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488. Bretschneider gives an approximate construction for $\pi$ accurate to the fifth decimal place: $\frac{13}{50} \sqrt{146}$, if the radius of the circle is taken as unity; construct this approximation.
## Problems of Zelena. | 488. Since $146=11^{2}+5^{2}$, the question reduces to constructing the expression
$$
\pi=\frac{13 \sqrt{11^{2}+5^{2}}}{50}
$$
We divide the diameter $A B$ (Fig. 113) into 10 equal parts and on its extension, we lay off $B C=\frac{1}{10} A B$ and $C D=\frac{2}{10} A B$. On the tangent at the end $A$ of the diameter, ... | MS=\frac{13}{25}\sqrt{246}=12\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,082 |
490. Five bandits took a wallet filled with ducats from a passerby. The strongest of them took 81 ducats, and each of the other four took a different amount. Due to the unequal division, a dispute arose, and the ataman who arrived at that time ordered the one who took the most to double the number of ducats each of the... | 490. Instruction. At the end of the section, each has $\frac{1}{5}$ of the total, so before doubling, four have $\frac{1}{10}$ each, and the fifth has $\frac{6}{10}$; before the second doubling, the first three have $\frac{1}{20}$ each, the fourth has $\frac{11}{20}$, and the fifth has $\frac{6}{20}$; before the third ... | 160 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,084 |
491. Find the ratio of the volumes obtained by the sequential rotation of a parallelogram about each of two adjacent sides. | 491. Answer. The ratio of volumes is equal to the ratio of adjacent sides. | Theofvolumesisequaltotheofadjacentsides. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,085 |
492. The volumes obtained by rotating a rectangle about each of its sides are respectively $a$ cubic meters and $b$ cubic meters. Find the length of the diagonal of the rectangle. | 492. If the measurements of a rectangle are $x$ and $y$, then the diagonal is
$$
\sqrt{x^{2}+y^{2}} .
$$
210
Since by condition $\pi x^{2} y=a$ and $\pi y^{2} x=b$, then
$$
\frac{y}{x}=\frac{b}{a}
$$
Multiplying the volumes gives $\pi^{2} x^{3} y^{3}=a b$, from which
$$
x y=\sqrt[3]{\frac{\overline{a b}}{\pi^{2}}... | \sqrt{\frac{^{2}+b^{2}}{}}\cdot\sqrt[6]{\frac{}{\pi^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,086 |
493. The height of a truncated cone is 3 meters, and the radii of its bases are 1 and 2 meters. Divide the volume into three parts, proportional to the numbers 2, 3, and 7, by two planes parallel to the bases.
## Problems from the course "Die Elemente der Mathematik" by R. Baltzer. | 493. Let the volumes of the sought parts be $v_{1}, v_{2}, v_{3}$, so that
$$
v_{1}: v_{2}: v_{3}=2: 3: 7
$$
The total volume $v=7 \pi$, hence,
$$
v_{1}=\frac{7}{6} \pi ; v_{2}=\frac{7}{4} \pi ; v_{3}=\frac{49}{12} \pi
$$
Triangles $A C B$ and $D E B$ (Fig. 114) are similar, therefore:
 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,087 |
496. System of equations:
$$
\begin{aligned}
x^{2}+y^{2} & =x y \\
x+y & =x y
\end{aligned}
$$ | 496. From the 1st equation $x^{2}+y^{2}+2 x y=3 x y$, or
$$
(x+y)^{2}=3 x y
$$
Substitution from the 2nd equation gives $(x+y)^{2}-3(x+y)=0$. Now we have the systems:
1) $x+y=0$
2) $x+y=3$,
$x y=0$.
$x y=3$.
The rest is clear. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,088 | |
497. Exponential equation:
$$
x^{a+b \lg x}=c
$$
## Sturm's Problem. | 497. $(a+b \lg x) \lg x=\lg c$; let $\lg x=y$, then
$$
b y^{2}+a y-\lg c=0 .
$$
Finding $y$, it is easy to find $x$.
Sturm, Jacques Charles (1803-1855), from Switzerland, the author of one important theorem of higher algebra, bearing his name. Author of a two-volume course of analysis, which has not lost its scienti... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,089 | |
498. A courier departs from $A$ and on the first day travels 10 li, and each subsequent day travels $\frac{1}{4}$ li more. After three days, another courier departs from city $B$, located 40 li behind city $A$, and travels in the same direction, traveling seven li on the first day, and each subsequent day traveling $\f... | 498. Sturm's Solution. The required number of days is $x$. The distance traveled by the first courier is the sum of the terms of an arithmetic progression, the extreme terms of which are 10 and $10+\frac{x-1}{4}$, i.e., it is equal to $\left(20+\frac{x-1}{4}\right) \frac{x}{2}$, or $\frac{(79+x) x}{8}$. The second cour... | 519 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,090 |
499. Through point $P$, a line is drawn intersecting the given perpendicular lines $O x$ and $O y$ at points $A$ and $B$. Then, from $P$, a perpendicular to $A B$ is drawn, intersecting the lines $O x$ and $O y$ at points $A_{1}$ and $B_{1}$. Perpendiculars are dropped from points $A_{1}$ and $B_{1}$ to $O P$, cutting ... | 499. The statement will be proven if it is shown that the midpoints of the lines $A B_{2}$ and $B A_{2}$ (Fig. 116) coincide,
Fig. 116.
$$
\text { that the perpendicular } M K \text { bei... | proof | Geometry | proof | Yes | Yes | olympiads | false | 38,091 |
500. In the plane of a triangle, find a point for which the sum of the squares of the distances from the sides of the triangle is minimal.
Problems from "Journal de Mathématiques elementaires," by M. Vuibert. | 500. Let $x, y, z$ be the distances from an arbitrary point (in the plane of the triangle) to its sides $a, b$, and $c$. We will use the Lagrange identity:
$$
\begin{gathered}
\left(x^{2}+y^{2}+z^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)=(a x+b y+c z)^{2}+(a y-b x)^{2}+ \\
+(b z-c y)^{2}+(c x-a z)^{2}
\end{gathered}
$$... | \frac{x}{}=\frac{y}{b}=\frac{z}{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,092 |
501. A trader has knowingly incorrect scales. To the first customer, he gives one pound of goods from one pan of the scales. To the second customer, he weighs out one pound of the same goods on the other pan, thinking that this compensates for the inaccuracy in weighing. The question is, did the trader actually gain or... | 501. The problem is based on the fact that the sum of two reciprocals is $>2$.
Indeed: let $a$ and $b$ be two unequal numbers, for example: $a>b$; then
$$
\begin{gathered}
a-b>0 \\
(a-b)^{2}>0 \\
a^{2}+b^{2}-2 a b>0 \\
a^{2}+b^{2}>2 a b \\
\frac{a}{b}+\frac{b}{a}>2
\end{gathered}
$$
*) C a se y, Géometrie élémentair... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 38,093 |
502. Prove that for any integer value of $n$ the number $n\left(n^{2}-1\right)\left(n^{2}-5 n+26\right)$ is divisible by 120
Problem from „Journal de Mathématiques speciales“. | 502. The given expression can be transformed as follows:
$$
\begin{gathered}
\left.n n^{2}-1\right)\left(n^{2}-5 n+26\right)=(n-1) n(n+1)\left[\left(n^{2}-5 n+6\right)+20\right]= \\
=(n-1) n(n+1)[(n-2)(n-3)+20]= \\
=(n-3)(n-2)(n-1) n(n+1)+20(n-1) n(n+1)
\end{gathered}
$$
The first term, being the product of five cons... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,094 |
503. Find a triangle with sides $a, b$ and $c$ and area $S$ expressed by four consecutive integers.
## Problem from "L'Education mathématique". | 503. Let the sides of the desired triangle be $x-1, x, x+1$, and the area $x+2$, then
$$
\begin{gathered}
\sqrt{3 x(x+2)(x-2) x}=4(x+2) \\
3 x^{2}\left(x^{2}-4\right)=16(x+2)^{2} \\
3 x^{2}(x-2)=16(x+2) \\
3 x^{3}-6 x^{2}-16 x-32=0 \\
3 x^{3}-12 x^{2}+6 x^{2}-24 x+8 x-32=0 \\
3 x^{2}(x-4)+6 x(x-4)+8(x-4)=0 \\
(x-4)\le... | 3,4,5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,095 |
504. Prove that for $p$ being a positive integer, the expression $4^{2 p}-3^{2 p}-7$ is divisible by 84.
## Problem from "Supplemento al Periodico di Matematica". | 504. Since the difference of even identical powers $4^{2 p}-3^{2 p}$ is divisible by the sum of the bases $(4+3)$, the given expression is divisible by 7. 216
In addition:
$\left.4^{2 p}-3^{2 p}-7=4^{2 p}-4-\left(3^{2 p}+3\right)=4^{1} 4^{2 p-1}-1\right)-3\left(3^{2^{p}-1}+1\right)$.
Since $4^{2 p-1}-1$ is divisible... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,096 |
505. Solve the system of equations:
$$
\begin{gathered}
x+y+z=9 \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \\
x y+x z+y z=27
\end{gathered}
$$
Problem from "Mathesis". | 505. The second of the given equations can be represented as:
$$
\frac{y z + x z + x y}{x y z} = 1
$$
Substitution from the third equation gives \( x y z = 27 \).
Now we have:
$$
\begin{gathered}
x + y + z = 9 \\
x y + x z + y z = 27 \\
x y z = 27
\end{gathered}
$$
Therefore, \( x, y \) and \( z \) are the roots o... | x=y=z=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,097 |
506. Solve the indeterminate equation in integers:
$x^{4}+y^{4}+z^{4}=u^{2}+v^{2}+w^{2}+t^{2}$.
Problems proposed by the Oxford Examination Board for university admission. | 506. Note. This equation is solved using the identity of E. N. Barisien:
$$
\begin{aligned}
& a^{4}+b^{4}+(a+b)^{4}=\left(a^{2}+a b+b^{2}\right)^{2}+ \\
& \left.\quad+a^{2} b^{2}+a^{2} a+b\right)^{2}+b^{2}(a+b)^{2}
\end{aligned}
$$
therefore, if
$$
\begin{gathered}
u=a^{2}+a b+b^{2} \\
v=a b \\
u=a^{\prime} a+b \\
t... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,098 |
1.1. Prove that if $x>0$, then $x+1 / x \geqslant 2$. | 1.1. For $x>0$ the given inequality is equivalent to the inequality $x^{2}-2 x+1 \geqslant 0$, i.e. $(x-1)^{2} \geqslant 0$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,100 |
1.2. a) Prove that $x(1-x) \leqslant 1 / 4$. b) Prove that $x(a-x) \leqslant a^{2} / 4$. | 1.2. a) The quadratic trinomial $x^{2}-x$ takes its minimum value at $x=1 / 2$; it is equal to $-1 / 4$.
b) The quadratic trinomial $x^{2}-a x$ takes its minimum value at $x=a / 2$; it is equal to $-a^{2} / 4$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,101 |
1.3. Prove that for numbers $a, b, c$, enclosed between 0 and 1, the inequalities $a(1-b)>1 / 4, b(1-c)>1 / 4$ and $c(1-a)>1 / 4$ cannot all be satisfied simultaneously. | 1.3. According to problem 1.2 b) $a(1-a) \leqslant 1 / 4, b(1-b) \leqslant 1 / 4, c(1-c) \leqslant$ $\leqslant 1 / 4$. Therefore
$$
a(1-b) b(1-c) c(1-a)=a(1-a) b(1-b) c(1-c) \leqslant(1 / 4)^{3}
$$ | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,102 |
1.4. For what $x$ does the function $f(x)=\left(x-a_{1}\right)^{2}+\ldots+\left(x-a_{n}\right)^{2}$ take its minimum value? | 1.4. It is clear that $f(x)=n x^{2}-2\left(a_{1}+\ldots+a_{n}\right) x+a_{1}^{2}+\ldots+a_{n}^{2}$. This quadratic trinomial takes its minimum value at $x=$ $=\frac{a_{1}+\ldots+a_{n}}{n}$. | \frac{a_{1}+\ldots+a_{n}}{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,103 |
1.5. Let $x, y, z$ be positive numbers whose sum is 1. Prove that $1 / x+1 / y+1 / z \geqslant 9$. | 1.5. By the condition $\frac{1}{x}=\frac{x+y+z}{x}=1+\frac{y}{x}+\frac{z}{x}$. Similar expressions can be written for $1 / y$ and $1 / z$. According to problem $1.2 y / x+x / y \geqslant 2$. Writing two more similar inequalities, we get the required result. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,104 |
1.6. Prove that the distance from the point $\left(x_{0}, y_{0}\right)$ to the line $a x+b y+c=0$ is $\frac{\left|a x_{0}+b y_{0}+c\right|}{\sqrt{a^{2}+b^{2}}}$. | 1.6. Let the point $(x, y)$ lie on the line $a x+b y+c=0$. Then $y=-\frac{a x+c}{b}$, so
$$
\begin{aligned}
\left(x-x_{0}\right)^{2}+(y & \left.-y_{0}\right)^{2}=\left(x-x_{0}\right)^{2}+\left(\frac{a x+c}{b}+y_{0}\right)^{2}= \\
& =\frac{a^{2}+b^{2}}{b^{2}} x^{2}+2\left(-x_{0}+\frac{a c}{b^{2}}+\frac{a y_{0}}{b}\righ... | \frac{|x_{0}+by_{0}+|}{\sqrt{^{2}+b^{2}}} | Geometry | proof | Yes | Yes | olympiads | false | 38,105 |
1.7. Let $a_{1}, \ldots, a_{n}$ be non-negative numbers, and $a_{1}+\ldots+a_{n}=a$. Prove that
$$
a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n} \leqslant a^{2} / 4
$$
## 1.2. Discriminant | 1.7. Let $x=a_{1}+a_{3}+a_{5}+\ldots$ Then $a-x=a_{2}+a_{4}+a_{6}+\ldots$ Therefore, according to problem 1.2 b)
$$
\begin{aligned}
& a^{2} / 4 \geqslant x(a-x)=\left(a_{1}+a_{3}+a_{5}+\ldots\right)\left(a_{2}+a_{4}+a_{6}+\ldots\right) \geqslant \\
& \geqslant a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}
\end{aligned}... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,106 |
1.8. a) Let $a, b, c$ be real numbers. Prove that the quadratic equation
$$
(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
$$
has a real root.
b) Prove that $(a+b+c)^{2} \geqslant 3(a b+b c+c a)$. | 1.8. a) Let, for definiteness, $a \leqslant b \leqslant c$. Then for $x=b$, the left-hand side of the equation takes the value $(b-c)(b-a) \leqslant 0$. And for very large $x$, the left-hand side is positive, so the equation has a real root.
b) It is sufficient to note that the discriminant of the equation considered ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,107 |
1.9. Let $a_{1}, \ldots, a_{n}, b_{1}, \ldots, b_{n}$ be real numbers. Prove the Cauchy inequality
$$
\left(a_{1} b_{1}+\ldots+a_{n} b_{n}\right)^{2} \leqslant\left(a_{1}^{2}+\ldots+a_{n}^{2}\right)\left(b_{1}^{2}+\ldots+b_{n}^{2}\right)
$$ | 1.9. The quadratic trinomial $\left(a_{1} x+b_{1}\right)^{2}+\ldots+\left(a_{n} x+b_{n}\right)^{2}$ is non-negative for all $x$, so its discriminant is non-positive. By calculating the discriminant, we obtain the required result. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,108 |
1.11. a) The golden section refers to the division of a segment into two parts, such that the whole segment is to the larger part as the larger part is to the smaller. What is the ratio of the smaller part to the larger in this case?
b) Let $a$ be the base of an isosceles triangle with a base angle of $72^{\circ}$, an... | 1.11. a) Answer: $\frac{\sqrt{5}-1}{2}$. Let the segment be divided into two segments of lengths $y$ and $z$, where $y<z$. We have the golden section if and only if $\frac{y}{z}=\frac{z}{y+z}=x$, where $x$ is the number we are looking for. For $x$, we get the equation $x(x+1)=\frac{y}{z} \frac{y+z}{z}=1$. Solving it an... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 38,110 |
1.12. Prove that the quadratic trinomial $a x^{2}+b x+c$ takes integer values for all integer $x$ if and only if the numbers $2 a, a+b$ and $c$ are integers. | 1.12. Suppose first that the quadratic trinomial $f(x)=$ $=a x^{2}+b x+c$ takes integer values for all integer $x$. Then, in particular, the number $f(0)=c$ is an integer. The numbers $f( \pm 1)-c=a \pm b$ are also integers. Therefore, the number $2 a=(a+b)+(a-b)$ is an integer.
Now suppose that the numbers $2 a, a+b$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,111 |
1.13. For the equations $x^{2}+a x+b=0$ and $x^{2}+c x+d=0$, there are no roots less than $x_{0}$. Prove that the equation $x^{2}+\frac{a+c}{2} x+$ $+\frac{b+d}{2}=0$ also has no roots less than $x_{0}$. | 1.13. It follows from the condition that if $x \geqslant x_{0}$, then $x^{2}+a x+b>0$ and $x^{2}+c x+d>0$. Therefore, if $x>x_{0}$, then $2 x^{2}+(a+c) x+(b+d)>0$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,112 |
1.14. Prove that if the equations with integer coefficients
\[
\begin{aligned}
& x^{2}+p_{1} x+q_{1}=0 \\
& x^{2}+p_{2} x+q_{2}=0
\end{aligned}
\]
have a common non-integer root, then \( p_{1}=p_{2} \) and \( q_{1}=q_{2} \). | 1.14. If the quadratic equation with integer coefficients $x^{2}+p_{1} x+q_{1}=$ $=0$ has a non-integer root $x_{1}$, then this root is irrational. Indeed, let $x_{1}=m / n$ be an irreducible fraction. Then $m^{2}+$ $+p_{1} m n+q_{1} n^{2}=0$, so $m^{2}$ is divisible by $n$. But by assumption, the numbers $m$ and $n$ a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,113 |
1.15. Prove that if for any positive $p$ all roots of the equation $a x^{2}+b x+c+p=0$ are real and positive, then the coefficient $a$ is equal to zero. | 1.15. Suppose that $a>0$. Then for large positive $p$ the discriminant $D=b^{2}-4 a c-4 a p$ is negative, so
the equation has no real roots at all. Suppose that $a<0$. Then for large positive $p$ the product of the roots, equal to $\frac{c+p}{a}$, is negative. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,114 |
1.16. a) The quadratic trinomial $a x^{2}+b x+c$ is a perfect fourth power for all integer $x$. Prove that then $a=b=0$.
b) The quadratic trinomial $a x^{2}+b x+c$ is a perfect square for all integer $x$. Prove that then $a x^{2}+b x+c=(d x+e)^{2}$. | 1.16. a) It is clear that $a \geqslant 0$ and $c \geqslant 0$. Consider the values of $x$ equal to $1, 2, \ldots, n$. If one of the numbers $a$ or $b$ is not zero, then the quadratic polynomial $a x^{2} + b x + c$ for such $x$ takes at least $n / 2$ different values. These values are between 0 and $a n^{2} + |b| n + c$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,115 |
1.17. Let $x_{1}$ and $x_{2}$ be the roots of the quadratic equation $x^{2} + a x + b = 0$, and let $s_{n} = x_{1}^{n} + x_{2}^{n}$. Prove that
$$
\frac{s_{n}}{n} = \sum_{m} (-1)^{n+m} \frac{(n-m-1)!}{m!(n-2m)!} a^{n-2m} b^{m}
$$
where the summation is over all integers $m$ such that $0 \leqslant m \leqslant n / 2$ (... | 1.17. According to Vieta's theorem, $x_{1}+x_{2}=-a$ and $x_{1} x_{2}=b$. For $n=1$ and $n=2$, the required equality takes the form $x_{1}+x_{2}=-a$ and $\frac{1}{2}\left(x_{1}^{2}+x_{2}^{2}\right)=$ $=\frac{1}{2} a^{2}-b$. The second equality is easily verified. Suppose that the required equality is proved for all nat... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,116 |
1.18. Prove the Intermediate Value Theorem for a quadratic trinomial. | 1.18. If a quadratic trinomial has no roots, then all its values are of the same sign. If a quadratic trinomial has exactly one root, then all its values are simultaneously non-negative or non-positive. If a quadratic trinomial has roots $x_{1}$ and $x_{2}$, where $x_{1}x_{2}$ have one sign, and for $x_{1}<x<x_{2}$ - t... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,117 |
1.19. Let $\alpha$ be a root of the equation $x^{2} + a x + b = 0$, and $\beta$ be a root of the equation $x^{2} - a x - b = 0$. Prove that between the numbers $\alpha$ and $\beta$ there is a root of the equation $x^{2} - 2 a x - 2 b = 0$. | 1.19. Let $f(x)=x^{2}-2 a x-2 b$. By the condition $\alpha^{2}=-a \alpha-b$ and $\beta^{2}=a \beta+\beta$, therefore $f(\alpha)=\alpha^{2}+2 \alpha^{2}=3 \alpha^{2}$ and $f(\beta)=\beta^{2}-2 \beta^{2}=-\beta^{2}$. Hence, $f(\alpha) f(\beta) \leqslant 0$. Therefore, on the interval $[\alpha, \beta]$ there lies at least... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,118 |
1.20. The quadratic trinomial $f(x)$ has two real roots, the difference between which is not less than a natural number $n \geqslant$ $\geqslant 2$. Prove that the quadratic trinomial $f(x)+f(x+1)+\ldots$ $\ldots+f(x+n)$ has two real roots. | 1.20. Let the quadratic trinomial $f(x)$ have real roots $x_{1}$ and $x_{2}=x_{1}+n+a$, where $a \geqslant 0$. For definiteness, we will assume that the coefficient of $x^{2}$ is positive. Let $x_{0}=x_{1}+a / 2$. Then $x_{0} \geqslant x_{1}$ and $x_{0}+n \leqslant x_{2}$. Therefore, $f\left(x_{0}\right)+f\left(x_{0}+1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,119 |
1.21. Given the equations $a x^{2}+b x+c=0$ and $-a x^{2}+b x+c=0$. Prove that if $x_{1}$ is a root of the first equation, and $x_{2}$ is a root of the second, then there exists a root $x_{3}$ of the equation $\frac{a}{2} x^{2}+b x+c=0$, for which either $x_{1} \leqslant x_{3} \leqslant x_{2}$, or $x_{1} \geqslant x_{3... | 1.21. When $x=x_{1}$ and $x=x_{2}$, the quadratic $\frac{a}{2} x^{2}+b x+c$ takes the values $-a x_{1}^{2} / 2$ and $3 a x_{2}^{2} / 2$. These values have different signs, so one root of the quadratic is located between $x_{1}$ and $x_{2}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,120 |
1.22. Prove that on the interval $[-1,1]$ the quadratic trinomial $f(x)=x^{2}+a x+b$ takes a value, the absolute value of which is not less than $1 / 2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 1.22. Suppose that $|f(x)|g(0)$. This means that the graphs of the functions $f$ and $g$ intersect at least at two points: one point of intersection lies on the segment $[-1,0]$, and the second one lies on the segment $[0,1]$.
Let's show that the graphs of the functions $f$ and $g$ intersect at only one point. From th... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,121 | |
1.24. Prove that if $\left|a x^{2}+b x+c\right| \leqslant 1$ for $|x| \leqslant 1$, then $\left|c x^{2}+b x+a\right| \leqslant 2$ for $|x| \leqslant 1$.
## 1.5. Equation of the tangent to a conic
## | 1.24. According to problem $1.23\left|a y^{2}+b y+c\right| \leqslant 2 y^{2}-1$ for $|y| \geqslant 1$. Let $y=1 / x$. Then
$$
\left|c x^{2}+b x+a\right|=\frac{1}{y^{2}}\left|a y^{2}+b y+c\right| \leqslant \frac{1}{y^{2}}\left(2 y^{2}-1\right) \leqslant 2
$$
for $|y| \geqslant 1$, i.e., for $0<|x| \leqslant 1$. For $x... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,123 |
1.25. Find the equation of the tangent to the conic $a x^{2}+$ $+b x y+c y^{2}+d x+e y+f=0$ at the point $\left(x_{0}, y_{0}\right)$. | 1.25. Suppose that the point $\left(x_{0}, y_{0}\right)$ belongs to the conic
$$
a x^{2}+b x y+c y^{2}+d x+e y+f=0
$$
Any line passing through the point $(x_{0}, y_{0})$ is given by the equation $y-y_{0}=k\left(x-x_{0}\right)$ (or the equation $x=x_{0}$, which corresponds to $k=\infty$). Let's find the second point o... | (2x_{0}+by_{0}+)x+(bx_{0}+2y_{0}+e)y+x_{0}+ey_{0}+2f=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,124 |
1.27. Prove that the quadratic equations $x^{2}+p_{1} x+q_{1}=$ $=0$ and $x^{2}+p_{2} x+q_{2}=0$ have a common root (possibly complex) if and only if
$$
\left(q_{2}-q_{1}\right)^{2}+\left(p_{1}-p_{2}\right)\left(p_{1} q_{2}-q_{1} p_{2}\right)=0
$$
The expression $\left(q_{2}-q_{1}\right)^{2}+\left(p_{1}-p_{2}\right)\... | 1.27. Let $x_{1}$ be a common root of the given equations. Subtracting one equation from the other, we get $\left(p_{1}-p_{2}\right) x_{1}=q_{2}-q_{1}$. If $p_{1}=p_{2}$, then $q_{1}=q_{2}$. If $p_{1} \neq p_{2}$, then $x_{1}=\frac{q_{2}-q_{1}}{p_{1}-p_{2}}$. Substituting this expression for $x_{1}$ into either of the ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,126 |
1.28. Prove that if the numbers $p_{1}, p_{2}, q_{1}, q_{2}$ satisfy the inequality $\left(q_{2}-q_{1}\right)^{2}+\left(p_{1}-p_{2}\right)\left(p_{1} q_{2}-q_{1} p_{2}\right)<0$, then the quadratic trinomials $x^{2}+p_{1} x+q_{1}$ and $x^{2}+p_{2} x+q_{2}$ have real roots and between the roots of each of them lies a ro... | 1.28. From the condition, in particular, it follows that $p_{1} \neq p_{2}$. Let $x_{1}=\frac{q_{2}-q_{1}}{p_{1}-p_{2}}$. Then
$$
x_{1}^{2}+p_{1} x_{1}+q_{1}=x_{1}^{2}+p_{2} x_{1}+q_{2}=\frac{\left(q_{2}-q_{1}\right)^{2}+\left(p_{1}-p_{2}\right)\left(p_{1} q_{2}-q_{1} p_{2}\right)}{\left(p_{1}-p_{2}\right)^{2}}<0
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,127 |
2.1. Solve the equation
$$
\left(x^{2}-x-1\right)^{3}+\left(x^{2}-3 x+2\right)^{3}=\left(2 x^{2}-4 x+1\right)^{3}
$$ | 2.1. Let $u=x^{2}-x-1$ and $v=x^{2}-3 x+2$. Then the equation in question can be written as $u^{3}+v^{3}=(u+v)^{3}$, i.e., $3 u v(u+v)=0$. It remains to solve the three quadratic equations $x^{2}-x-1=0, x^{2}-3 x+$ $+2=0$ and $2 x^{2}-4 x+1=0$. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,128 | |
2.5. Solve the equation $\frac{1}{x^{2}}-\frac{1}{(x+1)^{2}}=1$.
## 2.2. Guessing roots
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
2.5. Solve the equation $\frac{1}{x^{2}}-\frac{1}{(x+1)^{2}}=1$.
## 2.2. Guessi... | 2.5. This equation is equivalent to the equation $x^{4}+2 x^{3}+x^{2}-2 x-$ $-1=0$. The equation from problem 2.4 when $a=2$ and $b=-1$ takes exactly this form. | Algebra | proof | Yes | Yes | olympiads | false | 38,132 | |
2.6. Solve the equation $\frac{\left(x^{2}-x+1\right)^{3}}{x^{2}(x-1)^{2}}=\frac{\left(a^{2}-a+1\right)^{3}}{a^{2}(a-1)^{2}}$, where $a>1$. | 2.6. The rational function $R(x)=\frac{\left(x^{2}-x+1\right)^{3}}{x^{2}(x-1)^{2}}$ transforms into itself when $x$ is replaced by $1 / x$ or $1-x$. Therefore, the equation under consideration has roots $a, 1 / a, 1-a, 1 /(1-a), 1-1 / a$ and $a /(1-a)$.
For $a>1$, all these six numbers are distinct. The equation under... | ,\frac{1}{},1-,\frac{1}{1-},1-\frac{1}{},\frac{}{1-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,133 |
2.7. Solve the equation
$$
1-\frac{x}{1}+\frac{x(x-1)}{2!}-\ldots+(-1)^{n} \frac{x(x-1) \ldots(x-n+1)}{n!}=0
$$ | 2.7. Answer: $x=1,2, \ldots, n$. The equality
$$
1-\frac{k}{1}+\frac{k(k-1)}{2!}-\ldots+(-1)^{k} \frac{k(k-1) \cdot \ldots \cdot 2 \cdot 1}{k!}=(1-1)^{k}=0
$$
shows that the numbers $k=1,2, \ldots, n$ are roots of this equation. The equation cannot have more than $n$ roots, since its degree is $n$. | 1,2,\ldots,n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,134 |
2.8. Let $n>1$ be a natural number. Find all positive solutions of the equation $x^{n}-n x+n-1=0$.
## 2.3. Equations with Radicals
In problems $2.9-2.15$, it is assumed that the values of square roots are non-negative. We are interested only in the real roots of the equations. | 2.8. A n s w e r: $x=1$. It is clear that
$$
x^{n}-n x+n-1=\left(1+x+\ldots+x^{n-1}-n\right)(x-1)
$$
If $x>1$, then $1+x+\ldots+x^{n-1}-n>0$, and if $0<x<1$, then $1+x+\ldots$ $\cdots+x^{n-1}-n<0$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,135 |
2.9. Solve the equation $\sqrt{2 x-6}+\sqrt{x+4}=5$.
### 2.10. Solve the equation
$$
\sqrt[m]{(1+x)^{2}}-\sqrt[m]{(1-x)^{2}}=\sqrt[m]{1-x^{2}}
$$ | 2.9. Let $y=\sqrt{x+4}$. Then $x=y^{2}-4$, so $2 x-6=2 y^{2}-$ -14. Therefore, we get the equation $\sqrt{2 y^{2}-14}+y=5$. Move $y$ to the right side and square both sides. As a result, we get the equation $y^{2}+10 y-39=0$. Its roots are 3 and -13. But $y \geqslant 0$, so only the root $y=3$ remains, which correspond... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,136 |
2.11. Solve the equation
$$
3 \sqrt{x^{2}-9}+4 \sqrt{x^{2}-16}+5 \sqrt{x^{2}-25}=\frac{120}{x}
$$ | 2.11. The expression on the left side increases as $x$ increases, while the expression on the right side decreases. Therefore, the equation has no more than one solution. It is easy to verify that $x=5$ is a solution. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,137 |
2.12. Solve the equation $x+\sqrt{3+\sqrt{x}}=3$. | 2.12. If $x>1$, then $x+\sqrt{3+\sqrt{x}}>3$, and if $x<1$, then $x+$ $+\sqrt{3+\sqrt{x}}<3$. Only the root $x=1$ remains. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,138 |
2.13. Solve the equation $\sqrt{a-\sqrt{a+x}}=x$. | 2.13. By getting rid of the radicals, we first arrive at the equation $a-\sqrt{a+x}=x^{2}$, and then to the equation $\left(a-x^{2}\right)^{2}=a+x$, i.e.
$$
x^{4}-2 a x^{2}-x+a^{2}-a=0
$$
With respect to $a$, this is a quadratic equation. Solving it, we get two solutions:
$$
\begin{gathered}
a=x^{2}+x+1 \\
a=x^{2}-x... | -\frac{1}{2}+\sqrt{\frac{3}{4}}for\geq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,139 |
2.14. Solve the equation
$$
\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1
$$ | 2.14. Answer: $5 \leqslant x \leqslant 10$. Note that
\[
\begin{aligned}
& x+3-4 \sqrt{x-1}=(\sqrt{x-1}-2)^{2} \\
& x+8-6 \sqrt{x-1}=(\sqrt{x-1}-3)^{2}
\end{aligned}
\]
Therefore, the original equation can be written as
\[
|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1
\]
(we consider all roots to be positive). Let's consider all... | 5\leqslantx\leqslant10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,140 |
2.15. Solve the equation $\sqrt[3]{1-x}+\sqrt[3]{1+x}=p$, where $p$ is an arbitrary real number.
## 2.4. Different Equations
### 2.16. Solve the equation
$$
|x+1|-|x|+3|x-1|-2|x-2|=x+2
$$ | 2.15. Let's raise both sides of the equation to the third power:
$$
2+3 \sqrt[3]{1-x^{2}}(\sqrt[3]{1-x}+\sqrt[3]{1+x})=p^{3}
$$
Then substitute $p$ for $\sqrt[3]{1-x}+\sqrt[3]{1+x}$. As a result, we get the equation $2+3 p \sqrt[3]{1-x^{2}}=p^{3}$, from which
$$
x= \pm \sqrt{1-\left(\frac{p^{3}-2}{3 p}\right)^{3}}
$... | \\sqrt{1-(\frac{p^{3}-2}{3p})^{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,141 |
2.17. Solve the equation $x^{3}-[x]=3$, where $[x]$ denotes the greatest integer not exceeding $x$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2.17. Answer: $x=\sqrt[3]{4}$.
Let $[x]=n$ and $x=n+\alpha$, where $0 \leqslant \alpha<1$. The given equation can be written as $x^{3}-x+\alpha=3$. The inequality $0 \leqslant \alpha<1$ shows that $2<x^{3}-x \leqslant 3$. If $x \geqslant 2$, then $x\left(x^{2}-1\right) \geqslant 2 \cdot 3=6$, so in this case the inequ... | \sqrt[3]{4} | Number Theory | proof | Yes | Yes | olympiads | false | 38,142 |
3.1. $\left\{\begin{array}{l}x(y+z)=35 \\ y(x+z)=32 \\ z(x+y)=27 .\end{array}\right.$ | 3.1. The system is linear with respect to the unknowns $x_{1}=y z$, $y_{1}=x z, z_{1}=x y$. To find $x_{1}$, we need to add the last two equations and subtract the first equation from them. The result is $x_{1}=(32+27-35) / 2=12$. After this, we find $y_{1}=15$ and $z_{1}=20$.
Thus, $y z=12, x z=15, x y=20$. Therefore... | (5,4,3)(-5,-4,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,143 |
3.2. $\left\{\begin{array}{l}x+y+x y=19 \\ y+z+y z=11 \\ z+x+z x=14 .\end{array}\right.$ | 3.2. This system can be rewritten as $x_{1} y_{1}=20, y_{1} z_{1}=$ $=12, x_{1} z_{1}=15$, where $x_{1}=x+1, y_{1}=y+1, z_{1}=z+1$. Therefore, $\left(x_{1}, y_{1}, z_{1}\right)=(5,4,3)$ or $(-5,-4,-3)$, i.e., $(x, y, z)=(4,3,2)$ or $(-6,-5,-4)$. | (x,y,z)=(4,3,2)or(-6,-5,-4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,144 |
3.3. $\left\{\begin{array}{l}2 y=4-x^{2}, \\ 2 x=4-y^{2} .\end{array}\right.$ | 3.3. Subtracting the second equation from the first, we get $2(y-x)=$ $=y^{2}-x^{2}=(y-x)(y+x)$. Therefore, either $y=x$ or $y+x=2$. If $y=x$, then $2x=4-x^{2}$, i.e., $x=-1 \pm \sqrt{5}$. If $y+x=2$, then $2(2-x)=4-x^{2}$, i.e., $x^{2}=2x$. As a result, we obtain four solutions: $(-1+\sqrt{5},-1+\sqrt{5}),(-1-\sqrt{5}... | (-1+\sqrt{5},-1+\sqrt{5}),(-1-\sqrt{5},-1-\sqrt{5}),(0,2),(2,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,145 |
3.4. $\left\{\begin{array}{l}x+y+z=a, \\ x^{2}+y^{2}+z^{2}=a^{2}, \\ x^{3}+y^{3}+z^{3}=a^{3} .\end{array}\right.$ | 3.4. A n s w e r: $(0,0, a),(0, a, 0)$ and $(a, 0,0)$.
The identity $(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)=2(x y+y z+x z)$ shows that
$$
x y+y z+x z=0
$$
The identity $(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(y+z)(z+x)$ shows that $(x+y)(y+z)(z+x)=0$. Considering equation (1), we get $x y z=(x y+y z+x ... | (0,0,),(0,,0),(,0,0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,146 |
3.5. $\left\{\begin{array}{l}1-x_{1} x_{2}=0, \\ 1-x_{2} x_{3}=0, \\ 1-x_{3} x_{4}=0, \\ \ldots \ldots \ldots \ldots \ldots \ldots \\ 1-x_{n-1} x_{n}=0, \\ 1-x_{n} x_{1}=0 .\end{array}\right.$ | 3.5. Answer: $x_{1}=x_{2}=\ldots=x_{n}= \pm 1$ for odd $n$, $x_{1}=x_{3}=\ldots$ $\ldots=x_{n-1}=a$ and $x_{2}=x_{4}=\ldots=x_{n}=1 / a(a \neq 0)$ for even $n$.
Let $n$ be odd. It is clear that $x_{2} \neq 0$, so from the first and second equations we get $x_{1}=x_{3}$. From the second and third equations we get $x_{2... | x_{1}=x_{2}=\ldots=x_{n}=\1foroddn,x_{1}=x_{3}=\ldots=x_{n-1}=,x_{2}=x_{4}=\ldots=x_{n}=1/(\neq0)forevenn | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,147 |
3.6. $\left\{\begin{array}{l}x^{3}-y^{3}=26, \\ x^{2} y-x y^{2}=6 .\end{array}\right.$ | 3.6. Let $y=k x$. Immediately note that $k \neq 1$. From the equations
$$
\begin{aligned}
& x^{3}-k^{3} x^{3}=26 \\
& k x^{3} y-k^{2} x^{3}=6
\end{aligned}
$$
we get $x^{3}=\frac{26}{1-k^{3}}$ and $x^{3}=\frac{6}{k-k^{2}}$. Therefore,
$$
\frac{26}{1-k^{3}}=\frac{6}{k-k^{2}}
$$
This equation can be multiplied by $1-... | (-1,-3),(\frac{1\i\sqrt{3}}{2},\frac{3}{2}(1\i\sqrt{3})),(3,1),(\frac{-3\3i\sqrt{3}}{2},\frac{1}{2}(-1\i\sqrt{3})) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,148 |
3.9. $\left\{\begin{array}{l}x+y+x y=2+3 \sqrt{2}, \\ x^{2}+y^{2}=6 .\end{array}\right.$ | 3.9. Let $u=x+y$ and $v=x y$. Then $u+v=2+3 \sqrt{2}$ and $u^{2}-2 v=6$, so $u^{2}+2 u=6+2(2+3 \sqrt{2})=10+6 \sqrt{2}$. Therefore, $u=-1 \pm \sqrt{11+6 \sqrt{2}}=-1 \pm(3+\sqrt{2})$, i.e., $u=2+\sqrt{2}$ or $-4-\sqrt{2}$. In this case, $v=2+3 \sqrt{2}-u=2 \sqrt{2}$ or $6+4 \sqrt{2}$. If $u=-4-\sqrt{2}$ and $v=6+4 \sqr... | (x,y)=(2,\sqrt{2})or(\sqrt{2},2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,150 |
3.10. $\left\{\begin{array}{l}x^{3}+y^{3}=1, \\ x^{4}+y^{4}=1 .\end{array}\right.$ | 3.10. From the second equation, it follows that if $x=0$ or $\pm 1$, then $y= \pm 1$ or 0. It is also clear that $x \neq-1$ and $y \neq-1$. Therefore, there are exactly two solutions of this kind: $x=0, y=1$ and $x=1, y=0$. We will show that there are no other solutions.
We are interested in the case when $00$ and $y<... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,151 |
3.11. $\left\{\begin{array}{l}x+y=2, \\ x y-z^{2}=1 .\end{array}\right.$ | 3.11. From the second equation, it follows that $x y \geqslant 1$. The numbers $x$ and $y$ cannot both be negative, since their sum is 2. Therefore, the numbers $x$ and $y$ are positive and $x+y \geqslant 2 \sqrt{x y} \geqslant 2$, and the equality $x+y=2$ is possible only in the case when $x=y=1$. In this case, $z=0$. | 1,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,152 |
3.12. $\left\{\begin{array}{l}x+\frac{3 x-y}{x^{2}+y^{2}}=3, \\ y-\frac{x+3 y}{x^{2}+y^{2}}=0 .\end{array}\right.$ | 3.12. Multiply the first equation by $y$, the second by $x$ and add the resulting equations. As a result, we get $2 x y - 1 = 3 y$. In particular, $y \neq 0$, so $x = \frac{3}{2} + \frac{1}{2 y}$. Substituting this expression into the second equation, after some transformations, we get $4 y^{4} - 3 y^{2} - 1 = 0$. Cons... | x_1=2,y_1=1x_2=1,y_2=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,153 |
3.13. $\left\{\begin{array}{l}\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}, \\ \left(x_{4}+x_{5}+x_{1}\right)^{5}=3 x_{2}, \\ \left(x_{5}+x_{1}+x_{2}\right)^{5}=3 x_{3}, \\ \left(x_{1}+x_{2}+x_{3}\right)^{5}=3 x_{4}, \\ \left(x_{2}+x_{3}+x_{4}\right)^{5}=3 x_{5} .\end{array}\right.$
$. The function $f(x)=x^{5}$ is monotonically increasing, so $3 x_{2}=\left(x_{4}+x_{5}+x_{1}\right)^{5} \geqslant\left(x_{3}+x_{4}+x_{5}\right)^{5}=3 x_{1}$. Therefore, $x_{1}=x_{2}$ and $x_{3}=x_{1}$. Moreover, $3 x_{4... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,154 |
3.15. Find all positive solutions ($x_{1}>0, x_{2}>0$, $x_{3}>0, x_{4}>0, x_{5}>0$) of the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=x_{3}^{2} \\
x_{2}+x_{3}=x_{4}^{2} \\
x_{3}+x_{4}=x_{5}^{2} \\
x_{4}+x_{5}=x_{1}^{2} \\
x_{5}+x_{1}=x_{2}^{2}
\end{array}\right.
$$
## 3.4. The number of solutions of t... | 3.15. Let $x_{\min }=x_{i}$ be the smallest of the numbers $x_{1}, \ldots, x_{5}, x_{\max }=$ $=x_{j}$ be the largest. Then $x_{\min }^{2}=x_{i-2}+x_{i-1} \geqslant 2 x_{\min }$ (here it is assumed that $x_{0}=x_{5}$ and $x_{-1}=x_{4}$) and $x_{\max }^{2}=x_{j-2}+x_{j-1} \leqslant 2 x_{\max }$.
The numbers $x_{\min }$... | x_{\}=x_{\max}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,155 |
3.16. System of second-order equations
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=0 \\
(x-a)^{2}+y^{2}=1
\end{array}\right.
$$
generally has four solutions. For what values of $a$ does the number of solutions of the system decrease to three or to two?
## 3.5. Linear systems of equations | 3.16. Answer: the number of solutions decreases to three when $a= \pm 1$, and the number of solutions decreases to two when $a= \pm \sqrt{2}$.
From the first equation, we get $y= \pm x$. Substituting this expression into the second equation, we obtain
$$
(x-a)^{2}+x^{2}=1
$$
The number of solutions of the system dec... | \1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,156 |
3.17. Solve the system
$$
\left\{\begin{array}{l}
x_{1}+2 x_{2}+2 x_{3}+2 x_{4}+2 x_{5}=1 \\
x_{1}+3 x_{2}+4 x_{3}+4 x_{4}+4 x_{5}=2 \\
x_{1}+3 x_{2}+5 x_{3}+6 x_{4}+6 x_{5}=3 \\
x_{1}+3 x_{2}+5 x_{3}+7 x_{4}+8 x_{5}=4 \\
x_{1}+3 x_{2}+5 x_{3}+7 x_{4}+9 x_{5}=5
\end{array}\right.
$$ | 3.17. Answer: $x_{1}=x_{3}=x_{5}=1, x_{2}=x_{4}=-1$.
Let's first write down the first equation, then the second equation from which the first is subtracted, then the third equation from which the second is subtracted, and so on:
$$
\left\{\begin{aligned}
x_{1}+2 x_{2}+2 x_{3}+2 x_{4}+2 x_{5} & =1 \\
x_{2}+2 x_{3}+2 x... | x_{1}=x_{3}=x_{5}=1,x_{2}=x_{4}=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,157 |
3.18. Solve the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=6 \\
x_{2}+x_{3}+x_{4}=9 \\
x_{3}+x_{4}+x_{5}=3 \\
x_{4}+x_{5}+x_{6}=-3 \\
x_{5}+x_{6}+x_{7}=-9 \\
x_{6}+x_{7}+x_{8}=-6 \\
x_{7}+x_{8}+x_{1}=-2 \\
x_{8}+x_{1}+x_{2}=2
\end{array}\right.
$$ | 3.18. Answer: $x_{1}=-x_{8}=1, x_{2}=-x_{7}=2, x_{3}=-x_{6}=3, x_{4}=$ $=-x_{5}=4$.
By adding all the equations, we get $3\left(x_{1}+x_{2}+\ldots+x_{8}\right)=0$. Then, by adding the first, fourth, and seventh equations, we obtain $2 x_{1}+x_{2}+x_{3}+\ldots+x_{8}=1$, which means $x_{1}=1$. The other unknowns are fou... | x_{1}=-x_{8}=1,x_{2}=-x_{7}=2,x_{3}=-x_{6}=3,x_{4}=-x_{5}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,158 |
3.19. Let $a, b, c$ be pairwise distinct numbers. Solve the system of equations
$$
\left\{\begin{array}{l}
x + a y + a^{2} z + a^{3} = 0 \\
x + b y + b^{2} z + b^{3} = 0 \\
x + c y + c^{2} z + c^{3} = 0
\end{array}\right.
$$
34 | 3.19. Consider the polynomial $P(t)=t^{3}+t^{2} z+t y+x$. The pairwise distinct numbers $a, b, c$ are roots of the polynomial $P(t)$. Therefore, $P(t)=(t-a)(t-b)(t-c)$, which means $x=-a b c, y=a b+b c+c a$, $z=-(a+b+c)$. | -abc,++ca,-(+b+) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,159 |
3.20. Let $a_{1}, \ldots, a_{n}$ be pairwise distinct numbers. Prove that the system of linear equations
$$
\left\{\begin{array}{l}
x_{1}+\ldots+x_{n}=0 \\
a_{1} x_{1}+\ldots+a_{n} x_{n}=0 \\
7 a_{1}^{2} x_{1}+\ldots+a_{n}^{2} x_{n}=0 \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
a_{1}^{n-1} x_{... | 3.20. Using the fact that the numbers $a_{1}, \ldots, a_{n}$ are pairwise distinct, we construct a polynomial $P(t)$ which equals 0 for $t=a_{2}, \ldots, a_{n}$ and equals 1 for $t=a_{1}$. For this, we set $P(t)=\lambda\left(t-a_{2}\right) \ldots\left(t-a_{n}\right)$, where $\lambda\left(a_{1}-a_{2}\right) \ldots\left(... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,160 |
3.21. Find all solutions of the system of equations
$$
x\left(1-\frac{1}{2^{n}}\right)+y\left(1-\frac{1}{2^{n+1}}\right)+z\left(1-\frac{1}{2^{n+2}}\right)=0
$$
where $n=1,2,3,4, \ldots$ | 3.21. Answer: $y=-3 x, z=2 x(x-$ is an arbitrary number).
Let's prove that the given infinite system of equations is equivalent to the system of two equations:
$$
\begin{gathered}
x+y+z=0 \\
4 x+2 y+z=0
\end{gathered}
$$
First, if we subtract the second equation, multiplied by $\frac{1}{2^{n+2}}$, from the first equ... | -3x,2x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 38,161 |
3.22. Given 100 numbers $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$, satisfying the following conditions:
$$
\left\{\begin{array}{l}
a_{1}-3 a_{2}+2 a_{3} \geqslant 0 \\
a_{2}-3 a_{3}+2 a_{4} \geqslant 0 \\
a_{3}-3 a_{4}+2 a_{5} \geqslant 0 \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
a_{99}-3 a_{100}+2 a_... | 3.22. Let's add all these inequalities. The coefficient of $a_{k}$ will be $1-3+2=0$. Thus, we have a set of non-negative numbers $a_{1}-3 a_{2}+2 a_{3}, \ldots$, the sum of which is 0. Therefore, each of these numbers is 0, i.e., we have a system of equations instead of inequalities. These equations can be convenientl... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 38,162 |
3.23. Solve the system
$$
\left\{\begin{array}{rlrl}
10 x_{1}+3 x_{2}+4 x_{3}+x_{4}+x_{5} & & =0 \\
11 x_{2}+2 x_{3}+2 x_{4}+3 x_{5}+x_{6} & =0 \\
15 x_{3}+4 x_{4}+5 x_{5}+4 x_{6}+x_{7} & =0 \\
2 x_{1}+x_{2}-3 x_{3}+12 x_{4}-3 x_{5}+x_{6}+x_{7} & =0 \\
6 x_{1}-5 x_{2}+3 x_{3}-x_{4}+17 x_{5}+x_{6} & =0 \\
3 x_{1}+2 x_{... | 3.23. A n s w e r: $x_{1}=x_{2}=\ldots=x_{7}=0$.
Let's write the equations of the considered system in the form $\sum_{i=1}^{7} a_{i j} x_{i}=$ $=0$. The coefficients $a_{i j}$ have the following property: $\left|a_{j j}\right|>$ $>\sum_{i \neq i}\left|a_{i j}\right|$ for each $j$. Let $x_{1}, \ldots, x_{7}$ be a solut... | x_{1}=x_{2}=\ldots=x_{7}=0 | Algebra | proof | Yes | Yes | olympiads | false | 38,163 |
3.25. There is a system of equations
$$
\left\{\begin{array}{l}
* x+* y+* z=0 \\
* x+* y+* z=0 \\
* x+* y+* z=0
\end{array}\right.
$$
Two people take turns filling in the asterisks with numbers. Prove that the one who starts can always ensure that the system has a non-zero solution. | 3.25. The beginner, making the first move, writes down an arbitrary coefficient of $z$ in the first equation. Then, in response to the second move, he acts as follows. If the second writes some coefficient of $x$ or $y$, then the first writes the same coefficient of $y$ or $x$ in the same equation. If, however, the sec... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,164 |
4.2. Prove that the number of distinct divisors of a natural number $n$ (including 1 and the number itself) is odd if and only if the number is a perfect square. | 4.2. We associate with a divisor $d$ of the number $n$ the divisor $n / d$. If for all $d$ the numbers $d$ and $n / d$ are different (i.e., $n \neq d^{2}$), then the divisors of the number $n$ can be paired, so their number is even. If, however, $n=d^{2}$, then all divisors different from $d$ can be paired, so their nu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,166 |
4.3. Let $a_{1}, \ldots, a_{2 n+1}$ be integers, and $b_{1}, \ldots, b_{2 n+1}$ be the same numbers but in a different order. Prove that at least one of the numbers $a_{k}-b_{k}, k=1,2, \ldots, 2 n+1$, is even. | 4.3. Suppose all numbers $a_{k}-b_{k}$ are odd. Then the number $\left(a_{1}-b_{1}\right)+\ldots+\left(a_{2 n+1}-b_{2 n+1}\right)$ is also odd (the sum of an odd number of odd numbers is odd). But this number equals 0, since $a_{1}+\ldots$ $\ldots+a_{2 n+1}=b_{1}+\ldots+b_{2 n+1}$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,167 |
4.5. Prove that a polynomial with integer coefficients
$$
a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}
$$
taking odd values at $x=0$ and $x=1$, does not have integer roots. | 4.5. Let $P(x)=a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}$. By the condition, the numbers $a_{n}=P(0)$ and $a_{0}+a_{1}+\ldots+a_{n}=P(1)$ are odd. If $x$ is an even number, then $P(x) \equiv a_{n}(\bmod 2)$. If $x$ is an odd number, then $P(x) \equiv a_{0}+a_{1}+\ldots+a_{n}(\bmod 2)$. In both cases, we get that... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,169 |
4.6. Given two polynomials in the variable $x$ with integer coefficients. Their product is a polynomial in the variable $x$ with even coefficients, not all of which are divisible by 4. Prove that in one of the polynomials all coefficients are even, and in the other at least one is odd. | 4.6. From the fact that not all coefficients of the product are divisible by 4, it follows that one of the polynomials has an odd coefficient. We need to prove that the other polynomial has no odd coefficients. Suppose that both polynomials have odd coefficients. Replace each coefficient with its remainder when divided... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,170 |
4.8. Let $bc$ be divisible by $a$ and $\gcd(a, b)=1$. Prove that $c$ is divisible by $a$.
| 4.8. Let $m$ and $n$ be natural numbers for which $m a + n b = \gcd(a, b) = 1$. Then $m a c + n b c = c$, i.e., $a(m c + n_{1}) = c$, where the number $n_{1}$ is natural. This means that $c$ is divisible by $a$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,172 |
4.9. Prove the Fundamental Theorem of Arithmetic. | 4.9. Suppose that $a=p_{1} \ldots p_{r}=q_{1} \ldots q_{s}$, where $p_{1}, \ldots, p_{r}$, $q_{1}, \ldots, q_{s}$ are prime numbers. Clearly, either $\gcd \left(p_{1}, q_{1}\right)=1$, or $\gcd \left(p_{1}, q_{1}\right)=p_{1}$ and $p_{1}=q_{1}$. If $\gcd \left(p_{1}, q_{1}\right)=1$, then according to problem 4.8, the ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,173 |
4.10. Prove that the fraction $\frac{21 n+4}{14 n+3}$ is irreducible for any natural $n$. | 4.10. Let's find the greatest common divisor of the numbers $21 n+4$ and $14 n+3$ using the Euclidean algorithm. Dividing $21 n+4$ by $14 n+3$, we get a remainder of $7 n+1$. Dividing $14 n+3$ by $7 n+1$, we get a remainder of 1. Therefore, the numbers $21 n+4$ and $14 n+3$ are coprime. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,174 |
4.11. The sequence of natural numbers $a_{1}, a_{2}, \ldots$ is such that GCD $\left(a_{m}, a_{n}\right)=\text{GCD}\left(a_{m-n}, a_{n}\right)$ for any $m>n$. Prove that GCD $\left(a_{m}, a_{n}\right)=a_{d}$, where $d=$ GCD $(m, n)$. | 4.11. For any pair of natural numbers $\{m, n\}$, where $m>n$, successive operations $\{m, n\} \rightarrow\{m-n, n\}$ lead to a pair $\{d, d\}$, where $d=$ GCD $(m, n)$. Indeed, the operation of division of $m$ by $n$ with a remainder consists of several such operations. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,175 |
4.12. Prove that for any natural $a$ and any natural $m$ and $n$ the GCD $\left(a^{m}-1, a^{n}-1\right)=a^{d}-1$, where $d=$ GCD $(m, n)$
## 4.3. Prime Factorization | 4.12. According to problem 4.11, it is sufficient to check that GCD ( $a^{m}-$ $\left.-1, a^{n}-1\right)=$ GCD $\left(a^{m-n}-1, a^{n}-1\right)$ for any $m>n \geqslant 1$. But $a^{m}-1=$ $=a^{n}\left(a^{m-n}-1\right)+a^{n}-1$ and the numbers $a$ and $a^{n}-1$ are coprime. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,176 |
4.13. Let $p / q$ be an irreducible fraction, $p$ and $q$ be natural numbers. Set $f\left(\frac{p}{q}\right)=\frac{p^{2} q^{2}}{q_{1} \ldots q_{n}}$, where $q_{1}, \ldots, q_{n}$ are the distinct prime divisors of $q$. Prove that $f$ is a bijective mapping from the set of positive rational numbers to the set of all nat... | 4.13. Let $p=p_{1}^{a_{1}} \ldots p_{m}^{a_{m}}, q=q_{1}^{b_{1}} \ldots q_{n}^{b_{n}}$. Then $f(p / q)=p_{1}^{2 a_{1}} \ldots$ $\ldots p_{m}^{2 a_{m}} q_{1}^{2 b_{1}-1} \ldots q_{n}^{2 b_{n}-1}$. It is clear that each natural number can be represented in this form in exactly one way. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,177 |
4.14. Let $a_{n} a_{n-1} \ldots a_{1} a_{0}$ be the decimal representation of some number.
a) Prove that this number is divisible by 3 if and only if $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by 3.
b) Prove that this number is divisible by 9 if and only if $a_{0}+a_{1}+\ldots+a_{n}$ is divisible by 9.
c) Prove that th... | 4.14. a) The number 10 when divided by 3 gives a remainder of 1. Therefore, \(10^{k}\) when divided by 3 also gives a remainder of 1. Consequently, \(a_{k} 10^{k}\) when divided by 3 gives a remainder of \(a_{k}\).
b) The solution is analogous to the solution of part a).
c) The number 10 when divided by 11 gives a re... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,178 |
4.15. In the decimal representation of an integer, there are 300 ones, and the other digits are zeros. Can this number be a perfect square? | 4.15. This number is divisible by 3, but not by 9, so it cannot be a perfect square. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,179 |
4.16. There are seven tokens with the digits $1,2,3,4,5,6,7$. Prove that no seven-digit number formed using these tokens is divisible by another. | 4.16. Let $a$ and $b$ be seven-digit numbers formed using these tokens. Suppose that $a$ is divisible by $b$ and $a \neq b$. Then $a-b$ is also divisible by $b$. Clearly, $\frac{a-b}{b} \leqslant 7$. On the other hand, $a-b$ is divisible by 9, while $b$ is not divisible by 9. Therefore, $\frac{a-b}{b}$ is divisible
by ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,180 |
4.17. Let $a_{n} a_{n-1} \ldots a_{1} a_{0}$ be the decimal representation of some number. Replace this number with $a_{n} a_{n-1} \ldots a_{1} + 2 a_{0}$. Apply this transformation to the resulting number again and so on until a number not exceeding 19 is obtained. Prove that the original number is divisible by 19 if ... | 4.17. We replace the number $10 a+b$ with $a+2 b$. For any number $10 a+b$ greater than 19, the inequality $9 a>b$ holds, i.e., $10 a+b > a+2 b$. Therefore, the number always decreases under the specified transformations. It is clear that $10 a+b$ is divisible by 19 if and only if $20 a+2 b$ is divisible by 19, i.e., $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,181 |
4.18. Let $a_{n} a_{n-1} \ldots a_{1} a_{0}$ be the decimal representation of some number. Replace this number with $a_{n} a_{n-1} \ldots a_{1}-2 a_{0}$. Prove that the original number is divisible by 7 if and only if the resulting number is divisible by 7.
## 4.5. Greatest Common Divisor and Least Common Multiple | 4.18. Let's write the original number in the form $10 a + a_{0}$. We need to prove that it is divisible by 7 if and only if $a - 2 a_{0}$ is divisible by 7. It is clear that $10 a + a_{0}$ is divisible by 7 if and only if $20 a + 2 a_{0}$ is divisible by 7. It remains to note that when divided by 7, the number $20 a + ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,182 |
4.19. a) Prove that $\operatorname{GCD}(a, b)=\frac{a b}{\operatorname{LCM}(a, b)}$.
b) Prove that
$$
\operatorname{GCD}(a, b, c)=\frac{a b c \operatorname{LCM}(a, b, c)}{\operatorname{LCM}(a, b) \operatorname{LCM}(b, c) \operatorname{LCM}(c, a)}.
$$
(The general formula is given in problem 14.26.)
c) Prove that
$... | 4.19. a) Let $a=p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}}$ and $b=p_{1}^{\beta_{1}} \ldots p_{k}^{\beta_{k}}$. Then
$$
\begin{aligned}
& \text { GCD }(a, b)=p_{1}^{\min \left\{\alpha_{1}, \beta_{1}\right\}} \ldots p_{k}^{\min \left\{\alpha_{k}, \beta_{k}\right\}} \\
& \text { LCM }(a, b)=p_{1}^{\max \left\{\alpha_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,183 |
4.20. Prove that
GCD $(a$, GCD $(b, c))=$ GCD $($ GCD $(a, b), c)=$ GCD $(a, b, c)$. | 4.20. Let the highest power of a prime number $p$ that divides $a, b, c$ be $\alpha, \beta, \gamma$. It is required to prove that
$$
\min (\alpha, \min (\beta, \gamma))=\min (\min (\alpha, \beta), \gamma))=\min (\alpha, \beta, \gamma)
$$
where $\min (x, y)$ - the smallest of the numbers $x$ and $y$. This statement ab... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,184 |
4.21. Prove that $\frac{\operatorname{LCM}(a, a+b)}{\operatorname{LCM}(a, b)}=\frac{a+b}{b}$. | 4.21. According to problem 4.19 a)
$$
(a+b) \operatorname{LCM}(a, b)=\frac{(a+b) a b}{\operatorname{GCD}(a, b)}=\frac{(a+b) a b}{\operatorname{GCD}(a, a+b)}=b \text { LCM }(a, a+b) .
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,185 |
4.22. Natural numbers $a$ and $b$ are coprime. Prove that the GCD $\left(a+b, a^{2}+b^{2}\right)=1$ or 2. | 4.22. Suppose that the numbers $a+b$ and $a^{2}+b^{2}$ are divisible by $d$. Then the number $2 a b=(a+b)^{2}-\left(a^{2}+b^{2}\right)$ is also divisible by $d$. Therefore, the numbers $2 a^{2}=2 a(a+b)-2 a b$ and $2 b^{2}=2 b(a+b)-2 a b$ are also divisible by $d$. By the condition, the numbers $a$ and $b$ are coprime,... | 1or2 | Number Theory | proof | Yes | Yes | olympiads | false | 38,186 |
4.23. Prove that the greatest common divisor of the sum of two numbers and their least common multiple is equal to the greatest common divisor of the numbers themselves. | 4.23. The least common multiple of numbers $a$ and $b$ includes only those prime divisors that are present in $a$ and $b$. Only they can be included in the greatest common divisor of the sum and the least common multiple. Therefore, it is sufficient to track the degree of each prime factor separately. Let $a=p^{\alpha}... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,187 |
4.24. Prove that the least common multiple of $n$ natural numbers $a_{1}<a_{2}<\ldots<a_{n}$ is not less than $n a_{1}$. | 4.24. Let the least common multiple of the given numbers be $a$. Then $\frac{a}{a_{1}}>\frac{a}{a_{2}}>\ldots>\frac{a}{a_{n}}-$ are different natural numbers. Therefore, $\frac{a}{a_{1}} \geqslant n$, i.e., $a \geqslant n a_{1}$. | \geqslantna_{1} | Number Theory | proof | Yes | Yes | olympiads | false | 38,188 |
4.26. Given several natural numbers, each of which is less than a natural number $N \geqslant 4$. The least common multiple of any two of them is greater than $N$. Prove that the sum of the reciprocals of these numbers is less than 2.
## 4.6. Divisibility by an Integer | 4.26. Let $a_{1}, \ldots, a_{n}$ be given numbers. The number of terms in the sequence $1,2,3, \ldots, N$ that are divisible by $a_{k}$ is $\left[\frac{N}{a_{k}}\right]$. By the condition, the least common multiple of any two of the numbers $a_{1}, \ldots, a_{n}$ is greater than $N$, so there is no number among $1,2, \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,190 |
4.27. Prove that $7^{2 n}-5^{2 n}$ is divisible by 24. | 4.27. The number $a^{n}-b^{n}$ is divisible by $a-b$ (problem 5.1 a), therefore $7^{2 n}-5^{2 n}$ is divisible by $7^{2}-5^{2}=24$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 38,191 |
4.28. Prove that $\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7 n}{15}$ is an integer for any natural $n$. | 4.28. We need to prove that $3 n^{5}+5 n^{3}+7 n$ is divisible by 15, i.e., $5 n^{3}+7 n$ is divisible by 3 and $3 n^{5}+7 n$ is divisible by 5. It is clear that $5 n^{3}+7 n \equiv$ $\equiv-\left(n^{3}-n\right)(\bmod 3)$ and $3 n^{5}+7 n \equiv-2\left(n^{5}-n\right)(\bmod 5)$. By considering all different remainders, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 38,192 |
4.29. For which integers $n$ is the number $20^{n}+16^{n}-3^{n}-1$ divisible by 323? | 4.29. Answer: for even n. First of all, note that $323=17 \cdot 19$, so a number is divisible by 323 if and only if it is divisible by 17 and 19. The number $20^{n}-3^{n}$ is divisible by $20-3=17$. Furthermore, $16^{n} \equiv(-1)^{n}(\bmod 17)$, so the number $16^{n}-1$ is divisible by 17 if and only if $n$ is even. A... | forevenn | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 38,193 |
4.30. Prove that if for any natural $k \neq b$ the number $a-k^{n}$ is divisible by $b-k$, then $a=b^{n}$. (Here $a, b, n-$ are fixed natural numbers.) | 4.30. The polynomial $x^{n}-y^{n}$ is divisible by $x-y$ (problem 5.1 a), so $b^{n}-k^{n}$ is divisible by $b-k$. Therefore, the number $a-b^{n}=\left(a-k^{n}\right)-\left(b^{n}-k^{n}\right)$ is divisible by $b-k$ for any $k \neq b$. This is only possible if $a=b^{n}$. | b^{n} | Number Theory | proof | Yes | Yes | olympiads | false | 38,194 |
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