problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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6. Find a two-digit number if it is known that the sum of its digits is 13, and the difference between the sought number and the number written with the same digits but in reverse order is a two-digit number with 7 units. | 6. The desired number $N=10 x+y$, where $x=1,2,3,4,5,6,7,8,9 ; y=0,1,2$, $3,4,5,6,7,8,9$. Let's write down the two-digit numbers whose sum of digits is 13. Such numbers will be $49,58,67,76,85,94$. Let's write down the reversed numbers: $94,85,76,67,58,49$. The condition of the problem is satisfied only by the pair of ... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. The dividing circle of the dividing head is uniformly marked with 24 holes. What regular polygons can be marked using this instrument? | 10. By connecting all 24 holes sequentially, we get a twenty-four-sided polygon, connecting every other hole - a twelve-sided polygon,
connecting every two holes - an eight-sided polygon, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. In what ways can you give 78 rubles, having only five-ruble and three-ruble banknotes? | 13. Let $x$ be the number of 5-ruble stamps, and $y$ be the number of 3-ruble stamps, then $78 = 5x + 3y$, from which $x = 15 - \frac{3(y-1)}{5}$. Since $x$ is a positive integer, $15 - \frac{3(y-1)}{5} > 0$, and thus $y < 26$. Additionally, $\frac{3(y-1)}{5}$ is an integer, meaning $y-1$ is a multiple of 5. Let $y-1 =... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. The purchase costs 19 rubles. The buyer has only three-ruble bills, and the cashier has only five-ruble bills. Can the purchase be paid for with the available money, and if so, how exactly? (Limit the case to when the buyer and the cashier each have 15 bills of the specified denomination). | 14. 19 rubles can be paid in two ways:
1) the buyer gives 24 rubles and receives 5 rubles in change. 2) gives 39 rubles and receives 20 rubles in change. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
14. $\log _{\frac{\sqrt{3}}{3}}\left(\log _{8} \frac{\sqrt{2}}{2}-\log _{3} \frac{\sqrt{3}}{3}\right)$.
In № $15-27$ solve the equations: | 14. We should switch to the logarithm with base $\frac{\sqrt{3}}{3}$.
Then replace $\frac{1}{3}$ with $\left(\frac{\sqrt{3}}{3}\right)^{2}$. Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. $\log _{4}\left\{2 \log _{3}\left[1+\log _{2}\left(1+3 \log _{2} x\right)\right]\right\}=\frac{1}{2}$. | 16. Gradually potentiating, we arrive at the equation $1+3 \log _{2} x=4$. From which $x=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19. $\log _{5} 120+(x-3)-2 \log _{5}\left(1-5^{x-3}\right)=-\log _{5}\left(0.2-5^{x-4}\right)$. | 19. $1-5^{x-3} \neq 0, x \neq 3,0,2-5^{x-4}=\frac{1}{5}\left(1-5^{x-3}\right)$.
Raising the entire expression, we get
$$
\frac{120 \cdot 5^{x-3}}{\left(1-5^{x-3}\right)^{2}}=\frac{5}{1-5^{x-3}} ; \quad 5^{x-3}=5^{-2} . \quad \text { Hence } x=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
26. $5^{\lg x}-3^{\lg x-1}=3^{\lg x+1}-5^{\lg x-1}$. | 26. Grouping the powers with bases 5 and 3, we get
$$
5^{\lg x}\left(1+\frac{1}{5}\right)=3^{\lg x}\left(3+\frac{1}{3}\right) . \quad \text { Hence }\left(\frac{5}{3}\right)^{\lg \cdot x}=\left(\frac{5}{3}\right)^{2} ; \quad x=100
$$ | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Calculate the product $p=1 \cdot 2^{\frac{1}{2}} \cdot 4^{\frac{1}{4}} \cdot 8^{\frac{1}{8}} \cdot 16^{\frac{1}{16}} \cdot \ldots$
In № $15-20$ find the sums: | 14. Let's write all powers as powers with base 2
$$
p=1 \cdot 2^{\frac{1}{2}} \cdot 2^{\frac{2}{4}} \cdot 2^{\frac{3}{8}} \cdot 2^{\frac{4}{16}} \ldots=2^{\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots}
$$
The problem has been reduced to finding the sum $S=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\ld... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
21. Find the sum $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{n}{(n+1)!}$ and compute its limit as $n \rightarrow \infty$. | 21. By forming the sequence of partial sums, we get
$$
S_{1}=\frac{2!-1}{2!}, \quad S_{2}=\frac{3!-1}{3!}, \ldots \quad \text { Hence } S_{n}=\frac{(n+1)!-1}{(n+1)!}, \quad S=\lim _{n \rightarrow \infty} S_{n}=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. $y=\frac{|x|-2}{|x|}$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
15. $y=\frac{|x|-2}{|x|}$. | 15. (Fig. 15) The root of the submodular expression is 0.
For $x>0, \quad y=1-\frac{2}{x}$.
For $x<0, \quad y=1+\frac{2}{x}$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
31. How many planes of symmetry do the following have: a) a cube; b) a regular tetrahedron? | 31. a) A cube has 9 planes of symmetry; b) a regular tetrahedron has 6 planes of symmetry. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
32. How many axes of symmetry do the following have: a) a cube; b) a regular tetrahedron | 32. a) A cube has 13 axes of symmetry; b) a regular tetrahedron has 7 axes of symmetry. | 13 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
21. What is the maximum area that a triangle with sides \(a, b, c\) can have, given that the sides are within the following limits: \(0 \leqslant a \leqslant 1 \leqslant b \leqslant 2 \leqslant c \leqslant 3\) ? | 21. $S=\frac{1}{2} a b \sin \alpha$. Takes the maximum value when $a=1, b=2, \sin \alpha=1$,
$S=\frac{1}{2} \cdot 1 \cdot 2 \cdot 1=1$. Under these conditions, $c=\sqrt{1+4}=\sqrt{5}$, which satisfies the problem's condition ( $2 \leqslant c \leqslant 3$). | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
33. Determine $a$ so that the sum of the squares of the roots of the equation $x^{2}+(2-a) x-a-3=0$ is the smallest. | 33. $a=1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
35. Find the maximum value of the expression $\log _{2}^{4} x+12 \log _{2}^{2} x \log _{2} \frac{8}{x}$, assuming that $x$ varies from 1 to 64. | 35. Let's denote the given expression by $y$. Then it is easy to see that $y=\log _{2}^{2} x\left(6-\log _{2} x\right)^{2}$. But $1<x<64$, so $0<\log _{2} x<6$. The maximum value of $y$ coincides with the maximum value of the product. Since $\log _{2} x+(6-$ $\left.-\log _{2} x\right)=6$, the product will be the larges... | 81 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
43. Points $A$ and $B$ 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 per hour. At the same time, a motorcycle departs from $B$ in the same direction with a constant acceleration of 32 km/hour ${ }^{2}$. Determine... | 43. At time $t$, the car is at a distance of $40 t$ km from point $A$, and the motorcycle is at a distance of $\left(\frac{32 t^{2}}{2}+9\right)$ km from the same point. The distance between them is $\left|16 t^{2}+9-40 t\right|$ (Fig. 29). The greatest distance of 16 km will be reached in 1 hour 15 minutes. | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
30. Several families lived in one house. In total, there are more children in these families than adults: there are more adults than boys; more boys than girls; and more girls than families. There are no childless families, and no families have the same number of children. Each girl has at least one brother and at most... | 30. Three families lived in the house. In one of these families, there is a single child, a boy. In another family - two girls and a boy. In the third - two girls and three boys. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
32. During the lunch break, the members of the communist labor brigade started talking about how many newspapers each of them reads. It turned out that each member subscribes to and reads exactly two newspapers, each newspaper is read by five people, and any combination of two newspapers is read by one person. How many... | 32. 6 newspaper names, 15 members. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
51. There are three villages: $A, B$, and $C$. The inhabitants of $A$ always tell the truth, the inhabitants of $B$ always lie, and the inhabitants of $C$, depending on their mood, tell the truth or lie. Since the villages are located close to each other, the inhabitants visit each other. A tourist ended up in one of t... | 51. Four questions: 1) Am I in one of the settlements $A$ and $B$? 2) Am I in settlement C? 3) Do you live in settlement C? 4) Am I in settlement $A^{*}$? | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
52. My friend has thought of an integer between 10 and 19. To guess which number he has in mind, I can ask him questions that he will answer with "yes" or "no": What is the smallest number of questions and which specific questions can I ask to determine which number he thought of? | 52. The least number of questions is three. The first question: "Is the number you are thinking of among the first four numbers (11-14)?" If the answer is "Yes," then the second question can be: "Is the number you are thinking of among the numbers 11 and 12?" If the answer is "No," then the third question can be: "Is t... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. (7th grade) Several girls (all of different ages) were collecting white mushrooms in the forest. The collected mushrooms they distributed as follows: the youngest received 20 mushrooms and 0.04 of the remainder. The next oldest received 21 mushrooms and 0.04 of the remainder, and so on. It turned out that everyone r... | 1. Let the number of collected mushrooms be denoted by $x$. Then the youngest girl received $20+(x-20) \cdot 0.04$ mushrooms. The second girl received $19.392+0.0384 x$. Both received the same amount. By setting up the equation and solving it, we find $x=120$. There were 5 girls, each received 24 mushrooms. | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (9th grade) Find a natural number $n$, knowing that the sum $1+2+3+\ldots+n$ is a three-digit number consisting of identical digits. | 11. $1+2+3+\ldots+n=\frac{n+1}{2} n$. This sum equals a three-digit number $\overline{a a a}$, where $\overline{a a a}=100 a+10 a+a=111 a$. We get the equation $\frac{n+1}{2} n=111 a$. From this, $a=\frac{n(n+1)}{2 \cdot 3 \cdot 37}$. Since $a$ is an integer, $n(n+1)$ must be divisible by 2, 3, and 37. This is only pos... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21. (7th grade) (Newton's problem) Grass grows at the same rate and density everywhere in the corner. It is known that 70 cows would eat it in 24 days, and 30 cows in 60 days. How many cows would eat all the grass in 96 days? (It is assumed that the cows eat the grass uniformly). | 21. In one day, $x$ amount of grass grows. If the initial amount of grass is taken as 1, then over 24 days, the cows eat $1+24 x$ amount of grass, and one cow eats $\frac{1+24 x}{24 \cdot 70}$ amount of grass per day. But according to the second condition, a cow eats $\frac{1+60 x}{30 \cdot 60}$ amount of grass per day... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
29. (8th grade) When asked what his ticket number was, someone answered like this: "If all six two-digit numbers that can be formed from the digits of the number are added together, then half of the resulting sum will be exactly my ticket number." Determine the ticket number. | 29. Let's denote the ticket number as: $N=100 x+10 y+\boldsymbol{z}$. According to the problem,
$$
\begin{gathered}
2 N=10 x+y+10 x+z+10 y+x+10 y+z+10 z+x+10 z+y= \\
=22 x+22 y+22 z, \text { i.e. } N=11(x+y+z)
\end{gathered}
$$
By equating the first and second results, we get the following equation: $100 x+10 y+z=11(... | 198 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
37. (10-11 grades) How many planes are equidistant from four points that do not lie in the same plane? | 37. Taking these points as the vertices of a tetrahedron, it is easy to establish that only seven planes can be drawn equidistant from its vertices. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
41. (8th grade) Find the smallest natural number that has the following properties: a) its representation in the decimal system ends with the digit 6; b) if the last digit 6 is erased and this digit 6 is written in front of the remaining digits, the resulting number is four times the original number. | 41. The desired number can be written as: $N=\overline{a_{1} a_{2} a_{3} \ldots a_{n-1}}-1$. After rearranging the digit 6, we get the number $K=4 N=6 \bar{a}_{1} a_{2} a_{3} \ldots a_{n-1}$. Since the last digit of the number $N$ is 6, the last digit of the number $K$ is 4, i.e., $a_{n-1}=4$.
Substituting the value $... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
45. (9th grade) Find a four-digit number that is a perfect square, knowing that the first two digits, as well as the last two, are equal to each other. | 45. The desired number can be written as: $N=\overline{a a b} b$ or $N=1100 a+11 b=$ $=11(99 a+a+b)$. For the desired number to be a perfect square, it is necessary that the factor 11 appears in the number $N$ twice, i.e., the sum $a+b$ must be divisible by $11.1<a+b<18$, hence, $a+b=11$. Then $N=11^{2}(9 a+1)$. The bi... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
58. (8th grade) Given a thousand-sided polygon. Inside it, 500 points are taken, which are connected to each other and to the vertices of the thousand-sided polygon. How many triangles will be formed? (The sides of the triangles do not intersect). | 58. Let $x$ be the number of triangles, $2 d(n-2)$ be the sum of the interior angles of a polygon, $2 d x$ be the sum of the interior angles of the triangles, and $4 d \cdot 500$ be the sum of the full angles at the given 500 points. We can set up the equation $2 d x = 500 \cdot 4 d + 2 d(n-2)$, where $n=1000$. Answer:... | 1998 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
61. (9th grade) In a square table consisting of $8 \times 8$ cells, natural numbers from 1 to 64 are arranged in order. A number is selected, and the row and column in which it is located are crossed out. From the remaining numbers, another number is selected, and the row and column in which it is located are crossed o... | 61. The chosen number $N=M_{i}+8\left(n_{k}-1\right)$, where $M_{i}$ is the column number, and $n_{k}$ is the row number in which it stands. Then the total sum of the eight chosen numbers can be written as:
$$
S=\sum_{i=1, k=1}^{i=8, k=8} M_{i}+8\left(n_{k}-1\right)=\frac{8+1}{3} 8+8 \frac{7+1}{2} 7=260
$$ | 260 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
64. (9th grade) The number 7 is raised to the seventh power, the resulting number is again raised to the seventh power, and so on. This process is repeated 1000 times. What is the last digit of this number? | 64. Considering the powers of the number 7 in sequence, we notice that the last digits of these powers repeat every four, so the number given in the problem ends with the digit 7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
65. (10-11 grades) Into how many parts do the planes of the faces of a tetrahedron divide space? | 65. The calculation is easiest to perform as follows: inside the tetrahedron - 1 part, at the vertices of the trihedral angles - 4 parts, along the edges - 10 parts. In total, 15 parts. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
81. The difference $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$ is an integer. Find this number. | Solution. Since $40 \sqrt{2}-57<0$, then $|40 \sqrt{2}-57|=$ $=57-40 \sqrt{2}$. Then
$$
\begin{aligned}
A & =\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}= \\
& =\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{2}}
\end{aligned}
$$
Let $57-40 \sqrt{2}=(a+b \sqrt{2})^{2}$, where $a$ and $b$ are unknown coefficients. Then
$$
5... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
89. Simplify the expression:
a) $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$
b) $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$ | Solution. a) Let $\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{\overline{2}}}=x ; x<0$, since $\sqrt{57-40 \sqrt{2}}<\sqrt{57+40 \sqrt{2}}$.
Square both sides of the equation:
$$
x^{2}=57-40 \sqrt{2}-2 \sqrt{57^{2}-(40 \sqrt{2})^{2}}+57+40 \sqrt{2}
$$
from which
$$
x^{2}=114-2 \sqrt{49}, x^{2}=100, x=-10
$$
(Solution b... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
90. Prove that $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}=3$. | Solution. Let the value of the expression on the left side of the equality be denoted by $x$. Reasoning as in the solution of Example 89, b, we obtain the equation $x^{3}-3 x-18=0$, from which $x=3$.
## Solving Equations | 3 | Algebra | proof | Yes | Yes | olympiads | false |
91. Let's solve the equation
$$
(3 x-1) \sqrt{x-4}=17 \sqrt{2}
$$ | Solution. Method 1. The domain of expression (3) is the interval $[4 ;+\infty)$. Squaring both sides of the equation, we obtain the equivalent equation
$$
9 x^{3}-42 x^{2}+25 x-582=0
$$
We find the critical points of the function $f(x)=9 x^{3}-42 x^{2}+25 x-582$:
$$
f^{\prime}(x)=27 x^{2}-84 x+25=0
$$
from which $x... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
29. Space Division. Through a fixed point in space, we draw planes to divide the space into as many parts as possible. One plane will divide the space into two parts, two intersecting planes will divide it into four parts, and three planes intersecting at a point and having no other common point will divide the space i... | 29. Instead of the entire space, we will divide the tar, through the center of which we draw planes. On the surface of the sphere (bounded by its sphere), intersecting great circles will arise. We will take one of them as the әquator and project all other circles from the center of the sphere onto a plane tangent to th... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
31. Cube. Holding a model of a cube in your hand so that it can rotate around its longest axis (i.e., around the line connecting opposite vertices), you can wind black yarn around it without any gaps. The yarn will shade only half of the cube (why?). The same can be done with another axis; there are four in total, and ... | 31. The yarn wound around a cube rotating about one of its axes (Fig. 41) will remain only on those edges that do not have common points with the axis of rotation; the yarn will cover half of each face of the cube, i.e., half of the cube's surface.
Now, let's rotate the cube sequentially around each of the four axes, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
32. Geodesics. This problem does not require knowledge of mathematics. Let's place a rubber band (so-called "prescription", used in pharmacies for packaging medicines) on a stationary cube in such a way that it holds on the cube and does not cross itself.
The line along which this rubber band will lie is called a geod... | 32. We will prove that through each point on the surface of a cube, there pass four different geodesics, and in total, we have seven families of geodesic lines.
If we assume that the cube is smooth, then a rubber band wrapped around it will be arranged in such a way that the perimeter of the polygon it forms will reac... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
34. Unfolding a Cube. Models of polyhedra are made from flat nets. In a net, faces are adjacent to each other along edges, and the model is constructed by folding the cardboard net along the edges. A regular tetrahedron has two such different nets. How many does a cube have? | 34. All existing nets (a total of 11) are shown in Fig. 49. The first six solutions give nets in which four faces of the cube are arranged in one strip of the net. No other solutions of this type exist. The next four nets are those in which there are three faces in one
 can be formed from them. How many different tetrahedrons can result from this? | 37. The answer depends on whether we consider two tetrahedra different if one can be obtained from the other by a mirror reflection. We will show that in the first case, there are 60 different tetrahedra; in the second case, the number will obviously be reduced to 30.
Consider Fig. 55, which shows a tetrahedron with e... | 30 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
53. How old is Sofia Sergeyevna? Our acquaintance Sofia Sergeyevna is not old, as she was born after World War I, but she does not like to answer directly when asked how old she is.
When asked on July 27, 1950, how old she was, she replied: I am only one year old, because I only celebrate my birthday when it falls on ... | 53. Let's denote the days of the week as I, II, ..., VII and construct a table, agreeing that the day of the week that fell on a certain specific date in 1911 is denoted by the number I.
| $\boldsymbol{A}$ | B | c | $D$ | $A$ | $B$ | $c$ | D |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1911 | ... | 28 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
54. How many fish are in the pond? A certain ichthyologist wanted to determine how many fish in the pond were suitable for catching. For this purpose, he cast a net with a pre-selected mesh size and, pulling it out, found 30 fish, marked each of them, and threw them back into the pond. The next day, he cast the same ne... | 54. Let $n$ denote the number of fish in the pond suitable for catching. Then the ratio of the number of marked fish to the total number of fish is $30 / n$.
The second time, the ichthyologist caught 40 fish, two of which were marked. The ratio of the number of marked fish to the total number of fish caught is $1 / 20... | 600 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
72. Railway Network (I). There are five cities; no three of them lie on the same straight line. These cities need to be connected by a railway network consisting of four straight roads; at that, railway lines can be built one over another on viaducts.
How many such different railway networks exist? | 72. There can exist three types of railway networks, as shown in Fig. 149.
1) In the first case, each city can be a node where four lines converge, so there will be five different networks of this kind.
2) In the second case, a city that will be a node where three lines converge can be connected to three other cities i... | 125 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
87. Four dogs. Four dogs $A, B, C$, and $D$ stand at the corners of a square meadow and suddenly start chasing each other as indicated by the arrows in Fig. 2.
Each dog runs straight towards the next: $A$ towards $B$, $B$ towards $C$, $C$ towards $D$, and $D$ towards $A$. The side of the meadow is 100 m, and the speed... | 87. Since each dog runs at a right angle to the direction of the dog chasing it, and the one chasing runs straight toward the fleeing one, the chasing dog approaches the next dog at a speed of $10 \mathrm{m} /$ s and will catch it after 10 seconds. As a result, the path of each dog is $100 \mathrm{~m}$. At every moment... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
88. The Chase (I). Ship $P$ has spotted ship $Q$, which is sailing in a direction perpendicular to $P Q$, maintaining its course. Ship $P$ is chasing $Q$, always heading directly towards $Q$; the speed of both ships is the same at any moment (but can vary over time). Without calculations, it is clear that $P$ is sailin... | 88. Let $\alpha$ denote the instantaneous angle between the direction $P Q$ and the path of ship $Q$ (Fig. 163), and $v$ - the speed of ships $P$ and $Q$ at that moment. The mutual approach of the ships is influenced by the speed $v$ of ship $P$, directed towards $Q$, and the component $v \cos \alpha$ of the speed of s... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
95. Word game. Dr. Sylvester Sharadek announced that he can always guess the word you think of if he is allowed to ask 20 questions, to which the answers should be only "yes" or "no," and if the word is in the dictionary. Do you think he is boasting? | 95. There are no dictionaries containing more than a million words. For example, if a dictionary has 500 pages, we will open it in the middle and find the last word on page 250, let's say, "narcissus." The first question should be: is the word we are guessing located after the word "narcissus" in the dictionary? If the... | 20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
100. French cities. Dr. Sharadek, who knew strategy well, was interested in the latest war and in 1940 became acquainted with the map of the French theater of military operations. From this, the following problem probably arose. The distance (as the crow flies, as are all distances in this problem) from Chalon to Vitry... | 100. The five cities of Chalon, Vitry, Chaumont, Sän-Cantän, and Reims form a closed pentagon; one of its sides (the side Chaumont - Sän-Cantän) is the sum of the other four, since $236=86+40+30+80$. This is only possible when the vertices of the pentagon lie on a straight line. The cities are arranged on the line in t... | 150 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let's determine in which numeral system the following multiplication was performed: $352 \cdot 31=20152$. | Solution. Let $x$ be the base of the numeral system, then the given equality can be written in the form of an equation
$$
\left(3 x^{2}+5 x+2\right)(3 x+1)=2 x^{4}+x^{2}+5 x+2
$$
By performing the multiplication and combining like terms, we get:
$$
2 x^{4}-9 x^{3}-17 x^{2}-6 x=0
$$
It is clear that $x \neq 0$ and t... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let's find the sum of the cubes of the roots of the equation
$$
x^{3}+2 x^{2}+x-3=0
$$ | S o l u t i o n. This can be done in various ways, for example, by sequentially calculating the sum of the squares of the roots and then the sum of the cubes of the roots. However, we will use a frequently applied identity in mathematics:
$$
\begin{aligned}
a^{3}+b^{3}+c^{3}-3 a b c= & (a+b+c)\left(a^{2}+b^{2}+c^{2}-a... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1. Let's find pairs of real numbers $x$ and $y$ that satisfy the equation
$$
x^{2}+4 x \cos x y+4=0
$$ | S o l u t i o n. Several methods can be proposed to solve the given equation.
1st method. Since the equation is quadratic (if we do not consider $x$ under the cosine sign), we can express $x$ in terms of trigonometric functions of the angle $x y$:
$$
x=-2 \cos x y \pm \sqrt{4 \cos ^{2} x y-4}=-2 \cos x y \pm 2 \sqrt{... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Let's find all values of $a$ for which the system
$$
\left\{\begin{array}{l}
2^{b x}+(a+1) b y^{2}=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
has at least one solution for any value of $b,(a, b, x, y \in \mathbf{R})$. | S o l u t i o n. Suppose there exists some value $a$ for which the system has at least one solution, for example, for $b=0$.
In this case, the system will take the form:
$$
\left\{\begin{array}{l}
1=a^{2} \\
(a-1) x^{3}+y^{3}=1
\end{array}\right.
$$
It is clear that this system is consistent only if $a=1$ or $a=-1$.... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3. Compare the number $a$ with one, if
$$
a=0.99999^{1.00001} \cdot 1.00001^{0.99999} .
$$ | S o l u t i o n. Let's represent the numbers in the given expression as follows:
$$
0.99999=1-\alpha, \quad 1.00001=1+\alpha
$$
where $\alpha=0.00001$. Then
$$
a=(1-\alpha)^{1+\alpha}(1+\alpha)^{1-\alpha}=\left(1-\alpha^{2}\right)\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}
$$
Since $1-\alpha^{2}<1$, then $a<1$
... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
E x a m p l e 1. Let's find the limit of the sum
$$
S_{n}=\frac{3}{4}+\frac{5}{36}+\ldots+\frac{2 n+1}{n^{2}(n+1)^{2}}
$$
as $n \rightarrow \infty$. | S o l u t i o n. First, let's try to simplify $S_{n}$. Since the fraction 56
$\frac{2 k+1}{k^{2}(k+1)^{2}}$ can be represented as the difference $\frac{1}{k^{2}}-\frac{1}{(k+1)^{2}}$, then
$$
\begin{gathered}
\frac{3}{4}=1-\frac{1}{2^{2}} \\
\frac{5}{36}=\frac{1}{2^{2}}-\frac{1}{3^{2}} \\
\frac{2 n+1}{n^{2}(n+1)^{2}}=... | 1 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let $y \neq-1$. We set,
$$
x_{1}=\frac{y-1}{y+1}, \quad x_{2}=\frac{x_{1}-1}{x_{1}+1}, \quad x_{3}=\frac{x_{2}-1}{x_{2}+1}, \ldots
$$
What is $y$ if $x_{1978}=-\frac{1}{3}$? | Solution. Substituting the value of $x_{1}$ into the second equality, after simplifications we get:
$$
x_{2}=-\frac{1}{y}
$$
Further,
$$
x_{3}=\frac{y+1}{1-y}, \quad x_{4}=y, \quad x_{5}=\frac{y-1}{y+1}=x_{1}
$$
Therefore,
$$
x_{5}=x_{1}, \quad x_{6}=x_{2}, \quad x_{7}=x_{3}, \quad x_{8}=x_{4}, \ldots, \quad x_{19... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 2. Let's calculate the sum
$$
a^{2000}+\frac{1}{a^{2000}}
$$
if $a^{2}-a+1=0$. | S o l u t i o n. From the equality $a^{2}-a+1=0$, it follows that:
$$
a-1+\frac{1}{a}=0 \text { or } a+\frac{1}{a}=1
$$
Moreover,
$$
a^{3}+1=(a+1)\left(a^{2}-a+1\right)=0
$$
from which $a^{3}=-1$. Now we have:
$$
\begin{aligned}
& a^{2000}+\frac{1}{a^{2000}}=\left(a^{3}\right)^{666} a^{2}+\frac{1}{\left(a^{3}\righ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Points on a Line. If 10 points are placed at equal intervals on a line, they will occupy a segment of length s, and if 100 points are placed, the segment will have a length S. How many times greater is S than s? | 6. Between ten points there are nine intervals, and between a hundred points - ninety-nine. Therefore, $S$ is greater than $s$ by 11 times. | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Stereometric problem. How many faces does a hexagonal pencil have? | 7. It is important to ask: which pencil? If the pencil has not been sharpened yet, then 8, otherwise there may be variations...
Translating the text as requested, while preserving the original line breaks and formatting. | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12. The Musketeers' Journey. The distance between Athos and Aramis, riding on the road, is 20 leagues. In one hour, Athos travels 4 leagues, and Aramis - 5 leagues. What distance will be between them after an hour? | 12. Well, in which direction was each of the musketeers traveling? The problem statement does not mention this. If they were traveling towards each other, the distance between them would be 11 leagues. In other cases (make a diagram!), the possible answers are: 29 leagues; 19 leagues; 21 leagues. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16.On a circular route, two buses are operating, with an interval of 21 minutes. What would be the interval if three buses were operating on the route? | 16. Since the interval between buses is 21 minutes, the entire route is covered by a bus in 42 minutes. If there are 3 buses, the intervals between them will be 14 minutes. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
17. Economic forecast. Three chickens laid three eggs in three days. How many eggs will twelve chickens lay in twelve days? | 17. Of course not 12, as sometimes answered (inertial thinking!), but 48. | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12. Buying an album. To buy the album, Masha was short of 2 kopecks, Kolya was short of 34 kopecks, and Fedia was short of 35 kopecks. When they combined their money, it still wasn't enough to buy the album. How much does the album cost? | 12.Since Kolya has one kopek more than Fyodor, he has at least 1 kopek, and 1 kopek is added to Masha's money. But Masha was short of money for the album, so she was given less than 2 kopeks, which means Fyodor has no money at all and is short of the full cost of the album. Answer: the album costs 35 kopeks. | 35 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. An Arabic Tale. A Flock of Pigeons
$>\Delta \theta \Pi \mathrm{V} \oplus \theta\mathbf{v} \Pi \square \nabla \square \Lambda$
$\oplus \Lambda \nabla \theta \Pi \oplus \mathbf{V V} \varnothing \odot$
ロจ৫ఠ<>VIVOO flew up to a tall tree.
Some of the pigeons perched on the branches, while others settled under the t... | 13. From the condition of the problem, it is clear that the number of pigeons sitting on the branches is two more than those sitting below. Further, it follows from the condition that after one of the pigeons flew up to the branch, the number of pigeons sitting on the branch became twice as many as those sitting on the... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
14.Strike out a hundred digits from the number $12345678910111213 \ldots 5960 \quad$ so that the resulting number is the largest possible. | 14. The largest possible number should start with the maximum number of nines. We will "move" along the number from left to right, crossing out all digits except 9. We will cross out 27 digits: $12345678910111213141516171819 . . .5960$, 8 digits 19 digits
then 19 digits: $99 \underbrace{20212223242526272829 \ldots .59... | 99999785960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. In the government of the country of knights and liars, there are 12 ministers. Some of them are liars, and the rest are knights. Once, at a government meeting, the following opinions were expressed: the first minister said, “There is not a single honest person here,” the second said, “There is no more than one hones... | 4. Note that the number of true statements must match the number of honest people in the government. Further, if a statement from any minister is true, then the statements of each minister who spoke after him are also true. In this case, the only statement that would not lead to a contradiction is: "there are no more t... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Around a round table, eight people are sitting, each of whom is either a knight or a liar. When asked who their neighbors are, each of them answered: “My neighbors are a liar and a knight.” How many of them were liars? How would the answer change if nine people were sitting at the table? | 5. 6. At the table, there is at least one liar. Indeed, if only knights were sitting at the table, each knight's statement "next to me sits a knight and a liar" would be false, which is impossible. 2. The neighbors
=7$. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. What can you buy for a ruble? Nine boxes of matches cost 9 rubles and some kopecks, while ten such boxes cost 11 rubles and some kopecks. How much does one box cost? | 6. Note that a box of matches costs more than $\frac{11}{10}$ rubles, but less than $\frac{10}{9}$ rubles. That is, more than 1.10 rubles, but less than 1.111 rubles. Answer: 1 ruble 11 kopecks. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. The Big Laundry. After seven hours of washing, the length, width, and height of the soap piece were halved. For how many washes will the remaining soap last? | 12. Since the length, width, and height of the soap piece have been halved, its volume has decreased by 8 times, meaning that in 7 hours, the soap piece has decreased by $\frac{7}{8}$ of its volume (by $\frac{1}{8}$ of its volume per hour). Therefore, the soap will last for one more hour of heavy washing. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. How has the purchasing power of the population changed? The product has become cheaper by $20 \%$. By what percentage can more of the product be bought for the same amount of money? | 5. The product became cheaper by $20 \%$. Therefore, all the previously purchased goods could be bought by spending $80 \%$ of the money, and the remaining $20 \%$ could buy an additional $\frac{1}{4}$ of the goods, which constitutes $25 \%$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. At the conference. $85 \%$ of the delegates at the conference know English, and $75 \%$ know Spanish. What fraction of the delegates know both languages? | 7. Note that $85 \%+75 \%=160 \%$, which exceeds the total number of conference delegates by $60 \%$. The excess is due to those people who know both languages - we counted them twice. Thus, not less than $60 \%$ of the conference delegates know both languages. | 60 | Other | math-word-problem | Yes | Yes | olympiads | false |
9. Seawater contains $5 \%$ salt. How many kilograms of fresh water need to be added to 40 kg of seawater to make the salt content $2 \%$? | 9. In 40 kg of seawater, there is $40 \cdot 0.05=2$ (kg) of salt, which in the new solution constitutes $2\%$, therefore, the solution should be $2: 0.02=100$ (kg). Answer: 60 kg of fresh water should be added. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Vера and Аnya attend a math club, in which there are more than $91 \%$ boys. Find the smallest possible number of club participants. | 10. Let $x$ be the number of participants in the club, and $y$ be the number of girls. Then, according to the conditions of the problem, $0.09 x > y$ or $9 x > 100 y$, where $x$ and $y$ are natural numbers. Solving the problem by enumeration, we will verify that the smallest possible solution for $y=2$ is achieved at $... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. A painted cube with an edge of 10 cm was sawn into cubes with an edge of $1 \mathrm{~cm}$. How many of them will have one painted face? How many will have two painted faces? | 14. $K$ each face of the cube adjoins $8 \times 8=64$ cubes, painted only on one side (make a drawing!). There are 6 faces, so 384 cubes are painted on one side. Reasoning similarly, we get that 96 cubes are painted on two sides. | 384 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. To number the pages of an encyclopedia, 6869 digits were required. How many pages are there in it? | 12.Note that there are 9 single-digit numbers, 99-9=90 two-digit numbers, digits - 90 × 2=180; 999-99=900 three-digit numbers; digits - 900 × 3=2700. Let's calculate the number of four-digit numbers in the page numbering of the book: 6869-2700-180-9=3980, 3980: 4=995. Adding up the obtained results, we get: 9+90+900+99... | 1094 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. About the cat. A ribbon was stretched around the Earth along the equator. Then the length of the ribbon was increased by 1 meter and again evenly positioned around the equator. Will a cat be able to crawl through the resulting gap? | 13. Let the radius of the Earth be $R$, and the formed gap be $h$, then $2 \pi(R+h)-2 \pi R=1$ and $2 \pi h=1$. Therefore, the size of the gap $h=\frac{1}{2 \pi} \approx 15$ cm - a gap quite sufficient for a cat to squeeze through. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. $о$ about the fisherman and the fish. When asked how big the fish he caught was, the fisherman said: “I think that its tail is 1 kg, the head is as much as the tail and half of the body, and the body is as much as the head and the tail together.” How big is the fish? | 3. Let $2 x$ kg be the weight of the torso, then the head will weigh $x+1$ kg. From the condition that the torso weighs as much as the head and tail together, we get the equation: $2 x=x+1+1$. From this, $x=2$, and the whole fish weighs - 8 kg. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. A Herd of Elephants. Springs are bubbling at the bottom of the lake. A herd of 183 elephants could drink it dry in one day, while a herd of 37 elephants would take 5 days. How many days would it take for 1 elephant to drink the lake? | 6. Let's translate the problem into the language of algebra. Let the volume of the lake be $V$ liters, the elephant drinks $C$ liters of water per day, and $K$ liters of water flow into the lake from the springs per day. Then the following two equations hold: $183 C = V + K$ and $37 \cdot 5 C = V + 5 K$; from which we ... | 365 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. The Clock Hands Problem. At what time after 12:00 will the minute hand first catch up with the hour hand? | 11. At 13:00, the minute hand will lag behind the hour hand by 5 minute divisions. Before the "meeting," the hour hand will travel $x$ divisions, and the minute hand will travel $12x$ divisions. From the equation $x + 5 = 12x$, we get that $x = \frac{5}{11}$ minute divisions on the clock face. Answer: 1 hour $5 \frac{5... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
12.Cunning manipulations. With the number written on the board, it is allowed to perform the following operations: replace it with its double, or erase its last digit. How can you use these operations to get from the number 458 to the number 14? | 12.Let A be the doubling of a number, and B be the erasing of the last digit. Then: $458-\mathrm{B} \Rightarrow 45-\mathrm{A} \Rightarrow 90-\mathrm{B} \Rightarrow 9-\mathrm{A} \Rightarrow 18-\mathrm{A} \Rightarrow 36-\mathrm{A} \Rightarrow 72-\mathrm{B} \Rightarrow 7-\mathrm{A} \Rightarrow 14$. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Divide equally. There is milk in an eight-liter bucket. How can you measure out 4 liters of milk using a five-liter bucket and a three-liter jar? | 4. It is convenient to write the solution in the form of a table:
| 8-liter can | 8 | 3 | 3 | 6 | 6 | 1 | 1 | 4 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 5-liter can | 0 | 5 | 2 | 2 | 0 | 5 | 4 | 4 |
| 3-liter jar | 0 | 0 | 3 | 0 | 2 | 2 | 3 | 0 | | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. The Zmey Gorynych has 2000 heads. A legendary hero can cut off $1, 17, 21$ or 33 heads with one blow, but in return, $10, 14, 0$ or 48 heads grow back, respectively. If all heads are cut off, no new ones grow. Can the hero defeat the Zmey? | 5. We can propose the following tactic for cutting off the heads of the Snake: 1. Initially, we will cut off 21 heads (94 times); no new heads will grow, and the Snake will be left with 26 heads. 2. Next, we will cut off 17 heads three times (remember, 14 new heads will grow back each time) - after which 17 heads will ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. In a family, there are six children. Five of them are respectively 2, 6, 8, 12, and 14 years older than the youngest, and the age of each child is a prime number. How old is the youngest? | 4. Suppose the child's age does not exceed 35 years. Let's list all the prime numbers: $2,3,5,7,11,13,17,19,23$, 29,31. It is clear that the age of the younger child is an odd number. The numbers 29 and 31 also do not fit. Let's find the age of the younger child. He cannot be 1 year old, because $1+8=9$. His age cannot... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. The Eight Queens Problem. Can eight queens be placed on a chessboard so that they do not threaten each other? | 7. There are 92 solutions to this problem. The figure shows one of them. | 92 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. In the senate of the country of knights and liars - there are 100 senators. Each of them is either a knight or a liar. It is known that: 1. At least one of the senators is a knight. 2. Out of any two arbitrarily chosen senators, at least one is a liar. Determine how many knights and how many liars are in the senate... | 13. Only one of the senators is a knight. The fact that there is at least one knight among the senators follows from the first statement, the existence of a second knight contradicts the second statement. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Scientific organization of labor. There are logs of two types: 6 meters and 7 meters long. They need to be sawn into 1-meter logs. Which logs are more profitable to saw? | 1. To saw 42 one-meter logs from six-meter logs requires 35 cuts, while from seven-meter logs - 36. It can be considered that sawing six-meter logs is more profitable. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. To the Mechanics and Mathematics Faculty! Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks. How many books were bought if the price of one book is more than one ruble higher than the price of an album, and 6 more books were bought than albums? | 5. Since each book is more expensive than a ruble, no more than 10 books were bought. Moreover, it is clear that no fewer than 7 books were bought (since at least one album was bought). The number 1056 is divisible by 8 and not divisible by $7,9,10$. Therefore, 8 books were bought. (MSU, Mechanics and Mathematics, 1968... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. 175 shaltev cost more than 125 boltayev, but less than 126 boltayev. Prove that 80 kopecks will not be enough to buy 3 shaltev and 1 boltayev. | 6. Let's prove that 80 kopecks are not enough to buy 3 shaltays and 1 boltay. Let one shaltay cost $x$ kopecks, and one boltay $y$ kopecks, then $126 y > 175 x > 125 y$. From this, it follows that $y > 25(7 x - 5 y) > 0$ and $y \geq 26$. Further, $7 x \geq 5 \cdot 26 = 130$. Hence:
63 Natural numbers
$x \geq 19$. The... | 83 | Inequalities | proof | Yes | Yes | olympiads | false |
8. One or two? Let's take all natural numbers from 1 to 1000000 and for each of them, calculate the sum of its digits. For all the resulting numbers, we will again find the sum of their digits. We will continue this process until all the resulting numbers are single-digit. Among the million resulting numbers, 1 and 2 w... | 8. Let's use the statement: if a number when divided by 9 has a remainder of \( d \), then the sum of its digits will have the same remainder. Which numbers from 1 to 1000000 are more: those that have a remainder of 1 when divided by 9, or those that have a remainder of 2? In the range from 1 to 999999, there are an eq... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Centipedes and three-headed dragons. In a certain herd of centipedes and three-headed dragons, there are a total of 26 heads and 298 legs. Each centipede has one head. How many legs does a three-headed dragon have? | 10.Note: 1. the number of heads of dragons is divisible by 3, and the number of centipedes cannot exceed 7 (otherwise, there would be more than 298 legs), so the number of centipedes is either 2 or 5. 2. If there are 2 centipedes, then there are 8 dragons, and each dragon would have $\frac{218}{8}$ legs, which is impos... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Two detectives met. Here is their dialogue:
- Do you have two sons?
- Yes, they are young, they don't go to school.
- By the way, the product of their ages equals the number of pigeons near us.
- That's not enough information.
- And I named the older one after you.
- Now I know how old they are.
How old are the so... | 7. The ages of the sons could have been equal, hence the product of their ages - a perfect square. Upon receiving information about the product of the ages, the second detective could not answer the question, therefore, the product can be factored in two ways. Answer: 1 and 4 years. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Boxing Championship. In the Olympic boxing tournament, 247 people participated. How many matches were needed to determine the winner? | 1. In the Olympic tournament, when after each match one boxer is eliminated, and only the winner continues to compete, 246 matches will be required. | 246 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. In a round-robin chess tournament, 7 players participated. How many games did they play? | 2. Let's calculate the total number of games played in the tournament. Each player played 6 games - in total: $6 \times 7=42$. Notice that in this way, we counted each game twice (since two players participate in each game!), so the total number of games played in the tournament is $42: 2=21$. We can also solve this us... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. How many chess players participated in the round-robin tournament if a total of 190 games were played in this competition? | 3. Since $19 \times 20: 2=190$, there were 20 chess players in the tournament. | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In a chess tournament (played in a round-robin system) with 30 participants, to achieve the 4th category rating, a player needs to score $60 \%$ of the points. What is the maximum number of chess players who can achieve the category rating by the end of the tournament? | 5. In total, $30 \cdot 29: 2=435$ games are played in the tournament, and thus 435 points are contested. The number of participants who become rated players cannot exceed $435: 17.5=24$ people. If 24 participants draw all their games against each other and win the rest of their games, they will accumulate the required ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In a chess tournament, 8 players participated. They scored $7, 6, 4, 4, 3, 2, 1, 5$ and 0.5 points respectively. How many points did the players who took the first four places lose in matches against the others? | 7. From the condition of the problem, it follows that the tournament winner and the participant who took second place did not lose a single point in their matches against the others. Further, the participants who shared third and fourth places could have scored 8 points against the rest, but scored 7 (they played one p... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11. Homework. Fedya was supposed to divide a certain number by 4 and add 15 to it, but Fedya multiplied this number by 4 and subtracted 15, yet he still got the correct answer. What was this number? | 11. Solving the equation: $0.25 x + 15 = 4 x - 15$, we get the answer: 8. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Linear function. The distance between villages $\mathcal{A}$ and $B$ is 3 km. In village $\mathcal{A}-300$ students, and in village $B-$ 200 students. Where should a school be built to minimize the total distance traveled by students on their way to school? | 4. It is clear that the school should be built on the segment $A B$, but where exactly? Let the distance from village $A$ to the school be $x$, then the total distance traveled by all schoolchildren on the way
![](https://cdn.mathpix.com/cropped/2024_05_21_b6bdef2bf90cccc464adg-077.jpg?height=385&width=560&top_left_y=... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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