problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|
$7.327 \log _{\sqrt{x}}(x+12)=8 \log _{x+12} x, x$-integer. | ## Solution.
$\log _{x}(x+12)=\frac{4}{\log _{x}(x+12)} \Leftrightarrow \log _{x}(x+12)= \pm 2 \Leftrightarrow\left\{\begin{array}{l}x \geq 2, x-\text { integer, } \\ {\left[\begin{array}{l}x+12=x^{2} \\ x+12=\frac{1}{x^{2}}\end{array}\right.}\end{array} \Leftrightarrow\right.$ $\Leftrightarrow\left\{\begin{array}{l}x... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$7.333 \sqrt{\log _{2}\left(2 x^{2}\right) \cdot \log _{4}(16 x)}=\log _{4} x^{3}$. | ## Solution.
$$
\text { So, } \sqrt{\left(1+2 \log _{2} x\right)\left(2+\frac{1}{2} \log _{2} x\right)}=\frac{3}{2} \log _{2} x
$$
Let $y=\log _{2} x$, then $\sqrt{(2+4 y)(4+y)}=3 y \Leftrightarrow$ $\Leftrightarrow\left\{\begin{array}{l}y \geq 0, \\ (2+4 y)(y+4)=9 y^{2}\end{array} \Leftrightarrow\left\{\begin{array}... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$3.457 A=\operatorname{ctg}\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$
$3.457 A=\cot\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$ | Solution. Let $\alpha=\arccos \frac{3}{5}, \beta=\operatorname{arcctg}\left(-\frac{1}{2}\right) \Rightarrow \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$, $\operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=2, \operatorname{ctg} \beta=-\frac{1}{2}, \operatorname{ctg} 2 \beta=\frac{\operatorname{c... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.477 Find the minimum value of the expression
$$
A=\frac{\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha}{1+\sin \left(\frac{5 \pi}{2}-8 \alpha\right)} \text { for } 0<\alpha<\frac{\pi}{8}
$$ | ## Solution.
$\operatorname{ctg} 2 \alpha-\operatorname{tg} 2 \alpha=\frac{\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha}{\sin 2 \alpha \cdot \cos 2 \alpha}=\frac{2 \cos 4 \alpha}{\sin 4 \alpha} \Rightarrow A=\frac{2 \cos 4 \alpha}{\sin 4 \alpha \cdot(1+\cos 8 \alpha)}=$ $=\frac{2 \cos 4 \alpha}{2 \sin 4 \alpha \cdot \cos ^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.481 Find the minimum value of the expression
$A=\frac{\operatorname{ctg} \alpha-\operatorname{tg} \alpha}{\cos 4 \alpha+1}$ for $0<\alpha<\frac{\pi}{4}$. | ## Solution.
$\operatorname{ctg} \alpha-\operatorname{tg} \alpha=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\sin \alpha \cdot \cos \alpha}=\frac{2 \cos 2 \alpha}{\sin 2 \alpha} \Rightarrow A=\frac{2 \cos 2 \alpha}{\sin 2 \alpha \cdot 2 \cos ^{2} 2 \alpha}=\frac{2}{\sin 4 \alpha}$.
Since $0<\alpha<\frac{\pi}{4}$, the mi... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 3.485 Find the maximum value of the expression
$$
A=\frac{1}{\sin ^{6} \alpha+\cos ^{6} \alpha} \text { for } 0 \leq \alpha \leq \frac{\pi}{2}
$$ | ## Solution.
$\sin ^{6} \alpha+\cos ^{6} \alpha=\sin ^{4} \alpha-\sin ^{2} \alpha \cdot \cos ^{2} \alpha+\cos ^{4} \alpha=1-3 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=$ $=1-\frac{3}{4} \sin ^{2} 2 \alpha=\frac{1+3 \cos ^{2} 2 \alpha}{4} \Rightarrow A=\frac{4}{1+3 \cos ^{2} 2 \alpha}$.
From this, it is clear that $A$ t... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 3.486 Find the maximum value of the expression
$A=\frac{1}{\sin ^{4} \alpha+\cos ^{4} \alpha}$ for $0 \leq \alpha \leq \frac{\pi}{2}$. | Solution. $\sin ^{4} \alpha+\cos ^{4} \alpha=1-2 \sin ^{2} \alpha \cdot \cos ^{2} \alpha=1-\frac{\sin ^{2} 2 \alpha}{2}=\frac{1+\cos ^{2} 2 \alpha}{2} \Rightarrow$
$\Rightarrow A=\frac{2}{1+\cos ^{2} 2 \alpha}$. From this, it is clear that $A$ takes its maximum value when $1+\cos ^{2} 2 \alpha$ takes its minimum value,... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 9.303 Find integer values of $x$ that satisfy the inequality
$$
\log _{0.3}(\sqrt{x+5}-x+1)>0
$$ | Solution. The inequality is equivalent to the system:
$\left\{\begin{array}{l}\sqrt{x+5}-x+10, \\ x-\text { integer }\end{array} \Leftrightarrow\left\{\begin{array}{l}\sqrt{x+5}x-1, \\ x-\text { integer. }\end{array}\right.\right.$
From this, $x>0$ and, consequently, $x \geq 1$ ( $x$ - integer). Therefore,
$\left\{\... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
10.361 In triangle $A B C$, the measure of angle $A$ is twice the measure of angle $B$, and the lengths of the sides opposite these angles are 12 cm and 8 cm, respectively. Find the length of the third side of the triangle. | Solution.
A
Let $A K$ be the bisector of $\angle A$. Then
$\angle K A C = \angle A B C \Rightarrow \triangle A K C \sim \triangle A B C \Rightarrow$
$\Rightarrow \frac{A C}{B C} = \frac{K C}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.368 Calculate the length of the angle bisector of angle $A$ in triangle $A B C$ with side lengths $a=18$ cm, $b=15$ cm, $c=12$ cm. | Solution. Let $A K$ be the bisector of $\angle A$ of the given $\triangle A B C$. By Theorem 4, $\frac{C K}{B K}=\frac{15}{12} \Rightarrow \frac{18-B K}{B K}=\frac{15}{12} \Rightarrow B K=8, C K=10$.
Applying the Law of Cosines to $\triangle A B K$ and $\triangle A C K$, we have:
$\left\{\begin{array}{l}8^{2}=12^{2}+... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.369 In a triangle with a perimeter of 20 cm, a circle is inscribed. The segment of the tangent, drawn parallel to the base and enclosed between the sides of the triangle, measures 2.4 cm. Find the base of the triangle. | ## Solution.
Answer: 4 cm or 6 cm. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.373 In a triangle with sides 6, 10, and 12 cm, a circle is inscribed. A tangent is drawn to the circle such that it intersects the two larger sides. Find the perimeter of the cut-off triangle. | ## Solution.
Let $M N$ be the tangent to the circle inscribed in the given $\triangle A B C, A B=10, B C=12, A C=6$.
Then $B M+B N+M N=$ $=(10-A M)+(12-C N)+(A M+C N-6)=16$ (by theorem $2, ... | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.377 The bisector of a triangle's angle divides the opposite side into segments of length 4 and 2 cm, and the height drawn to the same side is $\sqrt{15}$ cm. What are the lengths of the sides of the triangle, given that they are expressed as integers? | Solution. Let the bisector of angle $A$ divide side $B C$ of the given $\triangle A B C$ into segments 2 and 4. Suppose $A Bh=\sqrt{15}(h$ - the height of $\triangle A B C) \Rightarrow \sqrt{15}<n<6 \Rightarrow 4 \leq n \leq 5$.
If $n=5$, then by Heron's formula $S_{A B C}=\sqrt{\frac{21}{2} \cdot \frac{11}{2} \cdot \... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
### 10.384 In triangle $A B C$, the bisectors $A D$ and $C E$ intersect at
point $F$. Points $B, D, E, F$ lie on the same circle. Show that angle $B$ is $60^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \angle E F D=180^{\circ}-(\angle F A C+\angle F C A)= \\
& =180^{\circ}-\frac{1}{2}(\angle A+\angle C) \Rightarrow 2 \angle B=\angle A+\angle C \Rightarro... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
10.391 On the segment $A C$ there is a point $B$, and $A B=14 \text{~cm}, B C=28 \text{~cm}$. On the segments $A B, B C$ and $A C$ as diameters, semicircles are constructed in the same half-plane relative to the boundary $\boldsymbol{A} \boldsymbol{B}$. Find the radius of the circle that is tangent to all three semicir... | ## Solution.
Let $O_{1}, O_{2}, O_{3}$ be the centers of semicircles with radii $r_{1}, r_{2}, r_{3}$ and diameters $A B, B C$, and $A C$ respectively; $P$ be the center of the circle with... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.403 In a right triangle $A B C$ ( $\angle C=90^{\circ}$ ), the altitude $CD$ is drawn. The radii of the circles inscribed in triangles $ACD$ and $BCD$ are 0.6 and 0.8 cm, respectively. Find the radius of the circle inscribed in triangle $\boldsymbol{A} \boldsymbol{\text { B }}$. | Solution. Let $x, r_{1}, r_{2}$ be the radii of the circles inscribed in $\triangle A B C$, $\triangle A C D, \triangle B C D$ respectively. $\triangle A C D \sim \triangle B C D \Rightarrow \frac{A C}{B C}=\frac{r_{1}}{r_{2}}=\frac{0.6}{0.8}$ (the radii of the circles inscribed in similar triangles are proportional to... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
### 10.417
In a right triangle $ABC\left(\angle C=90^{\circ}\right)$, a circle is inscribed, touching its sides at points $A_{1}, B_{1}, C_{1}$. Find the ratio of the area of triangle $ABC$ to the area of triangle $A_{1} B_{1} C_{1}$, if $AC=4 \text{ cm}, BC=3 \text{ cm}$. | ## Solution.
Let $O$ be the center of the circle with radius $r$ inscribed in the given $\triangle ABC$.
Then $AB=5, r=\frac{3+4-5}{2}=1$ (Theorem 1).
$S_{A_{1} B_{1} C_{1}}=S_{O A_{1} B_{1}}... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.198 The base of a right prism is a triangle with sides 6, 8, and 10 cm. A certain plane section of this prism cuts off segments of 12 cm each from the lateral edges passing through the vertices of the larger and medium angles of the base, and a segment of 18 cm from the edge passing through the vertex of the smalles... | Solution.
Let $A B C A_{1} B_{1} C_{1}$ be the given right prism with base $A B C, A C=6, B C=8, A B=10$. Let $L M N$ be the given section, $A M=C N=12, B L=18$.
By the condition, $A B^{2}... | 336 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.372 A consignment shop has accepted for sale cameras, watches, fountain pens, and receivers for a total of 240 rubles. The sum of the prices of a receiver and a watch is 4 rubles more than the sum of the prices of a camera and a fountain pen, and the sum of the prices of a watch and a fountain pen is 24 rubles less ... | Solution. Let $x$ be the price of a watch, $y$ be the price of a pen, $z$ be the price of a receiver, $t$ be the price of a camera, and $n$ be the number of cameras. Then, according to the problem, we have:
$\left\{\begin{array}{l}t=10 n, \\ y \leq 6, \\ z+x=y+t+4, \\ 3 n y+n x+n z+n t=240, \\ n, y \text { - integers ... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.373 A computing machine was given the task to solve several problems sequentially. Registering the time spent on the assignment, it was noticed that the machine spent the same multiple of time less on solving each subsequent problem compared to the previous one. How many problems were proposed and how much time did ... | Solution. Let $n$ be the number of problems, $b$ be the time to solve the first problem, and $q$ be the common ratio of the geometric progression that represents the times to complete the problems. Using the formula for the sum of $k$ terms of a geometric progression, we have:
$$
\begin{aligned}
& \left\{\begin{array}... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.374 Three candles have the same length but different thicknesses. The first candle was lit 1 hour earlier than the other two, which were lit simultaneously. At some point, the first and third candles had the same length, and 2 hours after this, the first and second candles had the same length. How many hours does it... | Solution. Let $x$ be the burning time of the first candle, and $l$ be the length of the candles. Then $\frac{l}{x}, \frac{l}{12}, \frac{l}{8}$ are the burning rates of the candles (i.e., the rates at which the lengths of the candles decrease). According to the problem, after $\frac{\frac{l}{x} \cdot 1}{\frac{l}{8}-\fra... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.376 The desired three-digit number ends with the digit 1. If it is erased and then written as the first digit of the number, the new three-digit number obtained will be less than the desired one by $10 a^{\log _{\sqrt{a}} 3}$. Find this number. | Solution. Let $\overline{x y 1}$ be the desired number. Then
$$
\begin{aligned}
& \overline{x y 1}-\overline{1 x y}=10 a^{\log _{\sqrt{a}} 3} \Leftrightarrow(100 x+10 y+1)-(100+10 x+y)=90 \Rightarrow \\
& \Rightarrow 10 x+y=21 \Leftrightarrow \overline{x y}=21 .
\end{aligned}
$$
Answer: 211 . | 211 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.377 The difference of the logarithms of the hundreds and tens digits of a three-digit number is equal to the logarithm of the difference of the same digits, and the sum of the logarithms of the hundreds and tens is equal to the logarithm of the sum of the same digits, increased by 4/3 times. If this three-digit numb... | Solution. Let $\overline{x y z}$ be the desired number. According to the condition, $\log _{a} x-\log _{a} y=\log _{a}(x-y) \Rightarrow x=y(x-y) \Rightarrow x: y \Rightarrow x=n y, n-$ integer. From here $n y=y(n y-y) \Rightarrow n=y(n-1) \Rightarrow n:(n-1)$
But $n$ and $n-1$ are coprime, hence $n-1=1, y=2, x=4$.
Ac... | 421 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.383 Several workers complete a job in 14 days. If there were 4 more people and each worked 1 hour longer per day, then the same job would be done in 10 days. If there were still 6 more people and each worked 1 hour longer per day, then this
job would be completed in 7 days. How many workers were there and how many h... | Solution. Let $x$ be the number of workers, $y$ be the duration (in hours) of the workday, and $t$ be the productivity of one worker per hour. Then
$$
\begin{aligned}
& 14 x y t=10(x+4)(y+1) t=7(x+10)(y+2) t \Leftrightarrow \text { (the volume of work remains constant) } \\
& \left\{\begin{array} { l }
{ 1 4 x y = 1 ... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.384 Five people perform a certain job. The first, second, and third, working together, can complete the entire job in 327.5 hours; the first, third, and fifth together - in 5 hours; the first, third, and fourth together - in 6 hours; and the second, fourth, and fifth together - in 4 hours. In what time interval will... | Solution. Let $x_{i}$ be the productivity of the $i$-th worker, $y$ be the volume of work. Then, according to the problem,
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=\frac{y}{7.5} \\
x_{1}+x_{3}+x_{5}=\frac{y}{5} \\
x_{1}+x_{3}+x_{4}=\frac{y}{6} \\
x_{2}+x_{4}+x_{5}=\frac{y}{4}
\end{array}\right.
$$
Multiplying the... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.389 From two pieces of alloy of the same mass but with different percentage content of copper, pieces of equal mass were cut off. Each of the cut pieces was melted with the remainder of the other piece, after which the percentage content of copper in both pieces became the same. How many times smaller is the cut pie... | Solution. Let $y$ be the mass of each alloy piece, $x$ be the mass of the cut-off piece, $p$ and $q$ be the concentration of copper in the first and second pieces of the alloy. Then $p x+q(y-x)$ and $q x+p(y-x)$ are the amounts of copper in the new alloys, respectively. According to the problem,
$$
\frac{p x+q(y-x)}{y... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.391 Two cars start simultaneously from points A and B and meet at 12 o'clock noon. If the speed of the first car is doubled while the speed of the second car remains the same, the meeting will occur 56 minutes earlier. If, however, the speed of the second car is doubled while the speed of the first car remains the s... | Solution. Let $x$ and $y$ be the initial speeds (km/min) of the cars, and $2z$ be the time (in minutes) after which they met after leaving from A and B. Then $2z(x+y)$ is the distance between A and B, hence
\[
\left\{
\begin{array}{l}
2z(x+y) = (2z-56)(2x+y) \\
2z(x+y) = (2z-65)(x+2y)
\end{array}
\right.
\]
\[
\left\{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.394 Two brothers had tickets to a stadium located 10 km from their home. At first, they planned to walk to the stadium, but they changed their mind and decided to use a bicycle, agreeing that one would go by bicycle while the other would walk at the same time. After covering part of the distance, the first brother w... | Solution. Since the brothers arrived at the stadium simultaneously (i.e., spent the same amount of time on the journey), they walked (and cycled) the same distance. But in total, they walked 10 km, so each of them walked 5 km. Therefore, the time gain will be $5 \cdot 12=60$ minutes.
Answer: 1 hour. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.395 A sportsman, walking along a highway, noticed that every 6 minutes a trolleybus catches up with him and every 3 minutes a trolleybus passes him in the opposite direction. Find the intervals at which trolleybuses depart from the terminal points and how many times slower the sportsman was walking compared to the t... | Solution. Let $x$ and $y$ be the speeds of the athlete and the trolleybuses, and $z$ be the interval of the trolleybuses' movement. Then $y z$ is the distance between the trolleybuses. According to the problem, $\left\{\begin{array}{l}y z=3(x+y), \\ y z=6(y-x),\end{array} \Rightarrow x=\frac{y}{3} ; z=4\right.$.
Answe... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.398 Two friends decided to go hunting. One of them lives 46 km from the hunting base, the other, who has a car, lives 30 km from the base between the base and his friend's house. They set off at the same time, with the car owner driving towards his friend who was walking. Upon meeting, they drove together to the bas... | Solution. Let $x$ and $y$ be the speeds (km/h) of the pedestrian and the car. Then $y \cdot 1$ is the total distance the car has traveled, from which $\frac{y-30}{2}$ is the distance the car traveled before meeting the pedestrian, $16-\frac{y-30}{2}=\frac{62-y}{2}$ is the distance the pedestrian traveled before the mee... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.399 The train was delayed at the departure station for 1 hour 42 minutes. Upon receiving the departure signal, the driver followed this schedule: on the section constituting 0.9 of the entire route from the departure station to the destination station, he maintained a speed 20% higher than the usual speed, and on 0.... | Solution. Let $z$ be the distance between stations, $x$ be the usual speed of the train.
$$
\begin{aligned}
& \text { Then } \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}-\text { is the travel time of the train, hence } \\
& \frac{0.9 z}{1.2 x}+\frac{0.1 z}{1.25 x}=\frac{z}{x}-1 \frac{42}{60} \Rightarrow \frac{z}{x}=10 .
\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.400 On the highway, points D, A, C, and B are arranged sequentially. A motorcyclist and a cyclist set off from A and B simultaneously, heading to C and D, respectively. Meeting at E, they exchanged vehicles and each continued their journey. As a result, the first one spent 6 hours on the trip from A to C, while the ... | Solution. Let $x=DA, y=AE, z=EC, u=CB$. Then $\frac{y}{60}+\frac{z}{25}$ is the total travel time of the first, $\frac{z+u}{25}+\frac{x+y}{60}$ is the total travel time of the second, $\frac{y}{60}=\frac{z+u}{25}$ is the time of their travel until they meet. In the end, we have:
$\frac{y}{60}+\frac{z}{25}=6$
$\left\{... | 340 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.402 The capacities of three vessels $A$, B, C, each of which has the shape of a cube, are in the ratio $1: 8: 27$, and the volumes of water poured into them are in the ratio 1:2:3. After transferring part of the water from vessel A to vessel B and from vessel B to vessel C, the water level in all three vessels becam... | Solution. Let $x$ be the initial volume of water in vessel $\mathrm{A}$, and $a$ be the linear size of vessel A. According to the problem, $2x$ and $3x$ are the volumes of water in B and C, and $2a$ and $3a$ are the linear sizes of B and C. Since after the first transfer, the water level in all vessels became the same,... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.405 Two pedestrians set out simultaneously from A to B and from B to A. When the first had walked half the distance, the second had 24 km left to walk to the end, and when the second had walked half the distance, the first had 15 km left to walk to the end. How many kilometers will the second pedestrian have left to... | Solution. Let $x$ be the distance between A and B, $y$ be the ratio of the speeds of the first and second pedestrians. Then
$\left(\frac{x}{2}\right):(x-24)=y,(x-15):\left(\frac{x}{2}\right)=y$, from which $\frac{x}{2(x-24)}=\frac{2(x-15)}{x} \Rightarrow x=40$ (the value $x=12$ is not suitable, as $x>24) \Rightarrow y=... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.406 Three motorcyclists travel the same section AB of the road at constant but different speeds. First, the first motorcyclist passed point A, and 5 s later, in the same direction, the second and third motorcyclists. After some time, the third motorcyclist overtook the first, and 10 s later, the second overtook him.... | Solution. Let $z$ be the distance between A and B, $x$ be the time (sec) it takes for the first motorcyclist to cover this distance. Then $\frac{z}{x}, \frac{z}{60}, \frac{z}{40}$ are the speeds of the motorcyclists, $\frac{5 z}{x}$ is the difference in distances between them at the moment the second and third pass poi... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.408 A material particle $m_{1}$ oscillates between points $A$ and $B$, which are 3.01 m apart. The particle's speed is constant, and it does not stop at the endpoints. After 11 s from the departure of particle $m_{1}$ from point $A$, another particle $m_{2}$ starts moving from point $B$ with a constant but lower spe... | Solution. Let $x$ and $y$ be the speeds (cm/s) of particles $m_{1}$ and $m_{2}$. By the time particle $m_{2}$ exits, the distance between $m_{1}$ and $m_{2}$ is $301-11 x$ cm, hence $301-11 x=10(x+y)$. Let $C$ be the point of the first meeting of the particles. Then $B C=10 y$, and therefore, by the time of the second ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.410 Along the sides of a right angle, towards the vertex, two spheres with radii of 2 and 3 cm are moving, with the centers of these spheres moving along the sides of the angle at unequal but constant speeds. At a certain moment, the center of the smaller sphere is 6 cm from the vertex, and the center of the larger ... | Solution. Let $x$ and $y$ be the speeds of the balls. After 1 second, the distances from the centers of the balls to the corner are $6-x$ and $16-y$, and after 3 seconds, they are $6-3x$ and $16-3y$. By the Pythagorean theorem, we have: $\left\{\begin{array}{l}(6-x)^{2}+(16-y)^{2}=13^{2}, \\ (6-3 x)^{2}+(16-3 y)^{2}=5^... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13.415 Two cyclists set out simultaneously from points $A$ and $B$ towards each other. Four hours after their meeting, the cyclist who had set out from $A$ arrived at $B$, and nine hours after their meeting, the cyclist who had set out from $B$ arrived at $A$. How many hours was each cyclist on the road? | Solution. Let $x$ and $y$ be the travel times of the first and second cyclists, respectively. Then $x-4=y-9$ is the time each of them traveled until they met. Since each of them traveled a distance after the meeting equal to the distance the other had traveled before the meeting, $\frac{4}{y-9}=\frac{x-4}{9}$.
$$
\tex... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.419 Two trains, 490 m and 210 m long, are moving towards each other at a constant speed on parallel tracks. The driver of one of them noticed the oncoming train at a distance of $700 \mathrm{~m}$; after this, the trains met after 28 s. Determine the speed of each train, given that the first train takes 35 s longer t... | Solution. Let $x$ and $y$ be the speeds (m/s) of the trains. According to the first condition, $700=28(x+y)$. According to the second condition of the problem, $\frac{490}{x}-\frac{210}{y}=35$.
In the end, we have: $\left\{\begin{array}{l}x+y=25, \\ \frac{14}{x}-\frac{6}{y}=1\end{array} \Rightarrow\left\{\begin{array}... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.421 If a two-digit number is divided by a certain integer, the quotient is 3 and the remainder is 8. If the digits in the dividend are swapped while the divisor remains the same, the quotient becomes 2 and the remainder is 5. Find the original value of the dividend. | Solution. Let $\overline{x y}$ be the desired number, $z$ be the divisor. Then, according to the condition,
\[
\left\{\begin{array}{l}
10 x+y=3 z+8 \\
10 y+x=2 z+5
\end{array}\right.
\]
By eliminating $z$ from these equations, we have: $17 x-28 y=1 \Leftrightarrow 28(x-y)=1+11 x$.
But $12 \leq 1+11 x \leq 100$. In t... | 53 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.423 In a brigade of earthworkers, each works the same number of hours daily. It is known that the labor productivity is the same for all workers in the brigade, and the brigade can dig a trench for laying a cable in 6 days. However, before the work began, it was found that the working day was reduced by 1 hour, and ... | Solution. Let $x$ be the original number of workers, $y+2$ be the originally planned duration of their working day. Then the volume of work, according to the first condition of the problem, is $6 x(y+2)$, according to the second condition - $9(x-5)(y+1)$; according to the third condition - $12(x-7) y$. Since the volume... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.426 Two pumps were installed to fill a swimming pool with water. The first pump can fill the pool 8 hours faster than the second one. Initially, only the second pump was turned on for a time equal to twice the amount of time it would take to fill the pool if both pumps were working simultaneously. Then, the first pu... | Solution. Let $x$ and $y$ be the pump efficiencies, and $z$ be the volume of work. Then $\frac{z}{x+y}$ is the time it takes to fill the pool when both pumps are working simultaneously. According to the conditions of the problem,
$\left\{\begin{array}{l}\frac{z}{y}-\frac{z}{x}=8, \\ \frac{2 z}{x+y} \cdot y+1.5(x+y)=z\e... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.427 After passing through the filter, the liquid is evenly poured into a 40-bucket barrel and can be drained through a tap at the bottom of the barrel. If this tap is open, the inflow and outflow of the liquid are such that every 4 minutes, one bucket is reduced in the barrel. How long will it take for the filtered ... | Solution. Let $x$ and $y$ be the rates (bucket/min) of liquid passing through the filter and the tap, respectively. Then, according to the conditions of the problem,
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 4 ( y - x ) = 1 , } \\
{ \frac { 6 6 } { y } - \frac { 4 0 } { x } = 3 }
\end{array} \Rightarrow \left... | 96 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.428 A batch of identical parts was processed on three machines of different designs in the following sequence: first, only the first machine worked for as many hours as it would take for the second and third machines to complete the entire job together; then, only the second machine worked for as many hours as it wo... | Solution. Let $x, y, z$ be the efficiencies of the machines, and $V$ be the volume of work. According to the problem, $\frac{V}{y+z}$ is the time taken by the first machine, $\frac{V}{x+z}$ is the time taken by the second machine, and $\frac{V}{x+y}$ is the time taken by the third machine.
From this, $\frac{V x}{y+z}+... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.430 A passenger can travel from Moscow to city $N$ by train. In this case, he will be on the way for 20 hours. If, however, he waits for the departure of the plane (and he will have to wait more than 5 hours after the train departs), the passenger will reach city $N$ in 10 hours, including the waiting time. How many... | Solution. Let $x$ and $y$ be the speeds (km/h) of the train and the airplane, respectively, and $z$ be the waiting time for the airplane to depart. Then $\frac{8}{9} y$ is the distance the airplane will fly until it meets the train. According to the problem, $\frac{8}{9} y = x \left(z + \frac{8}{9}\right)$. Additionall... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.435 The desired number is greater than 400 and less than 500. Find it, if the sum of its digits is 9 and it is equal to 47/36 of the number represented by the same digits but written in reverse order. | Solution. The condition that the desired number is within the range from 400 to 500 makes the problem trivial. Therefore, let's solve this problem by replacing this redundant condition with the condition that the desired number is a three-digit number.
Let $\overline{x y z}$ be the desired number. According to the con... | 423 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.437 What whole positive number should 180 be divided by, so that the remainder is $25\%$ of the quotient? | Solution. Let $n$ be the required number, $m$ be the remainder of the division of 180 by $n$, and $k$ be the quotient.
Then $\left\{\begin{array}{l}180=n \cdot k+m, \\ 4 m=k, \\ m4 m n>4 m^{2}$, i.e., $m^{2}<45 \Rightarrow m=4 \Rightarrow 4 n+1=45 \Rightarrow n=11$.
Answer: 11. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13.438 By mixing $2 \mathrm{~cm}^{3}$ of three substances, 16 g of the mixture was obtained. It is known that $4 \mathrm{r}$ of the second substance occupies a volume that is $0.5 \mathrm{~cm}^{3}$ larger than $4 \mathrm{r}$ of the third substance. Find the density of the third substance, given that the mass of the sec... | Solution. Let $x, y, z$ be the densities of three substances. Then, according to the condition $\left\{\begin{array}{l}2 x+2 y+2 z=16, \\ \frac{4}{y}-\frac{4}{z}=0.5, \quad \Rightarrow z=4 . \\ y=2 x\end{array}\right.$
Answer: 4 g $/ \mathrm{cm}^{3}$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.442 In the store, goods of the first and second grades arrived for a total amount of 450 rubles. An inspection determined that all the goods could only be sold at the price of the second grade, as a result of which the company would incur a loss of 50 rubles. The company eliminated the defects in the first-grade goo... | Solution. Let $x$ and $y$ be the cost of the first and second grade goods, respectively; $a$ and $b$ be the quantity of the first and second grade goods. Then $\frac{x}{a}, \frac{y}{b}$ are the cost per unit of the first and second grade goods, respectively.
According to the conditions of the problem, we have:
$$
\le... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.448 Two trains departed from station A with an interval of 12 min and almost immediately developed the same speed of 50 km/h. They are traveling in the same direction without stopping, maintaining the indicated speed unchanged. At what speed was the oncoming train traveling if it met these trains 5 min apart? | Solution. Let $x$ be the speed of the oncoming train. The distance between the other two trains is $50 \cdot \frac{1}{5}=10$ km. Therefore, the oncoming train met the second of these two trains after $\frac{10}{x+50}$ hours from meeting the first one.
$$
\text { According to the condition } \frac{10}{x+50}=\frac{1}{12... | 70 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.449 The desired three-digit number starts with the digit 1. If it is erased and then written as the last digit of the number, the new three-digit number obtained will be greater than the desired number by $9 a^{1 / \lg a}$. Find this number. | Solution. Let $\overline{1 x y}$ be the desired number. According to the condition
$$
\begin{aligned}
& \overline{1 x y}=\overline{x y 1}-9 \cdot a^{\log _{a} 10}=\overline{x y 1}-90 \text { or } 100+10 x+y=100 x+10 y+1-90 \Leftrightarrow \\
& \Leftrightarrow 21=10 x+y \Rightarrow x=2, y=1
\end{aligned}
$$
Answer: 12... | 121 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13.450 If at the beginning of the time measurement there were $m_{0}$ g of substance $A$ and $2 m_{0}$ g of substance $B$, then after any number $t$ years, as a result of the radioactive decay of these substances, there will remain respectively $\boldsymbol{m}=\boldsymbol{m}_{0} \cdot 2^{-\lambda_{1} t}$ and $M=2 m_{0}... | Solution. Let $x$ and $2x$ be the half-lives of the second and first substances, respectively. According to the conditions,
$$
\left\{\begin{array}{l}
m_{0} \cdot 2^{-\lambda_{1} \cdot 2x} = \frac{m_{0}}{2}, \\
2m_{0} \cdot 2^{-\lambda_{2} \cdot x} = m_{0}, \\
m_{0} \cdot 2^{-\lambda_{1} \cdot 20} + 2m_{0} \cdot 2^{-\... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
### 5.080 Find the largest term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$. | Solution. Let $T_{k}$ be the $k$-th term in the expansion of $(\sqrt{5}+\sqrt{2})^{20}$.
Then $\frac{T_{k+1}}{T_{k}}=\frac{C_{20}^{k} \cdot(\sqrt{5})^{20-k} \cdot(\sqrt{2})^{k}}{C_{20}^{k-1} \cdot(\sqrt{5})^{21-k} \cdot(\sqrt{2})^{k-1}}=\frac{\sqrt{2}}{\sqrt{5}} \cdot \frac{21-k}{k}$.
The ratio $\frac{T_{k+1}}{T_{k}}... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.082 If we expand all the brackets in the expression $(1+x)^{9}+(1+x)^{10}+\ldots$ $\ldots+(1+x)^{14}$ and combine like terms, we will get some polynomial. Determine the coefficient of $x^{9}$ in this polynomial without expanding the brackets. | Solution. The coefficient of $x^{9}$ is $A=C_{9}^{9}+C_{10}^{9}+C_{11}^{9}+C_{12}^{9}+C_{13}^{9}+C_{14}^{9}$. According to the condition of problem 5.076,
\[
\begin{aligned}
& A+C_{10}^{10}=1+C_{10}^{9}+C_{11}^{9}+\ldots+C_{14}^{9}+C_{10}^{10}=1+\left(C_{10}^{9}+C_{10}^{10}\right)+C_{11}^{9}+\ldots+C_{14}^{9}= \\
& =1... | 3003 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.089 "Early in the morning, smiling Igor was dashing barefoot to go fishing." How many different meaningful sentences can be formed using part of the words from this sentence without changing their order of succession?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | Solution. Let's break down the original sentence into phrases and number them with Roman numerals, and the words in these phrases with Arabic numerals:
"early in the morning" - I, "to go fishing" - II, "smiling Igor" - III, "ran barefoot" - IV.
In each meaningful sentence composed of phrases I-IV, either III or III. ... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7,8,9 |
| | Product Rule | |
Let $K(x)$ be the number of irreducible fractions $a / b$ such that $a<x$ and $b<x$ (where $a$ and $b$ are natural numbers).
For example, $K(5 / 2)=3$ (fractions $1, 2, 1 / 2$).
Calculate the sum $K(100)+K\left(100 / 2\right)+K\left(100 / 3\right)+\ldots+K\left(100 / 99\right)+K\left(1... | The irreducible fraction $a / b$ is counted several times in our sum. Indeed, it is counted once in all $K\left({ }^{100} /\right)$ where $l a<100$ and $l b<100$. Thus, we obtain that the desired sum is equal to the number of all fractions $a / b$ (including reducible ones) or, equivalently, pairs $(a, b)$ that satisfy... | 9801 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
A 9×9 square is divided into 81 unit cells. Some cells are shaded, and the distance between the centers of any two shaded cells is greater than 2.
a) Provide an example of a coloring where there are 17 shaded cells.
b) Prove that there cannot be more than 17 shaded cells. | b) Suppose it was possible to color 18 cells. Divide the $9 \times 9$ square into nine $3 \times 3$ squares. Label the cells of the square with horizontal coordinates as letters from $a$ to $i$, and vertical coordinates as numbers from 1 to 9. Denote the $3 \times 3$ squares with Greek letters from $\alpha$ to $\iota$,... | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Konagin S.
Find $x_{1000}$, if $x_{1}=4, x_{2}=6$, and for any natural $n \geq 3$, $x_{n}$ is the smallest composite number greater than $2 x_{n-1}-x_{n-2}$. | We will prove by induction that $x_{n}=1 / 2 n(n+3)$.
Base. For $n=3,4$, the formula is correct: $2 x_{2}-x_{1}=8$, so $x_{3}=9$; $2 x_{3}-x_{2}=12$, so $x_{4}=14$.
Induction step. $2 x_{n}-x_{n-1}=2 \cdot 1 / 2 n(n+3)-1 / 2(n-1)(n+2)=1 / 2(n+1)(n+4)-1$. By the condition, $x_{n+1}$ is the first composite number great... | 501500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Senderovv B.A.
Find the smallest natural number that cannot be represented in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are natural numbers. | $1=\frac{4-2}{4-2}, \quad 2=\frac{8-4}{4-2}, 3=\frac{8-2}{4-2}, 4=\frac{16-8}{4-2}, 5=\frac{32-2}{8-2}$
$6=\frac{16-4}{4-2}, 7=\frac{16-2}{4-2}, 8=\frac{32-16}{4-2}, 9=\frac{128-2}{16-2}, 10=\frac{64-4}{8-2}$.
Assume that $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$. Without loss of generality, let $a>b, c>d$. Denote $m=a-b,... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
M. Murraikin
In a racing tournament, there are 12 stages and $n$ participants. After each stage, all participants receive points $a_{k}$ depending on the place $k$ they occupy (the numbers $a_{k}$ are natural, and $a_{1}>a_{2}>\ldots>a_{n}$). For what smallest $n$ can the tournament organizer choose the numbers $a_{1}... | Evaluation. Suppose there are no more than 12 participants.
Let one of the participants (call him $A$) win all 11 stages, and each of the remaining participants at least once took the last place.
Then participant $A$ after 12 stages will score no less than $11 a_{1} + a_{n}$ points, while each of the remaining partic... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Solve the system in positive numbers:
$$
\begin{cases}x^{y} & =z \\ y^{z} & =x \\ z^{x} & =y\end{cases}
$$ | First, note that if one of the unknowns is equal to one, then the others are also equal to one.
Indeed, let $x=1$. Then $z=1^{\mathrm{y}}=1, y=z^{\mathrm{x}}=1^{1}=1$. Suppose there exists another solution besides $(1,1,1)$. Let's first consider $x>1$. Then $z=x^{\mathrm{y}}>1, y=z^{\mathrm{x}}>1$. Therefore, $z=x^{\m... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Numerical tables and their properties ]
In table $A$ of size $10 \times 10$, some numbers are written. Let the sum of all numbers in the first row be denoted by $s_{1}$, in the second row by $s_{2}$, and so on. Similarly, the sum of numbers in the first column is denoted by $t_{1}$, in the second column by $t_{2}$, ... | Example of table $A$, for which the sum of all numbers is 955.
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 91 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 92 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 93 |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 94 |
| 0 | 0 | 0 ... | 955 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Let's perform the following operation: between any two adjacent numbers, insert the number that results from subtracting the left number from the right one. We will perform the same operation on the new row, and so on. Find the sum of the numbers in the row that results after one hundred such operations.
# | Let's see how the sum of the numbers in a row changes after one operation. Let $a_{1}, a_{2}, \ldots, a_{\mathrm{n}}$ be the row to which the operation is applied. Then the new row has the form $a_{1}, a_{2}-a_{1}, a_{2}, a_{3}-a_{2}, \ldots, a_{\mathrm{n}-1}, a_{\mathrm{n}}-a_{\mathrm{n}-1}, a_{\mathrm{n}}$. The sum o... | 726 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$\left[\begin{array}{l}\text { Equations in integers } \\ {[\text { Case enumeration }}\end{array}\right]$
Let $S(x)$ be the sum of the digits of a natural number $x$. Solve the equation $x+S(x)=2001$. | If $x>1999$, then $x+S(x) \geq 2002$, if $x<1000$, then $x+S(x)<1000+3 \cdot 9=1027$. Therefore, $x-$ is a four-digit number with the first digit 1.
Let $a$ be the hundreds digit, $b$ be the tens digit, and $c$ be the units digit, then $(1000+100a+10b+c)+1=2001$, or $101a+11b+2c=1000$. If $a<9$ the last sum is less th... | 1977 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
In a bank, there are 2002 employees. All employees came to the anniversary and were seated at one round table. It is known that the salaries of those sitting next to each other differ by 2 or 3 dollars. What is the greatest possible difference between two salaries of the employees of this bank, given th... | Between the employees with the highest and lowest salaries, there are no more than 1000 people. Therefore, their salaries differ by no more than 3003 dollars. A difference of exactly 3003 dollars is not possible: in this case, there would be 1000 people on each side of the table, and the salaries of any two adjacent pe... | 3002 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Zhendarov R.G.
On the board, the product $a_{1} a_{2} \ldots a_{100}$ is written, where $a_{1}, \ldots, a_{100}$ are natural numbers. Consider 99 expressions, each of which is obtained by replacing one of the multiplication signs with an addition sign. It is known that the values of exactly 32 of these expressions are... | Consider the leftmost even number $a_{i}$ and the rightmost even number $a_{k}$. Note that the sums with indices from $i$ to $k-1$ are even and only they are (in sums with smaller indices, the first addend is odd and the second is even; in sums with larger indices, it is the opposite).
Thus, $k-i=32$. The number of ev... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
| |
a) The "Tower of Hanoi" puzzle consists of eight disks of decreasing size placed on one of three pegs. The goal is to move the entire tower to another peg, moving only one disk at a time and never placing a larger disk on a smaller one. Prove that the puzzle has a solution.
What method would be optimal (in terms... | a) Let the minimum number of steps to move a tower of $n$ disks be $K_{n}$. Express $K_{n+1}$ in terms of $K_{n}$. Consider a tower of $n+1$ disks. We can divide the entire procedure into three stages.
Stage 1 - from the beginning to the first move of the bottom disk. By the end of this stage, the tower of $n$ disks s... | 4373 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
[Methods for solving problems with parameters]
[Investigation of a quadratic trinomial]
For what positive value of $p$ do the equations $3 x^{2}-4 p x+9=0$ and $x^{2}-2 p x+5=0$ have a common root? | The common root of the given equations must also be a root of the equation $\left(3 x^{2}-4 p x+9\right)-3\left(x^{2}-2 p x+5\right)=0 \Leftrightarrow$ $2 p x-6=0$. Therefore, it equals $3 / p$. Substituting, for example, into the second equation, we get $9 / p^{2}=1$, from which $p=3$.
## Answer
For $p=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Dirichlet's Principle (continued).]
Every day, from Monday to Friday, the old man went to the blue sea and cast his net into the sea. Each day, the net caught no more fish than the previous day. In total, over the five days, the old man caught exactly 100 fish. What is the smallest total number of fish he could have... | Evaluation. On Tuesday and Thursday, the old man caught no more fish than on Monday and Wednesday, which means that over the specified three days, he caught no less than half of 100, that is, no less than 50 fish.
Example. If the old man caught 25 fish each of the first four days and caught nothing on Friday, the cond... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Initially on the computer screen - some prime number. Every second, the number on the screen is replaced by the number obtained from the previous one by adding its last digit, increased by 1. What is the maximum time it will take for a composite number to appear on the screen?
# | Let the initial number on the screen be 2, then we get the following sequence: $2-5-11-13-17-25$. The sixth number is composite, so in this case, it will take 5 seconds. We will prove that in other cases, it will take no more than 5 seconds.
Indeed, if the number on the screen was an odd prime number not ending in 9, ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In a certain class, in any distribution of 200 candies, there will be at least two students who receive the same number of candies (possibly none). What is the smallest number of students in such a class?
# | Let there be no more than 20 people in the class, then we will give nothing to the first student, one candy to the second student, two candies to the third student, and so on. In this case, no more than $0+1+2+\ldots+19=$ 190 candies will be distributed. The remaining candies will be given to the student with the most ... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The Vellum Skin grants wishes, but after each wish, its area decreases: either by 1 dm² in the usual case, or by half - if the wish was a heartfelt one. Ten wishes reduced the area of the skin threefold, the next few - sevenfold, and after a few more wishes, the skin disappeared completely. What was the original area o... | Let the original area of the skin be $S$ dm² $(S>0)$. After the area was reduced threefold and then sevenfold, it became $S / 21$ dm². This number must be an integer; otherwise, it would be impossible to get 0 by dividing by 2 and subtracting 1.
Consider the first 10 wishes. Among them, there cannot be more than one u... | 42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
The dentist forbade Sonya to eat more than ten caramels a day, and if on any day she eats more than seven caramels, then for the next two days she cannot eat more than five caramels per day. What is the maximum number of caramels Sonya will be able to eat in 25 days, following the dentist's instructions?
# | If Sonya eats more than 7 caramels on any day (except the last two days), then in the next two days she will eat no more than $5+5=10$, and over three days - no more than 20 candies. This means that it would be more beneficial for her to eat 7 candies per day over these three days.
Therefore, it is more beneficial for... | 178 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
What is the maximum number of queens that can be placed on the black squares of an $8 \times 8$ chessboard so that each queen is attacked by at least one of the others?
# | Consider some arrangement of queens that satisfies the condition. A queen standing at the edge of the board cannot be attacked by any other queen, which means that all the queens are in the "inner" $6 \times 6$ square. Note that in none of the $3 \times 3$ squares can there be more than four queens. Indeed, if there ar... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
On the board after the class, there was a note left:
"Calculate $t(0)-t(\pi / 5)+t\left((\pi / 5)-t(3 \pi / 5)+\ldots+t\left({ }^{8 \pi} / 5\right)-t(9 \pi / 5)\right.$, where $t(x)=\cos 5 x+* \cos 4 x+* \cos 3 x+* \cos 2 x+$ $*^{\cos x}+* "$.
Seeing it, a math student told a friend that he could calculate this sum e... | Let $t_{k}(x)=\cos k x$ for $k=0,1,2,3,4$. If we consider a similar sum for $t_{5}$ instead of $t$, we will get 10 as the result. We will check that for all other $t_{k}$ the sums are equal to 0. For $k=0$ this is obvious. For $k=1,2,3,4$ we arrive at the same equality
$\cos 0+\cos (2 \pi / 5)+\cos ((\pi / 5) \cos (6 ... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Coordinate method on the plane [ Ratio of areas of triangles with a common base or common height]
On the coordinate plane, points $A(1 ; 9)$, $C(5 ; 8)$, $D(8 ; 2)$, and $E(2 ; 2)$ are given. Find the area of the pentagon $A B C D E$, where $B$ is the intersection point of the lines $E C$ and $A D$. | If $y_{1} \neq y_{2}$ and $x_{1} \neq x_{2}$, then the equation of the line passing through the points $\left(x_{1} ; y_{1}\right)$ and $\left(x_{2} ; y_{2}\right)$ is
$$
\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}
$$
## Otвет
27.00 | 27 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[ Coordinate method on the plane [ The ratio of the areas of triangles with a common base or common height]
] On the coordinate plane, points $A(1 ; 3), B(1 ; 9), C(6 ; 8)$, and $E(5 ; 1)$ are given. Find the area of the pentagon $A B C D E$, where $D$ is the point of intersection of the lines $A C$ and $B E$. | If $y_{1} \neq y_{2}$ and $x_{1} \neq x_{2}$, then the equation of the line passing through the points ( $x_{1} ; y_{1}$ ) and $\left(x_{2} ; y_{2}\right)$ is
$$
\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}
$$
## Answer
21.00 | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Authors: Yashchenko I.V., Botin D.A.
Gulliver found himself in the land of the Lilliputians with 7,000,000 rubles. He immediately spent all his money on bottles of kefir, which cost 7 ruble... | Conduct calculations in "hard" currency - empty bottles. Then there will be no inflation!
## Solution
Note that if all transactions are conducted in hard currency - empty bottles, - there will be no inflation. That is, a bottle of kefir will always cost 7 empty bottles, and the kefir in it -6 empty bottles. Gulliver,... | 1166666 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
All natural numbers from 1 to 1000 inclusive are divided into two groups: even and odd.
In which of the groups is the sum of all digits used to write the numbers greater, and by how much?
# | The sum of the digits of the number 1 is equal to the sum of the digits of the number 1000; the remaining numbers will be divided into 499 pairs: $\{2,3\},\{4,5\}, \ldots$, $\{998,999\}$. In each pair, the sum of the digits of the odd number is 1 greater than the sum of the digits of the even number.
## Answer
The su... | 499 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Yatsenko I.v.
From the volcano station to the summit of Stromboli, it takes 4 hours to walk along the road, and then another 4 hours along the path. At the summit, there are two craters. The first crater erupts for 1 hour, then remains silent for 17 hours, then erupts again for 1 hour, and so on. The second crater eru... | The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the eruption of the first crater, this crater will not be dangerous.
Movement along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the eruption of... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogosnov I.I.
Numbers, either red or blue, are arranged in a circle. Each red number is equal to the sum of its neighboring numbers, and each blue number is equal to the half-sum of its neighboring numbers. Prove that the sum of the red numbers is zero.
# | Let $a, b, c$ - be three consecutive numbers. If $b$ is red, then $b=a+c$, and if $b$ is blue, then $2 b=a+c$. We will write such equations for all triples of consecutive numbers and add them. On the right side, we will get twice the sum of all numbers, and on the left side - the sum of red numbers plus twice the sum o... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
Rushkin C.
On the surface of a cube, a closed eight-segment broken line is drawn, the vertices of which coincide with the vertices of the cube.
What is the minimum number of segments of this broken line that can coincide with the edges of the cube?
# | The segments of the broken line are either edges of the cube or diagonals of its faces. To do this, let's color the vertices of the cube in two colors in a checkerboard pattern (see fig.). Note that an edge of the cube connects vertices of different colors, while a diagonal connects vertices of the same color. For the ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Rubanov I.S.
In five pots standing in a row, Rabbit poured three kilograms of honey (not necessarily into each and not necessarily equally). Winnie-the-Pooh can take any two adjacent pots. What is the maximum amount of honey that Winnie-the-Pooh can guarantee to eat?
# | Evaluation. Let's assume Winnie-the-Pooh cannot take at least a kilogram of honey. This means that in any pair of adjacent pots, there is less than a kilogram of honey. This is true for both the two rightmost pots and the two leftmost pots. However, then in the middle pot, there must be more than a kilogram of honey (o... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Auto: : Bogosnov I.I.
All integers from -33 to 100 inclusive were arranged in some order, and the sums of each pair of adjacent numbers were considered. It turned out that there were no zeros among them. Then, for each such sum, the number reciprocal to it was found. The obtained numbers were added. Could the result b... | Let's consider an example of such an arrangement. Consider the sequence $100, -33, 99, -32, \ldots, 34, 33$. Then, if the first number of the pair is in an odd position, the sum is 67. If it is in an even position, the sum is 66. The reciprocals will be $\frac{1}{67}$ and $\frac{1}{66}$, respectively, with the first oc... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
[ Pairing and grouping; bijections ] [ Counting in two ways ]
In a brigade of 7 people, their total age is 332 years. Prove that it is possible to select three people whose combined age is not less than 142 years.
# | Let's consider all possible triples of the work team. The sum of their ages, as is easy to calculate, is equal to 15 $\cdot$ 332, and there are 35 such triples in total. Therefore, there is a triple whose total age is not less than $15 \cdot 332 / 35$, which is more than 142. | 142 | Combinatorics | proof | Yes | Yes | olympiads | false |
[ Extremal Properties (other) ] [ Examples and Counterexamples. Constructions ]
The number of edges of a convex polyhedron is 99. What is the maximum number of edges that a plane, not passing through its vertices, can intersect? | The plane does not intersect at least one edge in each face.
## Solution
The answer is 66. We need to prove that this number is possible, and that a larger number is impossible. First, let's prove the latter. Each face of the polyhedron is intersected by the plane in no more than two edges, meaning that at least one ... | 66 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Yamenniko i.v.
The numbers 2, 3, 4, ..., 29, 30 are written on the board. For one ruble, you can mark any number. If a number is already marked, you can freely mark its divisors and numbers that are multiples of it. What is the minimum number of rubles needed to mark all the numbers on the board? | Let's mark the numbers $17, 19, 23$, and 29, spending four rubles. Then mark the number 2, spending another ruble. After this, we can freely mark all even numbers (since they are divisible by 2), and then all odd numbers not exceeding 15 - for any of them (let's say for the number $n$) the even number $2n$ is already m... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evdokimov M.A.
16 cards with integers from 1 to 16 are laid face down in a $4 \times 4$ table so that cards with consecutive numbers are adjacent (touching by a side). What is the minimum number of cards that need to be flipped simultaneously to definitely determine the location of all numbers (regardless of how the c... | Evaluation. Let's number the cells as shown in Figure 1.
| 1 | 2 | 3 | 4 |
| :--- | :--- | :--- | :--- |
| 2 | 1 | 4 | 3 |
| 5 | 6 | 7 | 8 |
| 6 | 5 | 8 | 7 |
Fig. 1
Fig. 2
 measure the distance in centimeters between two given points; b) compare two given numbers. What is the minimum number of operations this device needs to perform to definitely determi... | To determine whether $A B C D$ is a rectangle, it is sufficient to check the equalities $A B=C D$, $B C=A D$, and $A C=B D$ - a total of 9 operations (3 operations for each equality: two measurements and one comparison). The rectangle $A B C D$ will be a square if $A B=B C$ - for this, one more, the 10th, operation of ... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
[Sum of angles in a triangle. Theorem of the exterior angle.] Counting in two ways
In a square, 20 points were marked and connected by non-intersecting segments to each other and to the vertices of the square, such that the square was divided into triangles. How many triangles were formed? | We will consider the marked points and the vertices of the square as vertices, and the segments connecting them and the sides of the square as edges of a planar graph. For each piece into which this graph divides the plane, we will count the number of edges bounding it, and sum all the obtained numbers. Since each edge... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In a single-round-robin tournament, 10 chess players are participating. What is the minimum number of rounds after which a sole winner can be determined prematurely? (In each round, the participants are paired.
Win - 1 point, draw - 0.5 points, loss - 0).
# | Evaluation. After the sixth round, 30 points have been played, and the leader has no more than 6 points, while the other nine participants have collectively scored no less than 24 points. Therefore, among them, there is at least one who has no less than 3 points. Since there are still 3 rounds ahead, the winner is stil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In the language of the AU tribe, there are two letters - "a" and "u". Some sequences of these letters are words, and each word has no fewer than one and no more than 13 letters. It is known that if you write down any two words in a row, the resulting sequence of letters will not be a word. Find the maxim... | If all sequences, the number of letters in which is not less than 7 and not more than 13, are words, then, obviously, the condition of the problem is satisfied; in this case, the number of such words is $2^{7}+\ldots+2^{13}=2^{14}-2^{7}$. It remains to show that this number is the largest possible.
The first method. T... | 16256 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Marayev A.
Cubes of size $1 \times 1 \times 1$ are glued together to form a cube of size $3 \times 3 \times 3$. What is the maximum number of cubes that can be removed from it so that the remaining figure has the following two properties:
- from the side of each face of the original cube, the figure looks like a $3 \... | Example. The layers from bottom to top are shown in the figure. The 13 remaining cubes are marked in black. Each layer has cubes in all columns and rows. When the layers are overlaid, a black $3 \times 3$ square is formed. Therefore, the first condition is satisfied. The bottom and middle layers are connected and glued... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Bogdanov I.I.
In a store, 21 white and 21 purple shirts are hanging in a row. Find the smallest $k$ such that for any initial order of the shirts, it is possible to remove $k$ white and $k$ purple shirts so that the remaining white shirts hang together and the remaining purple shirts also hang together. | First, let's show that \( k \), equal to 10, is sufficient.
First method. We will walk along the row of shirts and count the white and purple shirts separately. As soon as we count 11 of one color - let's assume, without loss of generality, purple - shirts, we will stop. Now we will remove all the white shirts that we... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Gooovanovo A.C. Positive rational numbers $a$ and $b$ are written as decimal fractions, each of which has a minimal period consisting of 30 digits. The decimal representation of the number $a-b$ has a minimal period length of 15. For what smallest natural $k$ can the minimal period length of the decimal representation... | By multiplying, if necessary, the numbers $a$ and $b$ by a suitable power of ten, we can assume that the decimal representations of the numbers $a, b, a-b$, and $a+k b$ are purely periodic (i.e., the periods start immediately after the decimal point).
Then $a=\frac{m}{10^{30}-1}, b=\frac{n}{10^{30}-1}$. We also know t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Kozhevnikov P.A.
What is the minimum number of cells that need to be marked on a $15 \times 15$ board so that a bishop, from any cell on the board, attacks no fewer than two marked cells? (The bishop also attacks the cell it stands on.)
# | It is more convenient to solve the problem in general: we will show that on a board of size $(2 n+1) \times(2 n+1)(n>1)$, the answer is $4 n$. (The diagram corresponds to the case $n=3$.)
Example. Let's mark the cells along the perimeter of a rectangle $(2 n-1) \times 3$, as shown in the diagram. In this case, on the ... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Kuzneuvv A.
A square box of candies is divided into 49 equal square cells. In each cell lies a chocolate candy - either black or white. In one sitting, Sasha can eat two candies if they are of the same color and lie in adjacent cells, either by side or by corner. What is the maximum number of candies that Sasha can de... | Evaluation. For the layout shown in the left image, none of the 16 black candies can be eaten, and out of the 33 white candies, no more than 32 can be eaten due to parity.
Algorithm. We will... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Fomin D:
The numbers $1, 1/2, 1/3, \ldots, 1/100$ are written on the board. We choose two arbitrary numbers $a$ and $b$ from those written on the board, erase them, and write the number
$a + b + ab$. We perform this operation 99 times until only one number remains. What is this number? Find it and prove that it does ... | If $a_{1}, a_{2}, \ldots, a_{n}$ are numbers written on the board, then the value $\left(1+a_{1}\right)\left(1+a_{2}\right) \ldots\left(1+a_{n}\right)$ does not change under an allowable operation. Indeed, if $a$ and $b$ are replaced by $a+b+a b$, then the factors not containing $a$ and $b$ do not change, and the produ... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Bermov S.l.
Given a $15 \times 15$ board. Some pairs of centers of adjacent cells by side are connected by segments such that a closed non-self-intersecting broken line is formed, which is symmetric with respect to one of the diagonals of the board. Prove that the length of the broken line does not exceed 200. | Clearly, the broken line intersects the diagonal. Let $A$ be one of the vertices of the broken line lying on the diagonal. We will move along the broken line until we first reach another vertex $B$ lying on the diagonal. By symmetry, if we move along the broken line from $A$ in the other direction, $B$ will also be the... | 200 | Geometry | proof | Yes | Yes | olympiads | false |
Berlov S.L.
100 people came to a party. Then those who had no acquaintances among the attendees left. Then those who had exactly one acquaintance among the remaining people also left. Then, similarly, those who had exactly 2, 3, 4, ..., 99 acquaintances among the remaining people at the time of their departure also le... | It is not difficult to check that if all the attendees, except for two people $A$ and $B$, were acquainted with each other, then at the end, everyone except $A$ and $B$ should have remained, that is, 98 people.
Let's prove that 99 people could not have remained. It is clear that person $A$, who initially had the fewes... | 98 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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