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10.26. Prove that a polynomial $P(x)$ of degree $n$ cannot have more than $n$ distinct roots.
- A number $a$ is called a root of multiplicity $k$ ( $k \geqslant 1$ ) of the polynomial $P(x)$, if $P(x)$ is divisible by $(x-a)^{k}$ and is not divisible by $(x-a)^{k+1}$. | 10.26. Let $a$ be a root of the polynomial $P(x)$. Divide $P(x)$ by $x-a$ with a remainder. As a result, we get $P(x)=(x-a) R(x)+b$. In this case, $b=P(a)=0$. Therefore, the polynomial $P(x)$ is divisible by $x-a$. If $a_{1}, \ldots, a_{m}$ are the roots of the polynomial $P(x)$, then it is divisible by $\left(x-a_{1}\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,409 |
10.27. Prove that a polynomial $P(x)$ of degree $n$ cannot have more than $n$ roots, taking into account their multiplicities, i.e., if $a_{1}, \ldots, a_{m}$ are distinct roots with multiplicities $k_{1}, \ldots, k_{m}$, then $k_{1}+\ldots+k_{m} \leqslant n$ | 10.27. The polynomial $P(x)$ is divisible by $\left(x-a_{1}\right)^{k_{1}} \ldots\left(x-a_{m}\right)^{k_{m}}$, therefore $k_{1}+\ldots+k_{m} \leqslant n$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,410 |
10.28. Prove that the equation
$$
x^{n}-a_{1} x^{n-1}-a_{2} x^{n-2}-\ldots-a_{n-1} x-a_{n}=0
$$
where $a_{1} \geqslant 0, a_{2} \geqslant 0, a_{n} \geqslant 0$, cannot have two positive roots. | 10.28. Let's rewrite the given equation as
$$
1=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\ldots+\frac{a_{n}}{x^{n}}
$$
For $x>0$, the function $f(x)=\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\ldots+\frac{a_{n}}{x^{n}}$ is monotonically decreasing, so it cannot take the value 1 at two different positive values of $x$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,411 |
10.29. Let $f_{1}(x)=x^{2}-2, f_{n}(x)=f_{1}\left(f_{n-1}(x)\right)$. Prove that for any natural number $n$ the equation $f_{n}(x)=x$ has exactly $2^{n}$ distinct solutions.
## 10.8. Various Problems | 10.29. The function $f_{1}(x)$ monotonically decreases from 2 to -2 on the interval $[-2,0]$ and monotonically increases from -2 to 2 on the interval $[0,2]$. Therefore, the equation $f_{1}(x)=x$ has a root $x_{1} \in(-2,0)$ and a root $x_{2}=2$. Moreover, the equation $f_{1}(x)=0$ has two roots $\pm x_{1}^{\prime}$. T... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,412 |
10.31. What remainder does $x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}$ give when divided by $(x-1) ?$ | 10.31. Answer: 6. Let $x+x^{3}+x^{9}+x^{27}+x^{81}+x^{243}=P(x) \times$ $\times(x-1)+r$. By setting $x=1$, we get $r=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,414 |
10.32. In which of the expressions:
$$
\left(1-x^{2}+x^{3}\right)^{1000}, \quad\left(1+x^{2}-x^{3}\right)^{1000}
$$
after expanding the brackets and combining like terms, is the coefficient of \(x^{20}\) greater? | 10.32. Answer: in the expression $\left(1+x^{2}-x^{3}\right)^{1000}$. Let $P(x)=$ $=\left(1-x^{2}+x^{3}\right)^{1000}$ and $Q(x)=\left(1+x^{2}-x^{3}\right)^{1000}$. The coefficient of $x^{20}$ in the polynomial $P(x)$ is the same as in $P(-x)=\left(1-x^{2}-x^{3}\right)^{1000}$, and the coefficient of $x^{20}$ in the po... | in\the\expression\(1+x^{2}-x^{3})^{1000} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,415 |
10.33. a) Find the integer $a$ for which $(x-a) \times$ $\times(x-10)+1$ can be factored into the product $(x+b)(x+c)$ of two factors with integer $b$ and $c$.
b) Find such non-zero and distinct integers $a, b, c$ that the expression $x(x-a)(x-b) \times$ $\times(x-c)+1$ can be factored into the product of two polynomi... | 10.33. a) Let $(x-a)(x-10)+1=(x+b)(x+c)$. By setting $x=-b$, we get $(b+a)(b+10)=-1$. Since $a$ and $b$ are integers, $b+a$ and $b+10$ are also integers. The number -1 can be represented as the product of two integers in two ways. Accordingly, we have two cases: 1) $b+10=1$ and $b+a=-1$; 2) $b+10=-1$ and $b+a=1$. There... | =8or=12,(,b,):(1,2,3),(-1,-2,-3),(1,-2,-1),(2,-1,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,416 |
10.34. Let $x_{1}, \ldots, x_{n+1}$ be pairwise distinct numbers. Prove that there exists a unique polynomial $P(x)$ of degree not higher than $n$, which takes the given value $a_{i}$ at the point $x_{i}$ (the Lagrange interpolation polynomial). | 10.34. The uniqueness of the polynomial $P$ follows from the fact that the difference of two such polynomials vanishes at the points $x_{1}, \ldots, x_{n+1}$ and has a degree not higher than $n$. It is also clear that the following polynomial has all the required properties:
$$
\begin{aligned}
& P(x)=\sum_{k=1}^{n+1} ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,417 |
10.35. Prove that $\sum_{k=0}^{n}(-1)^{k} k^{m} C_{n}^{k}=0$ for $m<n$ ($m$ is a natural number) and $\sum_{k=0}^{n}(-1)^{k} k^{n} C_{n}^{k}=(-1)^{n} n!$. | 10.35. Let $P(x)$ be the Lagrange interpolation polynomial of degree $\leqslant n$, taking the values $x^{m}$ at $x=0,1,2, \ldots, n$. Then
$$
P(x)=\sum_{k=0}^{n} k^{m} \frac{x(x-1) \ldots(x-k+1)(x-k-1) \ldots(x-n)}{k(k-1) \cdot \ldots \cdot 1 \cdot(-1) \cdot(-2) \cdot \ldots \cdot(k-n)}
$$
The coefficient of $x^{n}$... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,418 |
10.36. Let $a_{1}, \ldots, a_{n}$ be pairwise distinct numbers. Prove that for any $b_{0}, b_{1}, \ldots, b_{n-1}$ the system of linear equations
$$
\left\{\begin{array}{l}
x_{1}+\ldots+x_{n}=b_{0} \\
a_{1} x_{1}+\ldots+a_{n} x_{n}=b_{1} \\
a_{1}^{2} x_{1}+\ldots+a_{n}^{2} x_{n}=b_{2} \\
\quad \ldots \ldots \ldots \ld... | 10.36. Let $P_{i}(t)=p_{0 i}+p_{1 i} t+p_{2 i} t^{2}+\ldots+p_{n-1, i} t^{n-1}$ be the Lagrange interpolation polynomial, which takes the value 1 at $t=a_{i}$ and the value 0 at $t=a_{j}$, where $j \neq i$. Multiply these equations by $p_{0 i}, p_{1 i}, \ldots, p_{n-1, i}$ and add them. As a result, we get $P_{i}\left(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,419 |
10.37. Given points $x_{0}, x_{1}, \ldots, x_{n}$. Prove that the polynomial $f(x)$ of degree $n$, taking values $f\left(x_{0}\right), \ldots, f\left(x_{n}\right)$ at these points, can be represented in the form
$$
\begin{aligned}
& f(x)=f\left(x_{0}\right)+\left(x-x_{0}\right) f\left(x_{0} ; x_{1}\right)+ \\
& \quad+... | 10.37. The following equalities must be satisfied:
\[
\begin{aligned}
& f\left(x_{1}\right)=f\left(x_{0}\right)+\left(x_{1}-x_{0}\right) f\left(x_{0} ; x_{1}\right) \\
& f\left(x_{2}\right)=f\left(x_{0}\right)+\left(x_{2}-x_{0}\right) f\left(x_{0} ; x_{1}\right)+\left(x_{2}-x_{0}\right)\left(x_{2}-x_{1}\right) f\left(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,420 |
10.38. Let $R(x)=P(x) / Q(x)$, where $P$ and $Q$ are coprime polynomials. Prove that $R(x)$ can be represented as
$$
R(x)=A(x)+\sum_{i, k} \frac{c_{i k}}{\left(x-a_{i}\right)^{k}}
$$
where $c_{i k}$ are some numbers, and $A(x)$ is some polynomial. | 10.38. By dividing $P$ by $Q$ with a remainder, we can move to the fraction $S / Q$, where the degree of $S$ is less than the degree of $Q$. Let $Q=Q_{1} Q_{2}$, where $Q_{1}$ and $Q_{2}$ are coprime polynomials. Then we can choose polynomials $a(x)$ and $b(x)$ such that $a(x) Q_{1}(x)+b(x) Q_{2}(x)=1$. Therefore,
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,421 |
10.40. Prove that the polynomial
$$
C_{x}^{k}=\frac{x(x-1) \ldots(x-k+1)}{k!}
$$
is integer-valued. | 10.40. For $k=1$ this is obvious. Suppose now that the polynomial $C_{x}^{k}$ is integer-valued. It is easy to verify that
$$
C_{x+1}^{k+1}-C_{x}^{k+1}=C_{x}^{k}
$$
Therefore, for all integers $m, n$ the difference
$$
C_{m}^{k+1}-C_{n}^{k+1}
$$
is an integer. It remains to note that $C_{0}^{k+1}=0$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,423 |
10.41. Let $p_{k}(x)$ be a polynomial of degree $k$ that takes integer values at $x=n, n+1, \ldots, n+k$ for some integer $n$. Then
$$
p_{k}(x)=c_{0} C_{x}^{k}+c_{1} C_{x}^{k-1}+c_{2} C_{x}^{k-2}+\ldots+c_{k}
$$
where $c_{0}, c_{1}, \ldots, c_{k}$ are integers.
### 10.12. Polynomials in Several Variables
A polynomi... | 10.41. By induction on $k$, it is easy to prove that any polynomial $p_{k}(x)$ of degree $k$ can be represented as
$$
p_{k}(x)=c_{0} C_{x}^{k}+c_{1} C_{x}^{k-1}+c_{2} C_{x}^{k-2}+\ldots+c_{k}
$$
where $c_{0}, c_{1}, \ldots, c_{k}$ are some numbers. Indeed,
$$
C_{x}^{0}=1, \quad C_{x}^{1}=x, \quad C_{x}^{2}=\frac{x^{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,424 |
10.42. Prove that the polynomial of the form $x^{200} y^{200}+1$ cannot be represented as a product of polynomials in only $x$ and only $y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10.42. Suppose there are polynomials $f(x)=a_{0} x^{n}+$ $+a_{1} x^{n-1}+\ldots+a_{n}$ and $g(y)=b_{0} y^{m}+b_{1} y^{m-1}+\ldots+b_{m}$, for which $f(x) g(y)=x^{200} y^{200}+1$. Setting $x=0$, we get $a_{n} g(y)=1$, i.e., $g(y)=1 / a_{n}$ for all $y$. Setting $y=0$, we similarly get that $f(x)=1 / b_{m}$ for all $x$. ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,425 |
10.43. a) Do there exist polynomials $P=P(x, y, z), Q=Q(x, y, z)$, and $R=R(x, y, z)$ in variables $x, y, z$ that satisfy the identity
$$
(x-y+1)^{3} P+(y-z-1)^{3} Q+(z-2 x+1)^{3} R=1 ?
$$
b) The same question for the identity
$$
(x-y+1)^{3} P+(y-z-1)^{3} Q+(z-x+1)^{3} R=1
$$
See also problem 16.10. | 10.43. a) The system of equations $x-y+1=0, y-z-1=0, z-2x+1=0$ has the solution $(x, y, z)=(1,2,1)$. For these values of the variables, the left side of the considered expression becomes zero. Therefore, such polynomials $P, Q, R$ do not exist.
b) Let $f=x-y+1, g=y-z-1, h=z-x+1$. Then $f+g+h=1$, so $(f+g+h)^{7}=1$. Th... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,426 |
11.1. Prove that $\sin \alpha<\alpha<\operatorname{tg} \alpha$ for $0<\alpha<\pi / 2$. | 11.1. Let $O$ be the center of a circle with radius $1, A$ and $B$ be points on this circle such that $\angle A O B=\alpha$. Consider also the point $C$, where the tangent to the circle at point $B$ intersects the ray $O A$. Let $S_{A O B}$ be the area of triangle $A O B, S$ be the area of the sector cut by radii $A O$... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,427 |
11.2. Prove that if $0<\alpha_{1}<\alpha_{2}<\ldots<\alpha_{n}<\pi / 2$ and $n \geqslant 2$, then
$$
\operatorname{tg} \alpha_{1}<\frac{\sin \alpha_{1}+\ldots+\sin \alpha_{n}}{\cos \alpha_{1}+\ldots+\cos \alpha_{n}}<\operatorname{tg} \alpha_{n}
$$ | 11.2. First, note that $\operatorname{tg} \alpha_{1}<\operatorname{tg} \alpha_{2}$. Therefore, according to problem 8.23,
$$
\operatorname{tg} \alpha_{1}<\frac{\sin \alpha_{1}+\sin \alpha_{2}}{\cos \alpha_{1}+\cos \alpha_{2}}<\operatorname{tg} \alpha_{2}<\operatorname{tg} \alpha_{3}
$$
Using problem 8.23 again, we ge... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,428 |
11.3. Compare the numbers $\operatorname{tg} 55^{\circ}$ and 1.4. | 11.3. It is clear that
$$
\operatorname{tg} 55^{\circ}=\operatorname{tg}\left(45^{\circ}+10^{\circ}\right)=\frac{\operatorname{tg} 45^{\circ}+\operatorname{tg} 10^{\circ}}{1-\operatorname{tg} 45^{\circ} \operatorname{tg} 10^{\circ}}=\frac{1+\operatorname{tg} 10^{\circ}}{1-\operatorname{tg} 10^{\circ}}
$$
Further, $\o... | 1.4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,429 |
11.4. Prove that if $0<\alpha, \beta \leqslant \pi / 4$, then
$$
\sqrt{\tan \alpha \tan \beta} \leqslant \tan \frac{\alpha+\beta}{2} \leqslant \frac{\tan \alpha+\tan \beta}{2}
$$ | 11.4. From the formula for the tangent of the difference of two angles, it follows that
\[
\begin{aligned}
\operatorname{tg} \alpha - \operatorname{tg} \frac{\alpha+\beta}{2} & = \operatorname{tg} \frac{\alpha-\beta}{2} \left(1 + \operatorname{tg} \alpha \operatorname{tg} \frac{\alpha+\beta}{2}\right) \\
\operatorname... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,430 |
11.5. Prove that the sum
$$
\cos 32 x+a_{31} \cos 31 x+a_{30} \cos 30 x+\ldots+a_{1} \cos x
$$
takes both positive and negative values. | 11.5. Suppose that the sum
$$
\cos 32 x + a_{31} \cos 31 x + a_{30} \cos 30 x + \ldots + a_{1} \cos x
$$
takes only positive values for all $x$. By replacing $x$ with $x + \pi$, we get that the expression
$$
\cos 32 x - a_{31} \cos 31 x + a_{30} \cos 30 x - \ldots + a_{2} \cos 2 x - a_{1} \cos x
$$
takes positive v... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,431 |
11.6. Prove that if for any angle $\varphi$ the inequality
$$
a_{1} \cos \varphi + a_{2} \cos 2 \varphi + \ldots + a_{n} \cos n \varphi \geqslant -1
$$
holds, then $a_{1} + a_{2} + \ldots + a_{n} \leqslant n$. | 11.6. According to problem 11.29 b) $\cos \varphi_{k}+\cos 2 \varphi_{k}+\ldots+\cos n \varphi_{k}=-1$ for $\varphi_{k}=\frac{2 k \pi}{n+1}$, where $k=1,2, \ldots, n$. Therefore, by adding $n$ inequalities
$$
a_{1} \cos \varphi_{k}+a_{2} \cos 2 \varphi_{k}+\ldots+a_{n} \cos n \varphi_{k} \geqslant-1
$$
we get $-a_{1}... | a_{1}+a_{2}+\ldots+a_{n}\leqslantn | Inequalities | proof | Yes | Yes | olympiads | false | 38,432 |
11.8. Let $A$ be an arbitrary angle, and $B$ and $C$ be acute angles. Does there always exist an angle $X$ such that
$$
\sin X=\frac{\sin B \sin C}{1-\cos A \cos B \cos C} ?
$$
## 11.2. Trigonometric Identities | 11.8. A n s w e r: yes, always. Po
Fig. 11.1 According to the condition $\cos B \cos C > 0$. Additionally, $\sin B \sin C + \cos B \cos C = \cos (B - C) \leqslant 1$ and $\cos A \leqslant... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,434 |
11.9. Simplify the expression
$$
\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}+\sqrt{3} \operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}
$$ | 11.9. Let's show that $\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}+\sqrt{3} \operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}=\sqrt{3}$, i.e., $\sqrt{3}=\frac{\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}}{1-\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}}$. Indeed, $\frac{\o... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,435 |
11.10. Prove that
$$
\operatorname{arctg} \frac{1}{3}+\operatorname{arctg} \frac{1}{5}+\operatorname{arctg} \frac{1}{7}+\operatorname{arctg} \frac{1}{8}=\frac{\pi}{4}
$$ | 11.10. Applying the formula $\operatorname{tg}(x+y)=\frac{\operatorname{tg} x+\operatorname{tg} y}{1-\operatorname{tg} x \operatorname{tg} y}$, we get
$$
\begin{gathered}
\operatorname{tg}\left(\operatorname{arctg} \frac{1}{3}+\operatorname{arctg} \frac{1}{5}\right)=\frac{\frac{1}{3}+\frac{1}{5}}{1-\frac{1}{15}}=\frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,436 |
11.11. Prove that
$$
4 \operatorname{arctg} \frac{1}{5}-\operatorname{arctg} \frac{1}{239}=\frac{\pi}{4}
$$ | 11.11. Let $\operatorname{tg} \varphi=1 / 5$. Using the identity $\operatorname{tg} 2 \varphi=\frac{2 \operatorname{tg} \varphi}{1-\operatorname{tg}^{2} \varphi}$ twice, we first get $\operatorname{tg} 2 \varphi=\frac{5}{12}$, and then $\operatorname{tg} 4 \varphi=\frac{120}{119}$. Therefore, $\operatorname{tg}\left(4 ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,437 |
11.12. Find the relationship between $\arcsin \cos \arcsin x \quad$ and $\quad \arccos \sin \arccos x$.
## 11.3. Equations | 11.12. Answer: $\arcsin \cos \arcsin x+\arccos \sin \arccos x=\frac{\pi}{2}$.
Let $\arcsin \cos \arcsin x=\alpha$ and $\arccos \sin \arccos x=\beta$. Then $0 \leqslant$ $\leqslant \alpha, \beta \leqslant \pi / 2$. Indeed, $0 \leqslant \cos \arcsin x \leqslant 1$, since $-\frac{\pi}{2} \leqslant$ $\leqslant \arcsin x \... | \alpha+\beta=\frac{\pi}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,438 |
11.13. Solve the equation
$$
3-7 \cos ^{2} x \sin x-3 \sin ^{3} x=0
$$ | 11.13. Answer: $x=\frac{\pi}{2}+2 k \pi$ and $x=(-1)^{k} \frac{\pi}{6}+k \pi$.
Let $t=\sin x$. Considering that $\cos ^{2} x=1-t^{2}$, we obtain the equation $4 t^{3}-7 t+3=0$. This equation has roots $t_{1}=1, t_{2}=1 / 2$ and $t_{3}=-3 / 2$. The last root does not fit. | \frac{\pi}{2}+2k\pi(-1)^{k}\frac{\pi}{6}+k\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,439 |
11.14. Solve the equation
$$
\frac{1+\operatorname{tg} x}{1-\operatorname{tg} x}=1+\sin 2 x
$$ | 11.14. Answer: $x=-\frac{\pi}{4}+k \pi$ and $x=k \pi$.
Let $t=\operatorname{tg} x$. Considering that $\sin 2 x=\frac{2 t}{1+t^{2}}$, we obtain the equation
$$
\frac{1+t}{1-t}=1+\frac{2 t}{1+t^{2}}
$$
i.e., $2(1+t) t^{2}=0$ (by the condition $t \neq 1$). This equation has roots $t_{1}=-1$ and $t_{2}=0$. | -\frac{\pi}{4}+k\pik\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,440 |
11.15. Find all real solutions of the equation
$$
x^{2}+2 x \sin (x y)+1=0
$$ | 11.15. Answer: $x= \pm 1, y=-\frac{\pi}{2}+2 k \pi$.
If we consider the given equation as a quadratic equation in terms of $x$, then its discriminant will be equal to $4\left(\sin ^{2}(x y)-1\right)$. The discriminant must be non-negative, so $\sin ^{2}(x y) \geqslant \geqslant 1$, i.e., $\sin (x y)= \pm 1$. The solut... | \1,-\frac{\pi}{2}+2k\pi | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,441 |
11.16. Solve the equation
$$
3 \sin \varphi \cos \varphi + 4 \sin \varphi + 3 \cos^2 \varphi = 4 + \cos \varphi
$$ | 11.16. Let $x=\sin \varphi$ and $y=\cos \varphi$. The considered equation is equivalent to the system of equations
$$
3 x y+4 x+3 y^{2}-y-4=0, \quad x^{2}+y^{2}=1
$$
Given the condition $x^{2}+y^{2}=1$, the first equation is equivalent to the equation
$$
3 x y+4 x+3 y^{2}-y-4+\lambda\left(x^{2}+y^{2}-1\right)=0
$$
... | (3\sin\varphi-1)(\cos\varphi-\sin\varphi+1)=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,442 |
11.17. How many roots does the equation $\sin x=\frac{x}{100}$ have?
## 11.4. Sums of Sines and Cosines Related to Regular Polygons
In solving the problems of this section, the following geometric problem is useful. | 11.17. Answer: 63. First, note that the number of positive roots is equal to the number of negative roots, and there is also a root 0. Therefore, it is sufficient to verify that the number of positive roots is 31. If $\sin x = x / 100$, then $|x| = 100|\sin x| \leqslant 100$. Consider the graphs of the functions $y = x... | 63 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 38,443 |
11.18. Prove that the sum of vectors going from the center of a regular polygon to its vertices is equal to the zero vector. | 11.18. The sum of vectors going from the center of a regular $n$-gon to its vertices, transforms into itself upon rotation by an angle of $2 \pi / n$. And the only vector that transforms into itself upon rotation by an angle of $2 \pi / n,$ is the zero vector. | proof | Geometry | proof | Yes | Yes | olympiads | false | 38,444 |
11.19. Prove the following equalities:
a) $\sin \frac{2 \pi}{n}+\sin \frac{4 \pi}{n}+\ldots+\sin \frac{2(n-1) \pi}{n}=0$;
b) $\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\ldots+\cos \frac{2(n-1) \pi}{n}=-1$;
c) $\sin \alpha+\sin \left(\alpha+\frac{2 \pi}{n}\right)+\sin \left(\alpha+\frac{4 \pi}{n}\right)+\ldots+\sin \... | 11.19. Let's take a regular $n$-sided polygon inscribed in a circle of radius 1 with the center at the origin. Rotate it so that one of its vertices falls on the point $(1,0)$. By considering the projections of the sum of vectors from the center of the regular $n$-sided polygon to its vertices onto the $x$ and $y$ axes... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,445 |
11.20. Prove the following equalities:
a) $\cos 36^{\circ}-\cos 72^{\circ}=1 / 2$
b) $\cos \frac{\pi}{7}-\cos \frac{2 \pi}{7}+\cos \frac{3 \pi}{7}=\frac{1}{2}$. | 11.20. According to problem 11.19 b) $\cos \frac{2 \pi}{5}+\cos \frac{4 \pi}{5}+\cos \frac{6 \pi}{5}+\cos \frac{8 \pi}{5}=$ $=1$. Here, $\cos \frac{8 \pi}{5}=\cos \frac{2 \pi}{5}=\cos 72^{\circ}$ and $\cos \frac{4 \pi}{5}=\cos \frac{6 \pi}{5}=-\cos 36^{\circ}$.
b) Let's take a regular 7-sided polygon with its center a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,446 |
11.21. Prove that
$\cos \frac{2 \pi}{2 m+1}+\cos \frac{4 \pi}{2 m+1}+\cos \frac{6 \pi}{2 m+1}+\ldots+\cos \frac{2 m \pi}{2 m+1}=-\frac{1}{2}$.
## 11.5. Calculation of sums and products | 11.21. For $n=2 m+1$ the sum from problem 11.19 b) consists of an even number of terms. These terms can be paired as $\cos \frac{2 k \pi}{n}=\cos \frac{2(n-k) \pi}{n}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,447 |
11.22. Prove that if $\alpha+\beta+\gamma=0$, then
$$
\sin \alpha+\sin \beta+\sin \gamma=-4 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\gamma}{2}
$$ | 11.22. We need to prove that
$$
\sin \alpha + \sin \beta - \sin (\alpha + \beta) = 4 \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \sin \frac{\alpha + \beta}{2}
$$
Replace \(\sin (\alpha + \beta)\) in the left part with \(\sin \alpha \cos \beta + \sin \beta \cos \alpha\). Further,
$$
\begin{aligned}
& 4 \sin \frac{\alp... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,448 |
11.23. Prove that if $\sin \alpha \neq 0$, then
$$
\cos \alpha \cos 2 \alpha \cos 4 \alpha \ldots \cos 2^{n} \alpha=\frac{\sin 2^{n+1} \alpha}{2^{n+1} \sin \alpha}
$$ | 11.23. First, note that $\cos \alpha=\frac{\sin 2 \alpha}{2 \sin \alpha}$. Therefore, $\cos \alpha \times$ $\times \cos 2 \alpha=\frac{\sin 2 \alpha \cos 2 \alpha}{2 \sin \alpha}=\frac{\sin 4 \alpha}{4 \sin \alpha}$ and so on. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,449 |
11.24. Prove that:
a) $\cos \frac{2 \pi}{7} \cos \frac{4 \pi}{7} \cos \frac{8 \pi}{7}=\frac{1}{8}$
b) $\cos \frac{2 \pi}{9} \cos \frac{4 \pi}{9} \cos \frac{8 \pi}{9}=-\frac{1}{8}$. | 11.24. a) We apply the formula from problem 11.23 for \( n=2 \) and \( \alpha = 2\pi / 7 \). As a result, we get that the required product is \( \frac{\sin (16\pi / 7)}{8 \sin (2\pi / 7)} \). But \( \sin (16\pi / 7) = \sin (2\pi / 7) \).
b) It is solved similarly to a). One only needs to notice that \( \sin (16\pi / 9... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,450 |
11.25. a) Prove that $\operatorname{ctg} \alpha - \operatorname{tg} \alpha = 2 \operatorname{ctg} 2 \alpha$.
b) Prove that
$\operatorname{tg} \alpha + 2 \operatorname{tg} 2 \alpha + 4 \operatorname{tg} 4 \alpha + \ldots + 2^{n} \operatorname{tg} 2^{n} \alpha = \operatorname{ctg} \alpha - 2^{n+1} \operatorname{ctg} 2^... | 11.25. a) It is clear that $\frac{\cos \alpha}{\sin \alpha}-\frac{\sin \alpha}{\cos \alpha}=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\cos \alpha \sin \alpha}=\frac{\cos 2 \alpha}{\frac{1}{2} \sin 2 \alpha}=$
$=2 \operatorname{ctg} 2 \alpha$.
b) According to part a)
$$
\begin{aligned}
& \operatorname{tg} \alpha=\opera... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,451 |
11.26. Prove that
$\frac{1}{\cos \alpha \cos 2 \alpha}+\frac{1}{\cos 2 \alpha \cos 3 \alpha}+\ldots+\frac{1}{\cos (n-1) \alpha \cos n \alpha}=\frac{\tan n \alpha-\tan \alpha}{\sin \alpha}$. | 11.26. It is clear that
$$
\begin{aligned}
\operatorname{tg}(k+1) \alpha-\operatorname{tg} k \alpha=\frac{\sin (k+1) \alpha}{\cos (k+1) \alpha} & -\frac{\sin k \alpha}{\cos k \alpha}= \\
& =\frac{\sin ((k+1) \alpha-k \alpha)}{\cos k \alpha \cos (k+1) \alpha}=\frac{\sin \alpha}{\cos k \alpha \cos (k+1) \alpha}
\end{ali... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,452 |
11.27. Prove that
$$
\cos \frac{\pi}{2 n+1} \cos \frac{2 \pi}{2 n+1} \ldots \cos \frac{n \pi}{2 n+1}=\frac{1}{2^{n}}
$$ | 11.27. Let's first prove that
$$
\begin{aligned}
\sin \frac{2 \pi}{2 n+1} \sin \frac{4 \pi}{2 n+1} \ldots \sin \frac{2 n \pi}{2 n+1} & = \\
\quad= & \sin \frac{\pi}{2 n+1} \sin \frac{2 \pi}{2 n+1} \ldots \sin \frac{n \pi}{2 n+1}
\end{aligned}
$$
We will consider the cases of even and odd \( n \) separately. If \( n =... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,453 |
11.28. Prove that
a) $\sin \alpha+\sin 2 \alpha+\sin 3 \alpha+\ldots+\sin n \alpha=\frac{\sin \frac{n \alpha}{2} \sin \frac{(n+1) \alpha}{2}}{\sin \frac{\alpha}{2}}$;
b) $\cos \alpha+\cos 2 \alpha+\cos 3 \alpha+\ldots+\cos n \alpha=\frac{\sin \frac{(2 n+1) \alpha}{2}}{2 \sin \frac{\alpha}{2}}-\frac{1}{2}$. | 11.28. a) Let $S$ be the desired sum of sines. Using the identity $2 \sin x \sin y = \cos (x-y) - \cos (x+y)$, we get $2 S \sin \frac{\alpha}{2} = 2 \sin \alpha \sin \frac{\alpha}{2} + 2 \sin 2 \alpha \sin \frac{\alpha}{2} + \ldots + 2 \sin n \alpha \sin \frac{\alpha}{2}=$
$$
\begin{aligned}
& \quad = \left(\cos \frac... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,454 |
11.29. a) Prove that $\cos \alpha+\cos (\alpha+x)+\cos (\alpha+2 x)+\ldots$ $\ldots+\cos (\alpha+n x)=\frac{\sin \left(\alpha+\left(n+\frac{1}{2}\right) x\right)-\sin \left(\alpha-\frac{1}{2} x\right)}{2 \sin \frac{1}{2} x}$.
b) Prove that if $\varphi=\frac{2 k \pi}{n+1}$, where the number $k$ is an integer and $1 \le... | 11.29. a) Add the identities
$$
\sin \left(\alpha+\left(k+\frac{1}{2}\right) x\right)-\sin \left(\alpha+\left(k-\frac{1}{2}\right) x\right)=2 \sin \frac{1}{2} x \cos (\alpha+k x)
$$
for $k=0,1,2, \ldots, n$. The result will be the required one.
b) For $\alpha=0$ and $x=\varphi$, the formula from part a) gives the eq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,455 |
11.30. a) Prove that $\cos n \varphi=T_{n}(\cos \varphi)$ and $\sin (n+1) \varphi=$ $=U_{n}(\cos \varphi) \sin \varphi$, where $T_{n}$ and $U_{n}$ are polynomials of degree $n$.
b) Prove that $\sin (2 k+1) \varphi=\sin \varphi P_{k}\left(\sin ^{2} \varphi\right)$, where $P_{k}$ is a polynomial of degree $k$. | 11.30. a) It is clear that $U_{0}(x)=1, U_{1}(x)=2 x$ and $T_{1}(x)=x$. Moreover, the formulas
$$
\begin{aligned}
& \sin (n+1) \varphi=\sin n \varphi \cos \varphi+\cos n \varphi \sin \varphi \\
& \cos (n+1) \varphi=\cos n \varphi \cos \varphi-\sin n \varphi \sin \varphi
\end{aligned}
$$
show that $U_{n}(x)=x U_{n-1}(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,456 |
11.31. Prove that
\[
\begin{aligned}
\frac{\sin (2 k+1) \varphi}{\sin \varphi}= & (-4)^{k}\left(\sin ^{2} \varphi-\sin ^{2} \frac{\pi}{2 k+1}\right) \times \\
& \times\left(\sin ^{2} \varphi-\sin ^{2} \frac{2 \pi}{2 k+1}\right) \ldots\left(\sin ^{2} \varphi-\sin ^{2} \frac{k \pi}{2 k+1}\right)
\end{aligned}
\]
## 11.... | 11.31. According to problem 11.30 b) $\frac{\sin (2 k+1) \varphi}{\sin \varphi}=P_{k}\left(\sin ^{2} \varphi\right)$, where $P_{k}$ is a polynomial of degree $k$. Further, if $\varphi=\frac{l \pi}{2 k+1}$, then $\sin (2 k+1) \varphi=$ $=0$. Therefore,
$$
P_{k}(x)=\lambda\left(x-\sin ^{2} \frac{\pi}{2 k+1}\right)\left(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,457 |
11.32. Solve the system of equations
$$
\left\{\begin{array}{l}
2 x+x^{2} y=y \\
2 y+y^{2} z=z \\
2 z+z^{2} x=x
\end{array}\right.
$$ | 11.32. The first equation can be rewritten as $y=\frac{2 x}{1-x^{2}}$ (it is clear that $x \neq \pm 1$). Let $x=\operatorname{tg} \varphi$. Then $y=\operatorname{tg} 2 \varphi$. Similarly, $z=\operatorname{tg} 4 \varphi$ and $x=\operatorname{tg} 8 \varphi$. Thus, $\operatorname{tg} 8 \varphi=\operatorname{tg} \varphi$.... | \operatorname{tg}\frac{k\pi}{7},\operatorname{tg}\frac{2k\pi}{7},\operatorname{tg}\frac{4k\pi}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,458 |
11.33. Solve the system of equations
$$
\left\{\begin{array}{l}
3\left(x+\frac{1}{x}\right)=4\left(y+\frac{1}{y}\right)=5\left(z+\frac{1}{z}\right) \\
x y+y z+z x=1
\end{array}\right.
$$ | 11.33. Answer: $(1 / 3,1 / 2,1)$ and ( $-1 / 3,-1 / 2,-1)$.
The equalities $\frac{x}{3\left(1+x^{2}\right)}=\frac{y}{4\left(1+y^{2}\right)}=\frac{z}{5\left(1+z^{2}\right)}$ show that the numbers $x, y, z$ have the same sign, and if $(x, y, z)$ is a solution to the system, then ( $-x,-y,-z$ ) is also a solution. Theref... | (1/3,1/2,1)(-1/3,-1/2,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,459 |
11.34. a) Let $x_{1}, \ldots, x_{n}$ be pairwise distinct positive numbers, $n \geqslant 3$. Prove that among them, one can choose two numbers $x_{i}$ and $x_{j}$, for which
$$
0<\frac{x_{i}-x_{j}}{1+x_{i} x_{j}}<\tan \frac{\pi}{2(n-1)}
$$
b) Let $x_{1}, \ldots, x_{n}$ be pairwise distinct numbers, $n \geqslant 3$. P... | 11.34. a) Let $x_{i}=\operatorname{tg} \varphi_{i}$, where $0<\varphi_{i}<\pi / 2$. Then $\frac{x_{i}-x_{j}}{1+x_{i} x_{j}}=$ $=\operatorname{tg}\left(\varphi_{i}-\varphi_{j}\right)$. We will assume that $x_{1}<\ldots<x_{n}$. Then the numbers $\varphi_{2}-\varphi_{1}$, $\varphi_{3}-\varphi_{2}, \ldots, \varphi_{n}-\var... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,460 |
11.35. Prove that for any positive number $x$ and any natural number $n$, the following inequality holds:
$$
\frac{x}{1+x^{2}}+\frac{x}{2^{2}+x^{2}}+\ldots+\frac{x}{n^{2}+x^{2}}<\frac{\pi}{2}
$$
## 11.8. Trigonometric Polynomials
A trigonometric polynomial of degree $n$ is a function $f(\varphi)=a_{0}+a_{1} \cos \va... | 11.35. Let's choose a point $A=(x, 0)$ on the $O x$ axis, and points $B_{k}=(0, k)$ on the $O y$ axis, where $k=0,1, \ldots$ Consider the triangle $B_{k-1} A B_{k}$. Let $\varphi=\angle B_{k-1} A B_{k}, k=1,2, \ldots$ By calculating the area of the triangle $B_{k-1} A B_{k}$ in two ways, we get $\frac{1}{2} x=\frac{1}{... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,461 |
11.36. Prove that for any trigonometric polynomial $f(\varphi)=a_{0}+a_{1} \cos \varphi+\ldots+a_{n} \cos n \varphi$ there exists a polynomial $P(x)$ of degree $n$ with the leading coefficient $2^{n-1} a_{n}$, such that $P(\cos \varphi)=f(\varphi)$. And conversely, any polynomial corresponds to a trigonometric polynomi... | 11.36. According to the definition of Chebyshev polynomials $\cos n \varphi=$ $=T_{n}(\cos \varphi)$, and $T_{n}(x)=2^{n-1} x^{n}+\ldots$ (Problem 32.30). Therefore, $f(\varphi)=a_{0}+a_{1} T_{1}(x)+a_{2} T_{2}(x)+\ldots+a_{n} T_{n}(x)$, where $x=\cos \varphi$.
Conversely, let a polynomial $P(x)=b_{0}+b_{1} x+\ldots+b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,462 |
11.37. Let $f$ be a trigonometric polynomial of degree $n$ with leading coefficient $a_{n}$.
a) Prove that $\frac{1}{2 n} \sum_{m=0}^{2 n-1}(-1)^{m} f\left(\frac{m \pi}{n}\right)=a_{n}$.
b) Prove that $\left|f\left(\frac{m \pi}{n}\right)\right| \geqslant\left|a_{n}\right|$ for some integer $m$. | 11.37. a) A trigonometric polynomial of degree $n$ is represented as a sum of terms of the form $\cos k \varphi, 0 \leqslant k \leqslant n$. Therefore, it is sufficient to verify that
$$
\frac{1}{2 n} \sum_{m=0}^{2 n-1}(-1)^{m} \cos \frac{k m \pi}{n}= \begin{cases}0 & \text { if } 0 \leqslant k \leqslant n-1 \\ 1 & \t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,463 |
11.38. Prove that if $P(x)=x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}$, then $\left|P\left(x_{0}\right)\right| \geqslant \frac{1}{2^{n-1}}$ for some $x_{0}$, where $-1 \leqslant x_{0} \leqslant 1$. | 11.38. According to problem 11.36, the polynomial $P(x)$ corresponds to a trigonometric polynomial $f(\varphi)$ with the leading coefficient $1 / 2^{n-1}$, for which $f(\varphi)=P(\cos \varphi)$. According to problem 11.37, $\left|f\left(\varphi_{0}\right)\right| \geqslant \frac{1}{2^{n-1}}$ for some $\varphi_{0}$. Let... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,464 |
11.39. Let $0<a_{0}<a_{1}<\ldots<a_{n}$. Prove that the trigonometric polynomial $f(\varphi)=a_{0}+a_{1} \cos \varphi+\ldots+a_{n} \cos n \varphi$ has exactly $n$ roots on the interval $[0, \pi]$. | 11.39. Consider the function $g(\varphi)=2 \sin \frac{\varphi}{2} f(x)$. Using the identity
$$
2 \sin \frac{\varphi}{2} \cos k \varphi=\sin \left(k+\frac{1}{2}\right) \varphi-\sin \left(k-\frac{1}{2}\right) \varphi
$$
we obtain $g(\varphi)=a_{n} \sin (n+1 / 2) \varphi-h(\varphi)$, where
$$
h(\varphi)=\left(a_{n}-a_{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,465 |
12.1. Prove that if $a, b, c$ is a Pythagorean triple, then one of these numbers is divisible by 3, another (or the same) is divisible by 4, and the third is divisible by 5. | 12.1. According to problem 4.42, the remainder of the square of an integer when divided by 3 and 4 is 0 or 1, and the remainder when divided by 5 is 0, 1, or 4. Using only the remainders 1 and 4, it is impossible to satisfy the equation \(a^{2}+b^{2}=c^{2}\). In the case of divisibility by 3 and 5, this completes the p... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,466 |
12.2. Let $a, b, c$ be a primitive Pythagorean triple. Prove that one of the numbers $a$ or $b$ is even, and the other is odd. | 12.2. The numbers $a$ and $b$ cannot both be even, because otherwise the number $c$ would also be even. The numbers $a$ and $b$ cannot both be odd, because otherwise the number $a^{2}+b^{2}$ would be divisible by 2, but not divisible by 4. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,467 |
12.3. Let $a, b, c$ be a primitive Pythagorean triple, and suppose the number $a$ is even. Prove that there exist coprime numbers $m$ and $n$ such that $a=2 m n, b=m^{2}-n^{2}$, $c=m^{2}+n^{2}$. | 12.3. The numbers $\frac{c-b}{2}$ and $\frac{c+b}{2}$ are coprime, so from the equation $\left(\frac{a}{2}\right)^{2}=\frac{c-b}{2} \cdot \frac{c+b}{2}$ it follows that $\frac{c-b}{2}=n^{2}$ and $\frac{c+b}{2}=m^{2}$, where $m$ and $n$ are coprime numbers. In this case, $c=m^{2}+n^{2}$ and $b=m^{2}-n^{2}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,468 |
12.5. Let $a, b, c$ be a primitive Pythagorean triple. Prove that the number $ab / 2$ (the area of the right triangle with legs $a$ and $b$) cannot be a perfect square.
## 12.2. Finding all solutions | 12.5. Suppose there exist coprime natural numbers $m$ and $n$ such that $m n(m+n)(m-n)=s^{2}$, where $s$ is a natural number. We will assume that $s$ is the smallest number for which such an equality holds. The numbers $m, n, m+n, m-n$ are pairwise coprime, so $m=x^{2}$, $n=y^{2}$, $m+n=z^{2}$, and $m-n=t^{2}$, where $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,470 |
12.6. Solve the equation $x+y=x y$ in natural numbers. 12.7. Solve the equation $2 x y+3 x+y=0$ in integers. | 12.6. It is clear that $(x-1)(y-1)=x y-x-y+1=1$, therefore $x-1=y-1=1$, i.e., $x=y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,471 |
12.8. Solve the equation in integers
$$
x y+3 x-5 y=-3
$$ | 12.8. The considered equation can be rewritten in the form $(x-5)(y+3)=-18$. Its solutions in integers correspond to the representations of the number -18 as the product of two integers. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,472 |
12.9. Solve the equation in integers
$$
x+y=x^{2}-x y+y^{2}
$$ | 12.9. Answer: $(0,0),(0,1),(1,0),(1,2),(2,1),(2,2)$. Let's consider the given equation as a quadratic equation in terms of $x$:
$$
x^{2}-(y+1) x+y^{2}-y=0 \text {. }
$$
The discriminant of this equation is $-3 y^{2}+6 y+1$. It is negative for $y \geqslant 3$ and for $y \leqslant-1$. Therefore, for $y$ we get three po... | (0,0),(0,1),(1,0),(1,2),(2,1),(2,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,473 |
12.10. Solve the equation $2^{x}+7=y^{2}$ in natural numbers. | 12.10. Answer: $x=1, y=3$. Let's check that there are no other solutions. It is clear that $2^{2}+7=11$ is not a square, so we can assume that $x \geqslant 3$. Then $2^{x}$ is divisible by 8, and therefore $y^{2} \equiv 7(\bmod 8)$. However, as it is easy to check, the square of an integer when divided by 8 can only gi... | 1,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,474 |
12.11. Let $p>2$ be a prime number. Prove that the number $2 / p$ can be uniquely represented in the form $1 / x+1 / y$, where $x$ and $y$ are different natural numbers. | 12.11. The equation $1 / x + 1 / y = 2 / p$ can be rewritten as $p(x + y) = 2xy$. This means that one of the numbers $x$ and $y$ is divisible by $p$. For example, $x = px'$. Then $px' + y = 2x'y$, i.e., $(2x' - 1)y = px'$. If $x' = 1$, then $y = p$ and $x = p$, but by the condition, the numbers $y$ and $x$ are differen... | y=\frac{p+1}{2},\,x=p\cdot\frac{p+1}{2} | Number Theory | proof | Yes | Yes | olympiads | false | 38,475 |
12.12. Let $n$ be a natural number. Prove that the number of solutions to the equation $1 / x + 1 / y = 1 / n$ in natural numbers is equal to the number of divisors of the number $n^{2}$. | 12.12. This equation can be written in the form $n^{2}=(x-n) \times$ $\times(y-n)$. Each divisor $d$ of the number $n^{2}$ corresponds to a solution $x=n+d, y=n+\frac{n^{2}}{d}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,476 |
12.13. a) Find all natural numbers \(x, y, z\) for which
\[
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1
\]
b) Find all natural numbers \(x, y, z>1\) for which
\[
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>1
\] | 12.13. a) Let's assume that $x \geqslant y \geqslant z$. Then $1=1 / x+1 / y+1 / z \leqslant 3 / z$, so $z \leqslant 3$. It is also clear that $z \neq 1$.
If $z=3$, then $1 / x+1 / y=2 / 3$ and in this case $1 / x+1 / y \leqslant 2 / y$, so $y \leqslant 3$. But $y \geqslant z=3$, so $y=3$ and $x=3$.
If $z=2$, then $1... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,477 | |
12.14. a) Find all solutions of the equation $x^{2}+y^{2}+z^{2}=$ $=2 x y z$ in natural numbers.
b) Find all solutions of the equation $x^{2}+y^{2}+z^{2}+u^{2}=2 x y z u$ in natural numbers. | 12.14. a) Let $x=2^{m} x_{1}, y=2^{n} y_{1}, z=2^{k} z_{1}$, where the numbers $x_{1}, y_{1}, z_{1}$ are odd. We can assume that $m \leqslant n \leqslant k$. Then both sides of the equation can be divided by $\left(2^{m}\right)^{2}$. As a result, we get
$$
x_{1}^{2}+2^{(n-m)} y_{1}^{2}+2^{(k-m)} z_{1}^{2}=2^{n+k-m+1} ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,478 |
12.15. Solve the equation in integers
$$
x^{3}-2 y^{3}-4 z^{3}=0
$$ | 12.15. Answer: $x=y=z=0$.
Let $x^{3}-2 y^{3}-4 z^{3}=0$, where $x, y, z$ are integers. Then the number $x$ is even. After substituting $x=2 x_{1}$, we get the equation $8 x_{1}^{3}-2 y^{3}-4 z^{3}=0$. Dividing by 2: $4 x_{1}^{3}-y^{3}-2 z^{3}=0$. Therefore, the number $y$ is even. After substituting $y=2 y_{1}$, we ge... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,479 |
12.16. Solve the equation $2^{n} + 1 = 3^{m}$ in natural numbers. | 12.16. Answer: $n=1, m=1$ or $n=3, m=2$.
If $n=1$, then $m=1$. Now consider the case when $n>1$. In this case $3^{m} \equiv 1(\bmod 4)$. Note that $3^{2 k+1}=3 \cdot 9^{k} \equiv 3(\bmod 4)$. Therefore, $m=2 k$, which means $\left(3^{k}-1\right)\left(3^{k}+1\right)=2^{n}$. Thus, the numbers $3^{k}-1$ and $3^{k}+1$ are... | n=1,=1orn=3,=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,480 |
12.17. Find all solutions to the equation $x^{y}=y^{x}$:
a) in natural numbers;
b) in rational numbers.
## 12.3. Finding some solutions | 12.17. There is always an obvious solution $x=y$, so it is sufficient to consider the case when $x>y$ (the case $x<y$ is analogous). Let $x = ky$ where $k>1$ is a rational number. Then $(ky)^y = (y^k)^y$, so $ky = y^k$, which means $y = k^{\frac{1}{k-1}}$. Let $\frac{1}{k-1} = \frac{p}{q}$ be an irreducible fraction. T... | 4,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,481 |
12.18. a) Prove that for any natural number $n$ the equation $a^{n}+b^{n}=c^{n+1}$ has infinitely many different solutions in natural numbers.
b) Prove that if $m$ and $n$ are coprime natural numbers, then the equation $a^{n}+b^{n}=c^{m}$ has infinitely many solutions in natural numbers.
## 12.4. Proof of the finitene... | 12.18. a) Rewrite the equation in the form $c=(a / c)^{n}+(a / c)^{n}$. We will look for solutions of the form $a=a_{1} c$ and $b=b_{1} c$, where $a_{1}$ and $b_{1}$ are natural numbers. Then $c=a_{1}^{n}+b_{1}^{n}, a=a_{1}\left(a_{1}^{n}+b_{1}^{n}\right)$ and $b=b_{1}\left(a_{1}^{n}+b_{1}^{n}\right)$. It is easy to ve... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,482 |
12.19. Prove that for any natural number $n$ the equation $x^{3}+y^{3}=n$ has a finite number of integer solutions.
## 12.5. Pell's Equation
The equation $x^{2}-d y^{2}=1$, where $d$ is a natural number that is square-free (i.e., a number that is not divisible by the square of any natural number other than 1), is cal... | 12.19. Let $x+y=a$ and $x^{2}-x y+y^{2}=b$. Then $a b=n$. The number $n$ can be factored into $a$ and $b$ in only a finite number of ways, so we get a finite number of systems of equations $x+y=a, x^{2}-x y+y^{2}=b$. Express $y$ from the first equation and substitute it into the second. As a result, we get the equation... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,483 |
12.20. Prove that the equation $x^{2}-2 y^{2}=1$ has infinitely many solutions in natural numbers. | 12.20. It is not difficult to find one solution to this equation, for example, $x_{1}=3$ and $y_{1}=2$. The equation $x_{1}^{2}-2 y_{1}^{2}=1$ can be written as
$$
\left(x_{1}-y_{1} \sqrt{2}\right)\left(x_{1}+y_{1} \sqrt{2}\right)=1
$$
It is clear that then
$$
\left(x_{1}-y_{1} \sqrt{2}\right)^{n}\left(x_{1}+y_{1} \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,484 |
12.21. Let $d$ be a square-free natural number. Prove that there exists a constant $C$ such that the inequality $\left|x^{2}-d y^{2}\right|<C$ has infinitely many integer solutions.
The solution of the Pell's equation in natural numbers with the smallest $y_{1}$ will be called the fundamental solution. | 12.21. Apply problem 17.12 for $\alpha=\sqrt{d}$. As a result, we get that there are infinitely many pairs of coprime numbers $x, y$ for which $|x-y \sqrt{d}|<1 / y$. In this case,
$$
|x+y \sqrt{d}|<|x-y \sqrt{d}|+2 \sqrt{d}|y|<\frac{1}{y}+2 \sqrt{d} y
$$
therefore,
$$
\left|x^{2}-d y^{2}\right|=|x+y \sqrt{d}| \cdot... | 2\sqrt{}+1 | Number Theory | proof | Yes | Yes | olympiads | false | 38,485 |
12.22. Prove that the Pell's equation for any natural $d$ (free of squares) has infinitely many solutions in natural numbers. Moreover, if $\left(x_{1}, y_{1}\right)$ is the fundamental solution, then any solution $\left(x_{n}, y_{n}\right)$ has the form
$$
x_{n}+y_{n} \sqrt{d}=\left(x_{1}+y_{1} \sqrt{d}\right)^{n}
$$ | 12.22. According to problem 12.21, for some integer $k$, the equation $x^{2}-d y^{2}=k$ has infinitely many natural solutions. We choose two different solutions $x_{1}^{2}-d y_{1}^{2}=k$ and $x_{2}^{2}-d y_{2}^{2}=k$, for which $y_{1} \equiv y_{2}(\bmod k)$. Then $x_{1} \equiv \pm x_{2}(\bmod k)$, so the solutions can ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,486 |
12.23. Prove that if $d \equiv 3(\bmod 4)$, then the equation $x^{2}-d y^{2}=-1$ has no solutions in natural numbers. | 12.23. It is clear that $d$ has a prime divisor $p=4k+3$. Suppose that $x^{2}-d y^{2}=-1$. Then $x^{2} \equiv-1(\bmod p)$, so the number -1 is a quadratic residue modulo $p$. On the other hand, if $p=4k+3$ is a prime number, then according to problem 31.36, the number -1 is not a quadratic residue modulo $p$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,487 |
12.24. Prove that if $d$ is a prime number, and $d \equiv 1(\bmod 4)$, then the equation $x^{2}-d y^{2}=-1$ has a solution in natural numbers. | 12.24. Let $\left(x_{1}, y_{1}\right)$ be the fundamental solution of the Pell equation $x^{2}-d y^{2}=1$. From the condition $d \equiv 1(\bmod 4)$, it follows that $x_{1}^{2}-y_{1}^{2} \equiv$ $\equiv 1(\bmod 4)$. Therefore, $x_{1} \equiv 1(\bmod 2)$ and $y_{1} \equiv 0(\bmod 2)$. Rewrite the equation $x_{1}^{2}-d y_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,488 |
12.25. Prove that if the equation $x^{2}-d y^{2}=-4$ has a solution with odd $x$ and $y$, then the equation $X^{2}-d Y^{2}=-1$ has a solution in natural numbers.
See also problems 31.37 and $35.11-35.13$.
## 12.6. Markov Equation
The Markov equation is defined as
$$
m^{2}+n^{2}+p^{2}=3 m n p
$$
where the numbers $... | 12.25. Let
$$
X+Y \sqrt{d}=\left(\frac{x+y \sqrt{d}}{2}\right)^{3}, \quad \text { i.e. } \quad X=\frac{x\left(x^{2}+3 d y^{2}\right)}{8}, Y=\frac{y\left(3 x^{2}+d y^{2}\right)}{8} .
$$
According to problem 4.42 g) $x^{2} \equiv 1(\bmod 8)$ and $y^{2} \equiv 1(\bmod 8)$. Therefore, $d^{2} \equiv 5(\bmod 8)$, which mea... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,489 |
12.27. Prove that all solutions of the equation $m^{2}+n^{2}+$ $+p^{2}=m n p$ in natural numbers have the form $3 m_{1}, 3 n_{1}, 3 p_{1}$, where $m_{1}, n_{1}, p_{1}$ are solutions of the Markov equation. | 12.27. It is sufficient to prove that if \(m^{2}+n^{2}+p^{2}=m n p\), then the numbers \(m, n\), and \(p\) are divisible by 3. If an integer is not divisible by 3, then its square, when divided by 3, gives a remainder of 1. Therefore, if \(m^{2}+n^{2}+p^{2}\) is not divisible by 3, then among the numbers \(m, n\), and ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,491 |
12.28. Prove that if $k \neq 1$ or 3, then the equation $x^{2}+y^{2}+z^{2}=k x y z$ has no solutions in natural numbers. | 12.28. The case $k=2$ is problem 12.14 a). Therefore, we will assume that $k>3$. Suppose that $x^{2}+y^{2}+z^{2}=k x y z$, where $x$, $y, z$ are natural numbers. First, we will show that these numbers are pairwise distinct. Let $y=z$. Then $x^{2}=k x y^{2}-2 y^{2}=(k x-2) y^{2}$, so $x=\lambda y$, where $\lambda$ is a ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,492 |
13.1. Calculate the sum
$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{n-1}{n!}
$$ | 13.1. We will prove by induction on $n$ that the considered sum equals $1-\frac{1}{n!}$. For $n=2$, we get the obvious equality $\frac{1}{2!}=1-\frac{1}{2!}$. Suppose that the required equality is proved for some $n$. Then
$$
\begin{aligned}
\frac{1}{2!}+\frac{2}{3!}+\ldots+\frac{n-1}{n!}+\frac{n}{(n+1)!}= & 1-\frac{1... | 1-\frac{1}{n!} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,494 |
13.2. Calculate the sum
$$
1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+n \cdot n!
$$ | 13.2. We will prove by induction on $n$ that the considered sum equals $(n+1)!-1$. For $n=1$, we get the obvious equality $1 \cdot 1! = 2! - 1$. Suppose the required equality is proven for some $n$. Then
$$
\begin{aligned}
1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n! & + (n+1) \cdot (n+1)! = \\
& = (n+1)! - 1 + (n+1)... | (n+1)!-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,495 |
13.3. Prove that $1^{3}+2^{3}+3^{3}+\ldots+m^{3}=\left(\frac{m(m+1)}{2}\right)^{2}$. | 13.3. The base case of induction is obvious, so we only need to check the equality
$$
\frac{m^{2}(m+1)^{2}}{4}+(m+1)^{3}=\frac{(m+1)^{2}(m+2)^{2}}{4}
$$
After dividing by $m+1$ and multiplying by 4, we get the obvious equality $m^{2}+4(m+1)=(m+2)^{2}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,496 |
13.4. Prove that
$$
1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2 n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2 n}
$$
## 13.2. Inequalities | 13.4. Let $A_{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{2 n}$ and $B_{n}=\frac{1}{n+1}+\frac{1}{n+2}+\ldots$ $\ldots+\frac{1}{2 n}$. Then $A_{n+1}=A_{n}+\frac{1}{2 n+1}-\frac{1}{2 n+2}$ and $B_{n+1}=B_{n}-\frac{1}{n+1}+$ $+\frac{1}{2 n+1}+\frac{1}{2 n+2}$. Therefore, $A_{n+1}-B_{n+1}=A_{n}-B_{n}+\frac{1}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,497 |
13.5. Prove the inequality
$$
1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+k \cdot k!<(k+1)!
$$ | 13.5. We apply induction on $k$. For $k=1$, the required inequality is obvious: $1 \cdot 1! < 2!$. Suppose that for some $k$ the required inequality holds. Then $1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \ldots + k \cdot k! + (k+1) \cdot (k+1)! < (k+1)! + (k+1) \cdot (k+1)! = (k+2)!$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,498 |
13.6. Prove that if $01-\left(x_{1}+\ldots+x_{n}\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 13.6. We apply induction on $n$. First, note that $\left(1-x_{1}\right) \times$ $\times\left(1-x_{2}\right)=1-\left(x_{1}+x_{2}\right)+x_{1} x_{2}>1-\left(x_{1}+x_{2}\right)$. Assume that $\left(1-x_{1}\right) \ldots\left(1-x_{n-1}\right)>1-\left(x_{1}+\ldots+x_{n-1}\right)$. Then $\left(1-x_{1}\right) \ldots\left(1-x_... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,499 |
13.9. a) Prove that for any $\alpha>0$ and any natural $n>1$ the inequality $(1+\alpha)^{n}>1+n \alpha$ holds.
b) Prove that if $0<\alpha \leqslant 1 / n$ and $n$ is a natural number, then the inequality $(1+\alpha)^{n}<1+n \alpha+n^{2} \alpha^{2}$ holds. | 13.9. a) Let's show that if $\alpha>0$ and $(1+\alpha)^{n}>1+n \alpha$, then $(1+\alpha)^{n+1}>1+(n+1) \alpha$. Indeed, $(1+\alpha)^{n+1}=(1+\alpha)(1+\alpha)^{n}>$ $>(1+\alpha)(1+n \alpha)=1+(n+1) \alpha+n \alpha^{2}>1+(n+1) \alpha$.
b) For $n=1$ the required inequality holds. Suppose that if $0<\alpha \leqslant 1 / n... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,502 |
13.10. Prove the inequality between the arithmetic mean and the geometric mean for $n$ positive numbers: $\left(a_{1} a_{2} \ldots a_{n}\right)^{1 / n} \leqslant\left(a_{1}+\ldots+a_{n}\right) / n$, with equality holding if and only if $a_{1}=\ldots=a_{n}$. | 13.10. First, we will prove the required inequality for numbers of the form \( n=2^{m} \) by induction on \( m \). For \( m=1 \), it follows from the fact that \( a-2 \sqrt{a b}+b=(\sqrt{a}-\sqrt{b})^{2} \geqslant 0 \); equality is achieved only if \( a=b \). Suppose the required inequality is proven for \( m \), and w... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,503 |
13.13. The differences between any two distinct odd natural numbers $a_{1}, a_{2}, \ldots, a_{n}$ are distinct. Prove that the sum $a_{1}+\ldots+a_{n}$ is not less than $\frac{n\left(n^{2}+2\right)}{3}$.
## 13.3. Proof of identities | 13.13. We apply induction on $n$. For $n=1$, the statement is obvious. Suppose that the required statement is true for any set of $n-1$ numbers with the given properties. We will prove it for the given set $a_{1}, a_{2}, \ldots, a_{n}$. We can assume that $a_{1}<a_{2}<\ldots<a_{n}$. By the induction hypothesis, $a_{1}+... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,505 |
13.14. Some of the numbers $a_{1}, a_{2}, \ldots, a_{n}$ are equal to +1, the others are equal to -1. Prove that
$$
\begin{aligned}
2 \sin \left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\right. & \left.\ldots+\frac{a_{1} a_{2} \ldots a_{n}}{2^{n-1}}\right) \frac{\pi}{4}= \\
& =a_{1} \sqrt{2+a_{2} \sqrt{... | 13.14. We apply induction on $n$. For $n=1$, we obtain an obvious identity. The equality
$$
\begin{aligned}
2\left(a_{1}+\frac{a_{1} a_{2}}{2}+\frac{a_{1} a_{2} a_{3}}{4}+\ldots+\right. & \left.\frac{a_{1} a_{2} \ldots a_{n} a_{n+1}}{2^{n}}\right) \frac{\pi}{4}= \\
& =a_{1} \frac{\pi}{2}+a_{1}\left(a_{2}+\frac{a_{2} a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,506 |
13.15. On a circular road, several identical cars are parked. It is known that if all the gasoline from the cars is drained into one of them, this car will be able to travel the entire circular road and return to its original position. Prove that at least one of these cars can travel the entire circular road in the giv... | 13.15. We will use induction on the number of cars $n$. For $n=1$, the statement is obvious. Suppose the statement is proven for $n$ cars. Now consider $n+1$ cars. Clearly, among them, there must be a car $A$ that can reach the next car $B$, since otherwise, the total amount of fuel in all cars would not be enough to t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 38,507 |
13.16. Prove that any fraction $m / n$, where $0<m / n<1$, can be represented in the form
$$
\frac{m}{n}=\frac{1}{q_{1}}+\frac{1}{q_{2}}+\frac{1}{q_{3}}+\ldots+\frac{1}{q_{r}}
$$
where $0<q_{1}<q_{2}<\ldots<q_{r}$, and the number $q_{k}$ is divisible by $q_{k-1}$ for $k=2,3, \ldots, r$.
See also problem 20.11. | 13.16. Let's consider that the fraction $m / n$ is irreducible. We will apply induction on $m$. For $m=1$, there is nothing to prove. Let $m>1$. We divide $n$ by $m$ with a remainder, and write the quotient as $d_{0}-1$, and the remainder as $m-k$, i.e., $n=m\left(d_{0}-1\right)+(m-k)=m d_{0}-k$, where $d_{0}>1$ and $0... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,508 |
14.1. In how many ways can $k$ items be selected from $n$ different items if the order in which the items are selected: a) matters; b) does not matter. | 14.1. a) The first item can be chosen in $n$ ways, the second item in $n-1$ ways, $\ldots$, the $k$-th item in $n-k+1$ ways. In total, we get $n(n-1) \ldots(n-k+1)=\frac{n!}{(n-k)!}$ ways.
b) Each set of $k$ items, where the order of items is not considered, corresponds to $k!$ sets where the order of items is conside... | \frac{n!}{k!(n-k)!} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,509 |
14.2. Prove that $(x+y)^{n}=\sum_{k=0}^{n} C_{n}^{k} x^{k} y^{n-k}$, where
$$
C_{n}^{k}=\frac{n!}{k!(n-k)!}
$$
| The number $C_{n}^{k}=\frac{n!}{k!(n-k)!}$ is called a binomial coefficient. It is convenient to assume that $C_{n}^{k}=0$ if $k<0$ or $k>n$. | 14.2. The coefficient of $x^{k} y^{n-k}$ in the expansion of $(x+y)^{n}$ is equal to the number of ways to mark $k$ factors in the product of $n$ factors $x+y$. Indeed, in each marked factor we take $x$, and in each unmarked factor we take $y$. It remains to use the result of problem 14.1 b). | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,510 |
14.3. Prove that $C_{n}^{k}+C_{n}^{k-1}=C_{n+1}^{k}$. | 14.3. Let's use the identity $(1+x)^{n}(1+x)=(1+x)^{n+1}$. The coefficient of $x^{k}$ on the left side is $C_{n}^{k}+C_{n}^{k-1}$, and on the right side it is $C_{n+1}^{k}$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,511 |
14.4. How many necklaces can be made from five white beads and two black ones? | 14.4. Answer: 3. Black beads divide the white beads into two groups (one of these groups may contain 0 beads). The appearance of the necklace is completely determined by the number of beads in the smaller group. This number can be 0, 1, or 2. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,512 |
14.5. a) Seven girls are forming a circle. In how many different ways can they stand in a circle?
b) How many necklaces can be made from seven different beads? | 14.5. a) Answer: 720. Seven girls can be arranged in a circle in $7!=5040$ ways. However, in a circle, arrangements that can be obtained by rotating the circle are not considered different. By rotating the circle, one arrangement can result in 7 other arrangements. Therefore, the total number of ways is $7!/ 7=6!=720$.... | 360 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,513 |
14.6. How many different paths exist on a plane leading from a point with coordinates ( $0, n$ ) to a point with coordinates ( $m, m$ ), if it is allowed to move each time either 1 up (coordinate $y$ increases by 1), or 1 left (coordinate $x$ decreases by 1)? | 14.6. Answer: $C_{n}^{m}$. In total, we need to make $m$ steps up and $n-m$ steps to the left. Therefore, out of $n$ steps, we need to choose $m$ steps up. | C_{n}^{} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,514 |
14.7. a) In how many ways can a natural number $n$ be represented as the sum of $m$ non-negative integers, if the representations $n=x_{1}+\ldots+x_{m}$ and $n=y_{1}+\ldots+y_{m}$ are considered the same if and only if $x_{1}=y_{1}, \ldots, x_{m}=y_{m} ?$
b) The same question for representation as the sum of natural n... | 14.7. a) Answer: $C_{n+m-1}^{n}$ ways. We can associate the decomposition $n=x_{1}+\ldots+x_{m}$ with a sequence of zeros and ones, where initially there are $x_{1}$ ones, followed by one zero, then $x_{2}$ ones, followed by one zero, and so on; at the end, there are $x_{m}$ ones. In total, this sequence contains $n+m-... | C_{n+-1}^{n} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 38,515 |
14.8. Prove that
$$
\left(x_{1}+\ldots+x_{k}\right)^{n}=\sum_{l_{1}+\ldots+l_{k}=n} \frac{n!}{l_{1}!\ldots l_{k}!} x_{1}^{l_{1}} \ldots x_{k}^{l_{k}}
$$
## 14.2. Identities for binomial coefficients | 14.8. To obtain the monomial $x_{1}^{l_{1}} \ldots x_{k}^{l_{k}}$ from the product of $n$ factors $x_{1}+\ldots+x_{k}$, one needs to first select $l_{1}$ factors from $n$ and choose $x_{1}$ in them, then select $l_{2}$ from the remaining $n-l_{1}$ factors and choose $x_{2}$ in them, then select $l_{3}$ from the remaini... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,516 |
14.9. Prove that:
a) $C_{n}^{0}+C_{n}^{1}+\ldots+C_{n}^{n}=2^{n}$.
b) $C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+C_{n}^{6}+\ldots=C_{n}^{1}+C_{n}^{3}+C_{n}^{5}+\ldots=2^{n-1}$. | 14.9. a) $2^{n}=(1+1)^{n}=C_{n}^{0}+C_{n}^{1}+\ldots+C_{n}^{n}$.
b) $0=(1-1)^{n}=C_{n}^{0}-C_{n}^{1}+C_{n}^{2}-C_{n}^{3}+\ldots$ | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 38,517 |
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