problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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4. On the board, the product of the three-digit numbers $\overline{K C I}$ and $\overline{\text { ICK }}$ is written, where the letters correspond to different decimal digits. The record of this product consists of three pairs of identical adjacent digits. What is written on the board? | Answer: $224455=385 \cdot 583$.
Solution. Since the problem does not change when I and K are swapped, we will assume I $<\mathrm{K}$. Let $\overline{\mathrm{KCИ}} \cdot \overline{\text { ИСK }}=\overline{x x y y z z}$. It is obvious that the right side is divisible by 11. Since
$$
\overline{\mathrm{KCU}} \bmod 11=(\m... | 224455=385\cdot583 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,198 |
5. Around a round table stand $n$ empty glasses ( $n \geqslant 3$ ). Petya and Vasya take turns (starting with Petya) to pour kvass into them. Petya, on his turn, pours kvass into a chosen empty glass, which has both of its neighboring glasses either empty or full. Vasya, on his turn, pours kvass into a chosen empty gl... | Answer: $n \neq 4$.
Solution. It is clear that $n=3$ works. The case $n=4$ does not work. Indeed, Vasya's first response move is to fill a glass next to the one Petya filled, after which Petya has no moves.
Let $n>4$. We will call a segment a block of several consecutive filled glasses, and a set of segments separate... | n\neq4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,199 |
6. On the table lie spheres with radii 2, 2, 1, touching each other externally. The vertex of the cone is located midway between the points of contact of the identical spheres with the table, and the cone itself touches all the spheres externally. Find the angle at the vertex of the cone. (The angle at the vertex of th... | Answer: $2 \operatorname{arcctg} 8$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A_{1}, A_{2}, A_{3}$ be the points of contact of the spheres with the table, $C$ be t... | 2\operatorname{arcctg}8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,200 |
1. Each cell of a $5 \times 5$ table is colored in one of three colors: blue, red, or yellow. In each row of the table, the number of yellow cells is not less than the number of red cells and not less than the number of blue cells, and in each column of the table, the number of red cells is not less than the number of ... | Answer: 5.
| $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| :---: | :---: | :---: | :---: | :---: |
| $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\mathrm{C}$ | $\mathrm{K}$ | $\mathrm{K}$ |
| $\mathrm{K}$ | $\mathrm{K}$ | $\m... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,201 |
2. Given numbers $x, y \in\left(0, \frac{\pi}{2}\right)$. Find the maximum value of the expression
$$
A=\frac{\sqrt[4]{\sin x \sin y}}{\sqrt[4]{\operatorname{tg} x}+\sqrt[4]{\operatorname{tg} y}}
$$ | Answer: $\frac{\sqrt[4]{8}}{4}$.
Solution. Note that $\sqrt[4]{a}+\sqrt[4]{b} \leqslant \sqrt[4]{8(a+b)}$ for any $a, b \geqslant 0$, since
$$
(\sqrt[4]{a}+\sqrt[4]{b})^{4} \leqslant(2(\sqrt{a}+\sqrt{b}))^{2} \leqslant 8(a+b)^{4}
$$
Applying the inequality for the harmonic mean and the arithmetic mean, we get
$$
\b... | \frac{\sqrt[4]{8}}{4} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,202 |
3. Quadrilateral $ABCD$ is inscribed in circle $\omega$, the center of which lies on side $AB$. Circle $\omega_{1}$ is externally tangent to circle $\omega$ at point $C$. Circle $\omega_{2}$ is tangent to circles $\omega$ and $\omega_{1}$ at points $D$ and $E$ respectively. Line $BD$ intersects circle $\omega_{2}$ agai... | Answer: $180^{\circ}$.
Solution 1. Since $\angle A C O=\angle Q C O_{1}$, the isosceles triangles $A O C$ and $Q O_{1} C$ are similar, from which $\angle A O C=\angle Q O_{1} C$. Similarly, ... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,203 |
4. There are four-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It is known that the number $m$ is divisible by 100, and its decimal representation consists of four pairs of identical adjacent digits. Find $m$ and $n$. | Answer: $6325 \cdot 5236=33117700$.
Solution. Let's write
$$
m=\overline{a b c d}, \quad n=\overline{d c b a}, \quad m n=\overline{x x y y z z 00}
$$
It is clear that one of the numbers $m$ and $n$ ends in 5 (let it be $m$). Then $d=5$, and the digit $a$ is even. Obviously, $m n$ is divisible by 11. Since
$$
m \bmo... | 6325\cdot5236=33117700 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,204 |
5. Around the edge of a round table, there are $n$ empty glasses ($n \geqslant 3$). Petya and Vasya take turns (starting with Petya) filling them with drinks: Petya - kvass, Vasya - morse. In one move, a player can fill one empty glass of their choice so that after their move, no two adjacent glasses contain the same d... | Answer: under no circumstances.
Solution. Vasya's strategy is as follows: at each step, he must fill the glass that is next in the clockwise direction to the one Petya filled on the previous move. We will show that this is always possible. We will use induction on the number of Petya's moves. Since $n \geqslant 3$, th... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,205 |
6. On the table lie spheres with radii 2, 2, 5, touching each other externally. The vertex of the cone is located midway between the points of contact of the identical spheres with the table, and the cone itself touches all the spheres externally. Find the angle at the vertex of the cone. (The angle at the vertex of th... | Answer: $2 \operatorname{arcctg} 72$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A_{1}, A_{2}, A_{3}$ be the points of tangency of the spheres with the table, $C$ be... | 2\operatorname{arcctg}72 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,206 |
1. In the cells of a $5 \times 5$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Answer: 48.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 5. Since all these sums are distinct, the minimal possible set of their values is $\{5,6, \ldots, 13,14\}$. By adding the sums of the rows and columns of the table, we get twice the... | 48 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,207 |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a+\frac{1}{b}\right)^{3}+\left(b+\frac{1}{c}\right)^{3}+\left(c+\frac{1}{d}\right)^{3}+\left(d+\frac{1}{a}\right)^{3}
$$ | Answer: 32.
Solution 1. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(2 \sqrt{\frac{a}{b}}\right)^{3}+\left(2 \sqrt{\frac{b}{c}}\right)^{3}+\left(2 \sqrt{\frac{c}{d}}\right)^{3}+\left(2 \sqrt{\frac{d}{a}}\right)^{3} \geqslant 32\left(\fr... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,208 |
3. Given an acute-angled triangle $A B C$. A circle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. Tangents drawn to the circle at points $D$ and $E$ intersect at point $K$. Find the angle between the lines $A K$ and $B C$. | Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle D B E, \beta=\angle B C D, \gamma=\angle C B E, O$ - the intersection point of segments $C D$ and $B E$. The angle between the tangent $D... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,209 |
4. On the board, the product of the numbers $\overline{\text{IKS}}$ and $\overline{\text{KIS}}$ is written, where the letters correspond to different non-zero decimal digits. The middle digits of this product are I, K, S, and o, written in some order, and the outermost digits are equal to each other and different from ... | Answer: $169201=269 \cdot 629$ or $193501=359 \cdot 539$.
Solution. Since the problem does not change when I and K are swapped, we can assume that I $=19$, which is impossible. In the case I $+K=8$, the pair (I, K) is $(2,6)$ or $(3,5)$. Since $269 \cdot 629=169201$ and $359 \cdot 539=193501$, both options are suitabl... | 169201=269\cdot629or193501=359\cdot539 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,210 |
5. Around a round table, there are $2 n$ empty glasses ( $n \geqslant 2$ ). Petya and Vasya take turns (starting with Petya) filling the empty glasses with orange and apple juice. In one move, each player chooses two empty glasses and pours orange juice into one and apple juice into the other. The game ends when all gl... | Answer: under no circumstances.
Solution. Let's divide the entire set of glasses into pairs of adjacent ones. To thwart Petya's plans, Vasya must adhere to the following strategy. If Petya fills two glasses from one pair, then Vasya fills two glasses from another pair. Suppose Petya fills glasses from two different pa... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,211 |
6. On the table lie spheres with radii 2, 2, 1, touching each other externally. The vertex of the cone \( C \) is on the table, and the cone itself touches all the spheres externally. Point \( C \) is equidistant from the centers of the two equal spheres, and the cone touches the third sphere along a generatrix perpend... | Answer: $2 \operatorname{arctg} \frac{\sqrt{5}-2}{3}$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A_{1}, A_{2}, A_{3}$ be the points of tangency of the spheres with ... | 2\operatorname{arctg}\frac{\sqrt{5}-2}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,212 |
1. In the cells of a $4 \times 6$ table, natural numbers are arranged such that all ten sums of these numbers in the rows and columns of the table are distinct. Find the smallest possible value of the sum of all the numbers in the table. | Answer: 43.
Solution. Since the elements of the table are natural numbers, the sums of the rows and columns of the table are no less than 4. Since all these sums are different, the minimal possible set of their values is $\{4,5, \ldots, 12,13\}$. By adding the sums of the rows and columns of the table, we get twice th... | 43 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,213 |
2. Given positive numbers $a, b, c, d$. Find the minimum value of the expression
$$
A=\left(a^{2}+\frac{1}{b c}\right)^{3}+\left(b^{2}+\frac{1}{c d}\right)^{3}+\left(c^{2}+\frac{1}{d a}\right)^{3}+\left(d^{2}+\frac{1}{a b}\right)^{3}
$$ | Answer: 32.
Solution. We will use the Cauchy inequality for means first in each parenthesis, and then for the entire sum. We get
$$
A \geqslant\left(\frac{2 a}{\sqrt{b c}}\right)^{3}+\left(\frac{2 b}{\sqrt{c d}}\right)^{3}+\left(\frac{2 c}{\sqrt{d a}}\right)^{3}+\left(\frac{2 d}{\sqrt{a b}}\right)^{3} \geqslant 32\le... | 32 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,214 |
3. Point $M$ is the midpoint of side $AB$ of triangle $ABC$. A circle $\omega_{1}$ is drawn through points $A$ and $M$, tangent to line $AC$, and a circle $\omega_{2}$ is drawn through points $B$ and $M$, tangent to line $BC$. Circles $\omega_{1}$ and $\omega_{2}$ intersect again at point $D$. Point $E$ lies inside tri... | Answer: $180^{\circ}$
Solution 1. The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,215 |
4. On the board, the product of the numbers $\overline{\text{IKS}}$ and $\overline{\text{KSI}}$ is written, where the letters correspond to different non-zero decimal digits. This product contains three digits $C$ and one each of $I$, $K$, and 0, and its leading digit is $C$. What is written on the board? | Answer: $112015=521 \cdot 215$.
Solution. Let $p=\overline{\text { UKC }} \cdot \overline{\mathrm{KCV}}$. The numbers $(\mathrm{U}+\mathrm{K}+\mathrm{C})^{2}$ and $(\mathrm{U}+\mathrm{K}+3 \mathrm{C})$ give the same remainders when divided by 9, from which $\mathrm{C} \bmod 9=\frac{(\text { U }+\mathrm{K}+\mathrm{C})(... | 112015=521\cdot215 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,216 |
5. Around the edge of a round table, there are $2 n$ empty glasses ( $n \geqslant 2$ ). Petya and Vasya take turns (starting with Petya) filling the empty glasses with orange and apple juice. In one move, each player chooses two empty glasses and fills them with the same type of juice (of their choice). The game ends w... | Answer: under no circumstances.
Solution. Let's divide the entire set of glasses into pairs of adjacent ones. To thwart Petya's plans, Vasya should follow this strategy. If Petya fills two glasses from one pair, then Vasya should fill two glasses from another pair. If Petya fills glasses from two different pairs, then... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,217 |
6. On the table lie spheres with radii $4,4,5$, touching each other externally. The vertex of the cone $C$ is on the table, and the cone itself touches all the spheres externally. Point $C$ is equidistant from the centers of the two equal spheres, and the cone touches the third sphere along a generatrix perpendicular t... | Answer: $2 \operatorname{arcctg} 7$.
Solution. Let $O_{1}, O_{2}, O_{3}$ be the centers of the spheres, $A_{1}, A_{2}, A_{3}$ be the points of tangency of the spheres with the table, $2 \alp... | 2\operatorname{arcctg}7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,218 |
1. What is the minimum number of cells that need to be marked in a $7 \times 7$ table so that every vertical or horizontal strip $1 \times 4$ contains at least one marked cell? | Answer: 12.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,219 |
2. Given numbers $x, y, z \in\left[0, \frac{\pi}{2}\right]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$ | Answer: 1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is positive a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,220 |
3. A circle $\omega$ is circumscribed around triangle $A B C$. Circle $\omega_{1}$ touches the line $A B$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $A C$ at point $A$ and passes through point $B$. A tangent is drawn from point $A$ to circle $\omega$, which intersects circle $... | Answer: $\frac{1}{2}$.
Solution 1. Let $O$ be the center of $\omega, \alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of the circle $\omega_{1}$ is equal to the... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,221 |
4. On the board, the product of the numbers $\overline{\text { IKS }}$ and $\overline{\text { ISK }}$ is written, where the letters correspond to different non-zero decimal digits, one of which is even. This product is a five-digit number and reads the same from left to right and from right to left. What is written on ... | Answer: $29392=167 \cdot 176$.
Solution. Since the problem does not change when swapping $\mathrm{K}$ and $\mathrm{C}$, we can assume that $\mathrm{K}<\mathrm{C}$. Let $p=\overline{\text { IKS }} \cdot \overline{\text { ISK }}$. The number $p$ is a five-digit number and $p \geqslant 10000 \cdot \text{I}^2$, from which... | 29392=167\cdot176 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,222 |
5. At the vertices of a regular $2 n$-gon, empty cups are placed. Petya and Vasya take turns (starting with Petya) pouring tea into the cups. In one move, a player can pour tea into either one empty cup or two cups that are symmetric with respect to the center of the $2 n$-gon, provided both are empty. The player who c... | Answer: for odd $n$.
Solution. Let $n$ be odd initially. Victory for Pete is ensured by the following strategy. His first move is to pour tea into two diametrically opposite cups. Then, on each step, he fills the cups that are symmetric relative to this diameter to the cups that Vasya filled on the previous move. Befo... | for\odd\n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,223 |
6. Two spheres with a radius of 4 are lying on a table, touching each other externally. A cone touches the side surface of the table and both spheres (externally). The distances from the vertex of the cone to the points where the spheres touch the table are 5. Find the angle at the vertex of the cone. (The angle at the... | Answer: $90^{\circ}$ or $2 \operatorname{arcctg} 4$.
Solution. Let $O_{1}, O_{2}$ be the centers of the spheres, $A_{1}, A_{2}$ be the points of tangency of the spheres with the table, $C$ b... | 90or2\operatorname{arcctg}4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,224 |
1. What is the minimum number of cells that need to be marked in a $9 \times 9$ table so that each vertical or horizontal strip $1 \times 5$ contains at least one marked cell? | Answer: 16.
Solution. Consider a more general problem when the table has size $(2 n-1) \times(2 n-1)$, and the strip is $-1 \times n$. Let's call the $n$-th row and the $n$-th column central, and the marked cells on them, except for the center of the board, - axial. Suppose there are $k$ axial cells in the central row... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,225 |
2. Given numbers $x, y, z \in [0, \pi]$. Find the minimum value of the expression
$$
A=\cos (x-y)+\cos (y-z)+\cos (z-x)
$$ | Answer: -1.
Solution. We can assume that $x \leqslant y \leqslant z$, since the expression $A$ does not change under pairwise permutations of the variables. Notice that
$$
\cos (x-y)+\cos (z-x)=2 \cos \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The first cosine in the right-hand side is non-negat... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,226 |
3. A circle $\omega$ with center at point $O$ is circumscribed around triangle $ABC$. Circle $\omega_{1}$ touches the line $AB$ at point $A$ and passes through point $C$, while circle $\omega_{2}$ touches the line $AC$ at point $A$ and passes through point $B$. A line through point $A$ intersects circle $\omega_{1}$ ag... | Answer: $90^{\circ}$.
Solution 1. Let $\alpha=\angle B A C$. The angle between the tangent $A B$ and the chord $A C$ of circle $\omega_{1}$ is equal to the inscribed angle that subtends $A ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,227 |
4. Given nine-digit numbers $m$ and $n$, obtained from each other by writing the digits in reverse order. It turned out that the product $mn$ consists of an odd number of digits and reads the same from left to right and from right to left. Find the largest number $m$ for which this is possible. | Answer: 220000001.
Solution. Let $m=\overline{a_{8} \ldots a_{0}}, n=\overline{a_{0} \ldots a_{8}}$. Since the number $m n$ contains an odd number of digits, it is a seventeen-digit number. Write $m n=\overline{b_{16} \ldots b_{0}}$. We will show by induction that
$$
b_{k}=a_{0} a_{8-k}+a_{1} a_{9-k}+\ldots+a_{k-1} a... | 220000001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,228 |
5. At the vertices of a regular $2 n$-gon, empty cups are placed. Petya and Vasya take turns (starting with Petya) pouring tea into the cups. In one move, a player can pour tea into two empty cups that are either adjacent or symmetric relative to the center of the $2 n$-gon. The player who cannot make a move loses. For... | Answer: for odd $n$.
Solution. Let's call a pair two cups that are symmetric relative to the center of the $2n$-gon. Suppose $n$ is even. This means the number of pairs is even. Vasya's winning strategy is as follows. If Petya pours tea into cups from one pair, then Vasya fills the cups from some other pair. Suppose P... | for\odd\n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,229 |
6. Two spheres with a radius of 12 are lying on a table, touching each other externally. A cone touches the table's surface and both spheres (externally). The distances from the vertex of the cone to the points where the spheres touch the table are 13. Find the angle at the vertex of the cone. (The angle at the vertex ... | Answer: $2 \operatorname{arcctg} \frac{4}{3}$ or $2 \operatorname{arcctg} 3$.
Solution. Let $O_{1}, O_{2}$ be the centers of the spheres, $A_{1}, A_{2}$ be the points of tangency of the sph... | 2\operatorname{arcctg}\frac{4}{3}or2\operatorname{arcctg}3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,230 |
1. In a $3 \times 3$ table, 9 numbers are arranged such that the six products of these numbers in the rows and columns of the table are all different. What is the maximum number of these numbers that can equal one? | Answer: 5.
Solution. Let's call the index of the table the total number of its rows and columns consisting of ones. According to the condition, the index does not exceed 1. Let $n$ be an element of the table different from 1. Then in one row or one column with $n$ there is another number not equal to 1 (otherwise the ... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,231 |
2. Given numbers $x, y, z \in\left[0, \frac{\pi}{2}\right]$. Find the maximum value of the expression
$$
A=\sin (x-y)+\sin (y-z)+\sin (z-x) .
$$ | Answer: $\sqrt{2}-1$.
Solution. We can assume that $x \leqslant y$ and $x \leqslant z$, since the expression $A$ does not change under cyclic permutation of the variables. Notice that
$$
\sin (x-y)+\sin (z-x)=2 \sin \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The argument of the sine on the right... | \sqrt{2}-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,232 |
3. Point $O$ is the center of the circumcircle of triangle $ABC$. On the circumcircle of triangle $BOC$, outside triangle $ABC$, point $X$ is chosen. On the rays $XB$ and $XC$ beyond points $B$ and $C$, points $Y$ and $Z$ are chosen respectively such that $XY = XZ$. The circumcircle of triangle $ABY$ intersects side $A... | Answer: $180^{\circ}$.
Solution. Note that $\angle B Y T=\angle B A T$ as inscribed angles subtending the same arc. Since quadrilateral $B O C X$ is cyclic, we get
$$
180^{\circ}-\angle B X C=\angle B O C=2 \angle B A C=2 \angle B Y T
$$
On the other hand, triangle $Y X Z$ is isosceles, so $180^{\circ}-\angle Y X Z=... | 180 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,233 |
4. On the board, the product of the three-digit numbers $\overline{\text{IKS}}$ and $\overline{\text{KSI}}$ is written, where the letters correspond to different non-zero decimal digits. This product reads the same from left to right and from right to left, two of its digits are 4, and the other four digits match I. Wh... | Answer: $477774=762 \cdot 627$ or $554455=593 \cdot 935$.
Solution. Let $p=\overline{\text { ИКС }} \cdot \overline{\mathrm{KCИ}}$. From the condition on $p$, it follows that $p \vdots 11$ and, therefore, one of the factors of $p$ is divisible by 11. Let's first assume that $\overline{\text { ИКС }} \vdots 11$. This m... | 477774=762\cdot627or554455=593\cdot935 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,234 |
5. On the table, there are empty glasses lined up. Petya and Vasya take turns (starting with Petya) to fill them with drinks: Petya - lemonade, Vasya - compote. In one move, a player can fill one empty glass of their choice so that after their move, no two adjacent glasses contain the same drink. If all the glasses are... | Answer: $n \notin\{1,2,4,6\}$.
Solution. Number the glasses from left to right with numbers from 1 to $n$. For $n=1$ and $n=2$, a draw is obvious. If $n$ is 4 or 6, Petya first fills the first glass with lemonade. For $n=4$, Petya can then fill one of the two last glasses and, thus, will not lose. For $n=6$, Petya fil... | n\notin{1,2,4,6} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,235 |
6. On the table lie two spheres with radii 4 and 1, with centers $O_{1}$ and $O_{2}$, touching each other externally. A cone touches the table's surface and both spheres (externally). The vertex $C$ of the cone is on the segment connecting the points of contact of the spheres with the table. It is known that the rays C... | Answer: $2 \operatorname{arctg} \frac{2}{5}$.
Solution. Let $A_{1}, A_{2}$ be the points of contact of the spheres with the table, and $2 \alpha$ be the angle at the vertex of the cone. By the problem's condition, the angles $O_{1} C A_{1}$ and $O_{2} C A_{2}$ are equal; denote their common value by $\varphi$. Note th... | 2\operatorname{arctg}\frac{2}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,236 |
1. In a $3 \times 4$ table, 12 numbers are arranged such that all seven sums of these numbers in the rows and columns of the table are distinct. What is the maximum number of numbers in this table that can be zero? | Answer: 8.
Solution. Let's call the index of the table the total number of its zero rows and columns. According to the condition, the index does not exceed 1. Let $n$ be a non-zero element of the table. Then in the same row or in the same column with $n$ there is another non-zero number (otherwise the sums in the row ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,237 |
2. Given numbers $x, y, z \in [0, \pi]$. Find the maximum value of the expression
$$
A=\sin (x-y)+\sin (y-z)+\sin (z-x)
$$ | Answer: 2.
Solution 1. We can assume that $x \leqslant y$ and $x \leqslant z$, since the expression $A$ does not change under cyclic permutation of the variables. Notice that
$$
\sin (x-y)+\sin (z-x)=2 \sin \left(\frac{z-y}{2}\right) \cos \left(\frac{z+y}{2}-x\right)
$$
The argument of the sine on the right-hand sid... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,238 |
3. Point $O$ is the center of the circumcircle of triangle $A B C$. Points $Q$ and $R$ are chosen on sides $A B$ and $B C$ respectively. Line $Q R$ intersects the circumcircle of triangle $A B R$ again at point $P$ and intersects the circumcircle of triangle $B C Q$ again at point $S$. Lines $A P$ and $C S$ intersect a... | Answer: $90^{\circ}$.
Solution. Note that $\angle C S Q=\angle C B Q$ and $\angle A P R=\angle A B R$ as inscribed angles subtending the same arc. Therefore, $\angle K S P=\angle K P S$, whi... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,239 |
5. Around a round table, there are $n$ empty glasses ( $n \geqslant 3$ ). Petya and Vasya take turns (starting with Petya) to fill them with compote or lemonade. In one move, a player can fill one empty glass with either of the two drinks of their choice. A player wins if, after their move, there is a glass with lemona... | Answer: for odd $n$.
Solution. Let $n$ be odd. Petya's winning strategy is as follows. He chooses the first glass arbitrarily. Suppose that Vasya's next move fills a glass that is $k$ positions away from the first, counting along the shortest arc. If Petya then has a winning move, he wins. Otherwise, Petya fills the s... | for\odd\n | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,240 |
6. Two spheres are lying on a table, touching each other externally. A cone touches the table with its lateral surface and both spheres (externally). The vertex of the cone is located on the segment connecting the points of contact of the spheres with the table. It is known that the rays connecting the vertex of the co... | Answer: $2 \operatorname{arcctg} 2$.
Solution. Let $O_{1}$ and $O_{2}$ be the centers of the spheres, $R$ and $r$ be the radii of the spheres, $A_{1}$ and $A_{2}$ be the points of tangency ... | 2\operatorname{arcctg}2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,241 |
1. (10 points) The lengths of the sides of a scalene triangle turned out to be consecutive terms of a certain geometric progression. The common ratio of this progression can be
1) 1.7 ;
2) 0.5 ;
3) 2.0 ;
4) another answer. | Answer: 4).
Solution: Let the sides of the triangle be $b, q b$, and $q^{2} b$, where $q$ is the sought common ratio of the geometric progression. For each side, the triangle inequality must hold:
$$
\left\{\begin{array}{l}
b+q b>b q^{2} \\
q b+q^{2} b>b \\
b+q^{2} b>b q .
\end{array}\right.
$$
From this, we have th... | (\frac{-1+\sqrt{5}}{2};\frac{1+\sqrt{5}}{2})\simeq(0.618;1.618) | Geometry | MCQ | Yes | Yes | olympiads | false | 8,242 |
2. (20 points) A unit segment is divided into 3 equal parts, and the middle part is discarded. Each of the remaining two segments is in turn divided into 3 equal parts, and its middle part is also discarded, after which the same operation is performed on each of the remaining segments, and so on. Suppose the operation ... | Answer: $2^{16}$.
Solution: Notice that after each operation, instead of one original segment, there remain two segments, each of which is one-third the length of the original. Thus, after each operation, the number of remaining segments doubles. Therefore, after sixteen operations, there will be $2^{16}$ segments rem... | 2^{16} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,243 |
3. (20 points) An equilateral triangle with unit sides is divided by three lines, parallel to its sides, into 4 equal triangles, and the middle triangle is discarded. Each of the remaining three triangles is in turn divided by three lines, parallel to its sides, into 4 equal parts, and its middle triangle is also disca... | Answer: $3^{12}$.
Solution: In this case, after each operation, the number of triangles triples (since one of the four triangles obtained from the division is discarded), and the side lengths of the resulting triangles are half the side lengths of the original triangle. Therefore, after twelve operations, there will b... | 3^{12} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,244 |
4. (20 points) Let $a_{1}, a_{2}, \ldots, a_{2015}$ be some permutation of the numbers 2015, 2016, 2017, ..., 4029. Prove that the product $(a_{1}-1)(a_{2}-2)(a_{3}-3) \cdot \ldots \cdot (a_{2015}-2015)$ is an even number. | Solution: Notice that among the numbers of the permutation, i.e., among the numbers from 2015 to 4029 inclusive, there is one more odd number than even numbers. For the considered product to be an odd number, it is necessary that all factors of this product be odd, i.e., from each number of the permutation, a number of... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,245 |
5. (20 points) Prove that the number $(1+1 / 1) \cdot(1+1 / 3) \cdot(1+1 / 5) \cdot \ldots \cdot(1+$ $+1 / 2015$ ) is not an integer. | Solution: We will bring the expressions in each of the parentheses to a common denominator. We will get the product of even numbers from 2 to 2016 inclusive in the numerator, and the product of odd numbers from 1 to 2015 inclusive in the denominator. Consider 2011 - the largest prime number among the factors of the den... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,246 |
7. (30 points) The bisector $Q K$ of triangle $P Q R$ intersects its circumcircle at point $M$ (different from $Q$). The circumcircle of triangle $P K M$ intersects the extension of side $P Q$ beyond point $P$ at point $N$. Prove that $N R$ and $Q M$ are perpendicular.
 A circle with center $O$, inscribed in angle $Q P R$, touches side $P R$ at point $L$. A tangent to the circle, parallel to $P O$, intersects ray $P Q$ at point $S$, and ray $L P$ at point $D$. Prove that $D S=2 P L$.
 Circles $K_{1}$ and $K_{2}$ touch a line at points $A$ and $B$ respectively and, in addition, intersect at points $X$ and $Y$, with point $X$ being closer to the line $AB$. Line $AX$ intersects $K_{2}$ again at point $P$. The tangent to $K_{2}$ at point $P$ intersects line $AB$ at point $Q$. Prove that a... | Solution: Let $Q Y$ intersect the second circle at point $S$, and the line $Y X$ intersect $A B$ at point $T$. From the similarity of triangles $Q S B$ and $Q B Y$, we have $B S: B Y=Q B: Q Y$. Similarly, $P S: P Y=Q P: Q Y$, hence $B S: S P=B Y: Y P$.
By the Law of Sines in triangle $X Y B$, we have: $B Y: X Y = \sin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,250 |
10. (40 points) On the board, there are 2015 pairwise distinct positive real numbers. It turns out that for any number $a>0$, the number of numbers on the board that are less than 2014/a, and the number of numbers that are greater than a, have the same parity. What can the product of all the numbers be? | Answer: $2014^{1007} \sqrt{2014}$.
Solution: We will prove that all numbers written on the board, except one, can be paired such that the product in each pair is 2014. Consider $a=\sqrt{2014}$. Since the number of numbers greater than it and the number of numbers less than it have the same parity, the number of number... | 2014^{1007}\sqrt{2014} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,251 |
11. (40 points) For the quadratic function $p(x)=a x^{2}+b x+c$, for some integers $n$, the equality $p(n)=p\left(n^{2}\right)$ holds. Provide an example of the function $p(x)$ for which the number of such numbers $n$ is the largest. What is this largest number of numbers $n$? | Answer: The maximum number of numbers $n$ is 4. An example of a function: $p(x)=x^{2}-6 x+1$.
Solution: Since $0=0^{2}$ and $1=1^{2}$, for any function $p(x)$ there are at least two such numbers $n$.
Write the equality $p(n)=p\left(n^{2}\right)$ and transform it:
$$
a n^{2}+b n+c=a n^{4}+b n^{2}+c \Leftrightarrow a ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 8,252 |
12. (40 points) Dasha, Masha, and Sasha are coming up with various five-digit numbers in the decimal system that can be obtained from each other by rearranging the digits. Can it happen that the sum of the numbers thought of by Sasha and Masha is equal to twice the number thought of by Dasha? | Answer: Yes, it could.
Solution: For example, if Dasha came up with the number 10061, Masha the number 10016, and Sasha the number 10106. | Yes,itcould | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,253 |
1. (20 points) On the board, all two-digit numbers divisible by 5, where the number of tens is greater than the number of units, were written down. There turned out to be $A$ such numbers. Then, all two-digit numbers divisible by 5, where the number of tens is less than the number of units, were written down. There tur... | Answer: 413.
Solution: We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's calculate what $A$ is. We need two-digit numbers divisible by 5, i.e., numbers where $y$ is either 0 or 5. Note that if $y=0$, then $x$ can take any value from 1 ... | 413 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,254 |
2. (20 points) Find the number of different four-digit numbers that can be obtained by rearranging the digits of the number 2021 (including this number itself). | Answer: 9.
Solution: The number of options can be found by enumerating the permutations of the digits: 2021, 2012, 2201, 2210, 2102, 2120, 1022, 1202, 1220.
The number of options can also be calculated using combinatorial methods. The position of zero can be chosen in three ways, as it should not be the first. Then, ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,255 |
3. (30 points) In math class, each of the seven dwarfs needs to find a two-digit number such that when 18 is added to it, the result is the number with its digits reversed. Can all the numbers found by the dwarfs be different? | Answer: Yes, they can.
Solution: We will write two-digit numbers in the form $\overline{x y}$, where $x$ is the number of tens, and $y$ is the number of units.
Let's find all valid numbers. According to the condition $10 x+y+18=10 y+x$. Transforming this:
$$
10 x+y+18=10 y+x \Leftrightarrow 18=9 y-9 x \Leftrightarro... | 13,24,35,46,57,68,79 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,256 |
4. (30 points) Let natural numbers $m$ and $n$ satisfy the equation $\frac{1}{m}+\frac{1}{n}=\frac{1}{2020}$. Prove that $m$ and $n$ cannot both be odd. | Solution: Transform the condition of the problem to the form
$$
2020(n+m)=m n
$$
The left side of this equality is an even number, therefore, its right side is also even. Hence, at least one of the numbers $m$ and $n$ must be even. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,257 |
5. (50 points) When distributing land plots, Farmer Novoselov was allocated 2 square plots of different areas, with integer side lengths. Is it possible to allocate Farmer Malinnikov 2 square plots with integer side lengths as well, so that the total area of Malinnikov's plots is twice the total area of Novoselov's plo... | Answer: Yes, it is possible.
Solution: Let $x$ and $y$ be the sides of the plots allocated to Farmer Novoselov. We will assume that $x>y$. We will show how to find the lengths of the sides of the plots that should be allocated to Farmer Malinnikov according to the problem's condition.
Let $x^{2}+y^{2}=a$. Then consid... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,258 |
6. (50 points) Let the sequence of numbers $A$ be such that $A_{1}=3, A_{2}=8, A_{3}=13, \ldots$. Prove that there exists an infinite number of numerical sequences $B$ with the following properties:
1) $B_{1}=7$
2) $B_{k}=B_{k-1}+d$, where $d-$ is some number $(k=2,3,4, \ldots)$;
3) have infinitely many matching terms ... | Solution: Note that if $B_{2}$ is in $A$, then the progression $B$ satisfies condition 3). Indeed, if $x$ is a common term of $A$ and $B$, then $x+5d$ is also. Consider the progressions $B$ with differences $A_{i}-7$, where $i \geq 3$. All of them are different, and in each $B_{2}$ is also a term of $A$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,259 |
1. (10 points) In the arithmetic class, children were studying some natural numbers and operations with them. The teacher gave the children many cards with the numbers 1, 2, 4, and 8 and asked them to arrange the cards in a circle using each of the numbers so that the sum of the numbers on any two adjacent cards is div... | Answer: a) and b).
First solution: It is sufficient to prove that the solutions to the problem are numbers of the form $6+2k$, where $k$ is a non-negative integer.
First, let's show that the minimum answer is 6. Next to the number 1, only the number 2 can be placed, and next to the number 8, only the number 4. Since ... | )b) | Number Theory | MCQ | Yes | Yes | olympiads | false | 8,260 |
2. (10 points) In triangle $A B C$, a point $D$ is marked on side $B C$, and a point $M$ is marked on segment $A D$ such that triangles $A M B$ and $A M C$ have equal area. Which of the following statements are true?
a) Point $D$ must be the midpoint of side $B C$.
б) Point $D$ may not be the midpoint of side $B C$.
... | Answer: a), g)
Solution: Rewrite the condition of the problem as
$$
1=\frac{S_{\triangle A M B}}{S_{\triangle A M C}}=\frac{\frac{1}{2} A B \cdot A M \cdot \sin \angle B A M}{\frac{1}{2} A C \cdot A M \cdot \sin \angle C A M}=\frac{A B \cdot \sin \angle B A M}{A C \cdot \sin \angle C A M}
$$
The right-hand side of t... | ),) | Geometry | MCQ | Yes | Yes | olympiads | false | 8,261 |
3. (20 points) The pensioners of one of the planets in the Alpha Centauri system enjoy spending their free time solving cybrow solitaire puzzles: they choose natural numbers from a certain interval $[A, B]$ in such a way that the sum of any two of the chosen numbers is not divisible by a given number $N$. Last week, th... | Answer: 356.
Solution: For $k=0,1, \ldots, 10$, let $I_{k}$ be the set of all numbers in $[A, B)$ that give a remainder of $k$ when divided by 11. Since $A$ and $B$ are multiples of 11, all sets $I_{k}$ contain an equal number of elements. Therefore, all numbers in $[A, B)$ that are not multiples of 11 can be paired a... | 356 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,262 |
4. (20 points) Before the geometry lesson, the teacher wrote on the board the values of all the angles (in degrees) of a certain convex polygon. However, the duty students erased one of the written numbers. When the lesson began, it turned out that the sum of the remaining numbers was 1703. What number did the duty stu... | Answer: 97.
Solution: Let the polygon have $n$ vertices. Since the $n$-gon is convex, each of its angles is less than $180^{\circ}$, and the sum of all angles is $(n-2) \cdot 180^{\circ}$. Therefore, the sum of all angles of the polygon minus one lies in the interval from $180(n-3)$ to $180(n-2)$. Then
$$
180(n-3)<17... | 97 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,263 |
5. (30 points) In an acute-angled triangle $S A P$, the altitude $A K$ is drawn. On the side $P A$, a point $L$ is chosen, and on the extension of side $S A$ beyond point $A$, a point $M$ is chosen such that $\angle L S P = \angle L P S$ and $\angle M S P = \angle M P S$. The lines $S L$ and $P M$ intersect the line $A... | Solution: From the equality of angles $L S P$ and $L P S$, it follows that triangle $L S P$ is isosceles with base $S P$, so its vertex $L$ lies on the perpendicular bisector of the base $S P$. Similarly, from the equality of angles $M S P$ and $M P S$, we get that $M$ also lies on the perpendicular bisector of $S P$. ... | 2ML=NO | Geometry | proof | Yes | Yes | olympiads | false | 8,264 |
6. (30 points) In triangle $K O I$, a point $M$ is marked on side $K O$ such that $K M = M I$, and a point $S$ is marked on side $I O$ such that $S I = S O$. On line $K S$, a point $U$ is marked such that line $M U$ is parallel to line $K I$. Prove that angle $KOI$ is equal to angle $MIU$. | Solution: Note that from the parallelism of $M U$ and $K I$, it follows that $\angle U M I = \angle M I K$,
and from the equality of $M I$ and $M K$, i.e., the isosceles nature of triangle $M... | proof | Geometry | proof | Yes | Yes | olympiads | false | 8,265 |
7. (40 points) In the thirtieth kingdom, there are 39 cities. It is known that from each city, there are no fewer than 21 one-way roads leading to other cities (no more than one road leads from one city to another). It is also known that exactly 26 cities are "transit" cities, meaning it is impossible to return directl... | Solution: Note that there are no more than $26 \cdot 25 / 2 = 325$ roads between the transit cities, since there can be no more than one road between any two transit cities. Since there are at least 21 roads leading from each city, including transit cities, there are at least $26 \cdot 21 - 325 = 546 - 325 = 221$ roads... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,266 |
8. (40 points) In a mental arithmetic competition, several teams participated. Each team received one natural number and was supposed to find the largest and the smallest, but not equal to 1, odd divisors of this number. All teams solved their tasks correctly. When reviewing the results, it turned out that any initial ... | Answer: 4 teams; 528, 880, 1232, 1936.
Solution: It is easy to see that $M$ is divisible by $m$, and $\frac{15 M+11 m}{M}$ is a power of two. Since $m \leqslant M$,
$$
15<\frac{15 M+11 m}{M} \leqslant \frac{15 M+11 M}{M}=26
$$
Among the integers in the interval $(15,26]$, only 16 is a power of two. Therefore, $15 M+... | 4teams;528,880,1232,1936 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,267 |
1. Prove the following divisibility rule: the number 1207 is divisible by a two-digit number x if and only if the sum of the cubes of the digits of x is 344. | Solution. Note that $1207=17 \cdot 71$. Since 17 and 71 are prime numbers, they are the only two-digit divisors of 1207. The sum of the cubes of the digits of both numbers is 344. Conversely, the sum of the cubes of two digits equals 344 only when these digits are 1 and 7. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,268 |
3. Points $A, B, C, D$ are located on a line (in that order). On the plane containing this line, a point $E$ is marked. It turns out that $A B = B E$ and $E C = C D$. Prove that angle $A E D$ is right.
^{2}=10^{18}+14 \cdot 10^{9}+49=1 \cdot 10^{18}+1 \cdot 10^{10}+4 \cdot 10^{9}+4 \cdot 10+9
$$
and the sum of its digits is 19. The square of the second number ... | doesnotlie | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,271 |
5. All three-digit numbers from 100 to 999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \) ... | Answer: $k=5$.
Solution. An example of a five-digit number that satisfies the condition of the problem is 22923. Indeed, Kostya could underline the fragment «22923» at the junction of the numbers 229 and 230, and Andrey - at the junction of the numbers 922 and 923.
Now we will show that it is not possible to choose a... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,272 |
6. There are p ministers, each of whom has issued a (unique) decree. To acquaint everyone with the decrees, the ministers send each other telegrams. No two telegrams can be sent simultaneously. When a minister receives a telegram, he reads it immediately. When sending a telegram, a minister includes his decree, as well... | Solution. If there are only two ministers, the statement is obvious: the minister who sent the telegram first will then receive the order from the second one along with his own. Suppose the statement is proven for $n-1$ ministers, and let's check it for $n$ ministers. Let the last of the telegrams be sent by minister $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,273 |
1. Prove the following divisibility criterion: the number 2701 is divisible by a two-digit number x if and only if the sum of the squares of the digits of the number x is 58. | Solution. Note that $2701=37 \cdot 73$. Since the numbers 37 and 73 are prime, they are the only two-digit divisors of 2701. The sum of the squares of the digits of both numbers is 58. Conversely, the sum of the squares of two digits equals 58 only when these digits are 3 and 7. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,274 |
4. Baron Munchausen claims that there exist two different 10-digit numbers, not divisible by 10, such that if you subtract the sum of the digits of their squares from each of these numbers, the results will be the same. Is the baron lying? | Answer: does not lie.
Solution. Examples of such numbers can be $10^{9}+8$ and $10^{9}+9$. Indeed, the square of the first number is
$$
\left(10^{9}+8\right)^{2}=10^{18}+16 \cdot 10^{9}+64=1 \cdot 10^{18}+1 \cdot 10^{10}+6 \cdot 10^{9}+6 \cdot 10+4
$$
and the sum of its digits is 18. The square of the second number ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,277 |
5. All four-digit numbers from 1000 to 9999 are written in a row without spaces. Kostya underlined \( k \) consecutive digits in this sequence, and Andrey underlined other \( k \) consecutive digits in this sequence. It turned out that the \( k \)-digit numbers underlined by the boys are equal. For what largest \( k \)... | Answer: $k=7$.
Solution. An example of a seven-digit number that satisfies the condition of the problem is 2229223. Indeed, Kostya could underline the fragment «2229223» at the junction of the numbers 2229 and 2230, and Andrey - at the junction of the numbers 9222 and 9223.
Now we will show that it is impossible to c... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,278 |
6. Each of the $n$ chemists in the laboratory has 1 kg of a certain chemical reagent (different chemists have different reagents). The chemists send packages to each other. No two packages can be sent simultaneously. Upon receiving a package, a chemist immediately adds all its contents to their laboratory. When a chemi... | Solution. If there are only two chemists, the statement is obvious: the chemist who first sent the parcel will then receive the reagent of the second one along with his own. Suppose the statement is proven for $n-1$ chemists, and let's check it for $n$ chemists. Let the last of the parcels be sent by chemist $A$ to che... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 8,279 |
1. Natural numbers $x$ and $y$ are not divisible by 59, and the number $3x + 28y$ is divisible by 59. Prove that $5x + 16y$ is not divisible by 59. | Solution. Suppose that $5 x+16 y$ is divisible by 59. Then the number
$$
3 \cdot(3 x+28 y)+10 \cdot(5 x+16 y)=59 x+244 y
$$
is also divisible by 59. Therefore, $244 y$ is divisible by 59. Since 244 and 59 are coprime, $y$ must be divisible by 59, which contradicts the condition.
We could have chosen other equations,... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,280 |
2. In a bag, there are 21 regular dice, each with numbers from 1 to 6 on their faces. All the dice are numbered. The audience member takes 3 dice from the bag, shows them to the magician, and puts one die in their pocket. The magician then places the two remaining dice on the table. The audience member writes down the ... | Solution. The spectator tells the second magician two numbers, each from 1 to 6, and the order in which the numbers are given carries no information. There are 21 such pairs of numbers:
| 11 | 12 | 13 | 14 | 15 | 16 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| | 22 | 23 | 24 | 25 | 26 |
| | | 33 | 34 | 35 | 36 |... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,281 |
3. Is it possible to arrange 8 different numbers from the range 1 to 220 at the vertices of a cube such that numbers in adjacent vertices have a common divisor greater than 1, and numbers in non-adjacent vertices do not have a common divisor greater than 1? | Answer: No.
Solution. For any two numbers connected by an edge, we write on this edge their common prime divisor (any one, if there is a choice). Then different prime numbers will be written on different edges. Indeed, if the number $p$ were written on two edges, then any two numbers at the vertices of these edges wou... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,282 |
4. In the maze diagram in Fig. 1, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, d3, a5.
## Solution.
Fig. 2: Parts of the maze
+8 \cdot(5 x+16 y)=61 x+230 y
$$
is also divisible by 61. Therefore, $230 y$ is divisible by 61. Since 230 and 61 are coprime, $y$ must be divisible by 61, which contradicts the condition.
We could have chosen other equations, ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,285 |
2. In the box, there are 23 ordinary dice, with numbers from 1 to 6 written on each face. The viewer takes several dice from the box (they can even take all of them, but at least three dice are mandatory), shows them to the magician, gives two dice to the magician, and puts the remaining dice in their pocket. The magic... | Solution. The spectator tells the second magician two numbers, each from 1 to 6, and the order in which the numbers are given carries no information. There are 21 such pairs of numbers:
| 11 | 12 | 13 | 14 | 15 | 16 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| | 22 | 23 | 24 | 25 | 26 |
| | | 33 | 34 | 35 | 36 |... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,286 |
3. Is it possible to arrange 8 different odd numbers from the range 1 to 600 at the vertices of a cube such that numbers in adjacent vertices have a common divisor greater than 1, while numbers in non-adjacent vertices do not? | Answer: No.
Solution. For any two numbers connected by an edge, we write on this edge their common prime divisor (any one, if there is a choice). Then, different prime numbers will be written on different edges. Indeed, if the number $p$ were written on two edges, then any two numbers at the vertices of these edges wo... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,287 |
4. In the maze diagram in Fig. 7, each segment (link) is a corridor, and each circle is a small room. In some rooms, there are beacons that hum - each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 8): part $\mathcal{K}$ is room $K$, the long dead-end corridor that exits from it to the left, and the dead-en... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,288 |
5. Numbers $0,1,2,3 \ldots$ are placed at the nodes of a grid in a spiral pattern. Then, in the center of each cell, the sum of the numbers at its nodes is written (see Fig. 10). Is it true that in the centers of the cells, numbers divisible by 68 will appear infinitely many times? Is it true that the numbers in the ce... | Answer: Yes, it is correct.
Solution. The cells of the plane form a spiral consisting of straight "corridors" that turn at right angles.
Fig. 11: Moving along the spiral
=77 x+352 y$ is divisible by 67, then $10 x+17 y=$ $77 x+352 y-67 x-5 \cdot 67 y$ is also divisible by 67. Therefore, the number one greater than this is not divisible by 67. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 8,290 |
2. In a safe, there are 20 wallets, each containing coins of 1, 2, 5, and 10 rubles. The wallets are numbered. The audience member takes 4 wallets from the safe, shows them to the magician, and puts one wallet in their pocket. The magician takes one coin from each of the three remaining wallets, gives them to the audie... | Solution. The second magician sees the denominations of the three coins lying on the table, but neither their order nor their position on the table, nor the sides of the coins (heads or tails) carry any information, since the coins were laid out by the audience member. In total, there are 20 triplets of coins:
| 111 |... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,291 |
3. Is it possible to arrange 8 different numbers, not divisible by 13, from the range 1 to 245 at the vertices of a cube such that numbers in adjacent vertices have a common divisor greater than 1, while numbers in non-adjacent vertices do not? | Answer: No.
Solution. For any two numbers connected by an edge, we write on this edge their common prime divisor (any one, if there is a choice). Then different prime numbers will be written on different edges. Indeed, if the number $p$ were written on two edges, then any two numbers at the vertices of these edges wou... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,292 |
4. In the maze diagram in Fig. 13, each segment (link) is a corridor, and each circle is a small room. Some rooms have beacons that hum, each with its own voice. When in any room, the robot hears the signal from each beacon and determines the distance to it by the attenuation, i.e., the number of links on the shortest ... | Answer: the minimum number of beacons is 3, for example, they can be placed in rooms a1, b3, d4.
## Solution.
b) Estimation. We will prove that two beacons are not enough. Consider three parts of our maze (Fig. 14): part $\mathcal{K}$ is room $K$ and the dead-end corridors that lead from it to the right and down, par... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 8,293 |
1. In each cell of a $100 \times 100$ table, a natural number is written. In each row, there are at least 10 different numbers, and in any four consecutive rows, there are no more than 15 different numbers. What is the maximum number of different numbers that can be in the table? | Answer: 175
Solution. In one line, there are no less than 10 different numbers, so in the next three lines together, there appear no more than 5 new numbers. Therefore, the first four lines contain no more than 15 different numbers, and each of the following three lines adds no more than 5 new numbers, making the tota... | 175 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,295 |
2. Find all primes $p$ for which the numbers $p+1$ and $p^{2}+1$ are double the squares of natural numbers.
# | # Answer: $p=7$
First solution. Let $p+1=2 x^{2}$ and $p^{2}+1=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Therefore, either $y-x$ or $y+x$ is divisible by $p$. From the inequality $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. No... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,296 |
3. The sum of positive numbers $a, b, c$ and $d$ does not exceed 4. Find the maximum value of the expression
$$
\sqrt[4]{a(b+2 c)}+\sqrt[4]{b(c+2 d)}+\sqrt[4]{c(d+2 a)}+\sqrt[4]{d(a+2 b)}
$$ | Answer: $4 \sqrt[4]{3}$
First solution. By the inequality of means for four numbers, we have
$$
\sqrt[4]{a(b+2 c)}=\frac{\sqrt[4]{3 a(b+2 c) \cdot 3 \cdot 3}}{\sqrt[4]{3^{3}}} \leqslant \frac{1}{\sqrt[4]{3^{3}}} \cdot \frac{3 a+b+2 c+6}{4}
$$
Summing this inequality with three similar ones, we get
$$
\begin{aligned... | 4\sqrt[4]{3} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,297 |
4. Point $O$ is the center of the circumscribed circle of an acute-angled triangle $ABC$, and $H$ is the intersection point of its altitudes. It turns out that the line $OH$ is parallel to side $BC$. On the plane, a point $K$ is marked such that $ABHK$ is a parallelogram. Segments $OK$ and $AC$ intersect at point $L$. ... | Answer: $1: 1$
First solution. Let $D$ be the foot of the altitude from point $A$, and point $E$ be the intersection of this altitude with the circumcircle of triangle $ABC$, point $A'$ be ... | 1:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,298 |
5. In a class, there are $n$ boys and $n$ girls ($n \geqslant 3$). They sat around a round table such that no two boys and no two girls sit next to each other. The teacher has $2 n$ cards, on which the numbers $1,2,3, \ldots, 2 n$ are written, each one appearing once. He distributed one card to each student such that t... | Answer: for odd $n$
Solution. According to the problem, the boys received cards with numbers from 1 to $n$, and the girls received cards with numbers from $n+1$ to $2 n$. Suppose that all the girls ended up with the number $m$ written on their sheets. Then the sum of all the numbers on the sheets is $m n$, which can a... | n | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,299 |
1. In each cell of a $75 \times 75$ table, a natural number is written. In each row, there are at least 15 different numbers, and in any three consecutive rows, there are no more than 25 different numbers. What is the maximum number of different numbers that can be in the table? | # Answer: 385
Solution. In one line, there are no less than 15 different numbers, so in the next two lines together, there appear no more than 10 new numbers. Therefore, the first three lines contain no more than 25 different numbers, and each of the following two lines adds no more than 10 new numbers, making the tot... | 385 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 8,300 |
2. Find all primes $p$ for which the numbers $p+7$ and $p^{2}+7$ are double the squares of natural numbers. | Answer: $p=11$
First solution. Let $p+7=2 x^{2}$ and $p^{2}+7=2 y^{2}$, then $2\left(y^{2}-x^{2}\right)=p(p-1)$. Since $p-$ is odd, $p \geqslant 3$ and $2 p^{2}>p^{2}+7$. Therefore, $xy-x \geqslant p$ and, thus, $2(y-x)(y+x) \geqslant 2 p^{2}>p(p-1)$, which is impossible. Therefore, $y+x$ is divisible by $p$. Note tha... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 8,301 |
3. The sum of positive numbers $a, b, c$ and $d$ does not exceed 4. Find the maximum value of the expression
$$
\sqrt[4]{a^{2}+3 a b}+\sqrt[4]{b^{2}+3 b c}+\sqrt[4]{c^{2}+3 c d}+\sqrt[4]{d^{2}+3 d a}
$$ | Answer: $4 \sqrt{2}$
First solution. By the inequality of means for four numbers, we have
$$
\sqrt[4]{a^{2}+3 a b}=\frac{\sqrt[4]{4 a \cdot(a+3 b) \cdot 4 \cdot 4}}{2 \sqrt{2}} \leqslant \frac{1}{2 \sqrt{2}} \cdot \frac{4 a+(a+3 b)+4+4}{4}
$$
Summing this inequality with three similar ones, we get
$$
\begin{aligned... | 4\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 8,302 |
4. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are dropped. A point $T$ is chosen on the plane such that the lines $T A$ and $T B$ are tangents to the circumcircle of triangle $A B C$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the li... | Answer: $90^{\circ}$
Solution. Let $\angle B A C=\alpha$ and $\angle A C B=\gamma$. Triangles $A O T$ and $B O T$ are congruent by three sides, and the inscribed angle $\angle A C B$ subten... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 8,303 |
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