problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
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25.15. Calculate $\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})$. | 25.15. Answer: 0. It is clear that
$$
0<\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}}
$$
Therefore, $\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})=0$. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
25.35. Find the first digit of the number $2^{400}$. | 25.35. Answer: 2. The first digit of the number $2^{400}=\left(2^{10}\right)^{40}=1024^{40}$ coincides with the first digit of the number $1,024^{40}$. On the one hand, according to problem 25.34, we get $1,024^{40}1+40 \cdot 0,024+\frac{40 \cdot 39}{2} 0,024^{2}=2,40928>2$. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
25.40. Does the sequence $a_{n}=\sin (2 \pi n!e)$ converge? | 25.40. A n s w e r: yes, converges to 0. According to problem 25.39, the fractional part of the number $n!e$ is between 0 and $1 / n$. Therefore, $0<\sin (2 \pi n!e)<$ $<\sin (2 \pi / n)$ for $n \geqslant 4$. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
27.13. Solve the equation $5^{2 x-1}+5^{x+1}=250$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 27.13. Answer: $x=2$. The function $f(x)=5^{2 x-1}+5^{x+1}$ is monotonically increasing, so it takes the value 250 at only one value of $x$. It is also clear that $f(2)=5^{3}+5^{3}=250$. | 2 | Algebra | proof | Yes | Yes | olympiads | false |
27.26. Provide an example of positive irrational numbers $a$ and $b$ such that the number $a^{b}$ is an integer.
## 27.6. Some Notable Limits | 27.26. Let $a=\sqrt{2}$ and $b=\log _{\sqrt{2}} 3$. The numbers $a$ and $b$ are irrational (problems 6.16 and 27.25). In this case, $a^{b}=(\sqrt{2})^{\log _{\sqrt{2}} 3}=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28.62. Calculate the limit $\lim _{x \rightarrow 0} \frac{\tan x - x}{x - \sin x}$.
### 28.10. The number of roots of the equation | 28.62. Answer: 2. Transform the ratio of derivatives:
$$
\frac{\frac{1}{\cos ^{2} x}-1}{1-\cos x}=\frac{1}{\cos ^{2} x} \cdot \frac{1-\cos x^{2}}{1-\cos x}=\frac{1+\cos x}{\cos ^{2} x}
$$
The last expression tends to 2 as $x \rightarrow 0$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
29.27. Calculate the area under the graph of the function $y=$ $=\sin x$ on the interval from 0 to $\pi$.
## 29.5. Calculation of Volumes
Let a body be located in space with rectangular coordinates $O x y z$, such that the projection of this body onto the $O x$ axis is the segment $[a, b]$. Suppose that a plane passi... | 29.27. The desired area is equal to $\int_{0}^{\pi} \sin x d x=-\left.\cos x\right|_{0} ^{\pi}=2$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
32.15. Prove that if $a+b+c+d=2$ and $1 / a+1 / b+1 / c+$ $+1 / d=2$, then
$$
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=2
$$
$$
\% * *
$$ | 32.15. Let $\sigma_{k}$ be the $k$-th elementary symmetric function of $a, b, c, d$. Given that $\sigma_{1}=2$ and $\sigma_{3}=2 \sigma_{4}$. Therefore,
$$
\begin{aligned}
\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d}=\frac{4-3 \sigma_{1}+2 \sigma_{2}-\sigma_{3}}{1-\sigma_{1}+\sigma_{2}-\sigma_{3}+\sigma_{4}... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
2. Write down and read all seven-digit numbers, the sum of the digits in each of which is equal to 2. How many such numbers are there? | 2. $2000000,1100000,1010000,1001000,1000100,1000010,1000001$. Bcero 7 numbers. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
27. Find the digit that has the property that if it is appended to the end of any natural number, the resulting number is equal to the sum of three terms, the first of which is the original number, the second is the number represented by the sought digit, and the third is the product of the first two terms. | 27. Let the natural number be $a$, and the appended digit be $x$. Then according to the condition $a \cdot 10 + x = a + x + a x$, from which $9 a = a x$, and therefore, $x = 9$. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
37. In the hundred-digit number $12345678901234567890 \ldots 1234567890$, all digits in odd positions were erased. In the resulting fifty-digit number, digits in odd positions were erased again. The erasing continued until nothing was left. What was the last digit to be erased? | 37. The first time, 50 odd digits were crossed out, leaving 50 even ones. The second time, 25 digits were crossed out, including 5 twos, 5 fours, 5 sixes, 5 eights, and 5 zeros. 25 remained in the following sequence: $4,8,2,6,0,4$, $8,2,6,0, \ldots, 2,6,0$. (The group of digits 4, 8, 2, 6,0 is written in sequence 5 tim... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. How many different products divisible by ten can be formed from the numbers $2,3,5,7,9 ?$ | 8. For the product of numbers to be divisible by 10, the factors must include the numbers 2 and 5. Taking this into account, the following products satisfy the condition of the problem:
1) $2 \cdot 5$
2) $2 \cdot 5 \cdot 3$
3) $2 \cdot 5 \cdot 7$
4) $2 \cdot 5 \cdot 9$;
5) $2 \cdot 5 \cdot 3 \cdot 7$
6) $2 \cdot 5 \cdo... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. A store needed to receive 185 kg of candies in closed boxes from the warehouse. The warehouse has boxes of candies weighing 16 kg, 17 kg, and 21 kg. What boxes and how many could the store receive? | 9. We notice that $185=37 \cdot 5$ and $16+21=37$. Therefore, the store could have received 5 boxes weighing 21 kg each and 5 boxes weighing 16 kg each. Since the number 185 has been factored into prime factors, the problem has a unique solution. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. A kilogram of one type of candy is 80 k more expensive than another. Andrey and Yura bought 150 g of candies, which included both types of candies, with Andrey having twice as much of the first type as the second, and Yura having an equal amount of each. Who paid more for their purchase and by how much?
14 | 10. Andrei has 100 g of one type of candy and 50 g of another, while Yura has 75 g of each type. Andrei's candies are more expensive than Yura's by the same amount that 25 g of one type is more expensive than 25 g of the other. Since 1 kg is more expensive by 80 k., then 25 g is more expensive by 2 k. $(80: 40=2)$. The... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
30. Think of a single-digit number: double it, add 3, multiply by 5, add 7, using the last digit of the resulting number, write down a single-digit number, add 18 to it, and divide the result by 5. What number did you get? No matter what number you thought of, you will always get the same number in the final result. Ex... | 30. Let the number thought of be $x$. Then we sequentially obtain:
1) $2 x ; 2) 2 x+3 ; 3) 10 x+15=10(x+1)+5$
2) $10(x+1)+12 ; 5) 2 ; 6) 2+18=20$; 7) $20: 5=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
31. The quotient of dividing one number by another is an integer that is two times smaller than one of them and six times larger than the other. Find this quotient. | 31. Let the dividend be $a$, and the divisor be $b$, then according to the condition, the quotient is $a: 2$ or $6 b$, i.e., $a: 2 = 6 b$, from which $a = 12 b$ and $a: b = 12$. Therefore, the quotient $a: b$ is 12. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
40. In a football tournament, each of the participating teams played against each other once. The teams scored $16,14,10,10,8,6,5,3$ points. How many teams participated in the tournament and how many points did the teams that played and finished in the top 4 positions lose? (A team gets 2 points for a win and 1 point f... | 40. If the number of teams is $n$, then the total number of games is $\frac{n(n-1)}{2}$, and the total number of points scored by them is $(n-1) n$. In our example, we have $16+14+10+10+8+$ $+6+5+3=72$. Therefore, $(n-1) \cdot n=72$, or $(n-1) \cdot n=8 \cdot 9$, so $n=9$, i.e., there were 9 teams. The maximum number o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
47. Over the past five years, 27 collective farmers have been awarded. Moreover, in each subsequent year, more were awarded than in the previous year. In the last year, three times as many were awarded as in the first year. How many collective farmers were awarded in the third year? | 47. If in the first year $x$ collective farm workers are awarded, then in the last year it is $3x$, and for these two years it is $4x$. The values $x=1$ and $x=2$ do not satisfy the condition, as in the second, third, and fourth years, 23 (27-4=23) and 19 (27-8=19) would be awarded, which does not meet the condition th... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
64. First, two numbers 2 and 3 are written, then they are multiplied and their product, the number 6, is recorded, then the product of the two last single-digit numbers $3 \cdot 6=18$, then $1 \cdot 8=8$, then $8 \cdot 8=64$ and so on. As a result, we get the following sequence of digits:
## $2361886424 \ldots$
Which... | 64. As a result of analyzing the sequence of digits $236188642483261224832612248 \ldots$, we notice that starting from the tenth digit, the group of digits 48326122 repeats (as in the previous problem, let's call it the period). Before the first period, there are 9 digits, and the period itself consists of 8 digits. Am... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Tourists from one of the groups bought various souvenirs, with each of them taking a set of souvenirs of the same cost and as many as the cost of one souvenir in rubles. All tourists paid with ten-ruble bills (one or several), and each received change that did not match the change received by any of the other touris... | 5. The amount in rubles that each tourist paid for a set of souvenirs is expressed by a number of the form $a^{2}$, where $a$ is the number of souvenirs or the price of one souvenir in rubles. Since the squares of natural numbers can only end in the digits $1,4,5,6,9$, there can only be 5 different changes, namely 9 p.... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. To the question: "How many two-digit numbers are there such that the sum of their digits is 9?" - Stepa Verkhoglyadkin started listing all two-digit numbers in a row, selecting the ones he needed. Show him a shorter way to solve the problem. | 6. For two-digit numbers, except for 99, the statement is true: “If a number is divisible by 9, then the sum of its digits is 9, and vice versa.” Up to 99 inclusive, there are 11 numbers divisible by 9 (99: 9=11). However, among them, two numbers (9 and 99) are not considered in this problem, so the number of numbers s... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
24. Mowgli asked his monkey friends to bring him some nuts. The monkeys gathered an equal number of nuts and carried them to Mowgli. But on the way, they quarreled, and each monkey threw a nut at every other monkey. As a result, Mowgli received only 35 nuts. How many nuts did the monkeys gather, if it is known that eac... | 24. Since the monkeys collected nuts equally and threw them away equally, they brought them equally. The number 35 is divisible by 5, so we have $35=5 \cdot 7$. There can be 2 cases:
1) There were 5 monkeys, each brought 7 nuts, and each threw away 4 nuts, so each collected $7+4=11$.
2) There were 7 monkeys, each broug... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. On New Year's Eve, Father Frost gave the children the following task: using all nine digits from 1 to 9 once each, insert "+" or "-" between every two adjacent digits so that after performing the operations, all possible two-digit prime numbers can be obtained. Help the children complete this task. How many such num... | 8. If we place only the "+" sign between the digits, we will get 45 after performing the operations. There are only 10 prime two-digit numbers less than 45. These numbers are $11,13,17,19,23,29,31,37,41,43$. They can be obtained as follows:
1) $1+2+3+4+5+6-8-9=11$
2) $1+2+3+4+5+6-7+8-9=13$
3) $1+2+3+4-5+6+7+8-9=17$
4) ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
20. Find $p$, if each of the numbers $p, p+10$, and $p+14$ is prime.
 | 20. All natural numbers can be divided into 3 groups:
1) divisible by 3, i.e., of the form $3 k$;
2) giving a remainder of 1 when divided by 3, i.e., of the form $3 k+1$;
3) giving a remainder of 2 when divided by 3, i.e., of the form $3 k+2$.
If $p=3 k$, then it is prime only when $k=1$; then $p+10=13$ and $p+14=17$.... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
21. Find $p$, if each of the numbers $p, 2p+1$, and $4p+1$ is prime. | 21. The number $p$ can be represented in one of three forms: $3 k, 3 k+1$ or $3 k+2$. For $p=3 k+1$, we have:
$2 p+1=2(3 k+1)+1=6 k+3=3(2 k+1)$ - a composite number. For $p=3 k+2$, we have:
$4 p+1=12 k+9=3(4 k+3)$ - a composite number.
Fig. 12
![](https://cdn.mathpix.com/cropped/2024_05_21_00604dc020e3721fc1f4g-145... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
23. Find all such prime numbers, so that each of them can be represented as the sum and difference of two prime numbers. | 23. Since the desired number $p$ must be the sum of two prime numbers, $p>2$. All such prime numbers are odd. To get odd numbers when adding and subtracting two numbers, one of the components must be odd and the other even. Since the only even prime number is 2, then $p=p_{1}+2$ and $p=p_{2}-2$. Therefore, the prime nu... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28. In all the carriages of a passenger train, 737 tourists were evenly distributed. How many carriages were there and how many tourists were in each carriage? | 28. Since $737=67 \cdot 11$ and the numbers 67 and 11 are prime, it is clear that there were 11 cars, and in each car there were 67 tourists. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
33. To answer the question: "Is the difference $9^{1986}-8^{1980}$ divisible by 5 and by 10?" - one could call upon a computer. But perhaps you would like to compete with it? Give it a try. | 33. Since $9^{1}=9, 9^{2}=81, 9^{3}=729$, it is clear that the odd powers of nine end in the digit 9, and the even powers end in the digit 1. Let's consider the powers of the number 8: $8^{1}=8 ; 8^{2}=64$-ends in the digit $4 ; 8^{3}$ ends in the digit $2 ; 8^{4}$ ends in the digit $6 ; 8^{5}$ ends in the digit 8, and... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
34. To calculate the product of all natural numbers from 1 to 50 inclusive, of course, it's better to use a computer. However, you can easily state the number of zeros at the end of this product without resorting to a computer. How many are there? | 34. Among the factors of the number $1 \cdot 2 \cdot 3 \cdot \ldots \cdot 50$, 10 numbers are divisible by 5 $(50: 5=10)$, and out of these, 2 numbers are divisible by $5^{2}=25$. Therefore, the product of the numbers from 1 to 50, when factored into prime factors, will contain 12 fives, and there will undoubtedly be m... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
40. The steamship has $a$ funnels, $b$ screws, and $c$ people on board. It set off on the $n$-th day, $k$-th month, of the year $(1900+p)$. The product of the six numbers $a$, $b$, $c$, $n$, $k$, $p$, increased by the number whose cube is the captain's age in years, equals 4752862. Determine the unknowns. | 40. Since $2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125$, it is clear that the captain is either 27 years old or 64 years old. If the captain is 64 years old, then from the number 4752862, 4 should be subtracted and the resulting number should be factored into 6 factors ( $4752862-4=$ $=4752858 ; 4752858=2 \cdot 3 \cdot 11 \cdo... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Find the greatest common divisor of all five-digit numbers written using the digits
$$
1,2,3,4,5
$$
without repetition. | 10. Each of such five-digit numbers is divisible by 3, since $1+2+$ $+3+4+5=15$, and $15: 3$. These numbers do not have any common divisors greater than 3, as, for example, $12345=3 \cdot 5 \cdot 823$, but the number 12354 does not divide evenly by either 5 or 823. Therefore, the greatest common divisor of these number... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. In one class, there are fewer than 50 students. On Gorky Street, $\frac{1}{7}$ of the students from this class live, on Pushkin Street - $\frac{1}{3}$, on Gaidar Street - $\frac{1}{2}$, and on Shipbuilding Street - the remaining students. How many students live on Shipbuilding Street? | 11. The number of students in the class must be a multiple of the numbers 7, 3, and 2. The least common multiple of these numbers is $42(7 \cdot 3 \cdot 2=42)$. It is less than 50, so it satisfies the condition. Then $42 \cdot \frac{1}{7}=6 ; 42 \cdot \frac{1}{3}=14 ; 42 \cdot \frac{1}{2}=21$; $42-(6+21+14)=1$. Therefo... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
13. At the lumber yard, there were logs 6 m and 8 m long of the same thickness. Which logs are more advantageous to take in order to get more pieces with fewer cuts when cutting them into 1-meter pieces? | 13. It is more profitable to take logs from which more pieces can be obtained with fewer or the same number of cuts. From a six-meter log, 6 pieces can be obtained with 5 cuts, while from an eight-meter log, 8 pieces can be obtained with 7 cuts. Let's compare the number of pieces obtained from these logs with the same ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. Half of the way the tourists walked, and the other half they traveled by bus, spending a total of 5.5 hours on the entire journey. If they had traveled the entire distance by bus, it would have taken them 1 hour. How much time in total will the tourists spend if they walk the entire distance? How many times faster i... | 9. Half of the way the tourists traveled by bus in 0.5 hours (1:2=0.5), then on foot they walked the other half in 5 hours (5.5-0.5). Therefore, the entire journey will take the tourists 10 hours (5 * 2=10). Traveling by bus is 10 times faster than walking. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. The bear was carrying pastries from the bazaar,
But on the forest edge
He ate half the pastries
And plus half a pastry.
He walked, walked, sat down to rest
And to the "coo-coo" of the cuckoo
He ate half the pastries again
And plus half a pastry.
It got dark, he quickened his pace,
But on the porch of the cottage
... | 11. On the porch of the little house, the bear ate all the remaining cookies, which were half a cookie and half a cookie. This means that half a cookie is half of the remainder, i.e., the third time he ate one cookie. Therefore, when he ate half the cookies and another half cookie the second time, one cookie was left. ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. A group of tourists was supposed to arrive at the station at 5 o'clock. By this time, a bus was supposed to arrive from the tourist base to pick them up. However, arriving at the station at 3:15 PM, the tourists, not waiting for the bus, started walking to the tourist base. Meeting the bus on the way, they got in a... | 18. Since the bus arrived at the tourist base 15 minutes earlier, it should have traveled from the meeting point with the tourists to the station for $\frac{1}{8}$ hours (15 minutes / 2 = $=7.5$ minutes $=\frac{1}{8}$ hours). In this time, it would travel 7.5 km $(60: 8=7.5)$, which means the tourists walked 7.5 km on ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
45. Calculate in the most rational way:
$$
x=\frac{225+375 \cdot 138}{375 \cdot 139-150}
$$ | 45. Since $375 \cdot 139-150=375 \cdot 138+375-150=375 \cdot 138+225$, then $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
48. On the sheet, several non-zero numbers are written, each of which is equal to half the sum of the others. How many numbers are written? | 48. Let there be $n$ numbers, then
$$
\begin{aligned}
a_{1} & =\frac{a_{2}+a_{3}+\ldots a_{n}}{2} ; \\
a_{2} & =\frac{a_{1}+a_{3}+\ldots+a_{n}}{2} ; \\
\ldots & \ldots \ldots \ldots+a_{n-1} \\
a_{n} & =\frac{a_{1}+a_{2}+\ldots+a_{n}}{2} ; \\
2\left(a_{1}\right. & \left.+a_{2}+\ldots+a_{n}\right)=(n-1) \cdot\left(a_{1}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
59. Find the non-negative integer values of $n$ for which $\frac{30 n+2}{12 n+1}$ is an integer. | 59. $\frac{30 n+2}{12 n+1}=\frac{(24 n+2)+6 n}{12 n+1}=2+\frac{6 n}{12 n+1}; \quad \frac{6 n}{12 n+1}$
will be an integer only when $n=0$, for other values of $n$ it is a proper fraction. Therefore, only when $n=0$ is $\frac{30 n+2}{12 n+1}$ an integer (2). | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
60. A barrel contains exactly 30 liters of linseed oil. For three construction brigades, 3 barrels were filled from it, each of which holds a whole number of liters, and the capacity of the first is $\frac{2}{3}$ of the capacity of the second or $\frac{3}{5}$ of the capacity of the third. How many liters of linseed oil... | 60. Let's express the capacities of the second and third buckets in terms of the capacity of the first one and schematically write it as: $\mathrm{II}=\frac{3}{2} \mathrm{I} ; \mathrm{III}=\frac{5}{3} \mathrm{I}$. Since each bucket contains a whole number of liters and the smallest number divisible by 2 and 3 is 6, the... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
66. After the tourists have walked 1 km and half of the remaining distance, they still have to walk a third of the entire distance and 1 km. What is the total distance? | 66. After the tourists have walked 1 km, half of the remaining distance is one third of the entire distance plus 1 km, which means the entire remaining distance is $\frac{2}{3}$ of the entire distance plus 2 km. Thus, the entire distance is $\frac{2}{3}$ of the entire distance plus $2+1=3$ (km), which means 3 km is $\f... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
70. A certain amount was spent on strawberries at 2 r. 40 k. per 1 kg of one variety and the same amount on another variety at 1 r. 60 k. Find the average price of 1 kg of the strawberries purchased. | 70. Let the amount spent on each type of strawberry be $a$ r. Then the first type purchased is $a: 2.4=\frac{a}{2.4}$ (kg), and the second type is $-\frac{a}{1.6}$ kg. In total, $\left(\frac{a}{2.4}+\frac{a}{1.6}\right)$ kg were purchased for a total of $2a$ r. The average price per 1 kg is
$$
2 a:\left(\frac{a}{2.4}+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
81. At exactly 24:00, the battery received an order to begin artillery preparation at the earliest time that will pass until the hour and minute hands overlap again. At what time should the artillery preparation begin? | 81. In 1 hour, the minute hand makes a full revolution and will point to 12, while the hour hand will have traveled $\frac{1}{12}$ of the circumference and will point to 1. This means that after 1 hour, the minute hand, following the hour hand, will be at a distance equal to $\frac{1}{12}$ of the circumference from it.... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
82. Andrey and Grandpa Grisha went mushroom picking sometime between six and seven in the morning, at the moment when the clock hands were aligned. They returned home between twelve and one in the afternoon, at the moment when the clock hands were pointing in exactly opposite directions. How long did their mushroom "hu... | 82. In 1 hour, the minute hand makes a full revolution (its end describes a circle), while the hour hand travels $\frac{1}{12}$ of the circumference. Therefore, in one hour, the minute hand will travel more than the hour hand by $\frac{11}{12}$ of the circumference. By the time the minute hand travels from 12 to 6, i.e... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
85. To which number can the expression
$$
\frac{(a+b)(b+c)(a+c)}{a b c}
$$
be equal if the numbers $a, b$ and $c$ satisfy the condition of problem 84? | 85. From the solution of the previous problem, we have $a+b=c n ; b+c=a n ; a+c=b n$. After subtracting, we get $a-c=n(c-a)$. The equality is possible when $a-c=0$, i.e., $a=c$. Similarly, $a=b$, so $a=b=c$, and from this, $\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=2$. Therefore, $\frac{(a+b)(b+c)(a+c)}{a b c}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. On a horizontal coordinate line, several integers are marked as points, the sum of which is equal to 25. If each point is moved 5 unit segments to the left, then the sum of the numbers corresponding to these points will become equal to -35. How many numbers were marked on the line? | 1. The sum of the numbers decreased by $25-(-35)=60$. Therefore, $60: 5=12$ numbers were marked. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. In one line, 19 numbers were written. The sum of any three consecutive numbers is positive. Can the sum of all 19 numbers be negative? | 9. Consider, for example, the sequence of numbers $\underbrace{-7,4,4, \ldots,-7,4,4,-7}_{19 \text { numbers }}$. The sum of any 3 consecutive numbers in this sequence is 1, i.e., positive, while the sum of all 19 numbers is -1, i.e., negative. Answer: It can. Find another similar example. | -1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
12. Based on the first three lines, establish the rule by which the number $c$ is obtained from the numbers $a$ and $b$, and fill in the missing numbers in the empty cells:
| $a$ | $b$ | $c$ |
| :---: | :---: | :---: |
| 5 | -11 | 6 |
| $-2.7$ | $-2.3$ | 5 |
| 32 | -18 | -14 |
| -17 | 5 | |
| 14 | -14 | | | 12. Obviously, $c=-(a+b)$, so the missing numbers are $-(-17+5)=12$; $-(14-14)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
19. On the number line, points $A, B, C, D$ were marked. It is known that the first three points correspond to the numbers 5, 8, and -10. What number corresponds to point $D$, if when the direction of the number line is reversed, the sum of the four numbers corresponding to these points does not change? | 19. Let the point $D$ correspond to the number $x$. The sum of the numbers before changing the direction of the line is $5+8-10+x$, and after changing the direction it is $-5-8+10-x$. According to the condition, $5+8-10+x=5-8+10-x$, from which $2x = -6$ and $x = -3$. The problem can be solved more simply without resort... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
21. Simplify the expression
$-1-(-1-(-1-(-1-(\ldots$, if the expression contains:
a) 1989 ones; b) 1990 ones. | 21. Observing
$$
\begin{aligned}
& -1-(-1)=-1+1=0 \\
& -1-(-1-(-1))=-1 ; \\
& -1-(-1-(-1-(-1)))=0 ; \\
& -1-(-1-(-1-(-1-(-1))))=-1
\end{aligned}
$$
we notice that with an even number of ones, the result is 0, and with an odd number of ones, the result is -1. Therefore, in case a) we have -1, and in case b) we have 0.... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
22. Given 5 numbers: $a_{1}=-1, a_{2}=0, a_{3}=1, a_{4}=2, a_{5}=3$. The sixth is the product of the first and the second, the seventh is the product of the second and the third, the eighth is the product of the third and the fourth, and so on. What is the last non-zero number in this sequence? | 22. Let's construct a table:
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
23. Given 5 numbers: $a_{1}=1, a_{2}=-1, a_{3}=-1, a_{4}=1, a_{5}=-1$. The following numbers are determined as follows: $a_{6}=a_{1} \cdot a_{2}, a_{7}=a_{2} \cdot a_{3}, a_{8}=a_{3} \cdot a_{4}$ and so on. What is $a_{1988}$? | 23. Let's construct a table:
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | -1 | -1 | 1 | -1 | -1 | 1 | -1 | -1 | 1 | -1 | -1 |
We notic... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Petya has as many sisters as brothers, while his sister Ira has twice as few sisters as brothers. How many boys and how many girls are there in this family? | 9. If there are $x$ girls, then there are $x+1$ boys. Without Irina, there will be $x-1$ girls, then from the condition we have
$$
2(x-1)=x+1
$$
from which $x=3$. Therefore, there are 3 girls and 4 boys in the family. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10. Aunt Masha is three years younger than Sasha and his same-aged friend Pasha combined. How old is Sasha when Aunt Masha was as old as Pasha is now? | 10. Let Sasha and Pasha be $a$ years old, then Aunt Masha is $(2 a-3)$ years old. Aunt Masha was $a$ years old $(a-3)$ years ago $(2 a-3-a=a-3)$. But so many years ago, Sasha was 3 years old $(a-(a-3)=3)$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. In the workshop, five clocks were checked. It turned out that clock No. 3 is 3 minutes behind clock No. 2 per day. Clock No. 1 is 1 minute behind clock No. 2 per day, clock No. 4 is behind clock No. 3, and clock No. 5 is behind clock No. 1. If the readings of all five clocks are added together and the sum is divide... | 14. Let clock No. 1 show $x$ min, then clock No. $2-(x+1)$, clock No. $3-(x-2)$, clock No. $4-(x-3)$ and clock No. $5-(x-1)$. The sum of the readings of all clocks $x+(x+1)+(x-2)+(x-3)+(x-1)=5 x-5$. Since $(5 x-5): 5=$ $=x-1$, and this is the reading of clock No. 5, then these clocks show the exact time. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
15. How many people are in the team if the average age of all team members is 25 years, the team leader is 45 years old, and the average age of the team members without the team leader is 23 years? | 15. Let there be $x$ people in the brigade, then we have $25 x=23(x-1)+45$, from which $2 x=22$ and $x=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
22. Yura left the house for school 5 minutes later than Lena, but walked twice as fast as she did. How many minutes after leaving will Yura catch up to Lena? | 22. If Lena walks $a$ m in 1 min, then Yura walks $2a$ m. Let him catch up with her after $x$ min from his start. Then we have $a \cdot 5 + a \cdot x = 2a \cdot x$, from which $x=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
24. Preparing his son for school, the father bought him a new uniform for 24 r. How did he pay for the purchase, if he only had 5 r bills, and the cashier only had 3 r bills? | 24. If the buyer gave the cashier $x$ bills and received $y$ bills as change, we have the equation $5 x-3 y=24$, from which $x=\frac{24+3 y}{5}$. Since $x$ must be an integer, $24+3 y$ must be divisible by 5. This is possible with the smallest value of $y=2$. Therefore, the buyer gave 6 bills (30 p.) and received 6 p. ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
26. In a competition of fun and ingenuity, 9 points were awarded for each correctly completed task, and 5 points were deducted for each uncompleted or incorrectly completed task. It is known that the team was offered no more than 15 tasks and scored 57 points. How many tasks did the team complete correctly? | 26. Let $x$ be the number of tasks completed correctly, and $y$ be the number of tasks completed incorrectly or not at all. Then we have $9 x-5 y=57$, from which $x=\frac{57+5 y}{9}$. Since $x$ is a natural number, $(57+5 y)$ must be divisible by 9. The smallest value of $y$ for which this condition is met is 3. In thi... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
27. If the team had scored 97 points in the last sport, the average number of points scored by it in one sport would have been 90; if it had scored 73 points in this last sport, the average number of points per sport would have been 87. In how many sports did the team compete? | 27. Let the team compete in $x$ sports and score $n$ points in all sports except the last one. Then we have
$$
(97+n): x=90 ;(73+n): x=87
$$
which means $97+n=90 x$ and $73+n=87 x$, so $x=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
30. Several girls were picking mushrooms. They divided the collected mushrooms among themselves as follows: one of them was given 20 mushrooms and 0.04 of the remainder, another - 21 mushrooms and 0.04 of the new remainder, the third - 22 mushrooms and 0.04 of the remainder, and so on. It turned out that everyone recei... | 30. Let the girls collected $x$ mushrooms. The first one received $20+0.04(x-20)$, the second one received $21+0.04(x-21-20-0.04(x-20))$ mushrooms. According to the condition $20+0.04 x-$ $-0.8=21+0.04(x-21-20-0.04(x-20))$, from which
$$
0.0016 x=0.192 \text { and } x=120 .
$$
Thus, 120 mushrooms were collected, each... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
37. Two people, who have one bicycle, need to get from point $A$ to point $B$, which is 40 km away from $A$. The first person walks at a speed of 4 km/h and rides the bicycle at 30 km/h, while the second person walks at a speed of 6 km/h and rides the bicycle at 20 km/h. What is the minimum time they can take to reach ... | 37. Let the first person travel $x$ hours by vehicle and $y$ hours on foot, then the second person travels $y$ hours by vehicle and $x$ hours on foot. We have the equation $30 x + 4 y = 20 y + 6 x$, from which $x = \frac{2}{3} y$. Therefore, the first person walked $y$ hours and traveled $\frac{2}{3} y$ hours by vehicl... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
42. Ten identical books cost no more than 11 p., while 11 such books cost more than $12 \mathrm{p}$. How much does one book cost? | 42. Let one book cost $x$ k. Then we have $x \cdot 10 \leqslant 1100$ and $x \cdot 11 > 1200$ or $x \leqslant 110$ and $x > 109 \frac{1}{11}$. Therefore, $x=110$ k., i.e., the book costs 1 p. 10 k. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Find the length of the side of a square whose area is numerically equal to its perimeter. | 2. Let the side length of the square be $a$, then the area $S=a \cdot a$, and the perimeter $P=4 a$. According to the condition $a \cdot a=4 a$, from which $a=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
25. A scout is in a house with four windows arranged in a rectangular shape. He needs to signal to the sea at night by lighting a window or several windows. How many different signals can he send? | 25. Let's schematically represent the windows and measure them:
a) Lighting all four windows gives one signal;
b) Lighting one of the windows is perceived as one signal, as in the dark, th... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10. Working in a collective farm orchard harvesting fruits, schoolchildren collected 22 crates, some of which contained apples, others pears, and still others plums. Can it be asserted that there are at least 8 crates, the contents of which are one of the specified types of fruit? | 10. Let's consider the most "unfavorable" case. Suppose there are 7 boxes of apples, pears, and plums each. This makes a total of 21 boxes. If the 22nd box is with apples, then there will be 8 boxes of apples; if this box is with pears, then there will be 8 boxes of pears; if this box is with plums, then there will be ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
18. The traffic inspector noticed that out of 20 cars that passed on the road to the airport, 14 were "Ladas," 15 were dark-colored, 17 were driven by men, and in 18 cars, there were passengers besides the driver. For what minimum number of cars could all 4 of these characteristics be true? | 18. Out of the total number of cars, 6 did not have the first feature (not "Zhiguli"), 5 did not have the second, 3 did not have the third, and 2 did not have the fourth. Therefore, the maximum number of cars that do not have at least one feature is 16 $(6+5+3+2=16)$. The remaining $4(20-16=4)$ have all four features. | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
22. The chain has broken into five parts,
But I hope you will
Join them as quickly as you can,
Doing the minimum work.
To cut and solder a link-
It takes two minutes...
Remember, the task is given
For a very thoughtful look.
Note. Each link has three rings. | 22. Cut 3 rings of one link and join the remaining 4 links with the three resulting rings (by welding). This will take 6 minutes. | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
28. A problem solved by Poisson (1781-1840) in his youth. Someone has 12 pints (a unit of volume) of honey and wants to pour out half of this amount, but he does not have a container with a capacity of 6 pints. He has 2 containers: one with a capacity of 8 pints, and the other with a capacity of 5 pints. How can he pou... | 28. Let's represent it with such a diagram:
| Vessel | | |
| :---: | :---: | :---: |
| 12 pints | 8 pints | 5 pints |
| 12 | - | - |
| 4 | 8 | - |
| 4 | 3 | 5 |
| 9 | - | - |
| 1 | 8 | 3 |
| 1 | 6 | 3 |
| 6 | 6 | - | | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
29. How can you use a balance scale and a 200g weight to divide 9 kg of sugar into two bags - 2 kg and 7 kg, if you are allowed to weigh no more than three times? | 29. Place a weight on one pan and distribute the available sugar across the pans so that the scales are in balance. Then, on one pan there will be 4 kg 600 g of sugar, and on the other (where the weight is) - 4 kg 400 g of sugar. The second time, distribute 4 kg 600 g of sugar across the pans. On one of them (without t... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
32. In which cases does a month have the maximum number of Saturdays? What is this number? | 32. The maximum number of Saturdays in one month is 5, since $7 \cdot 4=28 ; 28<30$ and $28<31$. In February, there can be 5 Saturdays if it is a leap year and the month starts on a Saturday. April, June, September, and November will have 5 Saturdays if they start on a Saturday or a Friday. The remaining months can hav... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
37. Football teams were holding a round-robin tournament. One of the fans of this game, returning from vacation, found the following situation: the total number of points scored by all teams was 44; one team, which scored the fewest points, received one point; the two teams occupying the first and second places scored ... | 37. Since $1+2+3+4+5+6+7+8+8=44$, there are 9 teams participating in the tournament. In total, they should score 72 points ( $8 \cdot 9=72$ ). Therefore, they still need to score 28 points $(72-44=28)$, which can be achieved in 14 games $(28: 2=14)$. | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
38. In a tournament where each pair of teams played each other twice, 4 teams participated. For each win, two points were awarded, for a draw - one, and for a loss - 0. The team that finished in last place scored 5 points. How many points did the team that finished in first place score? | 38. There were 12 games in total $(3 \cdot 4=12)$. In each match, -2 points were awarded, and the total points were -24. Since $5+6+7+8=26>24$, this case is impossible.
Therefore, $5+6+6+7=24$. This means the winning team scored 7 points. | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
39. In a chess tournament, each of the 8 participants plays one game with each other. All participants scored a different number of points (an integer), and the second-place finisher scored as many points as all the chess players who finished from fifth to eighth place combined. How did the players who finished in thir... | 39. The participant who took second place, out of a total of 28 points scored by all participants together, scored no more than six points, as otherwise, they would have taken either first place or shared the first and second places, which contradicts the condition. The only possible scenario is: $28=7+6+5+4+3+2+1+0$. ... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 22. Find the minimum value of the expression $(x+y) \cdot(x+z),$ if $x, y, z-$ are positive numbers and $x \cdot y \cdot z \cdot(x+y+z)=1$. | Solution. Let $a=y+z ; b=x+z ; c=x+y$. Consider a triangle with sides $a, b, c$ (it is not difficult to verify that such a triangle exists for any $x, y, z$). Then:
$$
\begin{gathered}
(x+y) \cdot(x+z)=b \cdot c \\
x y z(x+y+z)=(p-a)(p-b)(p-c) p=S^{2}=1
\end{gathered}
$$
But $b \cdot c=\frac{2 S}{\sin A}=\frac{2}{\si... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 101. The center of the inscribed circle divides the height of an isosceles triangle, drawn to the base, into segments of 5 cm and 3 cm, measured from the vertex. Find the sides of the triangle.
. Then $\frac{b+c}{a}=\frac{2 b}{a}=\frac{5}{3}$, or $a=\frac{6}{5} b$. Then $\quad C L_{1}=B L_{1}=\frac{a}{2}=\frac{3}{5} b . \quad$ By the Pythagorean theorem for triangle $A L_{1} C \quad A L_{1}=\frac{4}{5} b$. Thus:
76
$... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
19. Two natural numbers both give a remainder of 2 when divided by 3. What remainder does their product give when divided by 3? | 19. $(3 m+2)(3 n+2)=3(m n+m+n+1)+1$.
Translating the above text into English, while preserving the original text's line breaks and format, results in:
19. $(3 m+2)(3 n+2)=3(m n+m+n+1)+1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
65. What whole number is equal to the sum of all the whole numbers preceding it? Is there only one such number? | 65. $1+2=3$. No other number $>3$ has the required property: indeed, if we move forward in the sequence of natural numbers, then with each step taken, the sum of natural numbers increases by $3, 4, 5$, etc., while each subsequent natural number increases only by 1, and therefore the difference between the sum of natura... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
102. A certain number is written in the duodecimal system; for which divisor $m$ is the following divisibility rule valid: if the sum of the digits of the number is divisible by $m$, then the number is also divisible by $m$? | 102. $m=11$, because $12=11\cdot1+1, \quad 12^{2}=11\cdot13+1$ and so on. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
140. How to express with one equality sign that among three numbers $a, b, c$ at least one is equal to zero | 140. $a b c=0$.
The above text has been translated into English, maintaining the original text's line breaks and format. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
287. How many solutions does the system of equations generally have
$$
\begin{gathered}
a x^{2}+b x y+c y^{2}=d \\
a_{1} x^{2}+b_{1} x y+c_{1} y^{2}=d_{1} ?
\end{gathered}
$$
In particular, how many solutions does the system of question 280 have? | 287. Generally four. Excluding $d$ and $d_{1}$, we obtain a homogeneous equation of the second degree, which, generally, will determine two values of the ratio $\frac{x}{y}$ (see question 284) $\frac{x}{y}=k_{1}$ and $\frac{x}{y}=k_{2}$, from which $x=k_{1} y, x=k_{2} y$. Substituting each of these expressions sequenti... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
423. Can the sum of a finite number of terms of a geometric progression be zero? | 423. If all members of the progression are real numbers, then in the trivial case only, when the progression coefficient \( q = -1 \) and the number of terms is even. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
454. Prove that the third difference of the function $n^{3}$ is constantly equal to 6. Give the general form of a function that possesses this property. | 454. The first differences of the function $n^{3}$ are expressed by the function $3 n^{2}+3 n+1$, the second differences by $6 n+6$; consequently, the third differences are constant and equal to 6. If we add to $n^{3}$ an arbitrary function $\varphi(n)$, the third differences of which are zero, then the third differenc... | 6 | Algebra | proof | Yes | Yes | olympiads | false |
456. Prove that the third difference of the function
$$
f(n)=1^{2}+2^{2}+\ldots+n^{2}
$$
is constantly equal to 2. | 456. $f(n+1)-f(n)=(n+1)^{2}$; therefore, the first differences of $f(n)$ are $(n+1)^{2}$; the second differences are $(n+1)^{2}-n^{2}=2 n+1$, from which it is clear that the third differences are constantly equal to 2. | 2 | Algebra | proof | Yes | Yes | olympiads | false |
521. Through a point lying outside a plane, how many: a) lines perpendicular to the plane can be drawn? b) lines parallel to it? c) planes perpendicular to this plane? d) planes parallel to it? | 521. a) One, b) an infinite set - generating a plane parallel to the given one, c) an infinite set - passing through the perpendicular dropped from this point, d) one (see b). | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
551. On one of the sides of triangle $A B C$ there is a point $P$; draw a line through $\boldsymbol{P}$ that cuts off from triangle $\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}$ a triangle similar to it. How many solutions? | 551. In general, there are 4 solutions. If the triangle is isosceles or equilateral, the number of solutions decreases accordingly. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
641. Study the symmetric properties of an equilateral triangle. | 641. 3 axes of symmetry; no center of symmetry, for if such existed, it could only be (see question 634) the point of intersection of the altitudes. But it divides the altitudes in the ratio $1: 2$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
740. Show that a rectangle is the only convex polygon in which all angles are right angles. | 740. From the equation $\frac{180^{\circ}(n-2)}{n}=90^{\circ}$ it follows that $n=4$. | 4 | Geometry | proof | Yes | Yes | olympiads | false |
833. How many different values does the expression $\sin \frac{k \pi}{3}$ take when $k$ takes all integer values from 0 to $+\infty$? | 833. Let $q$ be the integer quotient of the division of $k$ by $3$, and $r$ the remainder, which lies between -2 and 2, then $k=3 q+r$.
We have $\sin \frac{k \pi}{3}=\sin \left(q \pi+\frac{r \pi}{3}\right)=\sin \left( \pm \frac{r \pi}{3}\right)$. Everything reduces to the number of different values of the seven member... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 11. Are the equations
$$
\frac{2(x-10)}{x^{2}-13 x+30}=1 \quad \text { and } \quad x^{2}-15 x+50=0
$$
equivalent? | Solution. We will solve the first equation. By getting rid of the denominator, i.e., multiplying both sides of the original equation by the expression $x^{2}-13 x+30$, we obtain the equation
$$
2 x-20=x^{2}-13 x+30
$$
The set of all roots of this equation consists of two numbers: $x_{1}=10$ and $x_{2}=5$. As a result... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 19. Solve the equation
$$
\sqrt{x+11}=x-1
$$ | Solution.
The first method. The domain of definition (DOD) of equation (7) is given by the condition $x \geqslant-11$. Considering that
$$
f(x)=g(x) \Rightarrow f^{2}(x)=g^{2}(x)
$$
we have
$$
\text { (7) } \begin{aligned}
\Rightarrow x+11=x^{2}-2 x+1 \Leftrightarrow x^{2}-3 x-10=0 \Leftrightarrow \\
\Leftrightarro... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Solve the equation
$$
\left|\frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}\right|=1
$$ | Solution. The given equation is equivalent to the system of equations
$\left[\begin{array}{l}\frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}=1, \\ \frac{x^{2}-6 \sqrt{x}+7}{x^{2}+6 \sqrt{x}+7}=-1\end{array} \Leftrightarrow\right.$
$$
\Leftrightarrow\left[\begin{array}{l}
\frac{-12 \sqrt{x}}{x^{2}+6 \sqrt{x}+7}=0 \\
\fr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7. Solve the equation
$$
|x-| 4-x||-2 x=4
$$ | Solution. The given equation is equivalent to the combination of two systems:
$$
\left\{\begin{array} { l }
{ 4 - x \geqslant 0 } \\
{ | x - ( 4 - x ) | - 2 x = 4 , }
\end{array} \quad \left\{\begin{array}{l}
4-x4 \\
-2 x=0
\end{array}\right.\right.
$$
The second system of the combination (4) has no solutions.
The ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 4. Solve the equation
$$
3 \sqrt{x+3}-\sqrt{x-2}=7
$$ | Solution. Isolate one of the roots on the left side:
$$
3 \sqrt{x+3}=\sqrt{x-2}+7
$$
By squaring both sides of the obtained equation, we have
$$
9(x+3)=x-2+14 \sqrt{x-2}+49
$$
Combining like terms and isolating the radical on the right side, we get the equation
$$
4 x-10=7 \sqrt{x-2}
$$
By squaring both sides of ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5. Solve the equation
$$
\sqrt{11 x+3}-\sqrt{2-x}-\sqrt{9 x+7}+\sqrt{x-2}=0
$$ | Solution. Isolate two radicals on each side so that after squaring we get the simplest equation:
$$
\sqrt{11 x+3}-\sqrt{2-x}=\sqrt{9 x+7}-\sqrt{x-2}
$$
Perform a chain of transformations:
$$
\begin{gathered}
11 x+3-2 \sqrt{(11 x+3)(2-x)}+2-x= \\
=9 x+7-2 \sqrt{(9 x+7)(x-2)}+x-2 \\
\sqrt{6+19 x-11 x^{2}}=\sqrt{9 x^{2... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6. Solve the equation
$$
\sqrt[3]{2 x-1}+\sqrt[3]{x-1}=1
$$ | Solution. Raising both sides of the equation to the third power, we get
$$
3 x-2+3 \sqrt[3]{(2 x-1)(x-1)}(\sqrt[3]{2 x-1}+\sqrt[3]{x-1})=1
$$
By the condition, the expression $\sqrt[3]{2 x-1}+\sqrt[3]{x-1}$ equals one. Substituting this expression with one in the obtained equation, we get the equation
$$
3 x-2+3 \sq... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 10. Solve the equation
$$
\sqrt{x+16}-x+4=0
$$ | Solution. The domain of the equation: $x \geqslant-16$. Isolating the radical, we get the equation $\sqrt{x+16}=x-4$. Squaring both sides of this equation and combining like terms, we obtain the equation $x^{2}-9 x=0$, the roots of which are: $x_{1}=0$ and $x_{2}=9$. Each of these roots belongs to the domain of the ori... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 11. Solve the equation
$$
\sqrt{x^{2}+3 x-4}=\sqrt{2 x+2}
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{l}
x^{2}+3 x-4 \geqslant 0 \\
2 x+2 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x \geqslant 1$.
By squaring both sides of the original equation, we obtain the equation
$$
x^{2}+3 x-4=2 x+2
$$
which is a consequence of i... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 12. Solve the equation
$$
\sqrt{2 x+5}+\sqrt{x-1}=8
$$ | Solution. Domain of the equation: $x \geqslant 1$. Squaring both sides of the equation and performing transformations, we obtain the equation
$$
2 \sqrt{2 x^{2}+3 x-5}=60-3 x
$$
After squaring both sides of this equation, we get the equation
$$
4\left(2 x^{2}+3 x-5\right)=(60-3 x)^{2}
$$
which is a consequence of t... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 15. Solve the equation
$$
\sqrt{x+7} \sqrt{3 x-2}=3 \sqrt{x-1} \sqrt{x+2}
$$ | Solution. Solving the system of inequalities
$$
\left\{\begin{array}{r}
x+7 \geqslant 0 \\
3 x-2 \geqslant 0 \\
x-1 \geqslant 0 \\
x+2 \geqslant 0
\end{array}\right.
$$
we find the domain of the equation: $x \geqslant 1$. Squaring both sides of the equation and combining like terms, we obtain the equation
$$
3 x^{2}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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