problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 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 |
|---|---|---|---|---|---|---|---|---|
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 |
5. There are two lead-tin alloys. In the first alloy, the mass of lead is to the mass of tin as $1: 2$; in the second - as $2: 3$. How many grams of the first alloy are needed to obtain 22 g of a new alloy with the mass ratio of lead to tin being 4:7? | Solution. Let the first alloy contain x g of lead and 2x g of tin. In the second alloy, there are 2y g of lead and 3y g of tin. Then $k \cdot 3x + n \cdot 5y = 22 ; \frac{kx + n \cdot 2y}{k \cdot 2x + n \cdot 3y} = \frac{4}{7} ; \quad$ we need to find $k \cdot 3x$ and $5ny$. Let $ny = b ; kx = a \cdot \frac{a + 2b}{2a ... | 12 | 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 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 |
1. The number $\overline{6 x 62 y 4}$ is divisible by 11, and when divided by 9, it gives a remainder of 6. Find the remainder when this number is divided by $13 .(15$ points) | # Solution
By the divisibility rule for 11, we get
$((x+2+4)-(6+6+y)) \vdots 11$ or $(x-6-y) \vdots 11$
Let's find suitable options: $(0 ; 5)(1 ; 6)(2 ; 7)(3 ; 8)(4 ; 9)(6 ; 0)(7 ; 1)(8 ; 2)$ $(9 ; 3)$.
If the number $\overline{6 x 62 y 4}$ gives a remainder of 6 when divided by 9, then by the divisibility rule for... | 6 | Number Theory | 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 |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(40 \cos 2 x+41)$ for $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]$ (10 points) | Solution: Since $\cos (\pi-\arcsin (-0.8))=\cos (\pi+\arcsin 0.8)=-\cos (\arcsin 0.8)=-0.6$, then $x \in[(5 / 3)(\operatorname{arctg}(1 / 5)) \cos (\pi-\arcsin (-0.8)) ; \operatorname{arctg} 3]=[-\operatorname{arctg}(1 / 5) ; \operatorname{arctg} 3]$. $2 x \in[-2 \operatorname{arctg}(1 / 5) ; 2 \operatorname{arctg} 3]$... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, $AE = \sqrt{3}$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, ... | # Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{A E}{\cos 30^{\circ}}=2, E C=A C \sin 30^{\circ}=1$
$, 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 |
6. Find the sum of the integers that belong to the set of values of the function $f(x)=\log _{3}(10 \cos 2 x+17)$ for $x \in[1,25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3] . \quad(10$ points $)$ | # Solution:
Since $\quad$ $\cos (\pi-\arcsin (-0.6))=\cos (\pi+\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg} 0.25) \cos (\pi-\arcsin (-0.6)) ; \operatorname{arctg} 3]=[-\operatorname{arctg} 0.25 ; \operatorname{arctg} 3] \quad$ Therefore, $2 x \in[-2 \operatorname{arctg} 0.25 ; 2 ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, $EC=1$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1:2$. Find the length of segment $BO$, where $O$... | # Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right triangle, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$,
$$
A C=\frac{E C}{\sin 30^{\circ}}=2, \quad A E=E C \operatorname{tg} 60^{\circ}=\sqrt... | 2 | 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
=\log _{2}(5 \cos 2 x+11)$ for $x \in[1,25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2] \quad$ (10 points) | # Solution:
Since $\quad$ $\cos (\pi+\arcsin (-0.6))=\cos (\pi-\arcsin 0.6)=-\cos (\arcsin 0.6)=-0.8, \quad$ then $x \in[1.25(\operatorname{arctg}(1 / 3)) \cos (\pi+\arcsin (-0.6)) ; \operatorname{arctg} 2]=[-\operatorname{arctg}(1 / 3) ; \operatorname{arctg} 2]$ Therefore, $2 x \in[-2 \operatorname{arctg}(1 / 3) ; 2 ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. On the side $AC$ of triangle $ABC$, a circle is constructed with $AC$ as its diameter, which intersects sides $AB$ and $BC$ at points $D$ and $E$ respectively. The angle $EDC$ is $30^{\circ}$, the area of triangle $AEC$ is $\sqrt{3} / 2$, and the area of triangle $DBE$ is to the area of triangle $ABC$ as $1: 2$. Fin... | Solution:
1) $\angle E D C=\angle E A C=30^{\circ}$ (inscribed angles subtending the same arc);
2) $A C$ - diameter of the circle $\Rightarrow \triangle A E C$ - right-angled, $\angle A E C=90^{\circ}, \angle E C A=60^{\circ}$, $A C=\frac{E C}{\sin 30^{\circ}}=2 E C, A E=E C \operatorname{tg} 60^{\circ}=\sqrt{3} E C ;... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. For chemical experiments, two identical test tubes were taken, each containing 200 ml of a liquid substance. From the first test tube, $1 / 4$ of the content was poured out and the same amount of water was added, then this procedure was repeated 3 more times, each time pouring out a quarter of the content of the tes... | Solution. The initial amount of the substance is $-V$. After pouring out $a$ part, the concentration of the substance in the test tube becomes $\frac{V-a}{V}$. After the second time, the concentration is $\left(\frac{V-a}{V}\right)^{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 |
4. Solve the equation $\quad \sqrt{x+\sqrt{x}-\frac{71}{16}}-\sqrt{x+\sqrt{x}-\frac{87}{16}}=\frac{1}{2}$. | Solution. Taking into account that $x$ is non-negative, we make the substitution $u=\sqrt{x+\sqrt{x}-\frac{71}{16}}, v=\sqrt{x+\sqrt{x}-\frac{87}{16}}, u \geq 0, v \geq 0$.
Then we obtain the system $\left\{\begin{array}{l}u-v=1 / 2, \\ u^{2}-v^{2}=1,\end{array} \Rightarrow\left\{\begin{array}{l}u-v=1 / 2, \\ u+v=2,\e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. What is the maximum possible number of consecutive terms of an increasing geometric progression that can be three-digit natural numbers? Provide an example of such a sequence. (16 points) | Solution. Let the required members of the progression be $a_{0}, a_{1}, \ldots, a_{n}, a_{k}=a_{0} q^{k}$, the common ratio - an irreducible fraction $q=r / s, r>s$. Then $a_{0}=b s^{n}, a_{n}=b r^{n}, b \in \mathbb{N}$, since $r^{n}$ and $s^{n}$ are coprime. We obtain the restriction
$$
r^{n}<1000 / b, \quad s^{n} \g... | 6 | Number Theory | 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 |
2. $f(-x)=3(-x)^{3}-(-x)=-3 x^{3}+x=-\left(3 x^{3}-x\right)=-f(x)$ $g(-x)=f^{3}(-x)+f\left(\frac{1}{-x}\right)-8(-x)^{3}-\frac{2}{-x}=-f^{3}(x)-f\left(\frac{1}{x}\right)+8 x^{3}+\frac{2}{x}=-g(x)$
Therefore, $g$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a roo... | Answer: 0.
Problem 8 (2nd version).
Find the sum of the roots of the equation $g^{3}(x)-g\left(\frac{1}{x}\right)=5 x^{3}+\frac{1}{x}$, where $g(x)=x^{3}+x$.
## Solution.
Consider the function $f(x)=g^{3}(x)-g\left(\frac{1}{x}\right)-5 x^{3}-\frac{1}{x}$, then the roots of the original equation are the roots of the... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. $g(-x)=(-x)^{3}+(-x)=-x^{3}-x=-\left(x^{3}+x\right)=-g(x)$
$$
f(-x)=g^{3}(-x)-g\left(\frac{1}{-x}\right)-5(-x)^{3}-\frac{1}{-x}=-g^{3}(x)+g\left(\frac{1}{x}\right)+5 x^{3}+\frac{1}{x}=-f(x)
$$
Therefore, $f$ is an odd function $\Rightarrow$ if $x_{0}$ is a root of the original equation, then $-x_{0}$ is also a roo... | # Answer: 0.
Task 9 (1st variant). In each vertex of an equilateral triangle with side $\sqrt{10}$, circles of radius $\sqrt{5}$ were constructed, the inner areas of which were painted gray, brown, and raspberry. Find the area of the gray-brown-raspberry region.
Solution. By symmetry, the areas of the figures formed ... | 0 | Algebra | proof | 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. Two trucks were transporting fertilizers, making the same number of trips. It turned out that 4 tons less could be loaded onto the first truck and 3 tons less onto the second truck than planned, so each truck had to make 10 extra trips. As a result, the first truck transported 60 tons more than the second, as planne... | # Solution:
Let $x, y$ - capacity, $t-$ number of trips as planned.
$$
\left\{\begin{array}{l}
x t=(x-4)(t+10), \\
y t=(y-3)(t+10), \Leftrightarrow\left\{\begin{array} { l }
{ 1 0 x - 4 t = 4 0 } \\
{ \quad x t - y t = 6 0 }
\end{array} \quad \left\{\begin{array}{l}
10 y-3 t=30, \\
(x-y) t=60
\end{array} \Rightarrow... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A batch of shoes, purchased for 180 thousand rubles, was sold in the first week at a price higher than the purchase price by $25 \%$, then the markup was reduced to $16 \%$ of the purchase price; and the entire batch of shoes was sold for $20 \%$ more than it was purchased for. For what amount was the shoes sold in ... | # Solution:
$x$ thousand rubles - the purchase cost of shoes sold in the first week, $y$ - the remainder.
$$
\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 |
1. The numerical sequence $\left\{a_{n}\right\}_{n=1}^{\infty}$ is defined such that $a_{1}=\log _{2}\left(\log _{2} f(2)\right), \quad a_{2}=$ $\log _{2}\left(\log _{2} f(f(2))\right), \ldots, a_{n}=\log _{2}(\log _{2} \underbrace{f(f(\ldots f}_{n}(2)))), \ldots$, where $f(x)=x^{x}$. Determine the index $n$ for which ... | # Solution.
If $\log _{2}\left(\log _{2} u\right)=t$, then $u=2^{2^{t}}, f(u)=\left(2^{2^{t}}\right)^{2^{2^{t}}}=2^{2^{t+2^{t}}}, \log _{2}\left(\log _{2} f(u)\right)=t+2^{t} . \quad$ If $u=2,2=2^{2^{t}}, t=0, a_{1}=\log _{2}\left(\log _{2} f(2)\right)=0+2^{0}=1$. If $u=f(2)$, then $t=\log _{2}\left(\log _{2} u\right)... | 5 | Algebra | 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 |
6. More and more countries are exploring space. The number of states that launch their satellites using their own launch vehicles is already 12. There are also countries that use the services of the main space powers to launch their satellites for economic purposes. Due to the increasing number of participants in space... | Solution. According to the condition, at the current moment, all satellites should be located on a sphere with radius $R+H$. Let $O$ be the center of the sphere, and its radius $R_{H}=R+H$. Denote the satellites by points $C_{i}, i=1, \ldots, n$. We need to determine the maximum value of $n$.
Let's find the distance $... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. Option I.
The decimal representation of a natural number $N$ contains 1580 digits. Among these digits are threes, fives, and sevens, and no other digits. It is known that the number of sevens is 20 less than the number of threes. Find the remainder when the number $N$ is divided by 3. | Solution. Let $x$ be the number of threes in the number $N$. The sum of the digits of the number $N$ is $S=3 x+7(x-20)+5(1580-(2 x-20))=7860$.
The remainder of $S$ divided by 3 is equal to the remainder of $N$ divided by 3 and is 0.
Answer: 0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 1. II variant.
The decimal representation of a 2015-digit natural number $N$ contains the digits 5, 6, 7 and no other digits. Find the remainder of the division of the number $N$ by 9, given that the number of fives in the representation of the number is 15 more than the number of sevens. | Solution. Let the number $N$ contain $x$ sevens. Then the sum of the digits of the number $N$ is $S=7x+5(x+15)+6(2015-(2x+15))=12075$.
$N \equiv S \equiv 6(\bmod 9)$.
Answer: 6. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. Option I.
Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}-a x+a+2=0$ is equal to -8.
# | # Solution.
1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=a\left(a^{2}-3(a+2)\right)=a^{3}-3 a(a+2)$.
2) $a^{3}-3 a(a+2)=-8 \Leftrightarrow\left[\begin{array}{l}a=-2, \\ a=1, \\ a=4 .\end{array}\right.$
3) $D=a^{2}-4 a-8$.
For $a=-2 \quad D=4+8-8=4>0$, therefo... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. II variant.
Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}+a x+a+1=0$ is equal to 1. | # Solution.
1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=-a\left(a^{2}-3(a+1)\right)=-a^{3}+3 a(a+1)$.
2) $-a^{3}+3 a(a+1)=1 \Leftrightarrow\left[\begin{array}{l}a=-1, \\ a=2 \pm \sqrt{3} \text {. }\end{array}\right.$
3) $D=a^{2}-4 a-4$.
For $a=-1 \quad D=1+4... | -1 | Algebra | 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 7. Option 1.
Given a triangle with sides 6, 8, and 10. Find the length of the shortest segment connecting points on the sides of the triangle and dividing it into two equal areas. | Solution. 1) Note that the given triangle is a right triangle: $6^{2}+8^{2}=10^{2}$.
2) Suppose first (and we will justify this later) that the ends of the desired segment $D E=t$ lie on the larger leg $A C=8$ and the hypotenuse $A B=10$ (see figure). Let $A D=x$, $A E=y, \angle B A C=\alpha$. Then
$S_{A D E}=\frac{1... | 4 | Geometry | 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 3.
One side of the parallelogram is $\sqrt{3}$ times larger than the other side. One diagonal of the parallelogram is $\sqrt{7}$ times larger than the other diagonal. How many times larger is one angle of the parallelogram than the other angle? | Solution. Let $x$ be the smaller side, then $\sqrt{3} x$ is the larger side. Let $y$ be the smaller diagonal, then $\sqrt{7} y$ is the larger diagonal. We have:
$2 x^{2}+2(\sqrt{3} x)^{2}=y^{2}+(\sqrt{7} y)^{2}$,
from which $x=y$.
We get: the acute angle of the parallelogram is $30^{\circ}$, the obtuse angle is $150^... | 5 | Geometry | 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. From point $A$ to point $B$, which are 8 km apart, a tourist and a cyclist set out simultaneously. The cyclist, who took no less than half an hour to travel from $A$ to $B$, without stopping, turned back and started moving towards point $A$, increasing his speed by $25 \%$. After 10 minutes from his departure from p... | Solution: Let $x$ (km/h) be the speed of the tourist, $y$ (km/h) be the initial speed of the cyclist, and $t$ (h) be the time spent by the cyclist traveling from $A$ to $B$. Then
$$
\left\{\begin{array}{c}
x(t+1 / 6)+5 y / 24=8, \\
y t=8, \\
t \geq 0.5,
\end{array} \Rightarrow x(8 / y+1 / 6)+5 y / 24=8, \Rightarrow 5 ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two numbers x and y satisfy the equation $280 x^{2}-61 x y+3 y^{2}-13=0$ and are the fourth and ninth terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression. | Solution: We factorize the expression $280 x^{2}-61 x y+3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(280\left(\frac{x}{y}\right)^{2}-61\left(\frac{x}{y}\right)+3\right)=280 y^{2}\left(\frac{x}{y}-\frac{3}{40}\right)\left(\frac{x}{y}-\frac{1}{7}\right)=(40 x-3 y)(7 x-y)$. This formula is also valid for all real numbers... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. From point $A$ to point $B$, which are 24 km apart, a pedestrian and a cyclist set out simultaneously. The cyclist, who spent no less than two hours on the journey from $A$ to $B$, without stopping, turned back and started moving towards point $A$ at a speed twice the initial speed. After 24 minutes from his departu... | Solution: Let $x$ (km/h) be the speed of the cyclist, $y$ (km/h) be the initial speed of the truck, and $t$ (h) be the time spent by the truck traveling from $A$ to $B$. Then
$$
\left\{\begin{array}{rl}
x(t+0.4)+0.8 y=24, \\
y t & =24, \\
t & \geq 2,
\end{array} \Rightarrow x(24 / y+0.4)+0.8 y=24, \Rightarrow 2 y^{2}+... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Two numbers \(x\) and \(y\) satisfy the equation \(26 x^{2} + 23 x y - 3 y^{2} - 19 = 0\) and are the sixth and eleventh terms, respectively, of a decreasing arithmetic progression consisting of integers. Find the common difference of this progression. | Solution: Factorize the expression $26 x^{2}+23 x y-3 y^{2}$. For $y \neq 0$, we have $y^{2}\left(26\left(\frac{x}{y}\right)^{2}+23\left(\frac{x}{y}\right)-3\right)=26 y^{2}\left(\frac{x}{y}+1\right)\left(\frac{x}{y}-\frac{3}{26}\right)=(x+y)(26 x-3 y)$. This formula is valid for all real numbers $y$. According to the ... | -3 | Algebra | 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\}
$$
We are interested in the number of integer points lying inside Π below its diagonal $x / 76 + y / 41=1$. There are no integer points on the diagonal itself, since 76 and 41 are coprime. So we get half the number of integer points insi... | 1500 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8. Indicate the smallest value of the parameter $a$ for which the equation has at least one solution
$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)+1=a+2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\frac{\pi x^{2}}{12}\right)$. | Solution.
Rewrite the equation as
$2 \sin \left(\pi-\frac{\pi x^{2}}{12}\right) \cos \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right)-2 \sin \left(\frac{\pi}{6} \sqrt{9-x^{2}}\right) \cos \left(\pi-\frac{\pi x^{2}}{12}\right)=a-1$, or
$\sin \left(\pi-\frac{\pi x^{2}}{12}-\frac{\pi}{6} \sqrt{9-x^{2}}\right)=\frac{a-1}{2}$.
... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Rhombus $ABCD$ is the base of a pyramid with vertex $S$. All its lateral faces are inclined to the base plane at the same angle of $60^{\circ}$. Points $M, N, K$, and $L$ are the midpoints of the sides of rhombus $ABCD$. A rectangular parallelepiped is constructed on rectangle $MNKL$ as its base. The edges of the up... | Solution.
The height of the pyramid $SO = h$. The diagonals of the rhombus $AC = d_1, BD = d_2$.
The height of the parallelepiped is $h / 2$, and the sides of the base of the parallelepiped are $d_1 / 2$ and $d_2 / 2$.
The volume of the parallelepiped is
$$
V_{\Pi} = \frac{d_1}{2} \cdot \frac{d_2}{2} \cdot \frac{h}... | 5 | Geometry | 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 |
8. Specify the greatest value of the parameter $p$ for which the equation has at least one solution
$2 \cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)-3=p-2 \sin \left(-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)$. (16 points) | # Solution.
Let's rewrite the equation as
$$
\cos \left(2 \pi-\frac{\pi x^{2}}{6}\right) \cos \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)+\sin \left(2 \pi-\frac{\pi x^{2}}{6}\right) \sin \left(\frac{\pi}{3} \sqrt{9-x^{2}}\right)=\frac{p+3}{2} \text {, or }
$$
$\cos \left(2 \pi-\frac{\pi x^{2}}{6}-\frac{\pi}{3} \sqrt{9... | -2 | Algebra | 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 |
3. (1b, 8-11) The probability that a purchased light bulb will work is 0.95. How many light bulbs need to be bought to ensure that with a probability of 0.99, there are at least 5 working ones among them? | # Solution.
Let's take 6 light bulbs. The probability that at least 5 of them will work is the sum of the probabilities that 5 will work and 1 will not, and that all 6 will work, i.e.,
$$
C_{6}^{5} \cdot 0.95^{5} \cdot 0.05 + C_{6}^{6} \cdot 0.95^{6} = 6 \cdot 0.95^{5} \cdot 0.05 + 0.95^{6} = 0.9672
$$
Let's take 7 ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. (3b,9-11) In the conditions of a chess match, the winner is declared as the one who outperforms the opponent by two wins. Draws do not count. The probabilities of winning for the opponents are equal. The number of decisive games in such a match is a random variable. Find its mathematical expectation. | # Solution.
Let $\mathrm{X}$ be the number of successful games. At the beginning of the match, the difference in the number of wins between the two participants is zero. Let's list the possible cases of two successful games, denoting a win by the first participant as 1 and a win by the second participant as 2: $11, 12... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
15. (4b, 8-11) In Anchuria, a checkers championship is being held in several rounds. The days and cities for the rounds are determined by a draw. According to the championship rules, no two rounds can take place in the same city, and no two rounds can take place on the same day. Among the fans, a lottery is organized: ... | # Solution.
In an $8 \times 8$ tour table, you need to select $k$ cells such that no more than one cell is chosen in any row or column. The value of $k$ should be chosen to maximize the number of combinations.
The number of combinations is given by $C_{8}^{k} A_{8}^{k}=\frac{8!\cdot 8!}{(8-k)!\cdot(8-k)!\cdot k!}$, w... | 6 | 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 |
9. Buratino the Statistician (from 7th grade, 2 points). Every month, Buratino plays in the "6 out of 45" lottery organized by Karabas-Barabas. In the lottery, there are 45 numbered balls, and in each draw, 6 random winning balls are drawn.
Buratino noticed that in each subsequent draw, there are no balls that appeare... | Solution. The probability that no numbers from the first draw will be repeated in the second draw is $a=\frac{39 \cdot 38 \cdot 37 \cdot \ldots \cdot 34}{45 \cdot 44 \cdot 43 \cdot \ldots 40}=0.40056 \ldots$, which is slightly more than 0.4. The probability that there will be no repetitions from the previous draw in bo... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. Winter Camp. In the winter camp, Vanya and Grisha live in a room. Every evening they draw lots to decide who will turn off the light before going to bed: the switch is near the door, and the loser has to go to bed in complete darkness, bumping into chairs.
Usually, Vanya and Grisha draw lots without any complicati... | Solution. a) Let's assume that heads in a coin toss give a one, and tails give a zero in the fractional part of a binary fraction. This results in some number $x$ represented by a binary fraction. For example, if the sequence of tosses starts with HTH, then the binary fraction is 0.101.
Obviously, $0 \leq x \leq 1$, a... | 2 | 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 |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | 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,
$$
\overline{x^{2}}-\bar{x}^{2}=\overline{y^{2}}-\bar{y}^{2}=\overline{z^{... | 800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Magic Pen (recommended for 8th grade, 1 point). Katya correctly solves a problem with a probability of $4 / 5$, while the magic pen correctly solves a problem without Katya's help with a probability of $1 / 2$. In the test, there are 20 problems, and to get a B, one needs to solve at least 13 of them correctly. How... | Solution. Let $x$ be the number of examples Katya solves herself, and $20-x$ be the number of examples solved by the pen. Then the expected number of correctly solved problems is
$$
\frac{4}{5} x+\frac{1}{2}(20-x)=0.3 x+10
$$
From the inequality $0.3 x+10 \geq 13$ we get that $x \geq 10$. Therefore, Katya needs to tr... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. Retro Collection (recommended from 8th grade, 2 points). Vitya collects toy cars from the "Retro" series. The problem is that the total number of different models in the series is unknown - it is the biggest commercial secret, but it is known that different cars are produced in the same edition, and therefore it ca... | Solution. Let there be a series of $n \geq 13$ cars. The probability that among the next $k$ offers there will only be the 12 models that Vitya has already seen is
$$
\left(\frac{12}{n}\right)^{k} \leq\left(\frac{12}{13}\right)^{k}
$$
We form the inequality: $\left(\frac{12}{13}\right)^{k}\log _{12 / 13} 0.01=\frac{\... | 58 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. New Year's Problem (recommended for 8th grade, 4 points). On the New Year's table, there are 4 glasses in a row: the first and third are filled with orange juice, while the second and fourth are empty. While waiting for guests, Vanya absent-mindedly and randomly pours the juice from one glass to another. In one mov... | Solution. We will encode full glasses with the digit 1 and empty ones with the digit 0. We will construct a graph of possible pourings (Fig. 4). This graph turns out to be the graph of an octahedron. From each state to any adjacent one, one can move with a probability of $1 / 4$, and each edge is "traversable" in both ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
15. Messengers (recommended for 9th grade, 3 points). Once, the beautiful Queen Guinevere, while staying at her parental castle, asked King Arthur to send her 20 pearls. The roads are not safe, and Arthur, just in case, decided to send 40 pearls, with different messengers, ordering them to ride on different roads. Band... | Solution. The probability of not saving at least 20 pearls if there are two messengers:
$$
\mathrm{P}_{2}=p^{2}
$$
The probability of not saving at least 20 pearls if there are three messengers:
$$
\mathrm{P}_{3}=p^{3}+2 p^{2}(1-p)=p^{2}(2-p) .
$$
The probability of not saving at least 20 pearls if there are four m... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. The Right Stars (from 8th grade, 3 points). Let natural numbers $k$ and $n$ be coprime, with $n \geq 5$ and $k < n / 2$. A proper $(n ; k)$-star is defined as a closed broken line that results from replacing every $k$ consecutive sides in a regular $n$-gon with a diagonal having the same endpoints. For example, the... | Solution. We will solve the problem in general for the $(n ; k)$-star. Let the vertices of the $n$-gon be denoted by $A_{0}, A_{1}$, and so on up to $A_{n-1}$. Take two consecutive segments of the star: $A_{0} A_{k}$ and $A_{1} A_{k+1}$. They intersect at a circle of radius $r_{1}$ with the center at the center of the ... | 48432 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12. Target (from 8th grade, 2 points). A target is hanging on the wall, consisting of five zones: a central circle (the bullseye) and four colored rings (see figure). The width of each ring is equal to the radius of the bullseye. It is known that the number of points for hitting each zone is inversely proportional to t... | Solution. Suppose, for definiteness, that the apple has a radius of 1. Then the blue zone is enclosed between circles with radii 3 and 4. The probability of hitting the blue zone relative to the probability of hitting the apple is the ratio of the areas of these zones:
$$
\frac{p_{g}}{p_{c}}=\frac{1}{4^{2}-3^{2}}=\fra... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18. Lonely Cars. (From 9th grade, 4 points.) On a very long narrow highway, where overtaking is impossible, $n$ cars are driving in a random order, each with its own preferred speed. If a fast car catches up to a slower one, the fast car has to slow down and drive at the same speed as the slower one. Thus, the cars for... | Solution. Let $I_{k}$ be the indicator of the event "the $k$-th car in line is alone." For $k \leq n$, this event consists of the slowest car among the first $k+1$ cars being the last, and the second slowest being the second to last. The probability of this is $\frac{1}{(k+1) k}$. If $k=n$, then this event consists of ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The smallest set. (From 6th grade, 2 points) In a numerical set, the median is 3, the arithmetic mean is 5, and the only mode of the set is 6. What is the smallest number of numbers that can be in a set with the given properties? | Solution. It is clear that the number 6 appears at least twice in the set, and in addition, there are at least two more numbers.
If the set contains exactly four numbers $a, b, 6, 6$, then we can assume that $a \leq b \leq 3$ and, moreover, the sum of all numbers is 20, so $a+b=20-6-6=8$. Contradiction. The set cannot... | 6 | Algebra | 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 |
9. Patrick and Slippers. Every day, the dog Patrick gnaws one slipper from the existing supply in the house. With a probability of 0.5, Patrick wants to gnaw a left slipper, and with a probability of 0.5 - a right slipper. If the desired slipper is not available, Patrick gets upset. How many pairs of identical slippers... | Solution. It is clear that if 7 pairs are bought, Patrick will definitely have enough of the desired, even if he chooses only left slippers every day. The question is about the smallest number of slippers that need to be bought so that with a probability of 0.8 or higher, Patrick will not be disappointed. Probability t... | 5 | 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 |
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 |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It i... | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then... | 3 | Logic and Puzzles | 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 |
13. There are fewer than 30 people in the class. The probability that a randomly chosen girl is an excellent student is $\frac{3}{13}$, and the probability that a randomly chosen boy is an excellent student is $\frac{4}{11} \cdot$ How many excellent students are there in the class? | # Solution.
According to classical probability theory, the probability that a randomly chosen girl is an excellent student is equal to the ratio of the number of excellent girl students to the total number of girls in the class. Accordingly,
$$
\frac{3}{13}=\frac{\text { number of girls-excellent students }}{\text { ... | 7 | Number Theory | 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 |
10. Disks (from 9th grade. 3 points). At a familiar factory, metal disks with a diameter of 1 m are cut out. It is known that a disk with a diameter of exactly 1 m weighs exactly 100 kg. During manufacturing, there is a measurement error, and therefore the standard deviation of the radius is 10 mm. Engineer Sidorov bel... | Solution. Given $\mathrm{E} R=0.5 \mathrm{m}, \mathrm{D} R=10^{-4}$ (sq.m). Let's find the expected value of the area of one disk:
$$
\mathrm{ES}=\mathrm{E}\left(\pi R^{2}\right)=\pi \mathrm{E} R^{2}=\pi\left(D R+\mathrm{E}^{2} R\right)=\pi\left(10^{-4}+0.25\right)=0.2501 \pi
$$
Thus, the expected value of the mass o... | 4 | Algebra | 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 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.