problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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|---|---|---|---|---|---|---|---|---|---|
25. Consider all pairwise non-congruent triangles with vertices at points dividing a circle into $N$ equal arcs ($N>2$). For which $N$ is exactly half of these triangles isosceles? | 79.25. Answer. For $N=10$ and for $N=11$. | N=10N=11 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,025 |
29. Solve the system of equations:
$$
\left\{\begin{array}{l}
X+\frac{X+2 Y}{X^{2}+Y^{2}}=2 \\
Y+\frac{2 X-Y}{X^{2}+Y^{2}}=0
\end{array}\right.
$$
## 10th grade | 79.29. Hint. Consider the sum of the squares of the left-hand sides of the equations. Answer. $X=0, Y=1$ or $X=2, Y=-1$. | X=0,Y=1orX=2,Y=-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,028 |
32. For which natural numbers $y$ is the number $y^{2}+3^{y}$ a perfect square | 79.32. Suppose that $y^{2}+3^{y}=x^{2}$, i.e., $x^{2}-y^{2}=3^{y}$. From this, we obtain that $x-y$ and $x+y$ are powers of three, specifically $3^{k}$ and $3^{y-k}$. Then $y=\left(3^{k}-3^{y-k}\right) / 2$. Since $\quad\left(3^{k}-3^{y-k}\right) / 2 \geqslant 3^{y-k-1} \geqslant 3^{(y-3) / 2}, \quad$ and for $y \geqsl... | 1or3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,029 |
33. In a circle, $n$ sectors are shaded such that the angular measure of each is less than $\pi /\left(n^{2}-n+1\right)$. Prove that the circle can be rotated so that all shaded sectors move to the unshaded part of the circle. | 79.33. Consider all possible angles of rotation of a circle from 0 to $2 \pi$. If one sector has a size of $\alpha$, and another has a size of $\beta$, then the angles of rotation at which the image of the first sector intersects with the second fill an interval of length $\alpha+\beta \leqslant$ $\leqslant 2 \pi /\lef... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,030 |
34. Given a convex polyhedron. Prove that if all its faces, except one, are centrally symmetric polygons, then the last face is also centrally symmetric. | 79.34. Instruction. Prove that for any side of the last face, you can find another side of this face that is parallel to it and equal in length.
Indication. Prove that for any side of the last face, you can find another side of this face that is parallel to it and equal in length. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,031 |
35*. In a circle, two diameters $A B$ and $C D$ are drawn. Prove that for any two points $E$ and $F$ lying on the circle, the intersection point of the lines $A E$ and $D F$, the center of the circle, and the intersection point of the lines $C E$ and $B F$ lie on the same line.
Comment. 1. In some archival versions of... | 79.35. Perhaps the shortest, but not the most elementary proof uses projective transformations. By removing the point of intersection of the lines $E A$ and $B F$ to infinity, we obtain a diagram symmetric with respect to the center of the circle, from which the required fact follows. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,032 |
1. Can all natural numbers from 1 to 30 be written in a $5 \times 6$ table so that the sums of the numbers in the columns are equal? | 80.1. Answer. No, it cannot be. Otherwise, the sum of the numbers in the table would be even. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,033 |
3. 200 points on the segment $A B$ are symmetrically arranged relative to the midpoint of this segment. Some 100 of them are painted red, the rest are painted blue. Prove that the sum of the distances from the red points to point $A$ is equal to the sum of the distances from the blue points to point $B$. | 80.3. Let's divide all points into pairs of symmetric points. All pairs can be divided into three types: both points are red; both points are blue; the points have different colors. It is clear that the number of pairs of the first type and pairs of the second type are equal. It remains to note that the sum of the dist... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,035 |
6. At the vertices and intersections of the diagonals of a convex polygon, there are tram stops, and it is known that no three diagonals intersect at the same point. On some of the diagonal-streets, tram traffic has been introduced such that at least one tram route passes through each stop. Prove that from any stop to ... | 80.6. Consider any two stops $A$ and $B$, through which tram routes $X Y$ and $M N$ pass, respectively. Either these two diagonals intersect, and then from $A$ to $B$ one can get with one transfer, or in the quadrilateral $X Y M N$ the diagonals intersect - say, $X N$ and $Y M$ intersect at point $O$. Then one of these... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,037 |
11. $2 n$ candies are somehow distributed among $n$ boxes. A girl and a boy take turns picking a candy: the girl picks first. Prove that the boy can choose the candies in such a way that the last two candies are from the same box. | 80.11. The boy must act as follows: if the girl takes a candy from a box containing the only candy, then the boy takes a candy from a box containing more than one candy. If the girl takes a candy from a box containing exactly two candies, then the boy takes the remaining candy from the same box. If the girl takes a can... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,039 |
13. Find all sets of integers $A, B$, and $C$, for which
$$
A^{2}-B^{2}-C^{2}=1, \quad B+C-A=3
$$ | 80.13. Answer. $A=9, B=8, C=4$ or $A=9, B=4, C=$ $=8$, or $A=-3, B=2, C=-2$, or $A=-3, B=-2, C=$ $=2$. | A=9,B=8,C=4orA=9,B=4,C=8orA=-3,B=2,C=-2orA=-3,B=-2,C=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,041 |
17. Prove that the number $53 \cdot 83 \cdot 109 + 40 \cdot 66 \cdot 96$ is composite. | 80.17. Note that $53+96=83+66=109+40=149$. Let's introduce the following notations: $A=53, B=83, C=109, x=149$. Then our number is
$$
\begin{gathered}
A B C+(x-A)(x-B)(x-C)= \\
=x\left(x^{2}-(A+B+C) x+(A B+A C+B C)\right)
\end{gathered}
$$
i.e., it is divisible by $x=149$ and the quotient is certainly different from ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,043 |
20. The sum of four positive numbers $a, b, c$ and $d$ is 1. Prove that
$$
\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}+\sqrt{4 d+1}<6
$$ | 80.20. Note. Use the obvious inequality $1 + 2 x > \sqrt{1 + 4 x}$ for any positive $x$.
Indication. Use the obvious inequality $1 + 2 x > \sqrt{1 + 4 x}$ for any positive $x$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,044 |
25. For what least $n$ in the decimal representation of the fraction $m / n$ can the digit sequence ...501... appear after the decimal point? | 80.25. We can consider that $m / n=0.501 \ldots$ Let's examine the difference $m / n-1 / 2$. On one hand, $m / n-1 / 2 < 0.001$. On the other hand, $m / n-1 / 2 = (2m-n) / 2n$. Since $2m-n$ is an integer, the fraction $(2m-n) / 2n$ can only take values of the form $1 / 2n, 2 / 2n, 3 / 2n, \ldots$. Therefore, $1 / 2n < ... | 251 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,046 |
26. Among nine coins, two are counterfeit. A genuine coin weighs 10 g, while a counterfeit coin weighs 11 g. How can you detect the counterfeit coins in five weighings using a single-pan balance? (The scale's division value is 1 g).
## 9th grade | 80.26. With one weighing, you can determine the number of counterfeit coins in any given set. Arrange nine coins in the cells of a $3 \times 3$ table. In four weighings, determine the number of counterfeit coins in the first two rows and in the first two columns of the table. If in one of these weighings, a row or colu... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,047 |
27. Angle $A$ of triangle $ABC$ is twice angle $B$. Prove that $|BC|^2 = (|AC| + |AB|)|AC|$. | 80.27. Let's draw the bisector $A K$ of angle $A$. From the condition, it follows that angles $C A K, K A B$, and $A B K$ are equal, and thus $A K = B K$ and triangles $A C K$ and $A B C$ are similar. From this, we get that $A C / B C = B K / A B$. But by the property of the bisector, we have $A C / (B C - B K) = A B /... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,048 |
29. Find two different natural numbers, the arithmetic mean and geometric mean of which are two-digit numbers, one of which is obtained from the other by rearranging the digits. | 80.29. Answer. These are the numbers 98 and 32. | 9832 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,049 |
40*. Two cockroaches are sitting on the contour of a convex polygon. At the same time, they start moving along the contour in the same direction at the same speed. For what initial arrangement of the cockroaches will the minimum distance between them during the movement be the greatest?
## SELECTION Round
## (8-9th g... | 80.40. The desired location of the cockroaches is when they divide the perimeter of the polygon in half. Suppose that in such a
Fig. 56 case, the minimum distance between the cockroaches ... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,054 |
44. In a country, a survey was conducted where people were asked to name their favorite writer, artist, and composer. It turned out that every artist mentioned at least once is a favorite of exactly $k$ people. Prove that all those surveyed can be divided into $3k-2$ groups such that in each group, any two people have ... | 80.44. Let's consider $3 k-2$ empty rooms and start distributing all people among them so that the given condition is satisfied. We will prove that at any moment, the next person can be added to one of the groups so that the condition still holds. Indeed, since the number of people who like the same composer, or the sa... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,058 |
46. The vertices of a convex polygon with an odd number of sides are colored in three colors such that any two adjacent vertices are colored differently. Prove that the polygon can be cut into triangles by non-intersecting diagonals so that the colors of all vertices of each triangle are different. | 80.46. We will prove a slightly more general fact by induction on $N$ - the number of vertices of the polygon - that such a cutting is possible if it is known that all three colors were used in the coloring, and $N$ is any natural number. The base case $N=3$ is obvious. Consider a vertex $A$ such that its neighbors $B$... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,059 |
2. Is it possible to form a nine-digit number from the digits $1,2, \ldots, 9$ such that there is an odd number of digits between 1 and 2, an odd number of digits between 2 and 3, ..., and an odd number of digits between 8 and 9? | 81.2. No, it cannot. Otherwise, all the digits would be in places of the same parity. | No | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,061 |
8. Point $C$ lies inside the right angle $X O Y$. On the ray $O X$, point $A$ is chosen, and on the ray $O Y$, point $B$ is chosen. Prove that the perimeter of triangle $A B C$ is greater than $2|O C|$. | 81.8. Reflect point $C$ with respect to both rays and obtain points $C^{\prime}$ and $C^{\prime \prime}$. Then the perimeter of triangle $A B C$ is equal to the length of the broken line $C^{\prime} A B C^{\prime \prime}$, which is obviously not less than the length of segment $C^{\prime} C^{\prime \prime}$, equal to $... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,064 |
11. A natural number $n$ is such that the number $n^{2}+1$ is a ten-digit number. Prove that its representation contains two identical digits. | 81.11. If all the digits of this number are distinct, then their sum is 45 and, therefore, this number is divisible by three. However, for no natural number $n$ is the number $n^{2}+1$ divisible by three. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,065 |
12. At the vertices of a cube, integers are written. It is allowed to simultaneously increase by 1 any two numbers located at the ends of one edge of the cube. Can we achieve, using such operations, that all numbers are divisible by three, if initially one of the vertices had a one written, and the others had zeros?
#... | 81.12. The vertices of a cube can be colored black and white so that the vertices at the ends of any edge have different colors. Now consider the difference between the sum of the numbers in the black vertices and the sum of the numbers in the white vertices. For any operation of the type described in the condition, th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,066 |
13. Does there exist a natural number that decreases by 1981 times when the first digit of its decimal representation is erased? | 81.13. Answer. No, such a number does not exist. | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,067 |
16. Let $M$ be an arbitrary natural number greater than 3, and let $S$ be the sum of all natural numbers $x$ not exceeding $M$ such that $x^2 - x + 1$ is divisible by $M$. Prove that $S$ is divisible by $M+1$. (In the case when there are no such numbers $x$, we set $S$ to be zero). | 81.16. It is not hard to see that along with the number $x$, the set also includes the number $M+1-x$. If $x=(M+1) / 2$, then it follows from this that $M=3$. Therefore, all numbers in the set can be divided into pairs with a sum of $M+1$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,068 |
19. Does there exist a set of 1981 consecutive integers whose sum is the cube of a natural number? | 81.19. The sum of 1981 consecutive integers starting from $n$ is $1981 n+990 \cdot 1981=1981(n+990)$. It remains to take $n=1981^{2}-990$. | 1981^2-990 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,069 |
20. A square is divided into several rectangles with sides parallel to the sides of the square. For each rectangle, the ratio of the length of the shorter side to the length of the longer side was calculated. Prove that the sum of these numbers is not less than 1. | 81.20. Since the lengths of all sides of the rectangles do not exceed one, each such ratio is not less than the area of the corresponding rectangle, and the sum of these areas is equal to 1. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,070 |
23. On an infinite grid paper with a cell side of 1, a triangle $ABC$ is constructed with vertices at the nodes. Prove that if $AB > AC$, then $AB - AC > 1 / p$, where $p$ is the perimeter of triangle $ABC$. | 81.23. $A B-A C=\left(A B^{2}-A C^{2}\right) /(A B+A C) \geqslant 1 /(A B+A C)$, since $A B^{2}$ and $A C^{2}$ are integers and not equal to each other. Since $1 /(A B+A C)>1 / p$, the required estimate is proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,071 |
24*. On an $8 \times 8$ chessboard, the squares are colored in two colors in a checkerboard pattern. A token moves on the black squares, and in one move, it can transition to any diagonally adjacent square. What is the minimum number of moves required to visit all the black squares?
## 9th grade | 81.24. We will paint all black cells below the diagonal connecting the top left and bottom right corners of the board in red and blue as follows: if the sum of the row number and column number (numbers are counted from the bottom and left) is divisible by four, we paint the cell red; if it is not divisible by four, we ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,072 |
28. $1<a_{1}<a_{2}<\ldots<a_{n}<\ldots$ - a sequence of natural numbers such that $a_{n+a_{n}}=2 a_{n}$ for any natural $n$. Prove that there exists a natural number $c$ such that $a_{n}=n+c$ for any $n$. | 81.28. The sequence $b_{n}=a_{n}-n$ is non-decreasing, since $a_{n+1} \geqslant a_{n}+1$. The condition of the problem can be rewritten as the equality $b_{n}=b_{2 n+b_{n}}$. But from this it follows that the number $b_{1}$, for example, occurs in the sequence $\left\{b_{n}\right\}$ infinitely many times and, therefore... | a_{n}=n+C | Number Theory | proof | Yes | Yes | olympiads | false | 37,074 |
29. Integers $a, b, c, d$ and $A$ are such that $a^{2}+A=b^{2}$, $c^{2}+A=d^{2}$. Prove that the number $2(a+b)(c+d)(a c+b d-$ $-A)$ is a square of a natural number. | 81.29 .
$$
\begin{gathered}
2(a+b)(c+d)(a c+b d-A)=(a+b)(c+d)(2 a c+2 b d-2 A)= \\
=(a+b)(c+d)\left(2 a c+2 b d-\left(b^{2}-a^{2}\right)-\left(d^{2}-c^{2}\right)\right)= \\
=(a+b)(c+d)\left((a+c)^{2}-(b-d)^{2}\right)= \\
=(a+b)(c+d)(a+b+c-d)(a-b+c+d)=
\end{gathered}
$$
230
$$
\begin{gathered}
=(a+b)(c+d)\left(a+b-\f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,075 |
31. In each of two congruent regular 16-gons, seven vertices are marked. Prove that it is possible to superimpose these polygons so that at least four of the marked vertices of one polygon coincide with the marked vertices of the other. | 81.31. Let's overlay one 16-sided polygon on another in an arbitrary manner and consider 16 rotations of the upper polygon by angles that are multiples of \(360^{\circ} / 16\). If at each of the corresponding overlays, no more than three marked vertices coincide, then the total number of coincidences will not exceed 48... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,076 |
36*. Does there exist a natural power of the number five, in the hundred least significant digits of whose decimal representation at least 30 zeros in a row will appear?
## QUALIFYING ROUND | 81.36. First, let's find the degree \( k \) of the number five such that \( 5^k - 1 \) is divisible by \( 2^{100} \). We can assume that the number of digits in the decimal representation of \( 5^k \) is greater than 100; otherwise, consider the exponents \( 2k, 3k \), etc.
Now, consider the number \( A = 5^{k+100} \)... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,078 |
40. Prove that there do not exist rational numbers $a, b$, $c$ and $d$ such that $(a+b \sqrt{3})^{4}+(c+d \sqrt{3})^{4}=1+\sqrt{3}$. | 81.40. Suppose such $a, b, c, d$ exist. Then the equality
$$
(a-b \sqrt{3})^{4}+(c-d \sqrt{3})^{4}=1-\sqrt{3}
$$
must also hold. Indeed, expanding the brackets in the given equality, we see that the left-hand side can be represented as $x + y \sqrt{3}$, where $x$ and $y$ are rational numbers, and the left-hand side o... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,080 |
44*. Can there be found in space five points, the pairwise distances between which are all different, such that the perimeter of any spatial pentagon with vertices at these five points is always the same? | 81.44. Yes, they will be found. Consider in space four pairwise tangent spheres with different radii and inscribe a fifth sphere between them, touching the other four. The centers of these five spheres are the five points we need. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,083 |
46*. Let $a_{1}, a_{2}, \ldots, a_{7} ; b_{1}, b_{2}, \ldots, b_{7} ; c_{1}, c_{2}, \ldots, c_{7}$ be three sets of integers. Prove that it is possible to discard several triples of numbers with the same indices (but not all!) so that the sum of the remaining numbers in each set is divisible by three. | 81.46. Consider the polynomial
$$
\begin{gathered}
F\left(x_{1}, x_{2}, \ldots, x_{7}\right)= \\
=\left[1-\left(\sum_{1}^{7} a_{i} x_{i}^{2}\right)^{2}\right]\left[1-\left(\sum_{1}^{7} b_{i} x_{i}^{2}\right)^{2}\right]\left[1-\left(\sum_{1}^{7} c_{i} x_{i}^{2}\right)^{2}\right]
\end{gathered}
$$
of degree 12 in seven... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,085 |
2. The grasshopper jumps 1 cm, then jumps 3 cm in the same or opposite direction, then 5 cm in the same or opposite direction, and so on. Can it end up at the starting point after the 25th jump? | 82.2. No, it cannot. After the 25th jump, the grasshopper will have moved by an odd number of centimeters relative to its initial position. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,087 |
4. Prove that if the sum of two natural numbers is 770, then their product does not divide 770. | 82.4. Since $770=2 \cdot 5 \cdot 7 \cdot 11$, for $x(770-x)$ to be divisible by 770, it is necessary that one of the factors is divisible by two, one by five, and so on. Notice now that if one of the factors is divisible by a divisor of 770, then the other factor is also divisible by this divisor. Therefore, if the pro... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,088 |
7. $A$ and $B$ are different two-digit numbers, the last digits of which are the same. It is known that the quotient of the division of $A$ by nine is equal to the remainder of the division of $B$ by nine, and the quotient of the division of $B$ by nine is equal to the remainder of the division of $A$ by nine. Find all... | 82.7. Answer. $(85,75),(25,65),(15,55)$. | (85,75),(25,65),(15,55) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,090 |
12. At a school dance, no boy danced with all the girls, but each girl danced with at least one boy. Prove that there will be two such pairs $M_{1}, D_{1}$ and $M_{2}, D_{2}$, that boy $M_{1}$ danced with girl $D_{1}$, and boy $M_{2}$ danced with girl $D_{2}$, but $M_{1}$ did not dance with $D_{2}$, and $M_{2}$ did not... | 82.12. Let's number the boys and denote by $A_{n}$ the set of girls who danced with the $n$-th boy $M_{n}$. Then, if we assume that the statement of the problem is false, we get that for any $m$ and $n$ either $A_{m} \subset A_{n}$ or $A_{n} \subset A_{m}$. Consequently, we obtain a chain
$$
A_{i_{1}} \subset A_{i_{2}... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,092 |
13. For two numbers $a$ and $b$, it is known that $\left(\frac{1+a b}{a+b}\right)^{2}<1$. Prove that one of the numbers $A$ and $B$ is greater than 1, and the other is less. | 82.13. Squaring, we get $1+a^{2} b^{2}<a^{2}+b^{2}$ or, equivalently, $\left(1-a^{2}\right)\left(1-b^{2}\right)<0$. | (1-^{2})(1-b^{2})<0 | Inequalities | proof | Yes | Yes | olympiads | false | 37,093 |
14. Prove that for any four angles of a convex polygon, the sum of the measures of two of them is greater than the difference of the measures of the other two. | 82.14. It is sufficient to prove that the sum of any three angles at vertices $A, B$, and $C$ is greater than the fourth angle at vertex $D$. Indeed, the sum of angles $A, B$, and $C$ is not less than the sum of the angles of triangle $ABC$, which is $180^{\circ}$, while angle $D$ is less than $180^{\circ}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,094 |
16. In a table tennis tournament, sixth-graders and seventh-graders participated, with the number of sixth-graders being twice the number of seventh-graders. The tournament was a round-robin. The number of matches won by seventh-graders was $40 \%$ more than the number of matches won by sixth-graders. How many particip... | 82.16. Let $x$ be the number of seventh graders. Then the number of points $N_{7}$, scored by the seventh graders, is no more than $x \cdot 2 x + (x^2 - x) / 2$, and also $N_{6} \geqslant 2 x(2 x-1) / 2$. We get
$$
2 x^{2} + (x^{2} - x) / 2 \geqslant \frac{7}{5} x(2 x-1)
$$
from which it follows that $x \leqslant 3$.... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,095 |
17. In the cells of a $3 \times 3$ square, place nine different natural numbers, not exceeding 40, such that the products of the numbers in any column, any row, and either of the two diagonals are the same. | 82.17. Here is an example of the arrangement:
$$
\left|\begin{array}{rrr}
12 & 9 & 2 \\
1 & 6 & 36 \\
18 & 4 & 3
\end{array}\right|
$$ | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,096 |
19. Let $p_{1}, p_{2}, p_{3}$ be quadratic trinomials with positive leading coefficients. Prove that if each of them has a common root, then the quadratic trinomial $p_{1} + p_{2} + p_{3}$ has a root. | 82.19. There are two cases. Either all the trinomials have a common root (and the proof is obvious), or there are numbers $a<b<c$ such that each of them is a root of two trinomials. Let $b$ be a root of the first and second trinomials. Then the value of the third trinomial at point $b$ is negative, and consequently, th... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,097 |
20. In triangle $ABC$, the measure of angle $C$ is twice the measure of angle $A$ and $|AC|=2|BC|$. Prove that this triangle is a right triangle. | 82.20. Instruction. Draw the bisector. CK of angle C. Then triangles $C K B$ and $A B C$ are similar, and triangle $A C K$ is isosceles. From this, the ratio $A B / B C$ can be found. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,098 |
21. A sequence of digits is written, the first four members of which are $1,9,8,2$, and each subsequent digit is the last digit of the sum of the four preceding ones. Will the sequence contain four consecutive digits $3,0,4,4$? | 82.21. Yes, it will meet. Indeed, sooner or later some sequence of four consecutive numbers will repeat. Moving backward along the sequence, it is not difficult to verify that the first to repeat will be the quartet $1,9,8,2$. Let's find the four digits that precede it. These will be the digits $3,0,4,4$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,099 |
23. The cells of a $5 \times 41$ rectangle are colored in two colors. Prove that it is possible to choose three rows and three columns such that all nine cells at their intersections have the same color. | 82.23. In any column of five cells, at least three have the same color. We will mark this triplet and the corresponding three rows. Since there are 41 columns in total and 20 possible triplets of rows (taking color into account), some triplet will be marked three times, which is what we need. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,100 |
24. The plane is divided into parts by $2 n$ lines ( $n>1$ ), no two of which are parallel and no three of which intersect at the same point. Prove that among these parts, there are no more than $2 n-1$ angles.
## 9th grade | 82.24. Let's construct a circle that contains all the points of intersection of the lines inside it. They intercept $4 n$ arcs on it, and two adjacent arcs cannot simultaneously belong to two angles - convince yourself of this. Therefore, there are no more than $2 n$ angles, and there can be exactly $2 n$ only in the c... | 2n-1 | Geometry | proof | Yes | Yes | olympiads | false | 37,101 |
29. Prove that for any natural $k$ there exists a natural $n$ such that $\sqrt{n+1981^{k}}+\sqrt{n}=(\sqrt{1982}+1)^{k}$. | 82.29. Let's raise the number $1+\sqrt{1982}$ to the $k$-th power. By combining like terms, we get $a+b \sqrt{1982}$, where $a$ and $b$ are integers. When raising the number $1-\sqrt{1982}$ to the $k$-th power, we get $a-$ $-b \sqrt{1982}$. Multiplying these powers, we have the equality $(-1981)^{k}=a^{2}-1982 b^{2}$, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,102 |
30*. Several chips are placed on a plane, and they do not all lie on the same straight line. You can move any chip to a point that is symmetric to it relative to any other chip. Prove that after several such operations, it is possible to achieve that the chips are located at the vertices of some convex polygon.
## 10t... | 82.30. If $A, B$ and $X$ are three chips, then reflecting $X$ sequentially with respect to $A$ and $B$ will move $X$ by the vector $2 \overrightarrow{A B}$. Now, let's choose chips $A, B$ and $C$, which do not lie on the same line. It is clear that by applying shifts by vectors $2 \overline{A B}, 2 \overline{A C}, 2 \o... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,103 |
35. Prove that in the vertices of a regular $n$-gon, non-zero numbers can be placed such that for any set of vertices of the original polygon, which are vertices of a regular $k$-gon $(k \leqslant n)$, the sum of the numbers standing in them is equal to zero. | 82.35. Let's number the vertices in order from 0 to $n-1$ and write the number $\cos (k \pi / n + \alpha)$ in the $k$-th vertex, where $k=0,1,2, \ldots, \ldots, n-1$, and $\alpha$ is some angle chosen so that there are no zeros among the specified numbers. This is the required set. Hint. These numbers are the projectio... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,105 |
$36 *$. On a circle, $4 n$ points are marked and colored alternately in red and blue. The points of each color are paired and the points of each pair are connected by a segment of the same color as these points (at the same time, no three segments intersect at one point). Prove that there will be at least $n$ points of... | 82.36. Suppose that two segments $A B$ and $C D$ of the same color intersect. Then replace them with segments $A C$ and $B D$ of the same color. In this case, the number of intersection points between blue segments and red segments will not increase. By performing such operations, we can achieve that no two segments of... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,106 |
40. Pete bought eight rice and cabbage buns and paid 1 ruble for them. Vasya bought nine buns and paid 1 ruble 1 kopek. How much does a rice bun cost, if it is known that it is more expensive than a cabbage bun and the buns cost more than 1 kopek? | 82.40. Answer. A rice-filled bun costs 13 kopecks, while a cabbage-filled bun costs 9 or 11 kopecks. | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,109 |
42. $\left(A_{n}\right)$ - an infinite sequence of natural numbers such that $A_{n+k}-A_{k}$ is divisible by $A_{n}$ for any $n$ and $k$. Denote by $B_{n}$ the product $A_{1} A_{2} \ldots A_{n}$. Prove that $B_{n+k}$ is divisible by $B_{n} B_{k}$ for any $n$ and $k$. | 82.42. We will prove this by induction on $n+k$. Thus, we need to check that $A_{n+1} A_{n+2} \ldots A_{n+k}$ is divisible by $A_{1} A_{2} \ldots A_{k}$. Since $A_{n+1} A_{n+2} \ldots A_{n+k}=A_{n+1} A_{n+2} \ldots A_{n+k-1}\left(A_{n+k}-A_{n}\right)+$ $+A_{n} A_{n+1} \ldots A_{n+k-1}$, and the second term is divisible... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,110 |
44*. Prove that the set of all natural numbers greater than one cannot be partitioned into two non-empty sets such that if numbers $a$ and $b$ lie in one set, then the number $a b-1$ also lies in the same set. | 82.44. Suppose such a partition into two sets $A$ and $B$ is possible and the number two belongs to set $A$. Let $n$ be the minimal number in set $B$, and $p$ be the minimal prime number that does not divide $n$. Clearly, $p < n$.
Lemma. If $x < p$, then for any number $b$ from $B$, the number $x b$ also belongs to se... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,111 |
5. A time machine allows traveling from March 1st to November 1st of any other year, from April 1st to December 1st, from May 1st to January 1st, and so on. Using the machine twice in a row is prohibited. Baron Munchausen set off on a time travel journey on April 1st. Returning after a moment, he reported that he had t... | 83.5. Using the time machine shifts the month number by four, if considering the remainders when dividing by 12. Since 26 is not divisible by four, such a journey is impossible. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,114 |
6. From four different digits, two four-digit numbers were formed - the largest of all possible and the smallest of all possible (digits in the number cannot be repeated). The sum of these numbers is 10477. What could the original digits have been?
## 6th grade | 83.6. Answer. These are the digits $7,4,3,0$.
保留源文本的换行和格式，翻译结果如下：
83.6. Answer. These are the digits $7,4,3,0$. | 7,4,3,0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,115 |
8. A straight rod 2 m long was cut into five pieces, each not less than 17 cm long. Prove that among them, there are three that can form a triangle. | 83.8. Let $a<b<c<d<e$ be the lengths of these pieces. Assuming the opposite of the statement of the problem, we get $c \geqslant a+b$, $d \geqslant b+c \geqslant a+2 b$, $e \geqslant c+d \geqslant 2 a+3 b$. Then the length of the rod is not less than $5 a+7 b \geqslant 12 \cdot 17=204$ - a contradiction. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,117 |
12. Can we mark 10 blue and 20 red points inside a triangle with blue vertices such that no three blue points lie on the same line and such that inside any triangle with blue vertices there is at least one red point?
## 7th grade | 83.12. Answer. No, it cannot. Hint. For any arrangement of blue points, the triangle can be divided into small triangles with blue vertices. The number of these small triangles does not depend on the method of division and is equal to $2 k+1$, where $k$ is the number of blue vertices inside the triangle. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,118 |
17. Prove that the number $2^{58}+1$ can be represented as the product of three natural numbers, each greater than 1. | 83.17.
$$
\begin{aligned}
2^{58}+1 & =2^{58}+1+2^{30}-2^{30}=\left(2^{29}+1\right)^{2}-2^{31}= \\
& =\left(2^{29}-2^{15}+1\right)\left(2^{29}+2^{15}+1\right)= \\
& =5\left(\frac{2^{29}-2^{15}+1}{5}\right)\left(2^{29}+2^{15}+1\right)
\end{aligned}
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,119 |
18. In each cell of a $24 \times 24$ table, a number is written, either +1 or -1. In one move, you can change the sign of one of these numbers, as well as simultaneously change the signs of all numbers that are in the same row or the same column with it. Prove that using several such moves, you can transform any initia... | 83.18. Let's perform such operations for all cells of an arbitrary cross formed by a row and a column. As it is not hard to see, the result will be precisely the change of the sign of the cell at the intersection of these lines. Therefore, we can change the sign of any number in the table. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,120 |
21. A point lying inside a given regular hexagon is connected to its vertices by segments. Prove that a hexagon can be formed from these segments, the area of which is not less than two thirds of the area of the original hexagon. | 83.21. Let $S$ be the area of the regular hexagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6}$ - see Fig. 57, a. For definiteness, assume that
$$
S\left(A_{1} O A_{2}\right)+S\left(A_{3} O A_{4}\right)+S\left(A_{5} O A_{6}\right) \geqslant \frac{S}{2}
$$
Consider the equilateral triangle $A B C$ with sides equal to the side... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,123 |
23. A regular 400-gon is cut into parallelograms. Prove that among these parallelograms, there are at least 100 rectangles. | 83.23. Consider two opposite sides of a 400-gon. It is clear that they are connected by a chain of adjacent parallelograms. Now consider another pair of opposite sides, perpendicular to the sides of the first pair, and the corresponding chain.
. Upon colliding, two points scatter, maintaining the same speed and remaining on the circle. Prove that there will be a moment when each point will return to its original position.
## ... | 83.24. Let's assume that upon collision, the points do not scatter but pass through each other, maintaining their speeds but swapping numbers. Then, after a time \( T \), which is the time needed for a point to complete a full revolution, the points will return to their original positions, but with different numbers. A... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,125 |
28. For which $n \geqslant 2$ is the inequality
$$
x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2} \geqslant x_{n}\left(x_{1}+x_{2}+\ldots+x_{n-1}\right)
$$
106
valid for all real values of the variables $x_{i}$? | 83.28. Answer. The inequality is true only for $n=2,3,4,5$. Hint. Rewrite the inequality as follows:
$$
\left(x_{1}-\frac{x_{n}}{2}\right)^{2}+\left(x_{2}-\frac{x_{n}}{2}\right)^{2}+\ldots+\left(x_{n-1}-\frac{x_{n}}{2}\right)^{2}+(5-n) \frac{x_{n}^{2}}{4} \geqslant 0
$$ | 2,3,4,5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 37,129 |
29. Two circles $K_{1}$ and $K_{2}$ have no common internal points. Lines $L_{1}$ and $L_{2}$ are externally tangent to these circles at four points, which are the vertices of a circumscribed quadrilateral. Prove that the circles $K_{1}$ and $K_{2}$ are tangent to each other. | 83.29. Let the radius of $K_{1}$ be no greater than the radius of $K_{2}$; denote the points of tangency of $K_{1}$ and $K_{2}$ with the lines $L_{1}$ and $L_{2}$ as $A, B, C$, and $D$ - see Fig. 58. It is clear that $A C \| B D$. Draw the tangents $A_{1} C_{1}$ and $B_{1} D_{1}$, parallel to the line $A C$ and tangent... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,130 |
34. Given a trihedral angle. It is known that the bisectors of two of its dihedral angles are perpendicular to each other. Prove that the plane passing through these bisectors is perpendicular to the plane of the third dihedral angle. | 83.34. Let's place unit vectors $\vec{a}, \vec{b}, \vec{c}$ on the edges of a trihedral angle. Suppose the bisectors along the vectors $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ are perpendicular. Then $(\vec{a}+\vec{b})(\vec{b}+\vec{c})=\overrightarrow{0}$, i.e., $\overrightarrow{a b}+\vec{b} \vec{c}+\overrightarrow{c a}... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,132 |
37. In a regular 20-gon, nine vertices are marked. Prove that there exists an isosceles triangle with vertices at the marked points. | 83.37. The set of vertices of a regular 20-gon can be represented as the union of the sets of vertices of four regular pentagons. It is clear that in one of them, no fewer than three vertices are marked. It only remains to convince ourselves by a small enumeration that any three vertices of a regular pentagon form an i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,134 |
39. Two players take turns placing crosses and noughts in the cells of an infinite sheet of squared paper. The first player aims to have four crosses form a square with sides parallel to the grid lines, while the second player tries to prevent this. Can the first player win? | 83.39. Answer. Yes, the first player wins with correct play. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,136 |
40. The angle $A$ at the vertex of the isosceles triangle $A B C$ is $100^{\circ}$. On the ray $A B$, a segment $A M$ is laid off, equal to the base $B C$. Find the measure of angle $B C M$. | 83.40. Answer. $10^{\circ}$. Indeed, let's lay off on the ray $A M$ a segment $B N$, equal in length to $B C$. Then, in the isosceles triangle $B N C$, the line $C M$ is the bisector of angle $C$. This follows from the fact that $B M / B N = C B / C N$. Since $\angle B C N = 20^{\circ}$, then $\angle B C M = 10^{\circ}... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,137 |
41. For which $n$ can $n$ numbers $a_{1}, a_{2}, \ldots, a_{n}$ (not all equal to zero) be placed on a circle so that for any $k \leqslant n$ the sum of $k$ consecutive numbers starting from $a_{k}$ is zero? | 83.41. Answer. Such an arrangement is possible if and only if $n$ is not a power of two. Hint. Let $s_{k}=a_{1}+a_{2}+\ldots+a_{k-1}$. Then the equalities specified in the condition can be rewritten as follows: for any $k$ we have $s_{k}=s_{2 k}$ (and there is also the equality $s_{1}=0$). Since there are $n$ independe... | n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 37,138 |
42. An infinite grid sheet of paper is covered with a layer of rectangular tiles of size $1 \times 2$, with sides aligned along the grid lines. Prove that it can be covered with three more layers of tiles in the same manner so that no tile lies exactly on top of another. | 83.42. We will lay the second layer as follows: divide the entire plane into 2x2 squares, each of which will be covered either by two horizontal or two vertical tiles. Then we will place the third layer as follows: connect with segments those adjacent cells that have not yet been covered by a single tile. Obviously, tw... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,139 |
43*. An increasing sequence of natural numbers $\left(a_{n}\right)$ is such that any natural number not equal to a term of the sequence can be represented as $a_{k}+2 k$ for some natural $k$. Prove that for all $k$ the inequality $a_{k}<\sqrt{2 k}$ holds. | 83.43. Suppose that $a_{k}$ is the first term of the sequence that does not satisfy the given inequality. Then it is clear that $a_{k}-1$ is not a term of the sequence, and therefore there exists an $s$ such that $a_{s}+2 s+1=a_{k}$. Let
$$
n=\left[\frac{a_{s}+1}{2+\sqrt{2}}\right]=a_{s}-\left[\frac{a_{s}+1}{\sqrt{2}}... | a_{k}\leqslant\sqrt{2}k | Number Theory | proof | Yes | Yes | olympiads | false | 37,140 |
44. In the Land of Fools, there are $N^{2}$ cities arranged in a square, with the distance between adjacent cities being 10 km. The cities are connected by a road system consisting of straight segments parallel to the sides of the square. What is the minimum length of such a road system, given that it is possible to tr... | 83.44. Answer. $10\left(N^{2}-1\right)$. Hint. Consider an arbitrary segment of such a system that starts and ends not in a city. Suppose that on one side it is adjacent to $a$ segments, and on the other $b$, with $a \geqslant b$. Moving it in the first direction, we will not increase the length of the system. Its move... | 10(N^{2}-1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 37,141 |
3. On a line, 45 points are marked, lying outside the segment $A B$. Prove that the sum of the distances from these points to point $A$ is not equal to the sum of the distances from these points to point $B$.
Translate the above text into English, please retain the original text's line breaks and format, and output th... | 84.3. The difference between the distance from any such point to $A$ and the distance from it to $B$ is equal to the length of the segment $A B$. It is clear that the sum of 45 numbers of the form $\pm|A B|$ is not equal to zero. | Combinatorics | MCQ | Yes | Yes | olympiads | false | 37,143 | |
5. A circle is divided into six sectors, and numbers are arranged in them as shown in Fig. 13. It is allowed to simultaneously increase by one any two adjacent numbers. Prove that it is impossible to achieve equality of all six numbers using such operations. | 84.5. Instruction. Let's paint the sectors in black and white in a checkerboard pattern. Then the difference between the sum of the numbers in the black sectors and the sum of the numbers in the white sectors does not change under the specified operations. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,144 |
6. The numbers from 1 to 100 are written in a row. Two players take turns inserting the signs +, - and $\times$ between these numbers (the players themselves choose where to place the next sign from the remaining positions). Prove that the player who makes the first move can ensure that the final result is an odd numbe... | 84.6. The first player places a multiplication sign between the one and the two, then mentally divides all the empty spaces, starting with the space between the three and the four, into pairs of consecutive spaces, and on each move of the second, places a multiplication sign in the other space of the same pair. If the ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,145 |
8. At point $O$, the diagonals $A C$ and $B D$ of quadrilateral $A B C D$ intersect. The perimeters of triangles $A B C$ and $A B D$ are equal. The perimeters of triangles $A C D$ and $B C D$ are also equal. Prove that $A O = B O$. | 84.8. From these equalities, it follows that $A C=B D, A D=B C$. But then triangles $A B C$ and $B A D$ are equal. The equality of angles $O C B, A C B, B D A$ and $O D A$ gives us the equality of triangles $O A D$ and $A B C$, from which the equality $O B=O A$ follows. | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,146 |
12. From a sheet of graph paper measuring $29 \times 29$ cells, 99 squares, each consisting of four cells, were cut out. Prove that it is possible to cut out one more square.
## 7th grade | 84.12. Let's mark all cells on the sheet for which both the column number and the row number give a remainder of 1 when divided by three (numbers are counted from the left and from the bottom). Now, for each marked cell, we will construct a marked square $2 \times 2$, with the lower left corner located in this cell. It... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,148 |
24. Set $A$ consists of natural numbers, and among any 100 consecutive natural numbers, there is a number from $A$. Prove that there exist four distinct numbers $a$, $b$, $c$, $d$ in $A$ such that $a+b=c+d$.
## 9th grade | 84.24. Let's divide the sequence of natural numbers into consecutive hundreds and select one element from the set $A$ in each of them. Consider the differences between pairs of consecutive selected numbers. They do not exceed 200 and, therefore, will eventually repeat. Consequently, there will be numbers $a, b, c, d$ f... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,156 |
30. The sum of the integers $A, B$, and $C$ is zero. Prove that the number $2\left(A^{4}+B^{4}+C^{4}\right)$ is a square of an integer.
## 10th grade | 84.30. Indeed,
\[
\begin{gathered}
A^{4}+B^{4}+C^{4}=A^{2}(B+C)^{2}+B^{2}(A+C)^{2}+C^{2}(A+B)^{2}= \\
=2\left(A^{2} B^{2}+B^{2} C^{2}+C^{2} A^{2}\right)+2 A B C(A+B+C)= \\
=\left(A^{2}+B^{2}+C^{2}\right)^{2}-A^{4}-B^{4}-C^{4}
\end{gathered}
\]
From here, \(2\left(A^{4}+B^{4}+C^{4}\right)=\left(A^{2}+B^{2}+C^{2}\right... | 2(A^{4}+B^{4}+C^{4})=(A^{2}+B^{2}+C^{2})^{2} | Algebra | proof | Yes | Yes | olympiads | false | 37,158 |
33. Prove that the difference of the squares of the lengths of adjacent sides of a parallelogram is less than the product of the lengths of its diagonals. | 84.33. Let us be given a parallelogram $A B C D$. Then
$|A C||B D|>|\overrightarrow{A C} \cdot \overrightarrow{D B}|=|(\overrightarrow{A B}+\overrightarrow{B C})(\overrightarrow{A B}-\overrightarrow{B C})| \geqslant|A B|^{2}-|B C|^{2}$. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 37,159 |
34. After several operations of differentiation and multiplication by $x+1$, performed in some order, the polynomial $x^{8}+x^{7}$ has turned into $a x+b$. Prove that the difference of the integers $a$ and $b$ is divisible by 49. | 84.34. Let's make the substitution $x=y-1$. The given polynomials will transform into $y^{8}-7 y^{7}+\ldots$ and $a y-(a-b)$, the operation of multiplication by $x+1$ - into multiplication by $y$, and differentiation with respect to $x$ - into differentiation with respect to $y$. It is clear that the constant term $a-b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 37,160 |
36. At the vertices of a 50-gonal prism, the numbers $1,2,3, \ldots, 100$ are arranged. Prove that the prism will have an edge such that the numbers at its ends differ by no more than 48.
## FINAL ROUND
## $(8-9$ th grades) | 84.36. Suppose that such an edge does not exist. Next to the number 50, there can only be three numbers: 1, 99, and 100, and the numbers 99 and 1 cannot have another common neighbor. Therefore, at the other end of the vertical edge where the number 50 is located, there is 100. Suppose the number 1 is to the right of 50... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,162 |
39. On the coordinate plane, there are four points with integer coordinates. It is allowed to replace any of these points with a point symmetric to it relative to any other of these points. Can one, after several such operations, transition from points with coordinates $(0 ; 0),(0 ; 1),(1 ; 0),(1 ; 1)$ to points with c... | 84.39. Answer. No, it cannot. Indeed, the final quadruple is such that for any point of this quadruple, the difference of its coordinates is divisible by three. This property is preserved under the transformations described in the problem, so the initial quadruple must have had this property as well. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 37,163 |
40. Given numbers: one and nine zeros. It is allowed to choose two numbers and replace each with their arithmetic mean. What is the smallest number that can end up in the place of one after a series of such operations? | 84.40. Let's consider the smallest non-zero number $A$. Under the averaging operation, this value either does not decrease or is halved if the averaging is done with zero. From this, it follows that the value of $A$ will not decrease by more than $2^{9}=$ $=512$ times. Answer. $1 / 512$. | \frac{1}{512} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,164 |
41. Positive numbers $a_{1}, a_{2}, \ldots, a_{k} ; b_{1}, b_{2}, \ldots, b_{n}$ are such that
$$
a_{1}+a_{2}+\ldots+a_{k}=b_{1}+b_{2}+\ldots+b_{n}
$$
Prove that in a rectangular table of size $k \times n$ it is possible to place non-negative numbers, among which at least $(k-1)(n-1)$ are zeros, so that the sums of t... | 84.41. Consider a segment of length $S=a_{1}+a_{2}+\ldots+a_{k}=$ $=b_{1}+b_{2}+\ldots+b_{n}$. After this, we will divide this segment with red points into segments of lengths $a_{1}, a_{2}, \ldots, a_{k}$, and then with blue points into segments of lengths $b_{1}, b_{2}, \ldots, b_{n}$. Now, place in the cell at the i... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 37,165 |
43. On a circle tangent to the sides of an angle with vertex $O$, two diametrically opposite points $A$ and $B$ (different from the points of tangency) are chosen. The tangent to the circle at point $B$ intersects the sides of the angle at points $C$ and $D$, and the line $O A-$ at point $E$. Prove that $B C = D E$. | 84.43. Consider the case when the circle is located outside the triangle \(OCD\) (see Fig. 59). Let \(C'\) and \(D'\) be the points of intersection of the sides of the angle with the tangent to the circle at point \(A\), so that the circle is inscribed in the triangle \(OC'D'\). Obviously, this triangle is homothetic t... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,166 |
44*. Prove that there exists a set $A$ of natural numbers such that any natural number not belonging to it is the arithmetic mean of two different numbers from $A$, and no number from $A$ has this property.
## 10th grade | 84.44. Note. As such a set, one can take the set of all natural numbers, in the ternary representation of which there is no digit 2. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,167 |
46. Prove that if the sum of the planar angles at the vertex of a pyramid is greater than $180^{\circ}$, then each lateral edge of the pyramid is less than half the perimeter of its base. | 84.46. Let $S A_{1} A_{2} \ldots A_{n}$ be a given pyramid. Cut its lateral surface along the longest edge, for example, $S A_{1}$, and unfold it onto a plane. From the condition, it follows that the point $S$ lies inside the polygon $A_{1} A_{2} \ldots A_{n} A_{1}^{\prime}$. Let $B$ be the second intersection point of... | proof | Geometry | proof | Yes | Yes | olympiads | false | 37,168 |
47. An integer $a$ is such that the number $3a$ can be expressed as $x^{2}+2y^{2}$, where $x$ and $y$ are integers. Prove that the number $a$ can also be expressed in this form. | 84.47. Consider the identity
$$
a=\frac{3 a}{3}=\frac{x^{2}+2 y^{2}}{3}=\left(\frac{x \pm 2 y}{3}\right)^{2}+2\left(\frac{x \mp 2 y}{3}\right)^{2}
$$
It is clear that the numbers $x$ and $y$ are either both divisible by three or not (in the first case, the proof is obvious). If their remainders when divided by three ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,169 |
51*. Integers $a, b, c, d$, $e$ are such that the numbers $a+b+$ $+c+d+e$ and $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}$ are divisible by an odd number $n$. Prove that the number $a^{5}+b^{5}+c^{5}+d^{5}+e^{5}-5 a b c d e$ is also divisible by $n$. | 84.51. Consider the polynomial $P(x)=(x-a)(x-b)(x-c) \times$ $\times(x-d)(x-e) . \quad$ Let $P(x)=x^{5}+p x^{4}+q x^{3}+r x^{2}+s x+t$. Then, by Vieta's theorem, we have $p=-(a+b+c+d+e), q=$ $=\left((a+b+c+d+e)^{2}-\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right)\right) / 2$, and both these numbers are divisible by $n$. Addi... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,170 |
52. In the sequence $1,0,1,0,1,0, \ldots$ each term, starting from the seventh, is equal to the last digit of the sum of the previous 112 six terms. Prove that in this sequence, the six numbers $0,1,0,1,0,1$ do not appear consecutively.
Comment. In 1984, the final rounds of the Olympiad were reformed: the qualifying r... | 84.52. Let $x_{1}, \ldots, x_{7}$ be seven consecutive terms of our sequence. Consider the function $S(a, b, c, d, e, f)=$ $=2 a+4 b+6 c+8 d+10 e+12 f$. The last digits of the numbers $S\left(x_{1}, \ldots, x_{6}\right)$ and $S\left(x_{2}, \ldots, x_{7}\right)$ coincide - check this. From this, it follows that for all ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 37,171 |
1. There are 68 coins, different in weight. Find the heaviest and the lightest coins using 100 weighings on a balance scale without weights. | 85.1. Let's divide 68 coins into 34 pairs and compare the coins in pairs. Then, among the 34 heavier coins, we will find the heaviest coin in 33 weighings, and among the 34 lighter coins, we will find the lightest coin in 33 weighings. | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,172 |
3. A traveler sets out from their hometown $A$ to the farthest city in the country - city $B$; then from $B$ to the farthest city from it - city $C$, and so on. Prove that if $C$ and $A$ are different cities, the traveler will never return home. | 85.3. It is clear that the distance between $B$ and $C$ is greater than the distance between $B$ and $A$ - otherwise, the traveler would have returned to $A$. Similarly, the next city is at least as far from $C$ as $B$ is, and so on. Therefore, the distance traveled by the traveler at each stage of the journey becomes ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 37,173 |
4. Find 1000 natural numbers whose sum equals their product. | 85.4. For example, you can take a set containing 998 ones, one two, and one number 1000. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,174 |
7. The number 1 is written on the board. Every second, the sum of its digits is added to the number on the board. Can the number 123456 appear on the board at some point? | 85.7. Answer. No, it cannot. Consider the remainders when divided by three. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 37,176 |
11. Three grasshoppers are sitting on a line. They start playing leapfrog, jumping over each other (but not over two at once). Can they end up in their original positions after 1985 jumps? | 85.11. Let's denote these grasshoppers as $A, B$, and $C$. There are six possible arrangements of them on a line (from left to right): $A B C, B C A, C A B, A C B, B A C, C B A$. We will call the first three arrangements correct, and the remaining three incorrect. Then, with each jump, the type of arrangement will chan... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 37,178 |
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