problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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3-5. Five people play several games of dominoes (two on two) so that each player has each other as a partner once and as an opponent twice. Find the number of games played and all possible ways of distributing the players. | Solution 5. Answer: 5 games; the distribution of players is unique (up to their numbering). First, let's see who the 1st player played with. We can assume that the 1st and 2nd played against the 3rd and 4th, and the 1st and 5th played against the 2nd and 3rd (this can be achieved by swapping the 3rd and 4th). The 3rd p... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-4. Find all numbers by which the fraction \(\frac{5 l+6}{8 l+7}\) can be reduced when \(l\) is an integer. | Solution 4. We will compute \(\text{GCD}(5l+6, 8l+7)\), using the fact that \(\text{GCD}(a, b) = \text{GCD}(a-b, b)\). As a result, we get \(\text{GCD}(5l+6, 8l+7) = \text{GCD}(5l+6, 3l+1) = \text{GCD}(2l+5, 3l+1) = \text{GCD}(2l+5, l-4) = \text{GCD}(l+9, l-4) = \text{GCD}(13, l-4)\). The number 13 is prime, so the fra... | 13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3-5. What is the smallest number of points that can be chosen on a circle of length 1956 so that for each of these points there is exactly one chosen point at a distance of 1 and exactly one at a distance of 2 (distances are measured along the circumference)? | Solve 5. Answer: 1304.
Let \(A\) be one of the selected points, \(B\) and \(C\) be the selected points, located at distances 1 and 2 from it, respectively. The arrangement \(A B C\) is impossible, since in this case, for point \(B\), there are two selected points at a distance of 1. Therefore, the points are arranged ... | 1304 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-2. Points \(A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}\) divide a circle of radius 1 into six equal parts. From \(A_{1}\), a ray \(l_{1}\) is drawn in the direction of \(A_{2}\), from \(A_{2}\) - a ray \(l_{2}\) in the direction of \(A_{3}, \ldots\), from \(A_{6}\) - a ray \(l_{6}\) in the direction of \(A_{1}\). From ... | Solve 2. Answer: \(B_{1} A_{1}=2\).
Let \(B_{2}, B_{3}, \ldots, B_{7}\) be the feet of the perpendiculars dropped from \(l_{6}, l_{5}, \ldots, l_{1} ; x_{1}=A_{2} B_{1}, x_{2}=\) \(A_{1} B_{2}, x_{3}=A_{6} B_{3}, \ldots, x_{7}=A_{2} B_{7}\). Then \(x_{k+1}=\frac{1}{2}\left(1+x_{k}\right)\). By the condition \(x_{1}=x_... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3-3. We took three numbers \(x, y, z\). We calculated the absolute values of the pairwise differences \(x_{1}=|x-y|\), \(y_{1}=|y-z|, z_{1}=|z-x|\). In the same way, from the numbers \(x_{1}, y_{1}, z_{1}\) we constructed the numbers \(x_{2}, y_{2}, z_{2}\) and so on. It turned out that for some \(n\), \(x_{n}=x\), \(y... | Solve 3. Answer: \(y=z=0\).
The numbers \(x_{n}, y_{n}, z_{n}\) are non-negative, so the numbers \(x, y, z\) are also non-negative. If all the numbers \(x\), \(y, z\) were positive, then the largest of the numbers \(x_{1}, y_{1}, z_{1}\) would be strictly less than the largest of the numbers \(x, y, z\), and then the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3-4. In a rectangular table composed of positive numbers, the product of the sum of the numbers in any column and the sum of the numbers in any row equals the number at their intersection. Prove that the sum of all numbers in the table is one. | Solution 4. First solution. Let \(x_{1}, \ldots, x_{n}\) be the sums of the numbers in the rows, and \(y_{1}, \ldots, y_{m}\) be the sums of the numbers in the columns. At the intersection of the \(i\)-th row and the \(j\)-th column, the number \(x_{i} y_{j}\) is placed. Therefore, the sum of the numbers in the \(i\)-t... | 1 | Algebra | proof | Yes | Yes | olympiads | false |
3-5. The distance from \(A\) to \(B\) is 999 km. Along the road, there are kilometer markers indicating the distances to \(A\) and to \(B: 0\) ।999, 1 ।998, \(\ldots, 999\) ।0. How many of these markers have only two different digits? | Solve 5. Answer: 40.
Assume that on the kilometer post, the number \(\overline{a b d a_{1} b_{1} c_{1}}\) is written. Then \(\overline{a b c} + \overline{a_{1} b_{1} c_{1}} = 999\), so \(a_{1} = 9 - a\), \(b_{1} = 9 - b\), and \(c_{1} = 9 - c\). If \(a = b = c\), then the required condition is satisfied. There will be... | 40 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-ча 1. In a chess tournament, 12 people participated. After the tournament, each participant made 12 lists. The first list includes only himself, the second list includes him and those he defeated, the third list includes all people from the second list and those they defeated, and so on. The 12th list includes all pe... | Solution 1. Answer: 54. If the \((k+1)\)-th list is the same as the \(k\)-th, then the lists numbered \(k+2, \ldots, 11, 12\) will also be exactly the same. However, according to the condition, the 11th list and the 12th list are different. Therefore, each participant's \(k\)-th list contains exactly \(k\) people. In p... | 54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3-6. All values of the quadratic trinomial \(a x^{2}+b x+c\) on the interval \([0,1]\) do not exceed 1 in absolute value. What is the maximum value that the quantity \(|a|+|b|+|c|\) can have in this case? | Solve 6.
\section*{9th Grade} | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.40. The leg $BC$ of the right triangle $ABC$ with a right angle at $C$ is divided by points $D$ and $E$ into three equal parts. Prove that if $BC = 3AC$, then the sum of the angles $AEC$, $ADC$, and $ABC$ is $90^{\circ}$. | 1.40. First solution. Consider the square $B C M N$ and divide its side $M N$ into three equal parts by points $P$ and $Q$ (Fig. 1.9). Then $\triangle A B C = \triangle P D Q$ and $\triangle A C D = \triangle P M A$. Therefore, triangle $P A D$ is an isosceles right triangle and $\angle A B C + \angle A D C = \angle P ... | 90 | Geometry | proof | Yes | Yes | olympiads | false |
2.15. A circle with a radius equal to the height of an equilateral triangle rolls along one of its sides. Prove that the angular measure of the arc cut off on the circle by the sides of the triangle is always $60^{\circ}$. | 2.15. Let the angular measure of the arc cut off by the sides of triangle $ABC$ on the circle be denoted by $\alpha$. Consider the arc cut off by the extensions of the sides of the triangle on the circle, and denote its angular measure by $\alpha^{\prime}$. Then $\left(\alpha+\alpha^{\prime}\right) / 2=\angle B A C=60^... | 60 | Geometry | proof | Yes | Yes | olympiads | false |
3.45*. On the diameter $A B$ of the circle $S$, a point $K$ is taken, and a perpendicular is erected from it, intersecting $S$ at point $L$. Circles $S_{A}$ and $S_{B}$ are tangent to the circle $S$, the segment $L K$, and the diameter $A B$, with $S_{A}$ touching the segment $A K$ at point $A_{1}$, and $S_{B}$ touchin... | 3.45. Let $\angle L A B=\alpha$ and $\angle L B A=\beta\left(\alpha+\beta=90^{\circ}\right)$. According to problem 3.42, b) $A B_{1}=A L$, so $\angle A B_{1} L=90^{\circ}-\alpha / 2$. Similarly, $\angle B A_{1} L=90^{\circ}-\beta / 2$. Therefore, $\angle A_{1} L B_{1}=(\alpha+\beta) / 2=45^{\circ}$. | 45 | Geometry | proof | Yes | Yes | olympiads | false |
5.33. In triangle $A B C$ with angle $A$ equal to $120^{\circ}$, the angle bisectors $A A_{1}, B B_{1}$ and $C C_{1}$ intersect at point $O$. Prove that $\angle A_{1} C_{1} O=30^{\circ}$. | 5.33. According to the solution of the previous problem, ray $A_{1} C_{1}$ is the bisector of angle $A A_{1} B$. Let $K$ be the point of intersection of the angle bisectors of triangle $A_{1} A B$. Then $\angle C_{1} K O=\angle A_{1} K B=90^{\circ}+\angle A / 2=120^{\circ}$. Therefore, $\angle C_{1} K O+\angle C_{1} A ... | 30 | Geometry | proof | Yes | Yes | olympiads | false |
6.90. What is the maximum number of acute angles a convex polygon can have
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 6.90. Let a convex $n$-gon have $k$ acute angles. Then the sum of its angles is less than $k \cdot 90^{\circ} + (n-k) \cdot 180^{\circ}$. On the other hand, the sum of the angles of an $n$-gon is $(n-2) \cdot 180^{\circ}$. Therefore, $(n-2) \cdot 180^{\circ} < k \cdot 90^{\circ} + (n-k) \cdot 180^{\circ}$, i.e., $k < 4... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
8.68. Using a compass and a ruler, divide an angle of $19^{\circ}$ into 19 equal parts. | 8.68. If there is an angle of magnitude $\alpha$, then angles of magnitude $2 \alpha, 3 \alpha$, etc., can be constructed. Since $19 \cdot 19^{\circ}=361$, an angle of $361^{\circ}$, which coincides with an angle of $1^{\circ}$, can be constructed. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.18*. How many sides can a convex polygon have if all its diagonals have the same length? | 9.18. We will prove that the number of sides of such a polygon does not exceed 5. Suppose that all diagonals of the polygon $A_{1} \ldots A_{n}$ have the same length and $n \geqslant 6$. Then the segments $A_{1} A_{4}, A_{1} A_{5}, A_{2} A_{4}$, and $A_{2} A_{5}$ have the same length, as they are diagonals of this poly... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.22. In a triangle, the lengths of two sides are 3.14 and 0.67. Find the length of the third side, given that it is an integer. | 9.22. Let the length of the third side be $n$. By the triangle inequality, $3.14-0.67<n<3.14+$ $+0.67$. Since $n$ is an integer, then $n=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.32. The perimeter of a convex quadrilateral is 4. Prove that its area does not exceed 1. | 9.32. According to problem $9.31 S_{A B C D} \leqslant(A B+C D)(B C+A D) / 4$. Since $a b \leqslant(a+$ $+b)^{2} / 4$, then $S_{A B C D} \leqslant(A B+C D+A D+B C)^{2} / 16=1$. | 1 | Geometry | proof | Yes | Yes | olympiads | false |
11.30. The diagonals of a convex quadrilateral $ABCD$ intersect at point $O$. What is the smallest area that this quadrilateral can have if the area of triangle $AOB$ is 4 and the area of triangle $COD$ is 9? | 11.30. Since $S_{A O B}: S_{B O C}=A O: O C=S_{A O D}: S_{D O C}$, then $S_{B O C} \cdot S_{A O D}=$ $=S_{A O B} \cdot S_{D O C}=36$. Therefore, $S_{B O C}+S_{A O D} \geqslant 2 \sqrt{S_{B O C} \cdot S_{A O D}}=12$, and equality is achieved if $S_{B O C}=S_{A O D}$, i.e., $S_{A B C}=S_{A B D}$, from which $A B \| C D$.... | 25 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.42. If five points are given on a plane, then by considering all possible triples of these points, one can form 30 angles. Let the smallest of these angles be $\alpha$. Find the maximum value of $\alpha$.
If five points are given on a plane, then by considering all possible triples of these points, one can form 30 ... | 11.42. First, assume that the points are the vertices of a convex pentagon. The sum of the angles of a pentagon is $540^{\circ}$, so one of its angles does not exceed $540^{\circ} / 5=108^{\circ}$. The diagonals divide this angle into three angles, so one of them does not exceed $108^{\circ} / 3=36^{\circ}$. In this ca... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.43*. In a city, there are 10 streets parallel to each other, and 10 streets intersecting them at right angles. What is the minimum number of turns a closed bus route can have, passing through all intersections? | 11.43. A closed route passing through all intersections can have 20 turns (Fig. 11.8). It remains to prove that such a route cannot have fewer than 20 turns. After each turn, there is a transition from a horizontal street to a vertical one or vice versa. Therefore, the number of horizontal segments in a closed route is... | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.44*. What is the maximum number of cells on an $8 \times 8$ chessboard that can be cut by a single straight line? | 11.44. A line can intersect 15 cells (Fig. 11.9). We will now prove that a line cannot intersect more than 15 cells. The number of cells intersected by a line is one less than the number of points of intersection with the segments that form the sides of the cells. Inside the square, there are 14 such segments. Therefor... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.45*. What is the maximum number of points that can be placed on a segment of length 1 so that on any segment of length $d$ contained in this segment, there are no more than $1+1000 d^{2}$ points?
See also problems $15.1,17.20$.
## §7. Extremal Properties of Regular Polygons | 11.45. First, let's prove that it is impossible to place 33 points in such a way. Indeed, if 33 points are on a segment of length 1, then the distance between some two of them does not exceed \(1 / 32\). The segment with endpoints at these points contains two points, but it should contain no more than \(1 + 1000 / 32^2... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12.52. In triangle $A B C$, the height $A H$ is equal to the median $B M$. Find the angle $M B C$.
Fig. 12.2 | 12.52. Drop a perpendicular $M D$ from point $M$ to line $B C$. Then $M D = A H / 2 = B M / 2$. In the right triangle $B D M$, the leg $M D$ is equal to half of the hypotenuse $B M$. Therefore, $\angle M B C = \angle M B D = 30^{\circ}$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.53. In triangle $ABC$, the angle bisectors $AD$ and $BE$ are drawn. Find the measure of angle $C$, given that $AD \cdot BC = BE \cdot AC$ and $AC \neq BC$. | 12.53. The quantities $A D \cdot B C \sin A D B$ and $B E \cdot A C \sin A E B$ are equal, as they are equal to twice the area of triangle $A B C$. Therefore, $\sin A D B = \sin A E B$. There are two possible cases.
1. $\angle A D B = \angle A E B$; in this case, points $A, E, D, B$ lie on the same circle, so $\angle ... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.54. Find the angle $B$ of triangle $A B C$, if the length of the height $C H$ is half the length of side $A B$, and $\angle B A C=75^{\circ}$. | 12.54. Let $B'$ be the point of intersection of the perpendicular bisector of segment $AC$ with line $AB$. Then $AB' = CB'$ and $\angle AB'C = 180^\circ - 2 \cdot 75^\circ = 30^\circ$. Therefore, $AB' = CB' = 2CH = AB$, i.e., $B' = B$ and $\angle B = 30^\circ$. | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.55*. In a right triangle $ABC$ with a right angle at $A$, a circle is constructed on the altitude $AD$ as a diameter, intersecting side $AB$ at point $K$ and side $AC$ at point $M$. Segments $AD$ and $KM$ intersect at point $L$. Find the acute angles of triangle $ABC$, given that $AK: AL = AL: AM$. | 12.55. It is clear that $A K D M$ is a rectangle and $L$ is the point of intersection of its diagonals. Since $A D \perp B C$ and $A M \perp B A$, then $\angle D A M = \angle A B C$. Similarly, $\angle K A D = \angle A C B$. Drop a perpendicular $A P$ from point $A$ to the line $K M$. Let's assume for definiteness that... | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18.5. On the sides $C B$ and $C D$ of the square $A B C D$, points $M$ and $K$ are taken such that the perimeter of triangle $C M K$ is equal to twice the side of the square. Find the measure of angle $M A K$. | 18.5. Rotate the given square around point $A$ by $90^{\circ}$ so that vertex $B$ moves to $D$. Let $M^{\prime}$ be the image of point $M$ under this rotation. Since by the condition $M K+M C+C K=(B M+M C)+(K D+C K)$, then $M K=B M+K D=D M^{\prime}+K D=K M^{\prime}$. Moreover, $A M=A M^{\prime}$, therefore $\triangle A... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
18.34*. In the circus arena, which is a circle with a radius of 10 m, a lion is running. Moving along a broken line, he ran $30 \mathrm{km}$. Prove that the sum of all his turning angles is not less than 2998 radians.
## §4. Compositions of Rotations | 18.34. Suppose the lion ran along the broken line $A_{1} A_{2} \ldots A_{n}$. Let's straighten the lion's trajectory as follows. Rotate the circus arena and the subsequent trajectory around point $A_{2}$ so that point $A_{3}$ falls on the ray $A_{1} A_{2}$. Then rotate the circus arena and the subsequent trajectory aro... | 2998 | Geometry | proof | Yes | Yes | olympiads | false |
21.7*. Each of the two disks is divided into 1985 equal sectors, and 200 sectors on each are painted in an arbitrary color (one color). The disks are placed on top of each other, and one is rotated by angles that are multiples of $360^{\circ} / 1985$. Prove that there are at least 80 positions in which no more than 20 ... | 21.7. Let's take 1985 disks colored the same as the second of our disks and place them on the first disk so that they occupy all possible positions. Then above each colored sector of the first disk, there are 200 colored sectors, i.e., there are a total of $200^{2}$ pairs of coinciding colored sectors. Let there be $n$... | 81 | Combinatorics | proof | Yes | Yes | olympiads | false |
21.9*. In the park, there are 10000 trees planted in a square grid pattern (100 rows of 100 trees each). What is the maximum number of trees that can be cut down so that the following condition is met: if you stand on any stump, you will not see any other stump? (The trees can be considered thin enough.) | 21.9. Let's divide the trees into 2500 quartets, as shown in Fig. 21.3. In each such quartet, no more than one tree can be cut down. On the other hand, all the trees growing in the upper left corners of the squares formed by our quartets of trees can be cut down. Therefore, the maximum number of trees that can be cut d... | 2500 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
22.4. Among all such numbers $n$ that any convex 100-gon can be represented as the intersection (i.e., common part) of $n$ triangles,
find the smallest. | 22.4. First, note that 50 triangles are sufficient. Indeed, let $\Delta_{k}$ be the triangle whose sides lie on the rays $A_{k} A_{k-1}$ and $A_{k} A_{k+1}$ and which contains the convex polygon $A_{1} \ldots A_{100}$. Then this polygon is the intersection of the triangles $\Delta_{2}, \Delta_{4}, \ldots, \Delta_{100}$... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
25.25*. A regular octagon with side 1 is cut into parallelograms. Prove that among them there are at least two rectangles, and the sum of the areas of all rectangles is 2.
## §5. Plane, cut by lines
Let $n$ pairwise non-parallel lines be drawn on the plane, with no three intersecting at the same point. Problems 25.26... | 25.25. Let us consider two mutually perpendicular pairs of opposite sides in a regular octagon and, as in problem 25.1, chains of parallelograms connecting opposite sides. At the intersections of these chains, there are rectangles. By considering two other pairs of opposite sides, we will obtain at least one more recta... | 2 | Geometry | proof | Yes | Yes | olympiads | false |
27.7. Ten points are marked on a circle. How many non-closed, non-self-intersecting nine-segment broken lines with vertices at these points exist? | 27.7. The first point can be chosen in ten ways. Each of the following eight points can be chosen in two ways, as it must be adjacent to one of the previously chosen points (otherwise, a self-intersecting broken line would result). Since the beginning and end are not distinguished in this calculation, the result must b... | 1280 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
43. It is obvious that any figure with a diameter of 1 can be enclosed within a square with a side of 2: for this, it is sufficient for the center of the square to coincide with any point of the figure. What is the side of the smallest square that can enclose any figure with a diameter of 1?
Note. One can also pose a ... | 43. It is obvious that a circle with a diameter of 1 cannot be enclosed in any square whose side is less than 1. On the other hand, it is almost equally clear that any figure with a diameter of 1 can be enclosed in a square with a side of 1. Indeed, any square that encloses some figure $\Phi$ can always be reduced so t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
49. How many circles of radius 1 can be applied ${ }^{1}$ ) to a given unit circle $\mathcal{S}$ so that no two of these circles intersect? So that no one of these circles contains the center of another circle inside itself? | 49. Since a unit circle tangent to $S$ is seen from the center $O$ of circle $S$ at an angle of $60^{\circ}$ (Fig. $132, a$), no more than $6\left(=\frac{360}{60^{\circ}}\right)$ non-overlapping unit circles can be applied to $S$. Six circles can obviously be applied (Fig. 132, b).
Further, if $O_{1}$ and $O_{2}$ are ... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
50*. What is the greatest number of circles of radius 1 that can be placed on a plane so that they all intersect a certain fixed unit circle $S$ and no one of them contains the center of $S$ or the center of another circle inside it? | 50. Fig. 134 shows that 18 circles satisfying the condition of the problem can be placed (six circles in Fig. 134 have centers at the vertices of a regular hexagon inscribed in $S$, the other 12 - at the vertices of squares constructed on the sides of the inscribed hexagon outside it; compare with Fig. 130). The fact t... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
51. What is the greatest number of squares with side 1 that can be placed ${ }^{1}$ ) next to a given unit square $K$ so that no two of them intersect?
$^{1}$) The superscript "1" is kept as is, since it might refer to a footnote or additional information in the original text. | 51. First solution. Let $O$ be the center of the main square $K$, and $O_{1}$ and $O_{2}$ be the centers of the non-overlapping squares $K_{1}$ and $K_{2}$ attached to it (Fig. $135, a$). Since the smallest distance from the center of a unit square to its boundary is $\frac{1}{2}$, the segments $O O_{1}$, $O O_{2}$, an... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
54. What is the smallest number of circles with which a circle of twice the radius can be completely covered? | 54. Since the diameter of the smaller circles is equal to the radius $R$ of the larger circle, each such circle intersects the circumference of the larger circle at points no more than $R$ apart, and thus covers an arc of this circumference no greater than $60^{\circ}$. It follows that at least six smaller circles are ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
102. Given a triangle $ABC$. A point $M$, located inside the triangle, moves parallel to side $BC$ until it intersects side $CA$, then moves parallel to $AB$ until it intersects side $BC$, then - parallel to $AC$ until it intersects $AB$, and so on. Prove that after a certain number of such steps, the point will return... | 102. If point $M$ lies on any of the medians of triangle $ABC$, for example, on $M_{3} M_{1}$, then it is obvious that after four steps it will return to its initial position (Fig. $227, a$). Now suppose that $M$ does not lie on any of the three medians of triangle $ABC$.
Let $M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6},... | 7 | Geometry | proof | Yes | Yes | olympiads | false |
31. a) The third term of an arithmetic progression is equal to 0. Find the sum of the first 5 terms.
b) The third term of a geometric progression is equal to 4. Find the product of the first 5 terms. | 31. a) We know that for an arithmetic progression $u_{1}=u_{3}-2 d ; u_{2}=u_{3}-d$; $u_{4}=u_{3}+d ; \quad u_{5}=u_{3}+2 d$. Therefore, the sum of the first five terms is
$$
\begin{aligned}
& \delta_{5}=u_{1}+u_{2}+u_{3}+u_{4}+u_{5}=\left(u_{3}-2 d\right)+ \\
& +\left(u_{3}-d\right)+u_{3}+\left(u_{3}+d\right)+ \\
& +... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
33. Someone arrives in the city with very interesting news and after 10 minutes reports it to two people. Each of the newly informed people, after 10 minutes, reports it to two more (who have not yet heard it) and so on. *). How long will it take for the entire city to know this news, if the city has three million resi... | 33. It is easy to see that the number of people who were told the news at the end of the 1st ten-minute period is 2, the number of people who were told the news after the 2nd ten-minute period is 4, and generally, the number of people who were told the news at the end of the $k$-th ten-minute period is $2^{k}$.
The nu... | 210 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
35. Find the sum of all positive odd numbers not exceeding a thousand. | 35. We need to find the sum $1+3+$ $+5+\ldots+997+999$. It is clear that this is the sum of the first 500 terms of an arithmetic progression with the first term 1 and common difference 2. Therefore, our sum is
$1 \cdot 500+\frac{500 \cdot 499}{2} \cdot 2=500+500 \cdot 499=$
$$
=500 \cdot 500=250000
$$ | 250000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
36. Find the sum of all positive three-digit numbers that are not divisible by 2 or 3. | 36. Subtract from the sum of all positive three-digit numbers $S^{(1)}$ the sum of positive three-digit numbers divisible by 2 ($S^{(2)}$), and the sum of positive three-digit numbers divisible by 3 ($S^{(3)}$). In this process, numbers divisible by both 2 and 3, i.e., divisible by 6, are subtracted twice.
Therefore, ... | 164700 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
44. In the kitchen, there are five light bulbs. How many lighting options exist?
Do you understand what lighting options mean? Each lamp can be on or off. Two options are considered different if they differ in the state of at least one lamp (see Fig. 7 and 8). | 44. First solution. The number of ways to illuminate with one lamp out of five is equal to the number of ways to illuminate with four lamps out of five (Fig. 32).
$$
\begin{aligned}
& 000 \sim 0000 \\
& 10000000 \\
& 0000 \text { } \\
& 1000 \\
& 0000 \sim 0000
\end{aligned}
$$
## Fig. 32
Similarly, the number of wa... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
55. What is the maximum number of bishops that can be placed on a chessboard so that they do not threaten each other? Prove that the number of ways to arrange the bishops in such a way is the square of some number.
To understand the condition, of course, you need to know how a bishop moves. A bishop moves diagonally. ... | 55. The instruction to the problem shows a specific example of placing 14 bishops. Therefore, it is possible to place 14 bishops. If we prove that it is impossible to place more than 14 bishops, the first part of the problem will be solved. Let's first consider the black-square bishops. How many can be placed on a ches... | 14 | Combinatorics | proof | Yes | Yes | olympiads | false |
56. Mom has two apples and three pears. Every day for five consecutive days, she gives out one fruit. In how many ways can this be done? | 56. Here is one way to distribute apples and pears: 000 'on the 1st and 3rd day - apples, on the 2nd, 4th, and 5th days - pears),
## 000
(on the first three days - pears, on the last two days - apples). Thus, we need to count all the tables with two light and three dark circles; but we have already done this when sol... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
81. Determine the sum of the coefficients of the polynomial that results from expanding and combining like terms in the expression $\left(1+x-3 x^{2}\right)^{1965}$. | 81. If we expand the expression $\left(1+x-3 x^{2}\right)^{1965}$ and combine like terms, we get a polynomial $a_{0}+a_{1} \cdot x+$ $+a_{2} \cdot x^{2}+a_{3} \cdot x^{3}+\ldots$. Note that the sum of its coefficients is equal to the value of the polynomial at $x=1$
$$
\begin{aligned}
& a_{0}+a_{1} \cdot 1+a_{2} \cdot... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
93. How many phone numbers contain the combination 12? (The number consists of six digits.) | 93. Let's consider the following five sets of phone numbers:
... | 49401 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
105. Is it true that there exists a number $C$ such that for all integers $k$ the inequality
$$
\left|\frac{k^{8}-2 k+1}{k^{4}-3}\right|<C ?
$$ | 105. Let's see how the expression $\left|\frac{k^{3}-2 k+1}{k^{4}-3}\right|$ behaves for large (in absolute value) values of $k$. Clearly, in the numerator, the term $k^{3}$ plays the main role, and in the denominator, $k^{4}$. Therefore, we can expect that for large values of $k$, our expression is approximately equal... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
148. To compute the square root of a positive number $a$, one can use the following method of successive approximations. Take any number $x_{0}$ and construct a sequence according to the following rule:
$$
x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)
$$
Prove that if $x_{0}>0$, then $\lim _{n \rightarrow \in... | 148. First, let's prove that if the limit of $\{x_{n}\}$ exists, then it equals $\pm \sqrt{a}$. Indeed, let $\lim _{n \rightarrow \infty} x_{n}=b$. Then $\lim _{n \rightarrow \infty} \frac{1}{2}\left(x_{n}+\frac{a}{x_{n}}\right)=\frac{1}{2}\left(b+\frac{a}{b}\right)$. We obtain the equation $b=\frac{1}{2}\left(b+\frac{... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
228. Given an angle MAN and a point $O$ not lying on the side of the angle. Draw a line through $O$ intersecting the sides of the angle at points $X$ and $Y$, such that the product $O X \cdot O Y$ has a given value $k$. | 228. Suppose the problem is solved. From OX. $O Y=k$, it follows that $X$ is obtained from point $Y$ by the inversion with center $O$ and power $k$; therefore, $X$ lies on the circle $S$, which is obtained from the line $A N$ by the inversion with center $O$ and power $k$ (i.e., $k$), i.e., $X$ is the intersection poin... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
298. The set of all cycles in Lobachevsky's non-Euclidean geometry that are perpendicular to two given cycles \( S_{1} \) and \( S_{2} \) is called a bundle of cycles. List all possible types of bundles of cycles in Lobachevsky's non-Euclidean geometry. Prove that for each bundle II, there are infinitely many cycles th... | 298. Obviously, the pencil of cycles of Lobachevsky's hyperbolic geometry does not differ from the pencil of circles of ordinary (Euclidean) geometry (see § 3 of this chapter); only in accordance with the definition of points in hyperbolic geometry, here one should consider not the entire circles but only arcs of them,... | 19 | Geometry | proof | Yes | Yes | olympiads | false |
2.12. (GDR, 77). How many pairs of values $p, q \in \mathbf{N}$, not exceeding 100, exist for which the equation
$$
x^{5}+p x+q=0
$$
has solutions in rational numbers? | 2.12. Let for some values
$$
p, q \in\{1 ; 2 ; \ldots ; 100\}
$$
the number $x \in \mathbf{Q}$ satisfies the equation
$$
x^{5}+p x+q=0
$$
Since all coefficients of the polynomial on the left side of the equation are integers, and the coefficient of the leading term is 1, by Theorem 60, any rational root of this pol... | 133 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4.2. (Belgium, 79). Find the sum of all $7!$ numbers that can be obtained by all possible permutations of the digits in the number 1234567. | 4.2. For any values of $i, i \in\{1 ; \ldots ; 7\}$, the number of numbers in which the $i$-th place is occupied by the digit $j$ is 61. Therefore, the sum of all numbers is
$$
\begin{aligned}
& (61 \cdot 1+\ldots+61 \cdot 7)+(61 \cdot 1+\ldots+61 \cdot 7) 10+ \\
& \quad+(61 \cdot 1+\ldots+61 \cdot 7) 10^{2}+\ldots+(6... | 22399997760 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4.17*. (NRP, 83). Find all values of $n \in \mathbf{N}$ for each of which there exists a permutation
$$
\left(a_{1} ; a_{2} ; \ldots ; a_{n}\right)
$$
of the numbers $0,1, \ldots, n-1$, such that all numbers
$$
a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \ldots, a_{1} a_{2} \ldots a_{n}
$$
give different remainders when... | 4.17. For $n=1$ the permutation $\left(a_{1}\right)=(0)$ satisfies the condition of the problem. For $n=4$, the permutation
$$
\left(a_{1} ; a_{2} ; a_{8} ; a_{4}\right)=(1 ; 3 ; 2 ; 0)
$$
possesses the required property. Let $n$ be a prime number. According to the Chinese "remainder theorem" (Theorem 23), for each v... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.6. (Yugoslavia, 83). Find all values of $n \in \mathbf{N}$ that have the following property: if the numbers $n^{3}$ and $n^{4}$ are written next to each other (in the decimal system), then in the resulting record each of the 10 digits $0,1, \ldots, 9$ will appear exactly once. | 5.6. Let $f(m)$ denote the number of digits in the decimal representation of the number $m \in N$, then for the desired number $n$ we have
$$
f\left(n^{3}\right)+f\left(n^{4}\right)=10
$$
Moreover, $f\left(n^{9}\right) \geqslant 4$, because otherwise the inequalities $n^{3}4$, then $n>10, n^{4}>10 n^{3}$, from which
... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.1. (GDR, 74). What is greater: $\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}$ or 0? | 7.1. Using Theorem 4, we get
$$
\begin{aligned}
& \sqrt{4+\sqrt{7}}=\sqrt{4-\sqrt{7}}-\sqrt{\overline{2}}= \\
& =\left(\sqrt{\frac{4+\sqrt{16-7}}{2}}+\sqrt{\frac{4-\sqrt{16-7}}{2}}\right)- \\
& -\left(\sqrt{\frac{4+\sqrt{16-7}}{2}}-\sqrt{\frac{4-\sqrt{16-7}}{2}}\right)-\sqrt{2}= \\
& =\sqrt{\frac{7}{2}}+\sqrt{\frac{1}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.2. (England, 75). Solve the equation
$$
[\sqrt[3]{1}]+[\sqrt[3]{2}]+\ldots+\left[\sqrt[3]{x^{3}-1}\right]=400
$$
in natural numbers. | 8.2. Note that the relation $[\sqrt[3]{m}]=k$, where $m, k \in \mathrm{N}$, is equivalent to the inequality $k^{3} \leqslant m \leqslant(k+1)^{3}-1$. The number of natural numbers $m$ satisfying this condition (for a fixed $k$) is $(k+1)^{3}-k^{3}=3 k^{2}+3 k+1$. Therefore, the left side of the equation is $\sum_{k=1}^... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.3. (Sweden, 82). Find all values of $n \in \mathbf{N}$ for each of which there exists a number $m \in \mathbf{N}$, a triangle $A B C$ with sides $A B=33, A C=21, B C=n$ and points $D$, $E$ on sides $A B, A C$ respectively, satisfying the conditions $A D=D E=E C=m$. | 9.3. Let the numbers $m, n \in \mathbf{N}$ satisfy the condition of the problem. Then $m = CE < AC = 21$ and from triangle $ADE$ (Fig. 3) we have $21 - m = AE < AD + DE = 2m$, hence
$$
7 < m < 21
$$
Furthermore, since $AD = DE$, for the angle $\alpha = \angle BAC$ we find
$$
\cos \alpha = AE / (2 \cdot AD) = (21 - m... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.20. (SFRY, 83). Inside triangle $A B C$, a point $M$ is taken, for which $\angle M B A=30^{\circ}, \angle M A B=10^{\circ}$. Find $\angle A M C$, if $\angle A C B=80^{\circ}$ and $A C=B C$. | 9.20. Let the height $CH$ of triangle $ABC$ intersect the line
Fig. 16
$BM$ at point $E$ (Fig. 16). Then $AE = BE$ and
\[
\begin{aligned}
& \angle EAM = \angle EAB - \angle MAB = 30^\circ... | 70 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.21. (England, 70). On the sides $B C$ and $A C$ of triangle $A B C$, points $D$ and $E$ are chosen respectively such that $\angle B A D=50^{\circ}, \angle A B E=30^{\circ}$. Find $\angle B E D$, if
$\angle A B C=\angle A C B=50^{\circ}$ | 9.21. Let $O$ be the point of intersection of lines $A D$ and $B E$ (Fig. 17), then $\angle A O B=180^{\circ}-30^{\circ}-50^{\circ}=100^{\circ}$, $\angle B D A=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}$, $\angle C B E=50^{\circ}-30^{\circ}=20^{\circ}$, $\angle A E B=\angle C B E+\angle E C B=70^{\circ}$, $\angle C A... | 40 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.22. (GDR, 64). On the side $BC$ of triangle $ABC$, a point $P$ is taken such that $PC = 2BP$. Find $\angle ACB$ if $\angle ABC = 45^\circ$ and $\angle APC = 60^\circ$. | 9.22. If point $C_{1}$ is symmetric to point $C$ with respect to line $A P$ (Fig. 18), then $C_{1} P = C P = 2 B P$ and $\angle C_{1} P B = 180^{\circ} - \angle A P C - \angle A P C_{1} = 180^{\circ} - 60^{\circ} - 60^{\circ} = 60^{\circ}$. Therefore, $\angle C_{1} B P = 90^{\circ}$ (since triangle $C_{1} P B$ is simil... | 75 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.16. (England, 66). Find the number of sides of a regular polygon if for four of its consecutive vertices \( A, B, C, D \) the equality
\[
\frac{1}{A B}=\frac{1}{A C}+\frac{1}{A D}
\]
is satisfied. | 11.16. Let a circle with center 0 and radius $R$ be circumscribed around a polygon (Fig. 58). Denote $\alpha=\angle A O B$, then $0<\alpha<$ $<120^{\circ}$ and
$A B=2 R \sin (\alpha / 2), A C=2 R \sin \alpha$,
$$
A D=2 R \sin (3 \alpha / 2)
$$
from which we have
$$
\frac{1}{\sin (\alpha / 2)}=\frac{1}{\sin \alpha}+... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.10. (SFRY, 76). Find all values of $n \in \mathbf{N}$, greater than 2, for which it is possible to select $n$ points on a plane such that any two of them are vertices of an equilateral triangle, the third vertex of which is also one of the selected points. | 12.10. We will prove that the condition of the problem is satisfied only by the value $n=3$ (in which case the points can be placed at the vertices of an equilateral triangle). Indeed, suppose that it is possible to arrange $n \geqslant 4$ points in the manner specified in the problem. We select two points $A$ and $B$,... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.11. (CSSR, 80). The set $M$ is obtained from the plane by removing three distinct points $A, B$, and $C$. Find the smallest number of convex sets whose union is the set $M$. | 12.11. Let points $A, B$, and $C$ initially lie on the same line. Then these points divide the line into 4 intervals, and no points from different intervals can lie in the same convex set. Therefore, the number of required sets cannot be less than 4. The number 4 is achieved if the set $M$ is divided into parts as show... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12.12. (Jury, SRP, 79). Prove that for any value of $n \in \mathbf{N}$, greater than some number $n_{0}$, the entire plane can be divided into $n$ parts by drawing several lines, among which there must be intersecting ones. Find the smallest such value of $n_{0}$. | 12.12. Among the lines drawn according to the condition, there are necessarily intersecting lines that already divide the plane into 4 parts. If another line is drawn, then, as a simple case analysis shows, the number of parts will increase by at least 2. Therefore, it is impossible to get exactly 5 parts, from which i... | 5 | Geometry | proof | Yes | Yes | olympiads | false |
12.17*. (CSSR, 73). In a square with side 50, a broken line is located. Prove that if the distance from any point of the square to at least one point of the broken line is no more than 1, then the length of the broken line is greater than 1248. | 12.17. Let $U(M)$ denote the union of all circles of radius 1, the centers of which belong to a set $M$ on the plane. We will prove by induction on $n \in \mathbb{N}$ that for any broken line $A_{0} A_{1} \ldots A_{n}$, the inequality
$$
S_{U\left(A_{0} \ldots A_{n}\right)} \leqslant 2 \sum_{i=1}^{n} A_{i-1} A_{i} + \... | 1248 | Geometry | proof | Yes | Yes | olympiads | false |
15.7. (NBR, 68). Inside the tetrahedron $A B C D$ is a point $O$ such that the lines $A O, B O, C O, D O$ intersect the faces $B C D, A C D, A B D, A B C$ of the tetrahedron at points $A_{1}, B_{1}$, $C_{1}, D_{1}$ respectively, and the ratios
$$
\frac{A O}{A_{1} O}, \frac{B O}{B_{1} O}, \frac{C O}{C_{1} O}, \frac{D O... | 15.7. Let \( V \) be the volume of the tetrahedron \( ABCD \), and \( k \) be the desired number. Then we have
\[
\frac{V}{V_{OBCD}} = \frac{AA_1}{OA_1} = \frac{AO}{A_1O} + \frac{OA_1}{OA_1} = k + 1
\]
\[
\frac{V}{V_{OACD}} = \frac{V}{V_{OABD}} = \frac{V}{V_{OABC}} = k + 1,
\]
from which
\[
k + 1 = \frac{4V}{V_{OBC... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16.15. (England, 68). Find the maximum number of points that can be placed on a sphere of radius 1 so that the distance between any two of them is: a) not less than $\sqrt{2} ;$ b) greater than $\sqrt{2}$. | 16.15. a) Let's prove that if points $A_{1}, A_{2}, \ldots, A_{n}$ are located on a sphere with center $O$ and radius 1 such that the distance between any points $A_{i}, A_{j} (i \neq j)$ is at least $\sqrt{2}$, then $n \leqslant 6$. Indeed, let $n>6$. By the cosine theorem, we have
$$
A_{i} A_{j}^{2}=2-2 \cos \angle ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16.21. (England, 70). Find the smallest number of planes that divide a cube into at least 300 parts.
16.22\%. (SRP, 78). On planes $\alpha$ and $\alpha^{\prime}$, intersecting along a line $l$, three points each are chosen: $A, B, C$ and $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. The plane $\alpha^{\prime}$ ro... | 16.21. We will prove by induction on $n$ that $n$ lines divide the plane into no more than
$$
p(n)=\frac{n(n+1)}{2}+1
$$
parts, and exactly $p(n)$ parts are obtained if no two lines are parallel and no three lines pass through the same point. Indeed, $p(0)=1$, and for any value of $n \in \mathbb{N}$, we have the ineq... | 13 | Geometry | proof | Yes | Yes | olympiads | false |
17.3. (New York, 74). Let
$$
a_{n}=\frac{1 \cdot 3 \cdot 5 \ldots(2 n-1)}{2 \cdot 4 \cdot 6 \ldots 2 n}, n \in \mathrm{N}
$$
Find $\lim a_{n}$.
$$
n \rightarrow \infty
$$ | 17.3. Since
$$
\begin{aligned}
a_{n}^{2}=\frac{1^{2} \cdot 3^{2} \cdot \ldots \cdot(2 n-1)^{2}}{2^{2} \cdot 4^{2} \cdot \ldots \cdot(2 n)^{2}} & = \\
& =\frac{1 \cdot 3}{2^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdots \frac{(2 n-1)(2 n+1)}{(2 n)^{2}} \cdot \frac{1}{2 n+1}<\frac{1}{2 n+1}
\end{aligned}
$$
for any $n \in \... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
21.5. (Austria, 83). Find all values of $a$ for which the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy the equation
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0
$$ | 21.5. Let's make the substitution $y=x-3$, then the numbers $y_{1}=x_{1}-3, y_{2}=$ $=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$
(y+3)^{3}-6(y+3)^{2}+a(y+3)+a=y^{3}+3 y^{2}+(a-9) y+4 a-27
$$
By Vieta's theorem, we have the equalities
$$
\begin{aligned}
& y_{1}+y_{2}+y_{3}=-3 \\
& y_{1} y_{2}+y_{... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
23.13*. (CSSR, 74). Let $M$ be the set of all polynomials of the form
$$
P(x)=a x^{3}+b x^{2}+c x+d \quad(a, b, c, d \in \mathbf{R})
$$
satisfying the inequality $|P(x)| \leqslant 1$ for $x \in[-1 ; 1]$. Prove that some number $k$ provides the estimate $|a| \leqslant k$ for all polynomials $P(x) \in M$. Find the smal... | 23.13. The polynomial $P_{0}(x)=4 x^{3}-3 x$ belongs to the set $M$, since $P_{0}(-1)=-1, P_{0}(1)=1$, and at its extremum points we have $P_{0}(-1 / 2)=1, P_{0}(1 / 2)=-1$. Let's prove that for any polynomial $P(x) \in M$ the estimate $|a| \leqslant 4$ holds. Suppose, on the contrary, that there exists a polynomial $P... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
25.2. (USA, 82). In a society consisting of 1982 people, among any four people, at least one can be chosen who is acquainted with the other three. What is the minimum possible number of people who are acquainted with everyone? | 25.2. If there are no strangers, then the number of people who know everyone is 1982. Let $A$ and $B$ not know each other. Then all other people know each other (if $C$ does not know $D$, then in the group $A, B, C, D$ no one knows the other three). If $A$ and $B$ know all the others, then 1980 people know everyone. If... | 1979 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
25.6. (USSR, 81; USA, 81). In a certain country, any two cities are directly connected by one of the following means of transportation: bus, train, or airplane. It is known that there is no city provided with all three types of transportation, and at the same time, there do not exist three cities such that any two of t... | 25.6. Suppose there are five cities connected in the manner indicated in the problem. First, let's prove that no city has three lines of the same type of transport leading out of it. Let city \( A \) be connected to cities \( B, C \), and \( D \), for example, by airplane. Then, according to the condition, no pair of c... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
27.5. (Jury, Brazil, 82; Australia, 83). An urn contains $n$ white and $m$ black balls, and next to the urn is a box with a sufficiently large number of black balls. The following operation is performed: a pair of balls is randomly drawn from the urn; if they are of the same color, a black ball from the box is moved to... | 27.5. Since the parity of the number of white balls contained in the urn does not change after each operation, the last ball will be white if and only if the number $n$ is odd. Therefore, the desired probability is either 1 (if $n$ is odd), or 0 (if $n$ is even). | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
27.11. (Belgium, 77). Three shooters $A, B, C$ decided to duel simultaneously. They positioned themselves at the vertices of an equilateral triangle and agreed on the following: the first shot is taken by $A$, the second by $B$, the third by $C$, and so on in a circle; if one of the shooters is eliminated, the duel con... | 27.11. Let's consider three events that may occur after the first shot of shooter $A$.
1) $C$ is hit. Then with probability 1, shooter $A$ will be hit by the first shot of $B$.
2) $B$ is hit. Then: either with probability 0.5, shooter $C$ will hit $A$ with his first shot, or with probability $0.5 \cdot 0.3$, shooter $... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10. In the city of "Diversity," 10,000 residents live, and any two of them are either enemies or friends. Each day, no more than one resident can "start a new life," which means breaking up with all their friends and befriending all their enemies; meanwhile, any three residents can become friends with each other. Prove... | 10. Let $A, B$ and $C$ be three residents of the city. It is clear that it is possible for all of them to be friends with each other; it is also possible that one of them (say, $A$) is not friends with either $B$ or $C$, while $B$ and $C$ are friends with each other: in this case, for $A, B$, and $C$ to all become frie... | 5000 | Combinatorics | proof | Yes | Yes | olympiads | false |
14. Some of the 20 metal cubes, identical in size and appearance, are aluminum, the rest are duralumin (heavier). How can you determine the number of duralumin cubes using no more than 11 weighings on a balance scale without weights?
Note. The problem assumes that all cubes can be aluminum, but they cannot all be dura... | 14. Let's place one cube on each pan of the scales (first weighing). In this case, two different scenarios may occur.
$1^{\circ}$. During the first weighing, one pan of the scales tips. In this case, one of the two weighed cubes is definitely aluminum, and the other is duralumin. Next, we place these two cubes on one ... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
16. a) One day, a guest came to the hotel owner, K, without money but with a silver chain consisting of seven links. The owner agreed to keep the guest for a week on the condition that the guest would give him one of the chain links as payment each day. What is the minimum number of links that need to be cut so that th... | 16. a) It is sufficient to saw off one third link; in this case, the chain will break into two parts, containing 2 and 4 links respectively, and one separate (sawed) link. On the first day, the guest will give this link; on the second day, he will take it back and give in exchange the part of the chain consisting of tw... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
27. Five friends, one of whom had a monkey, once bought a bag of nuts, which they intended to divide among themselves the next morning. However, one of the friends woke up at night and wanted nuts; he divided all the nuts in the bag into five equal parts, with one extra nut left over, which he gave to the monkey, and t... | 27. First solution. Let $n$ be the number of nuts each friend received in the morning; in this case, $5n + 1$ is the number of nuts that were in the bag in the morning. The last of those who woke up at night, obviously, took $\frac{5n + 1}{4}$ for himself, and before that, there were $5 \frac{5n + 1}{4} + 1 = \frac{25n... | 15621 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
28. Two brothers sold a flock of sheep that belonged to both of them, taking as many rubles for each sheep as there were sheep in the flock. The money received was divided as follows: first, the elder brother took ten rubles from the total amount, then the younger brother took ten rubles, after that the elder brother t... | 28. Let the number of sheep in the flock be denoted by $n$; in this case, the brothers received $n$ rubles for each sheep, and thus the total amount they received is $N=n \cdot n=n^{2}$ rubles. Let $d$ be the number of whole tens in the number $n$, and $e$ be the number of units; then $n=10 d+e$ and
$$
N=(10 d+e)^{2}=... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
34. a) Find the smallest integer starting with the digit 1 such that if this digit is moved to the end, the number triples. Find all such numbers.
b) What digits can non-zero integers start with if they triple when the first digit is moved to the end? Find all such numbers. | 34. a) First solution. Let's denote the (m-digit) number obtained by removing the first digit 1 from the desired number as \(X\). In this case, according to the problem, we have:
\[
\left(1 \cdot 10^{m} + X\right) \cdot 3 = 10 X + 1
\]
from which
\[
X = \frac{3 \cdot 10^{m} - 1}{7}
\]
From the last equation, it is ... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
35. Find the smallest natural number ending in the digit 6 that quadruples when its last digit is moved to the beginning of the number. | 35. First solution. Let the number $X$ satisfy the conditions of the problem:
$$
X=\widehat{a_{1} a_{2} \ldots a_{n-1} 6} \quad \text { and } \quad 4 X=\widehat{6 a_{1} a_{2} \ldots a_{n-1}},
$$
where $a_{1}, a_{2}, \ldots, a_{n-1}, 6$ are the digits of the number $X$. Since the last digit of $X$ is 6, the last digit... | 153846 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
39. Find the smallest integer starting with the digit 7 that decreases threefold when this digit is moved to the end. Find all such numbers. | 39. First solution. Let the unknown digits of the desired number be denoted by $x, y, \ldots, z, t$. In this case, using the notation from the second solution of problem 34 a), we have:
$$
\text { - } \overline{7 x y \ldots z t} \cdot \frac{1}{3}=\overline{x y \ldots z t 7}
$$
or
$$
\overline{x y \ldots z t 7} \cdot... | 7241379310344827586206896551 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
41. a) Find a six-digit number that increases by 6 times when the last three digits of the number, without changing their order, are moved to the beginning of the number.
b) Prove that there does not exist an eight-digit number that increases by 6 times when the last four digits are moved to the first four positions w... | 41. a) Let the number formed by the first three digits of the desired number $N$ be denoted by $p$, and the number formed by the last three digits of $N$ be denoted by $q$. In this case, the condition of the problem gives
$$
1000 q + p = 6(1000 p + q) \quad(=6 N)
$$
from which we get
$$
(1000 q + p) - (1000 p + q) =... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
42. Find a six-digit number, the products of which when multiplied by 2, 3, 4, 5, and 6 are written with the same digits as the number itself, but in a different order. | 42. Let $x$ be the number satisfying this condition. Since $6x$, like $x$, is a six-digit number, the first digit of the decimal representation of the number $x$ is 1. Therefore:
1) the first digits of the decimal representation of the numbers $x, 2x, 3x, 4x, 5x$, and $6x$ are all different, so they form a complete set... | 142857 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
51. What remainders can the hundredth power of an integer give when divided by 125? | 51. Any integer either is divisible by 5 or can be represented in one of the following four forms: $5k+1, 5k+2, 5k-2$, or $5k-1$. If a number is divisible by 5, then its hundredth power is clearly divisible by $5^3 = 125$. Further, using the binomial theorem, we get:
$$
(5k \pm 1)^{100} = (5k)^{100} \pm \ldots + \frac... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
54*. Let $N$ be an even number not divisible by 10. What will be the tens digit of the number $N^{20}$? What will be the hundreds digit of the number $N^{200}$? | 54. Let's find the last two digits of the number \(N^{20}\). The number \(N^{20}\) is divisible by 4 (since \(N\) is even). Further, the number \(N\) is not divisible by 5 (otherwise it would be divisible by 10) and, therefore, can be represented in the form \(5k \pm 1\) or in the form \(5k \pm 2\) (see the solution to... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
57. The number $123456789(10)(11)(12)(13)(14)$ is written in the base-15 numeral system, i.e., this number is equal to
(14) $+(13) \cdot 15+(12) \cdot 15^{2}+(11) \cdot 15^{3}+\ldots+2 \cdot 15^{12}+15^{13}$. What remainder does it give when divided by 7? | 57. The number 15 gives a remainder of 1 when divided by 7. Therefore, it follows that
$$
15^{2}=(7 \cdot 2+1) \cdot(7 \cdot 2+1)=7 n_{1}+1
$$
gives a remainder of 1 when divided by 7,
$$
15^{3}=15^{2} \cdot 15=\left(7 n_{1}+1\right) \cdot(7 \cdot 2+1)=7 n_{2}+1
$$
gives a remainder of 1 when divided by 7, and gene... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
65. Find the remainder when the number
$$
10^{10}+10^{\left(10^{0}\right)}+\ldots+10^{\left(10^{10}\right)}
$$
is divided by 7. | 65. First of all, note that \(10^{6}-1=999999\) is divisible by 7 (since \(999999=7 \cdot 142857\)). From this, it easily follows that \(10^{\mathrm{N}}\), where \(N\) is any integer, gives the same remainder when divided by 7 as \(10^{r}\), where \(r\) is the remainder from dividing \(N\) by 6. Indeed, if \(N=6k+r\), ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
68. a) What is the last digit of the number
$$
\left(\ldots\left(\left(\left(7^{7}\right)^{7}\right)^{7}\right)^{\ldots 7}\right)
$$
(raising to the power of 7 is repeated 1000 times)? What are the last two digits of this number?
b) What is the last digit of the number
$$
7\left(7^{\left(.7^{\left(7^{7}\right)}\righ... | 68. a) If you multiply two numbers, one of which ends in the digit $a$, and the second in the digit $b$, then their product will end in the same digit as the product $a b$. This observation allows us to easily solve the given problem. We will sequentially raise to powers, keeping track only of the last digit of the num... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
69*. Determine the five last digits of the number
$$
\left.N=9^{\left(9^{(} \cdot 9^{\left(9^{9}\right)}\right)} \ldots\right),
$$
written using 1001 nines in a similar manner to the number in problem 68). | 69. Let's consider the sequence of numbers:
$$
\begin{aligned}
& 1^{\circ} \cdot Z_{1}=9 \\
& \begin{aligned}
2^{\circ} \cdot Z_{2} & =9^{Z_{1}}=(10-1)^{Z_{1}}= \\
& =10^{Z_{1}}-C_{Z_{1}}^{1} \cdot 10^{Z_{1}-1}+\ldots+C_{Z_{1}}^{1} \cdot 10-1
\end{aligned}
\end{aligned}
$$
where the omitted terms in the expansion are... | 45289 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
70. For which natural numbers $n$ is the sum $5^{n}+n^{5}$ divisible by 13? What is the smallest $n$ that satisfies this condition? | 70. First, let's find the remainders of the division by 13 of the numbers \(5^n\) and \(n^5\) for the first few values of \(n = 0, 1, 2, \ldots\). It is more convenient to start with the numbers \(5^n\):
\(n\)
| 0 | 1 | 2 | 3 | 4 | . | . | . |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| 1 | 5 | 25 | 1... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
71. What are the last two digits of the number
$$
n^{a}+(n+1)^{a}+(n+2)^{a}+\ldots+(n+99)^{a}
$$
a) $a=4$
b) $a=8$ ? | 71. It is clear that the last two digits of the numbers $n, n^{2}, n^{3}, \ldots$, where $n$ is a non-negative integer, depend only on the last two digits of the number $n$ - this follows from the rule of multiplying multi-digit numbers "in a column". On the other hand, the last digits of 100 consecutive non-negative i... | 30 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
74. How many zeros does the product of all integers from 1 to 100 inclusive end with?
In the future, we will use the following notation:
$$
1 \cdot 2 \cdot 3 \cdot 4 \ldots(n-1) \cdot n=n!
$$
(read as: $n$ factorial). Thus, the previous problem can be more succinctly stated as: how many zeros does the number 100! en... | 74. The number of bullets at the end of a number indicates how many times 10 is a factor in that number. The number 10 is equal to the product of $2 \cdot 5$; in the product of all integers from 1 to 100, the factor 2 appears in a higher power than the factor 5. Therefore, the product $1 \cdot 2 \cdot 3 \ldots 100$ is ... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
81. The square of an integer ends with four identical digits. Which ones? | 81. A perfect square can only end in the digits 0, 1, 4, 9, 6, and 5. Furthermore, the square of any even number is clearly divisible by 4, while the square of any odd number gives a remainder of 1 when divided by 4 \(\left((2 k)^{2}=4 k^{2},(2 k+1)^{2}=4\left(k^{2}+k\right)+1\right)\); therefore, the square of no numb... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
86. a) Prove that the fraction $\frac{a^{3}+2 a}{a^{4}+3 a^{2}+1}$ is irreducible for any integer value of $a$.
b) Indicate all (natural) numbers by which the fraction $\frac{5 n+6}{8 n+7}$ can be reducible for integer $n$. | 86. The fraction $\frac{a^{3}+2 a}{a^{4}+3 a^{2}+1}$ is reducible or irreducible simultaneously with the fraction $\frac{a^{4}+3 a^{2}+1}{a^{3}+2 a} \Rightarrow a+\frac{a^{2}+1}{a^{3}+2 a}$, or - simultaneously with the fraction $\frac{a^{2}+1}{a^{3}+2 a}$. The fraction $\frac{a^{2}+1}{a^{3}+2 a}$ is reducible or irred... | 13 | Number Theory | proof | Yes | Yes | olympiads | false |
99. Find the smallest square that starts with six twos. | 99. Since the number 222222 is not a perfect square, the decimal representation of the unknown number has the form $222222 a_{7} a_{8} \ldots a_{n}$, where $a_{7}, a_{8}, \ldots, a_{n}$ are some unknown digits.
First, assume that the number of digits $n$ of the unknown number is even: $n=2 k$. We will now extract the ... | 222222674025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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