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Problem 5. There is a deck of 1024 cards, each with a different set of digits from 0 to 9, and all sets are distinct (including an empty card). We will call a set of cards complete if each digit from 0 to 9 appears exactly once on them.
Find all natural $k$ for which there exists a set of $k$ cards with the following condition: among them, no complete set can be selected, but adding any card from the deck violates this condition.
|
Solution. Answer: 512.
For each card, consider another card that complements it to a complete set (for example, for the card 3679, such a card would be 012458). It is clear that all 1024 cards can be divided into 512 non-overlapping pairs of cards that complement each other to a complete set. Next, we will prove that in any desired set, there must be exactly one card from each such pair, i.e., \( k = 512 \).
From the condition, it follows that at most one card from each pair can be among the selected ones. Now we will show that at least one card from each pair must be included.
Consider a pair of complementary cards, denoted as \( A \) and \( B \). Suppose neither of them is in the selected set. According to the condition, adding any card from the deck will result in a complete set. Adding card \( A \) to the set, we will find several cards that complement \( A \) to a complete set, i.e., all the digits on these cards simply match the set of digits on card \( B \). Similarly, adding card \( B \), we will find several cards from the set whose digits match the set of digits on card \( A \). Then, combining all these cards (which match the sets on cards \( A \) and \( B \)), we will get a complete set, which is a contradiction.
Now let's provide an example of a possible set for \( k = 512 \). Choose all cards that do not have the digit 9; there are exactly 512 such cards. It is clear that there is no complete set among them (the digit 9 is not present anywhere), and for each unselected card, its complement is among the selected ones, i.e., adding it will result in a complete set from these two cards.
## Criteria
## Points for the following achievements are summed up:
4 6. Proved that \( k \geqslant 512 \) (or that in each pair of cards that complement each other to a complete set, at least one card must be selected).
1 6. Proved that \( k \leqslant 512 \) (or that in each pair of cards that complement each other to a complete set, no more than one card should be selected).
1 6. Provided a correct example of a set of 512 cards.
1 6. Provided justification that the example set of 512 cards meets the conditions of the problem.
0 6. Only the correct answer.
|
512
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. In a certain company, there are 100 shareholders, and any 66 of them collectively own at least $50\%$ of the company's shares. What is the largest percentage of all shares that one shareholder can own? (The percentage of shares in the company owned by a shareholder can be non-integer.)
|
Answer: $25 \%$.
Solution. Consider any shareholder $A$. Divide the other 99 shareholders into three groups $B, C, D$ of 33 shareholders each. By the condition, $B$ and $C$ together have at least $50 \%$ of the company's shares, similarly for $C$ and $D$, and for $B$ and $D$. Adding all this up and dividing by two, we get that $B, C, D$ together have at least $((50+50+50) : 2) \% = 75 \%$ of the shares, so $A$ has no more than the remaining $25 \%$ of the shares.
Moreover, $A$ can have exactly $25 \%$ of the shares if all other shareholders have an equal number of shares, each with $\frac{75}{99} \% = \frac{25}{33} \%$. Clearly, in this case, any 66 shareholders of the company together have at least $66 \cdot \frac{25}{33} \% = 50 \%$ of the shares.
## Criteria
The largest applicable criterion is used:
The following criteria are summed:
5 p. It is proven that any shareholder has no more than $25 \%$ of the shares.
2 p. It is proven that some shareholder can have exactly $25 \%$ of the shares.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. In triangle $A B C$, the angle at vertex $B$ is $120^{\circ}$, point $M$ is the midpoint of side $A C$. Points $E$ and $F$ are chosen on sides $A B$ and $B C$ respectively such that $A E=E F=F C$. Find $\angle E M F$.
|
Answer: $90^{\circ}$.
Solution. Note that the sum of angles $A$ and $C$ is $60^{\circ}$. Let $A E=E F=F C=u$.
On the line $E M$, place a point $G$ such that point $M$ is the midpoint of segment $E G$. Triangles $A M E$ and $C M G$ are equal by two sides and the angle between them, so $\angle M C G=\angle M A E=\angle A$ and $C G=A E=u$.
Note that $\angle F C G=\angle F C M+\angle M C G=\angle C+\angle A=60^{\circ}$.
Since $F C=u=C G$, triangle $F C G$ is equilateral, and $F G=u$. Therefore, triangle $E F G$ is isosceles, $E F=F G=u$. Its median $F M$ is also an altitude, so $\angle E M F=90^{\circ}$.
## Criteria
One of the largest suitable criteria is used:
7 p. Any complete solution to the problem.
1 p. A point symmetric to $E$ or $G$ relative to $M$ is considered, but there is no further progress.
0 p. Only the correct answer is given.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8.3. How many lines exist that pass through the point $(0,2019)$ and intersect the parabola $y=x^{2}$ at two points with integer coordinates on the $y$-axis?
Answer: 9.
|
Solution. A vertical line obviously does not fit. All lines, different from the vertical one and passing through the point $(0,2019)$, are given by the equation $y=k x+2019$ for some $k$. Let such a line intersect the parabola at points $\left(a, a^{2}\right)$ and $\left(b, b^{2}\right)$, where $a^{2}$ and $b^{2}$ are integers. Without loss of generality, we will assume $a<b$. We write the system of equations
$$
\left\{\begin{array}{l}
a^{2}=k a+2019 \\
b^{2}=k b+2019
\end{array}\right.
$$
Subtracting one equation from the other, we get $a^{2}-b^{2}=k(a-b)$, and since $a \neq b$, we conclude that $k=a+b$. Substituting the obtained $k$ into any of the equations of the system, we arrive at the relation $-a b=2019$.
Note that this equality is equivalent to the fact that the points $\left(a, a^{2}\right)$ and $\left(b, b^{2}\right)$ lie on the same line. This can be verified by substituting $k=a+b$ and $2019=-a b$ into the system above.
Squaring the equation, we get $a^{2} b^{2}=2019^{2}$, where $a^{2}$ and $b^{2}$ are non-negative integers (but $a$ and $b$ are not necessarily integers!). Considering the prime factorization $2019^{2}=3^{2} \cdot 673^{2}$, we have 9 options for decomposing the number $2019^{2}$ into the product $a^{2} \cdot b^{2}$ (taking into account the order of the factors). Since $a$ and $b$ have different signs, and $a<b$, each such option corresponds to a unique solution $a=-\sqrt{a^{2}}, b=\sqrt{b^{2}}$ and, accordingly, one sought line.
Remark. A similar idea underlies the well-known geometric construction - the Matiyasevich-Steckin sieve.
## Criteria
The highest suitable criterion is used:
76 . Any correct solution to the problem.
3 6. The equation equivalent to the equality $-a b=2019$ from the solution is obtained, but the further calculation of the number of lines is done incorrectly for some reason.
2 6. The equation equivalent to the equality $k=a+b$ from the solution is obtained.
## 16 . The correct answer is given.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A cube with a side of 5 is made up of 125 smaller cubes with a side of 1. How many small cubes does a plane perpendicular to one of the cube's diagonals and passing through its midpoint intersect?
|
Answer: 55.
Let's introduce a coordinate system such that the cube is located in the first octant (the set of points with non-negative coordinates) and the mentioned diagonal extends from the origin $O$. The midpoint of the cube's diagonal has coordinates $(5 / 2, 5 / 2, 5 / 2)$, so the specified plane is given by the equation $x + y + z = 15 / 2$. Consider one of the 125 small cubes. Let the vertex closest to point $O$ have coordinates $(k, m, n)$, where the integers $k, m, n$ satisfy the inequalities $0 \leqslant k, m, n \leqslant 4$. The vertex farthest from point $O$ has coordinates $(k+1, m+1, n+1)$. Thus, the small cube intersects the plane if and only if $k + m + n \leqslant 15 / 2 \leqslant k + 1 + m + 1 + n + 1$, which is equivalent to the condition $k + m + n \in (9 / 2 ; 15 / 2)$. Considering the integer nature of the sum $k + m + n$, we get only three cases: $k + m + n = 5$, $k + m + n = 6$, or $k + m + n = 7$. Let's calculate the number of small cubes for each of the three cases.
(i) $k + m + n = 5$. List the ways to split 5 into three summands and indicate the number of different permutations of these summands: $0 + 1 + 4$ (6 solutions), $0 + 2 + 3$ (6 solutions), $1 + 1 + 3$ (3 solutions), $1 + 2 + 2$ (3 solutions) - a total of 18.
(ii) $k + m + n = 6. 0 + 2 + 4$ (6 solutions), $0 + 3 + 3$ (3 solutions), $1 + 1 + 4$ (3 solutions), $1 + 2 + 3$ (6 solutions), $2 + 2 + 2$ (1 solution) - a total of 19.
(iii) $k + m + n = 7$. All such small cubes are symmetric to those with the condition $k + m + n = 5$ relative to the center of the cube - a total of 18.
In total: $18 + 19 + 18 = 55$.
t The intersected small cubes are divided into three layers, and the number of small cubes in at least two layers is correctly calculated with justification. 6 points
± The intersected small cubes are divided into three layers, and the number of small cubes in at least one layer is correctly calculated with justification. 5 points
± The intersected small cubes are divided into three layers, and the number of small cubes in at least one layer is correctly calculated, but the justification is not strict or is missing. 4 points
Ғ The intersected small cubes are divided into three layers, but there is no further progress. 3 points
- The correct answer is given without justification. 2 points
- The equation of the plane is written. 1 point
|
55
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 1. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all possible options and prove that there are no others.
|
Answer: 30.
Solution. Consider any arbitrary quartet of consecutive people. If there were at least 3 knights in it, the leftmost of them would definitely lie, which is impossible. If there were at least 3 liars, the leftmost of them would definitely tell the truth, which is also impossible. Therefore, in every quartet of consecutive people, there are exactly 2 knights and 2 liars. Dividing all the people into 15 such quartets, we get that there are $15 \cdot 2=30$ knights.
Another solution. Notice that at the table, there cannot be only knights or only liars. Then there will be a pair of knight-liar sitting next to each other in this exact order. Consider 4 cases of who can sit to the right of this pair.
Case 1. Two knights sit to the right.
Then we get the sequence
$$
\text { knight-liar-knight-knight, }
$$
which contradicts the condition, as the first knight lied.
Case 2. Two liars sit to the right.
Then we get the sequence
knight-liar-liar-liar.
Whoever sits immediately after this sequence, the first liar will tell the truth, which contradicts the condition.
Case 3. A knight-liar pair sits to the right in this exact order.
Then we get the sequence
$$
\text { knight-liar-knight-liar. }
$$
Consider the first liar. He lies, so among the next three, there must be at least two knights. From this, we uniquely get
$$
\text { knight-liar-knight-liar-knight. }
$$
Now consider the second knight. He tells the truth, so the sixth person in the sequence will be a liar:
$$
\text { knight-liar-knight-liar-knight-liar. }
$$
Continuing these considerations, we get an alternating arrangement where there are an equal number of knights and liars.
Case 4. A liar-knight pair sits to the right in this exact order.
Then we get the sequence
$$
\text { knight-liar-liar-knight. }
$$
Consider the first liar. He lies, so among the next three, there must be at least two knights. From this, we uniquely get
$$
\text { knight-liar-liar-knight-knight. }
$$
Now consider the second knight. He tells the truth, so the sixth and seventh people in the sequence will be liars
knight-liar-liar-knight-knight-liar-liar.
Continuing these considerations, we get an arrangement where there are an equal number of knights and liars.
## Criteria
The highest applicable criterion is used:
7 p. Any complete solution to the problem.
7 6. The problem is correctly solved for the case where both knights and liars are present at the table.
4 p. Both possible cases of arranging knights and liars are considered in the solution, but there is no justification for why there are no other arrangements.
2 p. One of the two possible cases of arranging knights and liars is considered in the solution, but there is no justification for why there are no other arrangements.
1 6. Only the correct answer is present.
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On an island, there live knights who always tell the truth and liars who always lie. The population of the island is 1000 people, distributed across 10 villages (with no fewer than two people in each village). One day, every islander claimed that all their fellow villagers are liars. How many liars live on the island? (Two residents are fellow villagers if they live in the same village.)
|
Answer: 990.
In one village, at least two knights cannot live, because otherwise the knights would lie. Also, in the village, they cannot all be liars, since then these liars would tell the truth. Therefore, in each village there is exactly one knight, and there are 10 knights in total, and 990 liars.
|
990
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Petya climbed up a moving upward escalator, counting 75 steps, and then descended the same escalator (i.e., moving against the direction of the escalator), counting 150 steps. During the descent, Petya walked three times faster than during the ascent. How many steps are there on the stopped escalator?
|
Answer: 120.
For convenience, let's introduce a unit of time during which Petya took one step while ascending the escalator. We will measure all speeds in steps per unit of time. Petya's speed while ascending is 1 step per unit of time, and his speed while descending is 3 steps per unit of time. Let the speed of the escalator be $x$. The absolute speed of Petya was $1+x$ while ascending and $3-x$ while descending. Petya spent 75 units of time ascending and 50 units of time descending (he took steps three times faster while descending). The distance covered while ascending and descending was the same - the length of the escalator. Therefore, $75 \cdot(1+x)=50 \cdot(3-x)$, from which $125 x=75, x=0.6$.
The length of the escalator is $75 \cdot(1+0.6)=120$ steps. This is the number of steps on the stopped escalator.
|
120
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. For a convex quadrilateral $A B C D$, it is known that $A B=B C=C A=$ $C D, \angle A C D=10^{\circ}$. A circle $\omega$ is circumscribed around triangle $B C D$ with center $O$. Line $D A$ intersects circle $\omega$ at points $D$ and $E$. Find the measure of angle $E O A$, express your answer in degrees.
|
Answer: 65.
Fig. 2: to the solution of problem 3.
Fig. 2. Since triangle $A D C$ is isosceles, and we know the angle at its vertex, the angles at its base are $\angle D A C = \angle C D A = 85^{\circ}$. Therefore, $\angle B A E = 180^{\circ} - \angle D A C - \angle C A B = 35^{\circ}$. Angle $E B C$ complements angle $E D C$ to $180^{\circ}$ and is thus $95^{\circ}$, from which $\angle E B A = 35^{\circ}$. Consequently, triangle $A E B$ is isosceles, and line $C E$ is the perpendicular bisector of $A B$. But then $C B$ is the angle bisector of $\angle B C A$, so $\angle B C E = 30^{\circ}$, and the corresponding central angle $\angle B O E = 60^{\circ}$.
On the other hand, $\angle C O E = 2 \angle C D E = 170^{\circ}$, from which $\angle C O B = 110^{\circ}$.
Finally, from the symmetry of triangle $A B C$ with respect to $A O$, it is easy to see that the desired angle $\angle E O A$ complements to $180^{\circ}$ the sum of angle $\angle B O E$ and half of angle $\angle C O B$, i.e., $\angle E O A = 180^{\circ} - \angle B O E - \frac{1}{2} \angle C O B = 65^{\circ}$.
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Anya writes a natural number, and Boris replaces one of its digits with a digit differing by 1. What is the smallest number Anya should write to ensure that the resulting number is guaranteed to be divisible by 11?
|
Answer: 909090909.
According to the divisibility rule for 11, the remainder of a number when divided by 11 is the same as the remainder of the alternating sum of its digits when divided by 11. Therefore, changing a digit by 1 also changes the remainder by 1 when divided by 11. This means the original number gives a remainder of 10 or 1 when divided by 11.
If the original number gives a remainder of 10 when divided by 11, then any change in its digits should increase the remainder of the alternating sum of its digits by 1 when divided by 11. Thus, the digits in odd positions (counting from the end) can only be increased by 1, and the digits in even positions can only be decreased by 1, i.e., the number has the form ...909090. The smallest number of this form that gives a remainder of 10 when divided by 11 is 9090909090.
If the original number gives a remainder of 1 when divided by 11, then, similarly, it has the form ...90909, and the smallest number that fits the condition is 909090909.
|
909090909
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles.
|
Answer: 28.
Fig. 3: Solution to problem 5.
Example: The arrangement of 28 tiles can be seen in Fig. 3.
Estimation. Cells not covered by tiles will be called empty. First, let's prove that no two empty cells can be adjacent by side.
Assume the opposite: some two empty cells are adjacent. Consider the figure formed by the adjacent empty cells. Note that the figure cannot have concave angles (see the figure), because the tile covering the cell marked with a cross can be moved vertically if it is vertical, or horizontally if it is horizontal.
If the figure has no concave angles, then it is a rectangle. Denote its dimensions as $a \times b$, where $a \geqslant b$. The situation $a \geqslant 3$ is impossible, because at least one side of the rectangle of length $a$ must be adjacent to tiles (not the edge of the box), but then the tile adjacent to a non-edge cell of the considered side can be pushed into the rectangle. A short enumeration can prove that rectangles $2 \times 2$ and $2 \times 1$ of empty cells are also impossible.
Thus, all empty cells are isolated. No empty cell can be at the edge of the board, otherwise one of the tiles adjacent to this empty cell can be pushed onto this empty cell. Tiles adjacent to an empty cell can be placed in only two ways, as shown in the figure.
From this, it follows, in particular, that two empty cells cannot be in the same row or column at a distance of two, and they also cannot be adjacent diagonally.
Now we are ready to complete the estimation. Suppose that it is possible to use fewer than 28 tiles. Then more than 8 empty cells will remain. Since empty cells cannot be adjacent to the edge of the box, they will be concentrated in the central $6 \times 6$ square. Divide the $6 \times 6$ square into four $3 \times 3$ squares. One of the squares must contain at least three empty cells. However, a short enumeration of their potential placements shows that this is impossible. Contradiction.
|
28
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. On the parade ground, 2018 soldiers are lined up in one row. The commander can order either all soldiers standing in even positions or all soldiers standing in odd positions to leave the formation. After this order, the remaining soldiers close up into one row. In how many ways can the commander issue a series of 8 orders so that exactly 7 soldiers remain in the formation?
|
Answer: 30.
Let's add 30 imaginary people to the end of the line. We will number all the people in the line from 0 to 2047. We will write all these numbers in binary using 11 digits. This will result in sequences from 00000000000 to 11111111111. The imaginary soldiers correspond to numbers from 11111100010 to 11111111111 (from 2018 to 2047 in binary notation).
Notice that soldiers in even positions have 1 as the last digit in their binary representation, while soldiers in odd positions have 0. This means that after the order is executed, only soldiers with a fixed last digit remain in the line. After the next order, only soldiers with a fixed second-to-last digit remain, and so on.
After a series of 8 orders, the soldiers remaining in the line will have binary numbers of the form $\overline{x x a b c d e f g h}$, where the symbols $x$ represent digits that can take any value; $a, b, c, d, e, f, g, h$ are the digits fixed in the first eight orders. Clearly, soldiers with $\overline{x x x}=000,001,010,011,100,101,110$ were indeed present in the initial line, as the minimum number of an imaginary soldier starts with 111. We are interested in situations where the soldier with the number $\overline{111 a b c d e f g h}$ is imaginary (otherwise, 8 people remain). This happens exactly in 30 cases: from 11111100010 to 11111111111. In total, there are 30 possible groups of 7 people.
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. When a five-digit number is multiplied by 9, the result is a number composed of the same digits but in reverse order. Find the original number.
|
Answer: 10989.
Let $\overline{a b c d e}$ be the original number. The condition is written as the equation $9 \cdot \overline{a b c d e}=\overline{e d c b a}$. Note that $a=1$, because if $a \geqslant 2$, then $9 \cdot \overline{a b c d e} \geqslant 9 \cdot 20000>100000>\overline{e d c b a}$.
We have $9 \cdot \overline{1 b c d e}=\overline{e d c b 1}$. From this, the last digit $e=9$ is uniquely determined.
Working with $9 \cdot \overline{1 b c d 9}=\overline{9 d c b 1}$. Note that if $b \geqslant 2$, then $9 \cdot \overline{1 b c d 9}>$ $9 \cdot 12000>100000>\overline{e d c b a}$. Therefore, $b=0$ or $b=1$. Let's consider these two cases:
If $b=0$, the equation reduces to $9 \cdot \overline{10 c d 9}=\overline{9 d c 01}$. Considering the second-to-last digit, we conclude that $d=8$. By trying the possible values of the digit $c$, we find the unique solution $\overline{a b c d e}=10989$.
If $b=1$, the equation is rewritten as $9 \cdot \overline{11 c d 9}=\overline{9 d c 11}$. Analyzing the second-to-last digit, we get $d=7$. But $9 \cdot \overline{11 c 79}>9 \cdot 11000=99000>\overline{97 c 11}$, which is a contradiction. Therefore, there are no solutions in the case $b=1$.
The only number that satisfies the condition of the problem is 10989.
|
10989
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a pair of numbers magical if the numbers in the pair add up to a multiple of 7. What is the maximum number of magical pairs of adjacent numbers that can be obtained by writing down all the numbers from 1 to 30 in a row in some order?
|
Answer: 26.
Example: $1,6,8,13,15,20,22,27,29,2,5,9,12,16,19,23,26,30,3,4$, $10,11,17,18,24,25,7,14,21,28$. It is not hard to see that in this sequence, only the pairs $(29,2),(30,3),(25,7)$ are not magical.
Evaluation: Suppose it is possible to make no less than 27 pairs magical. Then there would be no more than two non-magical pairs. Let's color all numbers from 1 to 30 in four colors depending on the remainder when divided by 7: numbers with remainders 1 or 6 - in red; with remainders 2 or 5 - in blue; with remainders 3 or 4 - in purple; with remainder 0 - in orange.
Note that numbers in a magical pair must be of the same color. Therefore, if there were no more than two non-magical pairs, the color would have changed no more than twice when moving along the row from left to right, i.e., there would be no more than three colors. But there are four, a contradiction.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Dima has 25 identical bricks of size $5 \times 14 \times 17$. Dima wants to build a tower from all his bricks, each time adding one more brick on top (each new brick adds 5, 14, or 17 to the current height of the tower). We will call a number $n$ constructible if Dima can build a tower of height exactly $n$. How many constructible numbers exist?
|
Answer: 98.
In essence, we need to find the number of different towers by height that can be built from the given set of bricks.
Mentally reduce the length, width, and height of each brick by 5: from $5 \times 14 \times 17$ to $0 \times 9 \times 12$. Then the total height of a potential tower will decrease by $25 \cdot 5$, and the number of different towers by height, composed of the new bricks, will coincide with the number of different towers by height from the old ones. Now reduce the dimensions of the bricks by three times: from $0 \times 9 \times 12$ to $0 \times 3 \times 4$. The sizes of potential towers will also decrease by three times, and the number of different towers by height will remain unchanged.
Thus, the original problem is equivalent to the following: how many different values can be obtained by adding the numbers $0,3,4$ in a quantity of 25? Note that if the number 3 appears at least four times in the sum, then the set $3,3,3,3$ can be replaced by the set $4,4,4,0$; with this replacement, both the sum and the number of numbers remain the same. Therefore, we can consider only those sums where the number 3 appears no more than three times.
If the number 3 appears 0 times, then there are 26 different values of the sum: $0,4,8, \ldots, 4 \cdot 25$.
If the number 3 appears 1 time, then there are 25 different values of the sum: $3,3+4,3+8, \ldots, 3+4 \cdot 24$.
If the number 3 appears 2 times, then there are 24 different values of the sum: $6,6+4,6+8, \ldots, 6+4 \cdot 23$.
If the number 3 appears 3 times, then there are 23 different values of the sum: $9,9+4,9+8, \ldots, 9+4 \cdot 22$.
The values in the first, second, third, and fourth lists give remainders when divided by 4 of $0,3,2,1$, respectively, and therefore the values in different lists are different.
In total, $26+25+24+23=98$ possible values of the tower height.
|
98
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The bottom of the box is an $8 \times 8$ table. What is the smallest non-zero number of $2 \times 1$ or $1 \times 2$ tiles that can be placed on the bottom of the box so that no tile can be moved either horizontally or vertically? Each tile must occupy exactly two cells, not occupied by other tiles.
|
Answer: 28.
Fig. 1: to the solution of problem 6.
Example: the arrangement of 28 tiles can be seen in Fig. 1.
Estimate. Cells not covered by tiles will be called empty. To begin with, let's prove that no two empty cells can be adjacent by side.
Assume the opposite: some two empty cells are adjacent. Consider the figure formed by the adjacent empty cells. Note that the figure cannot have concave angles (see the figure), because the tile covering the cell marked with a cross can be moved vertically if it is vertical, or horizontally if it is horizontal.
If the figure has no concave angles, then it is a rectangle. Let its dimensions be $a \times b$, where $a \geqslant b$. The situation $a \geqslant 3$ is impossible, because at least one side of the rectangle of length $a$ must be adjacent to tiles (not the edge of the box), but then the tile adjacent to a non-edge cell of the considered side can be pushed into the rectangle. A short enumeration can prove that rectangles $2 \times 2$ and $2 \times 1$ of empty cells are also impossible.
Thus, all empty cells are isolated. No empty cell can be at the edge of the board, otherwise one of the tiles adjacent to this empty cell can be pushed onto this empty cell. Tiles adjacent to an empty cell can be placed in only two ways, as shown in the figure.
From this, in particular, it follows that two empty cells cannot be in the same row or column at a distance of two, and they also cannot be adjacent diagonally.
Now we are ready to complete the estimate. Suppose that it is possible to manage with fewer than 28 tiles. Then more than 8 empty cells will remain. Since empty cells cannot be adjacent to the edge of the box, they will be concentrated in the central $6 \times 6$ square. Divide the $6 \times 6$ square into four $3 \times 3$ squares. One of the squares must contain at least three empty cells. But a short enumeration of their potential placements shows that this is impossible. Contradiction.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. Through the point with coordinates $(10,9)$, lines (including those parallel to the coordinate axes) have been drawn, dividing the plane into angles of $10^{\circ}$. Find the sum of the x-coordinates of the points of intersection of these lines with the line $y=101-x$.
|
Answer: 867. Solution: Shift the entire picture to the left by 1. We get a set of lines passing through the point $(9,9)$ and intersecting the line $y=100-x$. The picture will become symmetric with respect to the line $y=x$, so the sum of the abscissas on it is equal to the sum of the ordinates. Through the point $(9,9)$, 18 lines are drawn, and the line $y=100-x$ intersects 17 of them. For each point on the line $y=100-x$, the sum of the coordinates is 100, so the total sum of the abscissas and ordinates is 1700, and the sum of the abscissas is half of that, which is 850. However, on the symmetric picture, each abscissa is 1 less than the original, so the desired sum is $850+17=867$.
Criterion: No less than 1 point for the correct answer without a correct justification. 1 point for the idea of shifting the picture so that it becomes symmetric with respect to the axes. 5 points if all ideas are found, but the answer is incorrect due to an incorrect count of the number of intersection points (for example, the student counts 18 or 36 points).
|
867
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. A pair of natural numbers is called good if one of the numbers is divisible by the other. Numbers from 1 to 30 were divided into 15 pairs. What is the maximum number of good pairs that could result?
|
Solution. Answer: 13.
First, let's prove that more than 13 good pairs could not have been formed. We will call a number bad if it is prime and not less than 15. There are only four bad numbers: \(17, 19, 23, 29\). None of the other numbers from 1 to 30 can be divisible by any bad number, and the bad numbers themselves can only be divisible by one (among the others). Consequently, at most one of the bad numbers can form a good pair with one, and the remaining at least three bad numbers will "ruin" at least 2 pairs.
Now let's consider the following 13 good pairs: \((1,27), (2,4), (3,6), (5,25), (7,21)\), \((8,16), (9,18), (10,20), (11,22), (12,24), (13,26), (14,28), (15,30)\). The remaining 4 numbers can be paired, for example, as follows: \((17,19), (23,29)\). In total, we have exactly 13 good pairs.
## Criteria
One of the largest suitable criteria is used:
7 points. Any complete solution to the problem.
4 points. An example consisting of 13 good pairs is constructed.
3 points. It is proven that there are no more than 13 good pairs.
1 point. The idea of considering "large" prime numbers (17, 19, 23, 29) is present, but there is no further progress.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.1. Jerry has nine cards with digits from 1 to 9. He lays them out in a row, forming a nine-digit number. Tom writes down all 8 two-digit numbers formed by adjacent digits (for example, for the number 789456123, these numbers are $78, 89, 94, 45$, $56, 61, 12, 23$). For each two-digit number divisible by 9, Tom gives Jerry a piece of cheese. What is the maximum number of pieces of cheese Jerry can get?
|
Answer: 4.
Solution. Note that among two-digit numbers, only 18, 27, 36, 45, and numbers obtained by swapping their digits are divisible by 9 (there are also 90 and 99, but we do not have the digit 0 and only one digit 9). Thus, only four pairs of digits from the available ones can form a number divisible by 9. To get an example, we need to arrange all these pairs in any order:
$$
182736459
$$
## Criteria
The highest applicable criterion is used:
## 76 . Any correct solution to the problem.
3 6. It is proven that it is impossible to get 5 two-digit numbers divisible by 9.
## 3 6. An example with 4 two-digit numbers divisible by 9 is provided.
## 16 . The correct answer is given.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6.3. A country has the shape of a square and is divided into 25 identical square counties. In each county, either a knight-count, who always tells the truth, or a liar-count, who always lies, rules. One day, each count said: "Among my neighbors, there are an equal number of knights and liars." What is the maximum number of knights that could have been? (Counts are neighbors if their counties share a common side.)
|
Answer: 8.
Solution. First, note that counties with exactly three neighboring counties must be governed by lying counts (Fig. ??). Therefore, the corner counties are also governed by liars, since both of their neighbors are liars.
| | $L$ | $L$ | $L$ | |
| :--- | :--- | :--- | :--- | :--- |
| $L$ | | | | $L$ |
| $L$ | | | | $L$ |
| $L$ | | | | $L$ |
| | $L$ | $L$ | $L$ | |
Fig. 1: to the solution of problem $? ?$
It remains to figure out the central $3 \times 3$ square. All cells cannot be occupied by knights, since then the central knight would have all neighbors as knights. Therefore, there are no more than 8 knights. An example with eight knights is shown in Fig. ??.
| | | | | |
| :--- | :--- | :--- | :--- | :--- |
| | $\mathrm{P}$ | $\mathrm{P}$ | $\mathrm{P}$ | |
| | $\mathrm{P}$ | | $\mathrm{P}$ | |
| | $\mathrm{P}$ | $\mathrm{P}$ | $\mathrm{P}$ | |
| | | | | |
Fig. 2: to the solution of problem $?$ ?
## Criteria
Any correct solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed:
4 6. It is proven that knights can only be in the central $3 \times 3$ square.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. In a $3 \times 3$ table, natural numbers (not necessarily distinct) are placed such that the sums in all rows and columns are different. What is the minimum value that the sum of the numbers in the table can take?
|
Answer: 17.
Solution. Note that in each row and column, the sum of the numbers is no less than 3. Then, the doubled sum of all numbers in the table, which is equal to the sum of the sums of the numbers in the rows and columns, is no less than \(3+4+\ldots+8=33\), so the simple sum of the numbers in the table is no less than 17.
Example of a table with a sum of 17:
| 1 | 1 | 1 |
| :--- | :--- | :--- |
| 1 | 2 | 2 |
| 2 | 3 | 4 |
## Criteria
The largest suitable criterion is used:
7 p. Any complete solution to the problem.
4 p. It is proven that the sum of the numbers in the table is no less than 17.
3 p. An example of a table with a sum of 17 is provided.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex?
|
Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, continuing this sequence in a cycle will include all numbers of the form $x_{3 k+1}$. Similar sequences can be constructed starting from the numbers $x_{2}$ and $x_{3}$. Thus, the numbers at the vertices that have the same remainder when divided by 3 are equal.
Since each set of nine consecutive numbers contains three sets of vertices with different remainders when divided by 3, the sum of the numbers in each such set is 100. Note that 19 gives a remainder of 1 when divided by 3, 20 gives a remainder of 2, and 2019 gives a remainder of 0, so the number at vertex 2019 is $100 - 19 - 20 = 61$.
## Criteria
## The largest suitable criterion is used:
## 7 p. Any complete solution to the problem.
4 p. It is proven that numbers that are 2 apart are equal.
1 p. It is noted that numbers that are 8 apart are equal.
|
61
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Polycarp has 2 boxes, the first of which contains $n$ coins, and the second is empty. In one move, he can either transfer one coin from the first box to the second, or remove exactly $k$ coins from the first box, where $k$ is the number of coins in the second box. For which $n$ can Polycarp make the first box empty in no more than 10 moves?
Answer: for $n$ from 0 to 30 inclusive.
|
Solution. Let's call a move where one coin is moved from the first box to the second box a move of the first type, and a move where coins are removed from the first box a move of the second type. Suppose a total of $x$ moves of the first type and $y$ moves of the second type were made. Then, the second box contains no more than $x$ coins. Therefore, when performing moves of the second type, we will remove no more than $x y$ coins from the first box, meaning we will remove no more than $x + x y$ coins in total. Since $x + y \leqslant 10$, then
$$
x + x y \leqslant x + x(10 - x) = -x^2 + 11x
$$
For integer values of $x$, the maximum of this expression is achieved at points $x = 5$ and $x = 6$ and is equal to $5 \cdot 6 = 30$. Thus, $n \leqslant 30$. We will prove that any $n \leqslant 30$ works.
If we make 5 moves of the first type and then 5 moves of the second type, we will remove exactly 30 coins from the first box. Write down the sequence of 1s and 2s, where 1 represents a move of the first type and 2 represents a move of the second type. Notice that if in this sequence we swap a neighboring pair 12 to a pair 21, the total number of coins removed from the first box will decrease by 1. We can do this until all 2s are on the left and all 1s are on the right. Then we will replace 1 with 2, and on each step, the number of coins removed using such a sequence of operations will also decrease by 1. Thus, for any $n$ from 0 to 30, we can provide the required sequence of operations.
## Criteria
The following single most appropriate criterion is used:
7 6. Any complete solution to the problem.
If in the solution $n$ is considered a natural number, and the number 0 is not mentioned in the answer, the score is not reduced.
4 b. It is proven that $n$ does not exceed 30, but it is not explained why all smaller $n$ also work.
3 6. All examples for $n$ from 1 to 30 are described.
For the following error in a solution that otherwise fits one of the criteria above, the score is reduced:
-1 6. The maximum value of the expression $-x^2 + 11x$ for integer $x$ is found incorrectly due to an arithmetic error.
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. Lёnya has cards with digits from 1 to 7. How many ways are there to glue them into two three-digit numbers (one card will not be used) so that their product is divisible by 81, and their sum is divisible by 9?
|
Answer: 36.
Solution. If one of the numbers is not divisible by 9, then the other is not either, since their sum is divisible by 9. But then the product cannot be divisible by 81, a contradiction. Therefore, both numbers are divisible by 9.
Then the sum of the digits in each number is divisible by 9, and thus the total sum of the used digits is also divisible by 9. The sum of all given digits $1+2+\ldots+7$ is 28. If we discard the digit 1, the remaining sum is 27, which is divisible by 9; discarding any other digit cannot result in a sum divisible by 9. Therefore, the digits used are 2, $3,4,5,6,7$.
Note that the smallest possible sum of three of these digits is $2+3+4=9$, and the largest is $5+6+7=18$. Other sums divisible by 9 cannot be obtained; 9 and 18 can only be obtained in one way. Thus, one number consists of the digits $2,3,4$, and the other of $5,6,7$.
There are six numbers that can be formed from the digits $2,3,4$, and six numbers that can be formed from the digits $5,6,7$. We need all possible pairs of these numbers.
## Criteria
Any complete solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed:
1 point. It is proven that both numbers must be divisible by 9.
2 points. Assuming (possibly unjustified) that both numbers are divisible by 9, it is proven that the digit 1 is not used.
3 points. Assuming (possibly unjustified) that both numbers are divisible by 9 and the digit 1 is not used, it is proven that one number consists of the digits 2,3,4, and the other of $5,6,7$.
1 point. The previous is not proven but stated.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\left(2017^{4}+4\right)}
$$
|
3. Calculate the value of the expression
$$
\frac{\left(3^{4}+4\right) \cdot\left(7^{4}+4\right) \cdot\left(11^{4}+4\right) \cdot \ldots \cdot\left(2015^{4}+4\right) \cdot\left(2019^{4}+4\right)}{\left(1^{4}+4\right) \cdot\left(5^{4}+4\right) \cdot\left(9^{4}+4\right) \cdot \ldots \cdot\left(2013^{4}+4\right) \cdot\left(2017^{4}+4\right)}
$$
Answer: $2020^{2}+1=4080401$.
Let's use the formula
$$
\begin{aligned}
a^{4}+4 & =a^{4}+4 a^{2}+4-4 a^{2}=\left(a^{2}+2\right)^{2}-(2 a)^{2}= \\
& =\left(a^{2}-2 a+2\right)\left(a^{2}+2 a+2\right)=\left((a-1)^{2}+1\right)\left((a+1)^{2}+1\right)
\end{aligned}
$$
Now the expression in the numerator of the desired fraction can be written as the product
$$
\left(2^{2}+1\right) \cdot\left(4^{2}+1\right) \cdot\left(6^{2}+1\right) \cdot \ldots \cdot\left(2018^{2}+1\right) \cdot\left(2020^{2}+1\right),
$$
while the expression in the denominator is written as
$$
\left(0^{2}+1\right) \cdot\left(2^{2}+1\right) \cdot\left(4^{2}+1\right) \cdot \ldots \cdot\left(2016^{2}+1\right) \cdot\left(2018^{2}+1\right)
$$
By canceling the common factors in the numerator and the denominator, we arrive at the answer.
## Criteria
+ Correct solution - 7 points.
+ Arithmetic error with the correct idea of the solution - 6 points.
+ The solution contains the idea of factoring $a^{4}+4$, but there is no further progress - 2 points.
- The problem is reduced to some infinite product or sum - this is not considered progress (0 points).
|
4080401
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal?
|
5. The natural numbers $1,2, \ldots, 64$ are written in the cells of an $8 \times 8$ table such that for all $k=1,2,3, \ldots, 63$ the numbers $k$ and $k+1$ are in adjacent cells. What is the maximum possible value of the sum of the numbers on the main diagonal?
Answer: 432.
Estimate. We color the cells of the table in a checkerboard pattern so that the cells on the chosen main diagonal are white. Without loss of generality, we can assume that the number 1 is not above the diagonal. We find the maximum value of the smallest number that falls on the diagonal. Since adjacent numbers are in cells of different colors, and there are only 12 white cells below the diagonal, one of the numbers from 1 to 26 must fall on the diagonal. The remaining numbers on the diagonal are guaranteed to have the same parity, so their sum does not exceed the sum of the even numbers from 52 to 64. Therefore, we conclude that the sum of the numbers on the diagonal has an upper bound:
$$
26+52+54+56+58+60+62+64=432
$$
An example of a suitable arrangement of numbers is shown in Fig. 6.
| 52 | 51 | 50 | 49 | 44 | 43 | 34 | 33 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 53 | 54 | 55 | 48 | 45 | 42 | 35 | 32 |
| 6 | 7 | 56 | 47 | 46 | 41 | 36 | 31 |
| 5 | 8 | 57 | 58 | 59 | 40 | 37 | 30 |
| 4 | 9 | 16 | 17 | 60 | 39 | 38 | 29 |
| 3 | 10 | 15 | 18 | 61 | 62 | 63 | 28 |
| 2 | 11 | 14 | 19 | 22 | 23 | 64 | 27 |
| 1 | 12 | 13 | 20 | 21 | 24 | 25 | 26 |
Fig. 6: to the solution of problem 5.
## Criteria
+ Correct solution - 7 points.
$\pm$ Correct upper bound on the sum of the numbers on the diagonal is proven - 4 points.
干 A correct example of an arrangement with the maximum sum of numbers on the diagonal is provided - 3 points.
- Correct answer without further progress - 1 point.
|
432
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. At a round table, 60 people are sitting. Each of them is either a knight, who always tells the truth, or a liar, who always lies. Each person at the table said: "Among the next 3 people sitting to my right, there is no more than one knight." How many knights could have been sitting at the table? List all possible options and prove that there are no others.
|
Answer: 30.
Solution. Consider any four consecutive people. If there were at least 3 knights among them, the leftmost knight would definitely lie, which is impossible. If there were at least 3 liars, the leftmost liar would definitely tell the truth, which is also impossible. Therefore, in every group of four consecutive people, there are exactly 2 knights and 2 liars. By dividing all the people into 15 such groups, we get that there are $15 \cdot 2 = 30$ knights.
Another solution. Notice that there cannot be only knights or only liars sitting at the table. Then there must be a pair of knight-liar sitting next to each other in that exact order. Consider the 4 cases of who can sit to the right of this pair.
Case 1. Two knights sit to the right.
Then we get the sequence
$$
\text { knight-liar-knight-knight, }
$$
which contradicts the condition, as the first knight lied.
Case 2. Two liars sit to the right.
Then we get the sequence
$$
\text { knight-liar-liar-liar. }
$$
Whoever sits immediately after this sequence, the first liar will tell the truth, which contradicts the condition.
Case 3. A knight and a liar sit to the right in that exact order.
Then we get the sequence
$$
\text { knight-liar-knight-liar. }
$$
Consider the first liar. He lies, so among the next three, there must be at least two knights. From this, we uniquely get
knight-liar-knight-liar-knight
Now consider the second knight. He tells the truth, so the sixth person in the sequence will be a liar:
knight-liar-knight-liar-knight-liar.
Continuing this reasoning, we get an alternating arrangement where there are an equal number of knights and liars.
Case 4. A liar and a knight sit to the right in that exact order. Then we get the sequence
$$
\text { knight-liar-liar-knight. }
$$
Consider the first liar. He lies, so among the next three, there must be at least two knights. From this, we uniquely get
knight-liar-liar-knight-knight
Now consider the second knight. He tells the truth, so the sixth and seventh people in the sequence will be liars
knight-liar-liar-knight-knight-liar-liar.
Continuing this reasoning, we get an arrangement where there are an equal number of knights and liars.
## Criteria
The highest applicable criterion is used:
7 6. Any complete solution to the problem.
7 6. The problem is correctly solved for the case where both knights and liars are present at the table.
4 6. Both possible cases of arranging knights and liars are considered in the solution, but there is no justification for why there are no other arrangements.
2 6. One of the two possible cases of arranging knights and liars is considered in the solution, but there is no justification for why there are no other arrangements.
1 6. Only the correct answer is present.
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Masha and the Bear ate a basket of raspberries and 60 pies, starting and finishing at the same time. At first, Masha was eating raspberries, and the Bear was eating pies, then (at some point) they switched. The Bear ate raspberries 6 times faster than Masha, and pies only 3 times faster. How many pies did the Bear eat, if he ate twice as many raspberries as Masha?
|
Answer: 54 pies. Solution: Divide the raspberries into 3 equal parts. The bear ate each part 6 times faster than Masha, but there are two parts, so he spent only 3 times less time on the raspberries than Masha. This means Masha ate pies in one-third of the time the bear did. Since she eats three times slower, she ate 9 times fewer pies than the bear. Dividing the pies in the ratio of 9:1, we see that Masha got the 10th part, which is 6 pies. The remaining 54 pies went to the bear.
Criterion: Correct answer - no less than 2 points, with full justification - 7 points.
|
54
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Buses from Moscow to Oryol depart at the beginning of each hour (at 00 minutes). Buses from Oryol to Moscow depart in the middle of each hour (at 30 minutes). The journey between the cities takes 5 hours. How many buses from Oryol will the bus that left from Moscow meet on its way?
|
Answer: 10.
It is clear that all buses from Moscow will meet the same number of buses from Orel, and we can assume that a bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Orel at $7:30, 8:30, \ldots, 15:30, 16:30$ and only them. There are 10 such buses.
$\pm$ Correct reasoning with an arithmetic error leading to an incorrect answer. 4-5 points
Ғ Frequency of encounters (every half hour) has been calculated, but the answer is incorrect. 3 points
- Correct answer without explanation. 1 point
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Nезнayka, Doctor Pilulkin, Knopochka, Vintik, and Znayka participated in a math competition. Each problem in the competition was solved by exactly four of them. Znayka solved strictly more than each of the others - 10 problems, while Nезнayka solved strictly fewer than each of the others - 6 problems. How many problems were there in the math competition?
|
Answer: 10.
Each of Dr. Pill, Knopochka, and Vintik, according to the condition, solved from 7 to 9 problems. Therefore, the total number of solved problems ranges from $10+6+3 \cdot 7=37$ to $10+6+3 \cdot 9=43$. Note that this number should be equal to four times the number of problems. Among the numbers from 37 to 43, only one is divisible by 4 - this number is 40. Therefore, the total number of solved problems is 40, and the total number of problems was 10.
## Criteria
+ Correct solution - 7 points.
$\mp$ Correct answer without explanations (or with incorrect explanations) - 2 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On the table, there are 2018 playing cards (2018 stacks, each with one card). Petya wants to combine them into one deck of 2018 cards in 2017 operations. Each operation consists of merging two stacks. When Petya merges stacks of $a$ and $b$ cards, Vasily Ivanovich pays Petya $a \cdot b$ rubles. What is the maximum amount of money Petya can earn by performing all 2017 operations?
|
Answer: $\frac{2017 \cdot 2018}{2}=2035153$
Let's mentally connect the cards in one stack with invisible threads, each with each. Then the operation of combining two stacks with $a$ and $b$ cards adds exactly $a \cdot b$ threads. In the end, each card will be connected by a thread to each other, and the number of threads will not depend on the way the operations are performed and will be equal to $\frac{2017 \cdot 2018}{2}$. This is exactly how many rubles Petya will earn.
Another solution. Notice that when combining stacks of $a$ and $b$ cards, we earn exactly $\frac{1}{2}\left((a+b)^{2}-a^{2}-b^{2}\right)$ rubles; therefore, the final amount we will earn is exactly
$$
\frac{1}{2}(2018^{2} \underbrace{-1^{2}-1^{2}-\ldots-1^{2}}_{2018 \text { times }})=\frac{1}{2}\left(2018^{2}-2018\right)
$$
## Criteria
+ Correct solution - 7 points.
+. In a logically correct solution, the answer is presented as a sum with an ellipsis - at least 1 point is deducted. For example, an answer in the form of $1+2+\ldots+2017$ - 6 points.
$\pm$ It is proven that the answer does not depend on the way the operations are performed - at least 5 points.
-. Correct answer without justification or with incorrect reasoning - 1 point.
Example of incorrect reasoning: in the work, the sum is calculated for a specific way of acting, but it is not proven that the maximum sum is achieved precisely with this method (and it is not proven that the sum does not actually depend on the order of operations).
|
2035153
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10.5. In each cell of a square table of size $200 \times 200$, a real number not exceeding 1 in absolute value was written. It turned out that the sum of all the numbers is zero. For what smallest $S$ can we assert that in some row or some column, the sum of the numbers will definitely not exceed $S$ in absolute value?
|
Answer: 100.
Solution. First, we show that $S<40000$.
$$
This means that one of the numbers $A$ or $D$ in absolute value exceeds 10000. However, each of the corresponding squares contains only 10000 cells, and the numbers in them do not exceed 1 in absolute value. Contradiction.
## Criteria
Any correct solution to the problem is worth 7 points. In the absence of such a solution, the following criteria are summed:
## 16. There is a correct answer.
1 6. "Example". It is proven that $S<100$ is impossible (i.e., an example is provided or its existence is proven).
5 6. "Estimate". It is proven that $S=100$ works for any arrangement of numbers satisfying the condition. In the absence of a complete proof of the "estimate", the following partial progress is credited:
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. If the width of a rectangle is increased by $30 \%$, and the height is decreased by $20 \%$, its perimeter will not change. Will the perimeter decrease or increase, and by what percentage, if instead the width of the original rectangle is decreased by $20 \%$, and the height is increased by $30 \%$?
|
Answer. It will increase by $10 \%$. Solution. Let the width of the original rectangle be $s$, and the height be $h$. In the first case, the modified width and height will be 1.3 and 0.8 $s$ respectively. According to the condition, $2(s+h)=2(1.3 s+0.8 h)$, from which $h=1.5 s$. This means that the original perimeter is $2(s+1.5 s)=5 s$. In the second case, the perimeter will be $2(0.8 s+1.3 h)=2(0.8 s+1.3 \cdot 1.5 s)=5.5$ s. This number is greater than $5 s$ by $0.5 s$, that is, by $10 \%$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Through the point with coordinates $(2,2)$, lines (including two parallel to the coordinate axes) are drawn, dividing the plane into angles of $18^{\circ}$. Find the sum of the abscissas of the points of intersection of these lines with the line $y=2016-x$.
|
Answer: 10080. Solution. The picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(2,2)$, 10 lines are drawn, and the line $y=2016-x$ intersects all of them. For each point on the line $y=2016-x$, the sum of the coordinates is 2016, so the total sum of the abscissas and ordinates is 20160, and the sum of the abscissas is half of that.
Criterion. 2 points for the correct answer without a correct justification. 5 points if all ideas are found, but the answer is incorrect due to a wrong count of the number of intersection points (for example, the student counts 9 or 18 points).
|
10080
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. In a school tic-tac-toe tournament, 16 students participated, each playing one game against every other student. A win was worth 5 points, a draw -2 points, and a loss -0 points. After the tournament, it was found that the participants collectively scored 550 points. What is the maximum number of participants who could have never played a draw in this tournament?
## Answer:
|
Solution. A total of $\frac{16 \cdot 15}{2}=120$ games were played in the tournament. In each game, either 5 points were awarded (in case of a win-loss) or 4 points (in case of a draw). If all games had ended in a draw, the total number of points for all participants would have been $120 \cdot 4=480$, which is 70 points less than the actual total points of all participants. In a non-draw game, the two participants collectively receive 1 point more than in a draw game. This means that exactly 70 games ended with one participant winning, and the remaining 50 games ended in a draw.
Suppose that at least 6 participants never played a draw. Then the draw games could only have been played between the remaining 10 participants, and they played a total of $\frac{10 \cdot 9}{2}=45$ games, which is less than 50. This is a contradiction. Therefore, no more than 5 participants never played a draw.
It is not difficult to describe an example for 5 participants. Fix 11 participants, they played $\frac{11 \cdot 10}{2}=55$ games among themselves. Choose any 50 of these games, let them be draws (it is clear then that each of the fixed 11 participants will play at least one draw), and all other games in the tournament ended with one participant winning. Therefore, $16-11=5$ people never played a draw. It is clear that all conditions of the problem are satisfied.
## Criteria
## The following criteria are cumulative:
2 6. A correct example is described where exactly 5 people never played a draw.
5 p. It is proven that no more than 5 people never played a draw.
If the proof that no more than 5 people never played a draw is missing:
2 p. It is proven that there were exactly 50 draws (or that there were exactly 70 decisive games).
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. Positive numbers $a, b, c, d$ are greater than 1. Find the smallest possible value of the expression
$$
\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)
$$
|
Answer: 67.
Solution. From the properties of logarithms, it follows that $\log _{a} b \cdot \log _{b} c \cdot \log _{c} d \cdot \log _{d} a=1$. Also, all these four factors are positive, since all numbers $a, b, c, d$ are greater than 1.
Transform and estimate the given expression
$$
\begin{gathered}
S=\log _{a}\left(a b^{2}\right)+\log _{b}\left(b^{2} c^{3}\right)+\log _{c}\left(c^{5} d^{6}\right)+\log _{d}\left(d^{35} a^{36}\right)= \\
=\left(\log _{a} a+\log _{a} b^{2}\right)+\left(\log _{b} b^{2}+\log _{b} c^{3}\right)+\left(\log _{c} c^{5}+\log _{c} d^{6}\right)+\left(\log _{d} d^{35}+\log _{d} a^{36}\right)= \\
=\left(1+2 \log _{a} b\right)+\left(2+3 \log _{b} c\right)+\left(5+6 \log _{c} d\right)+\left(35+36 \log _{d} a\right) \geqslant \\
\geqslant(1+2+5+35)+4 \sqrt[4]{2 \log _{a} b \cdot 3 \log _{b} c \cdot 6 \log _{c} d \cdot 36 \log _{d} a}=43+4 \cdot 6=67,
\end{gathered}
$$
here in the last transition, the inequality between the arithmetic and geometric means for four positive numbers $2 \log _{a} b, 3 \log _{b} c, 6 \log _{c} d, 36 \log _{d} a$ was used.
Also note that the value $S=67$ is achieved, for example, when $a=2, b=8, c=d=64$, since all four numbers $2 \log _{a} b, 3 \log _{b} c, 6 \log _{c} d, 36 \log _{d} a$ will be equal to 6.
## Criteria
Points for the estimate and example are summed.
Estimate. The largest suitable criterion is used:
5 p. The estimate $S \geqslant 67$ is proven.
1 p. The problem is reduced to finding the minimum of the sum $2 \log _{a} b+3 \log _{b} c+6 \log _{c} d+36 \log _{d} a$.
0 p. The positivity of the numbers $\log _{a} b, \log _{b} c, \log _{c} d, \log _{d} a$ is mentioned.
Example. The largest suitable criterion is used:
2 p. It is proven that the value $S=67$ is achieved for some $a, b, c, d$, greater than 1.
0 p. Only the correct answer is given.
|
67
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Several numbers were written on the board, their arithmetic mean was equal to $M$. They added the number 15, after which the arithmetic mean increased to $M+2$. After that, they added the number 1, and the arithmetic mean decreased to $M+1$. How many numbers were on the board initially? (Find all options and prove that there are no others.)
|
Answer: 4.
Let there be $k$ numbers in the original list with a sum of $S$. Then, by the condition,
$$
\frac{S+15}{k+1}-\frac{S}{k}=2, \quad \frac{S+15}{k+1}-\frac{S+16}{k+2}=1.
$$
By bringing to a common denominator and transforming in an obvious way, we get that these equations are equivalent to the following two:
$$
15 k-S=2 k(k+1), \quad S-k+14=(k+1)(k+2)
$$
Adding them, we arrive at the equation for $k$:
$$
14(k+1)=(k+1)(2 k+k+2),
$$
from which $k=4$.
GHz The correct system of equations is written, but solved incorrectly. 3 points
- One of the equations is obtained correctly, the second one incorrectly. 2 points
- The correct answer without justification. (An example of a list of four numbers satisfying the condition is not considered a justification.) 1 point
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Tetrahedron $ABCD$ with acute-angled faces is inscribed in a sphere with center $O$. A line passing through point $O$ perpendicular to the plane $ABC$ intersects the sphere at point $E$ such that $D$ and $E$ lie on opposite sides relative to the plane $ABC$. The line $DE$ intersects the plane $ABC$ at point $F$, which lies inside triangle $ABC$. It turns out that $\angle ADE = \angle BDE$, $AF \neq BF$, and $\angle AFB = 80^\circ$. Find the measure of $\angle ACB$.
|
Answer: $40^{\circ}$.
Note that point $E$ is equidistant from points $A, B, C$, so its projection onto the plane $A B C$ coincides with the projection of point $O$ onto this plane and is the center of the circumscribed circle of triangle $A B C$.
Consider triangles $A D E$ and $B D E$. They have a pair of equal sides $A E$ and $B E$, a common side $D E$, and equal angles $A D E$ and $B D E$. From the Law of Sines, it follows that these triangles are either equal or the angles $D A E$ and $D B E$ are supplementary to $180^{\circ}$. The first situation is impossible, as in the case of equality of triangles $A D E$ and $B D E$, points $A$ and $B$ are equidistant from any point on side $D E$, but by the condition $A F \neq B F$. Therefore, $\angle D A E + \angle D B E = 180^{\circ}$.
Consider the point $X$ of intersection of the ray $A F$ with the sphere $\Omega$ circumscribed around the tetrahedron $A B C D$. Note that the ray $A F$ lies in the planes $A B C$ and $A E D$, so point $X$ lies on the circumscribed circles of triangles $A B C$ and $A E D$. Point $E$ is equidistant from all points of the circumscribed circle of triangle $A B C$; in particular, $A E = X E$. From the inscribed quadrilateral $A E X D$, it follows that $\angle D A E + \angle D X E = 180^{\circ}$. Since $A E = X E$, $E$ is the midpoint of the arc $A X$ of the circumscribed circle of triangle $A D E$, and thus $\angle A D E = \angle X D E$.
Using the previously derived angle equalities, we conclude that triangles $D B E$ and $D X E$ are equal by the second criterion: $\angle D B E = 180^{\circ} - \angle D A E = \angle D X E, \angle X D E = \angle A D E = \angle B D E$, side $D E$ is common. Since triangles $D B E$ and $D X E$ are equal, vertices $B$ and $X$ are equidistant from any point on side $D E$; in particular, $B F = F X$.
It remains to calculate the angles in the plane $A B C$. Sequentially using the inscribed quadrilateral $A B X C$, the isosceles triangle $B F X$, and the exterior angle theorem for triangle $B F X$, we write
$$
\angle A C B = \angle A X B = \frac{1}{2} \cdot (\angle F X B + \angle F B X) = \frac{1}{2} \cdot \angle A F B = 40^{\circ}
$$
Another solution. Let the ray $A F$ intersect the sphere $\Omega$ circumscribed around the tetrahedron $A B C D$ at point $X$. By construction, the point $E$ satisfies the relation $E X = E A$, which implies that $\angle A D E = \angle E A F$. Similarly, we obtain that $\angle B D E = \angle E B F$, and thus $\angle E A F = \angle E B F$.
Denote the point of intersection of the line $O E$ with the plane $A B C$, which is the center of the circumscribed circle of triangle $A B C$, as $O_{1}$. Then $\angle O_{1} A E = \angle O_{1} B E$.
Consider the trihedral angles $A O_{1} E F$ and $B O_{1} E F$. In them, the planar angles $E A F$ and $E B F$, the planar angles $O_{1} A E$ and $O_{1} B E$, and the dihedral angles at the edges $A O_{1}$ and $B O_{1}$ are right. Therefore, the corresponding trihedral angles are equal. And thus the planar angles $\angle F A O_{1} = \angle F B O_{1}$ are equal. Note that this equality can also be derived from the Law of Cosines for trihedral angles.
The specified equality is possible in two cases: either point $F$ lies on the perpendicular bisector of $A B$ (points $A$ and $B$ are symmetric relative to $F O_{1}$), or point $F$ lies on the circumscribed circle of triangle $A B O_{1}$. The first case is forbidden by the condition $A F \neq B F$, so the second case must hold. Then $\angle A O B = \angle A F B = 80^{\circ}$ and is the central angle for the angle $A C B$ in the circumscribed circle of triangle $A C B$. As a result, we conclude that $\angle A C B = 40^{\circ}$.
## Criteria
The following points are summed.
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Buses from Moscow to Voronezh depart every hour, at 00 minutes. Buses from Voronezh to Moscow depart every hour, at 30 minutes. The trip between the cities takes 8 hours. How many buses from Voronezh will the bus that left from Moscow meet on its way?
|
Answer: 16.
It is clear that all buses from Moscow will meet the same number of buses from Voronezh, and we can assume that the bus from Moscow departed at 12:00. It is easy to understand that it will meet buses that left Oryol at 4:30, 5:30, ..., 18:30, 19:30 and only them. There are 16 such buses.
$\pm$ Correct reasoning with an arithmetic error leading to an incorrect answer. 4-5 points
干 The frequency of encounters (every half hour) was calculated, but the answer is incorrect. 3 points
- Correct answer without explanation. 1 point
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. How many solutions in natural numbers does the equation
$$
(2 x+y)(2 y+x)=2017^{2017} ?
$$
|
Answer: 0.
Note that the sum of the numbers $A=2x+y$ and $B=2y+x$ is divisible by 3. Since the number on the right side is not divisible by 3, neither $A$ nor $B$ are divisible by 3. Therefore, one of these two numbers gives a remainder of 2 when divided by 3, and the other gives a remainder of 1. Thus, their product gives a remainder of 2. However, the number 2017 gives a remainder of 1, and therefore $2017^{2017}$ also gives a remainder of 1.
$\pm$ It is established that the system of equations cannot be solved in integers due to divisibility by 3, but there is no strict justification for why this will happen in all cases. 5 points
Ғ In the solution, several different factorizations are analyzed, and for each, it is established that the corresponding system of equations has no solutions. 3 points
- Correct answer without justification. 1 point
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Dodson, Williams, and their horse Bolivar want to get from city A to city B as quickly as possible. Along the road, there are 27 telegraph poles that divide the entire journey into 28 equal segments. Dodson takes 9 minutes to walk one segment, Williams takes 11 minutes, and either of them can ride Bolivar to cover the distance in 3 minutes (Bolivar cannot carry both). They set out from city A simultaneously; the journey is considered complete when all of them arrive in city B.
They agreed that part of the way Dodson will ride, then tie Bolivar to one of the telegraph poles and continue on foot, while Williams will initially walk and then ride Bolivar. At which pole should Dodson tie Bolivar to minimize the travel time to city B?
Answer: at the 12th, counting from A.
|
Solution. Let the distance from A to B be taken as a unit, and time will be measured in minutes. Then Dodson's speed is $1 / 9$, Williams' speed is $1 / 11$, and Bolivar's speed is $-1 / 3$.
Let the desired post have the number $k$ (i.e., the distance from city A is $k / 28$). Then Dodson will arrive in time
$$
\frac{k}{28}: \frac{1}{3}+\frac{28-k}{28}: \frac{1}{9}=9-\frac{6 k}{28}
$$
while Williams will take
$$
\frac{k}{28}: \frac{1}{11}+\frac{28-k}{28}: \frac{1}{3}=3+\frac{8 k}{28}
$$
We need to find $k$ for which these values coincide:
$$
9-\frac{6 k}{28}=3+\frac{8 k}{28} \quad \Leftrightarrow \quad 6 \cdot 28=14 k \quad \Leftrightarrow \quad k=12
$$
It remains to understand why $k=12$ gives the best time. Indeed, for smaller $k$, Dodson's time will increase, and for larger $k$, Williams' time will increase. Thus, for all other $k$, the time of at least one of the characters will be greater than when $k=12$.
## Criteria
The highest suitable criterion is used:
7 p. Any complete solution to the problem.
4 p. The position of the post is found where the time of the characters coincides, but it is not proven that this is the best time.
2 p. The idea of equating the time of the characters is present, but an error was made in the formulation or solution of the equation.
1 p. The correct answer is given.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. In the vertices of a regular 2019-gon, numbers are placed such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the 19th vertex has the number 19, and the 20th vertex has the number 20. What number is in the 2019th vertex?
|
Answer: 61.
Solution. Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{2019}$. Since the sum of any nine consecutive numbers is the same, the numbers that are 8 apart are equal. Therefore, $x_{1}=x_{10}=x_{19}=\ldots=x_{1+9 k}=\ldots$. Since 2019 is not divisible by 9 but is divisible by 3, continuing this sequence in a cycle will include all numbers of the form $x_{3 k+1}$. Similar sequences can be constructed starting from the numbers $x_{2}$ and $x_{3}$. Thus, the numbers at the vertices that have the same remainder when divided by 3 are equal.
Since each set of nine consecutive numbers contains three sets of vertices with different remainders when divided by 3, the sum of the numbers in each such set is 100. Note that 19 gives a remainder of 1 when divided by 3, 20 gives a remainder of 2, and 2019 gives a remainder of 0, so the number written at vertex 2019 is $100 - 19 - 20 = 61$.
## Criteria
The largest suitable criterion is used:
7 points. Any complete solution to the problem.
4 points. Proven that numbers that are 2 apart are equal.
1 point. Noted that numbers that are 8 apart are equal.
|
61
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let's call a natural number an almost palindrome if it can be transformed into a palindrome by changing one of its digits. How many nine-digit almost palindromes exist? (20 points)
|
Solution: After replacing one digit in the almost palindrome, we get a number of the form $\overline{a b c d e d c b a}$. Let's divide the digits into pairs by place value: 1 and 9, 2 and 8, 3 and 7, 4 and 6. In any almost palindrome, the digits in three of the specified pairs of place values must be the same, and in exactly one pair they must be different.
International School Olympiad URFU "Emerald" 2022, 2nd stage
We can turn this number into a palindrome by replacing one of the two digits in the remaining pair with the other digit in that pair. We can choose three pairs of place values in 4 ways (3 of them include the pair 1 and 9, and exactly one way does not include this pair), in each pair (except for 1 and 9) the digits can be any and the same, in the pair 1 and 9 (if it is chosen) the digits are the same and not equal to zero, in the remaining pair the digits are different (any, if it is not the pair 1 and 9, and any except zero in the first place, if the pair 1 and 9 remains), and the middle digit is arbitrary. The total number of such numbers is
$$
3 \cdot 10^{3} \cdot 9 \cdot 10 \cdot 9+10^{4} \cdot 9 \cdot 9=3240000
$$
Answer: 3240000
|
3240000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. If in an acute scalene triangle three medians, three angle bisectors, and three altitudes are drawn, they will divide it into 34 parts.
|
Write the answer in digits in ascending order, without spaces (for example, 12345).
## Mathematics $8-9$ grade
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
34
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 2. (20 points)
Find the maximum possible value of the ratio of a three-digit number to the sum of its digits.
#
|
# Solution.
Let $N=\overline{a b c}$, where $a, b, c$ are the digits of the number. Clearly, for "round" numbers $N=$ $100, 200, \ldots, 900$, we have $\frac{N}{a+b+c}=100$. Furthermore, if the number $N$ is not round, then $b+c>0$ and $a+b+c \geq a+1$. Since the leading digit of the number $N$ is $a$, we have $N<(a+1) \cdot 100$ and
$$
\frac{N}{a+b+c}<\frac{(a+1) \cdot 100}{a+1}=100
$$
Thus, the maximum value of the considered ratio is 100. This value is achieved only for "round" numbers.
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. In a toy store, 125 plush bears are sold in $k$ different colors and six different sizes. For what largest $k$ can we assert that there will be at least three identical bears? (i.e., matching both in color and size) (20 points)
|
Solution: If $k=10$, then there are 60 varieties of teddy bears in the store. If there are no more than two of each type, then there are no more than 120 teddy bears in the store - a contradiction. Therefore, there will be at least three identical teddy bears among them.
If $k=11$, then there may not be three identical teddy bears. Indeed, there will be at least 66 varieties of teddy bears, among which 59 types can be sold with 2 copies each, and another 7 types - with one copy each. For $k \geq 11$, there will be more varieties of teddy bears, so as a counterexample, one can remove the second copies of the existing types and add one new copy of each added type.
Answer: 10
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the language of the "Tekimar" tribe, there are only 7 letters: A, E, I, K, M, R, T, but the order of these letters in the alphabet is unknown. A word is defined as any sequence of seven different letters from the alphabet, and no other words exist in the language. The chief of the tribe listed all existing words in alphabetical order and noticed that the word "Metrika" has the number 3634 in this list. What is the number of the word "Materik" in this list? (20 points)
|
Solution: The total number of words in the tribe's language is $7!=5040$. Note that the number of words starting with any particular letter is the same for any first letter, which is $7!\div 7=6!=720$. If a word starts with the first letter in the alphabet, the numbering of any such word starts from number 1 and ends at number 6!. If it starts with the second letter in the alphabet, the numbering starts from number $6!+1$ and ends at number $2 \cdot 6$!, and so on. In general, if a word starts with the $k$-th letter in the alphabet, its numbering varies from number $(k-1) \cdot 6!+1$ to number $k \cdot 6$!. The number 3634 satisfies the inequality $5 \cdot 6$! $<3634 \leq 6 \cdot 6$!, so M is the sixth letter in the alphabet. Now, we will repeat a similar process for the remaining six letters. Specifically, we will forget about the existence of the first letter by removing it from the word. Then the word's number will decrease by $5 \cdot 6!=3600$. If we now form all six-letter words and arrange them in alphabetical order, the word "Etrika" would have the number 34. The number of words starting with any particular letter is the same for any first letter, which is $6!\div 6=$ $5!=120$. If the first letter of the word among the remaining six letters is the first in alphabetical order, the numbering of any such word starts from number 1 and ends at number 5!. If it is the second letter in alphabetical order among the remaining, the numbering starts from number $5!+1$ and ends at number $2 \cdot 5$!, and so on. In general, if the first letter of the word is the $k$-th letter in alphabetical order among the remaining, its numbering varies from number $(k-1) \cdot 5$! to number $k \cdot 5$!. The number 34 satisfies the inequality above, which implies $34 \leq 5$!, so E is the first in alphabetical order among the remaining, and thus the first letter of the alphabet. Continuing this similar process, we get the order of the letters: E - first, A - second, T - third, R - fourth, I - fifth, M - sixth, K - seventh.
In the word "Materik," the letters are in the following order: $6,2,3,1,4,5,7$, so its number is 5 . $6!+1 \cdot 5!+1 \cdot 4!+0 \cdot 3!+0 \cdot 2!+0 \cdot 1!+0 \cdot 0!+1=3745$. (The multiplier before $k!$ indicates the number of digits to the right and less than the digit in the ($k+1$)-th position, counting from the end of the word. A one is added at the end because the numbering of words starts with it.
|
3745
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find four real numbers $x_{1}, x_{2}, x_{3}, x_{4}$, such that each, when added to the product of the others, equals two.
|
Solution. Let $x_{1} x_{2} x_{3} x_{4}=p$. Then our system of equations will take the form:
$$
x_{i}+\frac{p}{x_{i}}=2, \quad i=1,2,3,4
$$
The case when one of the sought numbers is zero leads to a contradiction. Indeed, substituting zero for $x_{1}$, we get
$$
x_{2} x_{3} x_{4}=2, x+2=x+3=x+4=2 \text {. }
$$
Thus, all $x_{i}$ are roots of the same quadratic equation $x_{i}^{2}-2 x_{i}+p=0$. Therefore, among them, there can only be two different ones. Let's consider three cases.
(a) $x_{1}=x_{2}=x_{3}=x_{4}=m, m+m^{2}=2$. Due to the monotonicity of the function $m+m^{2}$, we have only one real solution $m=1$.
(b) Three of the sought numbers are equal, and the fourth is not equal to them. Let $x_{1}=x_{2}=x_{3}=m$, $x_{4}=n, m \neq n$
$$
\left\{\begin{array}{l}
m+m^{2} n=2 \\
n+m^{3}=2
\end{array}\right.
$$
From this, $(m-n)\left(1-m^{2}\right)=0$. Since $m \neq n$, we get that $1-m^{2}=0$. Note that $m \neq 1$, because otherwise, from the system of relations, it immediately follows that $n=1$, which contradicts $m \neq n$. Therefore, $m=-1$, from which it follows that $n=3$. This gives us four more solutions.
(c) $x_{1}=x_{2}=m, x_{3}=x_{4}=n, m \neq n$,
$$
\left\{\begin{array}{l}
m+m n^{2}=2 \\
n+n m^{2}=2
\end{array}\right.
$$
From this, $(m-n)(1-m n)=0 \Rightarrow m n=1$. Then $m+n=2, m=n=1$. Contradiction.
Thus, either $x_{1}=x_{2}=x_{3}=x_{4}=1$, or one of the $x_{i}=3$, and the rest are equal to $-1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. On a piece of paper, 25 points are marked - the centers of the cells of a $5 \times 5$ square. The points are colored with several colors. It is known that no three points of the same color lie on any straight line (vertical, horizontal, or at any angle). What is the minimum number of colors that could have been used? (20 points)
|
Solution: If no more than two colors are used, then, in particular, in the first row there will be at least three points of the same color - they lie on the same straight line. An example of using three colors (identical numbers denote points painted in the same color):
## Answer: 3 colors
| 1 | 2 | 2 | 3 | 3 |
| :--- | :--- | :--- | :--- | :--- |
| 2 | 3 | 1 | 3 | 2 |
| 3 | 3 | 2 | 1 | 2 |
| 1 | 1 | 3 | 2 | 3 |
| 2 | 2 | 3 | 1 | 1 |
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Daria Dmitrievna is preparing a test on number theory. She promised to give each student as many problems as the number of addends they create in the numerical example
$$
a_{1}+a_{2}+\ldots+a_{n}=2021
$$
where all numbers $a_{i}$ are natural numbers, greater than 10, and are palindromes (do not change if their digits are written in reverse order). If a student does not find any such example, they will receive 2021 problems on the test. What is the minimum number of problems a student can receive? (20 points)
|
Solution: A student cannot receive one problem since 2021 is not a palindrome. Suppose he can receive two problems, then at least one of the numbers $a_{1}, a_{2}$ is a four-digit number. If it starts with 2, then the second digit is 0 and the number itself is 2002. In this case, the second number is 19, which is not a palindrome. If the number starts with 1, then its last digit is also 1, and the last digit of the second number must be zero, which is incorrect for palindromes. Therefore, the student could not have received two problems. An example of three problems exists, for instance, $1111+888+22=2021$.
Answer: 3
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find the number of triples of natural numbers $m, n, k$, which are solutions to the equation $m+$ $\sqrt{n+\sqrt{k}}=2023$. (20 points)
|
Solution: For the left side to be an integer, the numbers $k$ and $n+\sqrt{k}$ must be perfect squares, and since $n+\sqrt{k} \geq 2$, it follows that $\sqrt{n+\sqrt{k}} \geq 2$ and thus $m \leq 2021$. Since $1 \leq m \leq$ 2021, $\sqrt{n+\sqrt{k}}$ can take any value from 2 to 2022 - this value uniquely determines the number $m$. Let $k=x^{2}$ and $n+\sqrt{k}=y^{2}$, where $1 \leq x \leq y^{2}-1$ and $2 \leq y \leq 2022$, then the number $n$ is uniquely determined, specifically $n=y^{2}-x$. Therefore, we need to count the number of
admissible pairs $(x, y)$. There are a total of $\left(2^{2}-1\right)+\ldots+\left(2022^{2}-1\right)=1^{2}+2^{2}+\ldots+2022^{2}-2022$. The formula for the sum of the squares of the first $n$ natural numbers is known:
$$
1^{2}+2^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6}
$$
Applying this formula, we get $1^{2}+2^{2}+\ldots+2022^{2}-2022=\frac{2022 \cdot 2023 \cdot 4045}{6}-2022=$ 27575680773.
## Answer: 27575680773
|
27575680773
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. for $a<0.5$
## Correct answer: for $a<1$
## Question 3
Correct
Points: 1.00 out of a maximum of 1.00
Question 4
Correct
Points: 1.00 out of a maximum of 1.00
## Question 5
Correct
Points: 1.00 out of a maximum of 1.00
Worker $A$ and worker $B$ can complete the work in 7.2 days, worker $A$ and worker $C$ in 9 days, worker $B$ and worker $C$ in 12 days. How many days will they take to complete the work if they all work together?
|
Answer: 6
Correct answer: 6
A piece of copper-tin alloy has a total mass of 12 kg and contains 45% copper. How much pure tin should be added to this piece of alloy to obtain a new alloy containing 40% copper?
Answer: 1.5
Correct answer: 1.5
Find a two-digit number that is three times the product of its digits. If the digits of this number are reversed, the ratio of the resulting number to the given number is $\frac{7}{4}$.
Answer: 24
Correct answer: 24
## Question 7
Correct
Points: 1.00 out of a maximum of 1.00
## Question 8
Correct
Points: 1.00 out of a maximum of 1.00
Question 9
Correct
Points: 1.00 out of a maximum of 1.00
Answer: 8
Correct answer: 8
Three edges emanating from one vertex of a rectangular parallelepiped are equal to $a, b$, c. Find the length of the diagonal of this parallelepiped.
Choose one answer:
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A printing house determines the cost of printing a book as follows: it adds the cost of the cover to the cost of each page, and then rounds the result up to the nearest whole number of rubles (for example, if the result is 202 rubles and 1 kopeck, it is rounded up to 203 rubles). It is known that the cost of a book with 104 pages is 134 rubles, and a book with 192 pages costs 181 rubles. How much does the printing of the cover cost, if it is a whole number of rubles, and the cost of one page is a whole number of kopecks? (20 points)
|
Solution: Let's convert all costs into kopecks. Let one page cost $x$ kopecks, and the cover cost $100 y$ kopecks. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
13300<100 y+104 x \leq 13400 \\
18000<100 y+192 x \leq 18100
\end{array}\right.
$$
Subtracting the first inequality from the second, we get $4600<88 x<4800$. The integer solutions to the inequality are $x=53$ and $x=54$.
If $x=53$, then $7788<100 y \leq 7888$ and $y=78$. But then $100 y+192 x=17976$, which contradicts the second inequality.
If $x=54$, then $7684<100 y \leq 7784$ and $y=77$. Then $100 y+192 x=18068$.
## Answer: a page costs 54 kopecks, and the cover costs 77 rubles
|
77
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Maxim came up with a new way to divide numbers by a two-digit number $N$. To divide any number $A$ by the number $N$, the following steps need to be performed:
1) Divide $A$ by the sum of the digits of the number $N$;
2) Divide $A$ by the product of the digits of the number $N$;
3) Subtract the second result from the first.
For which numbers $N$ will Maxim's method give the correct result? (20 points)
|
Solution: Let $N=\overline{x y}=10 x+y$, where $x, y$ are digits. We need to find all pairs $(x, y)$ such that $\frac{A}{10 x+y}=\frac{A}{x+y}-\frac{A}{x y}$, which means $\frac{1}{10 x+y}=\frac{1}{x+y}-\frac{1}{x y}$. Note that $y \neq 0$. Transform the equation to the form
$$
\begin{gathered}
x y(x+y)=x y(10 x+y)-(x+y)(10 x+y) \\
(x+y)(10 x+y)=x y \cdot 9 x \\
10 x^{2}+11 x y+y^{2}=9 y x^{2}
\end{gathered}
$$
Solve the equation as a quadratic in terms of $x$. For this, rewrite the equation as
$$
\begin{gathered}
(10-9 y) x^{2}+11 y x+y^{2}=0 \\
D=121 y^{2}-4 y^{2}(10-9 y)=81 y^{2}+36 y^{3}=9 y^{2}(9+4 y)
\end{gathered}
$$
For the root to be an integer, the discriminant must be a perfect square, which means $9+4 y$ must also be a perfect square. Trying values of $y$ from 1 to 9, only $y=4$ works. Then the roots of the equation are $x_{1}=\frac{-11 \cdot 4+3 \cdot 4 \cdot 5}{2 \cdot(10-9 \cdot 4)}=-\frac{4}{13}$ and $x_{2}=\frac{-11 \cdot 4-3 \cdot 4 \cdot 5}{2 \cdot(10-9 \cdot 4)}=2$. Only $x=2$ is valid, so $N=24$. For this number, Maksim's method gives the correct result.
Answer: 24
|
24
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1. (20 points)
It is known that the only solution to the equation
$$
\pi / 4=\operatorname{arcctg} 2+\operatorname{arcctg} 5+\operatorname{arcctg} 13+\operatorname{arcctg} 34+\operatorname{arcctg} 89+\operatorname{arcctg}(x / 14)
$$
is a natural number. Find it.
|
# Solution.
Consider the equation:
$$
\operatorname{arcctg} a-\operatorname{arcctg} b=\operatorname{arcctg} y \text {. }
$$
Let $\alpha=\operatorname{arcctg} a, \beta=\operatorname{arcctg} b$. Now $y=\operatorname{ctg}(\alpha-\beta)=\frac{1+\operatorname{ctg} \alpha \operatorname{ctg} \beta}{\operatorname{ctg} \beta-\operatorname{ctg} \alpha}=\frac{1+a b}{b-a}$
$$
\begin{aligned}
\operatorname{arcctg} x / 14 & =\operatorname{arcctg} 1-\operatorname{arcctg} 2-\operatorname{arcctg} 5-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{2+1}{2-1}-\operatorname{arcctg} 5-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 3-\operatorname{arcctg} 5-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{15+1}{5-3}-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 8-\operatorname{arcctg} 13-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{104+1}{13-8}-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 21-\operatorname{arcctg} 34-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{814+1}{34-21}-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} 55-\operatorname{arcctg} 89 \\
& =\operatorname{arcctg} \frac{89 * 55+1}{89-55}=\operatorname{arcctg} 144
\end{aligned}
$$
From this, we have $x=144 * 14=2016$.
|
2016
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Before you is a segment-digit. For displaying time on electronic clocks, each digit uses seven segments, each of which can be lit or not; the lit segments form the digit as shown in the figure
## 8723456789
That is, to display zero, six segments are used, to display one - two segments, and so on. On electronic clocks, only hours and minutes are displayed. How many minutes in a day does it take to display a time that uses more lit segments than the time one minute later? (The day starts at 00:00 and ends at 23:59) (20 points)
|
Solution: Let's write down the number of segments required to display each digit:
\section*{8723456789
$$
\begin{array}{llllllllll}
6 & 2 & 5 & 5 & 4 & 5 & 6 & 3 & 7 & 6
\end{array}
$$
Consider several types of displayed time:
1) The last digit of the minutes is not 9. Then the next displayed time differs by only the last digit. To reduce the number of segments, the last digit of the minutes at this moment should be 0, 3, 6, or 8. In total, such time is displayed for $24 \cdot 6 \cdot 4 = 576$ minutes in a day.
2) The last digit of the minutes is 9, but the number of minutes is not 59. Then in the next minute, the last digit will become zero, and the number of segments for its display will not change, meaning this number should decrease in the tens of minutes. That is, the digit in the tens of minutes can only be 0 or 3. In total, such time is displayed for $24 \cdot 2 = 48$ minutes in a day.
3) The number of minutes is 59, and the number of hours does not end in 9. Then in the next minute, the number of minutes will be zero, which means the number of segments for displaying the minutes will increase by 1. Therefore, the number of segments for displaying the last digit of the hours should decrease by at least 2, meaning the last digit of the hours can only be 0 or 6. In total, such time is displayed for 5 minutes in a day.
4) The number of minutes is 59, and the number of hours ends in 9. Then in the next minute, one of the times 10:00, 20:00 will be displayed - only the time 09:59 fits.
In total, suitable time is displayed for $576 + 48 + 5 + 1 = 630$ minutes in a day.
## Answer: 630 minutes
|
630
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Masha chose a natural number $n$ and wrote down all natural numbers from 1 to 6 n on the board. Then, Masha halved half of these numbers, reduced a third of the numbers by a factor of three, and increased all the remaining numbers by a factor of six. Could the sum of all the resulting numbers match the sum of the original numbers? (20 points)
|
Solution: Yes, this could have happened. Let $n=2$, then the sum of all numbers from 1 to 12 is
$$
1+2+3+\ldots+12=78
$$
Suppose Masha halved the numbers 2, 3, 5, 9, 10, 11, reduced the numbers 4, 6, 8, 12 by a factor of three, and increased the numbers 1, 7 by a factor of six. Then the sum of the resulting numbers will be
$$
\frac{2+3+5+9+10+11}{2}+\frac{4+6+8+12}{3}+6(1+7)=78
$$
Answer: could have
|
78
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In how many ways can the numbers from 1 to 9 be arranged in a $3 \times 3$ table (each number appearing exactly once) such that in each column from top to bottom and in each row from left to right, the numbers are in increasing order? (20 points)
|
Solution: Let's number the cells of the table as shown in the figure. It is clear that the number 1 is in the upper left cell, and the number 9 is in the lower right cell.
| 1 | $a_{2}$ | $a_{3}$ |
| :---: | :---: | :---: |
| $a_{4}$ | $a_{5}$ | $a_{6}$ |
| $a_{7}$ | $a_{8}$ | 9 |
By the condition, $a_{5}>a_{2}, a_{5}>a_{4}, a_{5}<a_{6}, a_{5}<a_{8}$, so $4 \leq a_{5} \leq 6$. Let's consider the cases.
1) If $a_{5}=4$, then the numbers $a_{2}$ and $a_{4}$ are 2 and 3. There are 2 ways to arrange them. Now let's calculate the number of ways to choose the numbers $a_{3}$ and $a_{6}$. Any of the remaining pairs of numbers can be placed in their positions, and $a_{3}<a_{6}$, so the arrangement of each pair is unique. There are a total of $C_{4}^{2}=6$ such pairs. The remaining two numbers are arranged uniquely. In total, we have $2 \cdot 6=12$ ways of arranging.
2) If $a_{5}=6$, then the numbers $a_{6}$ and $a_{8}$ are 7 and 8, and the case is similar to the previous one. We get another 12 ways of arranging.
3) If $a_{5}=5$, then let's see which numbers can be in the cells $a_{3}$ and $a_{7}$. The numbers 2 and 8 cannot be placed in these cells, as these numbers must be neighbors of 1 and 9, respectively. If $a_{3}=3$, then $a_{2}=2$ and $a_{4}=4$. Any of the remaining numbers can be placed in the cell $a_{6}$ in 3 ways, and the remaining numbers are placed uniquely. The considered case is similar to the cases where $a_{3}=7, a_{7}=3$, and $a_{7}=7$ - in each we get 3 ways of arranging, but the cases where the numbers $a_{3}$ and $a_{7}$ are 3 and 7 were counted twice. There are a total of two such cases:
| 1 | 2 | 3 |
| :--- | :--- | :--- |
| 4 | 5 | 6 |
| 7 | 8 | 9 |
| 1 | 4 | 7 |
| :---: | :---: | :---: |
| 2 | 5 | 8 |
| 3 | 6 | 9 |
In the end, we get $3 \cdot 4-2=10$ ways.
If neither of the numbers in the cells $a_{3}$ and $a_{7}$ is 3 or 7, then the cells $a_{3}$ and $a_{7}$ can only contain the numbers 4 and 6 in any order. Then the cells $a_{2}$ and $a_{4}$ contain the numbers 2 and 3 in any order, and the cells $a_{6}$ and $a_{8}$ contain the numbers 7 and 8 in any order. In total, there are 8 ways of arranging.
All cases have been considered, and the desired number of ways is $24+10+8=42$.
Answer: 42
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Let $x$ be a natural number. Solve the equation
$$
\frac{x-1}{x}+\frac{x-2}{x}+\frac{x-3}{x} \cdots+\frac{1}{x}=3
$$
|
Solution. Multiply both sides of the equation by $x$. We get
$$
(x-1)+(x-2)+(x-3)+\ldots+1=3 x
$$
The left side of this equation is the sum of the terms of an arithmetic progression, where $a_{1}=x-1, d=-1$, and $a_{n}=1$. The progression has $(x-1)$ terms.
Since $S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n$, the equation can be rewritten as
$$
\frac{x-1+1}{2} \cdot(x-1)=3 x \quad \Rightarrow \quad x^{2}=7 x \quad \Rightarrow \quad x=7(x \neq 0)
$$
Answer. $x=7$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many units are in the number $S=9+99+999+\cdots+\overbrace{9 \ldots 90}^{1000}$?
|
$$
\begin{aligned}
& S=10-1+100-1+1000-1+\cdots+10^{1000}-1=10\left(1+10+100+\cdots+10^{999}\right)- \\
& -100=10 \frac{10^{1000}-1}{9}-1000=\underbrace{111 \ldots 10}_{999 \text { ones }}-1000 \Rightarrow
\end{aligned}
$$
the number has 998 ones. Answer: $\{998\}$.
|
998
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Solve the equation $\sqrt{6-x}+\sqrt{x-4}=x^{2}-10 x+27$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
3. Given $4 \leq x \leq 6$. Find the maximum of the function $f(x)=\sqrt{6-x}+\sqrt{x-4}$
$f^{\prime}(x)=-\frac{1}{\sqrt{6-x}}+\frac{1}{\sqrt{x-4}} \Rightarrow f^{\prime}(x)=0 ; \sqrt{4-x}=\sqrt{6-x} \Rightarrow x=5$. When $x=5$, $f(x)$ has a maximum value of 2. On the other hand, $x^{2}-10 x+27=(x-5)^{2}+2 \Rightarrow \min \varphi(x)=x^{2}-10 x+27=2$ when $x=5$. Therefore, equality is possible only when $x=5$. Answer: $\{5\}$.
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
4. Solve the inequality $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} \geq 0$. In the answer, indicate the sum of all natural numbers that are solutions to the inequality.
|
4. $\frac{\left|x^{2}-3 x\right|-\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|-\left|x^{2}-2 x\right|} * \frac{\left|x^{2}-3 x\right|+\left|x^{2}-2\right|}{\left|x^{2}-x-2\right|+\left|x^{2}-2 x\right|} \geq 0 \Leftrightarrow \frac{\left(x^{2}-3 x\right)^{2}-\left(x^{2}-2\right)^{2}}{\left(x^{2}-x-2\right)^{2}-\left(x^{2}-2 x\right)^{2}} \geq 0 \Leftrightarrow$ $\Leftrightarrow \frac{\left(x^{2}-3 x-x^{2}+2\right)\left(x^{2}-3 x+x^{2}-2\right)}{\left(x^{2}-x-2-x^{2}+2 x\right)\left(x^{2}-x-2+x^{2}-2 x\right)} \geq 0 \Leftrightarrow \frac{(2-3 x)\left(2 x^{2}-3 x-2\right)}{(x-2)\left(2 x^{2}-3 x-2\right)} \geq 0 \Leftrightarrow$ $\frac{x-\frac{2}{3}}{x-2} \leq 0 \Rightarrow \frac{2}{3} \leq x<2$
. Answer: $\{1\}$
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all solutions to the equation $2017^{x}-2016^{x}=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
5. It is obvious that $x=1$ will be a solution to the given equation. Let's show that there are no others. We have: $\left(\frac{2017}{2016}\right)^{x}=1+\frac{1}{2016^{x}} ;\left(\frac{2017}{2016}\right)^{x}-1=\frac{1}{2016^{x}}$ From this, it is clear that the function on the left side is increasing, while the function on the right side is decreasing. Therefore, the equation has no more than one solution. Answer: $\{1\}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Three circles with radii $1,2,3$ touch each other externally. Find the radius of the circle passing through the three points of tangency of these circles.
|
6. Let the centers of the given circles be denoted as $O_{1}, O_{2}, O_{3}$. Let $K, M, N$ be the points of tangency of the given circles. Then $O_{1} K=O_{1} N=1, O_{2} K=O_{2} M=2, O_{3} M=O_{3} N=3$.
We will show that the desired circle coincides with the incircle of the triangle $O_{1}, O_{2}, O_{3}$. Indeed, if we inscribe a circle in the triangle $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$, then the segments connecting its vertices to the points of tangency will be the radii of the corresponding circles. Let $r$ be the radius of this circle. We will write the area of the triangle $O_{1}, O_{2}, O_{3}$ in terms of the perimeter and $r$, and using Heron's formula.
By equating these expressions for the area, we get: $\frac{1}{2} \operatorname{Pr}=\sqrt{p(p-a)(p-b)(p-c)}$, where $\mathrm{P}$ is the perimeter, and $p=\frac{\mathrm{P}}{2} ; a, b, c$ are the sides of the triangle $\mathrm{O}_{1}, \mathrm{O}_{2}, \mathrm{O}_{3}$. We obtain: $\frac{1}{2} r(2+4+6)=\sqrt{6(6-3)(6-4)(6-5)} \Rightarrow 6 r=6 ; r=1$. Answer: $\{1\}$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. A raft departed from point $A$ to point $B$ downstream the river. After 2.4 hours, a motorboat (with its own speed of 20 km/h) set off in pursuit of the raft. The motorboat caught up with the raft and immediately turned back to point $A$. After 3.6 hours, the motorboat arrived at point $A$, while the raft reached point $B$. Determine the speed of the river current.
|
7. Let the speed of the river current be $x$ km/h, $t$-the time elapsed from the moment the boat left until the moment of the meeting. Then, by the condition: 1) the distance traveled by the raft until the meeting, 2) the distance traveled by the motorboat until the meeting, 3) the distance traveled by the motorboat after the meeting, are equal to each other. From this we get:
$(2.4+t) x=t(20+x)=(3.6-2.4-t)(20-x)$. From this we find: $t=\frac{2.4}{20} x \Rightarrow x=4$. Answer: 4 km/h.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the equation $n+S(n)=1964$, where $S(n)$ is the sum of the digits of the number $n$.
|
1. We have $S(n) \leq 9 * 4=36 \Rightarrow n \geq 1964-36=1928 \Rightarrow n=1900+10 k+l$, where $2 \leq k \leq 9$ $0 \leq l \leq 9 ; 1900+10 k+l+10+k+l=1964,11 k+2 l=54,2 \leq 2 l \leq 18,36 \leq 11 k \leq 54$ $3 \frac{3}{11} \leq k \leq 4 \frac{10}{11} \Rightarrow k=4, l=5, n=1945$. Answer: $\{1945\}$.
|
1945
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Calculate
the sum: $\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\cdots+\cos 160^{\circ}+\cos 180^{\circ}$
|
2.
$\cos 20^{\circ}+\cos 40^{\circ}+\cos 60^{\circ}+\cos 80^{\circ}+\cos 100^{\circ}+\cos 120^{\circ}+\cos 140^{\circ}+\cos 160^{\circ}+$ $\cos 180^{\circ}=\cos 20+\cos 40+\frac{1}{2}+\sin 10^{\circ}-\sin 10^{\circ}-\frac{1}{2}-\cos 40^{\circ}-1-\cos 20=-1$ Answer: $\{-1\}$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Two pedestrians set out simultaneously from $A$ to $B$ and from $B$ to $A$. When the first had walked half the distance, the second had 24 km left to walk, and when the second had walked half the distance, the first had 15 km left to walk. How many kilometers will the second pedestrian have left to walk after the first finishes his walk?
|
3. Let $v_{1}, v_{2}$ be the speeds of the first and second pedestrians, $S$ be the distance from $A$ to $B$, and $x$ be the distance the second pedestrian still needs to walk when the first pedestrian finishes the crossing. From the problem statement, we get the system $\frac{S}{2} \frac{v_{2}}{v_{1}}+24=S, \frac{S}{2} \frac{v_{1}}{v_{2}}+15=S$. Denoting $\frac{v_{1}}{v_{2}}=u$ and dividing the second equation by the first, we get $8 u^{2}-6 u-5=0 \Rightarrow u=\frac{5}{4} \Rightarrow S=40 \text{ km}$. Since $S \frac{v_{2}}{v_{1}}+x=S$, then $x=8 \text{ km}$. Answer: 8 km.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The denominator of an irreducible fraction is less than the square of the numerator by one. If 2 is added to both the numerator and the denominator, the value of the fraction will be greater than $\frac{1}{3}$. If 3 is subtracted from both the numerator and the denominator, the fraction will be less than $\frac{1}{10}$. Find this fraction, given that the numerator and denominator are single-digit numbers. Write the answer as a two-digit number, where the first digit is the numerator and the second digit is the denominator.
|
4. Let $\frac{m}{n}$ be the desired irreducible fraction, where $m, n$ are single-digit numbers. According to the problem, we have $n=m^{2}-1, \frac{m+2}{n+2}>\frac{1}{3} \Rightarrow 3 m+6>m^{2}+1, m^{2}-3 m-5<0$, then $\frac{3-\sqrt{29}}{2}<m<\frac{3+\sqrt{29}}{2} \Rightarrow$ $\Rightarrow 0<m<\frac{3+5.4}{2} \Rightarrow 0<m<4.2$. Therefore, $m \in\{1,2,3,4\}$. Then $m=1$ does not work because $n=0$; similarly, $m=4$ does not work because $n=15$ is a two-digit number. We check $m=2$, then $n=3$ and the fraction will be $\frac{2}{3}$, which is impossible according to the problem. The only option left is $m=3$, then $n=8$, The fraction $\frac{3}{8}$ satisfies all the conditions of the problem. Answer: $\{38\}$.
|
38
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Solve the inequality: $\frac{\left(\left|x^{2}-2\right|-7\right)(|x+3|-5)}{|x-3|-|x-1|}>0$. Write the largest integer that is a solution to the inequality in your answer.
|
5. $\frac{\left(\left|x^{2}-2\right|-7\right)\left(\left|x^{2}-2\right|+7\right)(|x+3|-5)(|x+3|+5)}{(|x-3|-|x-1|)(|x-3|+|x-1|)}>0 \Leftrightarrow \frac{\left(\left(x^{2}-2\right)^{2}-49\right)\left((x+3)^{2}-25\right)}{(x-3)^{2}-(x-1)^{2}}>0$ $\Leftrightarrow \frac{\left(x^{2}-2-7\right)\left(x^{2}-2+7\right)(x+3-5)(x+3+5)}{(x-3-x+1)(x-3+x-1)}>0 \Leftrightarrow \frac{(x-3)(x+3)(x-2)(x+8)}{(-2)(2 x-4)}>0$, $x \neq 2 \Leftrightarrow(x-3)(x+3)(x+8)<0$
, Solve using the interval method:
From here, the solution is: $x<-8 ;-3<x<3 ; x \neq 2$. Therefore, the answer is: $\{1\}$
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. By the property of absolute value, replacing $x$ with $-x$ does not change this relation. This means that the figure defined by the given inequality is symmetric with respect to the OY axis. Therefore, it is sufficient to find the area of half of the figure for $x \geq 0$. In this case, we obtain the inequality $\left|x-2 y^{2}\right|+x+2 y^{2} \leq 8-4 y$. By removing the absolute value sign, we get two regions: 1: $\left\{\begin{array}{c}x \geq 2 y^{2} \\ x-2 y^{2}+x+2 y^{2} \leq 8-4 y\end{array} \Leftrightarrow\left\{\begin{array}{l}x \geq 2 y^{2} \\ y \leq 2-\frac{x}{2}\end{array}\right.\right.$
Region II: $\left\{\begin{array}{c}x \leq 2 y^{2} \\ -x+2 y^{2}+x+2 y^{2} \leq 8-4 y\end{array} \Leftrightarrow\left\{\begin{array}{c}x \leq 2 y^{2} \\ y^{2}+y-2 \leq 0\end{array} \Leftrightarrow\left\{\begin{array}{c}x \leq 2 y^{2} \\ -2 \leq y \leq 1\end{array}\right.\right.\right.$
Next, on the coordinate plane xOy, we plot the graphs of the obtained inequalities for $x \geq 0$, taking into account that $x=2 y^{2}$ is the graph of a parabola with its vertex at the origin, and the branches of this parabola are directed along the positive direction of the Ox axis. The union of these regions gives a figure which is a trapezoid $M N A B$, the area of which is equal to $S=\frac{1}{2} * 3(8+2)=15$. Then the doubled area is 30 (see Fig. 2).
|
Answer: $\{30\}$.
Fig. 2
|
30
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Let the cost of a pound of rice be x coins, a pound of brown sugar be y coins, and a pound of refined sugar be z coins. We obtain the system:
$$
\left\{\begin{array}{l}
4 x+\frac{9}{2} y+12 z=6 ; \\
12 x+6 y+6 z=8
\end{array}\right. \text { Subtract the first equation from twice the second equation and }
$$
express y in terms of x. We get $y=\frac{4}{3}-\frac{8}{3} x$. Substitute into the second equation and express $\mathrm{z}$ in terms of $\mathrm{x} z=\frac{2}{3} x$. Find the cost of the purchase:
$$
4 x+3 y+6 z=4 x+4-8 x+4 x=4
$$
|
Answer: 4 pounds sterling.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Since you have to approach each apple and return to the basket, the number of meters walked will be equal to twice the sum of the first hundred numbers, or 101 taken a hundred times, i.e., $10100 \mathrm{M}$.
|
Answer: $\{10100$ meters $\}$
|
10100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Solve the inequality: $\log _{|x-1|}\left(\frac{x-2}{x}\right)>1$. In the answer, indicate the largest negative integer that is a solution to this inequality.
|
1. Find the domain: $(x-2) x>0$. From this, $x<0$ or $x>2$. If $|x-1|<1$, then $0<x<2$. But if $|x-1|>1$, then $x<0$ or $x>2$. Therefore, the inequality is equivalent to the system:
$\left\{\begin{array}{c}|x-1|>1 \\ \frac{x-2}{x}>|x-1|\end{array} \Leftrightarrow\right.$ a) $\left\{\begin{array}{c}x<0 \\ \frac{x-2}{x}>1-x\end{array}\right.$ or b) $\left\{\begin{array}{c}x>2 \\ \frac{x-2}{x}>x-1\end{array}\right.$
From this, the solution to the first system a) is $-\sqrt{2}<x<0$, and the second system b) has no solutions. Answer: $\{-1\}$
|
-1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Find the area of the region defined by the inequality: $|y-| x-2|+| x \mid \leq 4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3. Consider three intervals of change for $x$: 1) $x \leq 0$; 2) $0 \leq x \leq 2$; 3) $x \geq 2$.
1) $x \leq 0$; then $|y+x-2|-x \leq 4, |y+x-2| \leq 4+x \Rightarrow -4 \leq x \leq 0$. Squaring the inequality, we get $(y+x-2)^{2} \leq (4+x)^{2}$, from which $(y-6)(y+2x+2) \leq 0$
2) $0 \leq x \leq 2$, then $|y+x-2| \leq 4-x \Leftrightarrow (y+x-2)^{2} \leq (4-x)^{2} \Leftrightarrow (y+2x-6)(y+2) \leq 0$
3) $x \geq 2$, then $|y-x+2| \leq 4-x \Rightarrow 2 \leq x \leq 4$. Therefore, in this region $(y-x+2)^{2} \leq (4-x)^{2} \Leftrightarrow (y-2)(y-2x+6) \leq 0$. On the coordinate plane, we construct the corresponding regions taking into account the change in $x$. We obtain the region $A B C D E F$, which consists of two right triangles $A B F$ and $C D E$, as well as trapezoid $B C E F$. The coordinates of the points are easily found: $A(-4 ; 6), B(0 ; 6), C(2 ; 2), D(4 ; 2), E(2 ;-2)$ and $F(0 ;-2)$. Taking this into account, the area of the region will be 32. Answer: $\{32\}$
|
32
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find all roots of the equation
$1-\frac{x}{1}+\frac{x(x-1)}{2!}-\frac{x(x-1)(x-2)}{3!}+\frac{x(x-1)(x-2)(x-3)}{4!}-\frac{x(x-1)(x-2)(x-3)(x-4)}{5!}+$
$+\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{6!}=0 . \quad($ (here $n!=1 \cdot 2 \cdot 3 . . . n)$
In the Answer, indicate the sum of the found roots.
|
4. Note that by substituting the numbers $1,2,3,4,5,6$ sequentially into the equation, we will get the equality: $0=0$. This means these numbers are roots of the given equation. Since the equation is of the sixth degree, there are no other roots besides the numbers mentioned above. Answer: $\{21\}$
|
21
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Solve the equation in integers: $5 x^{2}-2 x y+2 y^{2}-2 x-2 y=3$. In the answer, write the sum of all solutions $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
5. Transforming the equation, we get: $(x+y)^{2}-2(x+y)+1+y^{2}-4 x y+4 x^{2}=4 \Leftrightarrow(x+y-1)^{2}+(y-2 x)^{2}=4$. Since $x$ and $y$ are integers, this equality is possible only in four cases:
$\left\{\begin{array}{c}x+y-1=0 \\ y-2 x=2\end{array}, \quad\left\{\begin{array}{c}x+y-1=2 \\ y-2 x=0\end{array}, \quad\left\{\begin{array}{c}x+y-1=0 \\ y-2 x=-2\end{array}, \quad\left\{\begin{array}{c}x+y-1=-2 \\ y-2 x=0\end{array}\right.\right.\right.\right.$
The first and fourth systems do not have integer solutions. The second system has the integer solution $x=1, y=2$, the third system has the solution $x=1, y=0$. $\underline{\text { Answer: }}\{4\}$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Two acute angles $\alpha$ and $\beta$ satisfy the condition $\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin}(\alpha+\beta)$. Find the sum of the angles $\alpha+\beta$ in degrees.
|
6. Using the formula for $\operatorname{Sin}(\alpha+\beta)$, we get:
$\operatorname{Sin}^{2} \alpha+\operatorname{Sin}^{2} \beta=\operatorname{Sin} \alpha \operatorname{Cos} \beta+\operatorname{Cos} \alpha \operatorname{Sin} \beta \Leftrightarrow \operatorname{Sin} \alpha(\operatorname{Sin} \alpha-\operatorname{Cos} \beta)=\operatorname{Sin} \beta(\operatorname{Cos} \alpha-\operatorname{Sin} \beta)$. Since $\alpha, \beta$ are acute angles, the brackets $\operatorname{Sin} \alpha-\operatorname{Cos} \beta$ and $\operatorname{Cos} \alpha-\operatorname{Sin} \beta$ have the same sign. Let them be positive, then:
$\left\{\begin{array}{l}\operatorname{Sin} \alpha-\operatorname{Cos} \beta>0 \\ \operatorname{Cos} \alpha-\operatorname{Sin} \beta>0\end{array} \Leftrightarrow \Leftrightarrow\left\{\begin{array}{l}\operatorname{Sin} \alpha>\operatorname{Cos} \beta \\ \operatorname{Cos} \alpha>\operatorname{Sin} \beta\end{array} \Leftrightarrow\left\{\begin{array}{l}\operatorname{Sin}^{2} \alpha>\operatorname{Cos}^{2} \beta \\ \operatorname{Cos}^{2} \alpha>\operatorname{Sin}^{2} \beta\end{array}\right.\right.\right.$ $\Rightarrow \operatorname{Sin}^{2} \alpha+\operatorname{Cos}^{2} \alpha>\operatorname{Cos}^{2} \beta+\operatorname{Sin}^{2} \beta=1$, which is impossible. Similarly, it is impossible that $\operatorname{Sin}^{2} \beta+\operatorname{Cos}^{2} \beta<1$. Therefore, $\operatorname{Sin} \alpha=\operatorname{Cos} \beta, \operatorname{Cos} \alpha=\operatorname{Sin} \beta \Rightarrow \operatorname{Sin} \alpha=\operatorname{Sin}(90-\beta) \Rightarrow \alpha+\beta=90^{\circ}$. A similar proof applies if the brackets $\operatorname{Sin} \alpha-\operatorname{Cos} \beta$ and $\operatorname{Cos} \alpha-\operatorname{Sin} \beta$ are both negative. Answer: $\left\{90^{\circ}\right\}$
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find $\max x^{2} y^{2} z$ subject to the condition that $x, y, z \geq 0$ and $2 x+3 x y^{2}+2 z=36$.
|
7. From the inequality for the arithmetic mean and geometric mean of three numbers, we have:
$\sqrt[3]{2 x \cdot 3 x y^{2} \cdot 2 z} \leq \frac{2 x+3 x y^{2}+2 z}{3}=12 \Rightarrow 3 x y^{2} \cdot 2 z \cdot 2 x \leq 12^{3}, x^{2} y^{2} z \leq 144$
This inequality becomes an equality if the numbers are equal, i.e., $2 x=3 x y^{2}=2 z=12 \Rightarrow z=6, x=6, y=\sqrt{2 / 3}$. Therefore, the maximum value is 144.
Answer: $\{144\}$
|
144
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Determine which of the numbers is greater: (1000!) $)^{2}$ or $1000^{1000}$. Write 1 in the answer if the first is greater or equal, 2 if the second is greater.
|
8. Let's show that the left number is greater than the right one. For this, we will write the first number as follows: $(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \ldots \cdot 1000)(1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot \ldots \cdot 1000)=(1 \cdot 1000)(2 \cdot 999)(3 \cdot 998)(4 \cdot 997) \ldots(k(1000-k+1)) \ldots$ In this product, in each pair in parentheses, the product of the numbers is obviously not less than 1000, and since there are exactly a thousand such pairs, the product of these pairs will be strictly
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. What is the maximum area that a triangle with sides $a, b, c$ can have, given that the sides are within the ranges: $0 < a \leq 1, 1 \leq b \leq 2, 2 \leq c \leq 3$?
|
9. Among the triangles with two sides $a, b$, which satisfy the conditions $0<a \leq 1, 1 \leq b \leq 2$. The right triangle with legs $a=1, b=2$ has the largest area. Its area $S=1$. On the other hand, the third side $c=\sqrt{5}$ as the hypotenuse of the right triangle satisfies the condition that $2 \leq c \leq 3$ and therefore among all the considered triangles, the right triangle with legs 1 and 2 has the largest area. Answer: $\{1\}$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Masha has an integer number of times more toys than Lena, and Lena has the same number of times more toys than Katya. Masha gave Lena 3 toys, and Katya gave Lena 2 toys. After that, the number of toys the girls had formed an arithmetic progression. How many toys did each girl originally have? Indicate the total number of toys the girls had.
|
1. Let initially Katya has $a$ toys, then Lena has $k a$, and Masha has $\kappa^{2} a$, where $\kappa-$ is a natural number $\geq 2$. After the changes: Katya has $a-2$, Lena has $a \kappa+5$, and Masha has $a \kappa^{2}-3$. Then these numbers form an arithmetic progression in some order. Let's consider the possible cases. If $a \kappa^{2}-3$ is the middle number in the arithmetic progression, then $2 a \kappa^{2}-6=a-2+a \kappa+5 \Rightarrow a\left(2 \kappa^{2}-\kappa-1\right)=9$. But $\kappa \geq 2$, so $2 \kappa^{2}-\kappa-1 \geq 5$ and therefore the possible equality: $2 \kappa^{2}-\kappa-1=9$. This equation does not have natural positive integer solutions. If $a \kappa+5$ is the middle number ($a-2$ is obviously not the middle number), then $2 a \kappa+10=a-2+a \kappa^{2}-3$. From this, $a(k-1)^{2}=15$. Therefore, $(\kappa-1)^{2} \in\{1,3,5,15\}$. If $(\kappa-1)^{2}=1$, then $\kappa=2, a=15 \Rightarrow 15,30,60$ are the required numbers. If $(\kappa-1)^{2}=3$, then $\kappa-$ is not an integer. The same statement is true in the other cases. Thus, Katya has 15 toys, Lena has 30, and Masha has 60. Therefore, the answer is: $\{105\}$
|
105
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given a triangle $\mathrm{PEF}$ with sides $\mathrm{PE}=3, \mathrm{PF}=5, \mathrm{EF}=7$. On the extension of side $\mathrm{FP}$ beyond point $\mathrm{P}$, a segment $\mathrm{PA}=1.5$ is laid out. Find the distance $d$ between the centers of the circumcircles of triangles $\mathrm{EPA}$ and $\mathrm{EAF}$. In the answer, specify the number equal to $2d$.
|
3. By the cosine theorem for angle EPF, we find that $\operatorname{Cos} E P F=-\frac{1}{2}$. Therefore, the angle $\mathrm{EPF}=120^{\circ}$; hence the adjacent angle will be $60^{\circ}$. Draw a perpendicular from point E to PF, meeting at point D. Since angle EPF is obtuse, point D will lie outside triangle EPF. In the right triangle PED, PD lies opposite the angle $30^{\circ}$, so $\mathrm{PD}=\frac{1}{2} \mathrm{EP}=1.5$. Therefore, point D coincides with point A. Thus, we have two right triangles: EAP and EAF. The centers of the circumcircles of these triangles lie on the midpoints of the corresponding hypotenuses. Therefore, the desired distance is equal to the midline of triangle EPF, i.e., $d=2.5$. Hence, the answer is: $\{5\}$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find the minimum value of the sum
$$
\left|x-1^{2}\right|+\left|x-2^{2}\right|+\left|x-3^{2}\right|+\ldots+\left|x-10^{2}\right|
$$
|
4. Grouping the terms, we get $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)+\left(\left|x-2^{2}\right|+\left|x-9^{2}\right|\right)+\left(\left|x-3^{2}\right|+\left|x-8^{2}\right|\right)+\left(\left|x-4^{2}\right|+\left|x-7^{2}\right|\right)+\left(\left|x-5^{2}\right|+\left|x-6^{2}\right|\right)$. It is known that the pair $|x-a|+|x-b|$ takes the minimum value when $a \leq x \leq b$ and this value is equal to $|b-a|$. Therefore, the minimum value of the first bracket $\left(\left|x-1^{2}\right|+\left|x-10^{2}\right|\right)$ will be $10^{2}-1^{2}$, the second bracket respectively $9^{2}-2^{2}$, the third $8^{2}-3^{2}$, and so on. However, the intervals $5^{2} \leq x \leq 6^{2}, 4^{2} \leq x \leq 7^{2}, \ldots$ are nested within each other, since the numbers $1^{2}<2^{2}<3^{2} \ldots<10^{2}$. Therefore, the interval $\left[1,10^{2}\right]$ contains all other intervals, on each of which all terms of the form $|x-a|+|x-b|$ simultaneously take the minimum value. Then our expression takes the minimum value, equal to $10^{2}-1^{2}+9^{2}-2^{2}+8^{2}-3^{2}+7^{2}-4^{2}+6^{2}-5^{2}$, (i.e., the sum of the lengths of the nested intervals). We find this sum: $11 \cdot 9+11 \cdot 7+11 \cdot 5+11 \cdot 3+11 \cdot 1=11(9+7+5+3+1)=$ $=11 \cdot 25=275$ Answer: $\{275\}$
|
275
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Find all integers $n$ for which the equality $\frac{n^{2}+3 n+5}{n+2}=1+\sqrt{6-2 n}$ holds.
|
5. We have: $\frac{n^{2}+3 n+5}{n+2}-1=\sqrt{6-2 n} ; \frac{n^{2}+2 n+3}{n+2}=\sqrt{6-2 n} ; n+\frac{3}{n+2}=\sqrt{6-2 n}$. Since for integer $n$, the right-hand side can be either an integer or irrational, while the left-hand side is either an integer or rational, the given equality is possible only if $\frac{3}{n+2}$ is an integer. Then $n+2=\{-1 ; 1 ;-3 ; 3\} \rightarrow n=\{-3 ;-1 ;-5 ; 1\}$. Then we find that $6-2 n$ will be the square of an integer only for $n=-5$ and $n=1$. Upon verification, we find that only $n=1$ is a solution to the original equation. Answer $\{1\}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Solve the system
$$
\left\{\begin{array}{l}
\operatorname{tg}^{3} x+\operatorname{tg}^{3} y+\operatorname{tg}^{3} z=36 \\
\operatorname{tg}^{2} x+\operatorname{tg}^{2} y+\operatorname{tg}^{2} z=14 \\
\left(\operatorname{tg}^{2} x+\operatorname{tg} y\right)(\operatorname{tg} x+\operatorname{tg} z)(\operatorname{tg} y+\operatorname{tg} z)=60
\end{array}\right.
$$
In the answer, indicate the sum of the minimum and maximum $\operatorname{tgx}$, which are solutions to the system.
|
6. Let $\operatorname{tg} x=x_{1}, \operatorname{tg} y=x_{2}, \operatorname{tg} z=x_{3}$.
Then $\left\{\begin{array}{l}x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=36 \\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=14 \\ \left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)=60\end{array}\right.$. Further, let $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=a \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=b \text {. } \\ x_{1} x_{2} x_{3}=c\end{array}\right.$.
We will use the identities: $(x+y+z)^{3}-3(x+y+z)(x y+x z+y z)=x^{3}+y^{3}+z^{3}-3 x y z$
$$
x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+x z+y z)
$$
Then the system can be written as
$\left\{\begin{array}{l}a^{3}-3(a b-c)=36 \\ a^{2}-2 b=14 \\ a b-c=60\end{array} \Rightarrow\left\{\begin{array}{l}a^{3}-3 \cdot 60=36 \\ a^{2}-2 b=14 \\ a b-c=60\end{array} \Rightarrow a^{3}=216, a=6 \Rightarrow\right.\right.$
Then $2 b=36-14=22, b=11$. And finally $c=66-60=6$.
Thus, we have the system: $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=6 \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=11 \\ x_{1} x_{2} x_{3}=6\end{array}\right.$.
Consider the cubic equation in $x$ with roots $x_{1}, x_{2}, x_{3}$. Then by Vieta's theorem, we have $x^{3}-x^{2}\left(x_{1}+x_{2}+x_{3}\right)+x\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)-x_{1} x_{2} x_{3}=0$.
Further, we have: $x^{3}-6 x^{2}+11 x-6=0$. Let's find the roots of this equation. The first root will be obviously $x=1$, the other roots $x=2, x=3$.
Since the system is symmetric with respect to the cyclic permutation of the variables $x_{1}, x_{2}, x_{3}$, the system has six solutions, which are obtained by cyclic permutation of the numbers $1,2,3$. Therefore, the minimum value of $\operatorname{tgx}$ is 1, and the maximum is 3.
Thus, the answer is: $\{4\}$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Solve the system
$\left\{\begin{array}{l}a+c=4 \\ a d+b c=5 \\ a c+b+d=8 \\ b d=1\end{array}\right.$
In the answer, write the sum of all solutions of the given system.
|
7. We have
$\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=x^{4}+(a+c) x^{3}+(a c+b+d) x^{2}+(a d+b c) x+b d=$
$=x^{4}+4 x^{3}+6 x^{2}+5 x+2=(x+1)(x+2)\left(x^{2}+x+1\right)$. Since the polynomial $x^{2}+x+1$ does not have
real roots, the identity $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=\left(x^{2}+3 x+2\right)\left(x^{2}+x+1\right)$
means that $\left\{\begin{array}{l}x^{2}+a x+b=x^{2}+3 x+2 \\ x^{2}+c x+d=x^{2}+x+1\end{array}\right.$, or $\left\{\begin{array}{l}x^{2}+a x+b=x^{2}+x+1 \\ x^{2}+c x+d=x^{2}+3 x+2\end{array}\right.$.
From this, we find that $a=3, b=2, c=1, d=1$ or $a=1, b=1, c=3, d=2$ (since the equality of polynomials means the equality of coefficients of corresponding powers). Answer: $\{14\}$
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Find the integer part of the number $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$
|
9. Let $a=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6}}}}}$. It is easy to see that $a>\sqrt{6}>2 \Rightarrow$ $a>2$. Let's find an upper bound for this number. At the end of the expression for $a$, replace $\sqrt{6}$ with $\sqrt{9}$. Then we get that $a<\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{9}}}}}=\sqrt{6+\sqrt{6+\sqrt{6+3}}}=\sqrt{6+\sqrt{6+3}}=\sqrt{6+3}=3$. Thus, $2<a<3$. From this, it follows that $[a]=2$. Answer: $\{2\}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1.
Since the pedestrian covers 1 km in 10 minutes, his speed is 6 km/h. There are more oncoming trams than overtaking ones because, relative to the pedestrian, the speed of the former is greater than that of the latter. If we assume the pedestrian is standing still, the speed of the oncoming trams is the sum of the tram's own speed $V$ and the pedestrian's speed. Therefore, the relative speed of the oncoming trams is $V+6$, and the relative speed of the overtaking trams is $V-6$. It is clear that the number of trams passing a given point in a certain time is proportional to their speed, from which we have:
$$
\frac{V+6}{V-6}=\frac{700}{300} \rightarrow 3 V+18=7 V-42 \text {, i.e. } V=15 \text {. }
$$
|
Answer: 15 km $/$ hour.
#
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1.
The pedestrian's speed is 5 km/h. There are more oncoming trams than overtaking ones, because the speed of the former relative to the pedestrian is greater than that of the latter. If we assume that the pedestrian is standing still, the speed of the oncoming trams is the sum of the tram's own speed $V+$ the pedestrian's speed. Therefore, the relative speed of the oncoming trams is $V+5$, and the relative speed of the overtaking trams is $V-5$. It is clear that the number of trams passing a given point in a certain time is proportional to their speed, from which we have the equation:
$$
\frac{V+5}{V-5}=\frac{600}{225} \rightarrow V=11 \text { km } / \text { hour. }
$$
|
Answer: $11 \mathrm{km} /$ hour.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A professor from the Department of Mathematical Modeling at FEFU last academic year gave 6480 twos, thereby overachieving the commitments taken at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a result, a new record was set for enclosed spaces: 11200 twos in a year. How many professors were there initially, if each professor gives the same number of twos as the others during the session?
|
1. The professors of the Department of Mathematical Modeling at DVFU in the previous academic year gave 6480 twos, thereby over-fulfilling the obligations they had taken on at the beginning of the year. In the next academic year, the number of professors increased by 3, and each of them started giving more twos. As a result, a new record for enclosed spaces was set: 11200 twos in a year. How many professors were there initially, if each professor gives the same number of twos in a session as the others?
Let the number of professors be $p$. Then the number of twos given by each of them is $6480 / p$. This number is less than the number of twos given in the next academic year $-11200 /(p+3)$. Therefore, $p \geq 5$. Each professor gives an integer number of twos, so $6480=2^{4} 3^{4} 5$ is divisible by $p$ and $11200=2^{6} 5^{2} 7$ is divisible by $p+3$. The number $p$ cannot be divisible by 3, since 11200 is not divisible by 3. Therefore, $p=5^{m} 2^{n} \geq 5$, where either $m=0 ; n=3,4$, or $m=1 ; n=0,1,2,3,4$. In the first case, we get values $p+3=11$ or $p+3=19$, which are not divisors of 11200. If $m=1$, then $p+3=13,23,43,83$ for $n=1,2,3,4$ and therefore the only option that satisfies the conditions is $m=1, n=0, p=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find natural numbers $n$ such that for all positive numbers $a, b, c$, satisfying the inequality
$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there exists a triangle with sides $a, b, c$.
|
2. Find natural numbers $n$ such that for all positive numbers $a, b, c$ satisfying the inequality
$$
n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there exists a triangle with sides $a, b, c$.
$\square$ Since $(a b+b c+c a) \leq\left(a^{2}+b^{2}+c^{2}\right)$, then $n>5$. If $n \geq 7$, then the numbers $a=1, b=1, c=2$ satisfy the inequality $n(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, but cannot be the lengths of the sides of a triangle. Therefore, it remains to check the value $n=6$. We will show that from the condition $6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)$, where $0<a \leq b \leq c$, it follows that $c<a+b$ and therefore the numbers $a, b, c$ are the lengths of the sides of some triangle.
Indeed, considering the condition as a quadratic inequality in terms of $c$, we get
$$
5 c^{2}+6(a+b) c+\left[5\left(a^{2}+b^{2}\right)-6 a b\right]<0 ; \quad D / 4=16\left(a b-(a-b)^{2}\right) \leq 16 a b
$$
Therefore, taking into account that $2 \sqrt{a b} \leq a+b$, we conclude
$$
c<\frac{3(a+b)+4 \sqrt{a b}}{5} \leq a+b
$$
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have?
|
3. The sequence of polynomials is defined by the conditions:
$$
P_{0}(x)=1, P_{1}(x)=x, P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x), n=1,2, \ldots
$$
How many distinct real roots does the polynomial $P_{2018}(x)$ have?
$\square$ We will prove by induction that the polynomial $P_{n}(x)=x^{n}+\ldots$ has exactly $n$ distinct real roots, and that between any two consecutive roots of $P_{n}(x)$ there is exactly one root of the polynomial $P_{n-1}(x)$. Indeed, $P_{2}(x)=x^{2}-1, P_{3}(x)=x\left(x^{2}-2\right)$, and this statement is true for $n=3$. Suppose $x_{1}<x_{2}<\ldots<x_{n}$ are the roots of $P_{n}(x)$, and $y_{1}<y_{2}<\ldots<y_{n-1}$ are the roots of $P_{n-1}(x)$. By the induction hypothesis, $x_{1}<y_{1}<x_{2}<y_{2}<\ldots<x_{n-1}<y_{n-1}<x_{n}$. Consider the interval $\left(x_{i}, x_{i+1}\right)$. On this interval, $P_{n}(x)$ changes sign, and $P_{n-1}(x)$ does not change sign. Therefore, $P_{n+1}(x)=x P_{n}(x)-P_{n-1}(x)$ changes sign on this interval, and hence has a root. Since $P_{n}(x)$ changes sign between $x_{n}$ and $a$ if $a$ is sufficiently large, there is a root of $P_{n+1}(x)$ on the interval $\left(x_{n}, a\right)$. Similarly, the interval $\left(-\infty, x_{1}\right)$ is considered. Thus, the polynomial $P_{n+1}(x)$ has exactly $n+1$ roots.
|
2018
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A rectangle is drawn on the board. It is known that if its width is increased by $30 \%$, and its length is decreased by $20 \%$, then its perimeter remains unchanged. How would the perimeter of the original rectangle change if its width were decreased by $20 \%$, and its length were increased by $30 \%$?
|
Answer. It will increase by $10 \%$.
Solution. Let's denote the width of the original rectangle as $a$, and the height as $b$. In the first case, the modified width and length will be $1.3a$ and $0.8b$ respectively. According to the condition, $2(a+b) = 2(1.3a + 0.8b)$, from which we get $b = 1.5a$. This means the original perimeter is $2(a + 1.5a) = 5a$. In the second case, the perimeter will be $2(0.8a + 1.3b) = 2(0.8a + 1.3 \cdot 1.5a) = 5.5a$. This number is greater than $5a$ by $0.5a$, which is $10 \%$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Misha, over the course of a week, picked an apple each day and weighed it. Each apple weighed a different amount, but the weight of each apple was a whole number of grams and ranged from 221 grams to 230 grams (inclusive). Misha also calculated the average weight of all the apples he picked, and it was always a whole number. The apple picked on the seventh day weighed 225 grams. How much did the apple picked on the sixth day weigh?
|
Answer: 230 grams.
Solution: Each apple weighed 220 grams plus an integer from 1 to 10. From the numbers 1 to 10, we need to choose 7 numbers such that their sum is divisible by 7. One of these numbers is 5, so we need to choose 6 from the numbers $1,2,3,4,6,7,8,9,10$. The smallest sum of six of these numbers is 23, and the largest is 44. Adding 5, we get numbers from 28 to 49. In this range, there are 4 numbers divisible by 7: 28, 35, 42, 49. Meanwhile, the sum of the weights for the first 6 days must be divisible by 6. From the numbers 23, 30, 37, 44, only the number 30 satisfies this condition. Thus, the total weight for the first 6 days was $220 \cdot 6 + 30$ (we will simply write 30, and add the multiple of 220 at the end). The sum of the weights for the first 5 days must be divisible by 5. The possible range for 5 days is from 16 to 40, and the numbers that fit are $20, 25, 30, 40$. However, the total weight for 5 days is less than that for 6 days, so it must be less than 30, i.e., 20 or 25. However, the difference between 30 and 25 is 5, and the number 5 is already used. Therefore, the weight of the apples for 5 days is $220 \cdot 5 + 20$, meaning the apple picked on the 6th day weighed $220 \cdot 6 + 30 - (220 \cdot 5 + 20) = 230$ grams. This value is valid, as the apples picked over the week could have weighed 221, 223, 222, 226, 228, 210, 205 grams.
|
230
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Several teams held a hockey tournament - each team played against each other once. 2 points were awarded for a win, 1 point for a draw, and no points for a loss. The team "Squirrels" won the most games and scored the fewest points. What is the minimum number of teams that could have participated in the tournament
|
Answer: 6 teams.
Solution. In a tournament with $n$ teams, the "Squirrels" scored less than the average number, i.e., no more than $n-2$ points. Therefore, some team must have scored more than the average, i.e., no less than $n$ points. For this, they had to win at least one match. Consequently, the "Squirrels" must have won at least two matches, i.e., scored no less than 4 points. Therefore, $n \geq 6$. An example for six teams is shown in the table.
| Team | $\mathbf{1}$ | $\mathbf{2}$ | $\mathbf{3}$ | $\mathbf{4}$ | $\mathbf{5}$ | Squirrels | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathbf{1}$ | | 1 | 1 | 1 | 1 | 2 | 6 |
| $\mathbf{2}$ | 1 | | 1 | 2 | 1 | 0 | 5 |
| $\mathbf{3}$ | 1 | 1 | | 1 | 2 | 0 | 5 |
| $\mathbf{4}$ | 1 | 0 | 1 | | 1 | 2 | 5 |
| $\mathbf{5}$ | 1 | 1 | 0 | 1 | | 2 | 5 |
| Squirrels | 0 | 2 | 2 | 0 | 0 | | 4 |
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. On the glade, there are 6 fewer bushes than trees. Birds flew in and sat both on the bushes and on the trees. They sat so that there were an equal number of birds on all the trees and an equal number of birds on all the bushes, but there were at least 10 more birds on a tree than on a bush. There were a total of 128 birds on the trees. How many bushes were there?
|
Answer: 2.
Solution: There are no fewer than 7 trees, as there are 6 more trees than shrubs. There are no fewer than 11 birds on one tree, as there are at least 10 more birds on one tree than on one shrub. There cannot be 12 or more trees, as then there would be $12 \cdot 11=$ 132 or more birds on the trees. Therefore, the number of trees can be 7 or 8 or 9 or 10 or 11. However, among these, only the number 8 is a divisor of the number 128. Therefore, there are 8 trees, and 6 fewer shrubs, which means 2 shrubs.
Comment: Correct solution - 20 points. There are gaps in the justification - 15 points. The problem is solved by trial and error, and it is not shown that there are no other options - 10 points. The solution is started, and there is some progress - 5 points. The solution is started, but the progress is insignificant - 1 point. The correct answer is given without explanation - 1 point.
## Variant 2
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Senya thought of two numbers, then subtracted the smaller from the larger, added both numbers and the difference, and got 68. What was the larger of the numbers Senya thought of?
|
Answer: 34.
Solution. The subtrahend plus the difference equals the minuend. Therefore, the doubled minuend equals 68.
Comment. Correct solution - 20 points. The result is obtained based on examples and the pattern is noticed but not explained - 15 points. The result is obtained based on one example - 10 points. The solution is started, but the progress is insignificant - 1 point. The correct answer is given without explanation - 1 point.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There are 28 students in the class. 17 have a cat at home, and 10 have a dog. 5 students have neither a cat nor a dog. How many students have both a cat and a dog?
|
Answer: 4.
Solution: The number of students with cats or dogs is $28-5=23$. If we add $17+10=27$, it exceeds 23 due to those who have both a cat and a dog, who were counted twice. The number of such students is $27-23=4$.
Comment: Correct solution - 20 points. Solution started, with some progress - 5 points. Solution started, but progress is insignificant - 1 point. Correct answer given without explanation - 1 point.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The little squirrel has several bags of nuts. In two bags, there are 2 nuts each, in three bags, there are 3 nuts each, in four bags, there are 4 nuts each, and in five bags, there are 5 nuts each. Help the little squirrel arrange the bags on two shelves so that there are an equal number of bags and nuts on each shelf.
|
Solution. For example, $5+5+5+4+4+2+2=27$ nuts in 7 bags - the first shelf, $5+5+4+4+3+3+3=27$ nuts in 7 bags - the second shelf.
Comment. Correct solution - 20 points. Solution started, some progress made - 5 points. Solution started, but progress insignificant - 1 point. Solution incorrect or absent - 0 points.
|
27
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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