problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
16.17. Four spheres touch each other at different points and their centers lie in the same plane P. Sphere $S$ touches all these spheres. Prove that the ratio of its radius to the distance from its center to the plane P is $1: \sqrt{3}$. | 16.17. Consider the inversion with the center at the point of tangency of some two spheres. Under this inversion, the plane P transitions into itself, as shown in
Fig. 116, the point of tan... | 1:\sqrt{3} | Geometry | proof | Yes | Yes | olympiads | false | 24,590 |
16.18. Three mutually tangent spheres touch a plane at three points lying on a circle of radius $R$. Prove that there exist two spheres tangent to the three given spheres and the plane, and if $r$ and $\rho(\rho>r)$ are the radii of these spheres, then $\frac{1}{r}-\frac{1}{\rho}=\frac{2 \sqrt{3}}{R}$.
## § 4. Stereog... | 16.18. Consider the inversion of degree $(2 R)^{2}$ with center $O$ of the family of tangent spheres with a plane; this inversion transforms the circle passing through the points of tangency of the spheres with the plane into a line $A B$, at a distance of $2 R$ from the point $O$ ( $A$ and $B$ are the images of the po... | \frac{1}{r}-\frac{1}{\rho}=\frac{2\sqrt{3}}{R} | Geometry | proof | Yes | Yes | olympiads | false | 24,591 |
16.19. a) Prove that the stereographic projection coincides with the restriction to the sphere of some inversion in space.
b) Prove that under stereographic projection, a circle on the sphere passing through point $B$ is mapped to a line, and a circle not passing through point $B$ is mapped to a circle.
c) Prove that... | 16.19. Let the plane П be tangent to the sphere $S$ with diameter $A B$ at point $A$. Let, further, $X$ be a point on the sphere $S$, and $Y$ be the point of intersection of the ray $B X$ with the plane П. Then $\triangle A X / 3$ is similar to $\triangle Y A B$, and therefore $A B : X B = Y B : A B$, i.e., $X B \cdot ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,592 |
16.20. In space, a circle \( S \) and a point \( B \) are given. Let \( A \) be the projection of point \( B \) onto the plane containing circle \( S \). For each point \( D \) on circle \( S \), consider the point \( M \) - the projection of point \( A \) onto the line \( DB \). Prove that all points \( M \) lie on on... | 16.20. Since $\angle A M B=90^{\circ}$, point $M$ belongs to the circle with diameter $A B$. Therefore, point $D$ is the image of point $M$ under the stereographic projection of the sphere with diameter $A B$ onto the plane containing circle $S$. Consequently, all points $M$ lie on one circle - the image of circle $S$ ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,593 |
16.21. Given a pyramid $S A B C D$, where its base is a convex quadrilateral $A B C D$ with perpendicular diagonals, and the plane of the base is perpendicular to the line $S O$, where $O$ is the point of intersection of the diagonals. Prove that the bases of the perpendiculars dropped from point $O$ to the lateral fac... | 16.21. Drop a perpendicular \( O A^{\prime} \) from point \( O \) to the trapezoid \( S A B \). Let \( A_{1} \) be the intersection point of the lines \( A B \) and \( S A^{\prime} \). Since \( A B \perp O S \) and \( A B \perp O A^{\prime} \), the plane \( S O A^{\prime} \) is perpendicular to the line \( A B \), and ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 24,594 |
16.22. Sphere $S$ with diameter $A B$ touches plane $\Pi$ at point $A$. Prove that the stereographic projection of the symmetry with respect to a plane parallel to $\Pi$ and passing through the center of sphere $S$, transforms into an inversion with center $A$ and power $A B^{2}$. More precisely, if points $X_{1}$ and ... | 16.22. Since points $X_{1}$ and $X_{2}$ are symmetric with respect to the plane perpendicular to segment $A B$ and passing through its midpoint, then $\angle A B X_{1} = \angle B A X_{2}$. Therefore, the right triangles $A B Y_{1}$ and $A Y_{2} B$ are similar. Consequently, $A B : A Y_{1} = A Y_{2} : A B$, i.e., $A Y_{... | AY_{1}\cdotAY_{2}=AB^{2} | Geometry | proof | Yes | Yes | olympiads | false | 24,595 |
3. The plane is divided into triangles. Any two triangles either have no common points, or share a common vertex, or share a common side 1). The vertices of these triangles are labeled with the digits $0,1,2$ such that vertices belonging to the same side are labeled with different digits (we will call such labeling cor... | 3. With such numbering, the vertices of any triangle will be numbered with all three digits $0,1,2$. We will place arrows on the sides of the triangles in the direction from 0 to 1, from 1 to 2, and from 2 to 0. Each triangle will thus have a certain orientation (direction of traversal). The triangles will be of two ty... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,598 |
6. Number the fields of a 25-cell chessboard in the order in which you traversed them with a knight in the previous task. Shade the fields that received even numbers. Indicate what kind of coloring would result if the same operations were performed not on a 25-cell board, but on a chessboard with an arbitrary number of... | 6. See Figs. 90 and 91. We have obtained a correct coloring of the 25-cell board. On any other board, we will also obtain a correct coloring. Indeed, suppose the chessboard is already correctly colored with two colors. Then, with each move, the knight moves from a square of one color to a square of the opposite color. ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 24,601 |
9. The knight made $n$ moves and returned to the starting field. Prove that $n$ is an even number. | 9. With each move, the knight changes the color of its square. Therefore, if it lands on a square of the same color as the starting one (in this case, the starting square itself), it has made an even number of moves. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,604 |
11. Is it possible to lay out all 28 dominoes in a chain so that one end of the chain has six dots, and the other end has five dots? | 11. No, it cannot. A domino tile consists of two halves, each of which displays one of 7 numbers: $0,1,2,3,4,5,6$. There are as many tiles as there are possible pairs of these numbers. Each number, therefore, appears six times in combination with other numbers; in addition, there is one tile - a double - in which this ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 24,606 |
13. There were 225 people present at the meeting. Acquaintances exchanged handshakes. Prove that at least one of the participants shook hands with an even number of acquaintances ${ }^{1}$ ).[^2] | 13. If all participants of the meeting shook hands with an odd number of acquaintances, it would turn out that each of the 225 people (225 is an odd number) made an odd number of handshakes. This contradicts the previous problem (if we set $N=225$). | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,608 |
23. The map is correctly colored with two colors. Prove that all its vertices have even multiplicity. | 23. If any vertex had an odd multiplicity, then the countries adjacent to it could not be properly colored with two colors (Fig. 101). | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,618 |
24. All vertices of the map have even degree. The rook toured a row of countries on this map, visiting none of them more than once, and returned to the starting country. Prove that it made an even number of moves.
. Now, let's eliminate all parts of the map that lie outside our path, and attach the path itself to the map. We will get a new map, as shown in Fig. 103. All internal vertices of this map, by condition, have even multiplicity, while the ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,619 |
25. All vertices on the map have even degree. The rook toured several countries on this map and returned to the starting country (it could have visited some countries more than once). Prove that it made an even number of moves. | 25. The case when the rook, having left $S_{0}$, visited each country no more than once, has already been discussed in problem 24. Let it now visit some country $S_{1}$ twice, i.e., cross its own path in this country (Fig. 104). Let the number of moves on the segment $S_{0} a S_{1}$ be $p$, on $S_{1} b S_{1}$ be $q$, a... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,620 |
27. All vertices of the map have even multiplicity. Prove that it can be properly colored with two colors (compare problem 83).
Problems 23 and 27 give the following theorem, completely solving the problem of proper two-coloring:
A map can be properly colored with two colors if and only if all its vertices have even ... | 27. First method. Let's move the rook out of some country $S_{0}$ and start visiting the countries on our map one by one, numbering them in the order of visit. It might happen that we visit the same country twice or even more times. Then this country will receive not one number, but two or several numbers. However, due... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,622 |
31. Number the fields of the hexagonal board in the order in which you walked them with the camel in problem 30. Paint the fields with numbers divisible by three black, and the fields with numbers that give a remainder of 1 when divided by three red. What coloring do you get? What coloring will you get if you perform t... | 31. We have obtained a correct three-color coloring of the board with a side of 3 (Figs. 106 and 107) ${ }^{2}$. The same will be true for any other board. Indeed, suppose the board is already correctly colored with three colors. In this case, the camel moves from a red field to a white one, from a white one to a black... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 24,626 |
32. The camel made $n$ moves and returned to the starting field. Prove that $n$ is divisible by 3. | 32. If, for example, the camel has moved from a black field, then the sequence of colors of the fields it passes has the form (see the previous problem)
bwb bwb bwb...
The period of this sequence consists of three members, so the camel can end up on a field of the same color as the starting one (in particular, on th... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,627 |
33. Prove that it is impossible for a camel to traverse all the fields of a hexagonal board with a side length of 3 exactly once if it starts from a corner field. | 33. To go around a board consisting of 19 fields (Fig. 107), the camel must make 18 moves, i.e., a number of moves that is a multiple of three, and therefore end up on a field of the same color as the starting one. Consequently, the camel passes more fields of this color than fields of each of the other colors. However... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,628 |
37. Prove that it is impossible for a camel to traverse each field of the board in Fig. 23 exactly once if the traversal starts from a corner field.
We can consider the camel on schemes that are significantly more general than the schemes depicted in Figs. 21 and 22. Let's recall that the triangulation of a polygon is... | 37. On our board, there are 25 fields. To cover it in the specified manner, 24 moves are required. Every third move, the camel lands on a field of the same color as the starting one (see Fig. 109). Thus, with the last move, the camel
13 Zak. 3398. E. Dunkin and V. Uspenskin
lands on a field of the starting color, and ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 24,631 |
39. The camel made $n$ steps and returned to the initial vertex. Prove that $n$ is divisible by 3 (see problem 4). | 39. First of all, it is clear that it is sufficient to prove this theorem for the case when the camel does not cross its own path anywhere, because loops can be dealt with in the same way as in solving problem 25. So, let's assume the camel visited each vertex only once and returned to the starting point. Let's elimina... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,633 |
54. The Five-Color Theorem. Prove that any map can be properly colored with five colors ${ }^{1}$ ) (use problem 51).
Methods similar to those used to solve the five-color problem can be applied to solve the four-color problem for a special case. This is done in problems 55 and 56. As in solving the five-color problem... | 54. Let's use the method of induction again. For maps with no more than 5 countries, the statement is obvious. Suppose the theorem is true for maps containing no more than \( n \) countries, and we will prove it for the case of \( n+1 \) countries.
By problem 52, there exists a country \( S \) with a number of borders... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,640 |
55. Prove that for a connected map, where the multiplicity of each vertex is at least 3, the following inequality holds:
$$
g \leqslant 3 s - 6
$$ | 55. The smallest possible vertex multiplicity is 8. Therefore, from problem 16 it follows that
$$
3 v \leqslant 2 g
$$
By Euler's theorem $v=g+2-s$, hence
$$
\begin{gathered}
3(g+2-s) \leqslant 2 g, \quad 3 g+6-3 s \leqslant 2 g \\
g \leqslant 3 s-6
\end{gathered}
$$
which is what we needed to prove. | \leqslant3-6 | Combinatorics | proof | Yes | Yes | olympiads | false | 24,641 |
59. What digit do the numbers
$$
6^{811}, 2^{1000}, 3^{999}
$$
end with?
In 7-arithmetic, there are seven numbers:
$$
0,1,2,3,4,5,6 .
$$
Addition and multiplication in 7-arithmetic are defined by the following rules: to add two numbers, find their sum in the usual arithmetic and then take the remainder of this su... | 59. To solve the problem, it is sufficient to compute the expressions $6^{811}$, $2^{1000}$, and $3^{999}$ in 10-arithmetic. In 10-arithmetic, the equality $6^{2}=6$ holds. Multiplying this equality successively by $6, 6^{2}, 6^{3}, \ldots$, we get
$$
6^{3}=6^{2}, 6^{4}=6^{3}, 6^{5}=6^{4}, \ldots
$$
From this, it fol... | 6,6,7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,644 |
60. Construct the addition table and the multiplication table for 7-arithmetic. List all factorizations of the numbers 0 and 1 into two factors. | 60. Addition Table
| | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 | 0 |
| 2 | 2 | 3 | 4 | 5 | 6 | 0 | 1 |
| 3 | 3 | 4 | 5 | 6 | 0 | 1 | 2 |
| 4 | 4 | 5 | 6 | 0 | 1 | 2 | 3 |
| 5 | 5 | 6 | 0 | 1 | 2 | 3 | 4... | 0=0\cdot0=1\cdot0=2\cdot0=3\cdot0=4\cdot0=5\cdot0=6\cdot0\\1=1\cdot1=2\cdot4=3\cdot5=6\cdot6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,645 |
61. Determine the remainder of the division by 7 of the number
$$
3^{100}
$$
The 10-arithmetic and 7-arithmetic we have discussed are special cases of residue arithmetic modulo $m$, or $m$-arithmetic. Let $m$ be any positive integer. The elements of $m$-arithmetic are the numbers $0,1,2, \ldots, m-1$. Addition and mu... | 61. Let's compute $3^{100}$ in 7-arithmetic:
$$
3^{2}=2,3^{8}=6,3^{4}=4,3^{5}=5,3^{6}=1
$$
Therefore,
$$
3^{100}=3^{6 \cdot 16+4}=\left(3^{6}\right)^{16} \cdot 3^{4}=3^{4} \Rightarrow 4
$$
$14^{*}$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,646 |
64. Determine the remainders of the number $2^{1000}$ when divided by 3, 5, 11, and 13.
Subtraction in $m$-arithmetic, like ordinary arithmetic, is defined as the inverse operation of addition: a number $x$ is called the difference of numbers $b$ and $a (x=b-a)$ if
$$
a+x=b
$$
For example, in 7-arithmetic,
$$
\begi... | 64. Answers: $1 ; 1 ; 1 ; 3$. | 1;1;1;3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,648 |
69. Let $a$ be any non-zero number from $p$-arithmetic. Prove that the multiplication scheme by $a$ has the following properties:
a) no number can have two arrows pointing to it;
b) every number has some arrow pointing to it. | 69. a) If the number $x$ were approached by two arrows, say from numbers $y$ and $z$, this would mean that $a y = x$ and $a z = x$. Subtracting one of these equations from the other, we get $a(y - z) = 0$, but according to problem 68, this is impossible because, by assumption, $a \neq 0$ and $y \neq z$.
b) From each n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,653 |
70. Let $a$ and $b$ be any numbers in $p$-arithmetic, with $a \neq 0$. Using the previous problem, prove that in $p$-arithmetic there exists and is unique a number $x$ such that $a x=b$. | 70. Let's consider the scheme of multiplication by $a$. According to problem 69b), an arrow points from some number $x$ to the number $b$. But this means that $a x = b$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,654 |
71. Let $a$ be a non-zero number from $p$-arithmetic. Prove that:
a) the multiplication scheme by $a$ splits into cycles;
b) all these cycles (excluding the cycle 0) have the same length.
From this, deduce that $a^{p-1}=1$. | 71. a) Let $b$ be an arbitrary number from $p$-arithmetic. We will show that $b$ belongs to some cycle. We will move, starting from $b$, according to our scheme, transitioning from number to number as indicated by the arrows. After several transitions (the number of transitions may, in particular, equal 1 and in any ca... | ^{p-1}=1 | Number Theory | proof | Yes | Yes | olympiads | false | 24,655 |
73. Construct tables of the reciprocals $\frac{1}{k}$ in 7-arithmetic, 11-arithmetic, and 13-arithmetic.
As an example, the table of reciprocals in 5-arithmetic is provided below:
| $k$ | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: |
| $\frac{1}{k}$ | 1 | 3 | 2 | 4 |
Show that in any $p$-arithmetic, only... | 73. Tables of inverse values:
in 7-arithmetic
| $k$ | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\frac{1}{k}$ | 1 | 4 | 5 | 2 | 3 | 6 |
in 11-arithmetic
| | 1 2 3 4 5 6 7 8 9 10 |
| :---: | :---: |
| | 1 6 4 3 9 8 7 |
in 13-arithmetic
| $k$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,657 |
74. a) Prove that the product of all elements of $p$-arithmetic, except zero, is equal to -1.
b) (Wilson's Theorem). Prove that if $p$ is a prime number, then $(p-1)! + 1$ is divisible by $p^2$. | 74. a) In the product $1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(p-1)$, each number, except for 1 and $p-1$, will cancel out with its inverse (see problem 73). Therefore,
$$
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(p-1)=1 \cdot(p-1)=-1
$$
b) Follows from problem a). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,658 |
78. Prove that if a squaring scheme is constructed in $p$-arithmetic ${ }^{1}$, then to each point (except for the point 0) either no arrow applies or two arrows apply. In other words: prove that for $a \neq 0$, the equation $x^{2}=a$ has in $p$-arithmetic either two distinct solutions or no solutions at all. | 78. Suppose the equation
$$
x^{2}=a
$$
is satisfied by some number $b$ from $p$-arithmetic, i.e., the identity
$$
b^{2}=a
$$
holds. Subtracting equation (2) from equation (1), we get
$$
x^{2}-b^{2}=0
$$
from which
$$
(x-b)(x+b)=0
$$
.
If in the diagrams constructed in the solution to problem 77 (see diagrams 134-136 in the solution to problem 77), the direction of all ar... | 79. Let's consider the squaring scheme in $p$-arithmetic and the numbers from which square roots are extracted, i.e., those points to which at least one arrow points. Let $k$ be the number of such points. Then, according to problem 78, one arrow points to one of them, specifically to the point 0, while two arrows point... | \frac{p+1}{2} | Number Theory | proof | Yes | Yes | olympiads | false | 24,663 |
80. a) Prove that the square root of -1 can be extracted in $p$-arithmetic if $p=4k+1$, and cannot be extracted if $p=4k+3$.
Hint. Use problems 71 and 74.
b) Prove that all odd prime divisors of the number $a^{2}+1$ (for any $a$) have the form $4k+1$. Every prime number of the form $4k+1$ appears in the prime factori... | 80. a) Let $p=4 k+1$. According to problem 74
$$
1 \cdot 2 \ldots(p-1)=-1
$$
Rewrite this equation as follows:
$$
\begin{aligned}
& -1=1.2 \ldots(p-1)=\left[1.2 \ldots \frac{p-3}{2} \cdot \frac{p-1}{2}\right]\left[\left(\frac{p-1}{2}+1\right) \times\right. \\
& \left.\times\left(\frac{p-1}{2}+2\right) \ldots(p-2)(p-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,664 |
81. Derive the formula for solving the quadratic equation in $p$-arithmetic
$$
a x^{2}+b x+c=0
$$
( $a, b$ and $c$ are numbers from $p$-arithmetic, $a \neq 0$ ).
Using this formula, prove that
if $\sqrt{b^{2}-4 a c}$ cannot be extracted in $p$-arithmetic, then the equation has no roots;
if $b^{2}-4 a c=0$, then th... | 81. The formula is derived exactly as in ordinary algebra.
We use the identity
$$
a x^{2}+b x+c=a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a}
$$
By this identity, the equation
$$
a x^{2}+b x+c=0
$$
is equivalent to the equality
$$
a\left(x+\frac{b}{2 a}\right)^{2}+\frac{4 a c-b^{2}}{4 a}=0
$$
or
$$
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,665 |
82. Solve the following quadratic equations in 7-arithmetic:
\[
\begin{gathered}
5 x^{2}+3 x+1=0 \\
x^{2}+3 x+4=0 \\
x^{2}-2 x-3=0
\end{gathered}
\] | 82. We calculate the discriminant $D=b^{2}-4 a c$ for each of the given equations. For the first equation, we have $D=3$, for the second $D=0$, and for the third $D=2$. Using, for example, the scheme from problem 77 (Fig. 134), we establish that in 7-arithmetic, $\sqrt{3}$ cannot be extracted, while $\sqrt{2}$ can be e... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,666 | |
83. Each quadratic equation can be reduced, by dividing it by the coefficient of the leading term, to the form
$$
x^{2}+c x+d=0
$$
The total number of different reduced quadratic equations in $p$-arithmetic is $p^{2}$. Calculate how many of them have no roots, how many have one root, and how many have two distinct ro... | 83. If $\alpha$ and $\beta$ are the roots of the equation $x^{2} + c x + d = 0$, then, as in ordinary algebra,
$$
x^{3} + c x + d = (x - \alpha)(x - \beta)
$$
(To convince yourself of this, it is sufficient to express $\alpha$ and $\beta$ in terms of $c$ and $d$ using the formulas for the solutions of a quadratic equ... | \frac{p(p-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,667 |
85. Let $p$ be a prime number of the form $3k + 2^1$. Prove that in $p$-arithmetic:
a) $\sqrt[3]{1}$ has only one value (equal to one) (use problem 71);
b) for no $a$, $\sqrt[3]{a}$ can have more than one value;
c) $\sqrt[8]{a}$ can be extracted for any $a$. | 85. a) Let
$$
b^{3}=1 \text {. }
$$
By problem 71
$$
b^{p-1}=1
$$
Since $p=3 k+2$, then $p-1=3 k+1$, and the equality takes the form
$$
b^{3 k+1}=1
$$
Raising equality (1) to the power of $k$, we find
$$
b^{3 k}=1
$$
and dividing (3) by (4), we get
$$
b=1 .
$$
Thus, (1) implies (5), which was to be proved. b)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,669 |
86. Using the equality $x^{3}-1=(x-1)\left(x^{2}+x+1\right)$, solve the equation
$$
x^{3}-1=0
$$
Applying the obtained formulas, find the three values of $\sqrt[3]{1}$ in 103 -arithmetic. | 86. From the equality $(x-1)\left(x^{2}+x+1\right)=0$ it follows that
$$
x-1=0
$$
or
$$
x^{2}+x+1=0
$$
From the first equation we find
$$
x_{1}=1
$$
from the second
$$
x_{2,3}=\frac{-1 \pm \sqrt{-3}}{2}
$$
In 103-arithmetic $-3=100$ and $\sqrt{-3}= \pm 10$. Substituting these values of $\sqrt{-3}$ into formula ... | x_{1}=1,\quadx_{2}=56,\quadx_{3}=46 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,670 |
87. Let $p>3$. Prove that in $p$-arithmetic $\sqrt[3]{1}$ has either three distinct values or only one value, depending on whether $\sqrt{-3}$ can be extracted or not. | 87. Any cube root of unity satisfies the equation $x^{3}-1=0$, and, therefore, if it is not equal to one, it satisfies the quadratic equation $x^{2}+x+1=0$. Depending on whether $\sqrt{-3}$ can be extracted in $p$-arithmetic, the latter equation either has two distinct roots, calculable by formula (1) from problem 86, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,671 |
89. Prove that if a polynomial has more roots than its degree, then all its coefficients are zeros ${ }^{1}$ ). | 89. The equation of the 1st degree $a x+b=0$ has only one root $x=-\frac{b}{a}$.
Let our theorem be proved for polynomials of degree $n$. We will prove that it is true for polynomials of degree $(n+1)$. Assume the opposite. Let the polynomial of degree $(n+1)$
$$
a_{0} x^{n+1}+a_{1} x^{n}+\ldots+a_{n} x+a_{n+1} \quad... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,673 |
90. Prove that if $x^{k}=1$ for all non-zero $x$, then $k$ is divisible by $p-1$. | 90. Let the remainder of dividing $k$ by $p-1$ be $r$:
$$
k=q(p-1)+r, \quad r0$, then the polynomial $x^{r}-1$ does not have all coefficients as zeros. At the same time, the number of its roots is $p-1>r(x=1,2, \ldots, \ldots, p-1)$, which contradicts the statement of problem 89. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,674 |
91. Let $a \neq 0, p \neq 2$. Prove that $a^{\frac{p-1}{2}}=1$ if $\sqrt{a}$ can be extracted in $p$-arithmetic, and $a^{\frac{p-1}{2}}=-1$ if $\sqrt{a}$ cannot be extracted.
${ }^{1}$ ) Strictly speaking, by our definition of the degree of a polynomial, a polynomial with all zero coefficients does not have any degree... | 91. We have
$$
\left(a^{\frac{p-1}{2}}\right)^{2}=a^{p-1} \Rightarrow 1
$$
(problem 71), therefore $a^{\frac{p-1}{2}}= \pm 1$. If $\sqrt{a}$ can be extracted, i.e., there exists $b$ such that $b^{2}=a$, then $a^{\frac{p-1}{2}}=b^{p-1}=1$. Thus, if $\sqrt{a}$ can be extracted, then $a$ is a root of the equation
$$
x^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,675 |
92. Prove that if a polynomial of degree $n$ has $n$ roots $x_{1}, x_{2}, \ldots, x_{n}$, then
$$
\begin{aligned}
& a_{0} x^{n}+a_{1} x^{n-1}+\ldots+a_{n-1} x+a_{n}= \\
& \quad=a_{0}\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)
\end{aligned}
$$ | 92. The proof of this theorem is contained in the solution to problem 89. For the sake of completeness, we will repeat our reasoning here. The polynomial
$$
\begin{aligned}
a_{0} x^{n}+a_{1} x^{n-1} & +\ldots+a_{n-1} x+a_{n}- \\
& -a_{0}\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{n}\right)
\end{aligned}
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,676 |
96. Prove that in the table placed below
where each number, starting from the second (from the top), is obtained by squaring the number above it:
a) the infinite number \( d = \ldots 90625... | 96. a) Let $x_{n}$ denote the n-th number in the table. We have $x_{1}=5, x_{2}=5^{2}, x_{3}=x_{2}^{2}=5^{2^{3}}, \ldots, x_{n}=5^{2^{n-1}}$. Let's factorize the difference $x_{n+1} - x_{n}$:
$$
\begin{aligned}
& x_{n+1}-x_{n}=5^{2^{n}}-5^{2^{n-1}}=5^{2^{n-1}}\left(5^{2^{n-1}}-1\right)= \\
& =5^{2^{n-1}}\left(5^{2^{n-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,678 |
97. Prove that the number $d$, constructed in the previous problem, does not have an inverse, i.e., it is impossible to find any infinite-valued number $x$ (neither integer nor fractional) such that $x d=1$. | 97. Assume the opposite and consider the product $x d e$. From the equality $x d=1$ it follows that $x d e=1 \cdot e=e$. From the equality $d e=0$ it follows that $x d e=x \cdot 0=0$. We have arrived at a contradiction, since $e \neq 0$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,679 |
99. Let $a$ be an arbitrary $p$-adic integer, the last digit of which is not zero. Prove that the last digit of the number $a^{p-1}-1$ is 0. (For the case when $a$ is a finite number, this problem coincides with problem 72.) | 99. Let $a=\ldots a_{3} a_{2} a_{1}$. The last digit of the product is determined by the product of the last digits of the factors, so the last digit of the number $a^{p-1}$ is $a_{1}^{p-1}$ (in the $p$-arithmetic sense). But in $p$-arithmetic, $a_{1}^{p-1}=1$ (problem 71). Therefore, the last digit of the number $a^{p... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,681 |
100. Let $a$ be an arbitrary $p$-adic integer, the last digit of which is not zero. Show that the number $a^{p^{k-1}(p-1)}-1$ ends with $k$ zeros. | 100. We will prove this by induction. The case $k=1$ was resolved in the previous problem. Suppose the theorem is proved for $k=l$. We will prove it for $k=l+1$. Denoting $a^{p^{l-1}(p-1)}$ by $y$, we have
$$
x^{p}(p-1)-1=y^{p}-1 \equiv (y-1)\left(y^{p-1}+y^{p-2}+\ldots+y+1\right) \pmod{p}
$$
Since $y-1$, by the indu... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,682 |
101. Prove that if in each term of the geometric progression $b_{0}, b_{1}, \ldots, b_{n}, \ldots$ (all $b_{1}$ are integers in the $p$-adic numbers; the denominator of the progression does not divide $p$) only the last $k$ digits are left, then a periodic sequence will be obtained, the length of which is a divisor of ... | 101. The general form of the $n$-th term of a geometric progression is $b_{n}=b_{0} q^{n}$. Therefore,
$$
\left.q^{k} \cdot 103 T 0 \mathrm{my}\right)=b_{0} q^{n+p^{k-1}(p-1)}=b_{n} q^{p^{k-1}(p-1)}
$$
From problem 100, it follows that the last $k$ digits of the number $q^{p^{k-1}(p-1)}$ are $000 \ldots 001$. Hence, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,683 |
106. Check that in 3-adic arithmetic 201 is a square root of the number ...112101 with an accuracy of three digits.
Let us assume we have already found $B_{n}$. We will show how to find $B_{n+1}$ - the square root of $A$ with an accuracy of $n+1$ digits. We seek $B_{n+1}$ in the form
$$
B_{n+1}=B_{n}+x \cdot 10^{n}
$... | 106. Indeed,
$$
\begin{aligned}
& \begin{array}{r}
201 \\
\times \frac{201}{201}
\end{array} \\
& 000 \\
& 1102 \\
& 111101
\end{aligned}
$$
The last three digits of the numbers 111101 and ... 112101 coincide. | 201 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,688 |
107. Let $u_{0}, u_{1}, u_{2}, \ldots, u_{n}, \ldots$ be some Fibonacci sequence. Prove that the $n$-th term of this sequence is expressed in terms of $u_{0}$ and $u_{1}$ by the formula
$$
u_{n}=a_{n-1} u_{0}+a_{n} u_{1}{ }^{\circ}
$$
(In order for this formula to remain valid for $n=0$, we must set $a_{-1}=1$.) | 107. Let us denote by $u_{n}^{\prime}$ the expression $a_{n-1} u_{0}+a_{n} u_{1}$. The sequence $u_{0}^{\prime}, u_{1}^{\prime}, u_{2}^{\prime}, \ldots, u_{n}^{\prime}, \ldots$ is a Fibonacci sequence. Indeed,
\[
\begin{aligned}
& u_{n-2}^{\prime}+u_{n-1}^{\prime}=\left(a_{n-8} u_{0}+a_{n-2} u_{1}\right)+\left(a_{n-2}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,689 |
108. Prove the formula
$$
a_{n+m-1}=a_{n-1} a_{m-1}+a_{n} a_{m}
$$ | 108. It is sufficient to apply the formula from problem 107 to the Fibonacci sequence
$$
a_{m-1}, a_{m}, a_{n+1}, a_{m+2}, \ldots, a_{m+(n-1)}, \ldots
$$
in which $u_{n}==a_{m+(n-1)}$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,690 |
110. Prove that if in the sequence $\varnothing^{0}$ any three consecutive terms $a_{n-1}, a_{n}, a_{n+1}$ are chosen, the product of the outer terms is taken and the square of the middle term is subtracted from it, the result will be 1 or -1. Prove that for any Fibonacci sequence (not necessarily starting with the num... | 110. Let
$$
u_{n-1} u_{n+1}-u_{n}^{2}=d_{n}
$$
We have
$$
\begin{aligned}
& d_{n+1}=u_{n} u_{n+2}-u_{n+1}^{2}=u_{n}\left(u_{n}+u_{n+1}\right)-u_{n+1}^{2}= \\
& =u_{n}^{2}+u_{n} u_{n+1}-u_{n+1}^{2}=u_{n}^{2}-u_{n+1}\left(-u_{n}+u_{n+1}\right)= \\
& \quad=u_{n}^{2}-u_{n-1} l_{n+1}=-d_{n}
\end{aligned}
$$
From this, i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,692 |
111. Prove that if you select four consecutive terms $a_{n-1}, a_{n}, a_{n+1}, a_{n+2}$ in the Fibonacci sequence and subtract the product of the middle terms $a_{n} a_{n+1}$ from the product of the outer terms $a_{n-1} a_{n+2}$, the result will be 1 or -1. Prove that for any Fibonacci sequence, the absolute value of t... | 111. Note that
$$
\begin{array}{r}
u_{n-1} u_{n+2}-u_{n} u_{n+1}=u_{n-1} u_{n}+u_{n-1} u_{n+1}-u_{n} u_{n+1}= \\
=u_{n-1} u_{n+1}-u_{n}\left(u_{n+1}-u_{n-1}\right)=u_{n-1} u_{n+1}-u_{n}^{2}= \\
=(-1)^{n-1}\left(u_{0} u_{2}-u_{1}^{2}\right)
\end{array}
$$
from which
$$
\left|u_{n-1} u_{n+2}-u_{n} u_{n+1}\right|=\left... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,693 |
112. Make sure that the last digits of the numbers in the Fibonacci sequence $^{\circ}$ repeat periodically. What is the length of the period?
Ensure that the last digits of the numbers in the Fibonacci sequence $^{\circ}$ repeat periodically. What is the length of the period? | 112. The last digits of the Fibonacci sequence $\varnothing^{0}$ themselves form a Fibonacci sequence in the sense of 10-arithmetic (sequence $\Phi_{10}^{0}$).
Let's write out the terms of this sequence:
$0,1,1,2,3,5,8,3,1,4,5,9,4,3,7$, $0,7,7,4,1,5,6,1,7,8,5,3,8,1,9$, $0,9,9,8,7,5,2,7,9,6,5,1,6,7,3$, $0,3,3,6,9,5,4,... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,694 |
113. (Generalization of problem 112.) Prove that all $m$-arithmetic Fibonacci sequences are periodic, and the length of the period does not exceed $m^{2}$. | 113. Let $v_{0}, v_{1}, v_{2}, \ldots, v_{n}, \ldots$ be some $m$-arithmetic Fibonacci sequence. We will prove that for some $r$ the equalities $v_{r}=v_{0}, v_{r+1}=v_{1}$ hold, after which the periodicity of the sequence $v_{0}, v_{1}, v_{2}, \ldots, v_{n}, \ldots$ is established in the same way as the periodicity of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,695 |
117. Let three elements $x, y, z$ of the circular sequence $\breve{\Phi}_{m}$ be arranged such that the distance between $x$ and $y$ is equal to the distance between $y$ and $z$ (Fig. 54). Prove that if $x=y=0$, then $z=0$. | 117. The number $y=0$ is at an equal distance from $x$ and $z$. Therefore, based on problem 116, either $x+z=0$ or $x-z=0$. Since $x=0$, in both cases $z=0$, | 0 | Number Theory | proof | Yes | Yes | olympiads | false | 24,699 |
120. Prove that if a circular sequence without repetitions $\Phi_{m}$ contains zero and consists of an odd number of elements, then the number of its elements is three. | 120. If the circular sequence $\breve{\Phi}_{m}$ contains zero and consists of an odd number of terms, then there will be a pair of adjacent elements $x, y$ located at the same distance from zero (Fig. 143). According to problem 116, either
) In the sense that the arcs $x 0$ and $y 0$ contain the same... | 121. Suppose a circular sequence contains at least three distinct zeros. Let us select three such consecutive zeros and number them $0_{1}, 0_{2}, 0_{3}$ in the order of their clockwise succession. The elements preceding $0_{1}, 0_{2}, 0_{3}$ will be denoted, respectively, as $u_{1}, u_{2}, u_{3}$, and the elements imm... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,703 |
122. The periodic sequence
$$
1,4,5,9,3,1,4,5,9,3,1,4,5,9,3, \ldots
$$
of elements in 11-arithmetic is simultaneously a geometric progression and a Fibonacci sequence. The sequence
$b, 4 b, 5 b, 9 b, 3 b, b, 4 b, 5 b, 9 b, 3 b, b, 4 b, 5 b, 9 b, 3 b, \ldots$, obtained from sequence (1) by multiplying by an arbitrary... | 122. a) A geometric progression with the initial term $b$ and common ratio $q$ has the form
$$
\left.\begin{array}{c}
b, b q, b q^{2}, b q^{3}, \ldots, b q^{n-1}, \\
b q^{n}, b q^{n+1}, \ldots
\end{array}\right\}
$$
For this sequence to be a Fibonacci sequence, it is necessary and sufficient that for all $n$ the rela... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,704 |
123. Decompose the sequence $\Phi_{11}^{0}$
$0,1,1,2,3,5,8,2,10,1,0,1,1,2,3,5,8,2,10,1, \ldots$
into the sum of two 11 -Fibonacci sequences, which are also geometric progressions ${ }^{1}$ ). | 123. In the previous problem, all 11-arithmetic sequences that are simultaneously geometric progressions and Fibonacci series were found. It was established that all such sequences fall into two families. Now, let's take a sequence from the first family with the initial term $x$ and a sequence from the second family wi... | c_{n}=3\cdot8^{n}+8\cdot4^{n} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 24,705 |
124. Prove that:
a) If in $p$-arithmetic $\sqrt{5}$ cannot be extracted, then in this arithmetic it is impossible to construct sequences that are simultaneously a geometric progression and a Fibonacci sequence.
b) If in $p$-arithmetic $\sqrt{5}$ can be extracted, then in this arithmetic there exist Fibonacci sequence... | 124. We repeat in general terms the reasoning that was already carried out for a particular case in solving problems 122-123. The denominator \( q \) of the geometric progression, which is a Fibonacci sequence, must satisfy the quadratic equation \( q^{2} - q - 1 = 0 \) (see formula (2) in the solution of problem 1.22)... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,706 |
125. Prove that if in some $p$-arithmetic $\sqrt{5}$ is extracted and if
$$
v_{0}, v_{1}, v_{2}, \ldots, v_{n}, \ldots
$$
- is an arbitrary Fibonacci sequence in this arithmetic, then
$v_{p-1}=v_{0}, \quad v_{p}=v_{1}, \quad v_{p+1}=v_{2}, \ldots, v_{k+p-1}=v_{k}, \ldots$ | 125. According to problem 71 in $p$-arithmetic $a^{p-1}=1$ for any $a \neq 0$. In particular, $q_{1}^{p-1}=1$ and $q_{2}^{p-1}=1$ and from formula (4) of the solution to problem 124 it follows
\[
\begin{gathered}
v_{p-1}=x q_{1}^{p-1}+y q_{3}^{p-1}=x+y=v_{0} \\
v_{p}=x q_{1}^{p}+y q_{3}^{p}=x q_{1}+y q_{2}=v_{1} \\
\c... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,707 |
126. If in $p$-arithmetic $\sqrt{5}$ can be extracted, then the number of elements in any circular Fibonacci series without repetitions is a divisor of the number $p-1$.
$p$-arithmetic Fibonacci series. Let $v_{0}, v_{1}$, $v_{2}, \ldots, v_{n}, \ldots$ be an arbitrary $p$-arithmetic Fibonacci series. Construct the se... | 126. Let's choose some element of the circular Fibonacci sequence as a starting point and move clockwise, passing the circle an unlimited number of times, with our sequence written around it. We will write down the elements of our sequence in the order we pass them. We will get an infinite periodic sequence $v_{0}, v_{... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,708 |
128. a) Prove that the consecutive terms of the sequence of ratios corresponding to an arbitrary Fibonacci sequence are related by the formula
$$
t_{n}=1+\frac{1}{t_{n-1}}
$$
Using this formula, prove that the sequence $t_{1}, t_{2}, \ldots, t_{n}, \ldots$ is periodic and that no number repeats twice within a period.... | 128. a) The numbers of the Fibonacci sequence are related by the relation
$$
v_{n}=v_{n-1}+v_{n-2}
$$
Dividing this relation by $v_{n-1}$:
$$
\frac{v_{n}}{v_{n-1}}=1+\frac{v_{n-2}}{v_{n-1}}.
$$
Replacing $\frac{v_{n}}{v_{n-1}}$ with $t_{n}$ and $\frac{v_{n-1}}{v_{n-2}}$ with $t_{n-1}$, we get the formula
$$
t_{n}=... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,710 |
129. Let
$$
v_{0}, v_{1}, v_{2}, \ldots, \quad v_{0}^{\prime}, v_{1}^{\prime}, v_{2}^{\prime}, \ldots
$$
be two arbitrary $p$-arithmetic Fibonacci sequences and
$$
\begin{aligned}
t_{1} & =\frac{v_{1}}{v_{0}}, \quad t_{2}=\frac{v_{2}}{v_{1}}, \ldots, t_{n}=\frac{v_{n}}{v_{n-1}}, \ldots, \\
t_{1}^{\prime} & =\frac{v_... | 129. First, let us make two remarks:
1) If the sequences of relations
$$
t_{1}, t_{2}, \ldots, t_{n}, \ldots
$$
and
$$
t_{1}^{\prime}, t_{2}^{\prime}, \ldots, t_{n}^{\prime}, \ldots
$$
have the same initial terms, then they coincide, i.e., for any $n$, $t_{n}=t_{n}^{\prime}$. This follows from the formula derived i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,711 |
130. Let
\[
\begin{aligned}
& t_{1}=\frac{v_{1}}{v_{0}}, \quad t_{2}=\frac{v_{2}}{v_{1}}, \ldots, t_{n}=\frac{v_{n}}{v_{n-1}}, \ldots \\
& t_{1}^{\prime}=\frac{v_{1}^{\prime}}{v_{0}^{\prime}}, \quad t_{2}^{\prime}=\frac{v_{2}^{\prime}}{v_{1}^{\prime}}, \ldots, t_{n}^{\prime}=\frac{v_{n}^{\prime}}{v_{n-1}^{\prime}}, \l... | 130. The formula of problem 107 can be rewritten for $p$-arithmetic rows in the form
$$
v_{n}=c_{n-1} v_{0}+c_{n} v_{1}
$$
(see the remark on page 82). Using this formula, we derive
$$
t_{r+1}=\frac{v_{r+1}}{v_{r}}=\frac{c_{r} v_{n}+c_{r+1} v_{1}}{c_{r-1} v_{0}+c_{r} v_{1}}
$$
Replace $c_{r+1}$ with the sum $c_{r-1... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,712 |
132. Prove that in the $n$-th row of the 2-arithmetic Pascal's triangle, all elements, except for the extreme ones (the zeroth and the $n$-th), are zeros if and only if $n=2^{k}$.
Let's construct the 3-arithmetic Pascal's triangle:
, looks like this:
Here, in the 2nd, 4th, 8th, and only in them, all members (except the extr... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,714 |
134. Let all elements in the $s$-th row of the $m$-ary Pascal triangle, except for the boundary elements, be zeros. Prove that the rows with numbers $s^{2}, s^{3}, \ldots, s^{k}, \ldots$ also have this property. | 134. Based on the previous task, the line with number $s^{2}$ - the first line of the $s$-strip - contains the upper vertices of the triangles $\Delta_{s}^{0}, \Delta_{s}^{1}, \ldots, \Delta_{s}^{s}$; all other members of this line are zeros. But $\Delta_{d}^{k}=P_{s}^{k} \Delta_{0}^{0}$, so the vertices of the triangl... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,716 |
135. Prove that if the binomial $1+x$ is raised to the $n$-th power and written in ascending powers of $x$, the result is the equality
$$
(1+x)^{n}=C_{n}^{0}+C_{n}^{1} x+C_{n}^{2} x^{2}+\ldots+C_{n}^{n} x^{n}
$$
or in $m$-arithmetic:
$$
(1+x)^{n}=P_{n}^{0}+P_{n}^{1} x+P_{n}^{2} x^{2}+\cdots+P_{n}^{n} x^{n}
$$ | 135. Let's conduct the proof by induction:
$$
\begin{aligned}
& \text { for } n=0 \quad(1+x)^{0}=1=C_{0}^{0} \\
& \text { for } n=1 \quad(1+x)^{1}=1+x=C_{1}^{0}+C_{1}^{1} x, \\
& \text { for } n=2 \quad(1+x)^{2}=1+2 x+x^{2}=C_{2}^{0}+C_{2}^{1} x+C_{2}^{2} x^{2} .
\end{aligned}
$$
Suppose this formula is proven up to ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,717 |
137. Construct schemes corresponding to the 7 arithmetic fractional-linear functions
$$
\frac{4 x+1}{2 x+3}, \quad \frac{2 x+1}{3 x+2}, \quad \frac{3 x-1}{x+1}
$$
Consider two fractional-linear functions
$$
f(x)=\frac{a x+b}{c x+d} \quad \text{and} \quad g(x)=\frac{-d x+b}{c x-a}
$$
It is not hard to see that if $f... | 137. The scheme of the function $\frac{4 x+1}{2 x+3}$ is shown in Fig. 149. It consists of two fixed points and three cycles.
Fig. 149.
 x-b=0
$$
Indeed,
$$
\begin{gathered}
\frac{a n+b}{c n+d}=n, \quad a n+b=c n^{2}+d n \\
c n^{2}+(d-a) n-b=0 .
\end{gathered}
$$
Assume that... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,721 |
141. Given two 7-arithmetic functions
$$
f(x)=\frac{x+5}{5 x+1}, \quad g(x)=\frac{4 x+3}{6 x+3}
$$
Calculate $f(g(x))$ and $g(f(x))$. | 141. We have
\[
\begin{aligned}
& f(g(x))=\frac{\frac{4 x+3}{6 x+3}+5}{5 \frac{4 x+3}{6 x+3}+1}=\frac{6 x+4}{5 x+4} \\
& g(f(x))=\frac{4 \frac{x+5}{5 x+1}+3}{6 \frac{x+5}{5 x+1}+3}=\frac{5 x+2}{5}=x+6
\end{aligned}
\] | f((x))=\frac{6x+4}{5x+4},\quad(f(x))=x+6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,722 |
142. Find a 7-arithmetic fractional-linear function $f(x)$ such that $f(0)=0, f(1)=4, f(4)=2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 142. The desired function is $f(x)=\frac{x}{2}$. We can find it in the following way. Let's write $f(x)$ in the general form
$$
f(x)=\frac{a x+b}{c x+d}
$$
and substitute $0, 1, 4$ for $x$ sequentially. Solving the equations for $a, b, c, d$
$$
\left\{\begin{array}{l}
\frac{a \cdot 0+b}{c \cdot 0+d}=0 \\
\frac{a \cd... | f(x)=\frac{x}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,723 |
144. Given a triple of distinct points $y_{1}, y_{2}, y_{3}$. Prove that there always exists a fractional-linear function $f(x)$ such that $f(0)=y_{1}, f(1)=y_{2}, f(\infty)=y_{3}$. | 144. Due to task 143, there exists a function $f(x)$ such that $f\left(y_{1}\right)=0, f\left(y_{2}\right)=1, f\left(y_{3}\right)=\infty$. The fractional-linear function $f_{-1}(x)$ meets our requirements
$$
f_{-1}(0)=y_{1}, \quad f_{-1}(1)=y_{2}, \quad f_{-1}(\infty)=y_{3}
$$ | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,725 |
145. Two triples of distinct points $x_{1}, x_{2}, x_{3}$ and $y_{1}, y_{2}, y_{3}$ are given. Prove that it is always possible to select, and in a unique way, a fractional-linear function $f(x)$ such that $f\left(x_{1}\right)=y_{1}, f\left(x_{2}\right)=y_{2}, f\left(x_{3}\right)=y_{\mathrm{g}}$. | 145. Let $f(x)$ map $x_{1}, x_{2}, x_{3}$ to $0,1, \infty$, and let $\varphi(x)$ map $0,1, \infty$ to $y_{1}, y_{2}, y_{3}$ (such fractional-linear functions exist based on problems 143 and 144). Form the function $\varphi(f(x))$. It is easy to see that this function maps $x_{1}, x_{2}, x_{3}$ to $y_{1}, y_{2}, y_{3}$,... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,726 |
146. Count how many different $p$-arithmetic fractional-linear functions there are.
The function $f(f(x))$ is denoted by $f_{2}(x)$. Similarly, we introduce fractional-linear functions $f_{\mathrm{B}}(x), f_{1}(x), \ldots, f_{2}(x)$, assuming
$$
\begin{aligned}
f_{3}(x) & =f\left(f_{2}(x)\right), \\
f_{4}(x) & =f\lef... | 146. Let's fix some triple of different points, for example $0,1, \infty$. Each fractional-linear function maps this triple of points to some other triple of different points $y_{1}, y_{2}, y_{3}$. Conversely, for each triple of different points $y_{1}, y_{2}, y_{3}$, one can find and uniquely determine a fractional-li... | (p+1)p(p-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 24,727 |
148. Let on the diagram of the function $f(x)$ the point $x_{0}$ belongs to a cycle of length $s$. Prove that if the number $k$ is divisible by $s$, then $f_{k}\left(x_{0}\right)=x_{0}$. Conversely, if $f_{k}\left(x_{0}\right)=x_{0}$, then $k$ is divisible by $s$. | 148. When applying the function $f_{k}(x)$, the point $x_{0}$ makes $k$ steps along its cycle, and since $k$ is divisible by $s$, it completes the cycle an integer number of times, i.e., it
... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,729 |
149. Let $f(x)$ be an arbitrary $p$-arithmetic fractional linear function. Prove that if any of the functions $f_{2}(x), f_{3}(x), \ldots, f_{k}(x)$ has at least one fixed point that is not a fixed point of $f(x)$, then it leaves all points unchanged. | 149. Let $n$ be a fixed point for $f_{k}(x)$, which is not a fixed point for $f(x)=\frac{a x+b}{c x+d}$, i.e., $f_{k}(n)=n$, $f(n) \neq n$. If $f_{k}(n)=n$, then all points of the cycle of $f(x)$ to which $n$ belongs are fixed points for $f_{k}(x)$. This immediately
$ be an arbitrary $p$-arithmetic fractional-linear function. Prove that the cycles into which its scheme decomposes all have the same length (excluding cycles consisting of a single point, i.e., fixed points of the function $f(x)$) (compare with problem 71). | 150. Let us discard the fixed points and among the remaining cycles consider the cycle of the smallest length $r$. The points of this cycle are fixed for the function $f_{r}(x)$. Therefore, by problem 149, $f_{r}(x)$ leaves all points fixed.
Now let us take any cycle different from a fixed point; let its length be $s$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,731 |
151. Let a $p$-arithmetic sequence be given:
$$
\begin{gathered}
x_{0}, \quad x_{1}=f\left(x_{0}\right)=\frac{a x_{0}+b}{c x_{0}+d} \\
x_{2}=f\left(x_{1}\right)=\frac{a x_{1}+b}{c x_{1}+d}, \ldots, x_{k}=f\left(x_{k-1}\right)=\frac{a x_{k-1}+b}{c x_{k-1}+d}, \ldots
\end{gathered}
$$
(or, equivalently, $x_{k}=f_{k}\le... | 151. If we discard the little interesting case when all points of $f(x)$ are fixed (in this case $x_{p+1}=$ $=x_{p}=x_{p-1}=x_{0}$), then the function $f(x)$ can have either no, one, or two fixed points (Problem 140). The other points are distributed into cycles of equal length (Problem 150); the length of each cycle i... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,732 |
153. a) Let $p$ be a prime number greater than three. Prove that in $p$-arithmetic $\sqrt{-3}$ can be extracted if and only if $p$ has the form $3k+1$.
b) Prove that for any integer $a$, all prime divisors of the number $a^{2}+3$, greater than three, have the form $3k+1$. Conversely, for any prime number $p$ of the fo... | 153. a) Consider the $p$-arithmetic function
$$
f(x)=\frac{x-3}{x+1}
$$
Here $(a-d)^{2}+4 b c=(1-1)^{2}+4(-3) \cdot 1=4(-3)$. Let $\sqrt{-3}$ be extracted. Then
$$
\sqrt{(a-d)^{2}+4 b c}=\sqrt{4(-3)}=2 \sqrt{-3}
$$
is also extracted. By the problem 151, $f_{p-1}\left(x_{0}\right)=x_{0}$ for any $x_{0}$, including $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,733 |
155. a) Let $p$ be a prime number greater than two. Prove that in $p$-arithmetic, $\sqrt{-1}$ can be extracted if and only if $p=4k+1$.
b) Prove that for any integer $a$, all odd prime divisors of the number $a^{2}+1$ have the form $4k+1$. Each prime number of the form $4k+1$ appears in the prime factorization of at l... | 155. a) Consider the $p$-arithmetic function
$$
f(x)=\frac{x-1}{x+1}
$$
Here $\quad(a-d)^{2}+4 b c=(1-1)^{2}+4(-1) \cdot 1=4(-1)$. Let $\sqrt{-1}$ be extracted. Then
$$
\sqrt{(a-d)^{2}+4 b c}=\sqrt{4(-1)}=2 \sqrt{-1}
$$
is extracted. By the problem 151, $f_{p-1}\left(x_{0}\right)=x_{0}$ for any $x_{0}$, including $... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,734 |
159. Prove that if $\sqrt{5}$ is extracted in $p$-arithmetic, then in the Fibonacci sequence of ordinary integers
$$
0,1,1,2,3,5,8, \ldots
$$
the term with the number $p^{k-1}(p-1)$ (i.e., $a_{p^{k-1}(p-1)}$) is divisible by $p^{k}$.
## § 2. The Connection Between Pascal's Triangle and the Fibonacci Sequence | 159. Let's write the sequence $0,1,1,2,3, \ldots$ in the $p$-adic number system. Based on the solution of the previous problem, the terms $a_{0}$ and $a_{p^{k-1}(p-1)}$ have the same last $k$ digits. But all the digits of $a_{0}$ are zeros ($\left.a_{0}=\ldots 000\right)$. Therefore, the last $k$ digits of $a_{p^{k-1}(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,737 |
160. Prove that the elements of Pascal's triangle can be calculated by the formula
$$
C_{n}^{k}=\frac{n!}{k!(n-k)!}
$$
( $n$ ! denotes the product $1 \cdot 2 \cdot 3 \ldots(n-1) \cdot n$ ; the symbol $0!$, which generally has no meaning, is conventionally understood to be 1$)$.
Note that if $k<p$ and $n-k<p$, then $... | 160. Let's check this formula for the first rows of the triangle
$$
C_{0}^{0}=\frac{a t}{0!0!}=1
$$
$$
C_{1}^{0}=\frac{11}{0!1!}=1 \quad C_{1}^{1}=\frac{11}{110!}=1
$$
$$
C_{3}^{0}=\frac{21}{0121}=1 \quad C_{2}^{1}=\frac{21}{111!}=2 \quad C_{2}^{2}=\frac{21}{2101}=1
$$.
Now, let's assume that our formula is true fo... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,738 |
161. Prove that in the $p$-arithmetic Pascal's triangle
$$
P_{p-1-k}^{k}=(-1)^{k} P_{2 k k}^{k} \quad\left(\text { for } k \leqslant \frac{p-1}{2}\right) .
$$ | 161. We have
$$
\begin{aligned}
& C_{p-k-1}^{k}=\frac{(p-k-1)!}{k!(p-2 k-1)!}= \\
& =\frac{(p-k-1)(p-k-2) \ldots(p-2 k)(p-2 k-1) \ldots 1}{k!(p-2 k-1)(p-2 k-2) \ldots 1} \\
& =\frac{(p-k-1)(p-k-2) \ldots(p-2 k)}{k!}= \\
& =\frac{[p-(k+1)][p-(k+2)] \ldots[p-2 k]}{k!}.
\end{aligned}
$$
Since $k \leqslant \frac{p-1}{2}<... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,739 |
164. Let $b_{n}$ be the sum of the numbers in the $n$-th diagonal of Pascal's triangle. Show that $b_{n}=a_{w}$ where $a_{n}$ is the $n$-th term of the Fibonacci sequence
$$
a_{0}=0, a_{1}=1, a_{2}=1, a_{3}=2, \ldots
$$ | 164. From Fig. 58 on p. 104, it is clear that $b_{n-1}+b_{n}=b_{n+1}$. Thus, the numbers $b_{n}$ form a Fibonacci sequence. But $b_{1}=1, b_{2}=1$, i.e., the initial terms of the Fibonacci sequences
$$
\begin{array}{lll}
a_{1}=1, & a_{2}=1, & a_{3}=2, \ldots \\
b_{1}, & b_{2}, & b_{3}, \ldots
\end{array}
$$
coincide,... | b_{n}=a_{n} | Combinatorics | proof | Yes | Yes | olympiads | false | 24,742 |
166. Let $n$ arbitrary numbers $d_{0}, \ldots, d_{n-1}$ be given. We form the sums
$$
d_{0}^{(1)}=d_{0}+d_{1}, \quad d_{1}^{(1)}=d_{1}+d_{2}, \ldots, d_{n-2}^{(1)}=d_{n-2}+d_{n-1}
$$
We will obtain $n-1$ numbers $d_{0}^{(1)}, d_{1}^{(1)}, \ldots, d_{n-2}^{(1)}$. Continue this process further
$$
d_{0}^{(2)}=d_{0}^{(1... | 166. We prove by induction. For $n=2$ everything is obvious.
$$
\begin{gathered}
d_{0} d_{1} \\
d_{0}^{(1)} \\
d_{0}^{(1)}=d_{0}+d_{1}=C_{1}^{0} d_{0}+C_{1}^{1} d_{1} .
\end{gathered}
$$
Suppose our theorem is proven for $n$ numbers. We will prove it for $n+1$ numbers. Consider the table
$$
\begin{aligned}
& d_{0} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,744 |
167. Show that the sum of the squares of a row of Pascal's triangle is again a number from Pascal's triangle. | 167. We will prove that $\left(C_{n}^{0}\right)^{2}+\left(C_{n}^{1}\right)^{2}+\ldots+\left(C_{n}^{n}\right)^{2}=C_{2 n}^{n}$. Apply to the table
$$
\begin{array}{cccccc}
C_{n}^{0} & C_{n}^{1} & & C_{n}^{2} & \ldots & \\
C_{n+1}^{1} & C_{n+1}^{2} & & \ldots & C_{n+1}^{n} \\
& & \ddots & & \ldots & \\
& & & C_{2 n}^{n}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 24,745 |
168. Prove that in any $p$-arithmetic Fibonacci sequence
$$
v_{0}, v_{1}, v_{2}, \ldots, v_{n}, \ldots
$$
the following equality holds:
$$
v_{k}+v_{k+p}=v_{k+2 p} \text{ (for any } k \text{ ). }
$$ | 168. Let's write the table
$$
\begin{array}{ccccccc}
v_{k} & v_{k+1} & v_{k+2} & v_{k+3} & \cdots & v_{k+p-2} & v_{k+p-1} \\
v_{k+2} & v_{k+3} & v_{k+1} & \cdots & & v_{k+p} \\
v_{k+1} & v_{k+5} & & \cdots & & v_{k+p+1} \\
& \cdot & \cdot & & & & v_{k+p+2} \\
& & & & & \cdots & \\
& & & & v_{k+2 p} & &
\end{array}
$$
... | v_{k+2p}=v_{k}+v_{k+p} | Number Theory | proof | Yes | Yes | olympiads | false | 24,746 |
169. Prove that in the $p$-arithmetic Fibonacci sequence, the terms whose indices are divisible by $p$ form again an arithmetic Fibonacci sequence.
## § 3. Members of the Fibonacci sequence divisible by a given number
Let us now study the distribution in the Fibonacci sequence of numbers divisible by an arbitrary num... | 169. The members of the $p$-arithmetic Fibonacci sequence whose indices are divisible by $p$ form the sequence $v_{0}, v_{p}$, $v_{2 p}, \ldots, v_{n p}, \ldots$ By setting in the formula of problem 168 $k=(n-1) p$, we get
$$
v_{(n-1) p}+v_{n p}=v_{(n+1) p}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,747 |
172. Prove that if $a_{k}$ is divisible by $m^{n}$, then the difference $a_{k+1}^{m}-a_{k-1}^{m}$ is divisible by $m^{n+1}$. | 172. Given $a_{k}=x m^{n}$, where $x$ is an integer. Hence,
$$
a_{k+1}=a_{k-1}+a_{k}=a_{k-1}+x m^{n}, \quad a_{k+1}^{m}=\left(a_{k-1}+x m^{n}\right)^{m}
$$
By problem 170,
$$
\begin{aligned}
\left(a_{k-1}+x m^{n}\right)^{m} & =a_{k-1}^{m}+m a_{k-1}^{m-1} x m^{n}+x^{2} m^{2 n} S= \\
& =a_{k-1}^{m}+m^{n+1}\left(a_{k-1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,749 |
173. Prove that if $a_{k}$ is divisible by $m^{n}$, then $a_{k m}$ is divisible by $m^{n+1}$. | 173. By task 171 (assuming $d=m^{n}$)
$$
\begin{aligned}
& a_{k m+1}=a_{k+1}^{m}+x m^{2 n} \\
& a_{k m-1}=a_{k-1}^{m}+y m^{2 n}
\end{aligned}
$$
According to the previous task, the difference $a_{k+1}^{m}-a_{k-1}^{m}$ is divisible by $m^{n+1}$, therefore, the difference $a_{k m+1}-a_{h m-1}=a_{k m}$ is also divisible... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 24,750 |
175. Prove that if $a+b \sqrt{5}=c+d \sqrt{5}$, then $a=$ $=c$ and $b=d$. | 175. Suppose that $b \neq d$. Then
$$
a-c=(d-b) \sqrt{5}, \quad \sqrt{5}=\frac{a-c}{d-b}
$$
i.e., $\sqrt{5}$ is equal to some rational number, which is impossible. Therefore, $b=d$, but then $a=c$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,752 |
176. Check that the product $(a+b \sqrt{5})(c+d \sqrt{5})$ can again be represented in the form $p+q \sqrt{5}$. Verify that if $a \geqslant 0, b \geqslant 0, c \geqslant 0, d \geqslant 0$, then $p \geqslant 0$ and $q \geqslant 0$. | 176. We have
$$
\begin{aligned}
(a+b \sqrt{5})(c+d \sqrt{5}) & =a c+b c \sqrt{5}+a d \sqrt{5}+5 b d= \\
& =(a c+5 b d)+(a d+b c) \sqrt{5}
\end{aligned}
$$
The second statement of the problem directly follows from the obtained formula. | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,753 |
177. Prove that if
$$
m+n \sqrt{5}=(a+b \sqrt{5})(c+d \sqrt{5})
$$
then
$$
m-n \sqrt{5}=(a-b \sqrt{5})(c-d \sqrt{5})
$$
Now let's consider the solutions of equation (1). | 177. Using the solution of problem 176, we write
$$
m+n \sqrt{5}=(a c+5 b d)+(a d+b c) \sqrt{5}
$$
By problem 175, \( m=a c+5 b d \), \( n=a d+b c \). Therefore,
$$
\begin{aligned}
m-n \sqrt{5} & =(a c+5 b d)-(a d+b c) \sqrt{5}= \\
& =[a c+5(-b)(-d)]+[a(-d)+(-b) c] \sqrt{5}= \\
& =(a-b \sqrt{5})(c-d \sqrt{5})
\end{a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,754 |
178. Let $a+b \sqrt{5}$ be a solution of equation (1). Show that
a) $a-b \sqrt{5}$ is also a solution of equation (1);
b) $\frac{1}{a+b \sqrt{5}}=a-b \sqrt{5}$. | 178. a) If $a^{2}-5 b^{2}=1$, then $a^{2}-5(-b)^{2}=1$, i.e.
$$
a+(-b) \sqrt{5}=a-b \sqrt{5}
$$
- solution of equation ( 1 ).
$$
\text { 6) } \begin{aligned}
\frac{1}{a+b \sqrt{5}} & =\frac{a-b \sqrt{5}}{(a+b \sqrt{5})(a-b \sqrt{5})}= \\
& =\frac{a-b \sqrt{5}}{a^{2}-5 b^{2}}=\frac{a-b \sqrt{5}}{1}=a-b \sqrt{5} .
\en... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,755 |
179. Let $a+b \sqrt{5}$ and $c+d \sqrt{5}$ be solutions of equation (1). Prove that
a) their product
$$
m+n \sqrt{5}=(a+b \sqrt{5})(c+d \sqrt{5})
$$
is a solution of equation ( 1 );
b) their quotient $\frac{a+b \sqrt{5}}{c+d \sqrt{5}}$ can be represented as $p+q \sqrt{5}$ and is also a solution of equation ( 1 ). | 179. a) Since by the condition
$$
m+n \sqrt{5}=(a+b \sqrt{5})(c+d \sqrt{5})
$$
then by problem 177
$$
m-n \sqrt{5}=(a-b \sqrt{5})(c-d \sqrt{5})
$$
Therefore
$$
\begin{aligned}
& m^{2}-5 n^{2}=(m+n \sqrt{5})(m-n \sqrt{5})= \\
&=(a+b \sqrt{5})(c+d \sqrt{5})(a-b \sqrt{5})(c-d \sqrt{5})= \\
&=(a+b \sqrt{5})(a-b \sqrt{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,756 |
180. Check that $9+4 \sqrt{5}$ is a solution to equation (1). Prove that equation (1) has an infinite number of different integer solutions. | 180. $9^{2}-5 \cdot 4^{2}=1$ and, therefore, $9+4 \sqrt{5}$ is a solution to equation (1). By problem 179a)
$$
p_{2}+q_{2} \sqrt{5}=(9+4 \sqrt{5})(9+4 \sqrt{5})=(9+4 \sqrt{5})^{2}
$$
is also a solution to equation (1). Similarly, the solutions will be
$$
p_{8}+q_{3} \sqrt{5}=\left(p_{2}+q_{2} \sqrt{5}\right)(9+4 \sq... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,757 |
181. Let $a+b \sqrt{5}$ and $c+d \sqrt{5}$ be the solutions of equation (1), where $a \geqslant 0, b \geqslant 0, c \geqslant 0, d \geqslant 0$. Prove that if
$$
a+b \sqrt{5}<c+d \sqrt{5}
$$
then $a<c$ and $b<d$. | 181. It cannot be that $a \geqslant c$ and $b \geqslant d$, because then it would be
$$
a+b \sqrt{5} \geqslant c+d \sqrt{5}
$$
Therefore, at least one of the two inequalities must hold: $a<c$ or $b<d$. We will show that the fulfillment of one of these inequalities implies the fulfillment of the other. Suppose $a<c$. ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 24,758 |
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