problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
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55.There were two positive numbers. One of them was increased by $1 \%$, the other by $4 \%$. Could their sum have increased by $3 \%$? | 55.Let the first number be $100 x$, and the second $100 y$. We can rephrase the problem as: "Can the equality $101 x + 104 y = 103(x + y)$ hold for some positive $x$ and $y$?", from which it follows that $y = 2x$. For $x = 1$ and $y = 2$, we get the numbers 100 and 200, which will satisfy the condition of the problem. | 2x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,813 |
56. 180 g of gold with a fineness of 920 was alloyed with 100 g of gold with a fineness of 752. What is the fineness of the resulting alloy? | 56. Assay of an alloy: $p=\frac{a}{b} \cdot 1000$, where $a$ is the weight of gold in the alloy, and $b$ is the weight of the alloy. That is: $p=\frac{100 \cdot 0.752 + 180 \cdot 0.920}{280} \cdot 1000=\frac{100 \cdot 752 + 180 \cdot 920}{280}=860$. | 860 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,814 |
57. After each wash, the piece of soap decreases by $20 \%$. After how many washes will it decrease by no less than a third | 57. Let the volume of a piece of soap be $x$, then after the second washing, its volume will become $0.64 x$, that is, it will decrease by more than a third. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,815 |
59.The distance between two cars driving on the highway is 200 km. The speeds of the cars are 60 km/h and 80 km/h. What will be the distance between them after one hour? | 59.There are four cases: (make a drawing!)
1) The cars are driving towards each other: $200-(60+80)=60$ km;
2) The cars are driving in opposite directions: $200+(60+80)=340$ km;
3) The second car is catching up to the first: $200+60-80=180 \text{ km}$; 4) The second car is ahead, the first is behind: $200+80-60=220$ k... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,817 |
60.Yura left the house for school 5 minutes later than Lena, but walked at twice her speed. How long after leaving will Yura catch up to Lena? | 60.Let Lena walk $s$ km in 5 minutes. Then in the next 5 minutes, Yura will walk $2 s$ km, and Lena will walk another $s$ km, that is, a total of $2 s$ km. Therefore, in 5 minutes, Yura will catch up with Lena. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,818 |
61.If a cyclist rides at a speed of 10 km/h, he will be 1 hour late. If he rides at a speed of 15 km/h, he will arrive 1 hour early. At what speed should he ride to arrive on time? | 61.If there were two cyclists, with the first one's speed being $10 \mathrm{km} / \mathrm{h}$ and the second one's speed being $15 \mathrm{km} / \mathrm{h}$. Then, according to the problem, if the first cyclist started 2 hours earlier than the second, they would arrive at the destination simultaneously. In this case, i... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,819 |
64.A motorboat travels 90 km downstream in the same time it takes to travel 70 km upstream. What distance can a raft drift in the same time? | 64.Let in $t$ hours the motorboat covers 90 km downstream and 70 km upstream. Then the speed of the motorboat downstream is $\frac{90}{t}$ km/hour, and the speed upstream is $\frac{70}{t}$ km/hour. From this, the doubled speed of the current will be $\frac{90}{t}-\frac{70}{t}=\frac{20}{t}$ km/hour. The speed of the cur... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,822 |
65. A cart is moving down the road, with a nail driven into one of its wheels. The diameter of the cart's wheels is 1 meter. Each time the nail hits the road, a click is heard. The clicks repeat every second. At what speed is the cart moving? | 65. In 1 second, the cart travels $\pi$ meters, therefore, in 1 hour $-3.6 \pi$ km $\approx 11.3$ km per hour. | 11.3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,823 |
66. Two beetles crawl from point $A$ to point $B$ and return back. The first beetle crawls both ways at the same speed. The second beetle crawls to $B$ 1.5 times faster than the first, but returns 1.5 times slower than the first. Which beetle returned to $A$ first? | 66. Let's write the condition of the problem in the form of a table:
| | speed | | path | | time | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| | there | back | there | back | there | back |
| 1st beetle | $v$ | $v$ | $s$ | $s$ | $s / v$ | $s / v$ |
| 2nd beetle | $3 v / 2$ | $2 v / 3$ | $s$ | $s... | t_{1}<t_{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,824 |
68. A sack contains 24 kg of nails. How, using only a balance scale without weights, can you measure out 9 kg of nails? | 68. The main operation available to us is dividing some (in general, arbitrary) amount of nails into two equal weight piles. We will record the results of the weighings in a table:
| | 1st pile. | 2nd pile. | 3rd pile | 4th pile |
| :--- | :--- | :--- | :--- | :--- |
| Initially: | 24 kg | | | |
| 1st step | 12 kg... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,826 |
69. Someone has 12 pints of wine and wants to pour out half, but he does not have a 6-pint container. However, he has two containers with capacities of 5 and 8 pints. How can he measure exactly 6 pints of wine? | 69. The solution is visible from the table:
| steps: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 12 l | 12 | 4 | 4 | 9 | 9 | 1 | 1 | 6 |
| 8 l | 0 | 8 | 3 | 3 | 0 | 8 | 6 | 6 |
| 5 l | 0 | 0 | 5 | 0 | 3 | 3 | 5 | 0 | | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,827 |
71. A biologist discovered an amazing variety of amoebas. Each of them divided into two every minute. The biologist puts an amoeba in a test tube, and exactly an hour later, the test tube is filled with amoebas. How much time would it take for the entire test tube to be filled with amoebas if, instead of one, two amoeb... | 71. At the beginning of the experiment, there is one amoeba in the test tube, after one minute there are already two, so 2 amoebas will fill the test tube in 59 minutes. | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,829 |
72.In a bottle, glass, jug, and jar, there are milk, lemonade, kvass, and water. It is known that water and milk are not in the bottle, the container with lemonade is between the jug and the container with kvass, the jar contains neither lemonade nor water. The glass stands next to the jar and the container with milk. ... | 72. Let's "enumerate" the information:
1) water and milk are not in the bottle; 2) the vessel with lemonade stands between the jug and the vessel with kvass, therefore, lemonade is not in the jug and kvass is not in the jug; 3) in the can - not lemonade and not water; 4) the glass stands near the can and the vessel wit... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,830 |
73. The police officer turned at the sound of breaking glass and saw four teenagers running away from the broken storefront window. Five minutes later, they were at the police station. Andrey claimed that Viktor broke the glass, Viktor insisted that Sergey was to blame. Sergey assured that Viktor was lying, and Yury in... | 73. Since only one of the boys told the truth, it is either - Viktor, or - Sergey (their statements contradict each other and cannot both be true or false at the same time). Therefore, all the others are lying. Yuri insisted that it was not him. If this is a lie, then Yuri is guilty. | Yuri | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,831 |
74. Three friends met: Belov, Chernov, and Ryzhov. One of them is a blonde, another is a brunette, and the third is a redhead. The brunette said to Belov: "None of us have hair color that matches our last name." What is the hair color of each of them? | 74. Belov is not a blonde, since his hair color cannot match his surname; he is also not a brunette, as he himself talked to a brunette. Therefore, he is a redhead. Chernov is neither a brunette nor a redhead. Therefore, he is a blonde. The only remaining possibility is: Ryzhov is a brunette. | Belov: | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,832 |
75. When Vanya was asked how old he was, he thought for a moment and said: “I am three times younger than Dad, but three times older than Seryozha.” At this point, little Seryozha ran up and reported that Dad is 40 years older than him. How old is Vanya? | 75. Dad is 3 times older than Vanya, who, in turn, is 3 times older than Seryozha, so Dad is 9 times older than Seryozha. Therefore, Dad is 8 times older than Seryozha, which is 40 years. Hence, Seryozha's age is 5 years. Vanya is three times older than Seryozha, so he is 15 years old.
109 Such problems! | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,833 |
76. In the box, there are 100 white, 100 red, 100 blue, and 100 black balls. What is the smallest number of balls that need to be pulled out, without looking into the box, to ensure that among them there are at least 3 balls of the same color? | 76. Answer: 9 balls. If we randomly draw 8 balls, there might not be three balls of the same color (2 white + 2 red + 2 blue + 2 black). If we add one more ball, then there will definitely be 3 balls of the same color. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,834 |
77. Ten teams participate in a football tournament. Prove that regardless of the game schedule, there are always two teams that have played the same number of matches. | 77. Since there are ten teams participating in the tournament, the number of games played by each team can be any integer from 0 to 9. If at any point in the tournament all teams had played a different number of games, then one team would not have played any matches, another would have played one match, a third would h... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,835 |
78. How many circles of radius 1 are needed to cover a square with a side length of $2$? | 78. A circle of radius 1 can only cover one vertex of a $2 \times 2$ square, so at least four circles are required. However, a circle of unit radius can completely cover a $1 \times 1$ square, so four circles are sufficient. Answer: 4 circles. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,836 |
79. The store received 25 boxes of apples of three varieties, and in each box, there are apples of only one variety. Can you find 9 boxes with apples of the same variety? | 79. Suppose it is impossible to find 9 boxes of apples of the same variety. This means that there are no more than eight boxes of the first variety, no more than eight boxes of the second variety, and no more than eight boxes of the third variety, which totals no more than 24 boxes. But we have 25 boxes. Therefore, the... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,837 |
80.In the school, there are 33 classes and 1150 students. Will there be a class in this school with at least 35 students? | 80. Let there be fewer than 35 students in each of the 33 classes of the school, then there are no more than $34 \times 33=1122$ students in the school, which contradicts the condition (there are 1150 students in the school). Therefore, there is a class in this school with no fewer than 35 students. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,838 |
81. In the box, there are pencils: 7 red and 5 blue. In the dark, pencils are taken. How many pencils need to be taken to ensure that there are at least two red and at least three blue among them? | 81. In order to definitely take no less than two red pencils, you need to take no less than 7 pencils, and to definitely take no less than 3 blue ones, you need to take no less than 10 pencils. Therefore, to definitely take no less than two red and no less than 3 blue, you need to take no less than 10 pencils. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,839 |
82.In a chess tournament, eight people participated, and all of them scored a different number of points. The chess player who took second place scored as many points as the last four together. How did the players who took third and seventh place play against each other? | 82. The chess players who took the last four places in the matches among themselves scored 6 points, that is, together they scored no less than 6 points. The chess player who took second place (and scored no less than 6 points) could not have scored 6.5 points, as he scored less than the one who took first place. Thus,... | The\chess\player\who\took\third\place\won\against\the\chess\player\who\took\seventh\place | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,840 |
83. A log with a nail in the middle was left in the cargo area of a truck. The log rolled freely during movement and scratched the bottom. Which part of the cargo area could be damaged? Dimensions of the cargo area: $3 \times 2$ m; length of the log 2 m. | 83. Four semicircles with a radius of 1, centered at the vertices of the rectangle, will not be damaged. The remaining areas of the body, with a total area of $6-\pi$, may be damaged. They constitute $1-\frac{\pi}{6}$ of the body. | 1-\frac{\pi}{6} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,841 |
84. We have 6 sticks 1 cm long, 3 sticks 2 cm long, 6 sticks 3 cm long, and 5 sticks 4 cm long. Can a square be formed from this set, using all the sticks without breaking them and not placing one on top of another? | 84. The sum of the lengths of all the sticks is 50; a square has 4 equal sides, so the perimeter of the square must be divisible by 4. The number 50 is not divisible by 4. Therefore, it is impossible to form a square with the sticks. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,842 |
86. The distance between points $A B=30, B C=80, C D=236$, $D E=86, E A=40$. What is the distance $E C ?$ | 86. Since $D E+E A+A B+B C=D C$, the points $E, A$ and $B$ lie on the segment $D C$ in the given order. Therefore, $E C=150$. | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,844 |
87. Prove that any amount exceeding seven rubles can be paid without change using banknotes of 3 and 5 rubles.
Prove that any amount exceeding seven rubles can be paid without change using banknotes of 3 and 5 rubles. | 87. Let $x>7$. If the number $x$ is divisible by 3, then it can be represented as: $x=6+3k$; if $x$ gives a remainder of 1 when divided by 3, then it can be represented as: $x=10+3k$; if $x$ gives a remainder of 2 when divided by 3, then it can be represented as $x$ $=8+3k$ ( $k$ is an integer; $k \geq 0$ ). From this,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,845 |
89.In a line, there are 1998 people. It is allowed to swap any two people standing one apart. Is it possible to arrange them by height? | 89. Note that people standing in "odd" positions can only move to "odd" positions, so it will not always be possible to arrange them by height.
111 Literature | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,847 |
P r o b l e m 5. Among all triangles sharing a common side $\boldsymbol{A} \boldsymbol{C}$ and with the sum of the lengths of the other two sides equal to $\boldsymbol{m}$, identify the one with the largest area. | ## S o l u t i o n.
This problem is the inverse of problem 4. We will prove that the condition is satisfied by an isosceles triangle where \( |AB| = |BC| = \frac{m}{2} \). In solving problem 4, it was established that among all triangles with a given area \( S \) and a given side \( AC \), the sum \( |AB| + |BC| \) at... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,851 |
Problem 8. Given the function $y=\frac{x+2}{x-3}$. Find $\boldsymbol{E}(f)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
---
Problem 8. Given the function $y=\frac{x+2}{x-3}$. Find $\boldsymbol{E}(f)$. | Rešenje.
$E(f)=\left\{y \mid\right.$ the equation $y=\frac{x+2}{x-3}$ has a solution $\}=\{y \mid$ the equation $y=\frac{(x-3)+5}{x-3}$ has a solution $\}=\left\{y \mid\right.$ the equation $y=1+\frac{5}{x-3}$ has a solution $\}=\left\{y \mid\right.$ the equation $y=1=\frac{5}{x-3}$ has a solution $\}=$ $=\{y \mid y \... | y\neq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,853 |
Problem 11. Find the maximum value of the function $y=3 x+4 \sqrt{1-x^{2}}$
$$ | \text { S o l u t i o n. }
$$
First, let's find the domain of the given function: $D(f)=$ $=\left\{x \mid 1-x^{2} \geqslant 0\right\}=\{x \mid-1 \leqslant x \leqslant 1\} ; \quad E(f)=\{y \mid \quad$ the equation $y=3 x+4 \sqrt{1-x^{2}}$ has a solution on $\left.[-1 ; 1]\right\}=\{y \mid$ the equation $y-3 x=4 \sqrt{1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,856 |
Problem 13. Determine the length and width of a rectangular plot of land that will maximize its area given a fixed perimeter. | ## S o l u t i o n.
Let $x$ and $y$ be the sides of the rectangle. Then the perimeter will be $p=2x+2y$, and the area will be $S=xy$. By their nature, $x$ and $y$ are positive, as they represent the lengths of segments. According to the condition, the perimeter $p$ is a constant, meaning it reaches its maximum value w... | \frac{p}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,857 |
Problem 15. A segment of a given length moves so that its ends slide along the sides of a right angle. In what position of this segment will the area of the cut-off triangle be the largest?
$$ | \text { Solution. }
$$
Let the ends of the segment $A B$ of length $a$ slide along the sides of the angle $A C B: \widehat{A C B}=90^{\circ}$ (Fig. 21). Let $|A C|=x, 0 \leqslant x \leqslant a$. Then $|C B|=\sqrt{a^{2}-x^{2}}$. The area of triangle $A B C$ is $S=$ $=\frac{1}{2}|A C| \cdot|C B|=\frac{1}{2} x \sqrt{a^{2... | \frac{1}{4}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,858 |
Task 16. Points $\boldsymbol{A}$ and $\boldsymbol{B}$ are located on a straight highway running from west to east. Point B is 9 km east of A. A car departs from point A heading east at a speed of 40 km/h. Simultaneously, a motorcycle departs from B in the same direction with a constant acceleration of 32 km/h².
$ km.
20
The distance between them is the absolute value of the difference between $16 t^{2}+9$ and $40 t$. Let's denote this distance by $y$:
$$
y=\left|16 t^{2}+9-4... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,859 |
Problem 17. A vessel of cylindrical shape is filled with water. The height of the water layer is h. Water flows out of an opening (its dimensions can be neglected) located at height $z$ (Fig. 23). At what $z$ will the range of the water jet be the greatest? | ## S o l u t i o n.
We introduce a rectangular coordinate system as shown in Figure 23. Let $v_{0}$ be the velocity of the water jet, $x$ be the distance from the $O x$ axis, and $y$ be the distance along the $O y$ axis. Then $v_{0}=\sqrt{2 g(h-z)}$, $x=v_{0} t, y=z-\frac{g t^{2}}{2}$. At the moment the water jet reac... | \frac{}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,860 |
Problem 18. Prove that among all quadrilaterals inscribed in a circle, the one with the greatest area is a square. | ## S o l u t i o n.
Denoting the length of side $A D$ of rectangle $A B C D$ by $x$, $(x<2 k)$, and the radius of the circle by $R$ (Fig. 24), we have:
$$
|C D|=\sqrt{|A C|^{2}-|A D|^{2}}=\sqrt{4 R^{2}-x^{2}}
$$
Then the area of the rectangle is $S=|A D| \cdot|D C|=$ $=x \sqrt{4 R^{2}-x^{2}}=\sqrt{x^{2}\left(4 R^{2}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,861 |
Problem 19. In triangle $\boldsymbol{A B C}$, draw a line $M N$ parallel to the base $A C$ so that the area of rectangle MNPQ is maximized. | S o l u t i o n.
Let $|A C|=a,|M N|=x,|Q M|=y,|D B|=H$ (Fig. 25). Since $\triangle A B C \sim \triangle M B N$, we have $\frac{|M N|}{|A C|}=\frac{|B O|}{|B D|}$, or $\frac{x}{a}=\frac{H-y}{H}$.
^{2} x
$$
Multiplying both sides of equation (1) by 4, we get:
$$
4 V=(60-2 x)(60-2 x) \cdot 4 x
$$
The sum of the (... | 16000(\mathrm{~}^{3}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,863 |
Problem 26. A company has 2 workshops and 3 warehouses. It is necessary to determine the most profitable organization of transportation.
| Warehouses | No. 1 | No. 2 | No. 3 | Produced |
| :---: | :---: | :---: | :---: | :---: |
| I | 3 | 3 | 2 | 10000 |
| II | 6 | 5 | 1 | 5000 |
| Required to deliver | 4000 | 8000 | ... | ## S o l u t i o n.
Let $x$ units be transported from workshop I to warehouse 1, and $y$ units to warehouse 2. Then, from workshop I to warehouse 3, $(10000-x-y)$ units will be transported, and from workshop II to warehouses 1, 2, and 3, $4000-x$, $8000-y$, and $x+y-7000$ units will be transported, respectively (Table... | 43000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,868 |
Problem 31. It is required to prepare a mixture containing three chemical substances $A, B, C$. It is known that the mixture to be prepared must contain substance $\boldsymbol{A}$ no less than 6 units, substance $B$ no less than 8 units, and substance $C$ no less than 12 units.
Substances $A, B, C$ are contained in th... | ## S o l u t i o n.
We will build a mathematical model of the problem of formulating a mixture.
Let the number of units of product I in the mixture be denoted by $x_{1}$, product II by $x_{2}$, and product III by $x_{3}$.
The mixture to be formulated must contain substance $A$. Substance $A$ is present in all three ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,870 |
Problem 33. Every day, 12 tons of potatoes are supplied to the city by one type of transport from three collective farms: from the I farm at a price of 4 rubles per ton, from the II - at a price of 3 rubles, from the III - 1 ruble. To ensure timely delivery of potatoes to the city, it is necessary to spend no more than... | S o l u t i o n.
First, let's construct a mathematical model of this problem. Let $x_{1}, x_{2}, x_{3}$ be the tons of potatoes delivered to the city from the I, II, and III collective farms, respectively. Then the problem can be formulated as follows: Minimize the objective function
$$
F=4 x_{1}+3 x_{2}+1 x_{3}
$$
... | (\frac{2}{3};5\frac{1}{3};6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,872 |
Problem 35. For the fattening of animals, two types of feed I and II are used. Each kilogram of feed I contains 5 units of nutrient $\boldsymbol{A}$ and 2.5 units of nutrient B, while each kilogram of feed II contains 3 units of nutrient A and 3 units of nutrient B. Experimental data has shown that the fattening of ani... | ## S o l u t i o n.
1. Let's construct a mathematical model of this problem. Let $x$ and $y$ be the number of kilograms of feed of types I and II, respectively, consumed daily. Then the system of constraints is:
$$
\left\{\begin{array}{l}
5 x+3 y \geqslant 30 \quad(a) \\
2.5 x+3 y \geqslant 22.5 \\
x \geqslant 0, y \... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,874 |
Problem 36. Find the maximum value of the objective function
$$
F=3-4 x+y
$$
on the set of solutions of the system of constraints
$$
\left\{\begin{array}{l}
1-x \geqslant 0 \\
1-x+y \geqslant 0 \\
x \geqslant 0, y \geqslant 0
\end{array}\right.
$$ | S o l u t i o n.
We construct the feasible region corresponding to system (2) (Fig. 34). From Figure 34, it is clear that, regardless of the value of $c$ (even arbitrarily large), the function will always intersect the feasible region, and therefore, there will exist a non-negative solution at which this value of the ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,875 |
Problem 37. On the set of solutions of the system of constraints
$$
\left\{\begin{array}{l}
2-2 x_{1}-x_{2} \geqslant 0 \\
2-x_{1}+x_{2} \geqslant 0 \\
5-x_{1}-x_{2} \geqslant 0 \\
x_{1} \geqslant 0, \quad x_{2} \geqslant 0
\end{array}\right.
$$
find the minimum value of the function $F=x_{2}-x_{1}$. | S o l u t i o n.
The set of feasible plans is the polygon $A B C D E$ (Fig. 35). The line $2-$ $-x_{1}-x_{2}=0$ is parallel to the level lines of the function $F=x_{2}$ $-x_{1}$. Therefore, all points on the segment $C D$ give the same minimum value of the objective function $F=x_{2}-x_{1}$ on the set of points of the... | -2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,876 |
Problem 39. Minimize $\boldsymbol{F}=\boldsymbol{x}_{2}-\boldsymbol{x}_{1}$ for non-negative $x_{1}$ and $x_{2}$, satisfying the system of constraints:
$$
\left\{\begin{aligned}
-2 x_{1}+x_{2}+x_{3} & =2 \\
x_{1}-2 x_{2}+x_{4} & =2 \\
x_{1}+x_{2}+x_{5} & =5
\end{aligned}\right.
$$
56 | ## S o l u t i o n.
These constraints can be considered as derived from inequalities, since each of the variables $x_{3}, x_{4}, x_{5}$ appears only in one equation.
1. Write the constraints as equations expressing the basic variables in terms of the non-basic variables:
$$
\left\{\begin{array}{l}
x_{3}=2+2 x_{1}-x_... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,878 |
For problem 40. Maximize $F=x_{1}-x_{2}+2 x_{3}-x_{4}$
$$
\text { s.t. }\left\{\begin{array}{l}
x_{1}+x_{2}=1 \\
x_{2}+x_{3}-x_{4}=1 \\
x_{1} \geqslant 0, x_{2} \geqslant 0, x_{3} \geqslant 0, \quad x_{4} \geqslant 0
\end{array}\right.
$$ | S o l u t i o n.
1. Rewrite the system of constraints (2) as:
$$
\left\{\begin{array}{l}
x_{1}=x_{3}-x_{4} \\
x_{2}=1-x_{3}+x_{4} \\
x_{1} \geqslant 0, x_{2} \geqslant 0, x_{3} \geqslant 0, x_{4} \geqslant 0
\end{array}\right.
$$
The basis \( B \) consists of variables \( x_{1} \) and \( x_{2} \). The basic solution... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,879 |
To find the maximum value of $F=4 x_{1}+6 x_{2}$
subject to
$$
\left\{\begin{array}{c}
x_{1}+x_{2} \leqslant 18 \\
0.5 x_{1}+x_{2} \leqslant 12 \\
0 \leqslant x_{1} \leqslant 12 \\
0 \leqslant x_{2} \leqslant 9
\end{array}\right.
$$ | S o l u t i o n.
This problem has already been solved graphically (see problem 32).
When solving by the simplex method, the system of constraints, consisting of inequalities, must be transformed into a system of linear equations by introducing additional variables:
$$
\begin{gathered}
x_{1}+x_{2}+x_{3}=18 \\
0.5 x_{... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,880 |
3(1244). In a six-digit number, the first digit coincides with the fourth, the second with the fifth, and the third with the sixth. Prove that this number is divisible by $7,11,13$. | Solution. Let the first digit of the number be denoted by the letter $a$, the second by the letter $b$, and the third by the letter $c$. Then the given number can be written as: $\overline{a b c a b c}$. We have:
$$
\overline{a b c a b c}=1000 \overline{a b c}+\overline{a b c}=1001 \overline{a b c}=7 \cdot 11 \cdot 13... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,884 |
5 (1246). By what percentage will the area of a rectangle increase if its length is increased by $20 \%$ and its width by $10 \%$? | Solution. Let the length of the rectangle be $x$, and the width be $y$, then its area is $x y$. After increasing the length and width of the rectangle by $20 \%$ and $10 \%$ respectively, its area became $1.2 x \cdot 1.1 y = 1.32 x y$, i.e., the area of the rectangle increased by $0.32 x y$, which is $32 \%$ of $x y$.
... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,885 |
6 (1247). Three boxes are filled with nuts. In the second box, there are $10 \%$ more nuts than in the first, and $30 \%$ more than in the third. How many nuts are in each box if there are 80 more nuts in the first box than in the third? | Solution. Let there be $x$ nuts in the first box and $y$ in the third. Then in the second box, there were $x+0.1 x=1.1 x$ or $y+0.3 y=1.3 y$. Considering that there were 80 more nuts in the first box than in the third, we form the system of equations:
$$
\left\{\begin{array}{l}
1.1 x=1.3 y \\
x-y=80
\end{array}\right.... | 520 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,886 |
7 (1248). The number a is $80 \%$ of the number b, and the number c is $140 \%$ of the number b. Find the numbers a, b, and c, given that c is 72 more than a. | Solution. According to the problem, we have $a=0.8 b, c=1.4 b$,
$$
\begin{gathered}
c-a=72, \text { from which } c-a=1.4 b-0.8 b=0.6 b \\
0.6 b=72, b=120, a=0.8 b=96, c=1.4 b=168
\end{gathered}
$$
Answer: $a=96, b=120, c=168$.
口 8(1249). The number $a$ is $75\%$ of the number $b$ and $40\%$ of the number $c$. The nu... | =96,b=120,=168 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,887 |
9(1250). When dividing a natural number $a$ by a natural number $b$, the quotient is $c$ and the remainder is $d$. Determine whether all numbers $a, b, c$, and $d$ can be odd. | Solution. According to the problem, we have $a=b c+d$.
Suppose that all numbers $a, b, c, d$ are odd. Then the number $b c$ is an odd number (the product of two odd numbers is an odd number), and the number $b c+d$ is an even number (the sum of two odd numbers is an even number). We have obtained that the number $a=b ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,888 |
12 (1253). Prove that the remainder of a prime number divided by 30 is a prime number or one. | Solution. Let the prime number $a$ when divided by 30 give a quotient $q$ and a remainder $r$. Then $a=30q+r$, where $r \in \boldsymbol{N}, r<30$. In this case, the remainder $r$ cannot be equal to 2, 3, or 5, because otherwise the number $a$ would be composite (if each term is divisible by some number, then the sum is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,890 |
13(1254). To a certain two-digit number, a one was added to the left and right. As a result, a number was obtained that is 23 times greater than the original. Find this two-digit number. | Solution. Let the desired two-digit number be $\bar{a} \bar{b}$. According to the problem, we form the equation: $\overline{\mid a b 1}=23 \cdot \overline{a b}$. We have $1000+$ $+10 \overline{a b}+1=23 \cdot \overline{a b}$, from which $13 \cdot \overline{a b}=1001, \overline{a b}=77$.
Answer: 77.
Remark 1. The unkn... | 77 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,891 |
3. If a zero is appended to the right of the thought number and the result is subtracted from 143, the result will be three times the thought number. What number was thought of? | Solution. Let the number be $x$. We have the equation $143-10 x=3 x$, from which $x=11$. A n s w e r: 11 . | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,892 |
4. If to a given number we append the digit 9 on the right and add to the resulting number twice the given number, the sum will be equal to 633. Find the given number. | Solution. The problem boils down to solving the equation $10 x+$ $+9+2 x=633$, where $x$ is the given number; $x=52$. A n s w e r: 52 . | 52 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,893 |
16(1257). Construct the graph of the equation:
a) $(x-2)(y+3)=0 ; \quad$ b) $x^{2}+x y=0$. | Solution. a) Since the product is equal to zero when at least one of the factors is equal to zero, the equation $(x-2) \times$ $\times(y+3)=0$ is satisfied by pairs of numbers $x=2, y$ - any number or $x$ - any number, $y=-3$. Therefore, the graph of the equation $(x-2)(y+3)=0$ consists of two lines $x=2$ and $y=-3$ (F... | 2-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,894 |
17(1258). Construct the graph of the equation:
a) $y+|y|=x ; \quad$ b) $y=x|y|$. | Solution. a) By the definition of the modulus of a number, we have:
Fig. 10
. Construct the graph of the function:
a) $y=|x|-3$; b) $y=4-|x|$. | Solution. a) By the definition of the absolute value of a number, we have:
$$
y=|x|-3=\left\{\begin{array}{l}
x-3, \text { if } x \geqslant 0, \\
-x-3, \text { if } x<0
\end{array}\right.
$$
The graph of the function $y=|x|-3$ is shown in Figure 14.
b) By the definition of the absolute value of a number, we have:
$... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,896 |
19(1260). Find the smallest natural number that, when multiplied by 2, becomes a square, and when multiplied by 3, becomes a cube of a natural number. | Solution. Let $x$ be the smallest natural number such that $2 x=b^{2}, 3 x=c^{3}$, where $b$ and $c$ are natural numbers. From the equation $2 x=b^{2}$, it follows that $x$ is divisible by 2. Since $3 x=c^{3}$, $x$ is divisible by $2^{3}=8$ and by $3^{2}=9$, i.e., $x=2^{3} \cdot 3^{2} a^{6}=72 a^{6}$, where $a$ is any ... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,897 |
20(1261). Prove that the value of the expression $96^{7}-22^{5}-$ $-48^{6}$ is divisible by 10. | Solution. Let's determine the last digit of each power in the given expression.
Since $6^{7}$ ends in the digit 6, then $96^{7}$ also ends in the digit 6. $2^{5}$ ends in the digit 2, therefore, $22^{5}$ also ends in the digit 2. $48^{6}$, like $8^{6}$, ends in the digit 4.
Considering the signs of the expression, we... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,898 |
21 (1263). Which is greater: $\frac{10^{11}+1}{10^{11}+1}$ or $\frac{10^{11}+1}{10^{12}+1}$ ? | Solution. To simplify the calculations, let's make the substitution $10^{10}=a$. Then we get:
$$
\frac{10^{11}+1}{10^{11}+1}=\frac{a+1}{10 a+1} ; \frac{10^{11}+1}{10^{12}+1}=\frac{10 a+1}{100 a+1}
$$
Comparing these fractions has been reduced to comparing the fractions
$$
\frac{a+1}{10 a+1} \text { and } \frac{10 a+... | \frac{10^{11}+1}{10^{11}+1}>\frac{10^{11}+1}{10^{12}+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,899 |
22(1264). Represent the expression $2 x^{2}+2 y^{2}$ as a sum of two squares. | Solution.
$$
\begin{aligned}
2 x^{2}+2 y^{2}= & \left(x^{2}+2 x y+y^{2}\right)+\left(x^{2}-2 x y+y^{2}\right)= \\
& =(x+y)^{2}+(x-y)^{2}
\end{aligned}
$$ | (x+y)^{2}+(x-y)^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,900 |
23(1265). If $a \neq 0$ or $b \neq 0$, then the value of the expression $5 a^{2}-6 a b+5 b^{2}$ is positive. Prove it. | Proof. Let's transform the given expression:
$$
\begin{aligned}
5 a^{2}-6 a b+5 b^{2} & =4 a^{2}-8 a b+4 b^{2}+a^{2}+2 a b+b^{2}= \\
& =(2 a-2 b)^{2}+(a+b)^{2}
\end{aligned}
$$
Under the given condition and for $a \neq \pm b$, each term of the obtained sum is positive. When $a=b$, the second term $(a+b)^{2}>0$, and w... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,901 |
24 (1266). Prove that for any value of $x$ the value of the expression $(x-3)(x-5)+2$ is a positive number. | Proof. Let's expand the brackets in the given expression and combine like terms:
$$
(x-3)(x-5)+2=x^{2}-8 x+15+2=x^{2}-8 x+17
$$
Now, let's complete the square:
$$
x^{2}-8 x+17=(x-4)^{2}+1>0
$$
Since $(x-4)^{2} \geqslant 0$, then $(x-4)^{2}+1>0$. The statement is proved. | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,902 |
25(1267). Factorize the polynomials:
a) $x^{8}+x^{4}-2$
b) $n^{4}+4$;
c) $a^{5}-a^{2}-a-1$
d) $n^{4}+n^{2}+1$. | Solution.
a) $x^{8}+x^{4}-2=\left(x^{4}-1\right)^{2}+3 x^{4}-3=\left(x^{4}-1\right)^{2}+3\left(x^{4}-1\right)=$ $=\left(x^{4}-1\right)\left(x^{4}-1+3\right)=\left(x^{2}-1\right)\left(x^{2}+1\right)\left(x^{4}+2\right)$.
4 Order 942
Another method of factoring this trinomial is given in $\$ 4, \mathbf{c} .28$.
6) $a^{... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,903 |
26 (1268). Prove that $p^{2}-1$ is divisible by 24 if $p$ is a prime number greater than 3. | Proof. Method I. $p^{2}-1=(p-1)(p+1)$. Since $24=3 \cdot 8$, where 3 and 8 are coprime numbers, to prove the divisibility of $p^{2}-1$ by 24, it is sufficient to prove its divisibility: a) by 3 and b) by 8 (the corresponding theorem can be considered in circle sessions).
a) To prove the divisibility of $p^{2}-1$ by 3,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,904 |
27 (1269). Prove that the sum of the squares of five consecutive natural numbers cannot be a square of a natural number. | Proof. Let $n-2, n-1, n, n+1, n+2$, where $n \in \boldsymbol{N}, n>2,$ be five consecutive natural numbers. Consider the sum of their squares:
50
$$
(n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}=5 n^{2}+10=5\left(n^{2}+2\right)
$$
For the expression $5\left(n^{2}+2\right)$ to be the square of a natural number, it is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,905 |
28 (1270). Prove that the difference between the square of a natural number not divisible by 3 and the number 1 is divisible by 3. | Proof. A natural number not divisible by 3 can be of one of two types: a) $n=3k+1$; b) $n=3k+2$, where $k=0,1,2,3, \ldots$.
In case a) according to the problem, we have $(3k+1)^2-1=9k^2+6k$.
In case b) we have $(3k+2)^2-1=9k^2+12k+3$.
In both cases, we obtained numbers divisible by 3.
$\triangle 29(1271)$. Simplify... | 2^{64}-1 | Number Theory | proof | Yes | Yes | olympiads | false | 35,906 |
31 (1273). Prove that there do not exist integer coefficients $a, b, c$ and $d$, such that the value of the polynomial $a x^{3}+b x^{2}+c x+d$ is 1 when $x=19$ and is 2 when $x=62$.
$4^{*}$ | Proof. Suppose there exist such coefficients $a, b, c$ and $d$ that the following equalities hold:
$$
\begin{aligned}
& a \cdot 19^{3}+b \cdot 19^{2}+c \cdot 19+d=1 \\
& a \cdot 62^{3}+b \cdot 62^{2}+c \cdot 62+d=2
\end{aligned}
$$
Subtracting the first equation from the second, we get:
$$
\begin{gathered}
a \cdot\l... | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,907 |
33 (1275). Find all prime numbers $p$ and $q$ for which $p^{2} - 2 q^{2}=1$. | Solution. From the equality $p^{2}-2 q^{2}=1$ we have $p^{2}-1=2 q^{2}$, from which $\frac{(p-1)(p+1)}{2}=q \cdot q$. Since in the right-hand side of the equality we have the product of equal prime numbers ( $q$ has no other divisors except $q$ and 1), then in the left-hand side of the equality we also have the product... | p=3,q=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,908 |
34 (1276). For what values of $a, b, c$ and $d$ is the equality $5 x^{3}-32 x^{2}+75 x-71=a(x-2)^{3}+b(x-2)^{2}+c(x-2)+d$ an identity | Solution. We will expand the brackets on the right side of the equality and combine like terms:
$$
\begin{gathered}
a(x-2)^{3}+b(x-2)^{2}+c(x-2)+d= \\
=a x^{3}-6 a x^{2}+12 a x-8 a+b x^{2}-4 b x+4 b+c x-2 c+d= \\
=a x^{3}+(b-6 a) x^{2}+(12 a-4 b+c) x-8 a+4 b-2 c+d
\end{gathered}
$$
52
For the equality
$$
\begin{ali... | =5,b=-2,=7,=-9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,909 |
35 (1277). Represent the polynomial $3 x^{3}+7 x^{2}+9 x+6$ as a polynomial $a y^{3}+b y^{2}+c y+d$, where $y=x+1$. | Solution. Method I. From the equality $y=x+1$, express $x$ and substitute the value $x=y-1$ into the given polynomial:
$$
\begin{gathered}
3 x^{3}+7 x^{2}+9 x+6=3(y-1)^{3}+7(y-1)^{2}+9(y-1)+6= \\
=3\left(y^{3}-3 y^{2}+3 y-1\right)+7\left(y^{2}-2 y+1\right)+9(y-1)+6= \\
=3 y^{3}-2 y^{2}+4 y+1 .
\end{gathered}
$$
We ob... | 3y^{3}-2y^{2}+4y+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,910 |
36(1278). For what natural values of $x$ and $y$ is the equation $3 x+7 y=23$ true? | Solution. Let's express one of the unknowns from the given equation, for example, $y$:
$$
y=\frac{23-3 x}{7}=\frac{21+2-3 x}{7}=3+\frac{2-3 x}{7}.
$$
By trial, we find the smallest natural value of $x$ for which $y$ is a natural number: $x=3$, from which $y=2$.
This value of $y$ is unique, since for $x>3$ ($x$ is a ... | 3,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,911 |
39(1281). Find two natural numbers whose sum is 168, and their greatest common divisor is 24. | Solution. Let $a$ and $b$ be two natural numbers such that $a+b=168, a=24 n_{1}, b=24 n_{2}$, where $n_{1}$ and $n_{2}$ are natural numbers. Then we have $24 n_{1}+24 n_{2}=168$, from which $n_{1}+n_{2}=7$. Considering that $n_{1}$ and $n_{2}$ are natural numbers summing to 7, it is easy to find the numbers $a$ and $b$... | 24144,or48120,or7296 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,912 |
40(1282). A schoolboy had several coins worth 15 k. and 20 k., with more 20 k. coins than 15 k. coins. The schoolboy spent one-fifth of all his money, paying two coins for movie tickets. He then spent half of the remaining money on lunch, paying with three coins. How many coins of each denomination did the schoolboy ha... | Solution. The problem reduces to solving the equation $20 x+15 y=n$ in natural numbers, where $x$ is the number of 20-kopeck coins, $y$ is the number of 15-kopeck coins, and $x>y$; the natural number $n$ is determined from the condition of the problem.
Since one fifth of all the money was paid with two coins, one fift... | 6,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,913 |
41(1283). The path from $A$ to $B$ goes 3 km uphill, 6 km downhill, and 12 km on flat ground. The motorcyclist covered this path in 1 hour and 7 minutes, and the return path in 1 hour and 16 minutes. Find the motorcyclist's speed uphill and downhill, if his speed on flat ground was 18 km/h. (Note that the motorcyclist ... | Solution. I method. On a flat surface in one direction, the motorcyclist traveled $\frac{2}{3}$ hours $\left(12: 18=\frac{2}{3}\right)$, or 40 minutes. Then, 3 km uphill and 6 km downhill, the motorcyclist traveled 27 minutes, and 6 km uphill and 3 km downhill, the motorcyclist traveled 36 minutes. If we denote the mot... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,914 |
42 (1284). How old is the brother and how old is the sister if 2 years ago the brother was twice as old as the sister, and 8 years ago - five times as old? | Solution. Method I. Let the current age of the sister be $x$ years, and the brother's age be $y$ years. According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
y-2=2(x-2) \\
y-8=5(x-8)
\end{array}\right.
$$
from which we find $x=10, y=18$.
Method II. Let 2 years ago the sister was $x$... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,915 |
43 (1285). From two cities $A$ and $B$, the distance between which is 180 km, a bus and a car set out towards each other at 6:20 AM. Their meeting occurred at 7:50 AM. If the bus had left 1 hour and 15 minutes earlier, and the car 15 minutes later, they would have met at 7:35 AM. What are the speeds of the bus and the ... | Solution. Let the speed of the bus be $v_{1}$ km/h, and the speed of the car be $v_{2}$ km/h. Since they met after 1.5 hours, we have the equation:
$$
1.5 v_{1} + 1.5 v_{2} = 180
$$
If the bus had left 1 hour and 15 minutes earlier, it would have been on the road for 2 hours and 30 minutes (7:35 - 5:05 = 2 hours 30 m... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,916 |
45 (1287). A rider and a pedestrian simultaneously set off from point $A$ to point $B$. The rider, arriving in $B$ 50 minutes earlier than the pedestrian, returned back to $A$. On the return trip, he met the pedestrian 2 kilometers from $B$. The rider spent 1 hour and 40 minutes on the entire journey. Find the distance... | S o l u t i o n. Since the rider spent 1 hour 40 minutes on the entire journey, he spent 50 minutes on the journey from $A$ to $B$ (1 hour 40 minutes: $2=50$ minutes). Then the pedestrian spent 1 hour 40 minutes on the journey from $A$ to $B$, as he arrived in $B$ 50 minutes later than the rider. Therefore, the pedestr... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,917 |
46 (1288). Freshly mined coal contains $2 \%$ water, and after two weeks of exposure to air, it contains $12 \%$ water. By how many kilograms has the mass of a ton of coal increased after the coal has been lying in the air for two weeks? | Solution. Since the freshly mined coal contains $2 \%$ water, in one ton of such coal, there are 20 kg of water and 980 kg of dry matter.
After two weeks of exposure to the air, the 980 kg of dry matter constitutes $88 \% (100 \% - 12 \% = 88 \%)$, meaning the total weight of the coal will be approximately 1114 kg $\l... | 114 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,918 |
47 (1289). Two brothers walk together from school to home at the same speed. One day, 15 minutes after leaving school, the first brother ran back to school and, upon reaching it, immediately set off to catch up with the second. Left alone, the second continued walking home at half his usual speed. When the first brothe... | Solution. Let the initial speed of the brothers be denoted by $v$ (m/min), and the segment of the path that the second brother traveled at a speed of $\frac{v}{2}$ be denoted by $s$. For this segment, he spent $\frac{2 s}{v}$ minutes $\left(s: \frac{v}{2}\right)$, which is 6 minutes more than the usual time spent (at s... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,919 |
50(1136). Before the swimming competition, each of the four participants $A, b, B$ and $\Gamma$ was asked what place they expected to finish in. $A$ said: “I will be first,” $B$ said: “I will not be last,” $V$ said: “I will not be first or last,” and $\Gamma$ said: “I will be last.” After the race, it turned out that o... | Solved. Let's create a table where we will indicate the assumed results with a "plus" sign.
| Swimmer | Place | | | |
| :---: | :---: | :---: | :---: | :---: |
| | 1 | 2 | 3 | 4 |
| $A$ | + | | | |
| 5 | + | + | + | |
| $B$ | | + | + | |
| | | | | + |
Let's assume that $A$ was wrong. Then he could have ... | A | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,920 |
51 (1137). Prove that any amount greater than seven kopecks can be paid using three-kopeck and five-kopeck coins without receiving change. | Solution. I with assistance. It is sufficient to check that 8, 9, and 10 k can be paid with 3 k and 5 k coins. $(8=3+5, 9=3+3+3, 10=5+5)$, and then add coins of $3 \mathrm{k}$.
II with assistance. Any natural number $m$, greater than 7, can be represented in one of the following forms:
\[
\begin{aligned}
& m=3 n, \te... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,921 |
52 (1138). In a box, there are multicolored balls: 5 white, 12 red, and 20 black. What is the smallest number of balls that need to be taken out of the box without looking inside, to ensure that among them there will be: a) at least one ball of each of the specified colors; b) 10 balls of one color? | Solution. a) In the worst case, 20 black and 12 red balls can be drawn, then the next ball drawn (white) will ensure the presence of balls of all the specified colors. Therefore, it is sufficient to draw 33 balls $(20+12+1=33)$ to ensure that there is at least one ball of each of the specified colors among them.
b) In... | 33 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,922 |
53(1139). I have thought of an integer not exceeding 1000. How can you find out what number I have thought of by asking no more than 10 questions, to which I will only answer “yes” or “no”? | Solution. When solving the problem, it is convenient to use a table of powers with base 2, as the number not exceeding 1000 is in the numerical interval $\left[2^{0}, 2^{10}\right)$. Questions should be asked in such a way that after each answer, the interval is reduced by half. (The found intervals will be marked on F... | 300 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,923 |
58(1154). Prove that the value of the expression $11^{6}+14^{6}-$ $-13^{3}$ is divisible by 10. | Solution. The number $11^{6}$ ends with the digit 1. The number $14^{6}$ ends with the digit 6, since $14^{6}=\left(14^{2}\right)^{3}$, and $14^{2}$ ends with the digit 6. The number $13^{3}$ ends with the digit 7. Therefore, the number $11^{6}+14^{6}-13^{3}$ ends with the digit 0, i.e., it is divisible by 10.
$59(115... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,925 |
60(1159). Prove that the value of the expression $3^{n+2}-$ $-2^{n+2}+3^{n}-2^{n}$ is divisible by 10 for any natural value of $n$. | Solution. $3^{n+2}-2^{n+2}+3^{n}-2^{n}=3^{n}\left(3^{2}+1\right)-2^{n}\left(2^{2}+1\right)=$ $=3^{n} \cdot 10-2^{n} \cdot 5=3^{n} \cdot 10-2^{n-1} \cdot 10=10 \cdot\left(3^{n}-2^{n-1}\right)$.
The value of the expression $3^{n}-2^{n-1}$ for any $n \in \boldsymbol{N}$ is a natural number, which means the value of the e... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,926 |
61 (1161). When the polynomial $2 x^{3}-5 x^{2}+7 x-8$ is multiplied by the polynomial $a x^{2}+b x+11$, the resulting polynomial does not contain either $x^{4}$ or $x^{3}$. Find the coefficients $a$ and $b$ and determine what polynomial results from the multiplication. | Solution. Let's multiply the given polynomials, writing their product (as well as the factors) in standard form:
$$
\begin{aligned}
& \left(2 x^{3}-5 x^{2}+7 x-8\right) \cdot\left(a x^{2}+b x+11\right)= \\
= & 2 a x^{5}-5 a x^{4}+7 a x^{3}-8 a x^{2}+2 b x^{4}-5 b x^{3}+ \\
+ & 7 b x^{2}-8 b x+22 x^{3}-55 x^{2}+77 x-88... | =4,b=10;8x^{5}-17x^{2}-3x-88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,927 |
64(1168). Can: a) the sum of five consecutive natural numbers be a prime number; b) the sum of the squares of five consecutive natural numbers be a prime number? | Solution. a) Let's check on specific examples whether the sum of five consecutive natural numbers is a prime or composite number: $1+2+3+4+5=15 ; 2+3+4+5+6=$ $=20 ; 3+4+5+6+7=25 ; \ldots ; 10+11+12+13+14=60$.
The numbers $15,20,25,60$ are composite numbers. We get the hypothesis: the sum of five consecutive natural nu... | 47 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,928 |
66(1182). Two containers with capacities of 144 liters and 100 liters contain some amount of water. If the larger container is filled to the top with water from the smaller one, then the latter will have $\frac{1}{5}$ of its initial amount of water left. If the smaller container is filled to the top from the larger one... | Solution. I with assistance. Let there be $x$ liters of water in the larger vessel, and $y$ liters in the smaller one. According to the condition, $\frac{4}{5} y$ is added to the larger vessel, filling it completely, i.e.,
$$
x+\frac{4}{5} y=144
$$
To the smaller vessel, $\frac{5}{12} x$ is added to fill it to the to... | 96,60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,929 |
67 (1183). One vessel contains 49 liters of water, and the other contains 56 liters. If the first vessel is filled to the top with water from the second vessel, the second vessel will be only half full. If the second vessel is filled to the top with water from the first, the first vessel will be only one-third full. Wh... | Solution. Method I. Let the capacity of the first vessel be $(49+x)$ liters, and the capacity of the second vessel be $(56+y)$ liters. If $x$ liters are poured from the second vessel into the first, then $(56-x)$ liters will remain in the second vessel, or $\frac{1}{2}(56+y)$. We have the equation:
$$
56-x=\frac{1}{2}... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,930 |
69 (1185). From $A$ to $B$, the distance between which is 37 km, two buses with the same speed left at 7:18 AM and 7:48 AM, respectively. A cyclist, who left $B$ for $A$ at 7:28 AM, met the first bus at 7:58 AM and the second bus at 8:19 AM. Find the speeds of the cyclist and the buses. | Solution. Let the speed of the buses be $v_{1}$ km/min, and the speed of the cyclist be $v_{2}$ km/min. Since the first bus traveled for 40 minutes (7:58 AM - 7:18 AM = 40 minutes) before meeting the cyclist, the second bus for 31 minutes (8:19 AM - 7:48 AM = 31 minutes), and the cyclist spent 30 minutes (7:58 AM - 7:2... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,931 |
70(1186). From points $M$ and $K$, which are 70 km apart, a bus and a cyclist set off towards each other simultaneously. They met after 1 hour and 24 minutes. Continuing their journey at the same speed, the bus arrived in $K$ and, after a 20-minute stop, set off back. Find the speeds of the bus and the cyclist, given t... | Solution. Let the speed of the bus be $v_{1}$ km/h, and the speed of the cyclist be $v_{2}$ km/h. The bus and the cyclist met after 1 hour and 24 minutes, i.e., after $\frac{7}{5}$ hours, so we have the equation $\frac{7}{5} v_{1}+\frac{7}{5} v_{2}=70$.
Since the cyclist had been riding for $4 \frac{1}{12}$ hours (1 h... | v_{1}=35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,932 |
71 (1187). A tourist left from $A$ to $B$. After 1 hour and 20 minutes, a cyclist left from $A$ in the same direction and overtook the tourist after 30 minutes. Upon arriving at $B$, the cyclist, without stopping, turned back and met the tourist 1.5 hours after the first meeting. Find the speeds of the tourist and the ... | Solution. Let the speed of the cyclist be $v_{1}$ km/h, and the speed of the tourist be $v_{2}$ km/h. Since the cyclist overtook the tourist 30 minutes $\left(\frac{1}{2}\right.$ hours) after departure and traveled $\frac{1}{2} v_{1}$ km, while the tourist walked 1 hour 20 minutes longer than the cyclist $\left(1 \frac... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,933 |
73 (1193). The least common multiple of two numbers, neither of which divides the other, is 90, and their greatest common divisor is 6. Find these numbers. | Solution. Knowing the rule for finding the LCM and GCD of numbers using their prime factorization, it is easy to find the solution method:
1) Represent the LCM - the number 90 - as a product of prime factors: $90=2 \cdot 3 \cdot 3 \cdot 5$.
2) Since the GCD of the sought numbers is 6, and they do not divide each other... | 1830 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,934 |
74(1195). Represent the expression $24 x y$ as the difference of squares of two polynomials.
$\mathrm{Pe}$ solution. Represent the expression $24 x y$ as the sum of two identically equal monomials:
$$
24 x y=12 x y+12 x y \text {. }
$$
Then the monomial $12 x y$ can be represented as the double product of one number... | Solution.
$$
\begin{gathered}
2 a\left(a^{2}+3 b^{2}\right)=2 a^{3}+6 a b^{2}= \\
=\left(a^{3}+3 a b^{2}+3 a^{2} b+b^{3}\right)+\left(a^{3}+3 a b^{2}-3 a^{2} b-b^{3}\right)= \\
=(a+b)^{3}+(a-b)^{3}
\end{gathered}
$$
74
$76(1197)$. Represent the expression $2 b \cdot\left(3 a^{2}+b^{2}\right)$ as the difference of cub... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,935 | |
77(1198). Find the coefficients $a, b$ and $c$ of the polynomial $a x^{2}+b x+c$, knowing that the equality
$$
2 x^{4}+x^{3}-41 x^{2}+83 x-45=\left(a x^{2}+b x+c\right)\left(x^{2}+4 x+9\right)
$$
is an identity. | Solution. By equating the coefficients of the fourth and third powers of $x$ on the left and right sides of the given equality, we will have $a=2, 4a+b=1$, from which $b=-7$.
By equating the free terms, we find $c: -45=-9c, c=5$. Answer: $a=2, b=-7, c=5$.
$\triangle 78(1199)$. Form a polynomial of the sixth degree wi... | =2,b=-7,=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,936 |
83(952). Prove that the value of the expression $16^{5}+2^{15}$ is divisible by 33. | Solution. $\quad 16^{5}+2^{15}=\left(2^{4}\right)^{5}+2^{15}=2^{20}+2^{15}=2^{15}\left(2^{5}+1\right)=$ $=33 \cdot 2^{15}$.
3 remark. Direct calculation of the value of the expression $16^{5}+2^{15}$ (for example, using a calculator) and obtaining an integer result from the division of the sum $16^{6}+2^{15}$ by 33 is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,937 |
84 (953). Prove that the value of the expression $333^{555}+555^{333}$ is divisible by 37. | Solution.
$$
\begin{aligned}
333^{555}+ & 555^{333}=3^{555} \cdot 111^{555}+5^{333} \cdot 111^{333}= \\
& =111^{333}\left(3^{555} \cdot 111^{222}+5^{333}\right) .
\end{aligned}
$$
Since $111=3 \cdot 37$, the statement is proved.
Note. It is advisable to draw students' attention to the fact that the given proof metho... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,938 |
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