but Andrey will leave the b... | Answer: 5 km before point $B$. Solution. Let $a=15$ (km), $u=5$ (km/h), $v=20$ (km/h). Let $x$ (km) be the distance from point $B$ to the place where the bicycle is left. Then Andrei's travel time is $t_{A}=\frac{2 a-x}{v}+\frac{x}{u}$, and Seva's travel time is $t_{C}=\frac{a-x}{u}+\frac{x}{v}$. We need to find such a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. From a rectangular table of $m \times n$ cells, several squares of different sizes need to be cut out along the grid lines. What is the maximum number of squares that can be cut out if: a) $m=8$, $n=11$; b) $m=8, n=12 ?$ | Answer: a) 5; b) 5. Solution. a) Note that the area of six different squares is no less than $1+4+9+16+25+36=91>88$. Therefore, it is impossible to cut out more than five different squares. A possible example for five squares (even in a rectangle $8 \times 9$) is shown in the figure. b) Suppose, for the sake of contrad... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. How many roots does the equation $\sqrt{14-x^{2}}(\sin x-\cos 2 x)=0$ have? | Answer: 6 roots. Solution. The domain of the equation: $14-x^{2} \geq 0 \Leftrightarrow|x| \leq \sqrt{14}$. On this domain, we solve the trigonometric equation $\sin x-\cos 2 x=0 \Leftrightarrow \sin x+2 \sin ^{2} x-1=0$; $\sin x=\frac{-1 \pm 3}{4}$, i.e., $\sin x=-1$ or $\sin x=\frac{1}{2}$. From this, we get three se... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.5. How many a) rectangles; b) right-angled triangles with integer sides exist, for which the area is numerically equal to the perimeter? (Congruent figures are counted as one). | Answer: a) two; b) two. Solution. a) Let the sides of the rectangle be integers $a$ and $b$ ( $a \leq b$ ). Then $a b=2(a+b)$, which is equivalent to the equation $(a-2)(b-2)=4$. The number 4 can be factored into the product of two integers in four ways: $(1 ; 4),(2,2),(-4 ;-1),(-2,-2)$, of which the first two give rec... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.1. Solve the equation $\cos ^{2}(\sqrt{2} x)-\sin ^{2} x=1$. | Answer: $x=0$. Solution. The left side of the equation does not exceed one, and it can equal one only if the following two conditions are simultaneously satisfied: $\cos ^{2}(\sqrt{2} x)=1$ and $\sin x=0$. Hence, $\sqrt{2} x=\pi n$ and $x=\pi k, k, n \in \mathbf{Z}$. Then $\frac{\pi n}{\sqrt{2}}=\pi k$, i.e., $n=\sqrt{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. $n$ vectors in space are such that any pair of them forms an obtuse angle. What is the largest value that $n$ can take? | Answer: 4. Solution. Let $\vec{a}_{1}, \vec{a}_{2}, \ldots, \vec{a}_{n}$ be the given vectors. Direct the $z$-axis of the coordinate space along $\vec{a}_{n}$. Then the $z$-coordinate of the other vectors must be negative (this follows from the formula for the cosine of the angle between vectors through the scalar prod... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.3. Find all values of the parameter $a$ for which the equation $a x^{2}+\sin ^{2} x=a^{2}-a$ has a unique solution. | Answer: $a=1$. Solution. Note that only $x=0$ can be the unique root of the equation, since due to the evenness of the functions involved, for any solution $x_{0} \neq 0$, $(-x_{0})$ will also be a solution. Therefore, we necessarily get $a^{2}-a=0 \Leftrightarrow a=0$ or $a=1$. Let's check these values. When $a=0$, we... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.4. On the coordinate plane, the parabola $y=x^{2}$ is drawn. On the positive $O y$-axis, a point $A$ is taken, and through it, two lines with positive slopes are drawn. Let $M_{1}, N_{1}$ and $M_{2}, N_{2}$ be the points of intersection with the parabola of the first and second line, respectively. Find the ordinate ... | Answer. 1. Solution. Let $a$ be the ordinate of point A. The line passing through point $A$ has the equation $\mathrm{y}=k \cdot x+a$, and the abscissas $x_{1}, x_{2}$ of points M and $N$ of intersection of the line with the parabola are the roots of the equation $x^{2}=k \cdot x+a$. From Vieta's theorem, we have $x_{1... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.2. Solve the equation $(\sqrt{2023}+\sqrt{2022})^{x}-(\sqrt{2023}-\sqrt{2022})^{x}=\sqrt{8088}$. | Answer: $x=1$.
Solution. Notice that $\sqrt{8088}=2 \sqrt{2022}$. This observation suggests that $x=1$ is a root of the equation. We will show that there are no other roots. Indeed, the left side represents the difference of two exponential functions: the base of the first is greater than one, and the base of the seco... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3 How many right-angled triangles with integer sides exist, where one of the legs is equal to 2021. | Answer: 4. Solution. Let the hypotenuse of a right triangle be $x$, one of the legs be $-y$, and the other be 2021. Then, by the Pythagorean theorem, $x^{2}-y^{2}=2021^{2}$, i.e., $(x-y) \cdot(x+y)=2021^{2}$. Considering that $x>y$, we have: $x+y>x-y>0$. Since the prime factorization of 2021 is $2021=43 \cdot 47$, the ... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.1. For what values of the parameter $a$ do the equations $a x+a=7$ and $3 x-a=17$ have a common integer root? | Answer: $a=1$.
Solution. Solving these two equations as a system with unknowns $x$ and $a$, express $a$ from the second equation and substitute it into the first. We get the quadratic equation $3 x^{2}-14 x-24=0$. It has two roots: 6 and (-4/3). For the integer root $x=6$, the corresponding value of $a=3 x-17=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.2. How many solutions in natural numbers $x, y$ does the equation $x+y+2 x y=2023$ have? | Answer: 6.
Solution. Multiply the equation by 2, add one to both sides, and factor the left side, while expressing the right side as a product of prime factors:
$$
2 x+4 x y+2 y+1=4047 \Leftrightarrow(2 x+1)(2 y+1)=4047 \Leftrightarrow(2 x+1)(2 y+1)=3 \cdot 19 \cdot 71
$$
Since \(x\) and \(y\) are natural numbers, e... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.2. Solve the equation $2 \cos (\pi x)=x+\frac{1}{x}$. | Answer: $x=-1$.
Solution. The left side of the equation is no more than two in absolute value, while the right side is no less than two in absolute value: for the right side, this follows from elementary inequalities, for example (for $x>0$), - from the inequality between the arithmetic mean and the geometric mean, an... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.1. Given the sequence $a_{n}=(-1)^{1+2+\ldots+n}$. Find $a_{1}+a_{2}+\ldots+a_{2017}$. | Answer: -1. Solution. We have $a_{n}=(-1)^{\text {sn }}$, where $S_{n}=\frac{n(n+1)}{2}$. It is easy to notice and prove that the parity of the number $S_{n}$ repeats with a period of 4: indeed, $S_{n+4}-S_{n}=\frac{(n+4)(n+5)-n(n+1)}{2}=\frac{8 n+20}{2}=4 n+10$, i.e., an even number. Therefore, $a_{1}+a_{2}+\ldots+a_{... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.7 Find all values of the parameter \(a\), for each of which the system of equations
\[
\left\{
\begin{array}{l}
x^{2}+y^{2}=2 a, \\
x+\log _{2}\left(y^{2}+1\right)=a
\end{array}
\right.
\]
has a unique solution. | Answer: $a=0$.
Solution. Note that if $(x, y)$ is a solution to the system, then $(x, -y)$ is also a solution to this system. Therefore, if for a parameter $a$ the system has a unique soluti... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.4. What is the minimum number of kings that need to be placed on a chessboard so that they attack all unoccupied squares? (A king attacks the squares that are adjacent to its square by side or corner).
 that there exists a rectangle whose vertices coincide with the centers of four cells of the same color? | Answer: $n=5$. Proof. We will prove that for $n=5$ (and thus for $\mathrm{n}>5$) such a rectangle will exist. Consider the bottom row of the table. In it, there are at least 3 cells of the same color. Let's assume for definiteness that these are white cells. Then consider the three columns with these cells at the base,... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.5. Girls stood in a circle, some of them wearing scarves. A circle is called correct if each girl without a scarf has a neighbor wearing a scarf. a) What is the minimum number of scarves in a correct circle of 25 girls? b) Prove that if in this correct circle of 25 girls there are more than 12 scarves, then some girl... | Answer: a) 9. Solution. a) Note that among three consecutive girls, there is at least one handkerchief: otherwise, the girl in the middle would not have a neighbor with a handkerchief. Fix one girl with a handkerchief in the circle, say Tanya, and consider three girls in sequence clockwise after Tanya. There are a tota... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.1. $\quad$ A cyclist planned to travel from point $A$ to point $B$ in 5 hours, moving at a constant speed. He traveled at the planned speed until the midpoint of the journey, then increased his speed by $25 \%$. He completed the journey to point $B$ at the new speed. How long did the entire trip take? | Answer: 4 hours 30 minutes. Solution. See problem 7.1. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.4. The sequence $a_{n}$ is defined as follows: $a_{1}=2^{20}, a_{n+1}=s\left(a_{n}\right)$ for all $n$, where $s(a)$ denotes the sum of the digits of the natural number $a$. Find $a_{100}$. | Answer: 5. Solution. The main fact used is that the sum of the digits of any number has the same remainder when divided by 9 as the number itself. This fact is proven in the same way as the well-known divisibility rule for 9. We will show that the sequence $a_{n}$ decreases rapidly until it becomes less than 10, and af... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.4. In triangle $A B C$, the angles $A$ and $C$ at the base are $20^{\circ}$ and $40^{\circ}$, respectively. It is known that $A C - A B = 5$ (cm). Find the length of the angle bisector of angle $B$. | Answer: 5 cm. Solution. Let $B M$ be the bisector of angle $B$. Mark a point $N$ on the base $A C$ such that $A N=A B$. Then triangle $A B N$ is isosceles and $\angle A B N=\angle A N B=80^{\circ}$. Since $\angle A B M=\frac{180^{\circ}-20^{\circ}-40^{\circ}}{2}=60^{\circ}$, then $\angle B M N=\angle A+\angle A B M=20^... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.2. Find all values of the parameter $a$ for which the equation $|x+a|=\frac{1}{x}$ has exactly two roots. | Answer: $a=-2$. Solution. The intersection of the graphs of the right and left parts can only be in the first
the graph $y=|x+a|$ in the first quadrant represents the line $y=x+a$, which inte... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5. Given a convex quadrilateral $A B C D$, where $A B=A D=1, \angle A=80^{\circ}$, $\angle C=140^{\circ}$. Find the length of the diagonal $A C$. | Answer: 1. Solution. We will prove that $A C=1$ by contradiction. If $A C>1$, then in triangle $A B C$ the larger side $A C$ is opposite the larger angle: $\angle B>\angle B C A$. Similarly, for triangle $A D C$ we have $\angle D>\angle D C A$. Adding these inequalities, we get $\angle B+\angle D>\angle C=140^{\circ}$.... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.2. From a three-digit number, the sum of its digits was subtracted and the result was 261. Find the second digit of the original number. | Answer: 7. Solution. Let $\overline{x y z}=100 x+10 y+z$ be the original number. According to the condition, $99 x+9 y=261$, i.e., $11 x+y=29$. Since $y \leq 9$, for $29-y$ to be divisible by 11, we get $y=7$ (and then $x=2$). | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.2. There are 11 kg of cereal. How can you measure out 1 kg of cereal using two weighings on a balance scale, if you have one 3 kg weight? | Solution. First weighing: place a weight (3 kg) on one scale pan, and on the other, initially 11 kg of grain, and keep pouring grain onto the first pan until equilibrium is achieved. We get 3 kg (weight) + 4 kg (grain) $=7$ kg (grain) (since $3+x=11-x=>x=4$). Second weighing: from the obtained 4 kg of grain, pour out 3... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.3. The numbers $a, b$, and $c$ satisfy the relation $\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}$. Find $a+b+c$, given that $b \neq c$. | Answer: 0. Solution. From the equality $\frac{a+b}{c}=\frac{a+c}{b}$, we get $a(b-c)=(c-b)(c+b)$. Dividing this equality by $b-c \neq 0$, we obtain $b+c=-a$. Therefore, $a+b+c=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.3. The sum $1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{45}$ is represented as a fraction with the denominator $45!=1 \cdot 2 \cdots 45$. How many zeros (in decimal notation) does the numerator of this fraction end with? | Answer: 8 zeros. Solution. The numerator of the fraction is the sum of numbers of the form $1 \cdot 2 \cdot \ldots \cdot(k-1)(k+1) \cdot \ldots .45$ (the product lacks one of the natural numbers from 1 to 45). Let's denote such a term as $c_{k}$. Note that $45!=5^{10} \cdot 2^{n} \cdot \boldsymbol{p}$, where $\boldsymb... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. There are 19 kg of cereal. Can 1 kg be measured with three weighings on a balance scale if there is one 3 kg weight? | Answer: It is possible. Solution. The first weighing can yield 8 kg of cereal. If a weight is placed on one (left) pan of the scales and, by pouring cereal from the right pan to the left, the scales are balanced. Indeed, from the equation $3+x=19-x$ we get $x=8$, i.e., there will be 8 kg of cereal on the pan with the w... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 4. (12 points)
Solve the equation $x^{2}+y^{2}+1=x y+x+y$.
# | # Solution
$$
\begin{gathered}
x^{2}+y^{2}+1=x y+x+y \Leftrightarrow(x-y)^{2}+(x-1)^{2}+(y-1)^{2}=0 \Leftrightarrow \\
x=y=1 .
\end{gathered}
$$
Answer: $x=y=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 1. (10 points)
How many natural numbers $n$ exist such that the equation $n x-12=3 n$ has an integer solution
# | # Solution.
Since $n x-12=3 n \Rightarrow x=3+\frac{12}{n}$, the condition of the problem is satisfied if 12 is divisible by $n$. Therefore, $n$ can take the values $1,2,3,4,6,12$.
Answer. 6
| Criterion | Evaluation | Points |
| :---: | :---: | :---: |
| Complete solution. | + | 10 |
| All main logical steps of the ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 5. (12 points)
On a plane, there are four different circles. We will call an intersection point a point where at least two circles intersect. Find the maximum possible number of intersection points of four circles.
# | # Solution
Any two circles can intersect at no more than two points. From four circles, we can choose 6 different pairs of circles. Therefore, the number of intersection points cannot exceed 12.
The figure below shows a case where there are exactly 12 intersection points.
How many distinct roots does the equation $f(f(f(x)))=1$ have, if $f(x)=x-\frac{2}{x}$. | # Solution.
Let $f(x)=k=x-\frac{2}{x}, f(f(x))=f(k)=k-\frac{2}{k}, f(f(f(x)))=f(f(k))=f(k)-\frac{2}{f(k)}$. The equation $f(f(f(x)))=f(k)-\frac{2}{f(k)}=1 \Leftrightarrow f^{2}(k)-f(k)-2=0$ has two solutions $f_{1}(k)=-1$ and $f_{2}(k)=2$.
We obtain $\left[\begin{array}{l}f(k)=k-\frac{2}{k}=-1 \\ f(k)=k-\frac{2}{k}=2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 11.6. (14 points)
Real numbers $a$ and $b$ are such that $\left(a+\sqrt{1+a^{2}}\right)\left(b+\sqrt{1+b^{2}}\right)=1$. Find the sum $a+b$.
# | # Solution.
$$
\begin{aligned}
& b+\sqrt{1+b^{2}}=\frac{1}{a+\sqrt{1+a^{2}}}=-a+\sqrt{1+a^{2}} \\
& a+\sqrt{1+a^{2}}=\frac{1}{b+\sqrt{1+b^{2}}}=-b+\sqrt{1+b^{2}}
\end{aligned}
$$
By adding the obtained equations, we get
$$
a+b+\sqrt{1+a^{2}}+\sqrt{1+b^{2}}=-a-b+\sqrt{1+a^{2}}+\sqrt{1+b^{2}} \Rightarrow a+b=0
$$
## ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 4. (12 points)
The angle bisectors of angles $A, B$, and $C$ of triangle $A B C$ intersect the circumcircle of this triangle at points $A_{1}, B_{1}$, and $C_{1}$, respectively. Find the distances between point $A_{1}$ and the center of the inscribed circle of triangle $A B C$, given that $\angle A_{1} B_{1} C_... | # Solution.
In the figure, identical numbers mark equal angles (this follows from the fact that $A A_{1}, B B_{1}, C C_{1}$ are the angle bisectors of triangle $ABC$, the angle marked "1+2" near point $O$ (which is the center of the inscribed circle) is equal to $\angle O A B + \angle O B A = \angle 1 + \angle 2$ by t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task 8. (16 points)
How many solutions in integers does the equation $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}=2031$ have. | # Solution.
Note that if the number $n$ is even, then $n^{4}$ is divisible by 16.
If $n$ is odd, then the number
$$
n^{4}-1=(n-1)(n+1)\left(n^{2}+1\right)
$$
is divisible by 16.
Therefore, the remainder of the left-hand side of the equation $x_{1}^{4}+x_{2}^{4}+\cdots+x_{14}^{4}$ when divided by 16 is equal to the... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. At the exchange office, there are two types of operations: 1) give 2 euros - get 3 dollars and a candy as a gift; 2) give 5 dollars - get 3 euros and a candy as a gift.
When the wealthy Pinocchio came to the exchange office, he only had dollars. When he left, he had fewer dollars, no euros appeared, but he ... | Answer: $10.
Solution: Since Buratino received 50 candies, he performed exactly 50 operations. At the same time, he exchanged all the euros he received back into dollars. Therefore, for every 3 operations of the first type, there were 2 operations of the second type. That is, Buratino received $3 30 times and gave awa... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Yura has a calculator that allows multiplying a number by 3, adding 3 to a number, or (if the number is divisible by 3) dividing the number by 3. How can one use this calculator to get from the number 1 to the number 11? $\quad[3$ points] (T. I. Golenishcheva-Kutuzova) | Answer. For example, $((1 \cdot 3 \cdot 3 \cdot 3)+3+3): 3=11$ or $(1 \cdot 3+3): 3+$ $3+3+3=11$.
Comment. Note that on Yura's calculator, any number can be increased by $1:(x \cdot 3+3): 3=x+1$. Therefore, in principle, any natural number can be obtained from one on it. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In a small town, there is only one tram line. It is circular, and trams run along it in both directions. There are stops on the loop: Circus, Park, and Zoo. The journey from Park to Zoo via Circus is three times longer than not via Circus. The journey from Circus to Zoo via Park is twice as short as not via ... | Answer: The path not through the Zoo is 11 times shorter.
Solution. Let's board the tram at the Zoo stop and travel through the Circus to the Park, and then, without leaving the tram, return to the Zoo. The second part of the journey is three times shorter than the first, meaning the first part takes up three quarters... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. On the surface of a planet shaped like a donut, two snails crawled, leaving trails behind: one along the outer equator, and the other along a spiral line (see figure). Into how many parts did the snails' trails divide the surface of the planet? (It is sufficient to write the answer.)
, then each rat ate $10 / k$ cheese heads on the first night. On the second night, each rat ate half as much, that is, $5 / k$ cheese heads. Thus, the 7 rats ate a total of $35 / k$ cheese heads. This is an integer. The only divisor of the ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Dima lives in a nine-story building. He descends from his floor to the first floor by elevator in 1 minute. Due to his small stature, Dima cannot reach the button for his floor. Therefore, when going up, he presses the button he can reach, and then walks the rest of the way. The entire journey upwards takes ... | Answer. Dima lives on the seventh floor.
First solution. Consider the part of the journey that Dima travels down by elevator and up on foot. On the one hand, the walk takes twice as long, and on the other, it is 10 seconds longer. Therefore, he traveled this part by elevator in 10 seconds and walked it in 20 seconds. ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. Bus stop $B$ is located on a straight highway between stops $A$ and $C$. After some time following departure from $A$, the bus found itself at a point on the highway such that the distance from this point to one of the three stops is equal to the sum of the distances to the other two. After the same amount o... | Answer: 3 hours.
Solution. At both moments in time mentioned in the problem, the sum of distances will obviously be the distance from the bus to the farthest stop from it. This cannot be $B$, as it is closer than $C$. Therefore, these were $C$ (before the bus had traveled halfway from $A$ to $C$) and $A$ (after this m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. In the park, there were lindens and maples. Maples among them were $60 \%$. In spring, lindens were planted in the park, after which maples became $20 \%$. In autumn, maples were planted, and maples became $60 \%$ again. By what factor did the number of trees in the park increase over the year?
[6 points] (... | Answer: 6 times.
First solution. Before the planting, lindens constituted $2 / 5$, and maples $-3 / 5$ of all the trees in the park. By summer, the number of maples did not change, but they began to constitute $1 / 5$ of all the trees. Therefore, the total number of trees in the park increased threefold. At the same t... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Children went to the forest to pick mushrooms. If Anya gives half of her mushrooms to Vitya, all the children will have the same number of mushrooms, and if instead Anya gives all her mushrooms to Sasha, Sasha will have as many mushrooms as all the others combined. How many children went mushroom picking? | Answer: 6 children.
Solution: Let Anya give half of her mushrooms to Vitya. Now all the children have the same number of mushrooms (this means that Vitya did not have any mushrooms of his own). For Sanya to now get all of Anya's mushrooms, he needs to take the mushrooms from Vitya and Anya. Then he will have the mushr... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Mom baked pies - three with rice, three with cabbage, and one with cherries - and laid them out in a circle on a plate (see fig.). Then she put the plate in the microwave to warm them up. On the outside, all the pies look the same.
 were either seals or otters, and Igor remembered that 5 were either otters or sea l... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. In an aquarium, there are three types of fish: gold, silver, and red. If the cat eats all the gold fish, the number of fish will be 1 less than $2 / 3$ of the original number. If the cat eats all the red fish, the number of fish will be 4 more than $2 / 3$ of the original number. Which fish - gold or silver ... | Answer: There are 2 more silver fish.
Solution: From the first condition, there is 1 more goldfish than a third. From the second condition, there are 4 fewer red fish than a third. Therefore, there are 3 more silver fish than a third. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Given an equilateral triangle $ABC$. On side $AB$, point $K$ is marked, and on side $BC$ - points $L$ and $M$ ( $L$ lies on segment $BM$ ) such that $KL=KM, BL=2, AK=3$. Find $CM$.
[7 points]
. Assign (draw with arrows) several more one-way flights so that from any city to any other, one could get there with no more than two transfers. Try to make the number of additional flights as small as possible.
 | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. Each face of a cube $6 \times 6 \times 6$ is divided into $1 \times 1$ cells. The cube is covered with $2 \times 2$ squares such that each square covers exactly four cells, no squares overlap, and each cell is covered by the same number of squares. What is the maximum value that this identical number can tak... | Answer: 3.
Solution: Estimation. A cell in the corner of a face can be covered in three ways (entirely within the face, with a fold over one edge of the corner, with a fold over the other edge of the corner). Therefore, each cell is covered by no more than three squares.
Example. Consider the usual covering of a cube... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Petya observes two ants crawling along a straight path at constant speeds. After 3 minutes of observation, the distance between the ants was 9 meters, after 5 minutes - 5 meters, after 9 minutes - 3 meters. What was the distance between the ants after 8 minutes of observation?
$[5$ points]
![](https://cdn.mathpix.c... | 4. The faster ant is ahead of the slower one and is moving away from it.
It is clear that situation 1 can over time turn into situation 2, and situation 3 can turn into situation 4. But if the ants at some point find themselves in situation 2 or 4, then from that moment on, they will always be moving away from each ot... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.