problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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## 19. Tram
On one tram line, trams depart regularly every 10 minutes throughout the day. A tram takes one hour to travel from one end of the line to the other. A passenger boards a tram at one terminal stop and rides to the last stop of the tram; out of boredom, he looks out the window and counts the oncoming trams o... | 19. The passenger will encounter all the tram cars that departed from the other end of the tram line less than an hour before his departure; generally speaking, there are six such trains. He will also encounter all the trains that will depart from the opposite end of the line within the next hour; generally speaking, t... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 20. Nicolas plays
## with tin soldiers
Nicolas has the same number of tin Indians, Arabs, cowboys, and Eskimos. After his cousin Sebastian's visit, he indignantly discovered the disappearance of a third of his soldiers. Suppose the number of Eskimos remaining is the same as the number of cowboys that disappeared, ... | 20. Let $x$ be the number of soldiers of each type that Nicolas had initially, and $y$ be the number of cowboys taken by Sebastian (or the number of remaining Eskimos); in this case, the number of Eskimos taken is $x-y$.
On the other hand, it is known that the number of Indians taken is $x / 3$. Let $z$ be the number ... | 0 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 22. General Cleaning
A solid joint-stock company occupies three floors in a tower, which has the shape of a parallelepiped and is located in Paris in the La Défense district: the 13th and 14th floors, where the company's offices are located, and the 25th floor, where its board of directors is situated.
To clean th... | 22. Let $n$ be the unknown number of cleaners. The number of man-hours required to clean two floors of the bureau is
$$
4 \cdot(n+n / 2)
$$
The number of man-hours required to clean the 25th floor is
$$
4 \cdot(n / 2)+8
$$
But this last number is half of the previous one, since the area of one floor is half the are... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
## 2. Marel
Take a piece of cardboard and draw the following diagram on it:
Also, take three silver and three gold coins (if you are on the beach, draw the diagram directly in the sand and... | 2. Let's renumber our nine circles (nine positions of the scheme):
Let $C$ be the player who starts, and $F$ be the player who finishes placing their coins.
I. The player who starts the ga... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
150*. Prove that if $a+b+c=0$, then
$$
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9
$$
where $a \neq 0, \quad b \neq 0, c \neq 0, a \neq b, \quad a \neq c, b \neq c$. | Proof.
$$
\begin{gathered}
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)= \\
=1+\frac{(b-c) c}{a(a-b)}+\frac{(c-a) c}{b(a-b)}+\frac{a(a-b)}{c(b-c)}+ \\
+1+\frac{(c-a) a}{b(b-c)}+\frac{b(a-b)}{c(c-a)}+\frac{(b-c) b}{a(c-a)}+1= \\
=3+\frac{c}{a-b}\left... | 9 | Algebra | proof | Yes | Yes | olympiads | false |
233. If $x_{1}$ and $x_{2}$ are the roots of the equation
$$
\frac{3 a-b}{c} x^{2}+\frac{c(3 a+b)}{3 a-b}=0
$$
then, without solving it, find $x_{1}^{117}+x_{2}^{117}$. | Solution.
$x_{1}^{117}+x_{2}^{117}=\left(x_{1}+x_{2}\right)\left(x_{1}^{116}-x_{1}^{115} x_{2}+\ldots+x_{2}^{116}\right), \quad$ but $\quad x_{1}+x_{2}=0$, so, $x_{1}^{117}+x_{2}^{117}=0$.
Note. It is useful to solve problem 233 when reviewing, after discussing the equality:
$$
a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
352. Find a natural number that is equal to the sum of all its preceding natural numbers. Does more than one such number exist, or is there only one? | Let $n$ be the desired natural number; the sum of the numbers preceding this natural number is expressed by the formula
$$
S_{n-1}=\frac{n(n-1)}{2}
$$
According to the problem, $\frac{n(n-1)}{2}=n$, from which $n^{2}-3 n=0$ or $n_{1}=3, n_{2}=0$. The desired number is 3. It is the only one, as the second root of the ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. How many four-digit numbers can be formed, the sum of whose digits is 3? List these numbers. | 2. For writing numbers, we use digits
$0,1,2,3,4,5,6,7,8,9$
Since the sum of the digits is three, we can exclude all digits starting from four.
$0,1,2,3,4,5,6,7,8,9$
A number cannot start with zero, so the first digit can be one of the three digits $1,2,3$. The further solution will be represented in the form of a ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Ducks were flying. One in front, two behind, one behind and two in front, one between two and three in a row. How many ducks were flying in total? | 3. Let's schematically depict how the ducks flew.
As can be seen from the diagram, the problem is about three ducks.
Answer: 3 ducks. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. There were 5 times more strawberry bushes on the first bed than on the second. When 22 bushes were transplanted from the first bed to the second, the number of strawberry bushes on each bed became the same. How many bushes were there on each bed? | ## 5. First method of solving.
Let's build a graphical model of the problem's condition
1) $22: 2=11$ (bushes) - this is $1/5$ of all the bushes (that were on the second bed);
2) $11 \ti... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let 2 cups and 2 pitchers weigh as much as 14 plates, 1 pitcher weighs as much as 1 cup and 1 plate. How many plates will balance a pitcher
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 4. First method of solution (graphical).
Reduce the number of items on each scale by half.
 $22725: 9=2525$ (rubles) — the amount the students paid for one month.
To determine how much each student paid monthly, we need to know the number of students in the class. This is unknown in the problem. However, from the problem's condition, it follows that this is a natural number that is a divisor of 2525.
... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Find the smallest number that gives a remainder of 1 when divided by 2, and a remainder of 2 when divided by 3. | 3. First, let's write down the numbers that give a remainder of 1 when divided by 2:
$$
3,5,7,9,11 \ldots
$$
Next, let's write down the numbers that give a remainder of 2 when divided by 3:
$$
5,8,11,14,17 \ldots
$$
We will choose from the obtained numbers the one that satisfies both conditions and is the smallest.... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. It is known that the perimeter of one rectangle is greater than the perimeter of another rectangle. Compare the areas of these rectangles.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. The solution to this problem can be obtained by students as a result of conducting a computational experiment with various rectangles, for example:
a) $a=3 \text{~cm}, b=4 \text{~cm}$;
$$
\begin{aligned}
& a=4 \text{~cm}, b=2 \text{~cm} ; \\
& p_{2}=12 \text{~cm} ; \\
& S_{2}=8 \text{~cm}^{2}
\end{aligned}
$$
$$
\... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. In the box, there are geometric shapes: triangles, squares, and circles. In total, there are 24 shapes. There are 7 times as many triangles as squares. How many of each shape could be in the box? | 3. The first method of solving. Let's represent the condition of the problem in the form of a drawing.
squares
triangles
 | 4. First method of solving. Since the machine is balanced by the ball and two cubes, the machine with a cube will be balanced by the ball and three cubes. From the second condition, we have that 2 balls balance a ball and 3 cubes, meaning one ball by mass is equal to 3 cubes. Thus, the machine can be balanced by 5 cube... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. Two numbers were first multiplied, and then the larger number was divided by the smaller one, resulting in equal outcomes. What are these numbers? How many such pairs of numbers exist? | 2. Let $a$ and $b$ be the two numbers mentioned in the problem, with $a > b$. According to the problem,
$$
a \times b = a : b
$$
or
$$
a b^{2} - a = 0
$$
Factoring out the common factor, we get
$$
a \left(b^{2} - 1\right) = 0
$$
From this, we can conclude that $b = 1$, and $a$ can be any number.
Students might r... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given a triangle with side lengths of 7 cm, 12 cm, and 9 cm respectively. Explain how to construct a segment connecting a vertex and the opposite side with a length of 9 cm so that the perimeters of the two resulting triangles are equal. | 6. First method of solution. Since in the obtained triangles one side will be common ( $A D$ ), in order for the perimeters to be equal, it is necessary that the side of 9 cm be divided into parts, the difference in lengths of which would equal the difference of the other two sides ( $12-7=5 \text{ cm}$ ). Based on thi... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Carlson has a runny nose. He uses square handkerchiefs measuring 25 cm $\times 25 \mathrm{~cm}$. Over eight days, Carlson used $3 \mathrm{~m}^{2}$ of fabric. How many handkerchiefs did Carlson use per day? | 2. First method of solving.
1) $25 \times 25=625\left(\mathrm{~cm}^{2}\right)-$ area of one handkerchief;
2) $3 \times 10000=30000\left(\mathrm{~cm}^{2}\right)$ - contained in $3 \mathrm{~m}^{2}$;
3) $30000: 625=48$ (handkerchiefs) - Carlson used over 8 days
4) $48: 8=6$ (handkerchiefs) - Carlson used per day.
Second ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a box, there are 7 blue and 5 red balls. What is the minimum number of balls that need to be taken out to ensure that among them there are at least 2 blue and 1 red? | 5. The worst-case scenario is if we draw 7 balls and they all turn out to be blue. To get another red ball, we need to draw one more ball. In this case, there will definitely be one red ball and 2 blue balls among the balls.
Answer: 8 balls. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. The boy caught a fish. When asked how much the caught fish weighed, he said: “I think the tail weighs 1 kg, and the head weighs as much as the tail and half the body, and the body weighs as much as the head and the tail together.” How much does the fish weigh? | 2. From the problem statement, it is known that the tail ( $X$ ) weighs 1 kg. The weight of the head is equal to the weight of the tail plus an additional $1 / 2$ of the body weight (T):
$$
\Gamma=\mathrm{X}+\mathrm{T} / 2 \text { or } 2 \Gamma=2 \mathrm{X}+\mathrm{T} .
$$
Since the tail weighs 1 kg, then $2 \Gamma=2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. If from each of two numbers you subtract half of the smaller one, the remainder of the larger one is three times the remainder of the smaller one. How many times larger is the larger number than the smaller one? | 1. The first way to solve the problem is to represent the condition of the problem in the form of a drawing.
From the drawing, it is clear that the first number is twice as large as the sec... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. One side of the square was increased by 5 times, and the other side was reduced by 2 times, resulting in a rectangle with an area of $160 \mathrm{~cm}^{2}$. What is the side length of the square | 6. First method of solution.
1) $160: 5=32\left(\mathrm{~cm}^{2}\right)$ - area of half the square;
2) $32 \times 2=64\left(\mathrm{~cm}^{2}\right)$ - area of the original square;
3) $64: ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. In 1996, a farmer bought two sheep. This year he did not get any offspring. The first sheep gave birth to 1 sheep every three years, and the second sheep gave birth to 1 sheep every two years. All the born sheep gave birth to 1 sheep annually. How many sheep will the farmer have in the year 2000? | 6. Let's draw a diagram according to the problem statement.
Answer: 9 sheep. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. From 16 m of fabric, 4 men's and 2 children's coats were sewn. How many meters of fabric are needed to sew one men's and one children's coat, if from 18 m of the same fabric, 2 men's and 6 children's coats can be sewn | 7. Let's make a brief record of the problem's condition:
$$
\begin{aligned}
& 4 \text { m, } 2 \text { d }-16 \text { m; } \\
& 2 \text { m, } 6 \text { d }-18 \text { m. }
\end{aligned}
$$
To ensure that the brief record contains the same quantities, we will double the second order:
$$
\begin{aligned}
& 4 \text { m... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. At the mathematics olympiad in two rounds, it is necessary to solve 14 problems. For each correctly solved problem, 7 points are given, and for each incorrectly solved problem, 12 points are deducted. How many problems did the student solve correctly if he scored 60 points? | 3. Let's determine the number of problems solved by the student, given that he received 7 points for each problem: $60: 7=8$ (remainder 4). Since 12 points were deducted for each incorrectly solved problem, the number of correctly solved problems was more than 8. Further, by trial and error, we can establish that there... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. One side of a rectangle is $5 \mathrm{~cm}$, and the other is the smallest two-digit number that is divisible by 3. For the second rectangle, one side is equal to the shorter side of the first. The area of one rectangle is $25 \mathrm{~cm}^{2}$ greater than the area of the other. Determine the unknown side of the se... | 6. First, let's determine the unknown side of the rectangle. For this, we will find the smallest two-digit number that is divisible by 3, which is 12. According to the problem, one side of the second rectangle is $5 \mathrm{~cm}$. Let's determine the area of the first rectangle $(12 \times 5=60)$. Since the problem doe... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A sheet of cardboard measuring 48 cm in length and 36 cm in width needs to be cut into cards measuring 16 cm in length and 12 cm in width. How can this be done to obtain the maximum number of cards? | 1. In the first step, it is necessary to determine how many cards will fit if the length of each is $16 (48: 16=3)$. Then determine how many cards will fit in the width (36:12 = 3).
Finally, find the number of cards $(3 \times 3=9)$. Then conduct similar reasoning for the case where the cards are laid out by width ( $... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (Ancient Indian problem 3-4th century AD). Of the four sacrificial donors, the second gave twice as many coins as the first, the third three times as many as the second, the fourth four times as many as the third, and all together gave 132 coins. How many did the first give? | 4. The first method of solving. By analyzing the condition of the problem, one can conclude that the number of coins given by the first person will not be more than 10. We will then refine this number through trial and error. Let's try the number 5.
$$
5+5 \times 2+5 \times 2 \times 3+5 \times 2 \times 3 \times 4=5+10... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Two boats departed from piers $A$ and $C$. The speed of the boat traveling from $A$ is 7 km per hour, and from $C$ is 3 km per hour. Determine after how much time the boats will meet, if the distance between the piers is 20 km, and the speed of the river current from $A$ to $C$ is 2 km per hour. | 4. Since the problem does not specify the direction in which the boats are moving, we need to consider various scenarios: they are moving towards each other; they are moving in the same direction downstream; they are moving in the same direction upstream; they are moving in opposite directions. Given that the boats mus... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (An old entertaining problem). Divide 46 rubles into 8 parts so that each part is 50 kopecks (50 cents) more than the previous one. | 5. First method of solution. Let the first part be $x$, then the second is $x+50$, the third is $-x+50 \times 2$, the fourth is $-x+50 \times 3$, the fifth is $x+50 \times 4$, the sixth is $-x+50 \times 5$, the seventh is $-x+50 \times 6$, and the eighth is $x+50 \times 7$. We will calculate the sum of all parts.
$$
\... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. How many numbers less than 2011 are divisible by 117 and 2? | 7. When solving this problem, one can reason as follows. First, list the numbers less than 2011 and divisible by 117 (2), then among them, select those that are divisible by 2 (117), and determine how many there are.
Since the number is divisible by both 2 and 117, it will also be divisible by their product $2 \times ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. What digits stand in the natural number sequence at the thirteenth and one hundred twentieth positions?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2. Let's write down the first fifteen numbers:
$$
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15
$$
To fill the first nine positions, 9 digits are required, and then two digits are needed to write each two-digit number. Starting from the tenth position, two-digit numbers are written. For each of them, two places are allocated. ... | 6 | Number Theory | proof | Yes | Yes | olympiads | false |
7. How many different products that are divisible by 10 (order does not matter) can be formed from the numbers $2,3,5,7,9$? Numbers in the product do not repeat! | 7. For a number to be divisible by 10, it must be divisible by 2 and 5.
Thus, the product must necessarily include 2 and 5. We get the first product $-2 \times 5$.
Now, let's add one more factor to this product:
$$
2 \times 5 \times 3, 2 \times 5 \times 7, 2 \times 5 \times 9
$$
Add one more factor to the obtained ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. In village $A$, there are 100 schoolchildren, and in village $B$, there are 50 schoolchildren. The distance between the villages is 3 kilometers. At what point on the road from $A$ to $B$ should a school be built to minimize the total distance traveled by all schoolchildren? | $\triangle$ It seems fair to build the school closer to $A$, as there are more students there. But how much closer? Sometimes it is suggested to build the school one kilometer from $A$ and two kilometers from $B$ (twice as many students, so the school is twice as close). However, from the perspective of the given probl... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. How many degrees does the minute hand turn in a minute? The hour hand? | $\Delta$ In one hour, the minute hand makes a full revolution, and in half an hour, it turns by $180^{\circ}$ (a straight angle). Half an hour is 30 minutes, so in one minute, it turns by $1 / 30$ of the angle in $180^{\circ}$, which is $180 / 30=6^{\circ}$.
The hour hand moves $1 / 12$ of a circle in one hour, meanin... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
449. Two points $A$ and $B$ are 8 units apart. How many lines exist that are 5 units away from point $A$ and 3 units away from point $B$? | $\triangleright$ The line $l$ is at a distance $d$ from the point $A$ when it is tangent to the circle of radius $d$ centered at $A$. Therefore, the problem can be restated as follows: given two circles with radii 3 and 5, and the distance between their centers is 8. How many common tangents exist?
We have already sol... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
521. By what factor should the side of a square be increased to make its area four times larger? | $\triangleright$ The area of a rectangle with sides $a$ and $b$ is $a b$, so the area of a square with side $a$ is $a \cdot a=a^{2}$. By doubling the side of the square, we increase its area by four times: $(2 a)^{2}=2^{2} a^{2}=4 a^{2} . \triangleleft$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
577. Draw an isosceles right triangle on graph paper with the legs equal to the side of a cell, and check the Pythagorean theorem for it: the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs. | $\triangleright$ The squares built on the legs have a unit area. The square built on the hypotenuse is divided into four triangles, each of which makes up half of a grid square, and thus has an area of 2 - as required. $\triangleleft$ | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
578. Check the Pythagorean theorem for a right triangle with legs 1 and 2. | $\triangleright$ The squares constructed on the legs have areas of 1 and 4. It remains to find the area of the square constructed on the hypotenuse. This can be done in two ways. First, it can be cut into 4 triangles and a central square. The triangles are equal to the original (legs 1 and 2) and have an area of 1. The... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
662. Point $A$ was moved 13 units to the right and resulted in a point with a coordinate of 8. What was the coordinate of point $A$? | $\triangleright$ Let $A$ have the coordinate $a$. Then $a+13=8$, and $a=8-13=(-5) . \triangleleft$
 | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
664. Given two points $A(-3)$ and $B(7)$ (the coordinates are given in parentheses). Find the coordinate of the midpoint of segment $A B$. | $\triangle$ The distance $A B$ is $7-(-3)=10$, half of the distance is 5. Shifting $A$ 5 units to the right (or $B$ 5 units to the left), we get the point with coordinate 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
694. Find the slope of a line that forms an angle of $45^{\circ}$ with the horizontal. | $\triangleright$ We already know that such a line is given by the equation $y=x$, so the slope is $1 . \triangleleft$ | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
735. Find the greatest divisor of the pair of numbers 123456789 and 987654321 (in other words, the greatest common measure of segments of such lengths). | $\triangleright$ In principle, this problem can be solved by brute force (and even in a reasonable time if you have a computer at hand), trying all numbers from 1 to 123456789 (larger numbers clearly do not fit).
However, it can also be done without a computer, using the Euclidean algorithm. The number 987654321 can b... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Find the minimum value of the expression $2 x+y$, defined on the set of all pairs $(x, y)$ satisfying the condition
$$
3|x-y|+|2 x-5|=x+1
$$ | 1.I.1. The following figure shows a set defined by the equation $3|x-y|+|2 x-5|=x+1$.
Let $C=2 x+y$. The problem requires finding the minimum value of $C$ for all points in the depicted set... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Find the minimum value of the function
$$
f(x)=4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5
$$
Solution: Since
$$
\begin{aligned}
4^{x}+4^{-x}-2^{x+1}-2^{1-x}+5=2^{2 x}+ & 2^{-2 x}-2\left(2^{x}+2^{-x}\right)+5= \\
=2^{2 x}+2 \cdot 2^{x} \cdot 2^{-x}+2^{-2 x} & -2-2\left(2^{x}+2^{-x}\right)+5= \\
& =\left(2^{x}+2^{-x}\right)^{2... | 1.I.8. Comment. The following error was made in the given solution. Since $2^{x}>0$, the range of the function $y=2^{x}+2^{-x}$ is the interval $[2 ;+\infty)$. Therefore, it is necessary to find the minimum value of the function $g(t)=t^{2}-2 t+3$ on the interval $[2 ;+\infty)$. On this interval, the function $g(t)$ is... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The radii of the excircles of a certain triangle are 2, 3, and 6 cm. Find the radius of the circle inscribed in this triangle. | 3.II.4. First, let's derive the formula for the radius of the excircle. Let $P$ be the center of the circle with radius $r_{3}$, which touches the side $AB$ and the extensions of the sides $CA$ and $CB$ of triangle $ABC$. From the equality $S_{PAB} + S_{ABC} = S_{PAC} + S_{PBC}$, it follows that
$$
S_{ABC} = S = \frac... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Compute $\operatorname{tg} \alpha$, if $3 \operatorname{tg} \alpha-\sin \alpha+4 \cos \alpha=12$. | 4.II.1. We have,
$$
\begin{aligned}
& 3 \operatorname{tg} \alpha - \sin \alpha + 4 \cos \alpha = 12 \\
& \Longleftrightarrow \operatorname{tg} \alpha (3 - \cos \alpha) - 4 (3 - \cos \alpha) = 0 \Longleftrightarrow \\
& (\operatorname{tg} \alpha - 4)(3 - \cos \alpha) = 0 \Longleftrightarrow \operatorname{tg} \alpha = 4... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Let $x$ and $y$ be positive numbers whose sum is 2. Find the maximum value of the expression $x^{2} y^{2}\left(x^{2}+y^{2}\right)$. | Solution 1. The idea is suggested by the formula for $(x+y)^{4}$. Since
$(x+y)^{4}-8 x y\left(x^{2}+y^{2}\right)=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}=(x-y)^{4} \geqslant 0$, then $8 x y\left(x^{2}+y^{2}\right) \leqslant(x+y)^{4}$. It is clear that $4 x y \leqslant(x+y)^{2}$. Therefore,
$$
32 x^{2} y^{2}\left... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Solve the equation $\frac{3}{\log _{2} x}=4 x-5$.
Answer: 2.
Solution: The function $y=\log _{2} x$ is increasing, therefore $y=\frac{3}{\log _{2} x}$ is a decreasing function. On the other hand, the function $y=4 x-5$ is increasing, therefore, the given equation has no more than one root. By trial, we find that $... | 5.I.8. Comment. The error in the given solution is that the function $y=\frac{3}{\log _{2} x}$ is decreasing not on its entire domain, but only on each of the intervals $(0 ; 1)$ and $(1 ;+\infty)$. Therefore, this equation has no more than one root in each of them. In addition to $x=2$, the solution to the equation is... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Solve the equation $27^{x}-7 \sqrt[3]{7 \cdot 3^{x}+6}=6$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution 1. Let $z=\sqrt[3]{7 y+6}$ and transition to the system
$$
\left\{\begin{array}{l}
y^{3}=7 z+6 \\
z^{3}=7 y+6
\end{array}\right.
$$
Then $y^{3}-z^{3}=7(z-y)$, from which it follows that $y=z$, since $y^{2}+y z+z^{2}$ cannot be equal to -7. As a result, we get the equation $y^{3}-7 y-6=0$, the roots of which ... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a row, all natural numbers less than a billion that have exactly 13 natural divisors (including one and the number itself) were written down. How many of them have an even sum of digits? | 6.I.3. If $n=p_{1}^{s_{1}} p_{2}^{s_{2}} \ldots p_{k}^{s_{k}}$, then the number of divisors of the number $n$ is $\left(s_{1}+1\right)\left(s_{2}+1\right) \ldots\left(s_{k}+1\right)$. Since the number of divisors of the given number is a prime number, the number itself is a power of a prime number; in this case $n=p^{1... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Solve the equation $3^{x^{2}+x-2}-3^{x^{2}-4}=80$. | 6.II.1. Let's rewrite the equation as $3^{x^{2}-4}\left(3^{x+2}-1\right)=80$ and set $f(x)=3^{x^{2}-4}\left(3^{x+2}-1\right)$. If $x \leqslant -2$, then $f(x) \leqslant 0$, hence the equation has no solutions on the interval $(-\infty; -2]$. If $-2 < x \leqslant 0$, then $0 < 3^{x+2} - 1 \leqslant 8$ and $3^{x^{2}-4} \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find the smallest prime number $p$ such that $n^{2}+n+11$ is divisible by $p$ for some integer $n$.
untranslated text is preserved in terms of line breaks and format, as requested. | 6.II.5. Since $n(n+1)$ is an even number, the number $n^{2}+n+11$ is odd, and therefore this number is not divisible by 2. Since $n^{2}+n+11=$ $(n-1)^{2}+1+3n+9$, the number $n^{2}+n+11$ is not divisible by 3, because no perfect square can have a remainder of 2 when divided by 3. Since $n^{2}+n+11=(n-2)^{2}+2+5n+5$, by... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. There are 9 sticks of different lengths from 1 cm to 9 cm. What are the side lengths of the squares that can be formed from these sticks, and in how many ways can they be formed? Ways of forming a square are considered different if different sticks are used and not necessarily all of them.
Can squares with side len... | 1. Solution. One way to form squares is to have side lengths of $7 \mathrm{~cm}, 8 \mathrm{~cm}, 10 \mathrm{~cm}, 11 \mathrm{~cm}$. In five ways, squares can be formed, each with a side length of $9 \mathrm{~cm}$. Thus, there are 9 ways to form the squares.
Hint. The side length of the square must be greater than 6 cm... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. The incident took place in 1968. A high school graduate returned from a written university entrance exam and told his family that he couldn't solve the following problem:
Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks, and the albums cost 56 kopecks. The number of book... | 4. S o l u t i o n. Since the 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. By checking, we find that the number 1056 is divisible by 8 and not divisible by $7,9,10$. A n s w e r: 8 books. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Twelve people are carrying 12 loaves of bread. Each man carries 2 loaves, each woman carries half a loaf, and each child carries a quarter of a loaf, and all 12 people are involved in the carrying. How many men, how many women, and how many children were there? | 7. 5 men, 1 woman, and 6 children. Show that the number of men cannot be, first, less than five, and second, more than five. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10. On the number line, four points are marked. Point $A$ corresponds to the number -3, point $B$ corresponds to the number -5, and point $C$ corresponds to the number 6. Find the fourth number corresponding to point $K$ under the following condition: if the direction of the number line is reversed, the sum of the new ... | 10. Solution. When the direction of the number line is changed, the sign of each number (except, of course, zero) changes to the opposite. Since the sum did not change in this case, it can only be equal to zero. Therefore, the fourth number sought is $0-(-5-3+6)=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. Several points are marked on a number line. The sum of the numbers corresponding to these points is $-1.5$. Each of the marked points was moved two units to the left on the number line, and therefore the sum of the numbers became $-15.5$. How many points were there? | 13. Solution. When a point is moved two units to the left on the number line, the number corresponding to this point decreases by 2 units. The sum of all numbers decreased by $-1.5-(-15.5)=$ $=14$, so there were a total of $14: 2=7$ points. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. What digits do the decimal representations of the following numbers end with:
1) $135^{x}+31^{y}+56^{x+y}$, if $x \in N, y \in N$
2) $142+142^{2}+142^{3}+\ldots+142^{20}$
3) $34^{x}+34^{x+1}+34^{2 x}$, if $x \in N$. | 12. 3) $\mathrm{Solution.} \mathrm{If} \mathrm{a} \mathrm{number} \mathrm{ends} \mathrm{in} \mathrm{four,} \mathrm{then}$ even powers of it end in 6, and odd powers end in 4. Therefore, one of the first two terms ends in four, and the other ends in six. The third term ends in six, so the decimal representation of the s... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2(!). Let the formula for the \(n\)-th term of the sequence be \(x=n^{2}\). Write down the first 10 terms of this sequence. Write down the sequence of differences between the second and first terms, the third and second terms, and so on. Write down another sequence of differences between the second and first, the third... | 2. The results of the calculations can be recorded in the form of the following table:
 | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. For what least natural \(m\) are the numbers of the form:
1) \(m^{3}+3^{m}\)
2) \(m^{2}+3^{m}\)
divisible by 7? | 10. Solution. Let \(r_{1}\) be the remainder of the division of \(m^{3}\) by 7, and \(r_{2}\) be the remainder of the division of \(3^{m}\) by 7. Clearly, the number \(m^{3}+3^{m}\) will be divisible by 7 when the sum of the remainders \(r_{1}+r_{2}\) equals 7. Let's construct the following table:
\begin{tabular}{|c|c... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. Simplify the expression:
\[
-(-(-(\ldots-(-1) \ldots)))
\]
which contains 200 pairs of parentheses.
If solving the problem is difficult, consider the numbers:
\[
\begin{aligned}
& -(-1)=\ldots, \\
& -(-(-1))=\ldots
\end{aligned}
\]
notice the pattern and draw a conclusion.
It is appropriate to inform students... | 12. -1. 13. 1) \(x=4n-1\); 2) \(y=(-1)^{n+1}\); 3) \(z=n^2+1\); 4) \(t=2^n-1\). | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16*. Among the first ten thousand numbers, how many of them end in 1 and can be represented in the following form: \(8^{m}+5^{n} ?(m \in N, n \in N)\) | 16. Five numbers. Hint. Show that \(m=4\). (KvN, 1972, No. 5, p. 81 .) | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Calculate at \(x=7\):
\((x-4)^{(x-5)^{(x-6)}}{ }^{(x+6)^{(x+5)}}\)
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6. Calculate at \(x=7\):
\((x-4)^{(x-5)^{(x-6)}}{ }^{(x+6)^{(x+5)}}\) | 6. When \(x=7\), the given expression equals \(3^{2^{1}}=9\). | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Given the monomial \((-1)^{n} a^{n-2} b^{9-n}\). Write in a row the set of all possible forms of this monomial for different permissible natural values of the exponents. | 14. Instruction. It is clear that \(n>2\) and \(n<9\), i.e., \(n=\{3 ; 4\); \(5 ; 6 ; 7 ; 8\}\). For these values of \(n\), six different monomials will be obtained. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Find \(k\), if it is known that for any \(x\):
\[
\begin{array}{r}
a x^{2}+b x+c \\
+b x^{2}+a x-7 \\
k x^{2}+c x+3 \\
\hline x^{2}-2 x-5
\end{array}
\] | 7. Solution.
\[
c=-5-(-7+3)=-1
\]
\[
\begin{aligned}
a+b & =-2-(-1) \\
k & =1-(a+b)=1-(-1)=2
\end{aligned}
\] | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Find the sum of the values of the polynomial \(x^{5}-1.7 \cdot x^{3}+2.5\) at \(x=19.1\) and \(x=-19.1\). | 11. Solution. If the sign of \(x\) is changed to the opposite, the given polynomial will take the form: \(-x^{5}+1.7 x^{3}+2.5\). The sum of the given polynomial and the obtained one is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13*. Calculate:
1) \(x^{4}-2 x^{3}+3 x^{2}-2 x+2\) given \(x^{2}-x=3\)
2) \(2 x^{4}+3 x^{2} y^{2}+y^{4}+y^{2} \quad\) given \(x^{2}+y^{2}=1\). | 13. 1) \( \mathrm{P} \) is a solution. \( x^{2}\left(x^{2}-x\right)-x\left(x^{2}-x\right)+2\left(x^{2}-x\right)+2= \) \( =3 x^{2}-3 x+2 \cdot 3+2=3\left(x^{2}-x\right)+8=3 \cdot 3+8=17 \);
2) Solution. \( 2 x^{4}+2 x^{2} y^{2}+x^{2} y^{2}+y^{4}+y^{2}=2 x^{2}\left(x^{2}+\right. \) \( \left.+y^{2}\right)+y^{2}\left(x^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Solve the equation: \(x^{3}+x^{2}+x+1=0\). | 9. Solution. Factoring the left side of the equation, we get:
\[
(x+1)\left(x^{2}+1\right)=0
\]
Since \(x^{2}+1>0\), then \(x+1=0, x=-1\). | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Find: 1) \(x^{6}+3 x^{2} y^{2}+y^{6}\), if \(x^{2}+y^{2}=1\); 2) \(x^{4}+\) \(+x^{2} y^{2}+y^{4}\), if \(x^{2}+y^{2}=a\) and \(x y=b\). | 5. 1) \(\mathrm{P}\) is the solution. First method. The given polynomial is equal to
\[
\left(x^{2}+y^{2}\right)\left(x^{4}-x^{2} y^{2}+y^{4}\right)+3 x^{2} y^{2}=\ldots=\left(x^{2}+y^{2}\right)^{2}=1
\]
Second method. Since \(y^{2}=1-x^{2}\), then, performing the substitution, we get that the given polynomial is ide... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Find the distance between the points of intersection of the three lines: \(y = 3x\), \(y = 3x - 6\), and \(y = 1975\). | 14. 2. Instruction. Show that the desired distance is equal to the distance between points \(A(0 ; 0)\) and \(B(2 ; 0)\). | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate in the most rational way:
\[
\frac{7^{16}-1}{2402000\left(49^{4}+1\right)}
\]
76 | 1. 2,4. Hint. Factor the numerator and reduce the fraction
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note I added for context, and it should not be part of the translation. Here is the requested translation:
1. 2,4. Hint. Factor the numerator and reduce the fraction | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. The price of a diamond is proportional to the square of its mass. If a diamond is broken into two parts, in which case will the total price of the two parts be the lowest? | 12. Solution. Let the price of a diamond be calculated by the formula \(y=a m^{2}\), where \(m\) is its mass. Let the mass of the first piece be \(\frac{m}{2}+x\). Then the mass of the second piece will be \(m-\frac{m}{2}-x=\frac{m}{2}-x\). The total cost of the two pieces will be:
\[
a\left(\frac{m}{2}+x\right)^{2}+a... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
297. The continuations of the medians of a triangle intersect the circumscribed circle at points $A_{1}, B_{1}, C_{1}$. Prove that
$$
\frac{A G}{G A_{1}}+\frac{B G}{G B_{1}}+\frac{C G}{G C_{1}}=3
$$
where $G$ is the centroid of triangle $A B C$. | 297. Using the concept of the power of a point, establish that
$$
\frac{A G}{G A_{1}}+\frac{B G}{G B_{1}}+\frac{C G}{G C_{1}}=\frac{A G^{2}+B G^{2}+C G^{2}}{R^{2}-d^{2}}
$$
But based on Leibniz's theorem, $A G^{2}+B G^{2}+C G^{2}=3 R^{2}-3 d^{2}$. 116 | 3 | Geometry | proof | Yes | Yes | olympiads | false |
45. Write the system of equations of a line passing through the origin and forming equal angles with the three coordinate axes. Determine the magnitude of these angles. How many solutions does the problem have? | 45. According to the problem, $\cos \alpha=\cos \beta=\cos \gamma$ and the equations take the form: $x=y=z$. Further, we have:
$$
\begin{gathered}
\cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1 ; \cos ^{2} \alpha=\frac{1}{3} \\
\cos \alpha= \pm \frac{1}{\sqrt{3}} ; \alpha \approx 54^{\circ} 44^{\prime} 8^{\prime... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given three pairwise intersecting lines, not parallel to one plane, and a point $P$ not belonging to any of them, construct a plane through this point so that it forms equal angles with the given lines.
保留源文本的换行和格式，这里直接输出翻译结果。 | 5. Draw through point $P$ the lines $A A^{\prime}, B B^{\prime}$, and $C C^{\prime}$, respectively parallel to the given lines $a, b$, and $c$. Point $P$ divides each of the lines $A A^{\prime}, B B^{\prime}, C C^{\prime}$ into two opposite rays $P A$ and $P A^{\prime}, P B$ and $P B^{\prime}, P C$ and $P C^{\prime}$. ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11*. From a point $M$ on the edge of a dihedral angle in one of its faces, a ray is drawn. Draw from the same point $M$ in the other face a ray that forms an angle of a given magnitude with the first ray. | 11. Let's take an arbitrary point $A$ on this ray; let $A^{\prime}$ be the projection of this point onto the other face. Suppose the desired ray is drawn, and we construct a segment $M B$ on it such that $|M B|=|M A|$. Then, in the isosceles triangle $A M B$, the two sides and the angle between them are known, and it c... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
58. Three spheres with different radii and centers not lying on the same straight line lie outside each other. What figure is formed by the lines of intersection of pairs of planes that are symmetric with respect to the plane of the centers of the spheres and simultaneously tangent to all three given spheres? | 58. The common tangent plane of two spheres passes through the center of their homothety, since the radii drawn to the points of tangency are parallel, and the line passing through the points of tangency also passes through the center of homothety (see the previous problem). The centers of homothety of three spheres be... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
36*. A polyhedron is called regular if all its faces are regular and congruent polygons to each other and all its polyhedral angles are also regular and congruent to each other ${ }^{2}$. Investigate the possibility of constructing simple regular polyhedra with $n$-sided faces and $m$-sided angles and determine the pos... | 36. Let the faces of a regular polyhedron be regular $n$-gons, and the polyhedral angles at the vertices be regular $m$-gonal angles; obviously, $m \geqslant 3, n \geqslant 3$. Divide the space into $f$ congruent regular $n$-gonal angles with a common vertex $S$ such that around each edge $[S A)$ there are $m$ angles. ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
46*. Inscribed a cube in a dodecahedron so that all eight vertices of the cube are vertices of the dodecahedron. How many solutions does the problem have? | 46. Consider a dodecahedron with center $S$ (Fig. 101). Due to the congruence of the faces, their diagonals are also congruent: $[A B] \cong|B C| \cong \cong[C D] \cong[D A]$. In the regular triangular pyramid $P A B Q$, the edge $[P Q]$ is perpendicular to the base edge $[A B]$ (see problem 8 from § 5). But $(P Q)\|(A... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. In a box, there are 7 red and 4 blue pencils. Without looking into the box, a boy randomly draws pencils. What is the minimum number of pencils he needs to draw to ensure that there are both red and blue pencils among them? | 1. Answer: 8 pencils. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
39. All odd numbers from 27 to 89 inclusive are multiplied together. What is the last digit of the resulting product? | 39. The product ends with the digit 5. Indeed, if in the multiplication of several odd numbers at least one factor ends with the digit 5, then the entire product ends with the digit 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
43. An adult needs 16 kg of salt for consumption over 2 years and 8 months. How much salt does an adult need for one year? | 43. 2 years 8 months make up 32 months. If 32 months require 16 kg of salt, then 12 months require 6 kg of salt. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
61. Insert arithmetic operation signs between the digits 12345 so that the result equals 1. | 61. There are several solutions. For example: $1+2-3-$ $-4+5=1 ; 1-2+3+4-5=1$. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
75. Can the sum of four consecutive integers be a prime number?
## Problems for the ninth grade | 75. The sum of four consecutive integers is an even number. There is only one even prime number. This number is 2. Since $-1+0+1+2=2$, the answer to the question of the problem is affirmative. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
82. There is a water tap and two containers: a three-liter and a five-liter. How can you get 4 liters of water in the larger one? | 82. Let's fill a five-liter container and pour 3 liters from it into a three-liter container. Then we will empty the three-liter container and transfer the 2 liters of water remaining in the five-liter container into it. Finally, we will fill the five-liter container from the tap and pour 1 liter from it into the three... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
86. Three numbers end with different digits. Can their pairwise products end with the same digit? | 86. They can. For example, the three numbers 2, 5, and 10 end in different digits, but their pairwise products all end in the same digit 0. | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
101. Write the number 100 using six identical non-zero digits in such a way that the method of writing does not depend on the digits used.
102 Calculate:
$\lg \operatorname{tg} 37^{\circ} \cdot \lg \operatorname{tg} 38^{\circ} \cdot \lg \operatorname{tg} 39^{\circ} \ldots \lg \operatorname{tg} 47^{\circ}$ | 101. Let $x$ be any non-zero digit. Then the expression $(\overline{x x x}-\overline{x x}): x=100$ is true for any digit $x$. 102. Since $\operatorname{tg} 45^{\circ}=1$, then $\lg \operatorname{tg} 45^{\circ}=0$. Therefore, the given product is also zero. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
108. The purchase of a book cost 1 ruble and one third of the book's cost. What is the cost of the book? | 108. Answer. 1 ruble 50 kopecks. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
109. What is the last digit of the difference
$$
1 \cdot 2 \cdot 3 \cdot 4 \ldots 13-1 \cdot 3 \cdot 5 \cdot 7 \ldots 13 ?
$$ | 109. The minuend ends in 0, and the subtrahend ends in 5. Therefore, the difference ends in 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
116. How many lines can divide a plane into 5 parts? | 116. Two lines can divide the plane only into 3 or 4 parts (see Fig. 15). Three lines can divide the plane only into 4, 6, and 7 parts (see Fig. 16). Four lines can divide the plane into 5 parts only if these lines are parallel (see Fig. 17). It is obvious that with any arrangement of five or more lines, the number of ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
118. A circle is inscribed in a square, and then a new square is inscribed in the circle. Find the ratio of the areas of these squares. | 118. It is clear that the second square can be placed so that its vertices fall on the points of tangency of the circle with the sides of the first square (see Fig. 18). After this, it is not difficult to verify that the desired ratio of the areas of the squares is 2.
 Prove that the square of an integer cannot end in any of the digits $2, 3, 7, 8$.
b) Prove that for no natural number $n$ can the numbers $5n+2$ and $5n+3$ be perfect squares.
c) Prove that for no integer $n$ is the number $n^2 + 3$ divisible by 5.
d) Find all natural values of $n$ for which the number $123 \l... | 7. a) Any integer can be represented in the form: \( n = 10k + r \), where \( k \) is an integer and \( r = 0, 1, 2, \ldots, 9 \). Based on the equality \( n^2 = 10(10k^2 + 2kr) + r^2 \), we conclude that the numbers \( n^2 \) and \( r^2 \) end with the same digit. It remains to verify directly that \( r^2 \) does not ... | 4 | Number Theory | proof | Yes | Yes | olympiads | false |
19. Find all prime numbers $p$ such that the numbers $p+4$ and $p+8$ are also prime. | 19. It is not hard to notice that the number 3 satisfies the requirements of the problem. Let's show that the problem has no other solutions. Indeed, every prime number not equal to 3 has one of two forms: $3k+1$ or $3k+2$, where $k$ is a non-negative integer. But in the case $p=3k+1$, the number $p+8$ is divisible by ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
22. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square. | 22. For $n=1$ the number $n^{2}+3 n$ equals 4, i.e., is a perfect square. Let now $n>1$. Obviously, in this case
$$
(n+1)^{2}<n^{2}+3 n<(n+2)^{2}
$$
Since a perfect square cannot exist between the squares of two consecutive natural numbers, the problem has a unique solution $n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
31. Find all prime numbers $p$ such that $p^{2}+14$ is also a prime number. | 31. It is easy to notice that the prime number $p=3$ satisfies the condition of the problem. If $p \neq 3$, then $p^{2}$ has the form $3 \mathrm{k}+1$ (see problem 5 a). Then $p^{2}+14=3 k+15=3(k+5)$. The obtained number is composite. Therefore, the problem has a unique solution $p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
18. This square should be cut by a broken line into two parts of equal area so that each segment of the broken line is parallel to the side or diagonal of the square, and the sum of the lengths of the segments parallel to the sides equals the length of the side, while the sum of the lengths of the segments parallel to ... | 18. It is obvious that each segment of the broken line must be shorter than the side or diagonal of the square parallel to it, since otherwise, this segment alone would already divide the square into two parts. From this, it follows that the number of segments of the broken line cannot be less than four. We will now sh... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. List all two-element subsets of a five-element set. | 7. Let's take the five-element set $\{A ; B ; C ; D ; E\}$. We will list its two-element subsets:
$\{A ; B\},\{A ; C\},\{A ; D\},\{A ; E\},\{B ; C\},\{B ; D\},\{B ; E\},\{C ; D\}$, $\{C ; E\},\{D ; E\}$. In total, we have 10 subsets. | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Given 5 points, no three of which lie on the same line. How many segments need to be drawn to connect each pair of points? | 8. The solution to the problem is shown in Figure 93. We will obtain 10 segments. It is useful to compare this problem with problem 7.
Fig. 93
 | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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