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class | __index_level_0__ int64 0 742k |
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231. Prove that the sum
$$
1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n}
$$
will exceed any predetermined number \( N \) if \( n \) is sufficiently large.
Note. The result of this problem can be significantly refined. Specifically, it can be shown that the sum \( 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldo... | 231. We will prove that the sum
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}
$$
can be made greater than any number $N$. We will consider $N$ to be an integer and take $n=2^{2 N};$ then
$$
\begin{aligned}
& 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n-1}+\frac{1}{n}=1+\frac{1}{2}+\left(\f... | proof | Calculus | proof | Yes | Yes | olympiads | false | 29,493 |
232. Prove that if in the sum
$$
1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{n}
$$
all terms where the denominator contains the digit 9 are removed, then the sum of the remaining terms for any $n$ will be less than 80. | 232. Let $n_{k}$ denote the number of unstruck terms between $\frac{1}{10^{k}}$ and $\frac{1}{10^{k+1}}$, including $\frac{1}{10^{k}}$ but not $\frac{1}{10^{k+1}}$. If the term $-\frac{1}{q}$, located between $\frac{1}{10^{h-1}}$ and $\frac{1}{10^{h}}$, is not struck out, then among the terms $\frac{1}{10 q}, \frac{1}{... | 80 | Number Theory | proof | Yes | Yes | olympiads | false | 29,494 |
233. a) Prove that for any $n$
$$
1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\ldots+\frac{1}{n^{2}}<2
$$
b) Prove that for any $n$
$$
1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\ldots+\frac{1}{n^{2}}<1 \frac{3}{4}
$$
It is clear that the inequality in part b) is a strengthening of the inequality in part a). An... | 233. a) Suppose that in the sum
$$
1+\frac{1}{4}+\frac{1}{9}+\ldots+\frac{1}{n^{2}}
$$
the number $n$ is less than $2^{h+1}$. Consider the sum $1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{\left(2^{h+1}-1\right)^{2}}$ and group the terms in the same way as in the solution to problem 231:
$$
\begin{aligned}
& 1+... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,495 |
235. Prove that
$$
(a+b+c)^{333}-a^{333}-b^{333}-c^{333}
$$
is divisible by
$$
(a+b+c)^{3}-a^{3}-b^{3}-c^{3}
$$ | 235. It is not hard to see that $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=$ $=3(a+b)(b+c)(c+a)$ (check this!); therefore, it is sufficient to prove that
$$
P(a, b, c)=(a+b+c)^{333}-a^{333}-b^{333}-c^{333}
$$
is divisible by $a+b$, by $b+c$, and by $a+c$. However, it is obvious that the polynomial (with literal coefficients) $P(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,497 |
236. Factor the expression
$$
a^{10}+a^{5}+1
$$ | 236. $a^{10}+a^{5}+1=\frac{\left(a^{5}\right)^{3}-1}{a^{5}-1}=\frac{a^{15}-1}{a^{5}-1}=$
$$
\begin{gathered}
=\frac{\left(a^{3}\right)^{5}-1}{(a-1)\left(a^{4}+a^{3}+a^{2}+a+1\right)}=\frac{\left(a^{3}-1\right)\left(a^{12}+a^{9}+a^{6}+a^{3}+1\right)}{(a-1)\left(a^{4}+a^{3}+a^{2}+a+1\right)}= \\
=\frac{\left(a^{2}+a+1\r... | ^{10}+^{5}+1=(^{2}++1)(^{8}-^{7}+^{5}-^{4}+^{3}-+1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,498 |
237. Prove that the polynomial
$x^{9999}+x^{8888}+x^{7777}+x^{6666}+x^{5555}+x^{4444}+x^{3333}+x^{2222}+$ $+x^{1111}+1$
is divisible by $x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+1$ | 237. First solution. Let's denote our polynomials as $B$ and $A$ respectively. In this case, we have:
$$
\begin{aligned}
& B-A=\left(x^{9999}-x^{9}\right)+\left(x^{8888}-x^{8}\right)+\left(x^{7777}-x^{7}\right)+\left(x^{66666}-x^{6}\right)+ \\
& +\left(x^{5555}-x^{5}\right)+\left(x^{444}-x^{4}\right)+\left(x^{3333}-x^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,499 |
238. a) Factorize the expression
$$
a^{3}+b^{3}+c^{3}-3 a b c
$$
b) Using the result from part a), find the general formula for solving the cubic equation
$$
x^{3}+p x+q=0
$$
$^{1}$ François Viète (1540-1603) - a prominent French mathematician, one of the creators of modern algebraic symbolism (and thus algebra).
... | 238. a) First solution. We have:
$a^{3}+b^{3}+c^{3}-3 a b c=$
$$
\begin{aligned}
& =a^{3}+3 a b(a+b)+b^{3}+c^{3}-3 a b c-3 a b(a+b)= \\
& =a^{3}+3 a^{2} b+3 a b^{2}+b^{3}+c^{3}-3 a b(c+a+b)= \\
& =(a+b)^{3}+c^{3}-3 a b(a+b+c)= \\
& =[(a+b)+c]\left[(a+b)^{2}-(a+b) c+c^{2}\right]-3 a b(a+b+c)= \\
& =(a+b+c)\left[(a+b)^... | x_{1}=-\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}-\sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,500 |
239. Solve the equation
$$
\sqrt{a-\sqrt{a+x}}=x
$$ | 239. First solution. Let $\sqrt{a+x}$ be denoted by $y$; we obtain a system of two equations.
$$
\sqrt{a+x}=y, \sqrt{a-y}=x
$$
Square these equations:
$$
a+x=y^{2}, a-y=x^{2}
$$
Subtract the second from the first:
$$
x+y=y^{2}-x^{2}
$$
$\mathrm{H} / \mathrm{H}$
$$
x^{2}-y^{2}+x+y=(x+y)(x-y+1)=0
$$
From this, th... | x_{1,2}=\frac{1}{2}\\sqrt{+\frac{1}{4}},\quadx_{3,4}=-\frac{1}{2}\\sqrt{-\frac{3}{4}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,501 |
240*. Find the real roots of the equation
$$
x^{2}+2 a x+\frac{1}{16}=-a+\sqrt{a^{2}+x-\frac{1}{16}} \quad\left(0<a<\frac{1}{4}\right) .
$$ | 240. First solution. Let
$$
x^{2}+2 a x+\frac{1}{16}=y, -a+\sqrt{a^{2}+x-\frac{1}{16}}=y_{1}
$$
In this case, our equation takes the form
$$
y=y_{1}
$$
Now let's express $x$ in terms of $y_{1}$. Simple calculations give
$$
x=y_{1}^{2}+2 a y_{1}+\frac{1}{16}
$$
Thus, we see that $x$ is expressed in terms of $y_{1}... | x_{1,2}=\frac{1-2}{2}\\sqrt{(\frac{1-2}{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,502 |
242. Solve the equation
$$
1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdot}} \cdot \cdot_{1+\frac{1}{x}}}=x
$$
(in the expression on the left, the fraction sign repeats $n$ times). | 242. We will sequentially simplify the fraction on the left:
$$
\begin{array}{rl}
1+\frac{1}{x}=\frac{x+1}{x} ; 1+\frac{1}{\frac{x+1}{x}}=1+\frac{x}{x+1} & =\frac{2 x+1}{x+1} \\
1 & 1+\frac{1}{\frac{2 x+1}{x+1}}=1+\frac{x+1}{2 x+1}=\frac{3 x+2}{2 x+1} ; \ldots
\end{array}
$$
Ultimately, we will obtain an equation of ... | x_{1}=\frac{1+\sqrt{5}}{2},x_{2}=\frac{1-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,504 |
243. Find the real roots of the equation
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$
(all square roots are considered positive), | 243. We have:
$$
\begin{aligned}
x+3-4 \sqrt{x-1}= & x-1-4 \sqrt{x-1}+4= \\
& =(\sqrt{x-1})^{2}-4 \sqrt{x-1}+4=(\sqrt{x-1}-2)^{2}
\end{aligned}
$$
and similarly
$$
x+8-6 \sqrt{x-1}=x-1-6 \sqrt{x-1}+9=(\sqrt{x-1}-3)^{2}
$$
Thus, our equation can be written as
$$
\sqrt{(\sqrt{x-1}-2)^{2}}+\sqrt{(\sqrt{x-1}-3)^{2}}=1... | 5\leqslantx\leqslant10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,505 |
244. Solve the equation
$$
|x+1|-|x|+3|x-1|-2|x-2|=x+2
$$ | 244. To solve this equation, we will first look for its roots lying on the interval from 2 to $\infty$, then on the intervals from 1 to 2, from 0 to 1, from -1 to 0, and from $-\infty$ to -1.
$1^{\circ}$. Let $x \geqslant 2$. Then $x+1>0, x>0, x-1>0, x-2 \geqslant 0$; therefore, $|x+1|=x+1,|x|=x,|x-1|=x-1 ;|x-2|=$ $=x... | -2x\geq2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,506 |
245. Solve the equation
$$
\begin{aligned}
& 1-\frac{x}{1}+\frac{x(x-1)}{1 \cdot 2}-\frac{x(x-1)(x-2)}{1 \cdot 2 \cdot 3}+\ldots \\
& \ldots+(-1)^{n} \frac{x(x-1)(x-2) \ldots(x-n)}{n!}=0
\end{aligned}
$$ | 245. Let us denote the right-hand side of the considered equation of the $n$-th degree by $f_{n}(x)$. It is easy to see that the equation $f_{1}(x)=0$, i.e., $1-x=0$, has the root $x_{1}=1$; the equation $f_{2}(x)=0$, i.e., $x(x-1)-2 x+2=0$, or $x^{2}-3 x+2=0$, has the roots $x_{1}=1$ and $x_{2}=2$. We will now prove t... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,507 |
246. Solve the equation $x^{3}-[x]=3$, where, as usual, $[x]$ is the integer part of the number $x$ (see p. 37). | 246. Let $\{x\}=x-[x]$ denote the so-called fractional part of the number $x$ (see p. 37); it is clear that $0 \leqslant\{x\}<1$; for $x<0$ and $x^{3}-x=x\left(x^{2}-1\right)<0<2$; for $x=-1$ we have $x^{3}-$ $-x=0<2$; for $-1<x \leqslant 0$ we have $x^{3}-x \leqslant-x<1$; for $0<x \leqslant 1$ we have $x^{3}-x<x^{3} ... | \sqrt[3]{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,508 |
247. System of two quadratic equations
$$
\left\{\begin{array}{r}
x^{2}-y^{2}=0 \\
(x-a)^{2}+y^{2}=1
\end{array}\right.
$$
generally has four solutions. For what values of \( a \) does the number of solutions of this system decrease to three or to two? | 247. From the first equation of the system, we immediately obtain:
$$
y^{2}=x^{2}, y= \pm x
$$
Substituting this value of $y^{2}$ into the second equation, we find:
$$
(x-a)^{2}+x^{2}=1
$$
- a quadratic equation which, generally speaking, gives two values of $x$. Since each value of $x$ corresponds to two values of... | \1,\\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,509 |
248. a) Solve the system of equations
$$
\left\{\begin{array}{c}
a x+y=a^{2} \\
x+a y=1
\end{array}\right.
$$
For which values of $a$ does this system have no solutions and for which values does it have infinitely many solutions?
b) The same question for the system
$$
\left\{\begin{array}{l}
a x+y=a^{3} \\
x+a y=1
... | 248. a) Solving the system, we obtain:
$$
x=\frac{a^{3}-1}{a^{2}-1}, y=\frac{-a^{2}+a}{a^{2}-1} .
$$
From this, it is clear that if $a+1 \neq 0$ and $a-1 \neq 0$, the system has a unique solution $x=\frac{a^{2}+a+1}{a+1}, y=\frac{-a}{a+1}$. If $a=-1$ or $a=+1$, our formulas lose their meaning; in the first case, we a... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,510 |
249. Find the conditions that the numbers $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ must satisfy for the system of 6 equations with 4 unknowns
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\alpha_{1} \alpha_{2} \\
x_{1}+x_{3}=\alpha_{1} \alpha_{3} \\
x_{1}+x_{4}=\alpha_{1} \alpha_{4} \\
x_{2}+x_{3}=\alpha_{2} \alpha_{... | 249. Subtracting the second equation from the first and the sixth from the fifth, and equating the two obtained expressions for \(x_{2}-x_{3}\), we get:
\[
\alpha_{1}\left(\alpha_{2}-\alpha_{3}\right)=\alpha_{4}\left(\alpha_{2}-\alpha_{3}\right), \text{ or }\left(\alpha_{1}-\alpha_{4}\right)\left(\alpha_{2}-\alpha_{3}... | \frac{\alpha^{2}}{2};\quad\alpha(\beta-\frac{\alpha}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,511 |
250. How many real solutions does the system have
$$
\left\{\begin{array}{c}
x+y=2 \\
x y-z^{2}=1 ?
\end{array}\right.
$$ | 250. From the first equation we get $x=2-y$; substituting into the second:
$$
2 y-y^{2}-z^{2}=1
$$
or
$$
z^{2}+y^{2}-2 y+1=0, \text { i.e. } z^{2}+(y-1)^{2}=0
$$
Each term of the last equality is non-negative and, consequently, equals zero. Hence,
$$
z=0, \quad y=1
$$
and, therefore,
$$
x=1 .
$$
Thus, the syste... | 1,1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,512 |
251. Find all real solutions of the system
$$
\left\{\begin{array}{c}
x^{3}+y^{3}=1 \\
x^{4}+y^{4}=1
\end{array}\right.
$$ | 251. If $x^{4}+y^{4}=1$, then either $x^{4}=1, y^{4}=0$, or $x^{4}=0$, $y^{4}=1$, or, finally, $0x^{4}+y^{4}=1
\end{gathered}
$$
from which it follows that two numbers $x, y$ of the same sign, such that $|x|<1$, $|y|<1$, cannot serve as a solution to our system. It is even clearer that
two numbers $x, y$ of different ... | 1,00,1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,513 |
253. Find all possible systems of four real numbers such that the sum of each of them with the product of the others is 2. | 253. Let $x, y, z, t$ be the desired numbers and $x y z t = A$; note that $A \neq 0$, since if, for example, $x=0$, then the conditions of the problem lead to contradictory equalities $y=z=t=2$ and $y z t=2$. Further, the equation $x + y z t = 2$ can be rewritten as:
$$
x + \frac{A}{x} = 2, \text{ or } x^2 - 2x + A = ... | x=y=z==1orx=y=z=-1,=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,515 |
254. Solve the system of four equations with four unknowns
$$
\begin{array}{rr}
|a-b| y+|a-c| z+|a-d| t & =1 \\
|b-a| x+|b-c| z+|b-d| t & =1 \\
|c-a| x+|c-b| y+|c-d| t & =1 \\
|d-a| x+|d-b| y+|d-c| z & =1
\end{array}
$$
where $a, b, c, d$ are some four pairwise distinct real numbers. | 254. To emphasize the complete symmetry of the equations of our system with respect to the unknowns and the coefficients of these unknowns, let us denote $x=x_{1}, y=x_{2}, z=x_{3}$ and $t=x_{4}; a=a_{1}, b=a_{2}, c=a_{3}, d=a_{4}$; in this case, our equations can be written as:
$$
\sum_{j=1}^{4}\left|a_{i}-a_{j}\righ... | \frac{1}{-},0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,516 |
255. Given a system of $n$ equations with $n$ unknowns $x_{1}, x_{2}, \ldots, x_{n}:$
$$
a x_{1}^{2}+b x_{1}+c=x_{2}, \quad a x_{2}^{2}+b x_{2}+c=x_{3}, \ldots
$$
$$
\ldots, a x_{n-1}^{2}+b x_{n-1}+c=x_{n}, \quad a x_{n}^{2}+b x_{n}+c=x_{1}
$$
(here $a \neq 0$). Prove that this system has no solutions if $(b-1)^{2}-... | 255. Let $x_{2}-x_{1}=X_{1}, x_{3}-x_{2}=X_{2}, \ldots, x_{n}-x_{n-1}=$ $=X_{n-1}, \quad x_{1}-x_{n}=X_{n}$. Then $X_{1}+X_{2}+\ldots+X_{n-1}+X_{n}=0$, and the given system of equations can be rewritten as:
$a x_{1}^{2}+(b-1) x_{1}+c=X_{1}, \quad a x_{2}^{2}+(b-1) x_{2}+c=X_{2}, \ldots$
$$
\ldots, a x_{n}^{2}+(b-1) x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,517 |
256. Let all numbers $a_{1}, a_{2}, \ldots, a_{n}$ (where $n \geqslant 2$) be positive; how many real solutions does the system of equations have:
$$
x_{1} x_{2}=a_{1}, x_{2} x_{2}=a_{2}, \ldots, x_{n-1} x_{n}=a_{n-1}, x_{n} x_{1}=a_{n} ?
$$ | 256. Let's consider two cases separately.
$1^{\circ} . n$ is even. By multiplying the "odd" (i.e., $1$-st, $3$-rd, ..., $(n-1)$-th) and "even" equations of our system, we get:
$$
x_{1} x_{2} x_{3} \ldots x_{n}=a_{1} a_{3} a_{5} \ldots a_{n-1} \text { and } x_{1} x_{2} x_{3} \ldots x_{n}=a_{2} a_{4} a_{6} \ldots a_{n}... | 0, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,518 |
257. a) How many roots does the equation
$$
\sin x=\frac{x}{100} ?
$$
b) How many roots does the equation
$$
\sin x=\lg x ?
$$ | 257. a) First of all, note that if $x_{0}$ is a root of the equation, then $-x_{0}$ is also a root. Therefore, the number of negative roots is the same as the number of positive roots. Furthermore, the number 0 is a root of the equation. Thus, it is sufficient to find the number of positive roots. Now, let's note that ... | 63 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 29,519 |
258. It is known that
$$
\begin{array}{r}
a_{1}-4 a_{2}+3 a_{3} \geqslant 0 \\
a_{2}-4 a_{3}+3 a_{4} \geqslant 0 \\
\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
\cdot \cdot \cdot \cdot \cdot \cdot \cdot \\
a_{98}-4 a_{99}+3 a_{100} \geqslant 0 \\
a_{99}-4 a_{100}+3 a_{1} \geqslant 0 \\
a_{100}-4 a_{1}+3 a_{2} \g... | 258. Adding the left parts of all our inequalities, we get the sum of the numbers $a_{1}, \alpha_{2}, \ldots, a_{99}, a_{100}$, each taken with the coefficient $1+(-4)+3=0$, i.e., the number 0. But if the sum of 100 non-negative numbers is zero, then all these numbers are zero; thus, our system of inequalities actually... | a_{2}=a_{3}=\ldots=a_{100}=1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 29,520 |
259. Let $a, b, c, d$ be any four positive numbers. Prove that the three inequalities
$$
\begin{gathered}
a+b<c+d \\
(a+b)(c+d)<a b+c d \\
(a+b) c d<(c+d) a b
\end{gathered}
$$
cannot hold simultaneously. | 259. First solution. Let's rewrite our inequalities as follows:
$$
\begin{aligned}
& A=-a-b+c+d>0 \\
& B=a b-a c-a d-b c-b d+c d>0 \\
& C=a b c+a b d-a c d-b c d>0
\end{aligned}
$$
and consider the equation
$$
\begin{aligned}
& P(x)=(x-a)(x-b)(x+c)(x+d)= \\
& \quad=x^{4}+A x^{3}+B x^{2}+C x+a b c d=0
\end{aligned}
$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,521 |
260. Prove that the fraction
$$
\frac{2-\sqrt{2+\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}}{2-\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}}
$$
where the numerator contains $n$ radicals, and the denominator contains $(n-1)$ radicals, is greater than $\frac{1}{4}$ for any $n \geqslant 1$. | 260. Since, obviously,
$$
\begin{aligned}
& (2-\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\ldots+\sqrt{2}})}}}_{\text {radicals }} \times \\
& \times\left(2+\sqrt{\frac{\sqrt{2+\sqrt{2+\sqrt{2+\ldots+V}})}}{\text { radicals }}}=2^{2}-\right. \\
& -(2+\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}})=2-\sqrt{\frac{2+\sqrt{2+\ldots+\sqrt{2}}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,522 |
261. The product of three positive numbers is 1; their sum is greater than the sum of the reciprocals of these numbers. Prove that one of the three numbers is greater than one, and the other two are less than one. | 261. Let $a, b, c$ be given numbers; since $a b c=1$, then $c=\frac{1}{a b}$. The second condition of the problem states that
$a+b+c>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, or $a+b+\frac{1}{a b}>\frac{1}{a}+\frac{1}{b}+a b$;
but the inequality ( $\left(^{\prime}\right.$ ) can be transformed as follows:
$$
a b-a-b+1(a-... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,523 |
262. The sum of 1959 positive numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{1959}\) is 1; prove that the sum of all possible products of 1000 different factors from our numbers is less than 1. [Among the considered products, all those that differ from each other by at least one factor are included, but not products that di... | 262. It is clear that the numbers 1959 and 1000 in the formulation of this problem are random - the fact that we need to prove is that if all $a_{i}>0 \quad$ and $\sum a_{i}=a_{1}+a_{2}+\ldots+a_{n}=1$, then the sum $S_{n, \xi}=\sum_{i_{1}, i_{2}, \ldots, i_{k}=1}^{n} a_{i_{1}} a_{i_{2} \ldots} a_{i_{i}}$ of all possib... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 29,524 |
264. 1973 positive numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{1973}$ are such that
$$
a_{1}^{a_{1}}=a_{2}^{a_{3}}=a_{3}^{a_{4}}=\ldots=\left(a_{1972}\right)^{a_{1973}}=\left(a_{1973}\right)^{a_{1}}
$$
Prove that $a_{1}=a_{1973}$. | 264. We claim that all our 1973 numbers are the same. Indeed, let it not be so, and, say, $a_{1}=a_{2}=a_{3}=\ldots=a_{i} \neq a_{i+1}$. For simplicity, we will renumber our numbers cyclically, assigning number 1 to the number $a_{i}$, number 2 to the number $a_{i+1}$, and so on, up to the number $a_{i-1}$, which will ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,526 |
265*. Prove that if $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}-6 x+1=0$, then $x_{1}^{n}+x_{2}^{n}$ for any integer $n$ is an integer not divisible by 5. | 265. The statement of the problem is true for $n=1$ and $n=2$, and
$$
\begin{gathered}
x_{1}^{0}+x_{2}^{0}=1+1=2, \quad x_{1}+x_{2}=6 \\
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(6)^{2}-2 \cdot 1=34
\end{gathered}
$$
Next, we have:
$$
\begin{aligned}
x_{1}^{n}+x_{2}^{n}=\left(x_{1}+x_{2}\right)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,527 |
266. Can the expression
$\left(a_{1}+a_{2}+\ldots+a_{999}+a_{1000}\right)^{2}=$ $=a_{1}^{2}+a_{2}^{2}+\ldots+a_{999}^{2}+a_{1000}^{2}+2 a_{1} a_{2}+2 a_{1} a_{3}+\ldots+2 a_{999} a_{1000}$, where some of the numbers $a_{1}, a_{2}, \ldots, a_{999}, a_{1000}$ are positive and others are negative, contain an equal number... | 266. Suppose the sum $a_{1}+a_{2}+\ldots+a_{1000}$ contains $n$ positive and 1000 - $n$ negative terms. In this case, all pairwise products of the $n$ positive terms (obviously, there will be $\frac{n(n-1)}{2}$ of them) and all pairwise products of the 1000 - $n$ negative terms (their number is $\frac{(1000-n)(1000-n-1... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,528 |
267. Prove that any integer power of the number $\sqrt{2}-1$ can be represented in the form $\sqrt{N}-\sqrt{N-1}$, where $N$ is an integer (for example, $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}=$ $=\sqrt{9} \div \sqrt{8}$, and $\left.(\sqrt{2}-1)^{3}=5 \sqrt{2}-7=\sqrt{50}-\sqrt{49}\right)$.
Note: The division symbol in the exa... | 267. First, we have:
$$
\begin{aligned}
& (\sqrt{2}-1)^{1}=\sqrt{2}-\sqrt{1} \\
& (\sqrt{2}-1)^{2}=3-2 \sqrt{2}=\sqrt{9}-\sqrt{8}
\end{aligned}
$$
Now let's prove that if
$$
(\sqrt{2}-1)^{2 h-1}=B \sqrt{2}-A
$$
can be represented as $\sqrt{N}-\sqrt{N-1}$, i.e., if $2 B^{2}-A^{2}=1$, then the number
$$
(\sqrt{2}-1)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,529 |
268. Prove that the expression $99999+111111 \sqrt{3}$ cannot be represented in the form $(A+B \sqrt{3})^{2}$, where $A$ and $B$ are integers. | 268. $\mathrm{If}(A+B \sqrt{3})^{2}=C+D \sqrt{3}$, then $C=A^{2}+3 B^{2}, D=2 A B$ and $(A-B \sqrt{3})^{2}=A^{2}+3 B^{2}-2 A B \sqrt{3}=C-D \sqrt{3}$. Therefore, if it were $(A+B \sqrt{3})^{2}=99999+111111 \sqrt{3}$, then it would also be $(A-B \sqrt{3})^{2}=99999-111111 \sqrt{3}$, which is impossible, since $99999-111... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,530 |
269. Prove that $\sqrt[3]{2}$ cannot be represented in the form $p+q \sqrt{r}$, where $p, q, r$ are rational numbers. | 269. Suppose that $\sqrt[3]{2}=p+q \sqrt{r}$, and raise both sides of this equation to the third power. In this case, we get:
$$
2=p^{3}+3 p^{2} q \sqrt{r}+3 p q^{2} r+q^{3} r \sqrt{r}
$$
or
$$
2=p\left(p^{2}+3 q^{2} r\right)+q\left(3 p^{2}+q^{2} r\right) \sqrt{r}
$$
Now let's show that if $\sqrt[3]{2}=p+q \sqrt{r}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,531 |
270. It is known that $A$ has the form $A=\left(\frac{n+\sqrt{n^{2}-4}}{2}\right)^{m}$ where $m, n \geqslant 2$ are natural numbers. Prove that $A$ can also be represented as $A=\frac{k+\sqrt{k^{2}-4}}{2}$, where $k$ is a natural number. | 270. Let $\frac{n+\sqrt{n^{2}-4}}{2}=x$; in this case
$$
\frac{1}{x}=\frac{2}{n+1-\sqrt{n^{2}-4}}=\frac{2\left(n-\sqrt{n^{2}-4}\right)}{4}=\frac{n-\sqrt{n^{2}-4}}{2}
$$
so $x$ satisfies the quadratic equation $x+\frac{1}{x}=n$. But if the number $x+\frac{1}{x}$ is an integer $(=n)$, then the number $x^{m}+\frac{1}{x^... | \frac{k+\sqrt{k^{2}-4}}{2} | Algebra | proof | Yes | Yes | olympiads | false | 29,532 |
271. Do there exist such rational numbers $x, y$, $z$ and $t$, that for some natural $n$
$$
(x+y \sqrt{2})^{2 n}+(z+t \sqrt{2})^{2 n}=5+4 \sqrt{2} ?
$$ | 271. First of all, note that a rational number $\alpha$ cannot be represented in two different ways as a sum $\alpha = x + y \sqrt{2}$, where $x$ and $y$ are rational, because if $\alpha = a + b \sqrt{2} = a_{1} + b_{1} \sqrt{2}$ (where $a, b, a_{1}, b_{1}$ are rational), then $\sqrt{2} = \frac{a - a_{1}}{b_{1} - b}$, ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,533 |
272. Given two barrels of infinite capacity. Is it possible, using two buckets of capacity $\sqrt{2}$ liters and $2-\sqrt{2}$ liters, to transfer exactly 1 liter of water from one of them to the other? | 272. No. Indeed, let us say we poured water from the 1st barrel to the 2nd barrel $k_{1}$ times using the first bucket and $k_{2}$ times using the same bucket to pour water from the 2nd barrel back to the 1st; in the end, we would have poured $\left(k_{1}-k_{2}\right) \sqrt{2}=k \cdot \sqrt{2}$ liters of water from the... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 29,534 |
273. For which rational $x$ will the expression $3 x^{2}-$ $-5 x+9$ represent the square of a rational number? | 273. First solution. The problem requires finding all rational solutions $(x, y)$ (where $y \geqslant 0$) of the equation $3 x^{2}-5 x+9 = y^{2}$ with two unknowns $x$ and $y$ (cf. problem cycle 5). Clearly, one solution is $x=0, y=3$. Let $x=x_{1}, y=y_{1}+3$; then we get:
$$
3 x_{1}^{2}-y_{1}^{2}-5 x_{1}-6 y_{1}=0
$... | \frac{5n^{2}+6n}{3n^{2}-^{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,535 |
274. The discriminant $\Delta=p^{2}-4 q$ of the quadratic equation $x^{2}+p x+q=0$ has a magnitude of the order of 10; prove that if, upon rounding the free term $q$ of the equation, we changed it by a magnitude of the order of 0.01, then the values of the roots of the equation will change by a magnitude of the order o... | 274. Let $x^{2}+p x+q=0$ and $y^{2}+p y+q_{1}=0$ be the original and "rounded" equations, where $\left|q_{1}-q\right|=|\varepsilon| \approx 0.01$. Subtracting the second equation from the first, we get: $\left(x^{2}-y^{2}\right)+p(x-y)=q_{1}-q=\varepsilon$, or $(x-y)(x+y+p)=\varepsilon$, from which, denoting by $x_{1}$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,536 |
275. We call the rounding of a number an integer that differs from the original by less than 1. Prove that any $n$ positive numbers can be rounded in such a way that the sum of any number of these numbers will differ from the sum of their roundings by no more than $\frac{n+1}{4}$. | 275. It is clear that if among our numbers there are several integers, then they can simply be discarded - since the difference between the sum of the roundings of any number of numbers and the sum of these numbers themselves will not change from the addition of some integers to the set of numbers, while the sum of the... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,537 |
276. In the decimal representation of a positive number $a$, all decimal places starting from the fourth digit after the decimal point are discarded, i.e., the number $a$ is rounded down to $a_{0}$ with an accuracy of 0.001. The obtained number $a_{0}$ is divided by the original number $a$, and the quotient is again ro... | 276. It is clear that for $a\frac{1}{2} \quad(\text { and } d \leqslant 1)
$$
- this estimate of the fraction $d$ is already final, because the ratio $\delta=\frac{\alpha}{a}$ can take any value within the range $0 \leqslant \delta < \frac{1}{2}$ (which means $d$ can take any value within the range $\frac{1}{2} < d \l... | 0;0.5;0.501;0.502;\ldots;0.999;1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,538 |
278. a) Prove that if $\propto$ is written as a decimal fraction $0.999 \ldots$, starting with 100 nines, then $\sqrt{\alpha}=0.999 \ldots$ is also written as a decimal fraction starting with 100 nines.
b)* Calculate the value of the root $\sqrt{\frac{\underbrace{111 \ldots .111}_{100}}{\text { ones }}}$ with an accur... | 278. a) If the number $\alpha$ is less than 1, then $\sqrt{\alpha}$ is also less than 1. Suppose now that the decimal fraction equal to $\sqrt{\alpha}$ starts with fewer nines than 100; this means that $\sqrt{\alpha}1-a$; therefore
$$
\sqrt{1-\left(\frac{1}{10}\right)^{100}} \\
& >\sqrt{1-\left(\frac{1}{10}\right)^{10... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,539 |
279. a) What is greater $\frac{2,00000000004}{1,0 \text { J)JJJ0JJ4) }}$ or
$$
\frac{2,00000000002}{(1,00000000002)^{2}+2,00000000002} ?
$$
b) Let $a>b>0$. What is greater
$$
\frac{1+a+a^{2}+\ldots+a^{n-1}}{1+a+a^{2}+\ldots+a^{n}} \text { or } \frac{1+b+b^{2}+\ldots+b^{n-1}}{1+b+b^{2}+\ldots+b^{n}} ?
$$ | 279. a) Let's denote 1.0000000004 as $\alpha$, and 1.00000000002 as $\beta$. In this case, the expressions in the problem will take the form $\frac{1+\alpha}{1+\alpha+\alpha^{2}}$ and $\frac{1+\beta}{1+\beta+\beta^{2}}$. Since $\alpha > \beta$, it is obvious that,
$$
\begin{aligned}
\frac{1+\alpha}{\alpha^{2}} & =\fra... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,540 |
282. a) Prove that for any real numbers $a_{1}, a_{2}, \ldots, a_{n} ; b_{1}, b_{2}, \ldots, b_{n}$, the following inequality always holds:
$$
\begin{aligned}
\sqrt{a_{1}^{2}+b_{1}^{2}} & +\sqrt{a_{2}^{2}+b_{2}^{2}}+\ldots+\sqrt{a_{n}^{2}+b_{n}^{2}} \geqslant \\
& \geqslant \sqrt{\left(a_{1}+a_{2}+\ldots+a_{n}\right)^... | 282. a) First of all, we can consider all \(a_{1}, a_{2}, \ldots, a_{n}; b_{1}, b_{2}, \ldots, b_{n}\) to be positive from the very beginning; otherwise, we would change the signs of the negative numbers to the opposite: the left side of the inequality would not change, while the right side could only increase. Now con... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,542 |
285. What is greater, $\cos \sin x$ or $\sin \cos x$? | 285. Since $\sin \cos x = -\cos \left(\frac{\pi}{2} + \cos x\right)$, then
$\cos \sin x - \sin \cos x = \cos \sin x + \cos \left(\frac{\pi}{2} + \cos x\right)=$
$$
=2 \cos \frac{\frac{\pi}{2} + \cos x + \sin x}{2} \cos \frac{\frac{\pi}{2} + \cos x - \sin x}{2}
$$
So
$$
\begin{aligned}
& |\cos x + \sin x| = \sqrt{\c... | proof | Calculus | math-word-problem | Yes | Yes | olympiads | false | 29,545 |
286. Prove, without using logarithmic tables, that
a) $\frac{1}{\log _{2} \pi}+\frac{1}{\log _{5} \pi}>2$;
b) $\frac{1}{\log _{2} \pi}+\frac{1}{\log _{\pi} 2}>2$. | 286. a) Let $\log _{2} \pi=a, \quad \log _{5} \pi=b$. From the equalities $2^{a}=\pi, 5^{b}=\pi$ we get $\pi^{1 / a}=2, \pi^{1 / b}=5, \pi^{1 / a} \cdot \pi^{1 / b}=2 \cdot 5=10$, $\pi^{1 / a+1 / b}=10$. But $\pi^{2} \approx 3.14^{2}2$, which is what we needed to prove.
b) Let $\log _{2} \pi=a, \log _{\pi} 2=b$. In th... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,546 |
287. Prove that if $\alpha$ and $\beta$ are acute angles and $\alpha < \beta$, then
a) $\alpha - \sin \alpha < \beta - \sin \beta$,
b) $\operatorname{tg} \alpha - \alpha < \operatorname{tg} \beta - \beta$. | 287. First solution. It is required to prove that if $\beta > \alpha$, then
a) $\sin \beta - \sin \alpha < \beta - \alpha$.
But, obviously, $\beta - \alpha > x^{1}$.
Second solution. Here we will consider only part a), as the solution to part b) is completely analogous.
Draw a unit circle with center at point $O$, a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,547 |
288*. Prove that if $\alpha$ and $\beta$ are acute angles and $\alpha<\beta$, then $\frac{\operatorname{tg} \alpha}{\alpha} > \frac{\operatorname{tg} \beta}{\beta}$. | 288. Let $A E$ and $A F$ be arcs of a unit circle with center at point $O$, equal to $\alpha$ and $\beta$ respectively, $B$ and $C$ - the points of intersection of the perpendicular, erected at point $A$ to the diameter $O A$, with the lines $O E$ and $O F$, $M$ and $N$ - the points of intersection of the perpendicular... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 29,548 |
289. Find the relationship between $\arcsin \cos \arcsin x$ and $\arccos \sin \arccos x$ | 289. Let
$$
\arcsin \cos \operatorname{arsin} x=\alpha
$$
The angle $\alpha$ is within the range $0 \leqslant \alpha \leqslant \frac{\pi}{2}$, because $0 \leqslant \cos \arcsin x \leqslant 1$ (since $-\frac{\pi}{2} \leqslant \arcsin x \leqslant \frac{\pi}{2}$). Further, $\sin \alpha=\cos \arcsin x$; therefore,
$$
\a... | \alpha+\beta=\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,549 |
290. Prove that regardless of the coefficients $a_{31}, a_{30}, \ldots, a_{2}, a_{1}$, the sum
$\cos 32 x + a_{31} \cos 31 x + a_{30} \cos 30 x + \ldots + a_{2} \cos 2 x + a_{1} \cos x$ cannot take only positive values for all $x$. | 290. Suppose that the sum $\cos 32 x+a_{31} \cos 31 x+a_{30} \cos 30 x+a_{29} \cos 29 x+\ldots$
$$
\ldots+a_{2} \cos 2 x+a_{1} \cos x
$$
takes only positive values for all values of $x$. Replace $x$ with $x+\pi$ in this sum; we will arrive at the expression
$$
\begin{array}{r}
\cos 32(x+\pi)+a_{31} \cos 31(x+\pi)+a_... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,550 |
291. Let some of the numbers $a_{1}, a_{2}, \ldots, a_{n}$ be +1, and the others be -1. Prove that
$$
\begin{aligned}
2 \sin \left(a_{1}+\frac{a_{1} a_{2}}{2}\right. & \left.+\frac{a_{1} a_{2} a_{3}}{4}+\ldots+\frac{a_{1} a_{2} \ldots a_{n}}{2^{n-1}}\right) 45^{\circ}= \\
& =a_{1} \sqrt{2+a_{2} \sqrt{2+a_{3} \sqrt{2+\... | 291. We will use the formula for the sine of a half-angle:
$$
2 \sin \frac{\alpha}{2}= \pm \sqrt{2-2 \cos \alpha}
$$
where the sign of the root is taken according to the known rule of signs for the sine. Using this formula, we will sequentially determine the sines of the angles
$$
\begin{aligned}
& a_{1} 45^{\circ} ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,551 |
293. In which of the expressions
$$
\left(1+x^{2}-x^{3}\right)^{1000} \text { and }\left(1-x^{2}+x^{3}\right)^{1000}
$$
will the coefficient of \(x^{20}\) be greater after expanding the brackets and combining like terms? | 293. By expanding the brackets and combining like terms in the two expressions under consideration, we will obtain two polynomials in terms of $x$. Now let's replace $x$ with $-x$ in these expressions. This means we will also have to replace $x$ with $-x$ in the resulting polynomials, i.e., keep the same coefficients f... | (1+x^{2}-x^{3})^{1000} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,552 |
294. Prove that in the product
$$
\begin{aligned}
\left(1-x+x^{2}-x^{3}+\ldots-x^{99}+x^{100}\right) & \times \\
& \times\left(1+x+x^{2}+\ldots+x^{99}+x^{100}\right)
\end{aligned}
$$
after expanding the brackets and combining like terms, there will be no terms containing $\boldsymbol{x}$ in an odd power. | 294. The statement of the problem directly follows from the following transformations:
$$
\begin{gathered}
\left(1-x+x^{2}-x^{3}+\ldots-x^{99}+x^{100}\right)\left(1+x+x^{2}+x^{3}+\ldots\right. \\
\left.\ldots+x^{99}+x^{100}\right)=\left[\left(1+x^{2}+x^{4}+\ldots+x^{100}\right)-\right. \\
\left.-x\left(1+x^{2}+x^{4}+\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,553 |
295. Find the coefficient of $x^{50}$ after expanding the brackets and combining like terms in the expressions:
a) $(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\ldots+x^{1000}$;
b) $(1+x)+2(1+x)^{2}+3(1+x)^{3}+\ldots+1000(1+x)^{1000}$. | 295. a) According to the formula for the sum of a geometric progression and the binomial formula of Newton, we have:
$(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+\ldots+x^{1000}=$
$$
\begin{gathered}
=\frac{\frac{x^{1001}}{1+x}-(1+x)^{1000}}{\frac{x}{1+x}-1}=\frac{x^{1001}-(1+x)^{1001}}{x-1-x}=(1+x)^{1001}-x^{1001}= \... | C_{1001}^{50} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,554 |
296 *. Determine the coefficient of $x^{2}$ after expanding the brackets and combining like terms in the expression
$$
\underbrace{\left(\ldots\left(\left((x-2)^{2}-2\right)^{n}-2\right)^{2}-\ldots-2\right)^{2}}_{k \text { times }}
$$ | 296. First, let's find the free term, which will result from the expression
$$
\underbrace{\left.\left(\ldots(x-2)^{2}-2\right)^{2}-\ldots-2\right)^{2}}_{k}
$$
if we expand the brackets and combine like terms. It is equal to the value of this expression at \( x=0 \), i.e.,
$$
\begin{aligned}
& \underbrace{\left(\ldo... | \frac{4^{2k-1}-4^{k-1}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,555 |
297. Find the remainder of the division of the polynomial
$$
x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}
$$
a) by $x-1$; b) by $x^{2}-1$. | 297. a) First solution. Since for any non-negative integer $k$ the binomial $x^{k}-1$ is divisible by $x-1$, then
$$
\begin{aligned}
& x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=(x-1)+\left(x^{3}-1\right)+ \\
&+\left(x^{9}-1\right)+\left(x^{27}-1\right)+\left(x^{81}-1\right)+\left(x^{243}-1\right)+6
\end{aligned}
$$
gives a... | 6x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,556 |
298. An unknown polynomial gives a remainder of 2 when divided by $x-1$, and a remainder of 1 when divided by $x-2$. What remainder does this polynomial give when divided by $(x-1)(x-2)$? | 298. Let $p(x)$ be our unknown polynomial, $q(x)$ the quotient from dividing this polynomial by $(x-1)(x-2)$, and $r(x)=a x+b$ the sought remainder:
$$
p(x)=(x-1)(x-2) q(x)+a x+b
$$
According to the problem, we have:
$$
\begin{aligned}
& p(x)=(x-1) q_{1}(x)+2, \text { hence } p(1)=2 \\
& p(x)=(x-2) q_{2}(x)+1, \text... | -x+3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,557 |
299. When dividing the polynomial $x^{1051}-1$ by $x^{4}+x^{3}+2 x^{2}+x+1$, a quotient and a remainder are obtained. Find the coefficient of $x^{14}$ in the quotient. | 299. The polynomial $x^{4}+x^{8}+2 x^{2}+x+1$ can be factored; it equals $\left(x^{2}+1\right)\left(x^{2}+x+1\right)$. From this, it is easy to see that this polynomial is a divisor of the polynomial
$$
x^{12}-1=\left(x^{6}-1\right)\left(x^{6}+1\right)=\left(x^{3}-1\right)\left(x^{3}+1\right)\left(x^{2}+1\right)\left(... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,558 |
300. Find all polynomials $P(x)$ for which the identity
$$
x P(x-1) \equiv (x-26) P(x)
$$
holds. | 300. From the identity given in the problem, it follows that the sought polynomial $P(x)=P_{n}(x)$ (where $n$ is the degree of the polynomial) is divisible by $x$, i.e., $P_{n}(x)=x P_{n-1}(x)$, where $P_{n-1}(x)$ is some new polynomial of degree $n-1$. Therefore,
$$
P(x-1)=(x-1) P_{n-1}(x-1)
$$
and, hence,
$$
x(x-1... | P(x)=(x-1)(x-2)\ldots(x-25) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,559 |
301. Given a polynomial $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+$ $4 a_{n} c$
a) natural; b) integer
coefficients $a_{0}, a_{1}, \ldots, a_{n}$; the sum of the digits of the number $P(n)$ is denoted by $s(n)$ (it is clear that the value $s(n)$ makes sense only if the number $P(n)$ is natural; otherwise, $s(n... | 301. a) If all coefficients of $P(x)$ are non-negative, then all numbers $s(1), s(2), s(3), \ldots$ make sense. Consider such a power of ten $N=10^{k}$ that $N$ is greater than all the coefficients $a_{0}, a_{1}, a_{2}, \ldots$ of the polynomial $P(x)$. Then the number $P(N)=P\left(10^{k}\right)$, obviously, starts wit... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,560 |
302. Prove that the polynomial $x^{200} y^{200}+1$ cannot be represented as a product $f(x) \cdot g(y)$ of two polynomials: one in the variable $x$ and one in the variable $y$. | 302. Let the polynomials $f(x)$ and $g(y)$ have free terms $a_{0}$ (i.e., $f(x) \equiv a_{0} + a_{1} x + \ldots + a_{n} x^{n}$) and $b_{0}$. Set the variable $x$ to 0 in the equation $x^{200} y^{200} + 1 \equiv f(x) g(y)$; then we get $a_{0} g(y) \equiv 1$, i.e., $g(y) = \frac{1}{a_{0}}$; thus, $g(y)$ equals $\frac{1}{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,561 |
303. The quadratic trinomial $p(x)=a x^{2}+b x+c$ is such that the equation $p(x)=x$ has no (real) roots. Prove that then the equation $p(p(x))=x$ also has no real roots. | 303. Since the (quadratic) equation $p(x)=a x^{2}+b x+c=x$ has no real roots, the quadratic trinomial $p(x)-x=$ $=a x^{2}+(b-1) x+c$ takes values of the same sign for all $x$, say, $p(x)-x>0$ for all $x$. In this case, for any $x=x_{0}$ we have $p\left(p\left(x_{0}\right)\right)-p\left(x_{0}\right)>0$, i.e., $p\left(p\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,562 |
304. The quadratic trinomial $p(x)=a x^{2}+b x+c$ is such that if $|x| \leqslant 1$, then $|p(x)| \leqslant 1$. Prove that in this case, from $|x| \leqslant 1$ it also follows that $\left|p_{1}(x)\right| \leqslant 2$, where $p_{1}(x)=$ $=c x^{2}+b x+a$. | 304. We will assume that $a \geqslant 0$ - otherwise, we will replace the polynomial $p(x)$ with (satisfying the same conditions) the polynomial $-p(x)=-a x^{2}-b x-c$. Similarly, we will assume that $b \geqslant 0$ - otherwise, we will replace $p(x)$ with $p(-x)=a x^{2}-$ $-b x+c$. Substituting now the values $x=1, x=... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,563 |
305. Prove that if $x_{1}$ is a root of the equation
$$
a x^{2}+b x+c=0
$$
and $x_{2}$ is a root of the equation
$$
-a x^{2}+b x+c=0
$$
then there exists an intermediate root $x_{3}$ of the equation
$$
\frac{a}{2} x^{2}+b x+c=0
$$
such that either $x_{1} \leqslant x_{3} \leqslant x_{2}$ or $x_{1} \geqslant x_{3} ... | 305. We will exclude from consideration the less interesting case $a=0$, when the three equations (1), (2), and (3) are linear equations, i.e., each has a unique root, and all coincide with each other (here $x_{1}=x_{2}=x_{3}$), as well as the case when, say, $c=0$ and our three equations have roots $\left(-\frac{b}{a}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,564 |
306. Let $\alpha$ and $\beta$ be the roots of the equation
$$
x^{2} + p x + q = 0
$$
and let $\gamma$ and $\delta$ be the roots of the equation
$$
x^{2} + P x + Q = 0
$$
Express the product
$$
(\alpha-\gamma)(\beta-\gamma) \quad, (\alpha-\delta)(\beta-\delta)
$$
in terms of the coefficients of the given equations... | 306. If $\alpha$ and $\beta$ are the roots of the equation
$$
x^{2}+p x+q=0
$$
then
$$
(x-\alpha)(x-\beta)=x^{2}+p x+q
$$
Therefore,
$$
\begin{aligned}
& (\alpha-\gamma)(\beta-\gamma)(\alpha-\delta)(\beta-\delta)= \\
& =[(\gamma-\alpha)(\gamma-\beta)][(\delta-\alpha)(\delta-\beta)]=\left(\gamma^{2}+p \gamma+q\righ... | Q^{2}+q^{2}-pP(Q+q)+^{2}+p^{2}Q-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,565 |
307. Given two equations:
$$
x^{2}+a x+1=0, \quad x^{2}+x+a=0
$$
Determine all values of the coefficient $a$, for which these equations have at least one common root. | 307. First solution. Let's find the coefficient $a$ from the second equation and substitute it into the first equation. Then we get:
$$
\begin{array}{r}
a=-(x^{2}+x) \\
x^{2}-(x^{2}+x) x+1=0 \\
x^{3}-1=0 \\
(x-1)\left(x^{2}+x+1\right)=0
\end{array}
$$
From this, we have:
$$
x_{1}=1, \quad x_{2,3}=\frac{-1 \pm i \sqr... | a_{1}=-2,a_{2,3}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,566 |
308. a) Find an integer \( a \) such that
\[
(x-a)(x-10)+1
\]
can be factored into the product \((x+b)(x+c)\) of two factors with integer \( b \) and \( c \).
b) Find such non-zero, distinct integers \( a, b, c \) that the polynomial of the fourth degree with integer coefficients
\[
x(x-a)(x-b)(x-c)+1
\]
can be re... | 308. a) Let $(x-a)(x-10)+1=(x+b)(x+c)$. By substituting $x=-b$ into both sides of the equation, we get:
$$
(-b-a)(-b-10)+1=(-b+b)(-b+c)=0
$$
From this,
$$
(b+a)(b+10)=-1
$$
Since $a$ and $b$ are integers, $b+a$ and $b+10$ are also integers. However, -1 can be represented as the product of two integers in only one w... | =8or=12,\quad(x-8)(x-10)+1=(x-9)^2,\quad(x-12)(x-10)+1=(x-11)^2,\quad=3,b=1,=2,\quad=-3,b=-1,=-2,\quad=1,b=2,=-1, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,567 |
309. For which distinct integers $a_{1}, a_{2}, \ldots, a_{n}$ do the polynomials with integer coefficients
a) $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right) \ldots\left(x-a_{n}\right)-1$,
b) $\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right) \ldots\left(x-a_{n}\right)+1$ factor into product... | 309. a) Suppose that
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right) \ldots\left(x-a_{n}\right)-1=p(x) q(x)
$$
where $p(x)$ and $q(x)$ are polynomials with integer coefficients, the sum of whose degrees is $n$; we can assume that the leading coefficients of both these polynomials are 1 (compare with t... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,568 |
310*. Prove that for any distinct integers $a_{1}, a_{2}, \ldots, a_{n}$, the polynomial
$$
\left(x-a_{1}\right)^{2}\left(x-a_{2}\right)^{2} \ldots\left(x-a_{n}\right)^{2}+1
$$
cannot be factored into the product of two other polynomials with integer coefficients. | 310. Similarly to the solution of the previous problem from the supposed equality
$$
\left(x-a_{1}\right)^{2}\left(x-a_{2}\right)^{2}\left(x-a_{3}\right)^{2} \ldots\left(x-a_{n}\right)^{2}+1=p(x) q(x)
$$
where $p(x), q(x)$ are some polynomials with integer coefficients (and with leading coefficients equal to 1), it f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,569 |
311. Prove that if the polynomial
$$
P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}
$$
with integer coefficients takes the value 7 at four integer values of $x$, then it cannot take the value 14 at any integer value of $x$. | 311. Let the polynomial $P(x)$ equal 7 when $x=a, x=b, x=c,$ and $x=d$. In this case, the equation $P(x)-7=0$ has four integer roots $a, b, c,$ and $d$. This means that the polynomial $P(x)-7$ is divisible by $x-a, x-b, x-c, x-d^{1}$, i.e.,
$$
P(x)-7=(x-a)(x-b)(x-c)(x-d) p(x)
$$
where $p(x)$ can equal 1.
Now suppose... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,570 |
313. Prove that if a polynomial with integer coefficients
$$
P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}
$$
takes odd values at $x=0$ and $x=1$, then the equation $P(x)=0$ has no integer roots. | 313. Let $p$ and $q$ be two integers, both even or both odd. Then the difference $P(p)-P(q)$ is even. Indeed, the expression
$$
\begin{gathered}
P(p)-P(q)=a_{0}\left(p^{n}-q^{n}\right)+a_{1}\left(p^{n-1}-q^{n-1}\right)+\ldots \\
\quad \ldots+a_{n-2}\left(p^{2}-q^{2}\right)+a_{n-1}(p-q)
\end{gathered}
$$
is divisible ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,571 |
314*. Prove that if the polynomial
$$
P(x)=a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n}
$$
with integer coefficients equals 1 in absolute value at two integer values $x=p$ and $x=q(p>q)$, and the equation $P(x)=0$ has a rational root $a$, then $p-q$ equals 1 or 2 and $a=\frac{p+q}{2}$. | 314. Suppose the equation $P(x)=0$ has a rational root $x=\frac{k}{l}$, i.e., $P\left(\frac{k}{l}\right)=0$. Let's expand the polynomial $P(x)$ in powers of $x-p$, i.e., write it in the form
$$
\begin{aligned}
& P(x)=\dot{c}_{0}(x-p)^{n}+c_{1}(x-p)^{n-1}+ \\
& \quad+c_{2}(x-p)^{n-2}+\ldots+c_{n-1}(x-p)+c_{n}
\end{alig... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,572 |
315*. Prove that the polynomials
a) $x^{2222}+2 x^{2220}+4 x^{2218}+6 x^{2216}+8 x^{2214}+\ldots$
$$
\ldots+2218 x^{4}+2220 x^{2}+2222
$$
b) $x^{250}+x^{249}+x^{248}+x^{247}+x^{240}+\ldots+x^{2}+x+1$
cannot be factored into polynomials with integer coefficients. | 315. a) Suppose that our polynomial can be factored into factors with integer coefficients:
$$
\begin{aligned}
& x^{2222}+2 x^{2220}+4 x^{2218}+\ldots+2220 x^{2}+2222= \\
&=\left(a_{n} x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots+a_{0}\right) \times \\
& \times\left(b_{m} x^{m}+b_{m-1} x^{m-1}+b_{m-2} x^{m-2}+\ldots+b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,573 |
316. Prove that if the product of two polynomials with integer coefficients is a polynomial with even coefficients, not all of which are divisible by 4, then in one of the multiplied polynomials all coefficients are even, and in the other not all are even. | 316. Let's write the polynomials in the form
$$
A=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}
$$
and
$$
B=b_{0}+b_{1} x+b_{2} x^{2}+\ldots+b_{m} x^{m}
$$
Since by the condition not all coefficients in the product are divisible by 4, the coefficients of both polynomials cannot all be even. Therefore, at least one o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,574 |
317. Prove that all rational roots of the polynomial
$$
P(x)=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n}
$$
with integer coefficients and with the coefficient of the leading term $x$ equal to 1, are integers. | 317. We will prove that for any rational but not integer value of $x$, the polynomial $P(x)$ cannot be an integer, and therefore cannot be zero, since zero is an integer.
Let $x=\frac{p}{q}$, where $p$ and $q$ are coprime. Then
\[
\begin{aligned}
P(x) & =x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+a_{n-1} x+a_{n}^{\prim... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,575 |
318*. Prove that there does not exist a polynomial
$$
P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}
$$
such that all numbers $P(0), P(1), P(2), \ldots$, are prime.
Note. The statement formulated in the problem was first proved by L. Euler¹). He also provided examples of polynomials whose values for many cons... | 318. Let $N$ be some integer and $P(N)=M$. For any integer $k$
\[
\begin{aligned}
P(N+k M) & -P(N)=a_{0}\left[(N+k M)^{n}-N^{n}\right]+ \\
& +a_{1}\left[(N+k M)^{n-1}-N^{n-1}\right]+\ldots+a_{n-1}[(N+k M)-N]
\end{aligned}
\]
is divisible by $k M$ (since $(N+k M)^{l}-N^{l}$ is divisible by $(N+k M)-N=$ $=k M$), and th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,576 |
319. Prove that if the polynomial
$$
P(x)=x^{n}+A_{1} x^{n-1}+A_{2} x^{n-2}+\ldots+A_{n-1} x+A_{n}
$$
takes integer values for all integer values of $x$, then it can be represented as a sum of polynomials
$$
\begin{aligned}
P_{0}(x)=1, P_{1}(x)=x, P_{2}(x) & =\frac{x(x-1)}{1 \cdot 2}, \ldots \\
\ldots, P_{n}(x) & =\... | 319. First of all, note that every polynomial $P(x)$ of degree $n$ can be represented as a sum of polynomials
$$
\begin{gathered}
P_{0}(x)=1, \quad P_{1}(x)=x \\
P_{2}(x)=\frac{x(x-1)}{1 \cdot 2}, \ldots, P_{n}(x)=\frac{x(x-1)(x-2) \ldots(x-n+1)}{1 \cdot 2 \cdot 3 \ldots n}
\end{gathered}
$$
taken with certain coeffi... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,577 |
320. a) Prove that if a polynomial $P(x)$ of degree $n$ takes integer values at $x=0,1,2, \ldots, n$, then it takes integer values for all integer values of $x$.[^4]b) Prove that any polynomial of degree $n$, which takes integer values at some $n+1$ consecutive integer values of $x$, takes integer values for any intege... | 320. a) From the solution of problem 319, it follows that a similar polynomial can be represented as the sum of polynomials $P_{0}(x), P_{1}(x), \ldots, P_{n}(x)$ with integer coefficients. From this and the fact that the polynomials $P_{0}(x), P_{1}(x), \ldots, P_{n}(x)$ take integer values for every integer $x$ (see ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,578 |
321. a) Prove that
$$
\begin{aligned}
& \cos 5 \alpha=\cos ^{5} \alpha-10 \cos ^{3} \alpha \sin ^{2} \alpha+5 \cos \alpha \sin ^{4} \alpha \\
& \sin 5 \alpha=\sin ^{5} \alpha-10 \sin ^{3} \alpha \cos ^{2} \alpha+5 \sin \alpha \cos ^{4} \alpha
\end{aligned}
$$
b) Prove that for each integer $n$
$$
\begin{array}{r}
\o... | 321. a) Using de Moivre's formula and the binomial formula of Newton, we have:
$$
\begin{aligned}
& \cos 5 \alpha+i \sin 5 \alpha=(\cos \alpha+i \sin \alpha)^{5}= \\
& =\cos ^{5} \alpha+5 \cos ^{4} \alpha i \sin \alpha+10 \cos ^{3} \alpha(i \sin \alpha)^{2}+ \\
& \quad+10 \cos ^{2} \alpha(i \sin \alpha)^{3}+5 \cos \al... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,579 |
322. Express $\operatorname{tg} 6 \alpha$ in terms of $\operatorname{tg} \alpha$. | 322. According to the formulas of problem 321 b), we have:
$$
\operatorname{tg} 6 \alpha=\frac{\sin 6 \alpha}{\cos 6 \alpha}=\frac{6 \cos ^{5} \alpha \sin \alpha-20 \cos ^{3} \alpha \sin ^{3} \alpha+6 \cos \alpha \sin ^{5} \alpha}{\cos ^{6} \alpha-15 \cos ^{4} \alpha \sin ^{2} \alpha+15 \cos ^{2} \alpha \sin ^{4} \alp... | \frac{6\operatorname{tg}\alpha-20\operatorname{tg}^{3}\alpha+6\operatorname{tg}^{5}\alpha}{1-15\operatorname{tg}^{2}\alpha+15\operatorname{tg}^{4}\alpha-\operatorname{tg}^{6}\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,580 |
323. Prove that if $x+\frac{1}{x}=2 \cos \alpha$, then $x^{n}+\frac{1}{x^{n}}=$ $=2 \cos n \alpha$. | 323. The equation $x+\frac{1}{x}=2 \cos \alpha$ can be rewritten as
$$
x^{2}+1=2 x \cos \alpha, \text { or } x^{2}-2 x \cos \alpha+1=0
$$
Thus,
$$
x=\cos \alpha \pm \sqrt{\cos ^{2} \alpha-1}=\cos \alpha \pm i \sin \alpha
$$
From this, it follows that
$$
\frac{1}{x^{n}}=\frac{1}{\cos n \alpha \pm i \sin n \alpha}=\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,581 |
325. Simplify
$$
\cos ^{2} \alpha+\cos ^{2} 2 \alpha+\ldots+\cos ^{2} n \alpha
$$
and
$$
\sin ^{2} \alpha+\sin ^{2} 2 \alpha+\ldots+\sin ^{2} n \alpha
$$ | 325. Let's use the fact that $\cos ^{2} x=\frac{1+\cos 2 x}{2}$. From this, using the result of the previous problem, we get:
$$
\begin{aligned}
& \cos ^{2} \alpha+\cos ^{2} 2 \alpha+\ldots+\cos ^{2} n \alpha= \\
& \quad=\frac{1}{2}[\cos 2 \alpha+\cos 4 \alpha+\ldots+\cos 2 n \alpha+n]= \\
& =\frac{1}{2}\left[\frac{\s... | \frac{\sin(n+1)\alpha\cosn\alpha}{2\sin\alpha}+\frac{n-1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,582 |
326. Simplify
$$
\begin{aligned}
\cos \alpha+C_{n}^{1} \cos 2 \alpha+C_{n}^{2} \cos 3 \alpha+\ldots+C_{n}^{n-1} & \cos n \alpha+ \\
& +\cos (n+1) \alpha
\end{aligned}
$$
and
$\sin \alpha+C_{n}^{1} \sin 2 \alpha+C_{n}^{2} \sin 3 \alpha+\ldots+C_{n}^{n-1} \sin n \alpha+$ $+\sin (n+1) \alpha$ | 326. It is required to compute the real part and the coefficient of the imaginary part of the sum
$(\cos \alpha+i \sin \alpha)+C_{n}^{1}(\cos 2 \alpha+i \sin 2 \alpha)+$
$$
+C_{n}^{2}(\cos 3 \alpha+i \sin 3 \alpha)+\ldots+(\cos (n+1) \alpha+i \sin (n+1) \alpha)
$$
Denoting $\cos \alpha+i \sin \alpha$ by $x$ and usin... | \begin{aligned}\cos\alpha+C_{n}^{1}\cos2\alpha+C_{n}^{2}\cos3\alpha+\ldots+\cos(n+1)\alpha&=2^{n}\cos^{n}\frac{\alpha}{2}\cos\frac{n+2}{2}\alpha\\\sin\alpha+C_{n}^{1}\sin2\alpha+C | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,583 |
327. Prove that if $m, n, p$ are arbitrary integers, then
$\sin \frac{m \pi}{p} \sin \frac{n \pi}{p} + \sin \frac{2 m \pi}{p} \sin \frac{2 n \pi}{p} + \sin \frac{3 m \pi}{p} \sin \frac{3 n \pi}{p} + \ldots$
$\ldots + \sin \frac{(p-1) m \pi}{p} \sin \frac{(p-1) n \pi}{p}=$
$$
=\left\{\begin{array}{l}
-\frac{p}{2}, \t... | 327. Let's use the formula
$$
\sin A \sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]
$$
From this, it follows that our sum can be represented as
$$
\begin{gathered}
\frac{1}{2}\left[\cos \frac{(m-n) \pi}{p}+\cos \frac{2(m-n) \pi}{p}+\cos \frac{3(m-n) \pi}{p}+\ldots\right. \\
\left.\ldots+\cos \frac{(p-1)(m-n) \pi}{p}\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,584 |
329. Form the equation whose roots are the numbers:
a) $\sin ^{2} \frac{\pi}{2 n+1}, \sin ^{2} \frac{2 \pi}{2 n+1}, \sin ^{2} \frac{3 \pi}{2 n+1}, \ldots, \sin ^{2} \frac{n \pi}{2 n+1}$;
b) $\operatorname{ctg}^{2} \frac{\pi}{2 n+1}, \operatorname{ctg}^{2} \frac{2 \pi}{2 n+1}, \operatorname{ctg}^{2} \frac{3 \pi}{2 n+1}... | 329. $\quad \cos \frac{2 \pi}{2 n+1}+i \sin \frac{2 \pi}{2 n+1}, \quad \cos \frac{4 \pi}{2 n+1}+i \sin \frac{4 \pi}{2 n+1}, \ldots$
$$
\ldots, \quad \cos \frac{4 n \pi}{2 n+1}+i \sin \frac{4 n \pi}{2 n+1}
$$
Since the coefficient of $x^{2 n}$ in the equation is zero, the sum of all these roots is zero:
$\left[1+\cos... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,586 |
330. Simplify the sums:
a) $\operatorname{ctg}^{2} \frac{\pi}{2 n+1}+\operatorname{ctg}^{2} \frac{2 \pi}{2 n+1}+\operatorname{ctg}^{2} \frac{3 \pi}{2 n+1}+\ldots+\operatorname{ctg}^{2} \frac{n \pi}{2 n+1}$;
b) $\operatorname{cosec}^{2} \frac{\pi}{2 n+1}+\operatorname{cosec}^{2} \frac{2 \pi}{2 n+1}+\operatorname{cosec... | 330. a) The sum of the roots of the equation of degree $n$
$$
x^{n}-\frac{C_{2 n+1}^{3}}{C_{2 n+1}^{1}} x^{n-1}+\frac{C_{2 n+1}^{5}}{C_{2 n+1}^{1}} x^{n-2}-\ldots=0
$$
(see the solution of problem 229 b)) is equal to the coefficient of $x^{n-1}$, taken with the opposite sign, i.e.
$\operatorname{ctg}^{2} \frac{\pi}{... | \frac{2n(n+1)}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,587 |
331. Simplify the products:
a) $\sin \frac{\pi}{2 n+1} \sin \frac{2 \pi}{2 n+1} \sin \frac{3 \pi}{2 n+1} \ldots \sin \frac{n \pi}{2 n+1}$
and
$$
\sin \frac{\pi}{2 n} \sin \frac{2 \pi}{2 n} \sin \frac{3 \pi}{2 n} \ldots \sin \frac{(n-1) \pi}{2 n}
$$
b) $\cos \frac{\pi}{2 n+1} \cos \frac{2 \pi}{2 n+1} \cos \frac{3 \p... | 331. a) First solution. The numbers $\sin ^{2} \frac{\pi}{2 n+1}, \quad \sin ^{2} \frac{2 \pi}{2 n+1}, \cdots$ $\ldots \sin ^{2} \frac{n_{\pi}}{2 n+1}$ are the roots of the equation of degree $n$, obtained in the solution of problem 329a). The coefficient of the leading term $x^{n}$ of this equation is equal to $(-1)^{... | \sin\frac{\pi}{2n+1}\sin\frac{2\pi}{2n+1}\cdots\sin\frac{n\pi}{2n+1}=\frac{\sqrt{2n+1}}{2^{n}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,588 |
332. From the results of problems 330 a) and b), deduce that for any positive integer $n$ the sum
$$
1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}
$$
is bounded between $\left(1-\frac{1}{2 n+1}\right)\left(1-\frac{2}{2 n+1}\right) \frac{\pi^{2}}{6}$ and $\left(1-\frac{1}{2 n+1}\right)\left(1+\frac{1}{2 n+1... | 332. Let's show that for any positive angle less than $\frac{\pi}{2}$,
$$
\sin \alpha < \alpha < \operatorname{tg} \alpha
$$
We have (Fig. 39):
$$
\begin{gathered}
S_{\triangle A O B} = \frac{1}{2} \sin \alpha \\
S_{\mathrm{Ce} T T A O B} = \frac{1}{2} \alpha \\
S_{\triangle A O C} = \frac{1}{2} \operatorname{tg} \a... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,589 |
333. a) On the circle circumscribed around a regular $n$-gon $A_{1} A_{2} \ldots A_{n}$, a point $M$ is taken. Prove that the sum of the squares of the distances from this point to all vertices of the $n$-gon does not depend on the position of the point on the circle and is equal to $2 n R^{2}$, where $R$ is the radius... | 333. a) Suppose that point $M$ is taken on the arc $A_{1} A_{\text {n }}$ of a circle (Fig. 40). Denote the arc $M A_{1}$ by $\alpha$; in this case, the arcs
$M A_{2}, M A_{3}, \ldots, M A_{n}$ are respectively equal to
$$
\alpha+\frac{2 \pi}{n}, \alpha+\frac{4 \pi}{n}, \ldots, \alpha+\frac{2(n-1) \pi}{n}
$$
But the ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,590 |
334. On the arc $A_{1} A_{n}$ of the circle circumscribed around the regular $n$-gon $A_{1} A_{2} \ldots A_{n}$, a point $M$ is taken. Prove that:
a) if $n$ is even, then the sum of the squares of the distances from point $M$ to the vertices of the $n$-gon with even indices is equal to the sum of the squares of the di... | 334. a) The statement of the problem immediately follows from the theorem of problem 333a), if we take into account that for even $n$, the even (and odd) vertices of the $n$-gon themselves serve as vertices of inscribed regular $\frac{n}{2}$-gons.
b) Let $n=2 m+1$. From the solution of problem $333 \mathrm{a}$), we de... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,591 |
335. The radius of the circle circumscribed around a regular $n$-gon $A_{1} A_{2} \ldots A_{n}$ is $R$. Prove that:
a) the sum of the squares of all sides and all diagonals of the $n$-gon is $n^{2} R^{2}$;
b) the sum of all sides and all diagonals of the $n$-gon is $n \operatorname{ctg} \frac{\pi}{2 n} R$
c) the pro... | 335. a) By problem 333 a), the sum of the squares of the distances from a point on the circumcircle of a regular $n$-gon to all its vertices is $2 n R^{2}$. Assuming that $M$ coincides with $A_{1}$, we obtain that the sum of all sides and diagonals of the $n$-gon emanating from one vertex is $2 n R^{2}$. If we multiply... | proof | Geometry | proof | Yes | Yes | olympiads | false | 29,592 |
337. It is known that $\left|z+\frac{1}{z}\right|=a$; what are the greatest and the least values that the modulus $|z|$ of the complex number $z$ can have? | 337. Since $|z|=|\bar{z}|=|-z|=|-\bar{z}|$ and $\left|z+\frac{1}{z}\right|=\left|\bar{z}+\frac{1}{\bar{z}}\right|=$ $\left|-z-\frac{1}{z}\right|=\left|-\bar{z}-\frac{1}{\bar{z}}\right|$, it is sufficient to consider only one of the numbers $\bar{z}, \bar{z}, -z$ and $-\bar{z}$ that lies in the first quadrant. If $|z|$ ... | \frac{1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 29,593 |
338. Let the sum of $n$ complex numbers be zero; prove that among them there are two numbers, the arguments of which differ by at least $120^{\circ}$.
Can the value $120^{\circ}$ be replaced by a smaller one here? | 338. It is clear that the three complex numbers $1(=1+i \cdot 0=\cos 0 + i \sin 0)$, $\frac{-1+\sqrt{3} i}{2}\left(=\cos 120^{\circ}+i \sin 120^{\circ}\right)$, and $\frac{-1-\sqrt{3} i}{2}\left(=\cos \left(-120^{\circ}\right)+i \sin \left(-120^{\circ}\right)\right)$ are such that the arguments of any two of them diffe... | proof | Algebra | proof | Yes | Yes | olympiads | false | 29,594 |
339. Let $c_{1}, c_{2}, \ldots, c_{n}, z$ be complex numbers such that
$$
\frac{1}{z-c_{1}}+\frac{1}{z-c_{2}}+\ldots+\frac{1}{z-c_{n}}=0
$$
Prove that if the numbers $c_{1}, c_{2}, \ldots, c_{n}$ represent the vertices of a convex $n$-gon on the complex plane, then the number $z$ represents a point lying inside this ... | 339. Suppose the point $A$ in the complex plane corresponding to the complex number $z$ does not belong to the convex polygon $C_{1} C_{2} \ldots C_{n}$, whose vertices correspond to the numbers $c_{1}, c_{2}, \ldots, c_{n}$
 - d).
Translated as requested, maintaining the original text's line breaks and format. | 340. First solution. Let $a$ not be divisible by $p$. In this case, the numbers $a, 2a, 3a, \ldots, (p-1)a$ will also not be divisible by $p$ and will all give different remainders when divided by $p$: indeed, if $ka$ and $la$ (where $p-1 \geqslant k > l$) gave the same remainder when divided by $p$, then the differenc... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,596 |
341. Euler's Theorem. Let $N$ be some integer and $r$ be the number of integers in the sequence $1,2,3, \ldots, N-1$ that are coprime with $N$. Prove that if $a$ is any integer coprime with $N$, then the difference $a^{r}-1$ is divisible by $N$.
Note. If the number $N$ is prime, then all the listed numbers are coprime... | 341. The proof of Euler's theorem is entirely analogous to the first proof of Fermat's theorem; $r$ numbers less than $N$ and coprime with $N$ we denote by $k_{1}, k_{2}, k_{3}, \ldots, k_{r}$. Consider $r$ numbers $k_{1} a, k_{2} a, \ldots, k_{r} a$, all of which are coprime with $N$ (since $a$ is coprime with $N$ by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,597 |
342*. According to Euler's theorem, the difference $2^{k}-1$, where $k=5^{n}-5^{n-1}$, is divisible by $5^{n}$ (see problem 341, in particular, the note to this problem). Prove that for no $k$, less than $5^{n}-5^{n-1}$, the difference $2^{k}-1$ is divisible by $5^{n}$. | 342. We will prove this by mathematical induction. First of all, it is clear that the statement of the problem is true for \( n=1 \): \( 2^{1}-1=1 \), \( 2^{2}-1=3 \), and \( 2^{3}-1=7 \) are not divisible by 5. We will also prove this statement for \( n=2 \). Let \( 2^{k} \) be the smallest power of 2 that gives a rem... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,598 |
343. Let's write down the consecutive powers of the number 2:
$2,4,8,16,32,64,128,256,512,1024,2048,4096, \ldots$
It is easy to notice that the last digit in this sequence of numbers repeats periodically with a period of 4:
$$
2,4,8,6,2,4,8,6,2,4,8,6, \ldots
$$
Prove that the last 10 digits in this sequence of numb... | 343. By Euler's theorem (see problem 341), the number $2^{5^{50}-5^{0}}-1=$ $=24 \cdot 5^{9}-1=2^{7812500}-1$ is divisible by $5^{10}$; therefore, for $n \geqslant$ $\geqslant 10$ the difference $2^{7812500+n}-2^{n}=2^{n}\left(2^{7812500}-1\right)$ is divisible by $10^{10}$, i.e., the last 10 digits of the numbers $2^{... | 7812500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 29,599 |
344.*. Prove that there exists such a power of the number 2, the last 1000 digits of which will all be ones and twos. | 344. We will prove an even more general statement, namely, that for any integer $N$, there always exists a power of the number 2 whose last $N$ digits are all ones and twos. Since $2^{5}=32$ and $2^{9}=512$, the statement is true for $N=1$ and $N=2$. We will further prove this by mathematical induction. Suppose that th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 29,600 |
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