problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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25. In the wallet, there are 71 kopecks in coins of 5, 2, and 1 kopeck. The number of 1 and 2 kopeck coins is the same. How many coins of each denomination are in the wallet if there are 31 of them in total? | 25. 1 kopeck - 12 coins, 2 kopecks - 12 coins, and 5 kopecks - 7 coins. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,730 |
1. Prove that for any natural $n$
a) $n^{4}+2 n^{3}+11 n^{2}+10 n \quad$ is divisible by 24,
b) $n^{7}-n$ is divisible by 42,
c) $7^{n}+3 n-1$ is divisible by 9,
d) $7+7^{2}+7^{3}+\ldots+7^{4 n}$ is divisible by 100,
e) $4^{2 n+2}-15 n-16$ is divisible by 225,
f) $n^{3}+11 n$
is divisible by 6,
g) $n^{5}-5 n^{3}... | 1. We apply the method of mathematical induction to prove the statement in part a) For $n=1$, the division by 24 is valid. Assume that $k^{4}+2 k^{3}+11 k^{2}+10 k$ is divisible by $24, k>1$. Consider the expression:
$$
\begin{gathered}
(k+1)^{4}+2(k+1)^{3}+11(k+1)^{2}+10(k+1)= \\
=\left(k^{4}+2 k^{3}+11 k^{2}+10 k\ri... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,732 |
2. Prove that $10^{n}+1$ is divisible by 11 for any odd natural value of $n$. | 2. First method. Let's write the given sum in the following form:
$$
10^{n}+1=(11-1)^{n}+1=11^{n}-n 11^{n-1}+C_{n}^{2} 11^{n-2}+\ldots+n 11-1+1
$$
The first $n$ terms of this sum contain the factor 11, therefore, the sum of these terms is divisible by 11.
Second method. We will use the method of mathematical inducti... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,733 |
4. Prove that if in a three-digit number the middle digit is equal to the sum of the extreme digits, then the number is divisible by 11. | 4. Let $100 x+10 y+z$ be the given three-digit number. By the condition $y=x+z$, i.e., $100 x+10 y+z=11 \cdot 10 x+11 z$, and the latter is divisible by 11 without a remainder. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,735 |
5. Prove that the sum of the cubes of three consecutive integers is divisible by 9. | 5. Let $x$ be the mean of the numbers $(x+1)^{3}+x^{3}+(x-1)^{3}=3 x\left(x^{2}+2\right)$. If $x$ is divisible by 3, then $3 x\left(x^{2}+2\right)$ will be divisible by 9. If $x$ is not divisible by 3, then either 1) $x=3 n+1$, or 2) $x=3 n+2$.
Then 1) $x^{2}+2=(3 n+1)^{2}+2=9 n^{2}+6 n+3$
2) $x^{2}+2=(3 n+2)^{2}+2=9 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,736 |
6. Prove that $27195^{8}-10887^{8}+10152^{8}$ is divisible by 26460. | 6. Note that $26460=2^{2} \cdot 3^{3} \cdot 5 \cdot 7^{2}$. The difference $27195^{8} - (10887^{8} - 10152^{8})$ is divisible by $3, 5, 7^{2}$, since $27195=3 \cdot 5 \cdot 7^{2} \cdot 37$, and $10887 - 10152 = 735 = 3 \cdot 5 \cdot 7^{2}$. On the other hand, the sum $(27195^{8} - 10887^{8}) + 10152^{8}$ is divisible b... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,737 |
7. The number $3^{105}+4^{105}$ is divisible by $13,49,181,379$, but not divisible by 5 and 11. How to verify this | 7. Instruction. $3^{105}+4^{105}=\left(3^{3}\right)^{35}+\left(4^{3}\right)^{35}=\left(3^{5}\right)^{21}+\left(4^{5}\right)^{21}+\left(3^{7}\right)^{15}+\left(4^{7}\right)^{15}$. When $3^{105}$ is divided by 5, the remainder is 3, and when $4^{105}$ is divided by 5, the remainder is 4. The sum $3^{105}+4^{105}$ is not ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,738 |
8. Prove that the difference between the cube of an odd number and the number itself is divisible by 24. | 8. If $2 n+1$ is an odd number, then the condition of the problem can be written as:
$$
(2 n+1)^{3}-(2 n+1)=4 n(n+1)(2 n+1)
$$
Prove that $n(n+1)(2 n+1)$ is divisible by 6 (use the method of mathematical induction). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,739 |
9. Prove that $n^{2}-8$ is not divisible by 5 for any integer value of $n$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 9. For $n^{2}-8$ to be divisible by $5, n^{2}$ must end in 8 or 3. But $n^{2}$ can only end in $1,4,9,6,5,0$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,740 |
10. Prove that for any natural number $n$ the number $\frac{10^{n}+8}{9}$ is an integer. | 10. $10^{n}+8=(9+1)^{n}+8=9^{n}+n 9^{n-2}+\ldots+9+1+8=$ $=9^{n}+n 9^{n-1}+\ldots+9+9$.
$\frac{10^{n}+8}{9}=9^{n-1}+n 9^{n-2}+\ldots+1+1$ is an integer. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,741 |
11. Prove that the product of $k$ consecutive natural numbers is divisible by $k!$ (where $k!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot k$). | 11. Consider the numbers $n+1, n+2, n+3, . . ., n+k$.
$\frac{(n+1)(n+2)(n+3) \ldots(n+k)}{1 \cdot 2 \cdot 3 \ldots k}=C_{n+k}^{k}$. This is always a natural number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,742 |
12. Find the remainder of the division of $67^{283}$ by 13. | 12. $65=13 \cdot 5$, then $67^{283}=(65+2)^{283}=65^{283}+283 \cdot 65^{282} \cdot 2+\ldots+2^{283}$. Each term, except the last one, contains a factor of 65, and therefore, is divisible by 13.
$$
2^{283}=16^{70} \cdot 2^{3}=(13+3)^{70} \cdot 2^{3}=\left(13^{70}+70 \cdot 13^{69} \cdot 3+\ldots+3^{70}\right) \cdot 2^{3... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,743 |
13. In a six-digit number, the first digit coincides with the fourth, the second with the fifth, and the third with the sixth. Prove that this number is divisible by $7, 13, 11$. | 13. Let the digits of the desired number be denoted by $a_{1}, a_{2}, a_{3}$. Then the number can be
$100100 a_{1} + 1010 a_{2} + 1001 a_{3} = 1001(100 a_{1} + 10 a_{2} + a_{3}) = 7 \cdot 1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,744 |
14. What digit does the number $777^{777}$ end with? | 14. Let's determine the last digits of the powers of 777. \(777^1\) ends in 7. \(777^2\) ends in 9; \(777^3\) ends in 3; \(777^4\) ends in 1; \(777^5\) ends in 7; \(777^6\) ends in 9; \(777^7\) ends in 3; \(777^8\) ends in 1, and so on. We establish that 777776 ends in 1, therefore, \(777^{777}\) ends in 7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,745 |
15. Prove that for any perfect square, the product of the last two digits is even. | 15. The last digits of any square can be $0,1,4,5,6,9$. Exclude all even numbers, as the statement of the problem is true for them. Consider squares that end in 1, 9, and 5. They have one of the following forms (for $x$ natural):
1) $(10 x \pm 1)^{2}=100 x^{2} \pm 10 \cdot 2 x+1, \quad 2 x \cdot 1$ is even;
2) $(10 x \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,746 |
16. If from a three-digit number, the digits of which are consecutive decreasing numbers, you subtract the number reversed, the difference is always 198. Prove it. | 16. Instruction. Denote the given number as $100(x+2)+10(x+1)+x$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,747 |
17. Can the sum of several consecutive natural numbers be a natural power of the number 2? | 17. Consider the sum of $n$ consecutive natural numbers from $k$ to $k+n-1$.
$k+(k+1)+(k+2)+\ldots+(k+n-1)=\frac{n(n+2 k-1)}{2}$. Let $\frac{n(n+2 k-1)}{2}=2^{x}$,
where $x$ is a natural number. $n(n+2 k-1)=2 \cdot 2^{x}$.
For all natural values of $k$, the factors on the left side of the equation differ from each o... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,748 |
18. How many digits does the number $2^{100}$ have? | 18. The number of digits in the representation of a number is always one more than the characteristic of the decimal logarithm of the number. In our case
$$
\lg 2^{100}=100 \lg 2 \approx 100 \cdot 0.3010=30.10
$$
Therefore, the number $2^{100}$ contains 31 digits. | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,749 |
19. Find the last two digits of the number $99^{99}-51^{51}$. | 19. Let's determine the last digits of the powers $99^{99}$ and $5151.99^{1}-99$, $99^{2}-01,99^{3}-99,99^{4}-01$ and so on. $99^{99} - 99$, i.e., $99^{99}=100 M+99$.
Similarly, we establish that $51^{51}=100 N+51$, then $99^{99}-5151=$ $=100(M-N)+99-51=100(M-N)+48$.
Answer: the last two digits are 4 and 8. | 48 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,750 |
20. Given a sequence $U_{n}$ of integers, for which $U_{1}=1, U_{k}=U_{k-1}+k$. Prove that $U_{n}+U_{n+1}=(n+1)^{2}$. | 20. If $U_{1}=1, U_{k}=U_{k-1}+k$, then $U_{2}=U_{1}+2=3, U_{3}=3+3=6$, and so on.
$\boldsymbol{U}_{\boldsymbol{n}}=\frac{\boldsymbol{n}(\boldsymbol{n}+1)}{2}$. The validity of this formula can be verified by mathematical induction. Then $U_{n}=\frac{n(n+1)}{2} ; U_{n+1}=\frac{(n+1)(n+2)}{2}$; $U_{n}+U_{n+1}=(n+1)^{2}... | U_{n}+U_{n+1}=(n+1)^{2} | Algebra | proof | Yes | Yes | olympiads | false | 34,751 |
21. Prove that the product of four consecutive integers, increased by one, is a perfect square. | 21. $n(n+1)(n+2)(n+3)+1=\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 n\right)+1$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,752 |
22. Prove that the sum of the squares of five consecutive integers is not a perfect square. | 22. Consider the consecutive numbers $n-2, n-1, n, n+1, n+2$. $(n-2)^{2}+(n-1)^{2}+n^{2}+(n+1)^{2}+(n+2)^{2}=5 n^{2}+10=5\left(n^{2}+2\right)$.
For the number $5\left(n^{2}+2\right)$ to be a perfect square, it is necessary that $n^{2}+2$ is divisible by 5, i.e., ends in 0 or 5. Then $n^{2}$ must end in 3 or 8, which i... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,753 |
25. For which natural values of $p$ will the numbers $p+10, p+14$ be prime simultaneously
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 note at the end is not part of the translation but is provided to clarify that the format and structure of the original text have been maintained in the translation. | 25. $p(p+10)(p+14)=p(p+1)(p+2)+9 p(p+2)+12 p(p+1)+9 \cdot 12 p$.
Each term is divisible by 3, so the entire sum is divisible by 3, i.e., the product $p(p+10)(p+14)$ is divisible by 3. This is only possible if $p=3$ (no other prime number is divisible by 3). | Number Theory | proof | Yes | Yes | olympiads | false | 34,756 | |
26. Prove that if $p$ and $8 p-1$ are prime numbers, then the number $8 p+1$ is composite. | 26. The product of three consecutive integers $(8p-1) 8p(8p+1)$ is divisible by 3. This means that one of the numbers $p, 8p-1, 8p+1$ is divisible by 3. But $p$ and $8p-1$ are prime, so $p(8p-1)$ is divisible by 3 only when $p=3$. Then $8p+1=25$ is a composite number. In all other cases, the number $8p+1$ must be divis... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,757 |
27. Prove that if $p \geqslant 5$ is a prime number, then $p^{2}-1$ is divisible by 24. | 27. $p^{2}-1=(p-1)(p+1)$. Since $p$ is a prime number, one of the factors is divisible by 2, and the other by 4, i.e., the entire product is divisible by 8. The product $(p-1) p(p+1)$ is divisible by 3. Since $p$ is prime, either $p-1$ or $p+1$ is divisible by 3. $p^{2}-1$ is divisible by 24.
Second method. We will pr... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,758 |
28. Let $p_{1}, p_{2}, p_{3}, \ldots, p_{n}$ be distinct prime numbers. How many divisors does the number $q=p_{1} p_{2} p_{3} \ldots p_{n}$ have, including 1 and $q$? | 28. The divisors of the number $q$ are, obviously, the numbers $p_{1}, p_{2}, p_{3}, \ldots, p_{n}$ and all possible products of them taken two at a time, three at a time, and so on. The number of such divisors is $C_{n}^{0}+C_{n}^{1}+C_{n}^{2}+\ldots+C_{n}^{n}=2^{n}$. (This is the sum of the binomial coefficients). | 2^{n} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,759 |
29. Prove that the square of a prime number, except for 2 and 3, when divided by 12, leaves a remainder of 1. | 29. A prime number greater than 3 can be written in the form $6 k+1$ or $6 k+5$, where $k=0,1,2,3, \ldots$ When the square of such a number is divided by 12, the remainder is 1. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,760 |
31. Prove that for any natural number $n$ there exists a number $N$ such that all numbers $N+1, N+2, N+3, \ldots, N+n$ are composite (not prime). | 31. Let $n$ be any natural number, $N=1 \cdot 2 \cdot 3 \ldots n(n+1)+1$.
Then $N+1=1 \cdot 2 \cdot 3 \ldots n(n+1)+2$, is divisible by 2;
$N+2=1 \cdot 2 \cdot 3 \ldots n(n+1)+3$, is divisible by 3.
$\dot{N}+\dot{n}=\dot{1} \cdot \dot{2} \cdot \dot{3} \ldots \dot{n}(\dot{n}+\dot{1}) \dot{+}(\dot{n}+\dot{1})$, is div... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,762 |
33. Prove that the smallest number that is coprime with each of the numbers $1,2,3, \ldots, n$ is a prime number. | 33. Let $N$ be the smallest number that is coprime with $1, 2, 3, \ldots, n$. If $N$ is composite, then $N = qs$, where $q$ is a prime number, $q < N$ and is coprime with $1, 2, 3, \ldots, n$. In addition to $N$, there is another number $q < N$ that is coprime with $1, 2, 3, \ldots, n$. This contradicts the condition. ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,764 |
34. Find the product of all divisors of the number $N$, if there are $n$ of them. | 34. Let the divisors of the number $N, a_{1}, a_{2}, a_{3}, \ldots, a_{n}$ form an increasing sequence. Then $a_{1}=1, a_{n}=N$. If $n$ is even, then $a_{i} a_{\gamma}=N$ (there will be pairs equidistant from the beginning and end), then $N \cdot N \cdot N \ldots N=\frac{n}{N^{2}}$ (a total of $\frac{n}{2}$ pairs). If ... | N^{\frac{n}{2}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,765 |
35. Prove that $(x-y)^{5}+(y-z)^{5}+(z-x)^{5}$ is divisible by $5(x-y)(y-z)(z-x)$, where $x, y, z$ are integers, pairwise distinct. | 35. The sum of odd powers is always divisible by the sum of the bases, which means that $(x-y)^{5}+(y-z)^{5}$ is divisible by $(z-x)$, i.e., $(x-y)^{5}+(y-z)^{5}+(z-x)^{5}$ is divisible by $(z-x)$. Similarly, we establish that $(x-y)^{5}+(y-z)^{5}+(z-x)^{5}$ is divisible by $(y-z)$ and $(x-y)$.
After expanding the bra... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,766 |
36. Prove that the polynomial $1+x+x^{2}+\ldots+x^{999}$ is divisible by $1+x+x^{2}+\ldots+x^{249}$. | 36. $1+x+x^{2}+\ldots+x^{999}=\frac{x^{1000}-1}{x-1} ; \quad 1+x+x^{2}+\ldots+x^{249}=\frac{x^{250}-1}{x-1}$. Their quotient $\frac{x^{1000}-1}{x^{250}-1}=x^{750}+x^{500}+x^{250}+1$. We obtained the polynomial. | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,767 |
37. Prove that there do not exist integers $a, b, c, d$ such that the expression $a x^{3}+b x^{2}+c x+d$ equals 1 when $x=19$ and equals 2 when $x=62$. | 37. Let $a x^{3}+b x^{2}+c x+d=1$ when $x=19$ and equals 2 when $x=62$. Substituting the values $x=19$ and $x=62$, we find the difference
$$
\begin{aligned}
& a \cdot 19^{3}+b \cdot 19^{2}+c \cdot 19+d=1 \\
& a \cdot 62^{3}+b \cdot 62^{2}+c \cdot 62+d=2 \\
& \hline a\left(62^{3}-19^{3}\right)+b\left(62^{2}-19^{2}\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,768 |
38. To find the product of three consecutive integers, it is sufficient to reduce the cube of the middle number by this number. Prove it. | 38. $n(n+1)(n+2)=n^{3}+n^{2}+2 n^{2}+2 n=(n+1)^{3}-(n+1)$.
Translate the above text into English, keeping the original text's line breaks and format:
38. $n(n+1)(n+2)=n^{3}+n^{2}+2 n^{2}+2 n=(n+1)^{3}-(n+1)$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,769 |
39. Find a two-digit number that has the following properties:
a) if the desired number is multiplied by 2 and 1 is added to the product, the result is a perfect square;
b) if the desired number is multiplied by 3 and 1 is added to the product, the result is a perfect square. | 39. The condition of the problem can be written as: $2 a+1=N^{2}, 3 a+1=M^{2}$. The squares of numbers can end with the digits $0,1,4,5,6,9$. Then $2 a$ can end with $0,4,8$ (since $2 a$ is an even number) and $a$ can end with 5,2, 4, and 0. $10 \leqslant a \leqslant 99$, i.e., $21<2 a+1<199,31<3 a+1<298$ or $21<N^{2}<... | 40 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,770 |
40. Find a five-digit number that has the following property: when multiplied by 9, the result is a number composed of the same digits but in reverse order, i.e., the number is reversed. | 40. Multiply the desired number by 10 and subtract the desired number from the product, which is equivalent to multiplying by 9, then $\overline{a b c d e o}-\overline{a b c d e}=\overline{e d c b a}$. It is easy to establish that $a=1$, then $e=9, b=0, d=8, c=9$. The desired number is 10989. | 10989 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,771 |
41. Find two numbers according to the following conditions:
a) their sum is 1244;
b) these numbers will become equal to each other if the digit 3 is appended to the end of the first number, and the digit 2 is removed from the end of the second number. | 41. The first number is $x$, the second number is $1244-x$, then according to the problem, we can form the equation $10 x+3=(1244-x-2): 10$. From this, $x=12$. The first number is 12, the second is 1232. | 12,1232 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,772 |
43. Find a four-digit number that is a perfect square, where the digit in the thousands place is the same as the digit in the tens place, and the digit in the hundreds place is 1 more than the digit in the units place. | 43. The desired number is $1000a + 100(b+1) + 10a + b = 1010a + 101b + 100$. $1010a + 101b + 100 = x^2$ or $101(10a + b) = (x-10)(x+10)$.
The left side has a factor of 101, so the right side must also have such a factor. Since $x$ is a two-digit number, $x-10 < 101$. Therefore, $x+10 = 101$, from which $x = 91$.
The ... | 8281 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,774 |
44. Find a two-digit number that is equal to three times the product of its digits. | 44. $x$ is the number of tens, $y$ is the number of units of the desired number, then $10 x+y=3 x y$, $x=\frac{y}{3 y-10}$. Since $x>0, y>0$, then $3 y-10>0, y>\frac{10}{3}$. But $x$ is an integer, therefore, $y \geqslant 3 y-10$, from which $y \leqslant 5$. Thus, $\frac{10}{3}<y \leqslant 5$. Therefore, $y=4, x=2$ or ... | 2415 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,775 |
45. Given an arithmetic progression consisting of four integers. It is known that the largest of these numbers is equal to the sum of the squares of the other three numbers. Find these numbers. | 45. Let $x$ be the largest term of the progression, then the second is $x-d$, the third is $x-2d$, and the fourth is $x-3d$. The condition of the problem can be written as:
$$
x=(x-d)^{2}+(x-2d)^{2}+(x-3d)^{2}
$$
From this, $d=\frac{6x \pm \sqrt{2x(7-3x)}}{14}, x>0, 7-3x \geqslant 0, x \leqslant 2 \frac{1}{3}$.
Thus... | 2,1,0,-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,776 |
46. Prove that the numbers $\sqrt{2}, \sqrt{3}$, and $\sqrt{5}$ cannot be members of the same arithmetic progression. | 46. Let $\sqrt{2}, \sqrt{3}, \sqrt{5}$ be terms of an arithmetic progression, then $\sqrt{3}-\sqrt{2}=m d, \sqrt{5}-\sqrt{2}=n d$. Expressing $d$ from the first and second equations, we get $5 m^{2}+3 n^{2}-2 m n \sqrt{15}=2(m-n)^{2}$. This equality is impossible, since $\sqrt{15}$ is an irrational number, while the ri... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,777 |
47. Can the numbers $2, \sqrt{6}$, and 4.5 be members of the same arithmetic or the same geometric progression? | 47. Let $2, \sqrt{6}$ and 4.5 be terms of one arithmetic progression, then 72
$\sqrt{6}=2+m d, 4.5=2+n d$. In the first case $d=\frac{\sqrt{6}-2}{m}$ is an irrational number. In the second case $d=\frac{5}{2 n}$ is a rational number. The numbers $2, \sqrt{6}$ and 4.5 are not terms of one arithmetic progression.
Let $2... | \sqrt{\frac{3}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,778 |
48. Prove that the number $\underbrace{3999 \ldots 9}_{n \text { digits }} \underbrace{000 \ldots 01}_{n \text { digits }}$ is a perfect square. | 48. $3 \underbrace{999 \ldots 9}_{n} \underbrace{600 \ldots .01}_{n}+3 \underbrace{3999 \ldots 9}_{n+1}-\frac{3999 \ldots 9}{n+1}=$
$$
=4 \cdot 10^{2 n+2}-4 \cdot 10^{n+1}+1=\left(2 \cdot 10^{n+1}-1\right)^{2}
$$ | (2\cdot10^{n+1}-1)^{2} | Number Theory | proof | Yes | Yes | olympiads | false | 34,779 |
49. Prove that the numbers $49, 4489, 444889, \ldots$, obtained by inserting 48 in the middle of the previous number, are perfect squares. | 49. $N=\left(\frac{2 \cdot 10^{n}+1}{3}\right)^{2}, \quad n$-th number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,780 |
50. The sequence of numbers $1,4,10,19, \ldots$ has the property that the differences between two consecutive numbers form an arithmetic progression. Find the $n$-th term and the sum of the first $n$ terms of this sequence. | 50. Let the terms of the arithmetic progression be $b_{1}, b_{2}, b_{3}, \ldots, b_{n}$, and the terms of the sequence be $a_{1}, a_{2}, a_{3}, \ldots$. According to the condition, $b_{1}=3, b_{2}=6$. Therefore, $d=b=3$. Then
$a_{2}-a_{1}=b$
$a_{3}-a_{2}=2 b$
$a_{4}-a_{3}=3 b$
$\dot{a}_{n}-\dot{a}_{n-1}=(n-1) b$
B... | a_{n}=1+\frac{3}{2}n(n-1),S_{n}=\frac{1}{2}n(n^{2}+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,781 |
51. The numbers $a, 1, b$ form an arithmetic progression, and the numbers $a^{2}, 1$ and $b^{2}$ form a geometric progression. Find the numbers $a$ and $b$. | 51. Given $\div a, 1, b$ and $\because a^{2}, 1, b^{2}$. Then $1-a=b-1, \frac{1}{a^{2}}=\frac{b^{2}}{1}$;
$$
\left\{\begin{array}{l}
a+b=2 \\
a b= \pm 1 .
\end{array} \quad a_{1}=b_{1}=1, \quad a_{2}=1 \pm \sqrt{2}, \quad b_{2}=1 \mp \sqrt{2}\right.
$$ | a_{1}=b_{1}=1,\quada_{2}=1\\sqrt{2},\quadb_{2}=1\\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,782 |
52. The numbers $a, b$, and $c$ form an arithmetic progression. The reciprocals of these numbers also form an arithmetic progression. Prove that $a=b=c$. | 52. Given $\div a, b, c$ and $\div \frac{1}{a}, \frac{1}{b}, \frac{1}{c}$.
Then $\quad b-a=c-b$ and $\frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b} ; \quad\left\{\begin{array}{l}a+c=2 b \\ 2 a b c=a b^{2}+b^{2} c\end{array}\right.$.
Solving the system of equations, we get $a=b=c$. | =b= | Algebra | proof | Yes | Yes | olympiads | false | 34,783 |
53. A student was given a problem to solve, which required dividing the first of the given numbers by the second and multiplying the resulting quotient by a number, $\frac{2}{3}$ of which equals 4. However, the student proceeded as follows: from the first given number, he subtracted the second and added $\frac{2}{3}$ o... | 53. 15 and 9. | 159 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,784 |
54. A student was asked to multiply two numbers, the product of which is 4500. While copying the numbers from the board, he wrote a 3 instead of the 5 at the end of one of the factors, and after multiplying, he obtained 4380. Find the given numbers. | 54. Let the required numbers be $x$ and $y$. Then $x y=4500, x(y-2)=4380$ (by replacing the digit 5 at the end of the multiplier with the digit 3, the student reduced the multiplier by 2 units). Answer: 60 and 75. | 6075 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,785 |
55. Find four such numbers that the sums of the numbers, taken three at a time, are $22, 24, 27, 20$. (This problem was given by the Greek mathematician Diophantus, who lived in the second century AD). | 55. Let the required numbers be $x, y, z, t$. Then
$$
\begin{cases}x+y+z=22 \\ x+y+t=24 & \text { Hence } x+y+z+t=31 \\ x+z+t=27 \\ y+z+t=20 & \text { Answer: } x=11, \quad y=4, \quad z=7, \quad t=9\end{cases}
$$ | 11,\quad4,\quad7,\quad=9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,786 |
56. If the thought number is multiplied by 3, 2 is appended to the right, the resulting number is divided by 19, and 7 is added to the quotient, the result is three times the thought number. What is this number? | 56. Let $x$ be the thought-of number. Then the condition of the problem can be written as: $(3 x \cdot 10 + 2) : 9 + 7 = 3 x$.
From which $x = 5$.
## $\S 3$ | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,787 |
1. The numbers $12, 14, 37$ and $65$ represent one of the solutions to the equation $x y - x z + y t = 182$. Determine which value each of these numbers represents for the variables. | 1. Let's represent the given equation in the form $x(y-z)+y t=2 \cdot 7 \cdot 13$.
If $x=14$, then the product $y t$, formed in any way from the numbers 12, 37, and 65, does not divide by 7, and therefore equation (1) will not be valid. Similarly, we can verify that $x \neq 65$. If we rewrite the given equation as $y(... | 12,37,65,=14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,788 |
2. The square root of a two-digit number is expressed as an infinite decimal fraction, the first four digits of which (including the integer part) are the same. Find this number without using tables. | 2. Let the desired number be $N_{\bullet}$ Then $10 \leqslant N \leqslant 100$. Therefore, $3, \ldots \leqslant$ $10,89$, i.e., $N=11 ; 12 ; \ldots$
But $\sqrt{11}=3.31662$. Therefore, there does not exist a two-digit number, the first four digits of the square root of which are written only with threes. By performing... | 79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,789 |
3. Find a six-digit number that, when multiplied by 2, 3, 4, 5, and 6, gives other six-digit numbers composed of the same digits but in a different order. | 3. Let the desired number be denoted as: \( N = \overline{a_{6} a_{5} a_{4} a_{3} a_{2} a_{1}} \), with \( a_{6} \neq 0 \), because otherwise \( N \) would not be a six-digit number. \( a_{6} \neq 2 \), because \( 5N \) would then be a seven-digit number. Therefore, \( a_{6} = 1 \).
It is clear that the first digits o... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,790 |
4. Find the smallest four-digit number that is equal to the square of the sum of the numbers formed by its first two digits and its last two digits. | 4. Let the desired number be $N=100 a+b$, where $a$ is the number written by the first two digits of the desired number, and $b$ is the last two digits. According to the problem, $100 a+b=(a+b)^{2}$. From this equation, we express $a+b$ in terms of $a$. By trial and error, we establish that $a+b$ will be an integer whe... | 2025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,791 |
5. Find all natural numbers ending in ...196, which, after erasing these digits, decrease by an integer factor.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. The desired number $N=1000 x+196$. Erasing the digits 196 in the number $N$ reduces it by 196 and also by 1000 times, so after erasing we get $x$. According to the condition, $1000 x+196=k x$, where $k$ is an integer. From this, $x=\frac{196}{k-1000}$. $k-1000=196 ; 98 ; 49 ; 28 ; 14 ; 4 ; 2 . \quad x=1,2,4,7,14,49,... | 1196,2196,4196,7196,14196,49196,98196 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,792 |
6. Find a two-digit number if it is known that the sum of its digits is 13, and the difference between the sought number and the number written with the same digits but in reverse order is a two-digit number with 7 units. | 6. The desired number $N=10 x+y$, where $x=1,2,3,4,5,6,7,8,9 ; y=0,1,2$, $3,4,5,6,7,8,9$. Let's write down the two-digit numbers whose sum of digits is 13. Such numbers will be $49,58,67,76,85,94$. Let's write down the reversed numbers: $94,85,76,67,58,49$. The condition of the problem is satisfied only by the pair of ... | 85 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,793 |
7. Represent the number 531441 as the sum of several addends such that the first one is 1, the second is $k$ times larger, the third is $k$ times larger than the sum of the first two, the fourth is $k$ times larger than the sum of all previous ones, and so on. Find the multiplier $k$ and the number of addends, given th... | 7. The condition of the problem can be written as:
$$
531441=3^{12}=1+k+(1+k) k=(1+k)^{2} k+\ldots+(1+k)^{n-3} k+(1+k)^{n-2} k
$$
Since the last term is $(1+k)^{n-2} k$, the sum of all previous terms is $(1+k)^{n-2}$. Therefore, $3^{12}=(1+k)^{n-2}+(1+k)^{n-2} k=(1+k)^{n-1}$. From this, $1+k=3^{\frac{12}{n-1}}$
The ... | k=80;26;8;20n=4;5;7;13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,794 |
8. The sum of several consecutive natural numbers,
starting from one, is a two-digit number, both digits of which are the same. How many numbers were taken? | 8. According to the condition of the problem, we form the equation $1+2+3+\ldots+n=10 a+a$, from which $\frac{n(n+1)}{2}=11 a, n^{2}+n=22 a, n=\frac{-1+\sqrt{1+88 a}}{2}$.
$1 \leqslant a \leqslant 9, a$-integer. By trial, we establish that the problem is possible for $a=5, n=10$ and $a=6, n=11$. | 1011 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,795 |
9. Find such four-digit numbers, each of which is divisible by 11, and the sum of the digits of each of them is 11. | 9. Let the desired number be denoted as: $1000 x+100 y+10 z+u$, then $x+y+z+u=11$. According to the divisibility rule for 11 (if the sum of the digits of a number at odd positions equals the sum of the digits at even positions or the difference between these sums is divisible by 11, then the number is divisible by 11),... | 2090,3080,4070,5060,6050,7040,8030,9020 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,796 |
10. The dividing circle of the dividing head is uniformly marked with 24 holes. What regular polygons can be marked using this instrument? | 10. By connecting all 24 holes sequentially, we get a twenty-four-sided polygon, connecting every other hole - a twelve-sided polygon,
connecting every two holes - an eight-sided polygon, ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,797 |
11. In chemical laboratories, floors are covered with tiles made of acid-resistant ceramic, which have the shape of regular polygons. Which polygons can fill the floor plane without gaps | 11. The internal angle of a regular polygon is equal to $\frac{2 d(n-2)}{n}, n$-the number of sides. Let $x$ polygons meet at one vertex, then $\frac{2 d(n-2)}{n} x=4 d$, from which $x=\frac{2 n}{n-2}=2+\frac{4}{n-2}$. This number must be an integer, i.e., $\frac{4}{n-2}$ must be an integer.
A polygon can have no fewe... | n_{1}=3;n_{2}=4;n_{3}=6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,798 |
12. Determine the sides of a triangle if they are expressed as integers forming an arithmetic progression, and if the perimeter of the triangle is 15. | 12. Let $a$ be the side of medium length in a triangle, the other two sides being $a-d$ and $a+d$. According to the problem, $3a=15, a=5$. The other two sides together amount to 10. We can construct a table of possible values for the sides.
The largest side of the triangle must be less than the sum of the other two si... | 5,5,5;4,5,6;3,5,7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,799 |
13. In what ways can you give 78 rubles, having only five-ruble and three-ruble banknotes? | 13. Let $x$ be the number of 5-ruble stamps, and $y$ be the number of 3-ruble stamps, then $78 = 5x + 3y$, from which $x = 15 - \frac{3(y-1)}{5}$. Since $x$ is a positive integer, $15 - \frac{3(y-1)}{5} > 0$, and thus $y < 26$. Additionally, $\frac{3(y-1)}{5}$ is an integer, meaning $y-1$ is a multiple of 5. Let $y-1 =... | 78 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,800 |
14. The purchase costs 19 rubles. The buyer has only three-ruble bills, and the cashier has only five-ruble bills. Can the purchase be paid for with the available money, and if so, how exactly? (Limit the case to when the buyer and the cashier each have 15 bills of the specified denomination). | 14. 19 rubles can be paid in two ways:
1) the buyer gives 24 rubles and receives 5 rubles in change. 2) gives 39 rubles and receives 20 rubles in change. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 34,801 |
15. From the book by A. Conan Doyle (English writer) "The Adventures of Sherlock Holmes".
The doctor and his guest Sherlock Holmes were sitting by the open window. Laughter and cheerful voices of a large group of children could be heard from the garden.
Guest. Tell me, please, how many children do you have?
Host. Th... | 15. The guest knew that all children were under 18, and he knew the house number $(N)$. Let the number of children in each family be $a, b, c, d$ respectively. Then
$$
a+b+c+db>c>d .
$$
The uncle should not have three children. Suppose $d=3$, then $c \geqslant 4, b \geqslant 5, a \geqslant 6$, so $a+b+c+d=6+5+4+3=18$... | N=120,\quad=5,\quadb=4,\quad=3,\quad=2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 34,802 |
1. Compute $\sin 18^{\circ}$ without tables. | 1. $\sin \left(18^{\circ} \cdot 2\right)=\cos \left(18^{\circ} \cdot 3\right)=$ $=4 \cos ^{3} 18^{\circ}-3 \cos 18^{\circ}$, but $\sin \left(18^{\circ} \cdot 2\right)=2 \sin 18^{\circ} \cos 18^{\circ}$. $2 \sin 18^{\circ} \cos 18^{\circ}=4 \cos ^{3} 18^{\circ}-$ $3 \cos 18^{\circ}$. From this, $\sin 18^{\circ}=$
$$
=\... | \frac{\sqrt{5}-1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,803 |
2. Prove that $\cos ^{2} 73^{\circ}+\cos ^{2} 47^{\circ}+\cos 73^{\circ} \cdot \cos 47^{\circ}=\frac{3}{4}$. | 2. We should perform the substitution $\cos ^{2} 73^{\circ}=\frac{1+\cos 146^{\circ}}{2}, \cos ^{2} 47^{\circ}=\frac{1+\cos 94^{\circ}}{2}$; $\cos 73^{\circ} \cos 47^{\circ}=\frac{1}{2}\left(\cos 120^{\circ}+\cos 26^{\circ}\right)$ | \frac{3}{4} | Algebra | proof | Yes | Yes | olympiads | false | 34,804 |
3. Prove that $\cos ^{2} \alpha+\cos ^{2}\left(\frac{2}{3} \pi+\alpha\right)+\cos ^{2}\left(\frac{2}{3} \pi-\alpha\right)=\frac{3}{2}$. | 3. Instruction. Use the formula $\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | proof | Yes | Yes | olympiads | false | 34,805 | |
4. Prove that $\operatorname{tg} \frac{\gamma+\alpha}{2} \cdot \operatorname{tg} \frac{\gamma-\alpha}{2}=\operatorname{tg}^{2} \frac{\beta}{2}$, if $\cos \gamma=\cos \alpha \cos \beta \cdot$ | 4. Expressing $\operatorname{tg} \frac{\gamma+\alpha}{2}$ and $\operatorname{tg} \frac{\gamma-\alpha}{2}$ in terms of sines and cosines of these angles and replacing the product with sums, we get
$$
\operatorname{tg} \frac{\gamma+\alpha}{2} \operatorname{tg} \frac{\gamma-\alpha}{2}=\frac{\cos \alpha-\cos \gamma}{\cos ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,806 |
5. Prove that $\alpha+2 \beta=45^{\circ}$, if $\alpha$ and $\beta$ are acute positive angles and $\operatorname{tg} \alpha=\frac{1}{7}, \sin \beta=\frac{1}{\sqrt{10}}$. | 5. Let's preliminarily calculate $\operatorname{tg} \beta$ and $\operatorname{tg} 2 \beta$, knowing that $\sin \beta=\frac{1}{\sqrt{10}}$. After this, it is easy to show that $\operatorname{tg}(\alpha+2 \beta)=1, \alpha+2 \beta=45^{\circ}$. | \alpha+2\beta=45 | Algebra | proof | Yes | Yes | olympiads | false | 34,807 |
8. Prove that $\sin \alpha+\sin \beta+\sin \gamma=4 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}$, if $\alpha+\beta+\gamma=180^{\circ}$. | 8. Replace $\sin \alpha+\sin \beta$ and $\sin \gamma=\sin (\alpha+\beta)$ with products. Factor the resulting sum.
Express the following in products and factor the resulting sum. | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,810 |
9. For what values of $\alpha$ is the equality
$$
\sqrt{\operatorname{tg}^{2} \alpha-\sin ^{2} \alpha}=\operatorname{tg} \alpha \cdot \sin \alpha ?
$$ | 9. Transforming the left side into the square of the right, we get $\sqrt{\operatorname{tg}^{2} \alpha-\sin ^{2} \alpha}=|\operatorname{tg} \alpha \sin \alpha|$. In this case, $|\operatorname{tg} \alpha \sin \alpha|=\operatorname{tg} \alpha \sin \alpha$. This equality is valid under the condition that $\operatorname{tg... | (4k-1)\frac{\pi}{2}<\alpha<(4k+1)\frac{\pi}{2};\quadk=0,\1,\2,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,811 |
10. For what values of $\alpha$ is the equality
$$
\sqrt{1+\sin 2 \alpha}=\sin \alpha+\cos \alpha ?
$$
Solve the equations from № 11 to 25. | 10. The equality is valid under the condition that $\cos \left(\frac{\pi}{4}-\alpha\right) \geqslant 0$,
i.e., $-\frac{\pi}{2}+2 \pi k<\frac{\pi}{4}-\alpha=\frac{\pi}{2}+2 \pi k$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,812 | |
11. $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
11. $\cos ^{2} x+\cos ^{2} 2 x+\cos ^{2} 3 x=1$. | 11. Using the formula $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$ and replacing the sum $\cos 2 x+\cos 4 x$ with a product, we get $4 \cos 3 x \cos 2 x \cos x=0$.
Answer: $x_{1}=\frac{\pi}{6}+\frac{\pi}{3} k, \quad x_{2}=\frac{\pi}{2}+\pi k, \quad x_{3}=\frac{\pi}{4}+\frac{\pi}{2} k, \quad k=0 ; \quad \pm 1, ... | x_{1}=\frac{\pi}{6}+\frac{\pi}{3}k,\quadx_{2}=\frac{\pi}{2}+\pik,\quadx_{3}=\frac{\pi}{4}+\frac{\pi}{2}k,\quadk=0;\quad\1,\2,\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,813 |
12. $\frac{a \sin x+b}{b \cos x+a}=\frac{a \cos x+b}{b \sin x+a}$,
where $a$ and $b$ are real numbers, different from zero. | 12. Given $(b \cos x + a)(b \sin x + a) \neq 0$. Multiplying both sides of the equation by $(b \cos x + a)(b \sin x + a)$ and grouping the terms, we get
$(\sin x - \cos x)\left[a b(\sin x + \cos x) + a^{2} + b^{2}\right] = 0. \sin x - \cos x = 0, \quad x_{1} = \frac{\pi}{4} + \pi k$. The equality $a b(\sin x + \cos x)... | x_{1}=\frac{\pi}{4}+\pik | Algebra | proof | Yes | Yes | olympiads | false | 34,814 |
13. $\sin 2 x+\cos 2 x+\sin x+\cos x+1=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
13. $\sin 2 x+\cos 2 x+\sin x+\cos x+1=0$. | 13. The equation is transformed into the form $(\sin x+\cos x)(2 \cos x+1)=0$.
From which $\quad x_{1}=\pi k-\frac{\pi}{4}, \quad x_{2}=2 \pi k \pm \frac{2}{3} \pi$. | x_{1}=\pik-\frac{\pi}{4},\quadx_{2}=2\pik\\frac{2}{3}\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,815 |
14. $2 \sin x-9 \cos x=7$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
14. $2 \sin x-9 \cos x=7$. | 14. Express $\sin x$ and $\cos x$ in terms of $\operatorname{tg} \frac{x}{2}$, then the equation will take the form $\operatorname{tg}^{2} \frac{x}{2}+2 \operatorname{tg} \frac{x}{2}-8=0$. From this, $x_{1}=2 \pi k+2 \operatorname{arctg}(-4), \quad x_{2}=2 \pi k+2 \operatorname{arctg} 2$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,816 | |
15. $\log _{\cos x} \sin x-2 \log _{\sin x} \cos x+1=0$. | 15. Transition to logarithms with base $\cos x, x \neq 0, x \neq \frac{\pi}{2}$ $\left(\log _{\cos x} x \sin x\right)^{2}+\log _{\cos x} \sin x-2=0, \quad \log _{\cos x} \sin x=1$.
From where $x_{1}=\frac{\pi}{4}+\pi k$. The roots of the equation $\log _{\cos } x \sin x=-2$ are extraneous. | x_{1}=\frac{\pi}{4}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,817 |
16. $\sin ^{2} 2^{\sqrt{-x}}=\frac{1}{2}$. | 16. Assuming $2^{\sqrt{-x}}=y$, we get $\sin ^{2} y=\frac{1}{2}, y=\frac{\pi}{4}+\frac{\pi}{2} k$, $2^{\sqrt{-x}}=\frac{\pi}{4}+\frac{\pi}{2} k$. Since $2^{\sqrt{-x}} \geqslant 1$, the possible values for $k$ are: $k=1 ; 2 ; 3 ; \ldots, \sqrt{-x}=\log _{2}\left(\frac{\pi}{4}+\frac{\pi}{2} k\right) . \quad$ Hence, $x=-\... | -\log_{2}^{2}(\frac{\pi}{4}+\frac{\pi}{2}k) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,818 |
17. $\sin ^{3} x+\cos ^{3} x+\sin \left(x+\frac{\pi}{4}\right)=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
17. $\sin ^{3} x+\cos ^{3} x+\sin \left(x+\frac{\pi}{4}\right)=0$. | 17. Factoring the sum of cubes and replacing the sum $\sin x+\cos x$ with a product, we can transform the equation into the following form:
$$
\cos \left(x-\frac{\pi}{4}\right)(\sqrt{2}-\sqrt{2} \sin x \cos x+1)=0
$$
From which $\cos \left(x-\frac{\pi}{4}\right)=0, \quad x=\frac{3}{4} \pi+\pi k$.
The second factor d... | \frac{3}{4}\pi+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,819 |
18. $\sin 3 x \sin ^{3} x+\cos 3 x \cos ^{3} x=\frac{1}{8}$. | 18. Moving on to the argument $x$. The equation will take the form $\cos ^{2} x=\frac{3}{4}$. From which $x= \pm \frac{\pi}{6}+\pi k$. | \\frac{\pi}{6}+\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,820 |
19. $(\operatorname{tg} x)^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1$.
19. $(\tan x)^{\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}}=1$. | 19. If $\operatorname{tg} x \neq 0$ and $\operatorname{tg} x \neq 1$, then $\sin ^{2} x-\frac{3}{2} \sin x+\frac{1}{2}=0$. This equation is equivalent to two equations $\sin x=1$ and $\sin x=\frac{1}{2}$. When $\sin x=1$, the function $\operatorname{tg} x$ is undefined. From the equation $\sin x=\frac{1}{2}$, we find $... | \pik+(-1)^k\frac{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,821 |
20. $2 \sin 17 x+\sqrt{3} \cos 5 x+\sin 5 x=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
20. $2 \sin 17 x+\sqrt{3} \cos 5 x+\sin 5 x=0$. | 20. Let's divide all terms of the equation by 2 and replace $\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$ with $\sin \frac{\pi}{3}$ and $\cos \frac{\pi}{3}$, respectively. We get the equation $\sin 17 x + \sin \left(\frac{\pi}{3} + 5 x\right) = 0$. By converting the sum into a product, we get $2 \sin \left(\frac{\pi}{6} + 11 ... | x_{1}=\frac{\pi}{66}(6k-1),\quadx_{2}=\frac{\pi}{18}(3k+2) | Algebra | proof | Yes | Yes | olympiads | false | 34,822 |
21. $\sin ^{6} x+\cos ^{6} x=\sin 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
21. $\sin ^{6} x+\cos ^{6} x=\sin 2 x$. | 21. By factoring, the left side of the equation is reduced to $1-\frac{3}{4} \sin ^{2} 2 x . \quad$ In this case, $3 \sin ^{2} 2 x+4 \sin 2 x-4=0 . \quad$ From which $\sin 2 x=\frac{2}{3}$, $x=(-1)^{k} \frac{1}{2}$ arc $\sin \frac{2}{3}+\frac{\pi}{2} k, \quad(\sin 2 x \neq-2)$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,823 | |
22. $\sin ^{9} x \cos x-\cos ^{9} x \sin x=\sin 4 x$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
22. $\sin ^{9} x \cos x-\cos ^{9} x \sin x=\sin 4 x$. | 22. By factoring the left side, the equation is reduced to $\sin 4 x\left(4+\sin ^{4} x+\cos ^{4} x\right)=0, \quad 4+\sin ^{4} x+\cos ^{4} x \neq 0, \sin 4 x=0, x=\frac{\pi}{4} k$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,824 | |
23. $x^{2}+2 x \sin x y+1=0$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
23. $x^{2}+2 x \sin x y+1=0$. | 23. Replace 1 with the sum $\sin ^{2} x y+\cos ^{2} x y$. The equation will take the form $(x+\sin x y)^{2}+\cos ^{2} x y=0$. Such an equality is possible only if $x+\sin x y=0$ and $\cos x y=0$. From this system, the values of $x$ and $y$ can easily be found. $x= \pm 1, y=-\frac{\pi}{2}+2 \pi k$. | \1,-\frac{\pi}{2}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,825 |
24. $\operatorname{ctg} x-\operatorname{tg} x=\sin x+\cos x$.
24. $\cot x - \tan x = \sin x + \cos x$. | 24. Expressing $\operatorname{tg} x$ and $\operatorname{ctg} x$ in terms of $\sin x$ and $\cos x$, we reduce the equation to the following form: $(\sin x+\cos x)(\cos x-\sin x-\sin x \cos x)=0$.
If $\sin x+\cos x=0$, then $\operatorname{tg} x=-1$, from which $x_{1}=-\frac{\pi}{4}+\pi k$.
If $\cos x-\sin x-\sin x \cos... | \frac{\pi}{4}(8k-1)\\arccos\frac{\sqrt{2}-1}{\sqrt{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,826 |
25. $1+\sin x+\cos 3 x=\cos x+\sin 2 x+\cos 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
25. $1+\sin x+\cos 3 x=\cos x+\sin 2 x+\cos 2 x$. | 25. Transferring all members of the equation to one side and grouping
$$
(1-\cos 2 x)+\sin x+(\cos 3 x-\cos x)-\sin 2 x=0
$$
By making the appropriate substitution in the groups, we get
$$
(2 \sin 2 x+1) \sin x(1-2 \cos x)=0
$$
If $\sin x=0$, then $x_{1}=\pi k$.
If $2 \sin 2 x+1=0$, then $x_{2}=(-1)^{k+1} \frac{\p... | x_{1}=\pik,\x_{2}=(-1)^{k+1}\frac{\pi}{6}+\pik,\x_{3}=\\frac{\pi}{3}+2\pik | Algebra | proof | Yes | Yes | olympiads | false | 34,827 |
26. Given the equation $a \cos ^{2} x+b=\cos x+c=0$, where $a, b$ and $c$ are given real numbers. Form a quadratic equation in terms of $a, b$, and $c$ whose roots are the corresponding values of $\cos 2 x$. Compare the given and the formed equations for the case $a=4, b=2, c=-1$. | 26. We solve the given equation with respect to $\cos x$. Then we calculate $\cos 2 x$, using the formula $\cos 2 x=2 \cos ^{2} x-1$,
$$
\cos 2 x=\frac{b^{2}-2 a c \pm b \sqrt{b^{2}-4 a c}-a^{2}}{a^{2}}
$$
Using these values as roots, we form a new equation
$$
a^{2} \cos ^{2} 2 x+2\left(a^{2}+2 a c-b^{2}\right) \cos... | ^{2}\cos^{2}2x+2(^{2}+2-b^{2})\cos2x+(+2)^{2}-2b^{2}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,828 |
27. Solve the system of equations: $\left\{\begin{array}{l}\sin x + a \cos y = \frac{1}{\sqrt{2}}(1 + a) \\ \cos x + a \sin y = \frac{1}{\sqrt{2}}(1 + a)\end{array}\right.$ | 27. For $a=0$ we get $\sin x=\frac{1}{\sqrt{2}}$, hence $x=\frac{\pi}{4}+2 \pi k, y$-any. For $a=-1$ we get the system $\sin x-\cos y=0, \cos x-\sin y=0$. Squaring both equations and adding them term by term, we get $\sin (x+y)=1$, hence $x+y=\frac{\pi}{2}+2 \pi k$, $x=\frac{\pi}{2}-y+2 \pi k, y$ - any, or $y=\frac{\pi... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,829 | |
28. Solve the system of equations: $\left\{\begin{array}{l}\operatorname{tg} x \operatorname{tg} y=\frac{1}{6} \\ \sin x \sin y=\frac{1}{5 \sqrt{2}}\end{array}\right.$ | 28. Dividing the second equation by the first, we get $\cos x \cos y=\frac{6}{5 \sqrt{\overline{2}}}$.
Subtracting the second equation from the result, we get
$$
\cos (x+y)=\frac{1}{\sqrt{2}}, \text { from which } x+y= \pm \frac{\pi}{4}+2 \pi k
$$
Similarly, $\cos (x-y)=\frac{7}{5 \sqrt{2}}$. From which $x-y= \pm \a... | x+\\frac{\pi}{4}+2\pik,\;x-\\arccos\frac{7}{5\sqrt{2}}+2\pik | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,830 |
29. Prove that the equality $\sin \alpha \cdot \sin 2 \alpha \cdot \sin 3 \alpha=\frac{4}{5} \quad$ is impossible for any value of $\alpha$. | 29. Let's use the equality $\sin \alpha \sin 2 \alpha=\frac{\cos \alpha-\cos 3 \alpha}{2}$.
Then $\frac{\cos \alpha-\cos 3 \alpha}{2} \sin 3 x=\frac{4}{5}$. Multiply each part by 4. The equation 80
will take the form $\sin 2 \alpha+\sin 4 \alpha-\sin 6 \alpha=3 \frac{1}{5}$. Since $|\sin n \alpha|<1$, this equality is... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,831 |
30. Prove that for any $x$ the inequality
$$
\cos ^{4} x+\sin ^{2} x-\sin 2 x \cos x \geqslant 0
$$
holds. | 30. By replacing $\sin 2 x$ with $2 \sin x \cos x$, the left part of the inequality can be written as: $\left(\cos ^{2} x-\sin x\right)^{2} \geqslant 0$. The inequality is obvious. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,832 |
31. Prove that $\sqrt{\sin x} \cdot \sqrt[4]{\tan x}+\sqrt{\cos x} \cdot \sqrt[4]{\cot x} \geqslant \sqrt[4]{8}$, if $0<x<\frac{\pi}{2}$. | 31. Squaring each part of the inequality and transitioning to the functions $\sin x$ and $\cos x$. Multiplying both sides by $\sqrt{\sin x \cos x}$, we get
$$
\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x \geqslant 2 \sqrt{\sin 2 x}
$$
Or $(1-\sqrt{\sin 2 x})^{2} \geqslant 0$. We obtain an obvious inequality. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,833 |
33. Prove that $\operatorname{ctg} \frac{\alpha}{2}>1+\operatorname{ctg} \alpha$ for all $\alpha$ satisfying the inequality $0<\alpha<\frac{\pi}{2}$. | 33. We transition to the argument $\frac{\alpha}{2}$. The inequality is transformed into $2 > 2 \operatorname{tg} \frac{x}{2} + 1 - \operatorname{tg}^{2} \frac{\alpha}{2}$ or $\left(\operatorname{tg} \frac{\alpha}{2} - 1\right)^{2} > 0$, which is obvious. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,835 |
34. Solve the inequality: $\sin x>\cos ^{2} x$. | 34. Let's move on to the function $\sin x$. The inequality takes the form $\sin ^{2} x+\sin x-$ $-1>0$. This inequality holds under the condition $\frac{\sqrt{5}-1}{2}<\sin x \leqslant 1$ or $\arcsin \frac{\sqrt{5}-1}{2}+2 \pi k<x<\pi-\arcsin \frac{\sqrt{5}-1}{2}+2 \pi k$. | \arcsin\frac{\sqrt{5}-1}{2}+2\pik<x<\pi-\arcsin\frac{\sqrt{5}-1}{2}+2\pik | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 34,836 |
36. Prove the identity
$$
\sin \alpha+\sin 2 \alpha+\ldots+\sin n \alpha=\frac{\sin \frac{n+1}{2} \alpha}{\sin \frac{\alpha}{2}} \cdot \sin \frac{n x}{2}
$$ | 36. The validity of the given formula can be verified by the method of mathematical induction or by applying an artificial trick.
Let $\sin \alpha + \sin 2 \alpha + \ldots + \sin n \alpha = x$. Multiplying both sides by $2 \sin \frac{\alpha}{2}$ and replacing the products of sines with the difference of the correspond... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,838 |
37. Simplify the expression $\sin ^{2}\left(\frac{\pi}{8}+\frac{\alpha}{2}\right)-\sin ^{2}\left(\frac{\pi}{8}-\frac{\alpha}{2}\right)$. | 37. Let's use the formula $a^{2}-b^{2}=(a+b)(a-b)$.
Then
$$
\begin{gathered}
\sin ^{2}\left(\frac{\pi}{8}+\frac{\alpha}{2}\right)-\sin ^{2}\left(\frac{\pi}{8}-\frac{\alpha}{2}\right)=\left[\sin \left(\frac{\pi}{8}+\frac{\alpha}{2}\right)+\sin \left(\frac{\pi}{8}-\frac{\alpha}{2}\right)\right] \times \\
\times\left[\s... | \frac{\sqrt{2}}{2}\sin\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,839 |
38. Prove the inequality $\sin ^{6} x+\cos ^{6} x \geqslant \frac{1}{4}$. | 38. Factorize the left part of the inequality
$$
\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x\right) \geqslant \frac{1}{4}
$$
or $\sin ^{4} x-\sin ^{2} x \cos ^{2} x+\cos ^{4} x \geqslant \frac{1}{4}$. Let's complete the square
$$
\left(\sin ^{2} x+\cos ^{2} x\right)^{2}-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 34,840 |
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