problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | 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 |
|---|---|---|---|---|---|---|---|---|
1. From point A to point B, which are 12 km apart, a pedestrian and a bus set out simultaneously. Arriving at point B in less than one hour, the bus, without stopping, turned around and started moving back towards point A at a speed twice its initial speed. After 12 minutes from its departure from point B, the bus met ... | Solution. Let $x$ be the pedestrian's speed (in km/h), $y$ be the car's speed (in km/h) on the way from $A$ to $B$, $2y$ be the car's speed on the way from $B$ to $A$, and $t$ be the time (in hours) the car spends traveling from $A$ to $B$.
$\left\{\begin{aligned} y t & =12, \\ x(t+0.2) & =12-0.4 y, \\ t & <1,\end{ali... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. How many roots does the equation
$\sqrt[3]{|x|}+10[x]=10 x$ ? ( $[x]$ - the integer part of the number $x$, i.e., $[x] \in Z,[x] \leq x<[x]+1$).
(5 points) | # Solution:
$\frac{\sqrt[3]{|x|}}{10}=\{x\},\{x\}=x-[x]$ Since $\{x\} \in[0 ; 1)$, then $x \in(-1000 ; 1000)$.
On the interval $[0 ; 1)$, the equation has 2 roots
$\sqrt[3]{x}=10 x, x=1000... | 2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can seven different items (3 weighing 2 tons each, 4 weighing 1 ton each) be loaded into two trucks with capacities of 6 tons and 5 tons, if the arrangement of the items inside the trucks does not matter? (12 points)
# | # Solution.
Solution. The load can be distributed as $6+4$ or $5+5$.
| 6 tons | 4 tons | number of ways | 5 tons | 5 tons | number of ways |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $2+1+1+1+1$ | $2+2$ | $C_{3}^{1}=3$ | $2+1+1+1$ | $2+2+1$ | $C_{3}^{1} \cdot C_{4}^{3}=12$ |
| $2+2+1+1$ | $2+1+1$ | $C_{3}^{... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In triangle $A B C$, with an area of $180 \sqrt{3}$, the angle bisector $A D$ and the altitude $A H$ are drawn. A circle with radius $\frac{105 \sqrt{3}}{4}$ and center lying on line $B C$ passes through points $A$ and $D$. Find the radius of the circumcircle of triangle $A B C$, if $B H^{2}-H C^{2}=768$.
# | # Solution.
Let $B C=a, A C=b, A B=c, \angle B A D=\angle C A D=\alpha, \angle A D C=\beta$, and the radius of the given circle is $r$. Note that $\beta>\alpha$ (as the external angle of tri... | 28 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Svetlana 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 on the 1580th step of applying this rule, if the initial triplet of numbers was $\{80 ; 71... | # 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 $b . B=80-20=60$.
Answer: 60 . | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In triangle $A B C$, side $B C$ is 19 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 $10 \, \text{cm}$. | # Solution:
Triangle $ABF (BF = AF)$ is isosceles, since $DF \perp AB$, and $D$ is the midpoint of $AB$. $P_{AFC} = AF + FC + AC = BF + FC + AC = BC + AC = 29 \text{ cm}$.
Answer: 29 cm. | 29 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Sasha bought pencils in the store for 13 rubles each and pens for 20 rubles each, in total he paid 350 rubles. How many items of pencils and pens did Sasha buy in total?
# | # Solution.
Let $x$ be the number of pencils, $y$ be the number of pens. We get the equation $13 x+20 y=355$
$13(x+y)+7 y=355$, let $x+y=t(1)$
$13 t+7 y=355$
$7(t+y)+6 t=355$, let $t+y=k(2)$
$7 k+6 t=355$
$6(k+t)+k=355$, let $k+t=n(3)$
$6 n+k=355$
$k=355-6 n$. Substitute into (3), $t=7 n-355$
Substitute into (... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. A boy wrote the first twenty natural numbers on a piece of paper. He did not like how one of them was written and crossed it out. It turned out that among the 19 remaining numbers, there is a number equal to the arithmetic mean of these 19 numbers. Which number did he cross out? If the problem has more than one solu... | # Solution:
The sum of the numbers on the sheet, initially equal to $1+2+3+\ldots+20=210$ and reduced by the crossed-out number, is within the range from 210-20=190 to 210-1=209. Moreover, it is a multiple of 19, as it is 19 times one of the addends. Since among the numbers $190,191,192, \ldots 209$ only 190 and 209 a... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A family of beekeepers brought containers of honey to the fair with volumes of $13,15,16,17,19,21$ liters. In August, three containers were sold in full, and in September, two more, and it turned out that in August they sold twice as much honey as in September. Determine which containers were emptied in August. In y... | # Solution:
A total of $13+15+16+17+19+21=101$ liters of honey were brought. The amount of honey sold is divisible by three. Therefore, the volume of the unsold container must give a remainder of 2 when divided by 3 (the same as 101), i.e., 17 liters. Thus, 101-17=84 liters were sold, with one-third of 84 liters, or 2... | 21 | 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. Find $\angle C_{1} B_{1} A_{1}$. | # Solution:
Extend side $A B$ beyond point $\mathrm{B}$, 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 $\angle B A C$, which means it is equidistant from its sides.
We get that $A... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. If a two-digit number is decreased by 36, the result is a two-digit number with the same digits but in reverse order. In the answer, specify the arithmetic mean of the resulting number sequence.
# | # 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 $10 x+y=10 y+x+36$
From the equation, it is clear that the two-digit number is greater than 36. Let's start the investigation with the ten... | 73 | Number Theory | 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 $\mathrm{O}$ is the center of the inscribed circle of triangle $A B C$. Find the angle between the lines $O_{1} O_{2}$ an... | # 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 |
3. If $a-$ is the first term and $d-$ is the common difference of an arithmetic progression, $\left\{\begin{array}{l}a+16 d=52, \\ a+29 d=13\end{array} \Leftrightarrow d=-3, a=100\right.$.
The sum of the first $n$ terms of an arithmetic progression $S_{n}$ reaches its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. ... | Answer: 1717
Consider the equation of the system $\sqrt{2} \cos \frac{\pi y}{8}=\sqrt{1+2 \cos ^{2} \frac{\pi y}{8} \cos \frac{\pi x}{4}-\cos \frac{\pi x}{4}}$. Given the condition $\sqrt{... | 1717 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 24$, passing through the point $M(4 ;-2 \sqrt{3})$.
# | # Solution:
$$
y=\frac{x^{2}}{8 \sqrt{3}}, M(4 ;-2 \sqrt{3}) \cdot y=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(x-x_{0}\right) ;-2 \sqrt{3}=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(4-x_{0}\right) ;
$$
$x_{0}^{2}-8 x_{0}-48=0 ; x_{0}=4 \pm 8 ;\left(x_{0}\right)_{1}=12,\left(x_{0}\righ... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 6$, passing through the point $M(1 ;-\sqrt{3} / 2)$. | Solution:
$$
y=\frac{x^{2}}{2 \sqrt{3}}, M\left(1 ;-\frac{\sqrt{3}}{2}\right) \cdot y=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(x-x_{0}\right) ;-\frac{\sqrt{3}}{2}=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(1-x_{0}\right) ; x_{0}^{2}-2 x_{0}-3=0
$$
$; x_{0}=1 \pm 2 ;\left(x_{0}\right)_{1}... | 90 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
# 5. Solution:
Let $x$ units of distance/hour be the speed of the bus, $y$ units of distance/hour be the speed of the tractor, and $S$ be the length of the path AB. Then the speed of the truck is $-2y$ units of distance/hour. We can set up a system of equations and inequalities:
$$
\left\{\begin{array}{c}
\frac{s}{x}... | Answer: 17 hours 45 minutes. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 2. (Option 2). Given an isosceles triangle $ABC (AB=BC)$ on the lateral side $BC$, points $M$ and $N$ are marked (M lies between B and $N$) such that $AN=MN$ and $\angle BAM = \angle NAC$. $MF$ is the distance from point M to the line $AC$. Find $\angle AMF$. | Solution: Let $\angle$ BAM
$=\angle \mathrm{NAC}=\alpha, \angle \mathrm{MAN}=\angle \mathrm{AMN}=\beta \prec=\angle \mathrm{MAC}=\alpha+\beta$ and $\angle \mathrm{MCA}=2 \alpha+\beta=\succ(\square \mathrm{AMC})$
$2 \beta+\alpha+2 \alpha+\beta=180^{\circ}=\succ \alpha+\beta=60^{\circ} \Rightarrow \succ \mathrm{MAF}=60... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Task 3. (Option 2)
Thirty clever students from 6a, 7a, 8a, 9a, and 10a grades were tasked with creating forty problems for the olympiad. Any two classmates came up with the same number of problems, while any two students from different grades came up with a different number of problems. How many people came up with ju... | Solution: 26 classmates solved 1 problem, the 27th person solved 2, the 28th solved 3, the 29th solved 4, and the 30th solved 5. This solution is immediately apparent. Let's prove that it cannot be otherwise.
Let $x$ be the number of people who solved one problem, $y$ be the number who solved two, $z$ be the number wh... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Two cyclists set off simultaneously from point $A$ to point $B$. When the first cyclist had covered half the distance, the second cyclist had 24 km left to travel, and when the second cyclist had covered half the distance, the first cyclist had 15 km left to travel. Find the distance between points $A$ and $B$. | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the cyclists. Then $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}}$ and $\frac{s-15}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24$; $s^{2}-52 s+480=0 ;... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the greatest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=327$ and the sum $S_{57}=57$? | # Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$\left\{\begin{array}{l}\frac{a+a+2 d}{2} \cdot 3=327, \\ \frac{a+a+56 d}{2} \cdot 57=57\end{array} \Leftrightarrow\left\{\begin{array}{l}a+d=109, \\ a+28 d=1\end{array} \Rightarrow 27 d=-108 ; d=-4, a=113\right.\ri... | 1653 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A pedestrian left point $A$ for point $B$. When he had walked 8 km, a second pedestrian set off from point $A$ after him. When the second pedestrian had walked 15 km, the first was halfway through his journey, and both pedestrians arrived at point $B$ at the same time. What is the distance between points $A$ and $B$... | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the pedestrians. Then,
$\frac{s-8}{v_{1}}=\frac{s}{v_{2}}$ and $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$. From this, $\frac{s}{2(s-8)}=\frac{s-15}{s} ; s^{2}=2 s^{2}-46 s+240 ; s^{2}-46 s+240=0$;
$s_{1,2}=23 \pm 17 \cdot... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the smallest value that the sum $S_{n}$ of the first $n$ terms of an arithmetic progression can take, given that the sum $S_{3}=-141$ and the sum $S_{35}=35$? | # Solution:
If $a-$ is the first term and $d-$ is the common difference of the arithmetic progression,
$$
\left\{\begin{array} { l }
{ \frac { a + a + 2 d } { 2 } \cdot 3 = - 141 , } \\
{ \frac { a + a + 34 d } { 2 } \cdot 35 = 35 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
a+d=-47, \\
a+17 d=1
\end{array}... | -442 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. (Option 2)
Calculate $x^{3}-3 x$, where $x=\sqrt[3]{7+4 \sqrt{3}}+\frac{1}{\sqrt[3]{7+4 \sqrt{3}}}$. | Solution: Let $\sqrt[3]{7+4 \sqrt{3}}=a$, then $x=a+\frac{1}{a}$,
$$
\begin{aligned}
& x^{3}-3 x=\left(a+\frac{1}{a}\right)^{3}-3\left(a+\frac{1}{a}\right)=a^{3}+\frac{1}{a^{3}}=7+4 \sqrt{3}+\frac{1}{7+4 \sqrt{3}}=\frac{(7+4 \sqrt{3})^{2}+1}{7+4 \sqrt{3}}= \\
& =\frac{98+56 \sqrt{3}}{7+4 \sqrt{3}}=14
\end{aligned}
$$
... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. (Option 1)
Find the sum of the squares of the roots of the equation $\left(x^{2}+4 x\right)^{2}-2016\left(x^{2}+4 x\right)+2017=0$. | Solution: Let's make the substitution: $x^{2}+4 x+4=t$, then $x^{2}+4 x=t-4$ and the equation will take the form:
$(t-4)^{2}-2016(t-4)+2017=0$
$t^{2}-2024 t+10097=0$
The discriminant of the equation is greater than zero, therefore, the equation has two roots. By Vieta's theorem: $t_{1}+t_{2}=2024, t_{1} \cdot t_{2}=... | 4064 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. (Option 2)
Find the sum of the squares of the roots of the equation $\left(x^{2}+6 x\right)^{2}-1580\left(x^{2}+6 x\right)+1581=0$. | Solution: Let's make the substitution: $x^{2}+6 x+9=t$, then $x^{2}+6 x=t-9$ and the equation will take the form:
$$
\begin{aligned}
& (t-9)^{2}-1580(t-9)+1581=0 \\
& t^{2}-1598 t+15882=0
\end{aligned}
$$
The discriminant of the equation is greater than zero, so the equation has two roots. By Vieta's theorem, $t_{1}+... | 3232 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Seeing a fox several meters away, the dog chased after it along a straight dirt road. The dog's jump is $23 \%$ longer than the fox's jump. There is a time interval during which both the fox and the dog make a whole number of jumps. Each time it turns out that the dog manages to make $t \%$ fewer jumps than the fox,... | Solution: Let $\mathrm{x}$ be the length of the fox's jump, and $y$ be the number of jumps it makes in some unit of time. Then $xy$ is the distance the fox covers in this time. The distance covered by the dog in the same time is $1.23 x\left(1-\frac{t}{100}\right) y$. The fox will escape from the dog if $1.23 x\left(1-... | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the arithmetic progression $\left(a_{n}\right) a_{1000}=150, d=0.5$.
Calculate: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cdot a_{1581}}+\frac{1}{a_{1581} \cdot a_{1582}}+\ldots+\frac{1}{a_{2019} \cdot a_{2020}}\right)$. | Solution: The expression in parentheses consists of several terms of the form $\frac{1}{x \cdot(x+d)}$, which can be decomposed into the sum of simpler fractions: $\frac{1}{x \cdot(x+d)}=\frac{1}{d}\left(\frac{1}{x}-\frac{1}{x+d}\right)$. Transform the original expression: $99 \cdot 100 \cdot\left(\frac{1}{a_{1580} \cd... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In $\triangle A B C$ with $\angle B=120^{0}$, the angle bisectors $A A_{1}, B B_{1}, C C_{1}$ are drawn.
## Find $\angle C_{1} B_{1} A_{1}$. | # Solution.
Extend side $A B$ beyond point $\mathrm{B}$, 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 is therefore equidistant from its sides, we get that $A_{... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. What is the minimum value that the function $F(x ; y)=x^{2}+8 y+y^{2}+14 x-6$ 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 $F(x ; y)=$ M, then $(x+7)^{2}+(y+4)^{2}=(M+71)$ - this is a circle with center $(-7 ;-4)$ and radius $(M+71)^{0.5}$.
Since the center of the second circle lies outside the first, ... | 29 | 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 35 cells (a $35 \times 35$ square, with a total of 1225 cells), so that among any four of its cells forming a corner (an "L" shape), there is at least one painted cell.
# | # Solution.
You should color every 3rd cell diagonally (see the figure). Thus, $\left[\frac{N^{2}}{3}\right]$ cells will be colored. This is the minimum possible number, as within any $3 \times 3$ square, at least three cells need to be colored. $\left[\frac{35^{2}}{3}\right]=408$.
. Find the minimum possible length of the text.
# | # Solution:
Let the length of the text be L. Let a character appear in the text $x$ times. The problem can be reformulated as: find the smallest natural number $\mathrm{L}$, for which there exists a natural number $x$ such that $\frac{10.5}{100}19$ when $x \geq 3$.
Answer: 19.
## Solution variant № 2 | 19 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. In the arithmetic progression $\left(a_{n}\right)$, $a_{1}=1$, $d=3$.
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:
$$
\begin{aligned}
& A=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots+\frac{1}{\sqrt{a_{1579}}+\sqrt{a_{1580}}}= \\
& =\frac{\sqrt{a_{2}}-\... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In rectangle $A B C D$, $A B: A D=1: 2$. Point $M$ is the midpoint of $A B$, and point $K$ lies on $A D$ and divides it in the ratio $3: 1$ starting from point $A$. Find the sum of $\angle C A D$ and $\angle A K M$.
# | # Solution:
Complete the rectangle $A B C D$ to a square $A E F D$ with side $A D$.
$$
\begin{aligned}
& \text { Let } L \in E F, E L: L F=1: 3, \\
& \triangle M E L=\triangle A K M \Rightarrow \angle E M L=\angle A K M \\
& N \in F D, F N=N C, M R\|A D \Rightarrow M N\| A C \Rightarrow \\
& \Rightarrow \angle N M R=... | 45 | Geometry | 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 35 cells (a total of $35 \times 35$ cells, which is 1225 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 shaded in a checkerboard pattern. Thus, $\left[\frac{N^{2}}{2}\right]$ cells will be shaded. 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 $4 \times 4$) such that the... | 612 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. A table consisting of 2019 rows and 2019 columns is filled with natural numbers from 1 to 2019 such that each row contains all numbers from 1 to 2019. Find the sum of the numbers on the diagonal that connects the top left and bottom right corners of the table, if the filling of the table is symmetric with respect to... | # Solution:
We will show that all numbers from 1 to 2019 are present on the diagonal. Suppose the number $a \in\{1,2,3 \ldots, 2019\}$ 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... | 2039190 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+2017^{2}$. | Solution: $1^{2}-2^{2}+3^{2}-4^{2}+\ldots+2017^{2}=$
$$
\begin{aligned}
& =\left(1^{2}-2^{2}\right)+\left(3^{2}-4^{2}\right)+\ldots+\left(2015^{2}-2016^{2}\right)+2017^{2}= \\
& =(1-2)(1+2)+(3-4)(3+4)+\ldots+(2015-2016)(2015+2016)+2017^{2}= \\
& =-(1+2)-(3+4)-\ldots-(2015+2016)+2017^{2}= \\
& =-(1+2+3+4+\ldots+2015+20... | 2035153 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The train is traveling at a speed of 60 kilometers per hour, making stops every 48 kilometers. The duration of each stop, except the fifth, is 10 minutes, and the fifth stop is half an hour. How far has the train traveled if it departed at noon on September 29 and arrived at its destination on October 1 at 10:00 PM? | Solution: The train was on the way for 58 hours.
The train covers a section of 48 kilometers in $\frac{4}{5}$ of an hour.
Let the train make $N$ stops during its journey, then the time of its movement will be $\left(\frac{4}{5}+\frac{1}{6}\right)(N-1)+\left(\frac{4}{5}+\frac{1}{2}\right)+t=58$, where $t$ is the time ... | 2870 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Calculate $f(2)$, if $25 f\left(\frac{x}{1580}\right)+(3-\sqrt{34}) f\left(\frac{1580}{x}\right)=2017 x$. Round the answer to the nearest integer. | Solution: Let's make the substitution: $\mathrm{t}=\frac{x}{1580}$, then the equation will take the form:
$25 f(t)+(3-\sqrt{34}) f\left(\frac{1}{t}\right)=2017 \cdot 1580 \cdot t \quad(1)$,
substitute $\frac{1}{t}$ for $t$ in the equation, we get
$$
25 f\left(\frac{1}{t}\right)+(3-\sqrt{34}) f(t)=2017 \cdot 1580 \cd... | 265572 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 1. Buratino, Karabas-Barabas, and Duremar are running along a path around a circular pond. They start simultaneously from the same point, with Buratino running in one direction and Karabas-Barabas and Duremar running in the opposite direction. Buratino runs three times faster than Duremar and four times faster ... | # Solution:
Let the length of the path be S.
Since Buratino runs three times faster than Duremar, by the time they meet, Buratino has run three-quarters of the circle ($3 / 4$ S), while Duremar has run one-quarter. Since Buratino runs four times faster than Karabas-Barabas, by the time they meet, Buratino has run fou... | 3000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. At the same time, a cyclist leaves point $A$ for point $B$ at a speed of $15 \mathrm{km} / \mathrm{u}$, and a tourist leaves point $B$ for point $C$ at a speed of 5 km/h. After 1 hour and 24 minutes from the start of their journey, they were at the minimum distance from each other. Find the distance between the poin... | Solution: Let $AB = BC = AC = S$. Denote the distance between the cyclist and the tourist as $r = r(t)$, where $t$ is the time from the start of the movement. Then, by the cosine theorem, we have: $r^{2} = (S - 15t)^{2} + (5t)^{2} - 5t(S - 15t)$. To find the time when the distance between the cyclist and the tourist wa... | 26 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. At the same time, a car departs from point $A$ to point $B$ at a speed of 80 km/h, and from point $B$ to point $C$ - a train at a speed of $50 \kappa м / ч$. After 7 hours of travel, they were at the shortest distance from each other. Find the distance between the points, if all three points are equidistant from eac... | Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 80t)^{2} + (50t)^{2} - 50t(S - 80t)$. To find the time at which the distance between the car and the train was the sm... | 860 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. At the same time, a car departs from point $A$ to point $B$ at a speed of $90 \mathrm{~km} / \mathrm{h}$, and from point $B$ to point $C$ - a train at a speed of $60 \mathrm{~km} /$ h. After 2 hours of travel, they found themselves at the minimum distance from each other. Find the distance between the points, if all... | Solution: Let $AB = BC = AC = S$. Denote the distance between the car and the train as $r = r(t)$, where $t$ is the time from the start of the motion. Then, by the cosine theorem, we have: $r^{2} = (S - 90t)^{2} + (60t)^{2} - 60t(S - 90t)$. To find the time at which the distance between the car and the train
^{2}$, and after the fourth time, it is $\left(\frac{V-a}{V}\right)^{4}$. Substituting th... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Find the smallest natural number $m$, for which the expression $148^{n}+m \cdot 141^{n}$ is divisible by 2023 for any odd natural $n$. (16 points) | Solution. $2023=7 \cdot 289$, GCD $(7 ; 289)=1$. Since $n-$ is an odd number, then $148^{n}+m \cdot 141^{n}=(289-141)^{n}+m \cdot 141^{n}=289 l+(m-1) 141^{n}, l \in \square$. Then $(m-1) 141^{n}$ must be divisible by 289. Since 289 and 141 are coprime, then $m-1=289 k, k \in\{0\} \cup \square$. On the other hand $148^{... | 1735 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. In triangle $\mathrm{KLM}$ with angle $\mathrm{L}=120^{\circ}$, the angle bisectors LA and $\mathrm{KB}$ of angles KLM and LKM are drawn respectively. Find the measure of angle KBA. | # Solution.
1). Let KS be the extension of KL beyond point L. Then LM is the bisector of angle MLS, since $\angle M L S = \angle M L A = \angle A L K = 60^{\circ}$. Point B is the intersecti... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A pedestrian left point $A$ for point $B$, and after some delay, a second pedestrian followed. When the first pedestrian had walked half the distance, the second had walked 15 km, and when the second pedestrian had walked half the distance, the first had walked 24 km. Both pedestrians arrived at point $B$ simultaneo... | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the pedestrians. Then $\frac{s}{2 v_{1}}=\frac{s-15}{v_{2}}$ and $\frac{s-24}{v_{1}}=\frac{s}{2 v_{2}}$. From this, $\frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; s^{2}=4 s^{2}-4 \cdot 39 s+60 \cdot 24 ;$ $s^{2}-52 s+480... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. The base of the pyramid is a rectangle with sides $AB=24$ and $BC=30$, and the lateral edge of the pyramid $TA=16$ is perpendicular to the plane of the base. What is the minimum area that the section of the pyramid by a plane passing through the center of symmetry of the base $O$, the vertex of the pyramid, and a p... | # Solution:
Regardless of the position of point $M$ on side $B C$, the face $T A B$ is the orthogonal projection of the section $T M N$. The area of the section will be the smallest if the angle between the cutting plane and the face $T A B$ is the smallest. Since the cutting plane passes through the center of symmetr... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Two cyclists set off simultaneously from point A to point B. When the first cyclist had covered half the distance, the second cyclist still had 24 km to go, and when the second cyclist had covered half the distance, the first cyclist still had 15 km to go. Find the distance between points A and B. | # Solution:
Let $s$ be the distance between points $A$ and $B$, and $v_{1}, v_{2}$ be the speeds of the cyclists. Then $\frac{s}{2 v_{1}}=\frac{s-24}{v_{2}} \quad$ and $\quad \frac{s-15}{v_{1}}=\frac{s}{2 v_{2}} . \quad$ From this, $\quad \frac{s}{2(s-24)}=\frac{(s-15) \cdot 2}{s} ; \quad s^{2}=4 s^{2}-4 \cdot 39 s+60... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A batch of shoes, purchased for 180 thousand rubles, was sold in the first week at a price higher than the purchase price by $25 \%$, then the markup was reduced to $16 \%$ of the purchase price; and the entire batch of shoes was sold for $20 \%$ more than it was purchased for. For what amount was the shoes sold in ... | # Solution:
$x$ thousand rubles - the purchase cost of shoes sold in the first week, $y$ - the remainder.
$$
\left\{\begin{array} { c }
{ x + y = 180 } \\
{ 25 x + 16 y = 20 ( x + y ) ; }
\end{array} \left\{\begin{array} { c }
{ 5 x = 4 y , } \\
{ x + 5 / 4 x = 180 ; }
\end{array} \left\{\begin{array}{l}
x=80 \\
y=... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. What is the greatest value that the sum of the first $n$ terms of the arithmetic progression $113,109,105, \ldots$ can take?
# | # Solution:
The sum of the first $n$ terms of an arithmetic progression $S_{n}$ takes its maximum value if $a_{n}>0$, and $a_{n+1} \leq 0$. Since $a_{n}=a_{1}+d(n-1)$, from the inequality $113-4(n-1)>0$ we find $n=[117 / 4]=29$.
Then $\max S_{n}=S_{29}=0.5 \cdot(113+113-4 \cdot 28) \cdot 29=1653$.
Answer: 1653. | 1653 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. In the laboratory, there are flasks of two sizes (volume $V$ and volume $V / 2$) in a total of 100 pieces, with at least three flasks of each size. The lab assistant randomly selects three flasks in sequence, and fills the first one with an 80% salt solution, the second one with a 50% salt solution, and the third on... | Solution. If $N$ is the number of large flasks in the laboratory, $N=3,4, \ldots, 97$, then $n=100-N$ is the number of small flasks in the laboratory, $n=3,4, \ldots, 97$. For the event $A=\{$ the salt content in the dish is between $45 \%$ and $55 \%$ inclusive $\}$, it is necessary to find the smallest $N$ such that ... | 46 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Clowns Plukha and Shmyaka have six pairs of valenki (traditional Russian felt boots) between them. Each pair of valenki is painted in a unique color, and the valenki in a pair are identical (they are not distinguished as left or right). In how many ways can both clowns be simultaneously wearing mismatched valenki? (... | Solution. We can use valenki (traditional Russian felt boots) from two, three, or four pairs.
1) We choose two pairs of valenki $C_{6}^{2}=15$ ways. Each clown puts on one valenok from different pairs, choosing which one for which foot. This gives us $15 \cdot 2 \cdot 2=60$ ways.
2) We use three pairs of valenki. We c... | 900 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. Option I.
Two different natural numbers are written on the board, the larger of which is 2015. It is allowed to replace one of the numbers with their arithmetic mean (if it is an integer). It is known that such an operation was performed 10 times. Find what numbers were originally written on the board. | Solution. After each iteration, the difference between the written numbers is halved. We get that the initial difference must be a multiple of $2^{10}=1024$. From this, we find the second number: $2015-1024=991$.
Answer: 991. | 991 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 5. II variant.
Two different natural numbers are written on the board, the larger of which is 1580. It is allowed to replace one of the numbers with their arithmetic mean (if it is an integer). It is known that such an operation was performed 10 times. Find what numbers were originally written on the board. | Solution. After each iteration, the difference between the written numbers is halved. We get that the initial difference must be a multiple of $2^{10}=1024$. From this, we find the second number: $1580-1024=556$.
Answer: 556. | 556 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1.
A librarian at a physics and mathematics school noticed that if the number of geometry textbooks in the school library is increased by several (integer) times and the number of algebra textbooks is added to the resulting number, the total is 2015. If the number of algebra textbooks is increased by the sam... | Solution. Let the number of geometry textbooks be $x$, and the number of algebra textbooks be $y$. We can set up the system
$$
\left\{\begin{array}{l}
x n+y=2015 \\
y n+x=1580
\end{array}\right.
$$
We can write an equivalent system, where the equations represent the sum and difference of the equations in the original... | 287 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6.
A $10 \times 10$ square was cut into rectangles, the areas of which are different and expressed as natural numbers. What is the maximum number of rectangles that can be obtained? | Solution. The area of the square is 100. If we represent 100 as the sum of natural numbers, the number of addends will be the largest if the difference between the numbers is one. Let's take rectangles with areas of $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$. Their total area is 55. Therefore, the sum of the areas of the remainin... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. The production of ceramic items consists of 3 sequential stages: forming a ceramic item on a potter's wheel for 15 minutes, drying for 10 minutes, and firing for 30 minutes. It is required to produce 75 items. How should 13 masters be distributed between molders and firers to work on stages 1 and 3 respectively for ... | Solution. Molders - 4, decorators - 8. The thirteenth master can work at any stage, or not participate in the work at all. In this case, the working time is 325 minutes $\left(55+\left(\left[\frac{75}{4}\right]+1\right) \cdot 15=325\right)$. We will show that with other arrangements, the working time is longer. Suppose... | 325 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Find all natural values of $n$ for which
$$
\cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9}, \text { and } \log _{2}^{2} n+45<\log _{2} 8 n^{13}
$$
In your answer, write the sum of the obtained values of $n$.
(6 points) | # Solution.
$$
\begin{aligned}
& \cos \frac{2 \pi}{9}+\cos \frac{4 \pi}{9}+\cdots+\cos \frac{2 \pi n}{9}=\cos \frac{\pi}{9} \quad \Leftrightarrow 2 \sin \frac{\pi}{9} \cos \frac{2 \pi}{9}+2 \sin \frac{\pi}{9} \cos \frac{4 \pi}{9}+\cdots+ \\
& 2 \sin \frac{\pi}{9} \cos \frac{2 \pi n}{9}=2 \sin \frac{\pi}{9} \cos \frac{... | 644 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In how many ways can a bamboo trunk (a non-uniform natural material) 4 m long be sawn into three parts, the lengths of which are multiples of 1 dm, and from which a triangle can be formed?
(12 points) | Solution.
Let $A_{n}$ be the point on the trunk at a distance of $n$ dm from the base. We will saw at points $A_{p}$ and $A_{q}, p<q$. To satisfy the triangle inequality, it is necessary and sufficient that each part is no longer than 19 dm:
$$
p \leq 19, \quad 21 \leq q \leq p+19
$$
Thus, the number of ways to choo... | 171 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. How many solutions in natural numbers $x, y$ does the inequality $x / 76 + y / 71 < 1$ have?
(12 points) | # Solution.
All solutions lie in the rectangle
$$
\Pi=\{0<x<76 ; \quad 0<y<41\}
$$
We are interested in the number of integer points lying inside Π below its diagonal $x / 76 + y / 41=1$. There are no integer points on the diagonal itself, since 76 and 41 are coprime. So we get half the number of integer points insi... | 1500 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
1. A workshop produces items of types $A$ and $B$. For one item of type $A$, 10 kg of steel and 23 kg of non-ferrous metals are used, and for one item of type $B-70$ kg of steel and 40 kg of non-ferrous metals are used. The profit from selling one item of type $A$ is 80 thousand rubles, and for type $B-100$ thousand ru... | Solution. Let $x$ be the number of items of type $A$, and $y$ be the number of items of type $B$. Then the profit per shift is calculated by the formula $D=80 x+100 y$, with the constraints $10 x+70 y \leq 700$, $23 x+40 y \leq 642$, and $x$ and $y$ are non-negative integers.
 | # Solution.
Factorize 210 into a product of prime numbers: $2 \cdot 3 \cdot 5 \cdot 7$. Let's see how 4 prime
#### Abstract
divisors can be distributed among the desired factors
1) $4+0+0+0-1$ way.
2) $3+1+0+0-C_{4}^{3}=4$ ways.
3) $2+2+0+0-C_{4}^{2} / 2=3$ ways.
4) $2+1+1+0-C_{4}^{2}=6$ ways.
5) $1+1+1+1-1$ way... | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The sequence is defined recursively:
$x_{0}=0, x_{n+1}=\frac{\left(n^{2}+n+1\right) x_{n}+1}{n^{2}+n+1-x_{n}} . \quad$ Find $x_{8453}$. (12 points) | Solution.
$$
\text { We calculate } x_{1}=\frac{1}{1}=1, x_{2}=\frac{4}{2}=2, x_{3}=\frac{15}{5}=3 \text {, a hypothesis emerges: } x_{n}=n \text {. }
$$
Let's verify by induction:
$$
x_{n+1}=\frac{\left(n^{2}+n+1\right) n+1}{n^{2}+n+1-n}=\frac{n^{3}+n^{2}+n+1}{n^{2}+1}=n+1
$$
## Answer: 8453 | 8453 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. What is the greatest area that a rectangle can have, the coordinates of whose vertices satisfy the equation $|y-x|=(y+x+1)(5-x-y)$, and whose sides are parallel to the lines $y=x$ and $y=-x$? Write the square of the found area in your answer. $\quad(12$ points $)$
# | # Solution.
Substitution: $x_{1}=x+y, y_{1}=y-x$. This substitution increases all dimensions by a factor of $\sqrt{2}$. We have $\left|y_{1}\right|=\left(x_{1}+1\right)\left(5-x_{1}\right), \quad S\left(x_{1}\right)=4\left(x_{1}-2\right)\left(x_{1}+1\right)\left(5-x_{1}\right), x_{1} \in(2 ; 5)$. $S^{\prime}\left(x_{1... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. The lateral face of a regular triangular pyramid $S A B C$ is inclined to the base plane $A B C$ at an angle $\alpha=\operatorname{arctg} \frac{3}{4}$. Points $M, N, K$ are the midpoints of the sides of the base $A B C$. Triangle $M N K$ is the lower base of a right prism. The edges of the upper base of the prism in... | # Solution.
The height of the pyramid $SO = h$. The side of the base of the pyramid $AC = a$.
The height of the prism is $3h/4$, and the sides of the base of the prism are $a/2$.
The area of triangle $MNK$:
$$
S_{MNK} = \frac{a^2 \sqrt{3}}{16}
$$
The area of triangle $FPR$:
$$
S_{FPR} = \frac{a^2 \sqrt{3}}{64}
$$... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. Solution. Let the correct numbers in the table be $a, b, c$ and $d$. Consider the events
$$
A=\{\text { There is a cat }\} \text { and } B=\{\text { There is a dog }\} .
$$
| | There is a dog | There is no dog |
| :--- | :---: | :---: |
| There is a cat | $a$ | $b$ |
| There is no cat | $c$ | $d$ |
These events ... | Answer: approximately 4272 people.
Note. Instead of approximate equalities, estimates can be made using inequalities.
## Grading Criteria
| Solution is complete and correct | 2 points |
| :--- | :--- |
| Answer is correct, but there are no arguments showing that other options are implausible (for example, there is a... | 4272 | Other | math-word-problem | Yes | Yes | olympiads | false |
8. Solution. Let the correct numbers in the table be $a, b, c$ and $d$. Consider the events $A=\{$ Has a card $\}$ and $B=\{$ Makes online purchases $\}$. These events are independent if the proportion of cardholders among online shoppers is the same as among those who do not make online purchases, that is,
| | Has a... | Answer: approximately 1796 people.
Note. Instead of approximate equalities, estimates can be made using inequalities.
## Grading Criteria
| Solution is complete and correct | 2 points |
| :--- | :--- |
| Answer is correct, but there are no arguments showing that other options are implausible (for example, there is a... | 1796 | Other | math-word-problem | Yes | Yes | olympiads | false |
15. Minimum Sum (9th grade, 4 points). There are a lot of symmetric dice. They are thrown simultaneously. With some probability $p>0$, the sum of the points can be 2022. What is the smallest sum of points that can fall when these dice are thrown with the same probability $p$? | Answer: 337.
Solution. The distribution of the sum rolled on the dice is symmetric. To make the sum $S$ the smallest possible, the sum of 2022 should be the largest possible. This means that the sum of 2022 is achieved when sixes are rolled on all dice. Therefore, the total number of dice is $2022: 6=337$. The smalles... | 337 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. The King's Path (from 7th grade, 2 points). A chess king is on the a1 square of a chessboard and wants to move to the h8 square, moving right, up, or diagonally up-right. In how many ways can he do this? | Solution. Instead of letter designations for the columns, we use numerical ones. Then, instead of al, we can write $(1,1)$. Let $S_{m, n}$ be the number of ways to get from the cell $(1,1)$ to the cell $(m, n)$. We are interested in $S_{8,8}$.
Obviously, $S_{1, n}=1$ and $S_{m, 1}=1$ for all $m$ and $n$. In particular... | 48639 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
17. Happy Sums (from 8th grade, 4 points). In the "Happy Sum" lottery, there are $N$ balls numbered from 1 to $N$. During the main draw, 10 balls are randomly drawn. During the additional draw, 8 balls are randomly selected from the same set of balls. The sum of the numbers on the drawn balls in each draw is announced ... | Solution. In the main draw, there are $C_{N}^{10}$ possible combinations, and in the additional draw, there are $C_{N}^{8}$ combinations. Let $S_{63,10}$ be the number of combinations in the first draw where the sum is 63. The addends are different natural numbers from 1 to $N$. If the smallest number is 2 or more, the... | 18 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In the village of Znoynoye, there are exactly 1000 residents, which exceeds the average population of villages in the valley by 90 people.
How many residents are there in the village of Raduzhny, which is also located in Sunny Valley? | Solution. Let $x$ be the total number of residents in all other villages except Znoynoye. Then the average population is
$$
\frac{1000+x}{10}=100+0.1 x=910, \text{ hence } x=8100
$$
Thus, the average population in 9 villages, except Znoynoye, is 900 people. If there are more than 900 residents in Raduzhny, there must... | 900 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Promotion. The store "Crocus and Cactus" announced a promotion: if a customer buys three items, the cheapest one of them is free. The customer took ten items, all of different prices: 100 rubles, 200 rubles, 300 rubles, and so on - the most expensive item cost 1000 rubles.
a) (from 6th grade. 1 point). What is the ... | Solution. a) The cheapest item in each trio should be as expensive as possible. If the buyer distributes the purchases into trios as follows (indicating only the prices):
$(1000,900,800),(700,600,500),(400,300,200)$ and a separate item for $100 \mathrm{p}.$, then he will pay $1000+900+700+600+400+300+100=4000$ rubles.... | 4583 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. Stubborn Squares (from 7th grade. 2 points). Given 100 numbers. 2 was added to each of them. The sum of the squares of the numbers did not change. 2 was added to each of the resulting numbers again. How did the sum of the squares change now? | Solution. Let the given numbers be $x_{1}, x_{2}, \ldots, x_{100}$. By adding 2 to each number, we get the set $y_{j}=x_{j}+2$. Adding 2 again, we get the set $z_{j}=y_{j}+2=x_{j}+4$. However, the variances of all three sets are equal, that is,
$$
\overline{x^{2}}-\bar{x}^{2}=\overline{y^{2}}-\bar{y}^{2}=\overline{z^{... | 800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. Retro Collection (recommended from 8th grade, 2 points). Vitya collects toy cars from the "Retro" series. The problem is that the total number of different models in the series is unknown - it is the biggest commercial secret, but it is known that different cars are produced in the same edition, and therefore it ca... | Solution. Let there be a series of $n \geq 13$ cars. The probability that among the next $k$ offers there will only be the 12 models that Vitya has already seen is
$$
\left(\frac{12}{n}\right)^{k} \leq\left(\frac{12}{13}\right)^{k}
$$
We form the inequality: $\left(\frac{12}{13}\right)^{k}\log _{12 / 13} 0.01=\frac{\... | 58 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. The Right Stars (from 8th grade, 3 points). Let natural numbers $k$ and $n$ be coprime, with $n \geq 5$ and $k < n / 2$. A proper $(n ; k)$-star is defined as a closed broken line that results from replacing every $k$ consecutive sides in a regular $n$-gon with a diagonal having the same endpoints. For example, the... | Solution. We will solve the problem in general for the $(n ; k)$-star. Let the vertices of the $n$-gon be denoted by $A_{0}, A_{1}$, and so on up to $A_{n-1}$. Take two consecutive segments of the star: $A_{0} A_{k}$ and $A_{1} A_{k+1}$. They intersect at a circle of radius $r_{1}$ with the center at the center of the ... | 48432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12. Target (from 8th grade, 2 points). A target is hanging on the wall, consisting of five zones: a central circle (the bullseye) and four colored rings (see figure). The width of each ring is equal to the radius of the bullseye. It is known that the number of points for hitting each zone is inversely proportional to t... | Solution. Suppose, for definiteness, that the apple has a radius of 1. Then the blue zone is enclosed between circles with radii 3 and 4. The probability of hitting the blue zone relative to the probability of hitting the apple is the ratio of the areas of these zones:
$$
\frac{p_{g}}{p_{c}}=\frac{1}{4^{2}-3^{2}}=\fra... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
19. How many airplanes? (11th grade, 8 points) The work of the Absent-Minded Scientist involves long business trips, and therefore he often flies on the same airline. This airline has many identical airplanes, and all of them have names. Since the Scientist does not fly every day or even every week, it can be assumed t... | Solution. Consider the random variable $X$ "the ordinal number of the flight when the Scientist first gets a plane he has flown before". We will make a point estimate of the number of planes using the method of moments. For this, we need to solve the equation $\mathrm{E} X=15$ at least approximately.
Let's find $\math... | 134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Ministers in Anchuria. In the government cabinet of ministers in Anchuria, there are 100 ministers. Among them, there are crooks and honest ministers. It is known that among any ten ministers, at least one minister is a crook. What is the smallest number of crook ministers that can be in the cabinet? | Solution. There are no more than nine honest ministers, otherwise, a group of ten honest ministers would be found, which contradicts the condition. Therefore, the number of minister-cheats is no less than $100-9=91$.
Answer: 91. | 91 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. For a procrastinator, the waiting time will increase by $a$ minutes.
Therefore, the total wasted time of all standing in the queue will decrease by $b-a$ minutes. We will thus swap people in pairs of "Procrastinator-Hurry" until we get a queue where all the hurries stand before all the procrastinators. In this queu... | Answer: a) 40 minutes and 100 minutes; b) 70 minutes.
Comment. The expected time turned out to be the arithmetic mean between the maximum and minimum possible. Try to prove the fact $\mathrm{E} T=\frac{T_{\min }+T_{\max }}{2}$ in general, that is, prove the combinatorial identity
$$
C_{n+m}^{2} \cdot \frac{b m+a n}{m... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Median. In a set of 100 numbers. If one number is removed, the median of the remaining numbers is 78. If another number is removed, the median of the remaining numbers is 66. Find the median of the entire set. | Solution. Arrange the numbers in ascending order. If we remove a number from the first half of the sequence (with a number up to 50), the median of the remaining numbers will be the $51-\mathrm{st}$ number in the sequence. If we remove a number from the second half, the median of the remaining numbers will be the numbe... | 72 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
, and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\ba... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false |
11. There are 25 children in the class. Two are chosen at random for duty. The probability that both duty students will be boys is $\frac{3}{25}$. How many girls are in the class? | # Solution.
Let there be $n$ boys in the class, then the number of ways to choose two duty students from them is $\frac{n(n-1)}{2}$, the number of ways to choose two duty students from the entire class is $25 \cdot 24$. The ratio of the two obtained fractions, i.e., $\frac{n(n-1)}{25 \cdot 24}$. It is also equal to $\... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young dr... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
19. How many airplanes? (11th grade, 8 points) The work of the Absent-Minded Scientist involves long business trips, and therefore he often flies on the same airline. This airline has many identical airplanes, and all of them have names. Since the Scientist does not fly every day or even every week, it can be assumed t... | Solution. Consider the random variable $X$ "the ordinal number of the flight when the Scientist first gets a plane he has flown before". We will make a point estimate of the number of planes using the method of moments. For this, we need to solve the equation $\mathrm{E} X=15$ at least approximately.
Let's find $\math... | 134 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. For a procrastinator, the waiting time will increase by $a$ minutes.
Therefore, the total wasted time of all standing in the queue will decrease by $b-a$ minutes. We will thus swap people in pairs of "Procrastinator-Hurry" until we get a queue where all the hurries stand before all the procrastinators. In this queu... | Answer: a) 40 minutes and 100 minutes; b) 70 minutes.
Comment. The expected time turned out to be the arithmetic mean between the maximum and minimum possible. Try to prove the fact $\mathrm{E} T=\frac{T_{\min }+T_{\max }}{2}$ in general, that is, prove the combinatorial identity
$$
C_{n+m}^{2} \cdot \frac{b m+a n}{m... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
, and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\ba... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false |
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
, and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\ba... | 40 | Other | math-word-problem | Yes | Yes | olympiads | false |
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young dr... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
. Ivan the Tsarevich is learning to shoot a bow. He put 14 arrows in his quiver and went to shoot at pine cones in the forest. He knocks down a pine cone with a probability of 0.1, and for each pine cone he knocks down, the Frog-Princess gives him 3 more arrows.... | Solution. First method. Let Ivan have $n$ arrows at the present moment. Let $X_{0}$ be the random variable "the number of shots needed to reduce the number of arrows by one." Ivan makes a shot. Consider the random variable - the indicator $I$ of a successful shot. $I=0$, if the shot is unsuccessful (probability of this... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. The diagram shows a track layout for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they like, returning to point $A$.
The young driver, Yu... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
. Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Solution. For clarity, let's assume that when a bite occurs, the Absent-Minded Scholar immediately reels in and casts the fishing rod again, and does so instantly. After this, he waits again. Consider a 5-minute time interval. During this time, on average, there are 5 bites on the first fishing rod and 1 bite on the... | Answer: 50 seconds.
## Grading Criteria
| Solution is correct and well-reasoned | 3 points |
| :--- | :---: |
| It is shown that on average there are 5 bites in 6 minutes, or an equivalent statement is proven | 1 point |
| Solution is incorrect or missing (including only the answer) | 0 points | | 50 | Other | math-word-problem | Yes | Yes | olympiads | false |
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young dr... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
. Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
![](https://cdn.mathpix.com/cropped/2024_05_06_cb774fbc1caae3aa402cg-05.jpg?height=634&width=1445&top_left_y=1268&top_left... | 190 | Other | 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.