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3. In the arithmetic progression $\left(a_{n}\right) a_{1}=1, d=4$.
$$
\text { Calculate } A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}
$$
In the answer, write the smallest integer greater than $A$. | Solution: We transform the expression by multiplying the numerator and denominator of each fraction by the expression conjugate to the denominator:
$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Right triangles $\triangle M D C$ and $\triangle A D K$ have a common right angle $\angle D$. Point $K$ lies on $C D$ and divides it in the ratio $2: 3$ from point $C$. Point $M$ is the midpoint of side $A D$. Find the sum of $\angle A K D$ and $\angle M C D$, if $A D: C D=2: 5$.
$ satisfies the condition $f(x+1)+1=f(x)+\frac{20}{(x+1)(x+2)}$. Calculate $\frac{2019}{f(2019)}$, if $\quad f(0)=2019$. | # Solution.
Notice that
$f(x+2019)-f(x)=(f(x+2019)-f(x+2018))+(f(x+2018)-f(x+2017))+\cdots+(f(x+$ 1) $-f(x))=\frac{20}{(x+2019)(x+2020)}-1+\frac{20}{(x+2018)(x+2019)}-1+\cdots+\frac{20}{(x+1)(x+2)}-1$. Therefore, $f(2019)-f(0)=20\left(\frac{1}{2019}-\frac{1}{2020}+\ldots+1-\frac{1}{2}\right)-2019=20\left(1-\frac{1}{2... | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. What is the minimum number of cells that need to be painted in a square with a side of 65 cells (a total of $65 \times 65$ cells, which is 4225 cells in the square), so that from any unpainted cell it is impossible to move to another unpainted cell with a knight's move in chess? | # Solution
Cells should be colored in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be colored. Since any "knight's move" lands on a cell of a different color, there is no move to a cell of the same color. A "knight's move" can traverse any square table larger than 4x4 such that the knight vi... | 2112 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Find the sum of all integer values of $\mathrm{h}$ for which the equation ||$r+h|-r|-4 r=9|r-3|$ in terms of $r$ has no more than one solution.
# | # Solution
Consider the function $\mathrm{f}(\mathrm{r})=9|\mathrm{r}-3|-|| \mathrm{r}+\mathrm{h}|-\mathrm{r}|+4 \mathrm{r}$. The coefficient of the first modulus in absolute value is greater than the sum of the other coefficients of $\mathrm{r}$. $9>1+1+4$. Therefore, on all intervals up to $r=3$, the coefficient of ... | -93 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. A table consisting of 1861 rows and 1861 columns is filled with natural numbers from 1 to 1861 such that each row contains all numbers from 1 to 1861. Find the sum of the numbers on the diagonal that connects the top left and bottom right corners of the table, if the table is filled symmetrically with respect to thi... | # Solution:
We will show that all numbers from 1 to 1861 are present on the diagonal. Suppose the number $a \in\{1,2,3 \ldots, 1861\}$ is not on the diagonal. Then, due to the symmetry of the table, the number $a$ appears an even number of times. On the other hand, since the number $a$ appears once in each row, the to... | 1732591 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Ilya takes a triplet of numbers and transforms it according to the rule: at each step, each number changes to the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 1989 applications of this rule, if the initial triplet of numbers was $\{70 ; 61 ; 20\}$? If... | # Solution:
Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b . \quad b=70-20=50$.
Answer: 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Two points are moving along a circle. When moving in the same direction, the faster point catches up with the slower point every 16 seconds. If they move in opposite directions at the same speeds, they meet every 4 seconds. It is known that when moving towards each other along the circle, the distance between the ap... | # Solution:
Let $v-$ be the speed of the faster point, and $u-$ be the speed of the slower point. According to the problem, we set up the system of equations: $\left\{\begin{array}{l}(u+v) \cdot 4=(v-u) \cdot 16 \\ (u+v) \cdot 2=64\end{array} ;\left\{\begin{array}{l}3 v=5 u \\ u+v=32\end{array} ;\left\{\begin{array}{l... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $A B C$, side $B C$ is 18 cm. The perpendicular $D F$, drawn from the midpoint of side $A B$ - point $D$, intersects side $B C$ at point $F$. Find the perimeter of triangle $A F C$, if side $A C$ is $9 \, \text{cm}$. | # Solution:
Triangle $ABF (BF = AF)$ is isosceles, since $DF \perp AB$, and $D$ is the midpoint of $AB$. $P_{AFC} = AF + FC + AC = BF + FC + AC = BC + AC = 27 \text{ cm}$.
Answer: 27 cm. | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. There are two types of containers: 27 kg and 65 kg. How many containers of the first and second type were there in total, if the cargo in the first type of containers exceeds the cargo in the second type of containers by 34 kg, and the number of 65 kg containers does not exceed 44 units?
# | # Solution:
Let $x$ be the number of containers weighing 27 kg, and $y$ be the number of containers weighing 65 kg. We get the equation $27 x - 65 y = 34$.
$27(x - 2y) - 11y = 34$, let $x - 2y = k$
$27k - 11y = 34$
$11(2k - y) + 5k = 34$, let $2k - y = t(2)$
$11t + 5k = 34$
$5(2t + k) + t = 34$, let $2t + k = n(3... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A farmer presented 6 types of sour cream in barrels of $9,13,17,19,20,38$ liters at the market. On the first day, he sold sour cream from three barrels completely, and on the second day, from two more barrels completely. The volume of sour cream sold on the first day was twice the volume of sour cream sold on the se... | # Solution:
A total of 116 liters were delivered; $116=3 \cdot 38+2$; therefore, the unsold barrel, when divided by 3, gives a remainder of 2.
1) If it is 17 liters, then $116-17=99$, then on the second day, one third of 99 liters were sold, which is $33=13+20$, then on the first day, barrels with capacities of $9, 1... | 66 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. In triangle $A B C$ with $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Segment $A_{1} B_{1}$ intersects the angle bisector $C C_{1}$ at point M. Find $\angle C B M$.
# | # Solution:
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$.
Considering that point $A_{1}$ lies on the bisector of $\angle B A C$, and therefore is equidistant from its sides.
We obtain that $A_{1}$ i... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. For what values of the parameter $\boldsymbol{a}$ does the equation $|f(x)-5|=p(x)$, where $f(x)=\left|\frac{x^{2}-10 x+25}{x-5}-\frac{x^{2}-3 x}{3-x}\right|$, $p(x)=a \quad$ have three solutions? If there is more than one value of the parameter, indicate their product in the answer. | # Solution:
Simplify $f(x)=\left|\frac{x^{2}-10 x+25}{x-5}-\frac{x^{2}-3 x}{3-x}\right|$, we get $f(x)=$ $|2 x-5|$, where $x \neq 5, x \neq 3$.
Solve the equation || $2 x-5|-5|=a$, where $x \neq 5, x \neq 3$ graphically in the system $x O a$.
The equation has three solutions when $a=4, a=5$.
The product is 20.
Ans... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If a two-digit number is decreased by 54, the result is a number written with the same digits but in reverse order. In the answer, specify the arithmetic mean of the resulting numbers.
# | # Solution:
$\overline{x y}=10 x+y-$ the original two-digit number, then $\overline{y x}=10 y+x$ - the number written with the same digits but in reverse order. We get the equation $10 x+y=10 y+x+54$.
From the equation, it is clear that the two-digit number is greater than 54. Let's start the investigation with the t... | 82 | Number Theory | 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 |
8. Given triangle $A B C . \angle A=\alpha, \angle B=\beta$. Lines $O_{1} O_{2}, O_{2} O_{3}, O_{1} O_{3}$ are the bisectors of the external angles of triangle $A B C$, as shown in the figure. Point O is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $\mathrm{CO}_{1}$ and $\mat... | # Solution:
We will prove that the bisectors of two external angles and one internal angle intersect at one point. Let \( O_{1} G, O_{1} H, O_{1} F \) be the perpendiculars to \( B C, A C \)... | 90 | 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 |
1. In how many ways can two knights - a white one and a black one - be placed on a chessboard consisting of $16 \times 16$ cells so that they threaten each other? (A knight moves in an "L" shape, i.e., it can move to one of
$ rectangular coordinate system to the board. Denote the coordinates of the two knights as $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$, where $x_{k}, y_{k} \in\{1,2, \ldots, n\}, k=1,2$. The knights threaten each other if 1) $\left|x_{1}-x_{2}\right|=1, \qua... | 1680 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest natural number that has exactly 70 natural divisors (including 1 and the number itself).
$(16$ points $)$ | Solution: Let $n$ be the required natural number, $n=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{m}^{k_{m}}$ - the prime factorization of the number $n$. Any natural divisor of this number has the form $d=p_{1}^{h_{1}} \cdot p_{2}^{l_{2}} \cdot \ldots \cdot p_{m}^{l_{m}}$, where $l_{i} \in\left\{0,1, \ldots... | 25920 | Number Theory | 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 |
3. A student wrote a program for recoloring a pixel into one of 128 different colors. These colors he numbered with natural numbers from 1 to 128, and the primary colors received the following numbers: white color - number 1, red - 5, orange - 13, yellow - 21, green - 45, blue - 75, indigo - 87, violet - 91, black - 12... | Solution. The final pixel color number is equal to $f^{[2019]}(5)$ where $f^{[k]}(n)=\underbrace{f(f(f(\ldots(f}_{k \text { times }}(n) \ldots)-k$-fold composition of the function $f(n)$, which is equal to $n+4$ when $n \leq 19$, and equal to $|129-2 n|$ when $n \geq 20$. Let's compute and write down the first few valu... | 75 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) On a tourist base, the number of two-room cottages is twice the number of one-room cottages. The number of three-room cottages is a multiple of the number of one-room cottages. If the number of three-room cottages is tripled, it will be 25 more than the number of two-room cottages. How many cottages are ... | Solution. Let the number of one-bedroom houses be $x$, two-bedroom houses $2 \mathrm{x}$, and three-bedroom houses nx.
$3 n x-25=2 x ; x(3 n-2)=25 ; \Rightarrow 25:(3 n-2) \Rightarrow 3 n-2=1 ; 5 ; 25$. If
$3 n-2=1 ; n=1 ; x=25 ; 25+50+25=100$ total houses.
If $3 n-2=5 ; 3 n=7$, which is impossible. If $3 n-2=25 ; 3... | 100 | 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 |
1. According to the theorem inverse to Vieta's theorem, we form a quadratic equation. We get $x^{2}-\sqrt{2019} x+248.75=0$.
Next, solving it, we find the roots $a$ and $b$: $a=\frac{\sqrt{2019}}{2}+\frac{32}{2}$ and $b=\frac{\sqrt{2019}}{2}-\frac{32}{2}$, and consequently, the distance between the points $a$ and $b$:... | Answer: 32
| 15 points | The correct answer is obtained justifiably |
| :---: | :---: |
| 10 points | The quadratic equation is solved, but an arithmetic error is made or the distance between the points is not found |
| 5 points | The quadratic equation is correctly formulated according to the problem statement. |
| 0... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Initially, the farmer placed his produce in boxes with a capacity of 8 kg each, but one box was not fully loaded. Then the farmer repackaged all the produce into boxes with a capacity of 6 kg each, which required 8 more boxes, but in this case, one box was also not fully loaded. When the produce was finally placed i... | Solution. Let $x$ kg be the weight of the farmer's produce. Then $\quad 8(n-1)<x<8 n$, $6(n+7)<x<6(n+8), \quad 5(n+13)=x, \Rightarrow 8(n-1)<5(n+13)<8 n, \quad 6(n+7)<5(n+13)<6(n+8)$, $\Rightarrow 21 \frac{2}{3}<n<23, \quad n=22, \quad x=35 \cdot 5=175$.
Answer: 175. | 175 | Number Theory | 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 |
4. In the country of Landia, which breeds an elite breed of horses, an annual festival is held to test their speed, in which only one-year-old, two-year-old, three-year-old, and four-year-old horses can participate. For each horse that meets the speed standard, the festival organizers pay a fixed amount of money to the... | Solution. A four-year-old horse can earn a maximum of 4 landricks over its entire participation in festivals. If the horse starts participating in festivals at 1 year old, it can participate for another 3 years after that. In the case of annual victories, it will earn $1+2+3+4=10$ landricks over 4 years. If the horse s... | 200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The number $N$ is written as the product of consecutive natural numbers from 2019 to 4036: $N=2019 \cdot 2020 \cdot 2021 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036$. Determine the power of two in the prime factorization of the number $N$.
(points) | Solution. The number $N$ can be represented as
$$
\begin{aligned}
& N=\frac{(2 \cdot 2018)!}{2018!}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4035 \cdot 4036}{2018!}=\frac{(1 \cdot 3 \cdot \ldots \cdot 4035) \cdot(2 \cdot 4 \cdot \ldots \cdot 4034 \cdot 4036)}{2018!}= \\
& =\frac{(1 \cdot 3 \cdot \... | 2018 | Number Theory | 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 |
5. Determine the smallest natural number $N$, among the divisors of which are all numbers of the form $x+y$, where $x$ and $y$ are natural solutions to the equation $6 x y-y^{2}-5 x^{2}=7$. points) | Solution. Transform the equation by factoring the right-hand side
$6 x y-y^{2}-5 x^{2}-x^{2}+x^{2}=7 \Rightarrow 6 x(y-x)-(y+x)(y-x)=7 \Rightarrow(y-x)(6 x-y-x)=7 \Rightarrow$ $(y-x)(5 x-y)=7$.
Considering that the variables are natural numbers, and 7 is a prime number, we get
$$
\left\{\begin{array} { l }
{ y - x ... | 55 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ of a circular track, a car and a motorcycle started simultaneously and in the same direction. The car drove two laps without stopping in one direction. At the moment when the car caught up with the motorcyclist, the motorcyclist turned around and increased his speed by 16 km/h, and after $3 / 8$ hours... | Solution. Let $x$ (km/h) be the speed of the motorcyclist, $x$ (km/h) be the speed of the car, $S$ (km) be the distance the motorcyclist travels before turning around, then the total length of the track is $2 S+5.25$. We have $\frac{S}{x}=\frac{3 S+5.25}{y}, \frac{3 x}{8}+6=S, \frac{3 y}{8}=S+5.25$. This leads to the q... | 21 | 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 |
5. Find the sum of all numbers of the form $x+y$, where $x$ and $y$ are natural number solutions to the equation $5 x+17 y=307$.
points) | Solution. We solve the auxiliary equation $5 x+17 y=1$. For example, its solutions can be 7 and 2. Multiplying them by 307 and considering linear combinations for integer $t$, we get values in natural numbers
$\left\{\begin{array}{l}x=7 \cdot 307-17 t, \\ y=-2 \cdot 307+5 t,\end{array} t \in Z, x>0, y>0 \Rightarrow t ... | 164 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. In 100 containers of three types with capacities of $1 \pi$, 10 liters, and 50 liters, 500 liters of oil have been distributed. How many containers of each type were used, if the amount of oil in each container corresponds to its capacity?
(12 points) | Solution. Let $n, m, k$ be the number of containers with capacities of 1 liter, 10 liters, and 50 liters, respectively. Then $\left\{\begin{array}{c}n+10 m+50 k=500, \\ n+m+k=100 .\end{array}\right.$ Since $n=500-10 m-50 k=10(50-m-5 k)$, $n$ is divisible by 10, i.e., $n=10 l, l \in N, n \leq 100, l \leq 10 .\left\{\beg... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Find all natural numbers $n \geq 2$, for which the equality $4 x_{n}+2 y_{n}=55 n^{2}+61 n-116$ 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
$$
\begin{aligned}
& x_{n}=1 \cdot 2+2 \cdot 3+\cdots+(n-1) \cdot n=(2+3+\cdots n)+(3+4+\cdots n)+\cdots((n-1)+n)+n= \\
& =\left(z_{n}-z_{1}\right)+\left(z_{n}-z_{2}\right)+\cdots+\left(z_{n}-z_{n-1}\right)=(n-1) z_{n}-\left(z_{1}+z_{2}+\c... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $ABC$, lines parallel to the sides of the triangle are drawn through an arbitrary point $O$. As a result, triangle $ABC$ is divided into three parallelograms and three triangles. The areas of the resulting triangles are $6 \text{ cm}^2$, $24 \text{ cm}^2$, and $54 \text{ cm}^2$. Find the area of triangle... | # Solution:
The triangles are similar. Let the first and third be similar with a similarity coefficient $k_{1}$, and the second and third with a similarity coefficient $k_{2}$. Then for the areas of these triangles, we have the ratios $\frac{S_{1}}{S_{3}}=k_{1}^{2}, \frac{S_{2}}{S_{3}}=k_{2}^{2}$.
^{2}}$ | Solution. Transform the equation $10 x-6+x^{2}=\sqrt{2 x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}}$
$10 x-6+x^{2}=\sqrt{x^{2}+x^{2}-2 x+1} \cdot \sqrt{36+(x+4)^{2}} \Rightarrow 6 x-6+x^{2}+4 x=\sqrt{x^{2}+(x-1)^{2}} \cdot \sqrt{6^{2}+(x+4)^{2}}$,
This equation can be interpreted as the scalar product of vectors $\vec{a}=(... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (15 points) Two medians of a triangle, equal to 18 and 24, are perpendicular to each other. Find the length of the third median of this triangle. | Solution. Let $\mathrm{AP}=18$ and $\mathrm{BH}=24$. We will find the length of the median $\mathrm{CM}$. By the property of medians in $\triangle A B C$ we have:
$A O: O P=B O: O H=2: 1 \Rightarrow A O=$
$=\frac{2}{3} A P=\frac{2}{3} \cdot 18=12 ; B O=\frac{2}{3} B H=\frac{2}{3} \cdot 24=16$.
Then, by the Pythagore... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (20 points) Calculate the value of the expression:
$$
1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots+2019 \cdot 2020 \cdot(2019+2020)
$$ | Solution. We will prove by mathematical induction for natural n that:
$$
\begin{gathered}
-0 \cdot 1 \cdot(0+1)+1 \cdot 2 \cdot(1+2)-2 \cdot 3 \cdot(2+3)+3 \cdot 4 \cdot(3+4)-\cdots- \\
-(2 n-2) \cdot(2 n-1) \cdot(4 n-3)+(2 n-1) \cdot 2 n \cdot(4 n-1)= \\
=(2 n-1) \cdot 2 n \cdot(2 n+1) .
\end{gathered}
$$
Base: \( n... | 8242405980 | 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 |
# Variant №1
№1. The number of male students in the university is $35 \%$ more than the number of female students. All students are distributed between two buildings, with $\frac{2}{5}$ of all male students in the first building and $\frac{4}{7}$ of all female students in the second building. How many students are the... | # Solution:
Let $n$ be the number of girls studying at the university, then the number of boys is $1.35 n = \frac{27}{20} n$, from which we can conclude that $n$ is a multiple of 20, i.e., $n = 20 m$.
Thus, the number of girls is $20 m$, and the number of boys is $27 m$.
In the first building, $\frac{2}{5}$ of all b... | 4935 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given the vertices of a regular 100-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{100}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points) | Solution. Let the vertices be numbered clockwise.
Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{100}(100-(m-k))>90^{\circ}, 0<m-k<50$.
The difference $m-k$ is taken modulo 100
(for example, $15-\left.70... | 117600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Given the vertices of a regular 120-gon $A_{1}, A_{2}, A_{3}, \ldots, A_{120}$. In how many ways can three vertices be chosen from them to form an obtuse triangle? (10 points) | Solution. Let the vertices be numbered clockwise.
Denote the selected vertices clockwise as $K, L, M$, where angle $K L M$ is obtuse. If $K=A_{k}, L=A_{l}, M=A_{m}$, then $\alpha=\angle K L M=\frac{180^{\circ}}{120}(120-(m-k))>90^{\circ}, 0<m-k<60$.
The difference $m-k$ is taken modulo 120
(for example, $15-\left.90... | 205320 | Combinatorics | 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 |
1. There are two types of containers: 27 kg and 65 kg. How many containers of the first and second type were there in total, if the cargo in the first type of containers exceeds the cargo in the second type of containers by 34 kg, and the number of 65 kg containers does not exceed 44 units? | Solution. Let $x$ be the number of containers weighing 27 kg, and $y$ be the number of containers weighing 65 kg. We obtain the equation $27 x - 65 y = 34$.
$27(x - 2y) - 11y = 34$, let $x - 2y = k$.
$27k - 11y = 34$,
$11(2k - y) + 5k = 34$, let $2k - y = t$.
$11t + 5k = 34$,
$5(2t + k) + t = 34$, let $2t + k = n$... | 66 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$ with angle $\angle B=120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn. Find the degree measure of angle $C_{1} B_{1} A_{1}$. | # Solution.
Extend side $A B$ beyond point $B$, then $B C$ is the bisector of angle $\angle B B_{1} B K$, which means point $A_{1}$ is equidistant from sides $B_{1} B$ and $B K$. Considerin... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Given a rectangular trapezoid $\mathrm{ABCE}$, the bases of which $\mathrm{BC}$ and $\mathrm{AE}$ are equal to 5 and 7, respectively. The smaller lateral side $\mathrm{AB}$ is equal to $\mathrm{BC}$. On $\mathrm{AB}$, a point $\mathrm{F}$ is marked such that $\mathrm{BF}: \mathrm{FA}=2: 3$, on $\mathrm{AC}$, a point... | Solution. Construct perpendiculars GI and GH.
1) $\Delta G I D = \Delta G F H$ - by two legs, since $G I = G H = 4; F H = I D = 1$, therefore $F G = G D$, $\angle F G H = \angle D G I = \alp... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. A chemistry student conducted an experiment: from a tank filled with syrup solution, he poured out several liters of liquid, refilled the tank with water, then poured out twice as much liquid and refilled the tank with water again. As a result, the amount of syrup in the tank decreased by $\frac{8}{3}$ times. Determ... | # Solution.
1) Let the syrup content in the original solution be $p \%$ and let $\mathcal{X}$ liters of the solution be poured out the first time.
2) Then after pouring out the liquid, there are $(1000-x)$ liters of solution left, and in it $(1000-x) \cdot \frac{p}{100}$ liters of syrup and $(1000-x) \cdot \frac{100-p... | 250 | 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 |
5. Elvira takes a triplet of numbers and transforms it according to the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and the smallest numbers in the triplet after 2019 applications of this rule, if the initial triplet was $\{100 ; 89 ; 60\}$? If the... | Solution. Let's denote the 3 numbers as $\{x ; x+a ; x+b\}$, where $0<a<b$. Then the difference between the largest and the smallest number at any step, starting from the zeroth step, will be an invariant, that is, unchanged and equal to $b$.
$b=100-60=40$.
Answer: 40. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B A C=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=30^{\circ}$. On the side $\mathrm{BC}$, a point М is chosen such t... | Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.
Then: 1) $\triangle T P M = \Delta K O... | 60 | Geometry | 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 |
2. Inside the square $A B C D$ with a side length of 5, there is a point $X$. The areas of triangles $A X B, B X C$, and $C X D$ are in the ratio $1: 5: 9$. Find the sum of the squares of the distances from point $X$ to the sides of the square. | Solution:
Let the side of the square be $a=5$, and the distances from point $X$ to sides $AB$, $BC$, $CD$, and $DA$ be $h_{1}$, $h_{2}$, $h_{3}$, and $h_{4}$, respectively. Since the area of a triangle is $S=\frac{1}{2} a h$, we conclude that $h_{1}: h_{2}: h_{3}=1: 5: 9$ or $h_{1}=x$, $h_{2}=5 x, h_{3}=9 x$. However,... | 33 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. The boat traveled 165 km upstream against the current and then returned. On the return trip, it took 4 hours less than the trip there. Find the boat's own speed if the river's current speed is 2 km/h. | # Solution:
$$
\begin{aligned}
& v_{\text {down}}=v_{\mathrm{c}}+v_{\text {current}} \\
& v_{\text {up}}=v_{\mathrm{c}}-v_{\text {current}}
\end{aligned}=>v_{\text {down}}-v_{\text {up}}=2 v_{\text {current}}=2 \cdot 2=4
$$
Then $\quad v_{\text {up}}=x$ km/h, $v_{\text {down}}=x+4$ km/h
Preliminary (online) stage of... | 13 | 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 |
9.3. How many solutions in integers $x, y$ does the inequality $|y-x|+|3 x-2 y| \leq a$ have a) for $a=2$; b) for $a=20$? | Answer: a) 13; b) 841. Solution. Let $m=y-x, n=3 x-2 y$, then (expressing $x$, y from these equations) we get $x=2 m+n$ and $y=3 m+n$ and thus, for any integers m, n there correspond integers $x, y$ (and vice versa), i.e., there is a one-to-one correspondence between ordered pairs (m.n) and ( $x, y$ ). Therefore, we ne... | 13 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
7.2. In 7a grade, there are 33 students. At the beginning of the school year, two clubs were organized in the class. According to school rules, a club can be organized if at least $70 \%$ of all students in the class sign up for it. What is the smallest number of students who could have signed up for both clubs simulta... | Answer: 15 students. Solution. In each club, there should be no less than $33 \cdot 0.7=23.1$ people, which means no less than 24 people. Let $n_{1}, n_{2}$ be the number of students who signed up for the first and second club, respectively, and $n$ be the number of students who signed up for at least one club. Obvious... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.3. In a three-digit number, the first digit was crossed out to obtain a two-digit number. If the original number is divided by the obtained number, the quotient is 8 and the remainder is 6. Find the original number. | Answer: 342. Solution. Let $a$ be the first digit of the original number, and $b$ be the two-digit number formed by the last two digits. According to the condition, $100 a+b=8 b+6 \Leftrightarrow 7 b=2(50 a-3)=2 \cdot 49 a+2(a-3)$. Thus, the number $a-3$ must be divisible by 7. Considering that $0<a \leq 9$, we get tha... | 342 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.5. There are 200 matches. How many ways are there to form, using all the matches, a square and (separately) an equilateral triangle? (Different ways differ in the sizes of the square and the triangle). | Answer: 16. Solution. Let $x$ (matches) be the side length of the square, and $y$ (matches) be the side length of the triangle. Then $4 x+3 y=200 \Leftrightarrow 3 y=4(50-x)$. Thus, for divisibility by 3, we need to consider all natural numbers up to 50 as $x$ that give a remainder of 2 when divided by 3 (i.e., the sam... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.3. How many points on the hyperbola $y=\frac{2013}{x}$ have integer coordinates $(x ; y)$? | Answer: 16. Solution. Integer points in the first quadrant correspond to the natural divisors of the number $2013=3 \cdot 11 \cdot 61$. The number of such divisors is 8 (they can be listed directly or the formula $\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \ldots\left(\alpha_{k}+1\right)$ for the number of natu... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.2. a) Given the quadratic equation $x^{2}-9 x-10=0$. Let $a$ be its smallest root. Find $a^{4}-909 a$. b) For the quadratic equation $x^{2}-9 x+10=0$, where $b$ is the smallest root, find $b^{4}-549 b$. | Answer: a) 910; b) -710. Solution. a) Let's solve the problem in general. Let $x^{2}-c x+d=0-$ be a quadratic equation, and $a-$ be its root. Then $a^{4}=(c a-d)^{2}=c^{2} a^{2}-2 a c d+d^{2}=$ $c^{2}(c a-d)-2 a c d+d^{2}=a\left(c^{3}-2 c d\right)+d^{2}-c^{2} d$. Therefore, $a^{4}-\left(c^{3}-2 c d\right) a=d^{2}-c^{2}... | 910 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.3. There are 12 matches, each 2 cm long. Can a polygon with an area of 16 cm $^{2}$ be formed from them? (The matches cannot be broken, and all matches must be used.) | Answer: It is possible. Solution. The result follows from the Pythagorean theorem (since $10^{2}=6^{2}+8^{2}$) and construction (see figure). The area of the polygon is $\frac{6 \cdot 8}{2}-8=16$.
 $\cdot \frac{1}{20}$ (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 |
9.2. In triangle $ABC$, side $BC$ is equal to segment $AM$, where $M$ is the point of intersection of the medians. Find the angle $\angle BMC$. | Answer: $90^{\circ}$. Solution. Let $A N-$ be the median. By the property of the intersection point of medians, $M N=\frac{1}{2} A M=\frac{1}{2} B C$. Thus, in triangle $B M C$, the median $M N$ is equal to half of side $B C$. Therefore, $\triangle B M N$ is a right triangle with a right angle at $B M N$ (this is a kno... | 90 | 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 |
9.2. Is the number
$$
\sqrt[3]{2016^{2}+2016 \cdot 2017+2017^{2}+2016^{3}}
$$
- rational or irrational? | Answer: the whole rational number 2017. Solution. Let $a=2017, b=2016$. Then $b^{2}+a b+a^{2}+b^{3}=\frac{a^{3}-b^{3}}{a-b}+b^{3}=a^{3}-b^{3}+b^{3}=a^{3}$ (since $a-b=1$ ). Thus, we get $\sqrt[3]{a^{3}}=a=2017$ | 2017 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.3. How many five-digit natural numbers exist, each of which has adjacent digits with different parity | Answer: 5625. Solution. If the first digit is even, then it can be chosen in four ways $(2,4,6,8)$. And all subsequent ones can be chosen in five ways $(1,3,5,7,9$ - for the second and fourth digits and $0,2,4,6,8$ - for the third and fifth). In the end (by the rule of product), we have a total of $4 \cdot 5^{4}=2500$ ... | 5625 | Combinatorics | 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 |
8.1. Find the value of the expression $\frac{\left(a^{2}+b^{2}\right)^{2}-c^{2}-4 a^{2} b^{2}}{a^{2}+c-b^{2}}$ for $a=2017, b=2016, c=2015$. Justify your result. | Answer: 2018. Solution. The numerator equals $a^{4}+2 a^{2} b^{2}+b^{4}-4 a^{2} b^{2}-c^{2}=\left(a^{2}-b^{2}\right)^{2}-c^{2}=$ $\left(a^{2}-b^{2}+c\right)\left(a^{2}-b^{2}-c\right)$, and after dividing by the denominator we get $a^{2}-b^{2}-c=$ $(a-b)(a+b)-c=2017+2016-2015=2018$. | 2018 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.4. Find the natural number $x$ that satisfies the equation
$$
x^{3}=2011^{2}+2011 \cdot 2012+2012^{2}+2011^{3}
$$ | Answer: 2012. Solution: Let $a=2011, \quad b=2012$. Then $a^{2}+a b+b^{2}=\frac{b^{3}-a^{3}}{b-a}=b^{3}-a^{3}$ (since $b-a=1$). Therefore, $x^{3}=b^{3}-a^{3}+a^{3}=b^{3}$. Hence, $x=b=2012$. | 2012 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.1. In a five-digit number, one of the digits was crossed out, and this four-digit number was subtracted from the original number. As a result, the number 54321 was obtained. Find the original number. | Answer: 60356. Solution. Let $x$ be the four-digit number obtained after crossing out a digit. Note that the crossed-out digit was the last digit of the five-digit number, because otherwise the last digit after subtraction would be zero. Let this digit be $y$. We have the equation $10 x+y-x=54321 \Leftrightarrow 9 x+y=... | 60356 | Number Theory | 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 |
7.2. Is it possible to place the 12 numbers $1,2, \ldots, 12$ on the edges of a cube such that the product of the four numbers on the top face equals the product of the four numbers on the bottom face? | Answer: It is possible. Solution. An example of such an arrangement can be as follows: on the top face, place the numbers $2,4,9,10$; on the bottom face, place the numbers $3,5,6,8$; the remaining numbers $1,7,11,12$ are placed on the side edges. The products on the top and bottom faces are the same and equal to 720. (... | 720 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.4. There are ten coins of different weights and a balance scale without weights. It is required to select the heaviest and the lightest coin. Can this be achieved in 13 weighings? | Answer: It is possible. Solution. First, divide all the coins into 5 pairs and in 5 weighings compare the weight of each pair. By selecting the heavier coin in each pair, we form a "heavy" group of 5 coins. The remaining 5 coins will form the "light" group. Now, in the heavy group, over 4 sequential weighings, select t... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. In a five-digit number, one digit was crossed out, and the resulting four-digit number was added to the original. The sum turned out to be 54321. Find the original number. | Answer: 49383. Solution. Note that the crossed-out digit must be the last digit of the number $N$, because otherwise the sum of the two numbers would have an even digit in the last place. Let's denote this crossed-out digit by $x$ and let $y$ be the four-digit number obtained after the crossing out. Then the condition ... | 49383 | Number Theory | 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 |
10.2. The number $a$ is a root of the quadratic equation $x^{2}-x-50=0$. Find the value of $a^{4}-101 a$. | Answer: 2550. Solution. We have $a^{2}=a+50$, therefore $a^{4}=(a+50)^{2}=a^{2}+100 a+2500=$ $a+50+100 a+2500=101 a+2550$. Hence $a^{4}-101 a=2550$. | 2550 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2. In triangle $A B C$, side $B C$ is equal to segment $A M$, where $M$ is the point of intersection of the medians. Find the angle $\angle B M C$. | Answer: $90^{\circ}$. Solution. Let $A N-$ be the median. By the property of the intersection point of medians, $M N=\frac{1}{2} A M=\frac{1}{2} B C$. Thus, in triangle $B M C$, the median $M N$ is equal to half of side $B C$. Therefore, $\triangle B M N$ is a right triangle with a right angle at $B M N$ (this is a kno... | 90 | Geometry | 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 |
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