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7. The decimal representation of a natural number $N$ consists only of ones and twos. It is known that by erasing digits from this number, any of the 10000 numbers consisting of 9999 ones and one two can be obtained. Find the smallest possible number of digits in the representation of $N$. (G. Chelnokov)
|
Answer: 10198. Solution: Example. The number 1...121...12...21...121...1, where there are 100 twos, 99 ones at the beginning and end, and 100 ones between adjacent twos. The number consisting of 9999 ones and a two, where before the two there are $100 m+n$ ones ($0 \leq m, n \leq 99$), is obtained by deleting all twos except the $(m+1)$-th, 99 ones before it, and $n$ ones after it. Estimation. Note that in the number $N$ there are no two consecutive twos - otherwise, it can be shortened by deleting one of these twos. Let the number $N$ have $k$ twos, with $a_{0}$ ones before the first two, $a_{1}$ ones between the first and second, ..., and $a_{k}$ ones after the last two. Let $s=a_{0}+\ldots+a_{k}$. To obtain a number where there is one one before the two, we will have to delete at least $a_{0}-1$ ones. Therefore, the number $s-\left(a_{0}-1\right)$ must be at least 9999, i.e., $s-a_{0} \geq 9998$. To obtain a number where there are $a_{0}+1$ ones before the two, we will have to delete the first two and at least $a_{1}-1$ ones, from which we get the inequality $s-a_{1} \geq 9998$. To obtain a number where there are $a_{0}+a_{1}+1$ ones before the two, we will have to delete the first two twos and at least $a_{2}-1$ ones, from which we get the inequality $s-a_{2} \geq 9998$. Reasoning similarly, we get that the inequality $s-a_{i} \geq 9998$ holds for all $i$ from 0 to $k-1$; in addition, to obtain a number where the two is the last, it is required that $s-a_{k} \geq 9999$. Adding all these inequalities, we get the inequality $(k+1) s-s \geq 9998(k+1)+1 \Rightarrow k s>9998(k+1) \Rightarrow$
$s>9998+9998 / k$. Since the desired number also has $k$ twos, the number of digits in it is greater than $9998+9998 / k+k \geq 9998+2 \sqrt{9998}>10197$, which is what we needed to prove.
|
10198
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On a $100 \times 100$ chessboard, 1975 rooks were placed (each rook occupies one cell, different rooks stand on different cells). What is the maximum number of pairs of rooks that could be attacking each other? Recall that a rook can attack any number of cells along a row or column, but does not attack a rook that is blocked by another rook. (I. Rubanov)
|
Answer: 3861. Solution: Sequentially remove from the $100 \times 100$ board the verticals and horizontals that do not contain rooks, each time gluing the edges of the removed strip. We will get a rectangle $\pi$, in each vertical and each horizontal of which there is at least one rook (obviously, the number of pairs of rooks attacking each other does not change in the process). Let there be $a$ horizontals and $b$ verticals in it. Note that if there are $k>0$ rooks in a horizontal or vertical, then there are exactly $k-1$ pairs of rooks attacking each other in it. Summing over all horizontals and verticals, we get that the number of pairs of rooks attacking each other is $1975 \cdot 2-(a+b)$ (*). At the same time, the area of the rectangle $\pi$ is not less than the number of rooks. We get that $a+b \geq 2 \sqrt{a b} \geq 2 \sqrt{1975}>2 \sqrt{1936}=88$, from which $a+b \geq 89$. Thus, the number of pairs of rooks attacking each other is no more than $1975 \cdot 2-89=3861$. To get an example where there are exactly 3861, it is sufficient to arbitrarily place 1975 rooks in some rectangle of size $45 \times 44$. This is possible since $45 \times 44=1980>1975$.
|
3861
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Zeus has scales that allow him to find out the weight of the load placed on them, and a bag with 100 coins, among which there are 10-gram and 9-gram coins. Zeus knows the total number $N$ of 10-gram coins in the bag, but it is unknown which ones weigh how much. He would like to make four weighings on the scales and as a result, guaranteed to find at least one 9-gram coin. For what largest $N$ is this possible? (K. Knop)
|
Answer. For $N=15$. Solution. First, let's outline Zeus's algorithm for $N=15$. By weighing a certain number of coins, he immediately determines the number of heavy coins among the weighed ones. Since he only needs to identify one light coin, he can weigh just 8 coins in the first weighing. If there are light coins among them, he continues to weigh these same coins (and forgets about all the others), otherwise, Zeus concludes that there are no more than seven heavy coins among the remaining coins, and he proceeds to weigh other coins. In any case, on the second weighing, he only needs to weigh four coins, on the third - two, and on the fourth - one.
Now let's prove that the task for Zeus is unsolvable when $N>15$. For this, consider Anti-Zeus, who, after Zeus places some coins on the scales during the first weighing, decides what the weight of these coins will be. Anti-Zeus tries to hinder Zeus, so his actions are as follows:
- if Zeus weighs no fewer than 8 and no more than $108-N$ coins, then Anti-Zeus makes exactly 8 of them heavy;
- if Zeus weighs fewer than 8 coins, then Anti-Zeus makes all these coins heavy,
- if Zeus weighs more than $108-N$ coins, then Anti-Zeus makes all the unweighed coins heavy.
Clearly, when $N>15$, Anti-Zeus achieves the following: no matter how Zeus divides the coins into two groups (weighed/unweighed), in each group consisting of at least 8 coins, there are at least 8 heavy coins.
On the second move, Anti-Zeus acts similarly: now each of the previous groups is divided into two parts (those that participated in the second weighing and those that did not), and Anti-Zeus ensures that in each of the four groups there are at least four heavy coins (or all are heavy if the group has fewer than 4 coins). After the third weighing, Anti-Zeus similarly tracks the 8 groups of coins that Zeus has, ensuring that in each group consisting of at least two coins, there are at least two heavy coins. Finally, on the fourth move, Anti-Zeus ensures that at least one heavy coin ends up in each non-empty group (of which there are no more than 16). Since Zeus knows nothing about each such group except its total weight, he cannot distinguish a heavy coin from a light one in this group, and therefore cannot reliably identify any light coin.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. What is the maximum number of white and black pawns that can be placed on a 9x9 grid (a pawn, regardless of its color, can be placed on any cell of the board) so that no pawn attacks any other (including those of the same color)? A white pawn attacks two diagonally adjacent cells on the next higher horizontal row, while a black pawn attacks two diagonally adjacent cells on the next lower horizontal row (see the figure). (A. Antropov)
|
Answer: 56. Solution: An example with 56 pawns is shown in the figure. Evaluation: Note that in each rectangle of three rows and two columns, there are no more than 4 pawns. Indeed, if there are at least 5, then on one of the colors, all three cells are occupied, and the pawn in the middle row attacks one of the two remaining ones. Therefore, in any rectangle of 9 rows and 8 columns, there are no more than 48 pawns (since it can be divided into 12 rectangles $3 \times 2$). Suppose we managed to place 57 pawns. Then in the ninth column, there must be at least 9 pawns, i.e., exactly 9. But then in the eighth column, there are no more than 2 pawns (otherwise, there would be a pawn not in the first and not in the last row that would attack some pawn in the ninth column). Then in the eighth and ninth columns together, there are no more than 11 pawns, and in the 2nd to 7th columns - no more than 36 pawns (they can be divided into 9 rectangles $3 \times 2$). It follows that in the first column, there should be 10 pawns - a contradiction.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. 200 people are standing in a circle. Each of them is either a liar or a conformist. Liars always lie. A conformist who stands next to two conformists always tells the truth. A conformist who stands next to at least one liar can either tell the truth or lie. 100 of those standing said: "I am a liar," and the other 100 said: "I am a conformist." Find the maximum possible number of conformists among these 200 people. (R. Zhendarov, S. Berlov)
|
Answer: 150. Solution: A liar cannot say, "I am a liar." Therefore, 100 people who said, "I am a liar," are conformists. All of them lied, so next to each of them stands a liar. Since next to a liar there can be a maximum of two conformists, there are no fewer than 50 liars. Thus, there are no more than 150 conformists. Example. Place 50 liars in a circle. In each of the 50 gaps between the liars, place three conformists. The middle one of these three tells the truth, the two at the ends lie that they are liars, and all liars lie that they are conformists.
|
150
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In pentagon $A B C D E A B=B C=C D=D E, \angle B=96^\circ$ and $\angle C=\angle D=108^\circ$. Find angle $E$.
|
Answer: $102^{\circ}$. Solution. Draw segments $B D$ and $C E$. Let them intersect at point $O$. Note that triangles $B C D$ and $C D E$ are isosceles with an angle of $108^{\circ}$ at the vertex, so the base angles are $36^{\circ}$ (they are marked on the diagram with one arc). Then $\angle B C E = \angle B D E = 72^{\circ}$. Angle $C O D$ is $108^{\circ}$ (since in triangle $C O D$ there are two angles of $36^{\circ}$). Therefore, $\angle C O B = 180^{\circ} - 108^{\circ} = 72^{\circ}$. Angles of $72^{\circ}$ are marked on the diagram with two arcs. We get that triangles $C B O$ and $D E O$ are isosceles. Thus, $A B = B O = B C = C D = D E = E O = x$. Note that $\angle O B A = 96^{\circ} - 36^{\circ} = 60^{\circ}$. Therefore, triangle $O B A$ is isosceles with an angle of $60^{\circ}$ at the vertex, i.e., equilateral. Hence, $A O = x$. Calculate angle $A O E$ $\angle A O E = \angle E O B - \angle A O B = 108^{\circ} - 60^{\circ} = 48^{\circ}$. Triangle $A O E$ is isosceles with an angle of $48^{\circ}$ at the vertex. Therefore, $\angle O E A = (180^{\circ} - 48^{\circ}) / 2 = 66^{\circ}$. We get that angle $E$ of the pentagon is $\angle A E D = \angle A E O + \angle O E D = 66^{\circ} + 36^{\circ} = 102$.
|
102
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. In triangle $ABC$, angle $C$ is three times larger than angle $A$, and side $AB$ is twice as long as side $BC$. Prove that angle $ABC$ is 60 degrees.
|
Solution. Let $D$ be the midpoint of side $A B$. Since $B D=B C$, triangle $B C D$ is isosceles. Let $\angle C A D=x, \angle A C D=y$. Then $\angle D C B=3 x-y$, and $\angle C D B=x+y$. Since $\angle D C B=\angle C D B$, we have $3 x-y=x+y$, from which $y=x$. Therefore, $D C=D A=D B=B C$, which means triangle $B C D$ is equilateral, and thus angle $B$ is 60 degrees.
|
60
|
Geometry
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
6. In the Thirtieth Kingdom, there are 100 cities, and no more than one road connects any two cities. One day, the tsar ordered that one-way traffic be introduced on each road, and at the same time, each road should be painted either white or black. The Minister of Transport proudly reported that after the order was carried out, from any city to any other, one could travel by alternating the colors of the roads, and in such a way that the first road in the journey would be white. What is the minimum number of roads that could have been in this country? When traveling from city to city, one can pass through intermediate cities any number of times. (M. Antipov)
|
Answer: 150. Solution: Example. Arrange the cities on a circle so that they divide it into equal arcs, and declare these arcs to be roads directed clockwise. Paint these 100 arcs in white and black colors so that the colors alternate on the circle. Direct another 50 white roads along the chords from the cities where black roads originate to the cities located one step away from them in the clockwise direction. Clearly, the described construction satisfies the emperor's order. Evaluation: We will prove a more general fact: if the kingdom has $2k$ cities, then there must be no fewer than $3k$ roads. Suppose this is not the case. Take the smallest natural $k$ such that in a kingdom of $2k$ cities, one can manage with fewer than $3k$ roads. Here $k>1$, since for two cities the order is clearly unfulfillable. Note that there are no fewer than $2k$ white roads among them, as a white road must leave each city. Therefore, there are no more than $k-1$ black roads, and hence there are at least two cities without black roads. Remove them along with the roads connected to them. In doing so, we will remove no fewer than three roads, since from each of the two cities one could leave and one could enter, and no more than one such road is counted twice. Thus, we now have $2(k-1)$ cities and fewer than $3(k-1)$ roads. Since no route with alternating road colors could pass through the removed cities, the remaining cities and roads still satisfy the emperor's condition, which contradicts the minimality of the number $k$.
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. At the vertices of a regular 100-gon, 100 chips numbered $1, 2, \ldots, 100$ were placed, in exactly that order clockwise. In one move, it is allowed to swap two adjacent chips if their numbers differ by no more than $k$. For what smallest $k$ can a series of such moves result in a configuration where each chip is shifted one position clockwise relative to its initial position? (S. Berlov)
|
Answer: 50. Solution: Example: The chip 50 is sequentially exchanged 99 times with the next one counterclockwise.
Evaluation. We reason by contradiction. Let $k<50$. First proof. We will consider the shifts of the chips relative to their initial positions, with shifts clockwise counted as positive and counterclockwise as negative. Then, when two chips are exchanged, 1 is added to the shift of one and 1 is subtracted from the shift of the other. Suppose after several moves all chips have shifted one position clockwise. Then the total shift of the chip numbered $k$ is $100 t_{k} + 1$, where $t_{k}$ is the number of complete revolutions of this chip (clockwise revolutions are counted with a plus sign, and counterclockwise with a minus sign). Since $k<50$, chips numbered 1 and 51 could not have exchanged places, and therefore they must have made the same number of complete revolutions, i.e., $t_{1} = t_{51}$. Similarly, $t_{2} = t_{52}, \ldots, t_{50} = t_{100}$. Therefore, the sum of all shifts of all chips is $100(2 t_{1} + \ldots + 2 t_{50} + 1)$. This sum must be 0, as the sum of shifts is 0 for each move. But it is not 0, since the sum in parentheses is odd. Contradiction.
Second proof. At any moment, we consider the \{lit painted\} arc from chip 100 to chip 1 clockwise. Since chips 100 and 1 cannot be exchanged in one move, any specific chip $\mathrm{m}$ $(2 \leq m \leq 99)$ could only enter the painted arc or leave the painted arc by exchanging with one of the chips 1 or 100. Since initially and at the end, chip $m$ was not on the painted arc, it must have made an equal number of \{lit entries\} to the painted arc and \{lit exits\} from the painted arc. For $m \leq 50$, chip $m$ could not have exchanged with chip 100, so it could only make \{lit entries\} or \{lit exits\} by exchanging with chip 1. On \{lit entry\}, chip 1 makes a shift of 1 clockwise, and on \{lit exit\}, 一 1 counterclockwise. We conduct similar reasoning for chips $m \geq 51$, which cannot exchange with chip 1. Thus, we find that chips 1 and 100 will make the same number of shifts clockwise and counterclockwise, so they will remain in their positions. Contradiction.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There is a cube, each face of which is divided into 4 identical square cells. Oleg wants to mark 8 cells with invisible ink so that no two marked cells share a side. Rustem has detectors. If a detector is placed in a cell, the ink on it becomes visible. What is the minimum number of detectors Rustem can place in the cells so that, no matter which cells Oleg marks afterward, all the marked cells can be identified? (R. Zhinodarov, O. Dmitriev)
|
Answer: 16. Solution: Example. Let's divide all 24 cells into eight triples, where each triple consists of three cells adjacent to one vertex of the cube. Any two cells in the same triple share a common side. Since the number of marked cells is the same as the number of triples, there must be exactly one marked cell in each triple. Place 16 detectors such that there are two detectors in each triple. If one of the detectors in a given triple triggers, we have found the marked cell in that triple; if neither detector triggers, the marked cell is the one without a detector. Evaluation: Suppose we place fewer than 16 detectors. Then there will be a triple with at least two cells without detectors (let's call them "free cells")—mark these cells on the unfolded cube diagram with a dark background. On the same unfolded diagram, mark 7 cells with the letter A as shown in the figure. Now note that if we mark the seven cells A and one of the free cells with invisible ink, the detectors will not allow us to determine which of the free cells is marked. Therefore, it is not possible to get by with fewer than 16 detectors.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Given an equilateral triangle ABC. Point $D$ is chosen on the extension of side $A B$ beyond point $A$, point $E$ is on the extension of $B C$ beyond point $C$, and point $F$ is on the extension of $A C$ beyond point $C$ such that $C F=A D$ and $A C+E F=D E$. Find the angle BDE. (A. Kuznetsov)
|
Answer: 60 - . Solution: Complete triangle $A C E$ to parallelogram $A C E G$. Since $C F=A D$, $C E=A G$ and $\cdot F C E=\cdot D A G=60 \cdot$, triangles $D A G$ and $F C E$ are equal, from which $G D=E F$. Therefore, $D E=A C+E F=G E+G D$. This means that point $G$ lies on segment $D E$, and thus $D E \| A C$, from which $\cdot B D E=\cdot B A C=60$
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. It is known that among 100 balls, exactly 51 are radioactive. There is a device into which two balls can be placed, and if both are radioactive, a light will turn on (if at least one of the two balls is not radioactive, the light will not turn on). Can all the radioactive balls be found using the device no more than 145 times?
|
Answer. Yes. Solution. Let's divide the balls into 50 pairs and test them. Consider two possible cases.
1) Exactly one of these tests revealed two radioactive balls. Then in each of the remaining 49 pairs, there is exactly one radioactive ball. Testing one of the found radioactive balls with one ball from each of the remaining pairs, we can identify all 98 remaining balls. In total, we conducted $50+49=99$ tests.
2) At least two tests revealed two radioactive balls each. Then we have already found 4 radioactive balls. Testing one of the found radioactive balls with 95 balls from the remaining 48 pairs, we will know about 99 balls whether they are radioactive or not. If there are 50 such balls, the remaining ball is radioactive, and if there are 51, it is not. We have found all the radioactive balls in $50+95=145$ tests.
Remark. More subtle reasoning can improve the estimate of 145.
|
145
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. On a white checkered board of size $25 \times 25$ cells, several cells are painted black, with exactly 9 cells painted black in each row and each column. What is the smallest $k$ such that it is always possible to repaint $k$ cells to white in such a way that it is impossible to cut out a black $2 \times 2$ square? (S. Berlov)
|
Solution. Evaluation. Note that if 9 cells are shaded in a row, then four of them can be repainted so that no two shaded cells are adjacent: it is enough to renumber the shaded cells from left to right and repaint the cells with even numbers. If such repainting is done with all even rows, then 48 cells will be repainted and there will be no shaded $2 \times 2$ squares, since there will be no two adjacent shaded cells in one even row.
Example. Let's shade non-overlapping squares along the main diagonal: the first with a side of 9 and two with a side of 8 - and the cells located along the unshaded main diagonal of the $16 \times 16$ square containing the $8 \times 8$ squares. Then on the board, it will be possible to highlight 48 non-overlapping $2 \times 2$ squares, all cells of which are shaded. Therefore, in this example, at least 48 cells need to be repainted.
|
48
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A two-digit number $N$ was multiplied by 2, the digits of the result were swapped, and then the number was divided by 2. The result was the same number $N$. How many such numbers $N$ exist?
Answers:
A) none (-) B) exactly 4 (-) C) at least 10 (+) D) at least 14 (+) E) at least 15 (-)
|
Solution. Since the result turned out to be the same number, two identical digits were swapped. This means that $2 \mathrm{~N}$ should have two such digits. Let's consider several cases:
1) When multiplying by 2, there was no carry-over to the next place value. Obviously, as $N$, the numbers 11, 22, 33, 44 fit and only they do.
2) When multiplying by 2, there was a carry-over to the tens place. Then the first digit of the number $2 \mathrm{~N}$ is 1. The last digit is always even. Note that if there was a carry-over in the units place during multiplication, then in $2 \mathrm{~N}$ only the last digit (the units digit) will be even. This means that in this case $2 \mathrm{~N}=11 *$, so $N$ can be 55, 56, 57, 58, or 59.
3) Now let's assume there was only a carry-over to the tens place. Then the number $2 \mathrm{~N}$ has two identical even last digits. That is, $2 \mathrm{~N}$ is equal to 100, 122, 144, 166, or 188. Then $N$ is equal to 50, 61, 72, 83, or 94.
Thus, there are a total of 14 different numbers.
|
14
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2021)$, if $f(1)=5, f(4)=2$.
|
# Solution.
Substituting the pairs of numbers \(a=4, b=1\) and \(a=1, b=4\) into the given equation, respectively, we get
If \(a=4, b=1\), then \(f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{2+2 \cdot 5}{3}=4, f(2)=4\).
If \(a=1, b=4\), then \(f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\frac{5+2 \cdot 2}{3}=3, f(3)=3\).
If we take \(a=0, b=3\), we get \(f\left(\frac{0+2 \cdot 3}{3}\right)=\frac{f(0)+2 f(3)}{3}, f(2)=\frac{f(0)+2 f(3)}{3}\).
Thus, \(f(0)=3 f(2)-2 f(3), f(0)=3 \cdot 4-2 \cdot 3, f(0)=6\).
Therefore, we have \(f(0)=6, f(1)=5, f(2)=4, f(3)=3, f(4)=2\).
We can form a chain of equalities
\[
\begin{aligned}
& f\left(\frac{2021+2 \cdot 2}{3}\right)=\frac{f(2022)+2 f(2)}{3}=f(675) \\
& f\left(\frac{675+2 \cdot 0}{3}\right)=\frac{f(675)+2 f(0)}{3}=f(225) \\
& f\left(\frac{225+2 \cdot 0}{3}\right)=\frac{f(225)+2 f(0)}{3}=f(75) \\
& f\left(\frac{75+2 \cdot 0}{3}\right)=\frac{f(75)+2 f(0)}{3}=f(25) \\
& f\left(\frac{25+2 \cdot 1}{3}\right)=\frac{f(25)+2 f(1)}{3}=f(9) \\
& f\left(\frac{9+2 \cdot 0}{3}\right)=\frac{f(9)+2 f(0)}{3}=f(3)
\end{aligned}
\]
Calculating in reverse order, we get:
\[
\begin{aligned}
& f(9)=3 f(3)-2 f(0) \text {, i.e., } f(9)=3 \cdot 3-2 \cdot 6=-3 ; \\
& f(25)=3 f(9)-2 f(1) \text {, i.e., } f(25)=3 \cdot(-3)-2 \cdot 5=-19 ; \\
& f(75)=3 f(25)-2 f(0) \text {, i.e., } f(75)=3 \cdot(-19)-2 \cdot 6=-69 ; \\
& f(225)=3 f(75)-2 f(0) \text {, i.e., } f(225)=3 \cdot(-69)-2 \cdot 6=-219 ; \\
& f(675)=3 f(225)-2 f(0) \text {, i.e., } f(675)=3 \cdot(-219)-2 \cdot 6=-669 ; \\
& f(2021)=3 f(675)-2 f(2) \text {, i.e., } f(2021)=3 \cdot(-669)-2 \cdot 4=-2015 \text {. }
\end{aligned}
\]
Answer. -2015.
|
-2015
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2021)$, if $f(1)=1, f(4)=7$.
#
|
# Solution.
Substituting the pairs of numbers $a=4, b=1$ and $a=1, b=4$ into the given equation, respectively, we get
If $a=4, b=1$, then $f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{7+2 \cdot 1}{3}=3, f(2)=3$.
If $a=1, b=4$, then $f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\frac{1+2 \cdot 7}{3}=5, f(3)=5$.
If we take $a=0, b=3$, we get $f\left(\frac{0+2 \cdot 3}{3}\right)=\frac{f(0)+2 f(3)}{3}, f(2)=\frac{f(0)+2 f(3)}{3}$.
Thus, $f(0)=3 f(2)-2 f(3), f(0)=3 \cdot 3-2 \cdot 5, f(0)=-1$.
Therefore, we have $f(0)=-1, f(1)=1, f(2)=3, f(3)=5, f(4)=7$.
Let's form a chain of equalities
$$
\begin{aligned}
& f\left(\frac{2021+2 \cdot 2}{3}\right)=\frac{f(2021)+2 f(2)}{3}=f(675) \\
& f\left(\frac{675+2 \cdot 0}{3}\right)=\frac{f(675)+2 f(0)}{3}=f(225) \\
& f\left(\frac{225+2 \cdot 0}{3}\right)=\frac{f(225)+2 f(0)}{3}=f(75) \\
& f\left(\frac{75+2 \cdot 0}{3}\right)=\frac{f(75)+2 f(0)}{3}=f(25) \\
& f\left(\frac{25+2 \cdot 1}{3}\right)=\frac{f(25)+2 f(1)}{3}=f(9) \\
& f\left(\frac{9+2 \cdot 0}{3}\right)=\frac{f(9)+2 f(0)}{3}=f(3)
\end{aligned}
$$
Calculating in reverse order, we get:
$$
\begin{aligned}
& f(9)=3 f(3)-2 f(0) \text {, i.e., } f(9)=3 \cdot 5-2 \cdot(-1)=17 ; \\
& f(25)=3 f(9)-2 f(1) \text {, i.e., } f(25)=3 \cdot 17-2 \cdot 1=49 ; \\
& f(75)=3 f(25)-2 f(0) \text {, i.e., } f(75)=3 \cdot 49-2 \cdot(-1)=149 ; \\
& f(225)=3 f(75)-2 f(0) \text {, i.e., } f(225)=3 \cdot 149-2 \cdot(-1)=449 ; \\
& f(675)=3 f(225)-2 f(0) \text {, i.e., } f(675)=3 \cdot 449-2 \cdot(-1)=1349 \\
& f(2021)=3 f(675)-2 f(2) \text {, i.e., } f(2021)=3 \cdot 1349-2 \cdot 3=4041
\end{aligned}
$$
Answer. 4041.
|
4041
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. (30 points) A regular triangular prism $A B C A_{1} B_{1} C_{1}$ with base $A B C$ and lateral edges $A A_{1}, B B_{1}, C C_{1}$ is inscribed in a sphere. Segment $C D$ is the diameter of this sphere, and point $K$ is the midpoint of edge $A A_{1}$. Find the volume of the prism if $C K=2 \sqrt{6}, D K=4$.
|
# Solution.
The planes of the bases $ABC$ and $A_1B_1C_1$ of the prism intersect the sphere along the circumcircles of the equilateral triangles $ABC$ and $A_1B_1C_1$; let their centers be points $O$ and $O_1$ respectively.
It is easy to show that the midpoint $M$ of the segment $OO_1$ is the center of the sphere (Fig. 3).
Fig. 3.
Draw the diameter $C_1D$ of the circle centered at point $O_1$ through point $C_1$. We will show that $CD$ is the diameter of the sphere. Indeed, the plane $CC_1D$ is perpendicular to the planes of the bases and, therefore, contains the segment $OO_1$ along with point $O_1$. Since $C_1D = 2DO_1$, the line $CD$ intersects the segment $OO_1$ at its midpoint, i.e., at the center $M$ of the given sphere.
Let $D_1$ be the projection of point $D$ onto the plane of the base $ABC$, the height of the prism is $h$, and the radii of the circles centered at $O$ and $O_1$ are $r$. Consider triangles $CAK$ and $KA_1D$. Given that $A_1D = AD_1 = r$ (triangle $A_1O_1D$ is equilateral), $AC = r\sqrt{3}$, $AK = KA_1 = \frac{h}{2}$, by the Pythagorean theorem, we obtain the system of equations:
$$
\left\{\begin{array}{l}
\frac{h^2}{4} + 3r^2 = (2\sqrt{6})^2 \\
\frac{h^2}{4} + r^2 = 4^2
\end{array}\right.
$$
Solving the system, we find that $r = 2$, $h = 4\sqrt{3}$. Then the side of the base is $2\sqrt{3}$, its area $S = 3\sqrt{3}$, and, consequently, the volume of the prism $V = S \cdot h = 36$.
Answer. 36.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 55 and 31, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
# Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of triangles $A O B$ and $D O C$, we have:
$\frac{A O}{O C}=\frac{55}{31}$ and $\overrightarrow{O C}=\frac{31}{55} \overrightarrow{A O}=\frac{31}{55} \vec{a}$
$\frac{B O}{O D}=\frac{55}{31}$ and $\overrightarrow{O D}=\frac{31}{55} \overrightarrow{B O}=\frac{31}{55} \vec{b}$.
Then
$\overrightarrow{A D}=\vec{a}+\frac{31}{55} \vec{b}$ and $\overrightarrow{B C}=\vec{b}+\frac{31}{55} \vec{a}$
Let's find the dot product
$\overrightarrow{A D} \cdot \overrightarrow{B C}=\left(\vec{a}+\frac{31}{55} \vec{b}\right) \cdot\left(\vec{b}+\frac{31}{55} \vec{a}\right)=\overrightarrow{a b}+\frac{31}{55} \vec{a} \cdot \vec{a}+\frac{31}{55} \vec{b} \cdot \vec{b}+\left(\frac{31}{55}\right)^{2} \vec{a} \vec{b}$.
Since $\vec{a} \perp \vec{b}$, then $\vec{a} \vec{b}=0$.
By the definition of the dot product of vectors, we get $\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cdot \cos 0^{0}=|\vec{a}|^{2}$ and $\vec{b} \cdot \vec{b}=|\vec{b}|^{2}$. Triangle $AOB$ is a right triangle, so $|\vec{a}|^{2}+|\vec{b}|^{2}=|A B|^{2}=55^{2}$.
Thus, $\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{31}{55}\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)=\frac{31}{55} \cdot 55^{2}=1705$.
Answer. 1705.
|
1705
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 41 and 24, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
Solution.
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of triangles $A O B$ and $D O C$, we have:
$\frac{A O}{O C}=\frac{41}{24}$ and $\overrightarrow{O C}=\frac{24}{41} \overrightarrow{A O}=\frac{24}{41} \vec{a}$.
$\frac{B O}{O D}=\frac{41}{24}$ and $\overrightarrow{O D}=\frac{24}{41} \overrightarrow{B O}=\frac{24}{41} \vec{b}$.
Then
$\overrightarrow{A D}=\vec{a}+\frac{24}{41} \vec{b}$ and $\overrightarrow{B C}=\vec{b}+\frac{24}{41} \vec{a}$.
Let's find the dot product
$\overrightarrow{A D} \cdot \overrightarrow{B C}=\left(\vec{a}+\frac{24}{41} \vec{b}\right) \cdot\left(\vec{b}+\frac{24}{41} \vec{a}\right)=\vec{a} \vec{b}+\frac{24}{41} \vec{a} \cdot \vec{a}+\frac{24}{41} \vec{b} \cdot \vec{b}+\left(\frac{24}{41}\right)^{2} \vec{a} \vec{b}$.
Since $\vec{a} \perp \vec{b}$, then $\vec{a} \vec{b}=0$.
By the definition of the dot product of vectors, we get $\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cdot \cos 0^{0}=|\vec{a}|^{2}$ and $\vec{b} \cdot \vec{b}=|\vec{b}|^{2}$.
Triangle $AOB$ is a right triangle, so $|\vec{a}|^{2}+|\vec{b}|^{2}=|A B|^{2}=41^{2}$.
Thus, $\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{24}{41}\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)=\frac{24}{41} \cdot 41^{2}=984$.
Answer. 984.
|
984
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds. The sequence must contain a term $a_{k}=2021$. Determine the maximum number of three-digit numbers, divisible by 25, that this sequence can contain.
#
|
# Solution.
The final sequence can contain all three-digit numbers, as it can consist of a given number of natural numbers starting from the chosen number $a_{i}$.
We will prove that for any term of the arithmetic progression $1,2,3, \ldots$ defined by the formula for the $n$-th term $a_{n}=n$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds.
Indeed, for any value of $k$, the equalities
$a_{k}=k, a_{k+1}=k+1, a_{k+2}=k+2$ are valid, from which it follows that
$3 a_{k+1}-2 a_{k}-1=3(k+1)-2 k-1=k+2=a_{k+2}$, i.e., the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds, which is what we needed to prove.
For example, the sequence containing 2021: 3,4,5,6,.., 2018, 2019,2020,2021.
Thus, the sequence can contain all three-digit numbers from 100 to 999. Among them, the numbers divisible by $25: 100,125,150,175,200,225,250,275, \ldots, 900,925$, 950,975 - 4 in each of the nine hundreds, i.e., 36 numbers.
Answer. 36.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2022)$, if $f(1)=1, f(4)=7$.
#
|
# Solution.
Substituting the pairs of numbers \(a=4, b=1\) and \(a=1, b=4\) into the given equation, respectively, we get
If \(a=4, b=1\), then \(f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{7+2 \cdot 1}{3}=3, f(2)=3\).
If \(a=1, b=4\), then \(f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\frac{1+2 \cdot 7}{3}=5, f(3)=5\).
If we take \(a=0, b=3\), we get \(f\left(\frac{0+2 \cdot 3}{3}\right)=\frac{f(0)+2 f(3)}{3}, f(2)=\frac{f(0)+2 f(3)}{3}\).
Thus, \(f(0)=3 f(2)-2 f(3), f(0)=3 \cdot 3-2 \cdot 5, f(0)=-1\).
Therefore, we have \(f(0)=-1, f(1)=1, f(2)=3, f(3)=5, f(4)=7\).
Let's form a chain of equalities
\[
\begin{aligned}
& f\left(\frac{2022+2 \cdot 0}{3}\right)=\frac{f(2022)+2 f(0)}{3}=f(674) \\
& f\left(\frac{674+2 \cdot 2}{3}\right)=\frac{f(674)+2 f(2)}{3}=f(226) \\
& f\left(\frac{226+2 \cdot 1}{3}\right)=\frac{f(226)+2 f(1)}{3}=f(76) \\
& f\left(\frac{76+2 \cdot 1}{3}\right)=\frac{f(76)+2 f(1)}{3}=f(26) \\
& f\left(\frac{26+2 \cdot 2}{3}\right)=\frac{f(26)+2 f(2)}{3}=f(10) \\
& f\left(\frac{10+2 \cdot 1}{3}\right)=\frac{f(10)+2 f(1)}{3}=f(4)
\end{aligned}
\]
Calculating in reverse order, we get:
\[
\begin{aligned}
& f(10)=3 f(4)-2 f(1) \text {, i.e., } f(10)=3 \cdot 7-2 \cdot 1=19 \\
& f(26)=3 f(10)-2 f(2) \text {, i.e., } f(26)=3 \cdot 19-2 \cdot 3=51 \\
& f(76)=3 f(26)-2 f(1) \text {, i.e., } f(76)=3 \cdot 51-2 \cdot 1=151 \\
& f(226)=3 f(76)-2 f(1) \text {, i.e., } f(226)=3 \cdot 151-2 \cdot 1=451 \\
& f(674)=3 f(226)-2 f(2) \text {, i.e., } f(674)=3 \cdot 451-2 \cdot 3=1347 \\
& f(2022)=3 f(674)-2 f(0) \text {, i.e., } f(2022)=3 \cdot 1347-2 \cdot(-1)=4043
\end{aligned}
\]
Answer. 4043.
|
4043
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 5. (20 points) At the first deposit, equipment of the highest class was used, and at the second deposit, equipment of the first class was used, with the highest class being less than the first. Initially, $40 \%$ of the equipment from the first deposit was transferred to the second. Then, $20 \%$ of the equipment that ended up on the second deposit was transferred back to the first, with half of the transferred equipment being of the first class. After this, the equipment of the highest class on the first deposit was 26 units more than on the second, and the total amount of equipment on the second deposit increased by more than $5 \%$ compared to the initial amount. Find the total amount of equipment of the first class.
|
# Solution.
Let there initially be $x$ units of top-class equipment at the first deposit and $y$ units of first-class equipment at the second deposit $(x1.05 y$, from which $y48 \frac{34}{67} .\end{array}\right.\right.$
This double inequality and the condition "x is divisible by 5" is satisfied by the unique value $x=50$. Then $y=130-70=60$. Thus, there were 60 units of first-class equipment.
Answer: 60.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. (10 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this sequence can contain.
|
# Solution.
Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that an arithmetic progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equation $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.
Indeed,
$a_{k}=2 k, \quad a_{k+1}=2 k+2, a_{k+2}=2 k+4$, or by the formula
$a_{k+2}=3 a_{k+1}-2 a_{k}-2=3(2 k+2)-2 \cdot 2 k-2=2 k+4$
The sequence containing 2022: 4, 6,
This finite sequence can contain all even three-digit numbers from 100 to 999. Among them, the numbers divisible by 4 are: 100, 104, 108, 112, ..., 992, 925, 950, 996 - 25 in each of the nine hundreds, i.e., 225 numbers.
Answer. 225.
|
225
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The function $f(x)$ satisfies the condition: for any real numbers $a$ and $b$, the equality $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$ holds. Find the value of the function $f(2022)$, if $f(1)=5, f(4)=2$.
|
# Solution.
Substituting the pairs of numbers $a=4, b=1$ and $a=1, b=4$ into the given equation, respectively, we get
If $a=4, b=1$, then $f\left(\frac{4+2}{3}\right)=\frac{f(4)+2 f(1)}{3}, f(2)=\frac{2+2 \cdot 5}{3}=4, f(2)=4$.
If $a=1, b=4$, then $f\left(\frac{1+2 \cdot 4}{3}\right)=\frac{f(1)+2 f(4)}{3}, f(3)=\frac{5+2 \cdot 2}{3}=3, f(3)=3$.
If we take $a=0, b=3$, we get $f\left(\frac{0+2 \cdot 3}{3}\right)=\frac{f(0)+2 f(3)}{3}, f(2)=\frac{f(0)+2 f(3)}{3}$.
Thus, $f(0)=3 f(2)-2 f(3), f(0)=3 \cdot 4-2 \cdot 3, f(0)=6$.
Therefore, we have $f(0)=6, f(1)=5, f(2)=4, f(3)=3, f(4)=2$.
Let's form a chain of equalities
$$
\begin{aligned}
& f\left(\frac{2022+2 \cdot 0}{3}\right)=\frac{f(2022)+2 f(0)}{3}=f(674) \\
& f\left(\frac{674+2 \cdot 2}{3}\right)=\frac{f(674)+2 f(2)}{3}=f(226) \\
& f\left(\frac{226+2 \cdot 1}{3}\right)=\frac{f(226)+2 f(1)}{3}=f(76) \\
& f\left(\frac{76+2 \cdot 1}{3}\right)=\frac{f(76)+2 f(1)}{3}=f(26) \\
& f\left(\frac{26+2 \cdot 2}{3}\right)=\frac{f(26)+2 f(2)}{3}=f(10) \\
& f\left(\frac{10+2 \cdot 1}{3}\right)=\frac{f(10)+2 f(1)}{3}=f(4)
\end{aligned}
$$
Calculating in reverse order, we get:
$$
\begin{aligned}
& f(10)=3 f(4)-2 f(1) \text {, i.e., } f(10)=3 \cdot 2-2 \cdot 5=-4 ; \\
& f(26)=3 f(10)-2 f(2) \text {, i.e., } f(26)=3 \cdot(-4)-2 \cdot 4=-20 \\
& f(76)=3 f(26)-2 f(1) \text {, i.e., } f(76)=3 \cdot(-20)-2 \cdot 5=-70 \\
& f(226)=3 f(76)-2 f(1) \text {, i.e., } f(226)=3 \cdot(-70)-2 \cdot 5=-220 \\
& f(674)=3 f(226)-2 f(2) \text {, i.e., } f(674)=3 \cdot(-220)-2 \cdot 4=-668 ; \\
& f(2022)=3 f(674)-2 f(0) \text {, i.e., } f(2022)=3 \cdot(-668)-2 \cdot 6=-2016
\end{aligned}
$$
Answer. -2016.
|
-2016
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 367 and 6, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
# Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of triangles $A O B$ and $D O C$, we have:
$\frac{A O}{O C}=\frac{367}{6}$ and $\overrightarrow{O C}=\frac{6}{367} \overrightarrow{A O}=\frac{6}{367} \vec{a}$.
$\frac{B O}{O D}=\frac{367}{6}$ and $\overrightarrow{O D}=\frac{6}{367} \overrightarrow{B O}=\frac{6}{367} \vec{b}$.
Then
$\overrightarrow{A D}=\vec{a}+\frac{6}{367} \vec{b}$ and $\overrightarrow{B C}=\vec{b}+\frac{6}{367} \vec{a}$.
Let's find the dot product
$\overrightarrow{A D} \cdot \overrightarrow{B C}=\left(\vec{a}+\frac{6}{367} \vec{b}\right) \cdot\left(\vec{b}+\frac{6}{367} \vec{a}\right)=\vec{a} \vec{b}+\frac{6}{367} \vec{a} \cdot \vec{a}+\frac{6}{367} \vec{b} \cdot \vec{b}+\left(\frac{6}{367}\right)^{2} \vec{a} \vec{b}$.
Since $\vec{a} \perp \vec{b}$, then $\vec{a} \vec{b}=0$.
By the definition of the dot product of vectors, we get $\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cdot \cos 0^{0}=|\vec{a}|^{2}$ and $\vec{b} \cdot \vec{b}=|\vec{b}|^{2}$
Triangle $AOB$ is a right triangle, so $|\vec{a}|^{2}+|\vec{b}|^{2}=|A B|^{2}=367^{2}$.
Thus, $\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{6}{367}\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)=\frac{6}{367} \cdot 367^{2}=2202$.
Answer. 2202.
|
2202
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are equal to 101 and 20, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of triangles $A O B$ and $D O C$, we have:
$\frac{A O}{O C}=\frac{101}{20}$ and $\overrightarrow{O C}=\frac{20}{101} \overrightarrow{A O}=\frac{20}{101} \vec{a}$
$\frac{B O}{O D}=\frac{101}{20}$ and $\overrightarrow{O D}=\frac{20}{101} \overrightarrow{B O}=\frac{20}{101} \vec{b}$.
Then
$\overrightarrow{A D}=\vec{a}+\frac{20}{101} \vec{b}$ and $\overrightarrow{B C}=\vec{b}+\frac{20}{101} \vec{a}$.
Let's find the dot product
$\overrightarrow{A D} \cdot \overrightarrow{B C}=\left(\vec{a}+\frac{20}{101} \vec{b}\right) \cdot\left(\vec{b}+\frac{20}{101} \vec{a}\right)=\vec{a} \vec{b}+\frac{20}{101} \vec{a} \cdot \vec{a}+\frac{20}{101} \vec{b} \cdot \vec{b}+\left(\frac{20}{101}\right)^{2} \vec{a} \vec{b}$.
Since $\vec{a} \perp \vec{b}$, then $\vec{a} \vec{b}=0$.
By the definition of the dot product of vectors, we get $\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cdot \cos 0^{0}=|\vec{a}|^{2}$ and $\vec{b} \cdot \vec{b}=|\vec{b}|^{2}$
Triangle $AOB$ is a right triangle, so $|\vec{a}|^{2}+|\vec{b}|^{2}=|A B|^{2}=101^{2}$.
Thus, $\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{20}{101}\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)=\frac{20}{101} \cdot 101^{2}=2020$.
Answer. 2020.
|
2020
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (6 points) In an ideal gas, a thermodynamic cycle consisting of two isochoric and two adiabatic processes is carried out. The ratio of the initial and final absolute temperatures in the isochoric cooling process is \( k = 1.5 \). Determine the efficiency of this cycle, given that the efficiency of the Carnot cycle with the same ratio of maximum to minimum temperature as in the considered cycle is \( 50\% \).
## Possible solution.
In the graph of the cyclic process, it is clear that the absolute temperatures \( T_{1} = T_{\max} \), and \( T_{3} = T_{\min} \). By definition, the efficiency is the ratio of the work done by the gas in the cycle to the amount of heat transferred to the gas. In the studied cycle, work is performed in adiabatic processes, while energy is transferred in the form of heat in isochoric processes. Using the first law of thermodynamics, we obtain for the work of
the gas in adiabatic processes:
$$
\begin{gathered}
A_{12} = -\Delta U_{12} = \frac{i}{2} v R \left( T_{\max} - T_{2} \right) > 0 \\
A_{34} = -\Delta U_{34} = \frac{i}{2} v R \left( T_{\min} - T_{4} \right) < 0
\end{gathered}
$$
where \(\Delta U\) is the change in internal energy of the gas, \(i\) is the number of degrees of freedom of the gas molecule, \(v\) is the amount of substance, and \(R\) is the universal gas constant.
The work done by the gas in the cycle:
$$
A = A_{12} + A_{34} = \frac{i}{2} v R \left[ \left( T_{\max} - T_{2} \right) - \left( T_{4} - T_{\min} \right) \right].
$$
The amount of heat transferred to the gas in the isochoric process \(4-1\):
$$
Q_{1} = \Delta U_{41} = \frac{i}{2} v R \left( T_{\max} - T_{4} \right)
$$
The efficiency
$$
\eta = \frac{\left( T_{\max} - T_{4} \right) - \left( T_{2} - T_{\min} \right)}{T_{\max} - T_{4}} = 1 - \frac{T_{2} - T_{\min}}{T_{\max} - T_{4}}
$$
According to Poisson's equation for adiabatic processes
$$
\begin{aligned}
p_{1} V_{1}^{\gamma} & = p_{2} V_{2}^{\gamma} \\
p_{4} V_{1}^{\gamma} & = p_{3} V_{1}^{\gamma} \\
\frac{p_{1}}{p_{4}} & = \frac{p_{2}}{p_{3}}.
\end{aligned}
$$
Here \(\gamma\) is the adiabatic index. By Charles's law
$$
\begin{aligned}
& \frac{p_{1}}{p_{4}} = \frac{T_{\max}}{T_{4}} \\
& \frac{p_{2}}{p_{3}} = \frac{T_{2}}{T_{\min}}
\end{aligned}
$$
We obtain that
$$
\frac{T_{\max}}{T_{4}} = \frac{T_{2}}{T_{\min}} = k
$$
Considering the obtained equality, we can write:
$$
\eta = 1 - \frac{T_{2} - T_{\min}}{T_{\max} - T_{4}} = 1 - \frac{T_{\min} (k - 1)}{T_{\max} \left( 1 - \frac{1}{k} \right)} = 1 - k \frac{T_{\min}}{T_{\max}}
$$
For the Carnot cycle
$$
\eta_{\mathrm{K}} = 1 - \frac{T_{\min}}{T_{\max}}
$$
Then
$$
\eta = 1 - k \left( 1 - \eta_{\mathrm{K}} \right) = 0.25 = 25\%.
$$
|
Answer: $\eta=1-k\left(1-\eta_{C}\right)=0.25=25 \%$.
## Evaluation Criteria
| Performance | Score |
| :--- | :---: |
| Participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for work in the cycle is written | $\mathbf{1}$ |
| Expression for the amount of heat received is written | $\mathbf{1}$ |
| Expression for the efficiency coefficient in terms of temperatures in the cycle states is written | $\mathbf{1}$ |
| Expression for the temperature ratios in isochoric processes is obtained | $\mathbf{1}$ |
| Expression for the efficiency of the Carnot cycle is written | $\mathbf{1}$ |
| Necessary transformations are performed and a numerical answer is obtained | $\mathbf{1}$ |
| :---: | :---: | :---: |
| Total points | $\mathbf{6}$ |
|
25
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 2. By how many units can the city's fleet of natural gas vehicles be increased in 2022, assuming that the capacity of each of the old CNG stations in the city is equal to the capacity of the new station on Narodnaya Street, and that the city's fleet constitutes only $70 \%$ of all vehicles refueling at CNG stations in St. Petersburg. Assume that the stations are used at $100 \%$ of their capacity.
|
Task 2. There are a total of 15 stations: 4 new ones and 11 old ones.
The throughput capacity of the old stations is 11 x $200=2200$ vehicles per day, and for the new ones: $200+700=900$ vehicles per day. In total, the stations can refuel: $2200+900=3100$ vehicles per day.
Vehicles from the city fleet account for only $70 \%$ of all vehicles refueled at the CNG stations in St. Petersburg, so $3100 x 0.7=2170$ vehicles can be refueled. Currently, there are 1000 vehicles in the city fleet, so the city fleet of gas-powered vehicles can be increased by 1170 units.
|
1170
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}+4096}{64 a^{6}}$, if $\frac{a}{2}-\frac{2}{a}=5$.
#
|
# Solution.
$$
\begin{aligned}
& \frac{a^{12}+4096}{64 a^{6}}=\frac{a^{6}}{64}+\frac{64}{a^{6}}=\frac{a^{6}}{64}-2+\frac{64}{a^{6}}+2=\left(\frac{a^{3}}{8}-\frac{8}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{8}-3 \cdot \frac{a}{2}+3 \cdot \frac{2}{a}-\frac{8}{a^{3}}+3\left(\frac{a}{2}-\frac{2}{a}\right)\right)^{2}+2= \\
& =\left(\left(\frac{a}{2}-\frac{2}{a}\right)^{3}+3\left(\frac{a}{2}-\frac{2}{a}\right)\right)^{2}+2=\left(5^{3}+3 \cdot 5\right)^{2}+2=19602
\end{aligned}
$$
Answer. 19602.
|
19602
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}+729^{2}}{729 a^{6}}$, if $\frac{a}{3}-\frac{3}{a}=4$.
#
|
# Solution.
$$
\begin{aligned}
& \frac{a^{12}+729^{2}}{729 a^{6}}=\frac{a^{6}}{729}+\frac{729}{a^{6}}=\frac{a^{6}}{729}-2+\frac{729}{a^{6}}+2=\left(\frac{a^{3}}{27}-\frac{27}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{27}-3 \cdot \frac{a}{3}+3 \cdot \frac{3}{a}-\frac{27}{a^{3}}+3\left(\frac{a}{3}-\frac{3}{a}\right)\right)^{2}+2= \\
& =\left(\left(\frac{a}{3}-\frac{3}{a}\right)^{3}+3\left(\frac{a}{3}-\frac{3}{a}\right)\right)^{2}+2=\left(4^{3}+3 \cdot 4\right)^{2}+2=5778
\end{aligned}
$$
Answer. 5778.
|
5778
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}+729^{2}}{729 a^{6}}$, if $\frac{a}{3}-\frac{3}{a}=2$.
|
Solution.
$$
\begin{aligned}
& \frac{a^{12}+729^{2}}{729 a^{6}}=\frac{a^{6}}{729}+\frac{729}{a^{6}}=\frac{a^{6}}{729}-2+\frac{729}{a^{6}}+2=\left(\frac{a^{3}}{27}-\frac{27}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{27}-3 \cdot \frac{a}{3}+3 \cdot \frac{3}{a}-\frac{27}{a^{3}}+3\left(\frac{a}{3}-\frac{3}{a}\right)\right)^{2}+2= \\
& =\left(\left(\frac{a}{3}-\frac{3}{a}\right)^{3}+3\left(\frac{a}{3}-\frac{3}{a}\right)\right)^{2}+2=\left(2^{3}+3 \cdot 2\right)^{2}+2=14^{2}+2=198
\end{aligned}
$$
Answer. 198.
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}+4096}{64 a^{6}}$, if $\frac{a}{2}-\frac{2}{a}=3$.
#
|
# Solution.
$$
\begin{aligned}
& \frac{a^{12}+4096}{64 a^{6}}=\frac{a^{6}}{64}+\frac{64}{a^{6}}=\frac{a^{6}}{64}-2+\frac{64}{a^{6}}+2=\left(\frac{a^{3}}{8}-\frac{8}{a^{3}}\right)^{2}+2= \\
& =\left(\frac{a^{3}}{8}-3 \cdot \frac{a}{2}+3 \cdot \frac{2}{a}-\frac{8}{a^{3}}+3\left(\frac{a}{2}-\frac{2}{a}\right)\right)^{2}+2= \\
& =\left(\left(\frac{a}{2}-\frac{2}{a}\right)^{3}+3\left(\frac{a}{2}-\frac{2}{a}\right)\right)^{2}+2=\left(3^{3}+3 \cdot 3\right)^{2}+2=1298
\end{aligned}
$$
Answer. 1298.
|
1298
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}-729}{27 a^{6}}$, if $\frac{a^{2}}{3}-\frac{3}{a^{2}}=4$.
#
|
# Solution.
$$
\begin{aligned}
& \frac{a^{12}-729}{27 a^{6}}=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}+1+\frac{9}{a^{4}}\right)=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}-2+\frac{9}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)^{2}+3\right)=4 \cdot\left(4^{2}+3\right)=76
\end{aligned}
$$
Answer. 76.
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}-4096}{64 a^{6}}$, if $\frac{a^{2}}{4}-\frac{4}{a^{2}}=3$.
|
# Solution.
$$
\begin{aligned}
& \frac{a^{12}-4096}{64 a^{6}}=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}+1+\frac{16}{a^{4}}\right)=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}-2+\frac{16}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\left(\frac{a^{2}}{16}-\frac{16}{a^{2}}\right)^{2}+3\right)=3 \cdot\left(3^{2}+3\right)=36
\end{aligned}
$$
Answer. 36.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}-729}{27 a^{6}}$, if $\frac{a^{2}}{3}-\frac{3}{a^{2}}=6$.
|
Solution.
$$
\begin{aligned}
& \frac{a^{12}-729}{27 a^{6}}=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}+1+\frac{9}{a^{4}}\right)=\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\frac{a^{4}}{9}-2+\frac{9}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)\left(\left(\frac{a^{2}}{3}-\frac{3}{a^{2}}\right)^{2}+3\right)=6 \cdot\left(6^{2}+3\right)=234
\end{aligned}
$$
Answer. 234.
|
234
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{12}-4096}{64 a^{6}}$, if $\frac{a^{2}}{4}-\frac{4}{a^{2}}=5$.
|
# Solution.
$$
\begin{aligned}
& \frac{a^{12}-4096}{64 a^{6}}=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}+1+\frac{16}{a^{4}}\right)=\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\frac{a^{4}}{16}-2+\frac{16}{a^{4}}+3\right)= \\
& =\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right)\left(\left(\frac{a^{2}}{16}-\frac{16}{a^{2}}\right)^{2}+3\right)=5 \cdot\left(5^{2}+3\right)=140
\end{aligned}
$$
Answer. 140.
|
140
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. (6 points) In an ideal gas, a thermodynamic cycle consisting of two isochoric and two adiabatic processes is carried out. The ratio of the initial and final absolute temperatures in the isochoric cooling process is \( k = 1.5 \). Determine the efficiency of this cycle, given that the efficiency of the Carnot cycle with the same ratio of maximum to minimum temperature as in the considered cycle is \( 50\% \).
## Possible solution.
In the graph of the cyclic process, it is clear that the absolute temperatures \( T_{1} = T_{\max} \), and \( T_{3} = T_{\min} \). By definition, the efficiency is the ratio of the work done by the gas in the cycle to the amount of heat transferred to the gas. In the studied cycle, work is performed in the adiabatic processes, while energy is transferred in the form of heat in the isochoric processes. Using the first law of thermodynamics, we obtain for the work of
the gas in the adiabatic processes:
$$
\begin{gathered}
A_{12} = -\Delta U_{12} = \frac{i}{2} v R \left( T_{\max} - T_{2} \right) > 0 \\
A_{34} = -\Delta U_{34} = \frac{i}{2} v R \left( T_{\min} - T_{4} \right) < 0
\end{gathered}
$$
where \(\Delta U\) is the change in internal energy of the gas, \(i\) is the number of degrees of freedom of the gas molecule, \(v\) is the amount of substance, and \(R\) is the universal gas constant.
The work done by the gas in the cycle:
$$
A = A_{12} + A_{34} = \frac{i}{2} v R \left[ \left( T_{\max} - T_{2} \right) - \left( T_{4} - T_{\min} \right) \right].
$$
The amount of heat transferred to the gas in the isochoric process \(4-1\):
$$
Q_{1} = \Delta U_{41} = \frac{i}{2} v R \left( T_{\max} - T_{4} \right)
$$
The efficiency
$$
\eta = \frac{\left( T_{\max} - T_{4} \right) - \left( T_{2} - T_{\min} \right)}{T_{\max} - T_{4}} = 1 - \frac{T_{2} - T_{\min}}{T_{\max} - T_{4}}
$$
According to Poisson's equation for adiabatic processes
$$
\begin{aligned}
p_{1} V_{1}^{\gamma} & = p_{2} V_{2}^{\gamma} \\
p_{4} V_{1}^{\gamma} & = p_{3} V_{1}^{\gamma} \\
\frac{p_{1}}{p_{4}} & = \frac{p_{2}}{p_{3}}.
\end{aligned}
$$
Here \(\gamma\) is the adiabatic index. By Charles's law
$$
\begin{aligned}
& \frac{p_{1}}{p_{4}} = \frac{T_{\max}}{T_{4}} \\
& \frac{p_{2}}{p_{3}} = \frac{T_{2}}{T_{\min}}
\end{aligned}
$$
We obtain that
$$
\frac{T_{\max}}{T_{4}} = \frac{T_{2}}{T_{\min}} = k
$$
Considering the obtained equality, we can write:
$$
\eta = 1 - \frac{T_{2} - T_{\min}}{T_{\max} - T_{4}} = 1 - \frac{T_{\min} (k - 1)}{T_{\max} \left( 1 - \frac{1}{k} \right)} = 1 - k \frac{T_{\min}}{T_{\max}}
$$
For the Carnot cycle
$$
\eta_{\mathrm{K}} = 1 - \frac{T_{\min}}{T_{\max}}
$$
Then
$$
\eta = 1 - k \left( 1 - \eta_{\mathrm{K}} \right) = 0.25 = 25\%.
$$
|
Answer: $\eta=1-k\left(1-\eta_{C}\right)=0.25=25 \%$.
## Evaluation Criteria
| Performance | Score |
| :--- | :---: |
| Participant did not start the task or performed it incorrectly from the beginning | $\mathbf{0}$ |
| Expression for work in the cycle is written | $\mathbf{1}$ |
| Expression for the amount of heat received is written | $\mathbf{1}$ |
| Expression for the efficiency coefficient in terms of temperatures in the cycle states is written | $\mathbf{1}$ |
| Expression for the temperature ratios in isochoric processes is obtained | $\mathbf{1}$ |
| Expression for the efficiency of the Carnot cycle is written | $\mathbf{1}$ |
| Necessary transformations are performed and a numerical answer is obtained | $\mathbf{1}$ |
| :---: | :---: | :---: |
| Total points | $\mathbf{6}$ |
|
25
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{8}+256}{16 a^{4}}$, if $\frac{a}{2}+\frac{2}{a}=5$.
|
Solution.
$$
\begin{aligned}
& \frac{a^{8}+256}{16 a^{4}}=\frac{a^{4}}{16}+\frac{16}{a^{4}}=\frac{a^{4}}{16}+2+\frac{16}{a^{4}}-2=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{4}+2+\frac{4}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{2}+\frac{2}{a}\right)^{2}-2\right)^{2}-2=\left(5^{2}-2\right)^{2}-2=527
\end{aligned}
$$
Answer. 527.
|
527
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) Laboratory engineer Sergei received an object for research consisting of about 200 monoliths (a container designed for 200 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequency (statistical probability) that a randomly selected monolith will be sandy loam is $\frac{1}{9}$. The relative frequency that a randomly selected
monolith will be marine clayey loam is $\frac{11}{18}$. How many monoliths of lake-glacial genesis does the object contain, if there are no marine sandy loams among the sandy loams?
|
# Solution.
Let's determine the exact number of monoliths. It is known that the probability of a monolith being loamy sand is $\frac{1}{9}$. The number closest to 200 that is divisible by 9 is 198. Therefore, there are 198 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands $(198: 9=22)$ and part of the clay loams. Let's find out how many monoliths of lacustrine-glacial origin are among the clay loams: The total number of clay loams is 198:9$\cdot$8=176. Among them, marine clay loams constitute $\frac{11}{18}$ of the total, which is 121. Therefore, the number of clay loams of lacustrine-glacial origin is 55. Thus, the total number of monoliths of lacustrine-glacial origin is: $22+55=77$.
Answer. 77.
|
77
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 5. (20 points)
Find $x_{0}-y_{0}$, if $x_{0}$ and $y_{0}$ are the solutions to the system of equations:
$$
\left\{\begin{array}{l}
x^{3}-2023 x=y^{3}-2023 y+2020 \\
x^{2}+x y+y^{2}=2022
\end{array}\right.
$$
|
# Solution.
Rewrite the system as
$$
\left\{\begin{array}{l}
x^{3}-y^{3}+2023 y-2023 x=2020 \\
x^{2}+x y+y^{2}=2022
\end{array}\right.
$$
Let $x_{0}$ and $y_{0}$ be the solution to the system of equations. Then
$\left\{\begin{array}{l}x_{0}{ }^{3}-y_{0}{ }^{3}+2023 y_{0}-2023 x_{0}=2020, \\ x_{0}{ }^{2}+x_{0} y_{0}+y_{0}{ }^{2}=2022 .\end{array}\right.$
Rewrite the first equation of the system as
$\left\{\begin{array}{l}\left(x_{0}-y_{0}\right)\left(x_{0}^{2}+x_{0} y_{0}+y_{0}{ }^{2}\right)-2023\left(x_{0}-y_{0}\right)=2020, \\ x_{0}{ }^{2}+x_{0} y_{0}+y_{0}{ }^{2}=2022 ;\end{array}\right.$
$\left\{\begin{array}{l}2022\left(x_{0}-y_{0}\right)-2023\left(x_{0}-y_{0}\right)=2020, \\ x_{0}^{2}+x_{0} y_{0}+y_{0}^{2}=2022 ;\end{array}\right.$
Then $x_{0}-y_{0}=-2020$.
Answer. -2020.
|
-2020
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}$, if $\frac{a}{2}-\frac{2}{a}=3$.
#
|
# Solution.
$$
\begin{aligned}
& \frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{4}}{16}-\frac{16}{a^{4}}\right) \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)\left(\frac{a^{2}}{4}-\frac{4}{a^{2}}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\frac{a^{2}}{4}-2+\frac{4}{a^{2}}+2\right)\left(\frac{a}{2}-\frac{2}{a}\right)\left(\frac{a}{2}+\frac{2}{a}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\left(\frac{a}{2}-\frac{2}{a}\right)^{2}+2\right)\left(\left(\frac{a}{2}-\frac{2}{a}\right)=\left(3^{2}+2\right) \cdot 3=33 .\right.
\end{aligned}
$$
Answer. 33.
|
33
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) Lab engineer Dasha received an object for research consisting of about 100 monoliths (a container designed for 100 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequency (statistical probability) that a randomly selected monolith will be sandy loam is $\frac{1}{7}$. The relative frequency that a randomly selected monolith will be marine clayey loam is $\frac{9}{14}$. How many monoliths of lake-glacial genesis does the object contain, if there are no marine sandy loams among the sandy loams?
#
|
# Solution.
Let's determine the exact number of monoliths. It is known that the probability of a monolith being loamy sand is $\frac{1}{7}$. The number closest to 100 that is divisible by 7 is 98. Therefore, there are 98 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands ($98: 7=14$) and part of the clay loams. Let's find out how many monoliths of lacustrine-glacial origin are among the clay loams: The total number of clay loams is $98: 7 \cdot 6=84$. Among them, marine clay loams constitute $\frac{9}{14}$ of the total, which is 63. Therefore, the number of clay loams of lacustrine-glacial origin is 21. Thus, the total number of monoliths of lacustrine-glacial origin is: $14+21=35$.
Answer. 35.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) At the research institute, a scientific employee, Ivan Ivanovich, received an object for research containing about 300 oil samples (a container designed for 300 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or high-sulfur, and density - either light or heavy. The relative frequency (statistical probability) that a randomly selected sample will be a heavy oil sample is $\frac{1}{8}$. The relative frequency that a randomly selected sample will be a light low-sulfur oil sample is $\frac{22}{37}$. How many high-sulfur oil samples does the object contain, if there were no low-sulfur samples among the heavy oil samples?
|
# Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a selected sample being a heavy oil sample is $\frac{1}{8}$, and the number closest to 300 that is divisible by $8-296$. Therefore, the total number of samples in the container is 296. The samples of high-sulfur oil consist of all heavy oil samples (296:8=37) and part of the light oil samples. Let's find out how many high-sulfur oil samples are among the light oil samples. The total number of light oil samples is 296:8.7=259. Among them, the samples of low-sulfur oil constitute $\frac{22}{37}$ of the total, which is 176. Therefore, the number of high-sulfur light oil samples is 83. Then the total number of high-sulfur oil samples is: $37+83=120$.
Answer. 120.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Calculate
$$
\left(\frac{10001}{20232023}-\frac{10001}{20222022}\right) \cdot 4090506+\sqrt{4092529}
$$
|
# Solution.
$$
\begin{aligned}
& \left(\frac{10001}{20232023}-\frac{10001}{20222022}\right) \cdot 4090506+\sqrt{4092529} \\
& =\left(\frac{10001}{2023 \cdot 10001}-\frac{10001}{2022 \cdot 10001}\right) \cdot 4090506+\sqrt{2023^{2}}= \\
& \quad=\left(\frac{1}{2023}-\frac{1}{2022}\right) \cdot 2022 \cdot 2023+2023=\frac{2022-2023}{2022 \cdot 2023} \cdot 2022 \cdot 2023+2023=-1+2023=2022
\end{aligned}
$$
Answer. 2022.
|
2022
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) At the research institute, a scientific employee, Tatyana Vasilyevna, received an object for research containing about 150 oil samples (a container designed for 150 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or high-sulfur, and density - either light or heavy. The relative frequency (statistical probability) that a randomly selected sample will be a heavy oil sample is $\frac{2}{11}$. The relative frequency that a randomly selected sample will be a light low-sulfur oil sample is $\frac{7}{13}$. How many high-sulfur oil samples does the object contain, if there were no low-sulfur samples among the heavy oil samples?
|
# Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{2}{11}$, and the number closest to 150 that is divisible by $11-143$. Therefore, the total number of samples in the container is 143. Samples of high-sulfur oil consist of all heavy oil samples ( $143: 11 \cdot 2=26$ ) and part of the light oil samples. Let's find out how many high-sulfur oil samples are among the light oil samples. The total number of light oil samples is 143:11.9=117. Among them, samples of low-sulfur oil constitute $\frac{7}{13}$ of the total, which is 77. Therefore, the number of high-sulfur light oil samples is 40. Then the total number of high-sulfur oil samples is: $26+40=66$.
Answer. 66.
|
66
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{8}+1296}{36 a^{4}}$, if $\frac{a}{\sqrt{6}}+\frac{\sqrt{6}}{a}=5$.
#
|
# Solution.
$$
\begin{aligned}
& \frac{a^{8}+1296}{36 a^{4}}=\frac{a^{4}}{36}+\frac{36}{a^{4}}=\frac{a^{4}}{36}+2+\frac{36}{a^{4}}-2=\left(\frac{a^{2}}{6}+\frac{6}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{6}+2+\frac{6}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{\sqrt{6}}+\frac{\sqrt{6}}{a}\right)^{2}-2\right)^{2}-2=\left(5^{2}-2\right)^{2}-2=527
\end{aligned}
$$
Answer. 527.
|
527
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) At the quality control department of an oil refinery, Engineer Pavel Pavlovich received a research object consisting of about 100 oil samples (a container designed for 100 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or high-sulfur, and density - either light or heavy. The relative frequency (statistical probability) that a randomly selected sample will be a heavy oil sample is $\frac{1}{7}$. The relative frequency that a randomly selected sample will be a light low-sulfur oil sample is $\frac{9}{14}$. How many high-sulfur oil samples does the object contain, if there were no low-sulfur samples among the heavy oil samples?
|
# Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{1}{7}$. The number closest to 100 that is divisible by $7$ is $98$. Therefore, there are 98 samples in total in the container. The samples of high-sulfur oil include all the heavy oil samples ($98: 7=14$) and a portion of the light oil samples. Let's find out how many high-sulfur oil samples are among the light oil samples. The total number of light oil samples is $98:7 \cdot 6=84$. Among them, the samples of low-sulfur oil constitute $\frac{9}{14}$ of the total, which is 63. Therefore, the number of high-sulfur light oil samples is 21. Thus, the total number of high-sulfur oil samples is: $14+21=35$.
Answer. 35.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300}$ holds. Find $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}-2023$.
|
# Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+298 x_{n}+x_{n+1}}{300} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{n}
\end{gathered}
$$
Let the common difference of this progression be $d, d \neq 0$ (as per the condition).
Then $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}-2023=\sqrt{\frac{x_{1}+2022 d-x_{1}-d}{2021} \cdot \frac{2022}{x_{1}+2022 d-x_{1}}}-2023=$ $=\sqrt{\frac{2021 d}{2021} \cdot \frac{2022}{2022 d}}-2023=1-2023=-2022$.
Answer. -2022.
|
-2022
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{8}-6561}{81 a^{4}} \cdot \frac{3 a}{a^{2}+9}$, if $\frac{a}{3}-\frac{3}{a}=4$.
|
Solution.
$\frac{a^{8}-6561}{81 a^{4}} \cdot \frac{3 a}{a^{2}+9}=\left(\frac{a^{4}}{81}-\frac{81}{a^{4}}\right) \cdot \frac{3 a}{a^{2}+9}=\left(\frac{a^{2}}{9}+\frac{9}{a^{2}}\right)\left(\frac{a^{2}}{9}-\frac{9}{a^{2}}\right) \cdot \frac{3 a}{a^{2}+9}=$
$=\left(\frac{a^{2}}{9}-2+\frac{9}{a^{2}}+2\right)\left(\frac{a}{3}-\frac{3}{a}\right)\left(\frac{a}{3}+\frac{3}{a}\right) \cdot \frac{3 a}{a^{2}+9}=$
$=\left(\left(\frac{a}{3}-\frac{3}{a}\right)^{2}+2\right)\left(\frac{a}{3}-\frac{3}{a}\right)=\left(4^{2}+2\right) \cdot 4=72$.
Answer. 72.
|
72
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) At the quality control department of an oil refinery, Engineer Valentina Ivanovna received a research object consisting of about 200 oil samples (a container designed for 200 samples, which was almost completely filled). Each sample has certain characteristics in terms of sulfur content - either low-sulfur or high-sulfur, and density - either light or heavy. The relative frequency (statistical probability) that a randomly selected sample will be a heavy oil sample is $\frac{1}{9}$. The relative frequency that a randomly selected sample will be a light low-sulfur oil sample is $\frac{11}{18}$. How many high-sulfur oil samples does the object contain, if there were no low-sulfur samples among the heavy oil samples?
|
# Solution.
Let's determine the exact number of oil samples. It is known that the relative frequency of a sample being a heavy oil sample is $\frac{1}{9}$. The number closest to 200 that is divisible by 9 is 198. Therefore, the total number of samples in the container is 198. The samples of high-sulfur oil consist of all heavy oil samples ($198: 9=22$) and a portion of the light oil samples. Let's find out how many high-sulfur oil samples are among the light oil samples. The total number of light oil samples is $198: 9 \cdot 8=176$. Among them, the samples of low-sulfur oil constitute $\frac{11}{18}$ of the total, which is 121. Therefore, the number of high-sulfur light oil samples is 55. Thus, the total number of high-sulfur oil samples is: $22+55=77$.
Answer. 77.
|
77
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+398 x_{n}+x_{n+1}}{400}$ holds. Find $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}+2021$.
|
# Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+398 x_{n}+x_{n+1}}{400} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{n}
\end{gathered}
$$
Let the common difference of this progression be $d, d \neq 0$ (as per the condition).
Then $\sqrt{\frac{x_{2023}-x_{2}}{2021} \cdot \frac{2022}{x_{2023}-x_{1}}}+2021=\sqrt{\frac{x_{1}+2022 d-x_{1}-d}{2021} \cdot \frac{2022}{x_{1}+2022 d-x_{1}}}+2021=$ $=\sqrt{\frac{2021 d}{2021} \cdot \frac{2022}{2022 d}}+2021=1+2021=2022$
Answer. 2022.
|
2022
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}$, if $\frac{a}{2}-\frac{2}{a}=5$.
|
Solution.
$$
\begin{aligned}
& \frac{a^{8}-256}{16 a^{4}} \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{4}}{16}-\frac{16}{a^{4}}\right) \cdot \frac{2 a}{a^{2}+4}=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)\left(\frac{a^{2}}{16}-\frac{16}{a^{2}}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\frac{a^{2}}{4}-2+\frac{4}{a^{2}}+2\right)\left(\frac{a}{2}-\frac{2}{a}\right)\left(\frac{a}{2}+\frac{2}{a}\right) \cdot \frac{2 a}{a^{2}+4}= \\
& =\left(\left(\frac{a}{2}-\frac{2}{a}\right)^{2}+2\right)\left(\frac{a}{2}-\frac{2}{a}\right)=\left(5^{2}+2\right) \cdot 3=81
\end{aligned}
$$
Answer. 81.
|
81
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) In the educational center "Young Geologist," an object consisting of about 150 monoliths (a container designed for 150 monoliths, which was almost completely filled) was delivered. Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequency (statistical probability) that a randomly selected monolith will be sandy loam is $\frac{2}{11}$. The relative frequency that a randomly selected monolith will be marine clayey loam is $\frac{7}{13}$. How many monoliths of lake-glacial genesis does the object contain, if there are no marine sandy loams among the sandy loams? #
|
# Solution.
Let's determine the exact number of monoliths. It is known that the relative frequency of a monolith being loamy sand is $\frac{2}{11}$. The number closest to 150 that is divisible by 11 is 143. Therefore, there are 143 monoliths in total. Monoliths of lacustrine-glacial origin make up all loamy sands ( $143: 11 \cdot 2=26$ ) and part of the loams. Let's find out how many monoliths of lacustrine-glacial origin are among the loams: The total number of loams is 143:11$\cdot$9=117. Among them, marine loams constitute $\frac{7}{13}$ of the total, which is 77. Therefore, the number of loams of lacustrine-glacial origin is 40. Thus, the total number of monoliths of lacustrine-glacial origin is: $26+40=66$.
Answer. 66.
|
66
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+98 x_{n}+x_{n+1}}{100}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2021$.
|
# Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+98 x_{n}+x_{n+1}}{100} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{n}
\end{gathered}
$$
Let the common difference of this progression be $d, d \neq 0$ (as per the condition).
Then $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2021=\sqrt{\frac{x_{1}+2022 d-x_{1}}{2022} \cdot \frac{2021}{x_{1}+2022 d-x_{1}-d}}+2021=$ $=\sqrt{\frac{2022 d}{2022} \cdot \frac{2021}{2021 d}}+2021=1+2021=2022$.
Answer. 2022.
|
2022
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 1. (5 points) Find $\frac{a^{8}+256}{16 a^{4}}$, if $\frac{a}{2}+\frac{2}{a}=3$.
|
Solution.
$$
\begin{aligned}
& \frac{a^{8}+256}{16 a^{4}}=\frac{a^{4}}{16}+\frac{16}{a^{4}}=\frac{a^{4}}{16}+2+\frac{16}{a^{4}}-2=\left(\frac{a^{2}}{4}+\frac{4}{a^{2}}\right)^{2}-2= \\
& =\left(\frac{a^{2}}{4}+2+\frac{4}{a^{2}}-2\right)^{2}-2=\left(\left(\frac{a}{2}+\frac{2}{a}\right)^{2}-2\right)^{2}-2=\left(3^{2}-2\right)^{2}-2=47
\end{aligned}
$$
Answer. 47.
|
47
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) An educational center "Young Geologist" received an object for research consisting of about 300 monoliths (a container designed for 300 monoliths, which was almost completely filled). Each monolith has a specific name (sandy loam or clayey loam) and genesis (marine or lake-glacial deposits). The relative frequency (statistical probability) that a randomly selected monolith will be sandy loam is $\frac{1}{8}$. The relative frequency that a randomly selected monolith will be marine clayey loam is $\frac{22}{37}$. How many monoliths of lake-glacial genesis does the object contain, if there are no marine sandy loams among the sandy loams?
|
# Solution.
Let's determine the exact number of monoliths. It is known that the relative frequency of a monolith being loamy sand is $\frac{1}{8}$. The number closest to 300 that is divisible by $8-296$. Therefore, there are 296 monoliths in total. Monoliths of lacustrine-glacial origin consist of all loamy sands $(296: 8=37)$ and part of the clays. Let's find out how many monoliths of lacustrine-glacial origin are among the clays: The total number of clays is 296:8$\cdot$7=259. Among them, marine clays constitute $\frac{22}{37}$ of the total number, which is 176. Therefore, the number of clays of lacustrine-glacial origin is 83. Thus, the total number of monoliths of lacustrine-glacial origin is: $37+83=120$.
Answer. 120.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) For the numerical sequence $\left\{x_{n}\right\}$, all terms of which, starting from $n \geq 2$, are distinct, the relation $x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200}$ holds. Find $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2022$.
|
# Solution.
From the given relations in the problem, it is easily deduced that for all $n \geq 2$, $x_{n}-x_{n-1}=x_{n+1}-x_{n}$, which implies that the sequence is an arithmetic progression. Indeed,
$$
\begin{gathered}
x_{n}=\frac{x_{n-1}+198 x_{n}+x_{n+1}}{200} \\
2 x_{n}=x_{n-1}+x_{n+1} \\
x_{n}-x_{n-1}=x_{n+1}-x_{n}
\end{gathered}
$$
Let the common difference of this progression be $d, d \neq 0$ (as per the condition).
Then $\sqrt{\frac{x_{2023}-x_{1}}{2022} \cdot \frac{2021}{x_{2023}-x_{2}}}+2022=\sqrt{\frac{x_{1}+2022 d-x_{1}}{2022} \cdot \frac{2021}{x_{1}+2022 d-x_{1}-d}}+2022=$ $=\sqrt{\frac{2022 d}{2022} \cdot \frac{2021}{2021 d}}+2022=1+2022=2023$.
Answer. 2023.
|
2023
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) In a sequence of natural numbers, each subsequent number, starting from the third, is equal to the absolute difference of the two preceding ones. Determine the maximum number of elements such a sequence can contain if the value of each of them does not exceed 2022.
#
|
# Solution.
To maximize the length of the sequence, the largest elements should be at the beginning of the sequence. Let's consider the options:
1) $n, n-1,1, n-1, n-2, n-3,1, n-4, n-5,1, \ldots, 2,1,1$;
2) $n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
3) $1, n, n-1,1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
4) $n-1, n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
Thus, the length of the sequence that satisfies the condition of the problem will be the greatest if the largest member is the second, and the first is one less. This sequence will look like:
$$
n-1, n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1
$$
For $1<n \leq 4$, the length of such a sequence can be determined by direct counting $\left(l_{2}=4, l_{3}=6, l_{4}=7\right)$, and for larger $n$, we will prove that $l_{n+2}=l_{n}+3$.
Let's form a sequence with the largest member $n+2$:
$$
n+1, n+2,1, n+1, n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1 .
$$
That is, the length of such a sequence will be:
$$
\begin{gathered}
l_{n}=\frac{3 n}{2}+1=\frac{3 n+2}{2} \text { for even } n, \\
l_{n}=\frac{3(n-1)}{2}+3=\frac{3 n+3}{2} \text { for odd } n .
\end{gathered}
$$
Then, for $n=2022$
$$
l_{2022}=\frac{3 \cdot 2022+2}{2}=3034
$$
Answer. 3034.
|
3034
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) In a sequence of natural numbers, each subsequent number, starting from the third, is equal to the absolute difference of the two preceding ones. Determine the maximum number of elements such a sequence can contain if the value of each of them does not exceed 2021.
#
|
# Solution.
To maximize the length of the sequence, the largest elements should be at the beginning of the sequence. Let's consider the options:
1) $n, n-1,1, n-1, n-2, n-3,1, n-4, n-5,1, \ldots, 2,1,1$;
2) $n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
3) $1, n, n-1,1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
4) $n-1, n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1$.
Thus, the length of the sequence that satisfies the condition of the problem will be the greatest if the largest member is the second, and the first is one less. Then this sequence will look like:
$$
n-1, n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1
$$
For $1<n \leq 4$, the length of such a sequence can be determined by direct counting $\left(l_{2}=4, l_{3}=6, l_{4}=7\right)$, and for larger $n$, we will prove that $l_{n+2}=l_{n}+3$. Let's form a sequence with the largest member $n+2$:
$$
n+1, n+2,1, n+1, n, 1, n-1, n-2,1, n-3, n-4,1, \ldots, 2,1,1 .
$$
That is, the length of such a sequence will be:
$$
\begin{gathered}
l_{n}=\frac{3 n}{2}+1=\frac{3 n+2}{2} \text { for even } n, \\
l_{n}=\frac{3(n-1)}{2}+3=\frac{3 n+3}{2} \text { for odd } n .
\end{gathered}
$$
Then, for $n=2021$
$$
l_{2021}=\frac{3 \cdot 2021+3}{2}=3033
$$
Answer. 3033.
|
3033
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. (20 points) A finite increasing sequence $a_{1}, a_{2}, \ldots, a_{n}$ ( $n \geq 3$ ) of natural numbers is given, and for all $k \leq n-2$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds. The sequence must contain a term $a_{k}=2021$. Determine the maximum number of three-digit numbers, divisible by 25, that this sequence can contain.
|
# Solution.
The final sequence can contain all three-digit numbers, as it can consist of a given number of natural numbers starting from the chosen number $a_{i}$.
We will prove that for any term of the arithmetic progression $1,2,3, \ldots$ defined by the formula for the $n$-th term $a_{n}=n$, the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds.
Indeed, for any value of $k$, the equalities
$a_{k}=k, a_{k+1}=k+1, a_{k+2}=k+2$ are valid, from which it follows that
$3 a_{k+1}-2 a_{k}-1=3(k+1)-2 k-1=k+2=a_{k+2}$, i.e., the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-1$ holds, which is what we needed to prove.
For example, the sequence containing 2021: 3,4,5,6,..., 2018, 2019,2020,2021.
Thus, the sequence can contain all three-digit numbers from 100 to 999. Among them, the numbers divisible by $25: 100,125,150,175,200,225,250,275, \ldots, 900,925$, 950,975 - 4 in each of the nine hundreds, i.e., 36 numbers.
Answer. 36.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 6. (30 points) At the first deposit, equipment of the highest class was used, and at the second deposit, equipment of the first class was used, with the highest class being less than the first. Initially, $40 \%$ of the equipment from the first deposit was transferred to the second. Then, $20 \%$ of the equipment that ended up on the second deposit was transferred back to the first, with half of the transferred equipment being of the first class. After this, the equipment of the highest class on the first deposit was 26 units more than on the second, and the total amount of equipment on the second deposit increased by more than $5 \%$ compared to the initial amount. Find the total amount of equipment of the first class.
|
# Solution.
Let there initially be $x$ units of top-class equipment at the first deposit and $y$ units of first-class equipment at the second deposit $(x1.05 y$, from which $y48 \frac{34}{67} .\end{array}\right.\right.$
This double inequality and the condition “x is divisible by 5” is satisfied by the unique value $x=50$. Then $y=130-70=60$. Thus, there were 60 units of first-class equipment.
Answer: 60.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4. (20 points) A finite increasing sequence of natural numbers $a_{1}, a_{2}, \ldots, a_{n}(n \geq 3)$ is given, and for all $\kappa \leq n-2$ the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$ holds. The sequence must contain $a_{k}=2022$. Determine the maximum number of three-digit numbers, divisible by 4, that this sequence can contain.
|
# Solution.
Since it is necessary to find the largest number of three-digit numbers that are multiples of 4, the deviation between the members should be minimal. Note that the arithmetic
progression with a difference of $d=2$, defined by the formula $a_{k}=2 k$, satisfies the equality $a_{k+2}=3 a_{k+1}-2 a_{k}-2$.
Indeed,
$$
a_{k}=2 k, a_{k+1}=2 k+2, a_{k+2}=2 k+4, \text { or by the formula }
$$
$$
a_{k+2}=3 a_{k+1}-2 a_{k}-2=3(2 k+2)-2 \cdot 2 k-2=2 k+4
$$
The sequence containing 2022: 4, 6,
This finite sequence can contain all even three-digit numbers from 100 to 999. Among them, the numbers divisible by 4 are: 100, 104, 108, 112, ..., 992, 925, 950, 996 - 25 in each of the nine hundreds, i.e., 225 numbers.
Answer. 225.
|
225
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3. (15 points) The bases $AB$ and $CD$ of trapezoid $ABCD$ are 41 and 24, respectively, and its diagonals are perpendicular to each other. Find the scalar product of vectors $\overrightarrow{AD}$ and $\overrightarrow{BC}$.
|
# Solution.
Fig. 1
Let the point of intersection of the diagonals be $O$ (Fig. 1).
Consider the vectors $\overrightarrow{A O}=\bar{a}$ and $\overrightarrow{B O}=\bar{b}$.
From the similarity of triangles $A O B$ and $D O C$, we have:
$\frac{A O}{O C}=\frac{41}{24}$ and $\overrightarrow{O C}=\frac{24}{41} \overrightarrow{A O}=\frac{24}{41} \vec{a}$
$\frac{B O}{O D}=\frac{41}{24}$ and $\overrightarrow{O D}=\frac{24}{41} \overrightarrow{B O}=\frac{24}{41} \vec{b}$
Then
$\overrightarrow{A D}=\vec{a}+\frac{24}{41} \vec{b}$ and $\overrightarrow{B C}=\vec{b}+\frac{24}{41} \vec{a}$
Let's find the dot product
$\overrightarrow{A D} \cdot \overrightarrow{B C}=\left(\vec{a}+\frac{24}{41} \vec{b}\right) \cdot\left(\vec{b}+\frac{24}{41} \vec{a}\right)=\vec{a} \vec{b}+\frac{24}{41} \vec{a} \cdot \vec{a}+\frac{24}{41} \vec{b} \cdot \vec{b}+\left(\frac{24}{41}\right)^{2} \vec{a} \vec{b}$
Since $\vec{a} \perp \vec{b}$, then $\vec{a} \vec{b}=0$.
By the definition of the dot product of vectors, we get $\vec{a} \cdot \vec{a}=|\vec{a}| \cdot|\vec{a}| \cdot \cos 0^{0}=|\vec{a}|^{2}$ and $\vec{b} \cdot \vec{b}=|\vec{b}|^{2}$.
Triangle $AOB$ is a right triangle, so $|\vec{a}|^{2}+|\vec{b}|^{2}=|A B|^{2}=41^{2}$.
Thus, $\overrightarrow{A D} \cdot \overrightarrow{B C}=\frac{24}{41}\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)=\frac{24}{41} \cdot 41^{2}=984$.
Answer. 984.
|
984
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Calculating Samson loves to have lunch at the Italian restaurant "At Pablo's". During his next visit to the restaurant, Samson was offered to purchase a loyalty card for a period of 1 year at a price of 30000 rubles, which gives the client a $30 \%$ discount on the bill amount.
(a) Suppose that during the week Samson visits the restaurant three times (regardless of whether he has a loyalty card or not) with an average bill amount of 900 rubles without the discount. Is it profitable for him to buy the loyalty card?
(b) According to data from the information portal, the average bill amount per person in this restaurant is 600 rubles. On average, how many visits per year does a customer need to make to be indifferent about buying the restaurant's card or not?
(c) Recently, many stores and organizations operating in the service sector have been issuing so-called customer cards. Some of these cards are cumulative and provide discounts only after accumulating a certain amount on the card's account, while others immediately provide the customer with a discount that can be used with any purchase. Provide at least two strong reasons and substantiate why it is profitable for trading and service enterprises to issue such cards and use them in their work with customers. (10 points)
|
# Solution:
(a) After purchasing the card, one lunch will cost the client $900 * 0.3 = 270$ rubles less. Since there are 52 full weeks in a year $(365 / 7 = 52.14)$, Samson will save $270 * 3 * 52 = 42120$ rubles on discounts over the year, which is more than the cost of the card. Therefore, it is beneficial for him to buy the card.
(b) On average, a lunch with the card will be $600 * 0.3 = 180$ rubles cheaper. Then, from the condition of equal expenses, we determine that one needs to visit $30000 / 180 = 166$,(6), i.e., it will be beneficial to purchase the card if the restaurant is visited 167 times a year or more, and it will be unprofitable to purchase the card if the restaurant is visited 166 times or less.
(c) The use of cards with a cumulative discount system allows firms to "retain" regular customers, thereby preventing them from "switching" to the services of other firms.
The use of cards that provide discounts without a cumulative system also helps to attract new and "retain" old customers with discounts.
Often, providing such cards to customers requires their authorization in the company's databases. In this case, companies can inform their customers about planned promotions and also provide customer databases to their partners, albeit not without a cost. Additionally, issuing such cards requires additional expenses from the company owners. Therefore, it can signal to customers about the stability of the company, its concern for its customers, and its willingness to provide them with discounts.
## Grading Criteria:
(a) Correct calculations allowing for a comparison of expenses - 3 points, Correct conclusion about the profitability of purchasing the card
(b) Correctly found number of visits - 3 points, (c) 2 points each (total 4 points) for the reason and corresponding explanation.
|
167
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 1. Maximum 15 points
A company that produces educational materials for exam preparation incurs average costs per textbook of $100+\frac{100000}{Q}$, where $Q$ is the number of textbooks produced annually. What should be the annual production volume of the textbook to reach the break-even point if the planned price of the textbook is 300 monetary units?
#
|
# Solution
At the break-even point $\mathrm{P}=\mathrm{ATC}=\mathrm{MC}$
Form the equation $100+10000 / Q=300$
$100 \mathrm{Q}+100000=300 \mathrm{Q}$
$100000=200 \mathrm{Q}$
$\mathrm{Q}=100000 / 200=500$
## Evaluation Criteria
1. The correct answer is justified: 15 points
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. Maximum 15 points
The great-grandfather-banker left a legacy to his newborn great-grandson. According to the agreement with the bank, the amount in the great-grandson's account increases. Every year, on the day after the birthday, the current amount is increased by 1 million rubles more than in the previous year. Thus, if the initial amount was zero rubles, after one year +1 million rubles, after 2 years $1+2$ million, after 3 years $1+2+3$ and so on. According to the terms of the agreement, the process will stop, and the great-grandson will receive the money when the amount in the account becomes a three-digit number consisting of three identical digits.
How old will the great-grandson be when the terms of the agreement are met?
|
# Solution:
Since the number consists of identical digits, it can be represented as 111 multiplied by a. According to the problem, the same number should be obtained as the sum of an arithmetic progression. The first element of the progression is 1, the last is \( \mathrm{n} \), and the number of elements in the progression is \( \mathrm{n} \). Since \( 111 = 3 \times 37 \):
\[
n(n+1) = 2 \times 3 \times 37 \times a
\]
Since 37 is a prime number, and \( \frac{n(n+1)}{2} \) should be less than 1000, then either \( \mathrm{n}+1=37 \) or \( \mathrm{n}=37 \). By checking, we find that \( \mathrm{n}=36 \), and the required three-digit number consisting of identical digits is 666.
## Criteria:
Correctly formulated equation in integers: 5 points;
Correct answer: 5 points;
Analysis of the equation and proof of the absence of other options: 5 points.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. Maximum 20 points
At the conference "Economics of the Present," an intellectual tournament was held, in which more than 198 but fewer than 230 scientists, including doctors and candidates of sciences, participated. Within one match, participants had to ask each other questions and record correct answers within a certain time. Each participant played against each other exactly once. The winner of the match received one point, the loser received no points; in the event of a draw, both participants received half a point. At the end of the tournament, it turned out that in matches against doctors of sciences, each participant scored half of all their points. How many candidates of sciences participated in the tournament? Provide the smallest of all possible answers.
|
Solution: Let there be $\mathrm{n}$ scientists participating in the tournament, of which $\mathrm{m}$ are doctors and $\mathrm{n}-\mathrm{m}$ are candidates of science. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the doctors of science competed in $\mathrm{m}(\mathrm{m}-1) / 2$ matches and scored the same number of points. The candidates conducted $(n-m)(n-m-1)/2$ matches among themselves and scored $(n\mathrm{m})(\mathrm{n}-\mathrm{m}-1) / 2$ points when playing against each other.
Since each participant scored half of their points against the doctors of science, the doctors of science scored half of their points against each other, meaning they scored twice as much in total: $\mathrm{m}(\mathrm{m}-1)$.
Since the candidates of science scored half of their points against the doctors of science, they also scored half of their points when competing among themselves, and in total they scored $(n-m)(n-m-1)$ points. The total number of points scored by all participants is: $m(m-1)+(n-m)(n-m-1)=n(n-1) / 2$.
After a series of transformations, we get:
$n^{2}-4 n m+4 m^{2}=n$
From which $(\mathrm{n}-2 \mathrm{~m})^{2}=\mathrm{n}$, that is, $\mathrm{n}$ is the square of a natural number.
Given $198<\mathrm{n}<230$, we find that $\mathrm{n}=225$, then $(225-2 \mathrm{~m}) 2=225$, from which $\mathrm{m}=$ 105 or $\mathrm{m}=120$.
The number of participants in the tournament is 225; the number of candidates of science is 105, or 120. The smallest value is 105.
## Criteria.
- A correct solution to the problem and the correct answer are provided - 20 points (maximum),
- A correct solution to the problem, but an incorrect answer or no answer - 8 points (rounded to the nearest whole number, half the maximum),
- An error in the solution, but the correct answer is provided - 7 points (rounded to the nearest whole number, half the maximum minus one),
- The solution is incorrect or absent, but the correct answer is provided - 2 points (under any circumstances),
- The solution is incorrect or absent, and an incorrect answer or no answer is provided - 0 points.
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. Maximum 20 points
At the "Economics and Law" congress, a "Tournament of the Best" was held, in which more than 220 but fewer than 254 delegates—economists and lawyers—participated. Within one match, participants had to ask each other questions within a limited time and record the correct answers. Each participant played against each other exactly once. The winner of the match received one point, the loser received no points; in the event of a draw, both participants received half a point. At the end of the tournament, it turned out that in matches against economists, each participant scored half of all their points. How many lawyers participated in the tournament? Provide the smallest of all possible answers.
|
Solution: Let there be $\mathrm{n}$ delegates participating in the tournament, of which $\mathrm{m}$ are economists and $\mathrm{n}-\mathrm{m}$ are lawyers. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the economists competed in $\mathrm{m}(\mathrm{m}-1) / 2$ matches and scored the same number of points. The lawyers conducted $(n-m)(n-m-1)/2$ matches among themselves and scored $(n\mathrm{m})(\mathrm{n}-\mathrm{m}-1) / 2$ points while playing against each other.
Since each participant scored half of their points against economists, the economists scored half of their points against each other, meaning they scored twice as many points in total: $\mathrm{m}(\mathrm{m}-1)$.
Since the lawyers scored half of their points against economists, they also scored half of their points against each other, meaning they scored $(n\mathrm{m})(\mathrm{n}-\mathrm{m}-1)$ points in total. The total number of points scored by all participants is:
$m(m-1)+(n-m)(n-m-1)=n(n-1) / 2$.
After a series of transformations, we get:
$n^{2}-4 n m+4 m^{2}=n$
From which $(\mathrm{n}-2 \mathrm{~m})^{2}=\mathrm{n}$, meaning $\mathrm{n}$ is a perfect square of a natural number.
Given $220<\mathrm{n}<254$, we find that $\mathrm{n}=225$, then $(225-2 \mathrm{~m})^{2}=225$, from which $\mathrm{m}=$ 105 or $\mathrm{m}=120$.
The number of participants in the tournament is 225; the number of lawyers is 105, or 120. The smallest value is 105.
## Criteria.
- Correct solution and the right answer - 20 points (maximum),
- Correct solution but the wrong answer or no answer -
8 points (rounded to the nearest whole number, half the maximum),
- Error in the solution but the right answer - 7 points (rounded to the nearest whole number, half the maximum minus one),
- Incorrect or no solution, but the right answer - 2 points (under any circumstances),
- Incorrect or no solution, and the wrong answer or no answer - 0 points.
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Find the minimum loss, which is EXPENSE - INCOME, where the letters $\boldsymbol{P}, \boldsymbol{A}, \boldsymbol{C}, \boldsymbol{X}, \boldsymbol{O}, \boldsymbol{D}$ represent digits forming an arithmetic progression in the given order. (2 points).
#
|
# Solution:
The difference in the progression is 1; otherwise, 6 digits will not fit (if the "step" is 2, then 6 digits will exceed the field of digits). The smaller the first digit, the smaller the loss. Therefore, $\boldsymbol{P}=1, \boldsymbol{A}=2, \boldsymbol{C}=3, \boldsymbol{X}=4, \boldsymbol{O}=5, D=6$. Thus, 123456-65456=58000.
Answer: 58000.
| Points | Criteria for evaluating the performance of task № 1 |
| :---: | :--- |
| $\mathbf{9}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{6}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification for the choice of correspondence between letters and digits or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{3}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the difference in the progression is found); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the choice of correspondence between letters and digits is incorrect). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the above criteria for scoring 3 and 6 points. |
|
58000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a certain state, only liars and economists live (liars always lie, while economists tell the truth). At a certain moment, the state decided to carry out monetary and credit, as well as budgetary and tax reforms. Since it was unknown what the residents expected, everyone was asked several questions (with only "yes" and "no" answers). The questions were answered affirmatively as follows:
"Will taxes be raised?" $40 \%$,
"Will there be an increase in the money supply?" $30 \%$,
"Will bonds be issued?" $50 \%$,
"Will the gold and foreign exchange reserves be squandered?" $\mathbf{0} \%$.
What percentage of the population actually expects taxes to be raised, if each of them realistically expects only one measure? (14 points).
|
# Solution:
Let $\boldsymbol{x}$ be the proportion of liars in the country, then (1-x) is the proportion of economists. Each economist answers affirmatively to one question, and each liar answers affirmatively to three. Therefore, we can set up the equation:
$3 x+1-x=0.4+0.3+0.5+0 ; 2 x=0.2 ; x=0.1$. Thus, in the country, $10\%$ are liars and $90\%$ are economists. Since no one said they expect a depletion of gold and foreign exchange reserves, this means that all liars are expecting it. Therefore, the actual percentage of the population expecting a tax increase is $40-10=30\%$.
Answer: $30\%$.
| Points | Criteria for evaluating the performance of task № $\mathbf{3}$ |
| :---: | :--- |
| $\mathbf{14}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are correctly performed. The correct answer is obtained. |
| $\mathbf{10}$ | A correct sequence of all steps of the solution is provided. There are gaps in justifying the percentage of the population that actually expects a tax increase, or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{5}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is correctly performed, possibly inaccurately (for example, an equation is set up); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the percentage of the population that actually expects a tax increase is found incorrectly). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the criteria for scoring 5 and 10 points. |
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There are 2015 coins on the table. Two players play the following game: they take turns; on a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. How many coins should the first player take on the first move to then guarantee a win with an unchanging strategy? (20 points)
#
|
# Solution:
The strategy of the first player: he takes 95 coins, and then on each move, he takes (101-x) coins, where $\boldsymbol{x}$ is the number of coins taken by the second player. Since $\boldsymbol{x}$ is even (by the condition), 101-x is odd. Then $2015-95=1920$, since 101-x+x=101 coins will be taken per move, 1920/101=19 (remainder 1). In total, there will be 19 moves, in which 101 coins will be taken each time. The second player will be left with 1 coin and will not be able to make a move.
Answer: 95.
| Points | Criteria for evaluating the solution to problem № 4 |
| :---: | :--- |
| $\mathbf{2 0}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{1 4}$ | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the first player's strategy or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| $\mathbf{7}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the first move of the first player is indicated); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the first player's strategy is incorrectly determined). At the same time, the solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the criteria for scoring 7 and 14 points. |
|
95
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The bank issued a loan to citizen $N$ on September 9 in the amount of 200 mln rubles. The repayment date is November 22 of the same year. The interest rate on the loan is $25 \%$ per annum. Determine the amount (in thousands of rubles) that citizen N will have to return to the bank. Assume that there are 365 days in a year, the bank accrues interest daily on the loan amount, and no interest is charged on the days of loan issuance and repayment. (14 points).
|
# Solution:
Number of days of the loan: September - 21 days, October - 31 days, November - 21 days, i.e., $21+31+21=73$ days. The accrued debt amount is calculated using the formula:
$\boldsymbol{F V}=\boldsymbol{P V} \cdot(\boldsymbol{1}+\boldsymbol{t} \cdot \mathbf{Y}$ ), where $\boldsymbol{F} \boldsymbol{V}$ - the accrued (future - future value) amount of money after a certain period; $\boldsymbol{P} \boldsymbol{V}$ - the initial (present - present value) amount of money, $\boldsymbol{t}$ - the term of the operation in days, $\mathbf{Y}$ - the duration of the year in days, $\boldsymbol{i}$ - the interest rate for the period. We have $t=73, Y=365, i=0.25 F V=200 *(1+73 * 0.25 / 365)=1.05 * 200=210$ thousand rubles.
Answer: 210 thousand rubles.
## Grading Criteria:
1) The number of days of the loan is correctly calculated - 3 points.
2) The correct formula for calculating the debt amount is obtained - 7 points
3) The debt amount is correctly calculated - 4 points. If an arithmetic error is made while using the correct formula for calculating the debt amount, 3 points are given.
|
210
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. In the Tumba-Yumba tribe with a population of 30 people, a trader arrives. After studying the customs of the tribe, the trader proposes to play a game. For each natural exchange of goods conducted in the market by two tribespeople, the trader gives each participant one gold coin. If at the end of the day, two different tribespeople end up with the same number of coins, all goods and coins go to the trader. A tribesperson trades goods only with another known tribesperson, and the cunning trader knows this. By the end of the day, the trader finds a group of tribespeople with the same number of coins. The tribespeople are shocked, the trader is happy, and the offending tribespeople are expelled from the market. Then the Chief appears and offers the trader a rematch. Let the trader distribute 270 coins among the remaining tribespeople so that no one has the same number of coins (i.e., each tribesperson should have a unique number of coins). If the trader cannot do this, all the coins and goods remain in the tribe.
- Is it true that in the original game, the tribespeople had no chance?
- Is it true that the Chief can return all the goods and gold?
- Is it true that the Chief knew how many tribespeople were expelled from the market?
- What is the maximum number of tribespeople that needed to be expelled for the Chief to win?
- Is it true that if the trader had lost the first game, the tribespeople could have lost the second game?
Explain your answers.
|
Solution: In any company, there are at least two people who have the same number of acquaintances. Therefore, the natives had no chance. The chief proposed to distribute 270 coins, which means he knew that some natives would be removed. Let's say x people were removed. To distribute a different number of coins to everyone, you need to take $\frac{0+30-x-1}{2}(30-x)$ coins, and the Chief should make sure that 270 coins are not enough. Then, at least 24 natives should remain, so $24 * 23 / 2=276$ coins are needed to avoid repetition. Since $23 * 11=253$ contains an arithmetic error, which does not affect the result. The correct answer to 2, 3, or 4 questions | 5 |
| The correct answer and justification for only one of the questions in the problem | 2 |
| The answer is either incorrect or guessed, with no justification | 0 |
|
24
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In how many ways can two knights, two bishops, two rooks, a queen, and a king be arranged on the first row of a chessboard so that the following conditions are met:
1) The bishops stand on squares of the same color;
2) The queen and the king stand on adjacent squares. (20 points).
|
# Solution:
Let's number the cells of the first row of the chessboard in order from left to right with numbers from **1** to **8** ( **1** - the first white cell, **8** - the last black cell). Since the queen and king are standing next to each other, they can occupy one of 7 positions: 1-2, 2-3, ..., 7-8. Additionally, on each position, we can swap their places (for example, on position 1-2, we can place the king first, then the queen, and vice versa). In total, we have 2$\cdot$7=14 ways to place the queen and king on the first row of the chessboard so that they stand next to each other. Now let's see how we can place the bishops. Suppose we have placed the queen and king somewhere (for example, on positions 4-5). Notice that no matter how we place the queen and king, we will have 3 cells of each color left: 3 white and 3 black. To place the bishops, we need to choose the color of the cell - this gives us 2 options. Then we have 3 positions on which we want to place two bishops (the number of combinations of 3 taken 2 at a time). This can be done in 3 ways. Therefore, we have $2 \cdot 3=6$ ways to place the two bishops on cells of the same color given a fixed position of the king and queen. We have 4 cells left, on which we need to place two knights and two rooks. That is, we need to choose 2 positions out of 4 for the rooks, and the knights will take the remaining 2 positions. We can choose 2 cells out of 4 in 6 ways (the number of combinations of 4 taken 2 at a time). Therefore, we have 6 ways to place the two rooks and two knights on 4 positions in any order. Since we fixed the positions sequentially, the multiplication rule applies. The total number of ways to place the figures is $N=14 \cdot 6 \cdot 6=504$.
## Answer: 504.
| Points | Criteria for evaluating the completion of task № 4 |
| :---: | :---: |
| 20 | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| 14 | A correct sequence of all steps of the solution is provided. There are gaps in the justification of the number of ways to arrange the figures (for example, rooks and knights), or the distinguishability of figures in one pair, or a computational error or typo that does not affect the further course of the solution. As a result of this error or typo, an incorrect answer may be obtained. |
| 7 | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, the number of ways to arrange the king, queen, and either only the bishops or only the rooks and knights; the problem is solved separately only with condition №1 and only with condition №2); - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, the number of ways to arrange only the rooks and knights or only the bishops). The solution may be incomplete. |
| 0 | All cases of the solution that do not meet the above criteria for scoring 7 and 14 points. |
|
504
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. In his country of Milnlandia, Winnie-the-Pooh decided to open a company that produces honey. Winnie-the-Pooh sells honey only in pots, and it costs him 10 milnovs (the monetary units of Milnlandia) to produce any pot of honey. The inverse demand function for honey is given by $\boldsymbol{P}=310-3 \boldsymbol{Q}$ (where $\boldsymbol{Q}$ is the number of pots, and $\boldsymbol{P}$ is the price of one pot). There are no other suppliers of honey.
a) How many pots of honey will Winnie-the-Pooh produce if his main goal is to maximize profit?
b) The government of Milnlandia decided to introduce a tax $\boldsymbol{t}$ milnovs on each pot of honey sold by Winnie-the-Pooh's company. What should be the amount of the tax to maximize the tax revenue for the budget of Milnlandia? (17 points).
|
# Solution:
a) Profit $=P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q = 310Q - 3Q^2 - 10Q = 300Q - 3Q^2$.
Since the graph of the function $300Q - 3Q^2$ is a parabola opening downwards, its maximum is achieved at the vertex: $Q = -b / 2a = -300 / (-6) = 50$.
b) Winnie-the-Pooh maximizes the quantity $P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q - tQ = (300 - t)Q - 3Q^2$, $Q = (300 - t) / 6$. The state receives a monetary amount in the form of tax revenue $Q \cdot t = (300 - t) / 6 \cdot t = (50 - 1/6 \cdot t) \cdot t = -1/6 \cdot t^2 + 50t$. The graph of this function is a parabola opening downwards, its maximum is achieved at $t = -50 / (-2/6) = 50 \cdot 6 / 2 = 50 \cdot 3 = 150$.
Answer: a) 50; b) 150.
## Grading Criteria:
a)
1) The problem of profit maximization for a single producer is correctly formulated: 3 points
2) The solution to the profit maximization problem is correctly found and justified: 4 points. The absence of justification for the maximum value at the extremum point results in a 1-point deduction.
b)
1) The consequences of introducing a profit tax for a single producer are correctly accounted for: 3 points
2) The tax amount is correctly found: 7 points. If there is an arithmetic error in calculating the tax, 6 points are given.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. To climb from the valley to the mountain top, one must walk 4 hours on the road, and then -4 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 17 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour, then sleeps for 9 hours, then spews fire for 1 hour again, and so on. It is dangerous to walk on both the road and the path during the first dragon's fire-spewing, while it is dangerous to walk only on the path during the second dragon's fire-spewing. Both dragons start spewing fire simultaneously at midnight. Is it possible to safely climb from the valley to the mountain top and return back? (8 points)
|
# Solution:
The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the second dragon's eruption, this dragon will not be dangerous. For a safe ascent, it is sufficient for the first dragon to stop erupting by the start of the journey along the road, and for the second dragon to stop erupting 4 hours later, by the start of the journey along the trail. The first dragon erupts in hours numbered $18 x+1$, and the second dragon in hours numbered $10 y+1$, where $x, y$ are natural numbers. We need them to erupt with a 4-hour shift. We get the following Diophantine equation:
$$
10 y+1=(18 x+1)+4,5 y=9 x+2 .
$$
The smallest solution in natural numbers is $x=2, y=4$. Therefore, for a safe ascent, you need to start at the beginning of the 38th hour ( $18 \cdot 2+1=37)$.
Answer: You need to start at the beginning of the 38th hour.
| llы | Я№ 1 |
| :---: | :---: |
| 8 | Contains all steps of the solution. and explanation. |
| 5 | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is correctly performed, possibly inaccurately (for example, a Diophantine equation is formulated, or a graph of eruptions is constructed, or the possibility of a safe path is demonstrated in principle); - another part is either not performed or performed incorrectly, possibly even with logical errors (for example, the solution to the Diophantine equation is not found, or there are errors in the graph of eruptions, or a specific moment for the start of a safe path is not indicated, or the possibility of stopping and waiting is allowed). The solution may be incomplete. |
| 0 | All cases of the solution that do not correspond to the above criteria for scoring 5 and 8 points. |
|
38
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. To climb from the valley to the mountain top, one must walk 6 hours on the road, and then - 6 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 25 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour, then sleeps for 13 hours, then spews fire for 1 hour again, and so on. During the first dragon's fire-spewing, it is dangerous to walk both on the road and on the path, while during the second dragon's fire-spewing, it is dangerous to walk only on the path. At midnight, both dragons start spewing fire simultaneously. Is it possible to safely climb from the valley to the mountain top and return back? (6 points)
|
# Solution:
The path along the road and the trail (there and back) takes 24 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 12 hours. Therefore, if you start moving along the trail immediately after the second dragon's eruption, this dragon will not be dangerous. For a safe ascent, it is necessary that by the start of movement along the road, the first dragon has stopped erupting, and 6 hours later, by the start of movement along the trail, the second dragon has stopped erupting. The first dragon erupts in hours numbered $26 x+1$, and the second dragon in hours numbered $14 y+1$, where $x, y$ are natural numbers. We need them to erupt with a 6-hour shift. We get the following Diophantine equation:
$$
14 y+1=(26 x+1)+6, \quad 7 y=13 x+3
$$
The smallest solution in natural numbers is $x=3, y=6$. Therefore, for a safe ascent, you need to start at the beginning of the 80th hour $(26 \cdot 3+1=79)$.
Answer: You need to start at the beginning of the 80th hour.
| Points | Criteria for evaluating the performance of task No. 1 |
| :---: | :--- |
| $\mathbf{6}$ | A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are performed correctly. The correct answer is obtained. |
| $\mathbf{3}$ | The problem is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, a Diophantine equation is formulated); - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, the solution to the Diophantine equation is not found). The solution may be incomplete. |
| $\mathbf{0}$ | All cases of the solution that do not meet the above criteria for scoring 3 and 6 points. |
|
80
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In all other cases - o points.
## Task 2
Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$.
How many solutions of this equation satisfy the condition:
$0 \leq x \leq 2020 ?$
|
Solution. Let $t=\frac{\pi x}{4}$. Then the equation takes the form
$2 \sqrt{2} \sin ^{3} t=\cos \left(\frac{\pi}{4}-t\right)$.
$2 \sqrt{2} \sin ^{3} t=\cos \frac{\pi}{4} \cos t+\sin \frac{\pi}{4} \sin t$
$2 \sqrt{2} \sin ^{3} t=\frac{\sqrt{2}}{2} \cos t+\frac{\sqrt{2}}{2} \sin t$
$4 \sin ^{3} t=\cos t+\sin t ; 4 \sin ^{3} t-\sin t=\cos t$
Case 1: $\sin t=0 \rightarrow \cos t=0 \rightarrow \emptyset$
Case 2: $\sin t \neq 0$. Dividing both sides of the equation by $\sin t$, we get:
$4 \sin ^{2} t-1=\cot t$. Considering that $\sin ^{2} t=\frac{1}{1+\cot ^{2} t}$ and denoting $y=\cot t$, we obtain
$\frac{4}{1+y^{2}}-1=y$, or after transformation $y^{3}+y^{2}+y-3=0$
$(y-1)\left(y^{2}+2 y+3\right)=0$, the only root of which is $y=1$.
Making reverse substitutions: $\cot t=1 \rightarrow t=\frac{\pi}{4}+\pi n, n \in \mathbb{Z}$.
$\frac{\pi x}{4}=\frac{\pi}{4}+\pi n, n \in \mathbb{Z}$, from which $x=1+4 n, n \in \mathbb{Z}$.
$0 \leq x \leq 2020 ; 0 \leq 1+4 n \leq 2020, -0.25 \leq n \leq 504.75$.
Since $n$ is an integer, $0 \leq n \leq 504$. The number of integers $n$ is 505.
Answer: 505.
## Evaluation Criteria:
|
505
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In all other cases - $\mathbf{0}$ points.
## Task 2
## Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi}{4}(1+x)\right)$.
How many solutions of this equation satisfy the condition: $2000 \leq x \leq 3000$?
|
# Solution:
$\sin \left(\frac{\pi}{4}(1+x)\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$. The equation becomes
$2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$, i.e., we get problem 2 from option 1, the solution of which is:
$x=1+4 n, n \in Z . \quad 2000 \leq x \leq 3000,2000 \leq 1+4 n \leq 3000,499.75 \leq n \leq 749.75$.
Since $n$ is an integer, $500 \leq n \leq 749$. The number of integers $n$ is 250.
Answer: 250.
## Grading Criteria:
|
250
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 3.
## Maximum 10 points
In the Country of Wonders, a pre-election campaign is being held for the position of the best tea lover, in which the Mad Hatter, March Hare, and Dormouse are participating. According to a survey, $20 \%$ of the residents plan to vote for the Mad Hatter, $25 \%$ for the March Hare, and $30 \%$ for the Dormouse. The rest of the residents are undecided. Determine the smallest percentage of the undecided voters that the Mad Hatter must attract to ensure he does not lose to the March Hare and the Dormouse (under any distribution of votes), knowing that each of the undecided voters will vote for one of the candidates. The winner is determined by a simple majority of votes. Justify your answer.
|
# Solution:
Let the number of residents in Wonderland be $N$, then $0.2 N$ residents are going to vote for Dum, $0.25 N$ residents for the Rabbit, and $0.3 N$ residents for Sonya. The undecided voters are $0.25 N$ residents. Let $\alpha$ be the fraction of the undecided voters who are going to vote for Dum. Dum will not lose the election if the number of votes for him is not less than the number of votes for his opponents. He will receive $0.2 N + 0.25 N \alpha$ votes. In the "worst" case, the remaining fraction $1 - \alpha$ of the undecided voters will vote for the Rabbit or Sonya. Therefore, for the condition of the problem to be met, the following must hold:
$$
\left\{\begin{array} { c }
{ 0.2 N + 0.25 N \alpha \geq 0.25 N + 0.25 N (1 - \alpha), } \\
{ 0.2 N + 0.25 N \alpha \geq 0.3 N + 0.25 N (1 - \alpha). }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
0.2 + 0.25 \alpha \geq 0.25 + 0.25(1 - \alpha) \\
0.2 + 0.25 \alpha \geq 0.3 + 0.25(1 - \alpha)
\end{array}\right.\right.
$$
It is clear that it is sufficient to solve the second inequality of the system:
$$
0.2 + 0.25 \alpha \geq 0.3 + 0.25 - 0.25 \alpha \Leftrightarrow 0.5 \alpha \geq 0.35 \Leftrightarrow \alpha \geq 0.7
$$
The smallest value of $\alpha = 0.7$, which corresponds to $70\%$ of the undecided voters.
Answer: $70\%$.
## Grading Criteria:
- 10 points - the correct answer is obtained with proper justification.
- 5 points - there is a correct sequence of reasoning leading to the right answer, but an arithmetic error has been made.
- 0 points - the solution does not meet any of the criteria provided.
|
70
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Assignment 7.
## Maximum 10 points
In the modern world, every consumer often has to make decisions about replacing old equipment with more energy-efficient alternatives. Consider a city dweller who uses a 60 W incandescent lamp for 100 hours each month. The electricity tariff is 5 rubles/kWh.
The city dweller can buy a more energy-efficient lamp with a power of 12 W, which costs 120 rubles and provides the same light output as the aforementioned incandescent lamp. Alternatively, he can approach an investor (an energy service company) that will install such an energy-saving lamp itself, but in return, the city dweller must pay the company 75% of the resulting cost savings on electricity for 10 months (the tariff payment is made regardless of the method of lamp installation).
(a) If the city dweller plans his expenses only for the next 10 months, will he install the energy-saving lamp himself or approach the investor company?
(b) If the city dweller plans his expenses for the entire service life of a typical energy-saving lamp, what decision will he make?
|
# Solution and Grading Scheme:
a) Expenses for 10 months when installing an energy-saving lamp independently:
$$
120 \text { rub. }+12 \text { (W) * } 100 \text { (hours) / } 1000 \text { * } 5 \text { (rub./kW*hour) * } 10 \text { (months) = } 180 \text { rub. }
$$
Expenses for 10 months when turning to an energy service company:
$$
(12+(60-12) * 0.75)(W) * 100 \text { (hours) / } 1000 \text { * } 5 \text { (rub./kW*hour) * } 10 \text { (months) = } 240 \text { rub. }
$$
Thus, the most cost-effective of the alternative options considered is to install the energy-saving lamp independently.
Full correct solution - 6 points. Only one option correctly calculated - 3 points.
b) The expected service life of a typical lamp is more than 10 months. Since after 10 months, you only need to pay for electricity consumption, as in part a), it will be more cost-effective to install the lamp independently.
Justification (including calculations) 4 points
#
|
180
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. 25 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the amount of tax revenue were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit commodity tax on the consumer in the amount of 30 monetary units per unit of the good. It is known that the market supply is given by $Q_s = 6P - 312$, and the market demand is linear. In the situation where no taxes are imposed, at the equilibrium point, the price elasticity of market supply is 1.5 times higher than the absolute value of the price elasticity of the market demand function. After the tax was introduced, the consumer price increased to 118 monetary units.
1) Restore the market demand function.
2) Determine the amount of tax revenue collected at the chosen rate.
3) Determine the rate of the quantity tax that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. For linear demand functions, using the definition of elasticity, we get: $1.5 \cdot \frac{b P_{e}}{Q_{e}}=\frac{6 P_{e}}{Q_{e}}$. From this, we find that $b=4$.
If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118$; from which $a=688$.
The market demand function is $Q_{d}=688-4 P$.
2) It is known that $P_{d}(t=30)=118$. Therefore, $Q_{d}=688-472=216, T=216 \cdot 30=6480$.
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $\quad T=Q \cdot t=288 t-2.4 t^{2}$. The graph of the function is a parabola opening downwards, the maximum of the function is achieved at $t^{*}=60$.
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$.
## Grading Criteria:
1) Correctly determined the slope of the demand curve - 3 points.
Correctly determined the selling price with the incorrect tax - 1 point.
Correctly determined the sales volume with the incorrect tax - 2 points.
Correctly determined the intercept of the demand curve - 3 points.
Correctly found the demand function - 1 point.
If an error/typo is made in the solution that leads to results contradicting economic logic, only 3 points are awarded for this item even if the rest of the solution is correct.
2) Correctly found the tax revenues - 3 points.
3) Explained how to find the rate that maximizes tax revenues - 4 points.
Found the rate that maximizes tax revenues - 4 points. If the rate found contradicts logic, even if the value of $t$ is correct, these 4 points are not awarded.
4) Found the quantity of goods that will be sold at the rate that maximizes tax revenues - 2 points. If the quantity found contradicts logic, even if the value of $t$ is correct, these 2 points are not awarded.
Found the tax revenues at this tax rate - 2 points. If the tax revenues are found but their value does not align with the condition in the problem about the mixed-up digits, these 2 points are not awarded.
## Penalties:
An arithmetic error was made in calculations that did not lead to results contradicting economic logic - 1 point.
The fact that the tax rate found in item 3) gives a maximum, not a minimum of the objective function, is not justified.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4. 25 points
A novice economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. In the cryptogram, the amount of tax revenue to be collected was specified. It was also emphasized that it was impossible to collect a larger amount of tax revenue on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the amount of tax revenue were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the producer in the amount of 90 monetary units per unit of the product. It is known that the market demand is given by $Q d=688-4 P$, and the market supply is linear. In the situation where there are no taxes, the price elasticity of market supply at the equilibrium point is 1.5 times higher than the absolute value of the price elasticity of the market demand function. After the tax was introduced, the producer's price decreased to 64 monetary units.
1) Restore the market supply function.
2) Determine the amount of tax revenue collected at the chosen rate.
3) Determine the rate of the quantity tax that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot\left|E_{p}^{d}\right|=E_{p}^{s}$. Using the definition of price elasticity for linear demand functions, $1.5 \cdot$ $\frac{4 P_{e}}{Q_{e}}=\frac{d P_{e}}{Q_{e}}$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64$. Therefore, $688-4\left(P_{s}+90\right)=6 P_{s}+c ; 0.1 c+32.8=P_{s}=64 ; c=-312$.
The market supply function is $Q_{s}=6 P-312$.
2) It is known that $P_{s}(t=90)=64$. Therefore, $Q_{s}=6 P-312=72$. Then the government revenue is $T=72 \cdot 90=6480$.
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenue will be $T=Q \cdot t=288 t-2.4 t^{2}$. The graph of the function is a parabola opening downwards, and the maximum of the function is achieved at $t^{*}=60$.
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$.
## Grading Criteria:
1) Correctly determined the slope of the supply curve - 3 points.
Correctly determined the purchase price with the incorrect tax - 1 point.
Correctly determined the sales volume with the incorrect tax - 2 points.
Correctly determined the intercept of the supply curve - 3 points.
Correctly found the supply function - 1 point.
If an error/typo is made in the solution that leads to results contradicting economic logic, only 3 points are awarded for this section even if the rest of the solution is correct.
2) Correctly found the tax revenue - 3 points.
3) Explained how to find the rate that maximizes tax revenue - 4 points.
Found the rate that maximizes tax revenue - 4 points. If the rate found contradicts logic, even if the value of $t$ is correct, these 4 points are not awarded.
4) Found the quantity of goods sold at the rate that maximizes tax revenue - 2 points. If the quantity found contradicts logic, even if the value of $t$ is correct, these 2 points are not awarded. Found the tax revenue at this tax rate - 2 points. If the tax revenue is found but does not align with the condition in the problem about mixed-up digits, these 2 points are not awarded.
## Penalties:
An arithmetic error was made in calculations that did not lead to results contradicting economic logic -1 point.
The fact that the tax rate found in section 3) gives a maximum, not a minimum, of the objective function is not justified.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 2. Maximum 16 points
Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from point $B$ to the road connecting $A$ and $C$, and from point $C$ to the road connecting $A$ and $B$ is 300 km. It is known that point $D$ is equidistant from the roads connecting points $A, B, C$ and lies within the area bounded by these roads. Any resident of all settlements spends no more than 1 liter of fuel for every 10 km of road. What is the maximum amount of fuel that would be needed by a motorist who needs to get from settlement $D$ to any of the roads connecting the other settlements?
#
|
# Solution
The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of triangle $\mathrm{ABC}$ is maximal, or, equivalently, when the radius $r$ of the inscribed circle in the given triangle is maximal under the conditions of the problem. The conditions of the problem can be easily reformulated as follows (see the figure):
$\mathrm{ha}=100, \mathrm{hb}+\mathrm{hc}=300$,
where $\mathrm{ha}, \mathrm{hb}, \mathrm{hc}$ are the heights of triangle ABC. It is not difficult to establish that the following equality holds:
1ha+1hb+1hc=1r.
Indeed, let $S$ be the area of the triangle, then
$$
\mathrm{S}=12 \mathrm{aha}=12 \mathrm{bhb}=12 \mathrm{chc}=\text { pr. }
$$
From this, we have:
$$
h a=2 S a, h b=2 S b, h c=2 S c, r=S p .
$$
And we get:
$$
1 \mathrm{ha+}+1 \mathrm{hb}+1 \mathrm{hc}=\mathrm{a} 2 \mathrm{~S}+\mathrm{b} 2 \mathrm{~S}+\mathrm{c} 2 \mathrm{~S}=\mathrm{a}+\mathrm{b}+\mathrm{c} 2 \mathrm{~S}=\mathrm{pS}=1 \mathrm{r} \text {. }
$$
From the condition, it follows that
$$
1 \mathrm{r}=1100+1 \mathrm{hb}+1 \mathrm{hc}=1100+1 \mathrm{hb}+1300-\mathrm{hb}=1100+300 \mathrm{hb} 300-\mathrm{hb} .
$$
The radius $\mathrm{r}$ is maximal when the sum on the right side of the equation is minimal. But this is equivalent to the maximality of the expression
hb300-hb.
This is a quadratic function, and its maximum is achieved at its vertex, i.e., as it is not difficult to understand, when $\mathrm{hb}=150$. Therefore, the maximum value of the radius of the inscribed circle is found from the equation
$$
1 \mathrm{r}=1100+300150300-150=1100+175=7300 \Leftrightarrow \mathrm{r}=3007 \text { km. }
$$
From this, the maximum fuel consumption is:
$$
\text { r10=307 l. }
$$
Answer: 307 l.
|
307
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. Maximum 20 points
The commander of a tank battalion, in celebration of being awarded a new military rank, decided to invite soldiers to a tank festival, where the main delicacy is buckwheat porridge. The commander discovered that if the soldiers are lined up by height, there is a certain pattern in the change of their demand functions. The demand for buckwheat porridge of the shortest soldier is given by $Q d=510-5.1 P$, the next tallest soldier's demand is $Q d=520-5.2 P$, the next one's is $Q d=530-5.3 P$, and so on. The commander's individual demand for buckwheat porridge is $Q d=500-5 P$. The battalion commander always tries to feed as many soldiers as possible. Apart from the commander and the invited soldiers, no one else demands buckwheat porridge. At the festival, 25 perfectly competitive firms offer buckwheat porridge, with each firm's supply function being $Q s=302 P$. The battalion commander, after consulting with his family, decided to make a surprise - initially, the guests and the commander will make decisions about consuming buckwheat porridge, assuming that everyone will pay for it themselves, but at the end of the celebration, the commander's family will pay the total bill from their savings, which amount to 2,525,000 monetary units. If this amount is insufficient, the commander will ask all guests to equally share the remaining bill. It is known that the equilibrium price is set at 20 monetary units. Assume that if guests, like the commander, have to make a certain fixed payment, this payment does not affect their demand for buckwheat porridge.
(a) How is the individual demand of the 60th soldier invited to the celebration described? How will the aggregate demand for buckwheat porridge be described if the commander plans to invite 40 soldiers and does not refuse the porridge himself? Explain the construction of the demand function in this case.
(b) How many soldiers did the commander invite to the celebration? Did the guests have to pay part of the bill? If so, how much did each of them pay?
|
# Solution
(a) The commander tries to feed as many soldiers as possible, which means he will invite relatively short soldiers first - all other things being equal, their consumption of porridge is less.
Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n- (5+0.1 n) P$, where $n$ is the ordinal number of the soldier. Also note that all consumers are willing to purchase the product at $P \in [0 ; 100]$, so to find the market demand, it is sufficient to sum the individual demand functions.
The individual demand of the 60th soldier is $Q_{d}=1100-11 P$. To find the demand of 40 soldiers, we can use the formula for the arithmetic progression:
$Q_{d}=\frac{510-5.1 P+500+10 n-(5+0.1 n) P)}{2} \cdot n$
$Q_{d}=\frac{1410-14.1 P}{2} \cdot 40$
$Q_{d}=28200-282 P$
Adding the commander's demand, we get the market demand: $Q_{d}=28700-287 P$.
Answer: $Q_{d}=1100-11 P, Q_{d}=28700-287 P$.
(b) Knowing the individual supply of one firm, we can find the market supply:
$Q_{s}=302 P \cdot 25=7550 P$
It is known that $Q_{s}(20)=Q_{d}(20)$, so $Q_{d}=7550 \cdot 20=151000$.
From the previous part, we know that
$Q_{d}=5 n^{2}+505 n+500-\left(0.05 n^{2}+5.05 n+5\right) P$
$Q_{d}(20)=4 n^{2}+404 n+400=151000$
$(4 n+1004)(n-150)=0$
$n_{1}=-251, n_{2}=150$.
Obviously, $n>0$. The commander invited 150 guests.
$T R=151000 \cdot 20=3020000>2525000$ - the expenses for buckwheat porridge exceed the family budget, which means the soldiers will help pay part of the bill. Each soldier will contribute an amount of $\frac{495000}{150}=3300$ monetary units.
Answer: the commander invited 150 guests, each of whom paid 3300 monetary units.
## Grading Criteria
(a) Individual demand is formulated - (3p), market demand is formulated - (5p).
(b) Market supply function is formulated - (2p), equilibrium quantity is determined - (2p), market demand function in terms of the number of guests is formulated - (2p), maximum number of guests is determined - (3p), behavior of guests and their payment of the bill is determined - (3p).
A penalty of 1 point for arithmetic errors that did not lead to significant distortion of the results. Alternative solutions may be awarded full points if they contain a correct and justified sequence of actions.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The number 2458710411 was written 98 times in a row, resulting in a 980-digit number. From this number, it is required to erase 4 digits. What is the number of ways this can be done so that the newly obtained 976-digit number is divisible by 6?
|
Solution. If a number is divisible by 6, then it is divisible by 3 and 2. A number is divisible by 2 if and only if its last digit is even. The 980-digit number given in the condition ends with 2458710411, i.e., it has the form ...2458710411. For the number to be divisible by 2, it is necessary to strike out the last two ones so that the number ends with an even digit. The resulting number will have the form 2458710411...245871041124587104. Now it is necessary for this number to be divisible by 3. A number is divisible by 3 if and only if the sum of its digits is divisible by 3. The sum of the digits of the resulting number is:
$(2+4+5+8+7+1+0+4+1+1) * 97 + (2+4+5+8+7+1+0+4) = 33 * 97 + 31 = 3232$.
The remainder of the division of the number 3232 by 3 is 1. This means that the numbers must be struck out in such a way that their sum is divisible by 3 and leaves a remainder of 1. Then, upon subtraction, a number will be obtained whose sum of digits will be exactly divisible by 3. The maximum sum that can be obtained by striking out digits is $16 (8+8)$. Therefore, it is necessary to consider all numbers from 0 to 16 that are divisible by 3 and leave a remainder of 1, and all possible combinations of their formation. These numbers are: $1, 4, 7, 10, 13, 16$. The combinations by which they can be formed are as follows:
$1 (0+1); 4 (0+4, 2+2); 7 (2+5, 7+0); 10 (5+5, 2+8); 13 (8+5); 16 (8+8)$.
It remains to calculate the number of ways in which each combination can be obtained. The first combination can be obtained by striking out one 1 and one 0. There are $97 * 3 + 1 = 292$ ones in our number, and 98 zeros. Therefore, the number of ways in which this can be done is $(97 * 3 + 1) * 98 = 28616$. For subsequent combinations, we get:
$(0+4): 98 * 98 = 9604$
$(2+2): 98 * 97 / 2 = 4753$
$(2+5): 98 * 98 = 9604$
$(7+0): 98 * 98 = 9604$
$(5+5): 98 * 97 / 2 = 4753$
$(2+8): 98 * 98 = 9604$
$(5+8): 98 * 98 = 9604$
$(8+8): 98 * 97 / 2 = 4753$.
Summing all the obtained combinations, we get the number of ways 90895. However, it should be noted that one of the proposed ways is not suitable. If we strike out the last digits 0 and 4, then our number will end with 1, i.e., it will not be divisible by 2, and therefore, it will not be divisible by 6. Therefore, the number of combinations is $90895 - 1 = 90894$ (ways).
Answer: 90894 ways.
| Criteria for evaluating the performance of the task | Points |
| :--- | :---: |
| A correct and justified sequence of all steps of the solution is provided. All transformations and calculations are correctly performed. The correct answer is obtained. | **10** |
| A correct sequence of all steps of the solution is provided. Gaps in the justification of the choice of number combinations (for example, one extra solution (0+4) is not taken into account, pairs of digits are not divided by 2, etc.) or a computational error or typo that does not affect the further course of the solution are allowed. As a result of this error or typo, an incorrect answer may be obtained. | **6** |
| The task is not solved, but its solution is significantly advanced, i.e.: - a significant part of the solution is performed correctly, possibly inaccurately (for example, part of the combinations is taken into account and calculated correctly); |
| - the other part is either not performed or performed incorrectly, possibly even with logical errors (for example, most combinations are not taken into account or extra combinations are taken into account, combinations are calculated incorrectly, cases of striking out the same digit (2+2, |
| 5+5, 8+8) are not taken into account, etc.); |
All cases of solutions that do not meet the above criteria for scoring 3 and 6 points.
|
90894
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the tax revenue amount were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the consumer in the amount of 30 monetary units per unit of the good. The market supply is given by \( Q_s = 6P - 312 \), and the market demand is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the consumer price increased to 118 monetary units.
1) Restore the market demand function.
2) Determine the amount of tax revenue collected at the chosen tax rate.
3) Determine the per-unit tax rate that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler specified to be collected?
|
# Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 b=6$. We find that $b=$ 4. If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118 ; a=688$. The market demand function is $Q_{d}=688-4 P$. (8 points).
2) It is known that $P_{d}(t=30)=118$. Therefore, $Q_{d}=688-472=216, T=216 \cdot 30=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, the maximum of the function is reached at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: an arithmetic error was made - 2 points, lack of justification for the maximum - 5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the tax revenue amount were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the producer in the amount of 90 monetary units per unit of the product. It is known that the market demand is given by $Q_d = 688 - 4P$, and the market supply is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the producer's price decreased to 64 monetary units.
1) Restore the market supply function.
2) Determine the amount of tax revenue collected at the chosen rate.
3) Determine the rate of the quantity tax that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot 4=d$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64.688-4\left(P_{s}+90\right)=6 P_{s}+c$; $0.1 c+32.8=P_{s}=64 ; c=-312$. The market supply function is $Q_{s}=6 P-312$. (8 points).
2) It is known that $P_{s}(t=90)=64$. Therefore, $Q_{s}=6 P-312=72, T=72 \cdot 90=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, and the maximum of the function is achieved at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: An arithmetic error was made -2 points, lack of justification for the maximum -5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In all other cases - 0 points
## *Important: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.
## Assignment 2 (12 points)
Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest in the 1st project an amount equal to $p_{1}>0$ thousand rubles, and after a year receive $x_{1}$ thousand rubles such that
$$
4 x_{1}-3 p_{1}-44=0
$$
and at the same time Gena agreed to invest in the 2nd project an amount equal to $p_{2}>0$ thousand rubles, and after a year receive $x_{2}$ thousand rubles such that
$$
p_{2}^{2}-12 p_{2}+x_{2}^{2}-8 x_{2}+43=0
$$
According to the terms of the contract, the parameters $p_{1}, x_{1}, p_{2}, x_{2}$ were chosen so that the value
$$
d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(p_{1}-p_{2}\right)^{2}}
$$
was minimized. Determine:
1) the minimum value of the quantity $d$;
2) is the deal profitable for Gena? Find the amount of profit or loss for Gena if he agrees to this deal.
|
# Solution:
1) It is clear that the expression $4 x_{1}-3 p_{1}-44=0 \Leftrightarrow x=\frac{3}{4} p+11$ defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression
$$
\begin{aligned}
p^{2}-12 p+x^{2}-8 x+4 & =0 \Leftrightarrow(p-6)^{2}+(x-4)^{2}-36-16+43=0 \\
\Leftrightarrow & (p-6)^{2}+(x-4)^{2}=3^{2}
\end{aligned}
$$
It defines in the plane $x O p$ the equation of a circle $\Phi$ with center at the point $(6 ; 4)$ and radius 3. Then the expression of the form
$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(p_{1}-p_{2}\right)^{2}}
$$
defines the distance between points lying on $l$ and $\Phi$. Therefore, according to the problem's condition, we need to find the minimum distance between points lying on
the line and the circle. Let $O_{1} H$ be the perpendicular from the center of the circle $O_{1}$ to the line $l$, then $O_{1} P$ is the radius of the circle. We will show that $H P=O_{1} H-O_{1} P$ is the desired value. Indeed, let $H_{1} \in l, F \in \Phi$. Draw the tangent $a$ to the circle at point $P$. The radius $O_{1} P$, drawn to the point of tangency, is perpendicular to $a$. Therefore, $a \| l$. And thus, $HP$ is equal to the distance between the parallel lines. Now we have:
$$
H_{1} F=H_{1} P_{1}+P_{1} F \geq H_{1} P_{1} \geq H P .
$$
Thus, $H P$ is the shortest distance between the points of the circle and the line $l$. Let's find $O_{1} H$ using the formula for the distance from a point to a line (one of the ways to find it):
$$
O_{1} H=\frac{|4 \cdot 4-3 \cdot 6-44|}{\sqrt{3^{2}+4^{2}}}=\frac{46}{5}=9.2 .
$$
Therefore, $H P=9.2-3=6.2$.
2) Let's define the equation of the line $O_{1} H$. Since it is perpendicular to the line $l$ and passes through the point $O_{1}(6 ; 4)$, its equation in the plane $x O p$ has the form
$$
x=-\frac{4}{3}(p-6)+4 \Leftrightarrow x=-\frac{4}{3} p+12
$$
Let's find the coordinates of the point $H$ as the coordinates of the intersection of the lines $l$ and $O_{1} H$. Form the equation:
$$
-\frac{4}{3} p+12=\frac{3}{4} p+11 \Leftrightarrow \mathrm{p}=\frac{12}{25}=0.48
$$
Thus, $p_{1}=0.48$ and $x_{1}=\frac{3}{4} p_{1}+11=11.36$.
Let's find the coordinates of the point $P$ as the coordinates of the intersection of the line $O_{1} H$ and the circle $\Phi$ :
$$
(p-6)^{2}+\left(-\frac{4}{3} p+12-4\right)^{2}=3^{2}
$$
We get two solutions $p=\frac{39}{5}$ and $p=\frac{21}{5}$. Since the point $P$ lies to the left of the point $O_{1}$, we finally have $p_{2}=\frac{21}{5}=4.2$. Thus, $x_{2}=6.4$.
The advantage of the deal will be:
$$
x_{1}+x_{2}-p_{1}-p_{2}=11.36+6.4-0.48-4.2=13.08
$$
Thus, the deal is profitable for Crocodile Gena, and the profit will be 13080 rubles.
Answer: 1) 6.2. 2) profitable, profit 13080 rubles.
## Criteria:
|
13080
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 3. Maximum 14 points
Settlements $A, B$, and $C$ are connected by straight roads. The distance from settlement $A$ to the road connecting settlements $B$ and $C$ is 100 km, and the sum of the distances from settlement $B$ to the road connecting $A$ and $C$, and from settlement $C$ to the road connecting $A$ and $B$ is 300 km. It is known that settlement $D$ is equidistant from the roads connecting settlements $A, B, C$ and lies within the area bounded by these roads. Any resident of all settlements spends no more than 1 liter of fuel for every 10 km of road. What is the maximum amount of fuel that would be needed by a motorist who needs to get from settlement $D$ to any of the roads connecting the other settlements?
#
|
# Solution
The settlements form a triangle $\mathrm{ABC}$, and point $\mathrm{D}$, being equidistant from the sides of the triangle, is the incenter of the triangle (i.e., the center of the inscribed circle). Note that the fuel consumption will be maximal when the distance from point $\mathrm{D}$ to the sides of triangle $\mathrm{ABC}$ is maximal, or, equivalently, when the radius $\mathrm{r}$ of the inscribed circle in the triangle is maximal under the conditions of the problem. The conditions of the problem can be easily reformulated as follows (see the figure):
$$
\text { ha=100, hb+hc=300, }
$$
where $\mathrm{ha}, \mathrm{hb}, \mathrm{hc}$ are the heights of triangle ABC. It is not difficult to establish that the following equality holds:
$$
1 \mathrm{ha} + 1 \mathrm{hb} + 1 \mathrm{hc} = 1 \mathrm{r} \text {. }
$$
Indeed, let $S$ be the area of the triangle, then
$$
\text { S=12aha=12bhb=12chc=pr. }
$$
From this, we have:
$$
\text { ha=2Sa, hb=2Sb, hc=2Sc, r=Sp. }
$$
And we get:
$$
1 \mathrm{ha} + 1 \mathrm{hb} + 1 \mathrm{hc} = \mathrm{a} / 2 \mathrm{~S} + \mathrm{b} / 2 \mathrm{~S} + \mathrm{c} / 2 \mathrm{~S} = (\mathrm{a} + \mathrm{b} + \mathrm{c}) / 2 \mathrm{~S} = \mathrm{p} / \mathrm{S} = 1 \mathrm{r} \text {. }
$$
From the condition, it follows that
$$
1 \mathrm{r} = 1 / 100 + 1 / \mathrm{hb} + 1 / \mathrm{hc} = 1 / 100 + 1 / \mathrm{hb} + 1 / (300 - \mathrm{hb}) = 1 / 100 + 300 / (\mathrm{hb} (300 - \mathrm{hb})).
$$
The radius $\mathrm{r}$ is maximal when the sum on the right side of the equation is minimal. This is equivalent to the maximality of the expression
$$
\mathrm{hb} / (300 - \mathrm{hb}).
$$
This is a quadratic function and its maximum is achieved at its vertex, i.e., as it is easy to understand, when $\mathrm{hb} = 150$. Therefore, the maximum value of the radius of the inscribed circle is found from the equation
$$
1 \mathrm{r} = 1 / 100 + 300 / (150 * (300 - 150)) = 1 / 100 + 1 / 75 = 7 / 300 \Leftrightarrow \mathrm{r} = 300 / 7 \text { km. }
$$
From this, the maximum fuel consumption is:
$$
\text { r10 = 30 / 7 = 307 liters. }
$$
Answer: 307 liters.
|
307
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 5. Maximum 20 points
The commander of a tank battalion, in celebration of being awarded a new military rank, decided to organize a mass celebration, inviting subordinate soldiers. Only the soldiers whom the commander personally invites can attend. The main delicacy at the celebration is buckwheat porridge. However, after consulting with his family, the tank battalion commander decided to make a surprise - initially, guests, as well as the commander himself, will make decisions about consuming buckwheat porridge, assuming that they will pay for it themselves, but at the end of the celebration, the commander will pay the total bill from his family's savings. He discovered that if the soldiers are lined up by height, there is a certain pattern in the changes in their demand functions. The demand for buckwheat porridge of the shortest soldier is given by \( Q_d = 510 - 5.1P \), the next tallest \( Q_d = 520 - 5.2P \), the next \( Q_d = 530 - 5.3P \), and so on. The commander will first invite the shortest tankers. The individual demand of the commander for buckwheat porridge is given by \( Q_d = 500 - 5P \).
(a) How is the individual demand of the 45th soldier invited to the celebration defined? How will the aggregate demand for buckwheat porridge be defined if the commander invites 45 soldiers and does not refuse the porridge himself?
(b) The commander can invite the soldiers to a local cafe, the only one in the region. The marginal costs of producing buckwheat porridge are 0. The commander's family is willing to spend no more than 2,525,000 monetary units on the celebration. What is the maximum number of guests whose demand for buckwheat porridge will be satisfied at the celebration if the commander eats the porridge with everyone?
(c) After estimating how many soldiers he can feed in the cafe, the battalion commander decided to invite the military personnel to a tank festival. In addition to the commander and the soldiers he invites, no one else demands porridge. Buckwheat porridge at the festival is offered by 25 perfectly competitive firms, each with a supply function \( Q_s = 302P \). The commander's family is willing to spend the entire amount allocated for the celebration, but if it is not enough, the commander will ask all guests to equally share the remaining bill. It is known that the equilibrium price was set at 20 monetary units. How many soldiers did the commander invite to the festival? Did the guests have to pay part of the bill? If so, how much did each of them pay? Assume that if the guests, like the commander, have to make a certain fixed payment, the payment does not affect their demand for buckwheat porridge.
(d) Which option (b or c) would the soldiers prefer if each of them made a decision about the amount of buckwheat porridge to consume according to their demand function as described in the problem? Justify your answer.
|
# Solution
(a) The commander tries to feed as many soldiers as possible, which means he will primarily invite relatively short soldiers - all other things being equal, their porridge consumption is less.
Note that the individual demand of soldiers is determined by the formula $Q_{d}=500+10 n-(5+0.1 n) P$, where $n$ is the ordinal number of the soldier. Also note that all consumers are willing to purchase the product at $P \in [0; 100]$, which means that to find the market demand, it is sufficient to sum the individual demand functions.
The individual demand of the 45th soldier is $Q_{d}=950-9.5 P$. To find the demand of 45 soldiers, we can use the formula for the arithmetic progression:
$Q_{d}=\frac{(510-5.1 P + 500 + 10 n - (5 + 0.1 n) P)}{2} \cdot n$
If $n=45$, then:
$Q_{d}=32850-328.5 P$
Adding the commander's demand, we get the market demand: $Q_{d}=33350-333.5 P$.
Answer: $Q_{d}=950-9.5 P, Q_{d}=33350-333.5 P$.
(b) For the canteen, the profit maximization problem coincides with the revenue maximization problem, since marginal costs are zero. If the commander invites $n$ soldiers, then the market demand will be:
$Q_{d}=500-5 P + \frac{1010 n + 10 n^2 - (10.1 n + 0.1 n^2) P}{2}$
$Q_{d}=5 n^2 + 505 n + 500 - (0.05 n^2 + 5.05 n + 5) P$
$T R=(5 n^2 + 505 n + 500) P - (0.05 n^2 + 5.05 n + 5) P^2$
The graph of $T R$ is a parabola opening downwards, since for $n \geq 0$ it holds that $0.05 n^2 + 5.05 n + 5 > 0$. The maximum value is at the vertex of the parabola:
$$
\begin{aligned}
& P^{*}=\frac{-5 n^2 - 505 n - 500}{-2(0.05 n^2 + 5.05 n + 5)}=50 \\
& T R_{\max }=T R(50)=125 n^2 + 12625 n + 12500 \\
& T R_{\max } \leq 2525000 \\
& 125 n^2 + 12625 n + 12500 - 2525000 \leq 0 \\
& n^2 + 101 n - 20100 \leq 0 \\
& (n + 201)(n - 100) \leq 0 \\
& n_{1}=-201, n_{2}=100
\end{aligned}
$$
Obviously, the number of guests cannot be negative or a non-integer, so the maximum number of guests the commander can invite is 100.
Answer: 100.
(c) Knowing the individual supply of one firm, we can find the market supply: $Q_{s}=302 P \cdot 25=7550 P$
It is known that $Q_{s}(20)=Q_{d}(20)$, which means $Q_{d}=7550 \cdot 20=151000$.
From the previous part, we know that
$Q_{d}=5 n^2 + 505 n + 500 - (0.05 n^2 + 5.05 n + 5) P$
$Q_{d}(20)=4 n^2 + 404 n + 400=151000$
$(4 n + 1004)(n - 150)=0$
$n_{1}=-251, n_{2}=150$
The commander invited 150 guests.
$T R=151000 \cdot 20=3020000>2525000$ - the expenses for buckwheat porridge exceed the family budget, which means the soldiers will help pay part of the bill. Each soldier will contribute an amount of $\frac{495000}{150}=3300$ monetary units.
Answer: the commander invited 150 guests, each of whom paid 3300 monetary units.
(d) Let's find the consumer surplus (CS) for the soldier with number $i$ in each case. For a linear demand function, $C S=\frac{(P_{\max }-P_{e}) Q}{2}$. The demand function of each soldier is linear, and $P_{\max }=100$.
In part (b), consumers do not pay for the buckwheat porridge, i.e., $P_{e}=0$.
$Q_{i}^{1}=500+10 i-(5+0.1 i) \cdot 50=250+5 i$
$C S_{i}^{1}=(P_{\max }-P_{e}) \cdot 0.5 \cdot Q_{1}^{i}=(100-0) \cdot 0.5(250+5 i)=12500+250 i$.
In part (c), consumers do not pay for the buckwheat porridge directly, but spend 3300 monetary units.
$Q_{i}^{2}=500+10 i-(5+0.1 i) \cdot 20=400+8 i$
$C S_{i}^{2}=(100-0) \cdot 0.5(400+8 i)-3300=16700+400 i$
It is clear that even with the expenses for partial payment of the bill, for any $i \geq 0$ it holds that $C S_{i}^{2}>C S_{i}^{1}$.
Answer: all soldiers prefer the option (c).
## Grading Criteria
(a) Individual demand is formulated - (1p), market demand is formulated - (2p).
(b) The firm's problem is formulated - (1p), the optimal price is determined - (2p), the maximum number of guests is determined - (3p).
(c) The market supply function is formulated - (1p), the equilibrium quantity is determined - (1p), the market demand function in terms of the number of guests is formulated - (2p), the maximum number of guests is determined - (2p), the behavior of guests and their payment of the bill is determined - (2p).
(d) A justified comparison of benefits is formulated, from which a conclusion is drawn about the preference of a certain option - (3p).
A penalty of 1 point for arithmetic errors that did not lead to significant distortion of the results. Alternative solutions may be evaluated for the full number of points if they contain a correct and justified sequence of actions.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that it was impossible to collect a larger amount of tax revenue on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the amount of tax revenue were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the consumer in the amount of 30 monetary units per unit of the good. The market supply is given by $Q_s = 6P - 312$, and the market demand is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the consumer price increased to 118 monetary units.
1) Restore the market demand function.
2) Determine the amount of tax revenue collected at the chosen tax rate.
3) Determine the per-unit tax rate that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler specified to be collected?
|
# Solution:
1) Let the demand function be linear $Q_{d}=a-b P$. It is known that $1.5 b=6$. We find that $b=$ 4. If a per-unit tax $t=30$ is introduced, then $P_{d}=118 . a-4 P_{d}=6\left(P_{d}-30\right)-312 ; 0.1 a+$ $49.2=P_{d}=118 ; a=688$. The market demand function is $Q_{d}=688-4 P$. (8 points).
2) It is known that $P_{d}(t=30)=118$. Therefore, $Q_{d}=688-472=216, T=216 \cdot 30=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, the maximum of the function is reached at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: an arithmetic error was made - 2 points, lack of justification for the maximum - 5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. 20 points
A beginner economist-cryptographer received a cryptogram from the ruler, which contained another secret decree on the introduction of a commodity tax on a certain market. The cryptogram specified the amount of tax revenue to be collected. It was also emphasized that a larger amount of tax revenue could not be collected on this market. Unfortunately, the economist-cryptographer decrypted the cryptogram with an error - the digits in the tax revenue amount were determined in the wrong order. Based on the incorrect data, a decision was made to introduce a per-unit tax on the producer in the amount of 90 monetary units per unit of the product. It is known that the market demand is given by $Q_d = 688 - 4P$, and the market supply is linear. Additionally, it is known that a change in price by one unit results in a change in the quantity demanded that is 1.5 times smaller than the change in the quantity supplied. After the tax was introduced, the producer's price decreased to 64 monetary units.
1) Restore the market supply function.
2) Determine the amount of tax revenue collected at the chosen tax rate.
3) Determine the per-unit tax rate that would allow the ruler's decree to be fulfilled.
4) What are the tax revenues that the ruler indicated to collect?
|
# Solution:
1) Let the supply function be linear $Q_{s}=c+d P$. It is known that $1.5 \cdot 4=d$. We find that $d=6$. If a per-unit tax $t=90$ is introduced, then $P_{s}=64.688-4\left(P_{s}+90\right)=6 P_{s}+c$; $0.1 c+32.8=P_{s}=64 ; c=-312$. The market supply function is $Q_{s}=6 P-312$. (8 points).
2) It is known that $P_{s}(t=90)=64$. Therefore, $Q_{s}=6 P-312=72, T=72 \cdot 90=6480$. (4 points).
3) Let $P_{s}=P_{d}-t, 688-4 P_{d}=6 P_{d}-6 t-312, P_{d}=100+0.6 t ; \quad Q_{d}=288-2.4 t$. Tax revenues are $T=Q \cdot t=288 t-2.4 t^{2}$. This is a downward-opening parabola, and the maximum of the function is achieved at $t^{*}=60$. (4 points).
4) $T_{\max }=288 \cdot 60-2.4 \cdot 60 \cdot 60=8640$. (4 points).
Penalties: an arithmetic error is made -2 points, lack of justification for the maximum -5 points.
|
8640
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Maximum 15 points. On side AB of an equilateral triangle $\mathrm{ABC}$, a right triangle $\mathrm{A} H \mathrm{~B}$ is constructed ( $\mathrm{H}$ - the vertex of the right angle), such that $\angle \mathrm{HBA}=60^{\circ}$. Let point K lie on ray $\mathrm{BC}$ beyond point $\mathrm{C}$ and $\angle \mathrm{CAK}=15^{\circ}$. Find the angle between line HK and the median of triangle $\mathrm{AHB}$, drawn from vertex $\mathrm{H}$.
|
# Solution:
Extend NB and NA beyond points B and A respectively (H-B-B1, H-A-A1)
$\angle \mathrm{B} 1 \mathrm{BC}=60^{\circ}$
$\angle$ KAA1 $=75^{\circ}$, so BK is the bisector of $\angle \mathrm{ABB} 1$
AK is the bisector of $\angle \mathrm{A} 1 \mathrm{AC}$, therefore, HK is the bisector of $\angle \mathrm{AHB}$, so we need to find the angle between the bisector and the median of triangle AHB, drawn from the vertex of the right angle. Answer: $15^{\circ}$
| Criteria | Score |
| :--- | :---: |
| The solution algorithm is provided, the drawing is correct, and the correct answer is obtained | 15 |
| The solution algorithm is provided, the drawing is correct, but due to an arithmetic error, the answer is incorrectly calculated or the answer is correct, but the logic of the proof is violated | 6 |
| The solution algorithm has significant logical errors, but some points of the proof are correct | 2 |
| The solution does not meet any of the presented criteria | 0 |
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Maximum 15 points. Masha was given a chest with multicolored beads (each bead has a unique color, there are a total of $\mathrm{n}$ beads in the chest). Masha chose seven beads for her dress and decided to try all possible combinations of them on the dress (thus, Masha selects from a set of options to sew one, two, three, four, five, six, or seven beads, and the order of the beads does not matter to her). Then she counted how many options she got and was very surprised that the number turned out to be odd.
1) What number did Masha get?
2) Is it true that if Masha had chosen from an even number of beads, she could have gotten an even number of options?
3) Is it true that if the order of the beads sewn on the dress mattered to Masha, she could have gotten both even and odd numbers?
|
# Solution:
1) Consider one bead. Before Masha sews it onto the dress, there are two options: to take the bead or not. If we now choose two beads, the number of options becomes four, which can be obtained by multiplying the first option by two. By increasing the number of beads, we conclude that the total number of all possible options for p beads is $2^{\text {n }}$. According to the problem, Masha chose 7 beads, but the option of not sewing anything is not considered, so the number of options is $2^{7}-1=127$.
2) The number of options is odd (see point 1).
3) This is correct; it is sufficient to consider the case of one and two beads. In the first case, the number of options is 1, and in the second case, it is 4.
| Criterion | Score |
| :--- | :---: |
| The correct answer is obtained in points 1), 2), and 3) through a logically justified algorithm | 15 |
| The correct answer is obtained in only one of the points through a logically justified algorithm | 2 |
| The solution does not meet any of the presented criteria | 0 |
|
127
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Maximum 15 points. The company "Intelligence, Inc" has developed a robot with artificial intelligence. To manufacture it, a special machine is required, which can produce 1 robot in 1 hour. The company owns a large number of such machines, but the created robot is so intelligent that it can produce an exact copy of itself in exactly 2 hours. "Intelligence, Inc" has received an urgent order for 3250 robots, which need to be manufactured in just 10 hours. The company's expenses for creating one robot (regardless of the time of its production and whether it was created by another robot or a special machine) are constant and amount to 100 monetary units, and the maintenance of one special machine during its operation to produce a robot costs the company 70 monetary units. "Intelligence, Inc" does not incur any other expenses for creating robots. What are the minimum costs the company will incur in fulfilling the order?
|
# Solution:
To minimize the company's costs, it is necessary to find the minimum number of machines that will allow the company to complete the order within the specified time frame.
Let $x$ be the number of machines. Then, in the first hour of operation, they will produce $x$ robots. These $x$ robots will start manufacturing new robots, and after 3 hours, there will be $2x$. Meanwhile, the machines continue their work and will produce a new batch of robots by the end of the 2nd hour, which will also start manufacturing their own kind. Thus, after 3 hours of the company's operation, there will be $2x + x$ robots. If we consider that after 9 hours from the start of the work on the order, the robots manufacturing their own kind will not have time to produce new ones by the deadline, then after 10 hours of operation, $62x$ robots will be produced.
The calculation of the number of robots is also presented in the table:
| 1 hour | 2 hour | 3 hour | 4 hour | 5 hour | 6 hour | 7 hour | 8 hour | 9 hour | 10 hour | Total |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{x}$ | - | $2x + x$ | - | $6x + x$ | - | $14x + x$ | - | $30x + x$ | - | $60x + 2x$ |
| - | $\mathrm{x}$ | - | $2x + x$ | - | $6x + x$ | - | $14x + x$ | - | $30x + x$ | |
It remains to find $x$ and round it up to the nearest whole number (otherwise, there won't be enough machines to produce the required number of robots within the specified time).
$$
x = \frac{3250}{62} \approx 52.42
$$
Thus, the company's costs for fulfilling the order will be $100 * 3250 + 70 * 53 = 328710$ monetary units.
## Grading Scheme:
Full correct solution 15 points.
Correctly found the number of robots - 10 points out of 15.
Correctly found the company's expenses for fulfilling the order - 5 points out of 15.
The number of robots is found incorrectly, but there is a correct logic for finding the costs - 2 points out of 15.
There is progress in finding the number of robots, but there are minor logical errors and the number of robots is found incorrectly - 5 points out of 15.
|
328710
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4. Maximum 20 points
## Option 1
At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved problem. A complete solution to a planimetry problem is worth 7 points, and a problem in stereometry is worth 12 points. The tournament winner is the team that scores the highest number of points. Andrey is organizing his team of 3 people, where he will be the captain. He is considering whether to invite Volodya and Zhanna or Petya and Galina. Therefore, he asked all the candidates to honestly provide information about their capabilities in solving problems within the allotted time for this stage of the tournament. It is known that the opportunity costs for each student in solving planimetry and stereometry problems are always constant.
| Name | Maximum number of stereometry problems if solving only them | Maximum number of planimetry problems if solving only them |
| :--- | :---: | :---: |
| Andrey | 7 | 7 |
| Volodya | 6 | 3 |
| Zhanna | 3 | 18 |
| Petya | 12 | 3 |
| Galina | 7 | 14 |
Help Andrey decide which pair of students to take into his team if the team's only goal is to win the tournament.
|
# Solution:
Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve.
Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points for the team.
Volodya, instead of solving 1 problem in planimetry, can solve 2 problems in stereometry. Since 1 problem in planimetry is less valuable than 2 problems in stereometry, he should specialize in stereometry problems, earning $12 * 6 = 72$ points for the team.
Zhanna, instead of solving 1 problem in stereometry, can solve 6 problems in planimetry. Since 1 problem in stereometry is less valuable than 6 problems in planimetry, she should specialize in planimetry problems, earning $7 * 18 = 126$ points for the team. In total, the team of Andrey, Volodya, and Zhanna can earn a maximum of 84 + 72 + 126 = 282 points.
Let's find out what the maximum result the team of Andrey, Petya, and Galina could achieve.
Petya, instead of solving 1 problem in planimetry, can solve 4 problems in stereometry. Since 1 problem in planimetry is less valuable than 4 problems in stereometry, he should specialize in stereometry problems, earning $12 * 12 = 144$ points for the team. Galina, instead of solving 1 problem in stereometry, can solve 2 problems in planimetry. Since 1 problem in stereometry is less valuable than 2 problems in planimetry, she should specialize in planimetry problems, earning $7 * 14 = 98$ points for the team.
In total, the team of Andrey, Petya, and Galina can earn a maximum of $84 + 144 + 98 = 326$ points.
Thus, Andrey should invite Petya and Galina to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Volodya, Zhanna, Petya, and Galina.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Andrey will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
## Variant 2
A school is holding the remote stage of a team tournament in physics, where the results of the participants are evaluated based on the number of points earned for a fully solved problem. A fully solved problem in kinematics is worth 14 points, and a problem in thermodynamics is worth 24 points. The team that scores the most points wins the tournament. Volodya is forming his team of 3 people, in which he will be the captain. He is considering whether to invite Andrey and Tatyana, or Semyon and Maria. Therefore, he asked all the candidates to honestly indicate in the table their capabilities for solving problems during the allocated time for this stage of the tournament. It is known that the opportunity costs of each student for solving problems in kinematics and thermodynamics are always constant.
| Name | Maximum number of kinematics problems if solving only them | Maximum number of thermodynamics problems if solving only them |
| :--- | :---: | :---: |
| Volodya | 7 | 7 |
| Semyon | 4 | 8 |
| Maria | 8 | 2 |
| Andrey | 2 | 12 |
| Tatyana | 2 | 1 |
Help Volodya decide which pair of students to invite to his team if the team's only goal is to win the tournament.
## Solution:
Let's find out what the maximum result the team of Volodya, Semyon, and Maria could achieve.
Volodya, instead of solving 1 problem in thermodynamics, can solve 1 problem in kinematics. Since a problem in thermodynamics is more valuable, he should specialize in thermodynamics problems, earning $24 * 7 = 168$ points for the team.
Semyon, instead of solving 1 problem in kinematics, can solve 2 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 2 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 8 = 192$ points for the team. Maria, instead of solving 1 problem in thermodynamics, can solve 4 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 4 problems in kinematics, she should specialize in kinematics problems, earning $14 * 8 = 112$ points for the team. In total, the team of Volodya, Semyon, and Maria can earn a maximum of $168 + 192 + 112 = 472$ points.
Let's find out what the maximum result the team of Volodya, Andrey, and Tatyana could achieve.
Andrey, instead of solving 1 problem in kinematics, can solve 6 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 6 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 12 = 288$ points for the team.
Tatyana, instead of solving 1 problem in thermodynamics, can solve 2 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 2 problems in kinematics, she should specialize in kinematics problems, earning $14 * 2 = 28$ points for the team.
In total, the team of Volodya, Andrey, and Tatyana can earn a maximum of $168 + 288 + 28 = 484$ points.
Thus, Volodya should invite Andrey and Tatyana to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Semyon, Maria, Andrey, and Tatyana.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Volodya will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
|
326
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 20 points
In the city of Eifyadl, runic stones are sold. It is known that the first merchant offers a fixed discount of $\mathrm{n} \%$ for every 5th stone purchased, while the second merchant increases the discount by $1 \%$ for each subsequent stone purchased (0% for the 1st stone, 3% for the 4th stone, and so on), but not exceeding $20 \%$. An order for 100 runic stones has been placed. Assume that all stones must be purchased from only one merchant.
a) Find the minimum discount rate for the first merchant so that it is more advantageous for the buyer to purchase all the stones from him.
b) What is the maximum number of runic stones that would be more advantageous to buy from the first merchant with the discount found in part a)?
c) Suppose now that the buyer has increased the order to 175 stones. Upon learning this, the second merchant increased the discount limit to $30 \%$ and the discount increment to $2 \%$. Can the first merchant compete with the second?
|
# Solution and Evaluation Criteria:
a) Let's assume the cost of one rune stone without a discount is 1 unit of currency. We find the average cost of a rune stone from the first merchant:
$(20(1-n)+80) / 100=(100-20 n) / 100$
For the second merchant:
$(1+0.99+0.98+\ldots+0.81+0.8 * 80) / 100=82.1 / 100$
Then:
$100-20 n>82.1$
$17.9 / 20 \quad n>0.895$
Thus, $n=90\%$
Answer: $90\%$.
## Evaluation Criteria:
Logically justified solution - 5 points, logically justified solution with arithmetic error - 0 points
b) We find the average cost of a rune stone from the second merchant:
$(1+0.98+\ldots+0.72+0.7 * 159) / 175=125.06 / 175$
And from the first merchant:
$(35 *(1-n)+140) / 175=(175-35 n) / 175$
Then:
$125.06>175-35 n$
$n>(175-125.06) / 35$
$n>49.94 / 35$
$n>1.427$, which is impossible, as the maximum discount is $100\%$, and paying extra for the purchase of a stone significantly affects the merchant's costs. Answer - no, $n>100\%$
## Evaluation Criteria:
Logically justified solution - 7 points,
justified solution with arithmetic error - 3.5 points, incorrect answer - 0 points
c) Let the number of stones be X. We write the general equation for the average cost of a stone from the first and second merchants:
$1: (x / 5 * (1-0.9) + (x - x / 5)) / x$
$2: (18.1 + 0.8 * (x - 20)) / x$
We get:
$0.1 x / 5 + x - x / 5 < 18.1 + 0.8 * (x - 20)$
$x - 0.18 x - 0.8 x < 2.1$
$0.02 x < 2.1$
$x < 105$
Then: $x=104$ stones.
Answer: 104.
Evaluation Criteria:
Logically justified solution - 8 points,
solution with arithmetic error - 0 points, unsolved part a) - 0 points.
|
104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 4. Maximum 20 points
## Option 1
At a school, the remote stage of a team geometry tournament is taking place, where participants' results are evaluated based on the number of points earned for a fully solved problem. A complete solution to a planimetry problem is worth 7 points, and a problem in stereometry is worth 12 points. The team that scores the highest number of points wins the tournament. Andrey is organizing his team of 3 people, where he will be the captain. He is considering whether to invite Volodya and Zhanna or Petya and Galina. Therefore, he asked all the candidates to honestly provide information about their capabilities in solving problems within the allotted time for this stage of the tournament. It is known that the opportunity costs for each student in solving planimetry and stereometry problems are always constant.
| Name | Maximum number of stereometry problems if solving only them | Maximum number of planimetry problems if solving only them |
| :--- | :---: | :---: |
| Andrey | 7 | 7 |
| Volodya | 6 | 3 |
| Zhanna | 3 | 18 |
| Petya | 12 | 3 |
| Galina | 7 | 14 |
Help Andrey decide which pair of students to take into his team if the team's only goal is to win the tournament.
|
# Solution:
Let's find out what the maximum result the team of Andrey, Volodya, and Zhanna could achieve.
Andrey, instead of solving 1 problem in planimetry, can solve 1 problem in stereometry. Since a problem in stereometry is more valuable, he should specialize in stereometry problems, earning $12 * 7 = 84$ points for the team.
Volodya, instead of solving 1 problem in planimetry, can solve 2 problems in stereometry. Since 1 problem in planimetry is less valuable than 2 problems in stereometry, he should specialize in stereometry problems, earning $12 * 6 = 72$ points for the team.
Zhanna, instead of solving 1 problem in stereometry, can solve 6 problems in planimetry. Since 1 problem in stereometry is less valuable than 6 problems in planimetry, she should specialize in planimetry problems, earning $7 * 18 = 126$ points for the team. In total, the team of Andrey, Volodya, and Zhanna can earn a maximum of 84 + 72 + 126 = 282 points.
Let's find out what the maximum result the team of Andrey, Petya, and Galina could achieve.
Petya, instead of solving 1 problem in planimetry, can solve 4 problems in stereometry. Since 1 problem in planimetry is less valuable than 4 problems in stereometry, he should specialize in stereometry problems, earning $12 * 12 = 144$ points for the team. Galina, instead of solving 1 problem in stereometry, can solve 2 problems in planimetry. Since 1 problem in stereometry is less valuable than 2 problems in planimetry, she should specialize in planimetry problems, earning $7 * 14 = 98$ points for the team.
In total, the team of Andrey, Petya, and Galina can earn a maximum of $84 + 144 + 98 = 326$ points.
Thus, Andrey should invite Petya and Galina to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Volodya, Zhanna, Petya, and Galina.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Andrey will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
## Variant 2
A school is holding the remote stage of a team tournament in physics, where the results of the participants are evaluated based on the number of points earned for a fully solved problem. A fully solved problem in kinematics is worth 14 points, and a problem in thermodynamics is worth 24 points. The team that scores the most points wins the tournament. Volodya is forming his team of 3 people, in which he will be the captain. He is considering whether to invite Andrey and Tatyana, or Semyon and Maria. Therefore, he asked all the candidates to honestly indicate in the table their capabilities for solving problems during the allocated time for this stage of the tournament. It is known that the opportunity costs of each student for solving problems in kinematics and thermodynamics are always constant.
| Name | Maximum number of kinematics problems if solving only them | Maximum number of thermodynamics problems if solving only them |
| :--- | :---: | :---: |
| Volodya | 7 | 7 |
| Semyon | 4 | 8 |
| Maria | 8 | 2 |
| Andrey | 2 | 12 |
| Tatyana | 2 | 1 |
Help Volodya decide which pair of students to invite to his team if the team's only goal is to win the tournament.
## Solution:
Let's find out what the maximum result the team of Volodya, Semyon, and Maria could achieve.
Volodya, instead of solving 1 problem in thermodynamics, can solve 1 problem in kinematics. Since a problem in thermodynamics is more valuable, he should specialize in thermodynamics problems, earning $24 * 7 = 168$ points for the team.
Semyon, instead of solving 1 problem in kinematics, can solve 2 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 2 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 8 = 192$ points for the team. Maria, instead of solving 1 problem in thermodynamics, can solve 4 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 4 problems in kinematics, she should specialize in kinematics problems, earning $14 * 8 = 112$ points for the team. In total, the team of Volodya, Semyon, and Maria can earn a maximum of $168 + 192 + 112 = 472$ points.
Let's find out what the maximum result the team of Volodya, Andrey, and Tatyana could achieve.
Andrey, instead of solving 1 problem in kinematics, can solve 6 problems in thermodynamics. Since 1 problem in kinematics is less valuable than 6 problems in thermodynamics, he should specialize in thermodynamics problems, earning $24 * 12 = 288$ points for the team.
Tatyana, instead of solving 1 problem in thermodynamics, can solve 2 problems in kinematics. Since 1 problem in thermodynamics is less valuable than 2 problems in kinematics, she should specialize in kinematics problems, earning $14 * 2 = 28$ points for the team.
In total, the team of Volodya, Andrey, and Tatyana can earn a maximum of $168 + 288 + 28 = 484$ points.
Thus, Volodya should invite Andrey and Tatyana to his team.
## Grading Criteria:
To compare the benefits of forming each team, it is sufficient to compare the contributions of only four participants: Semyon, Maria, Andrey, and Tatyana.
A correct calculation of the points each of them can bring to the team, conducted in any correct way - 4 points.
A correct calculation of the total number of points (for the team as a whole or for the pair of participants that Volodya will invite to the team) - 2 points.
A comparison of the benefits of each pair of participants and the correct answer - 2 points.
Any arithmetic error that did not lead to a significant distortion of the results is penalized by 1 point.
Any arithmetic error that led to a significant distortion of the results is penalized by 5 points.
|
326
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 5. Maximum 20 points
In the city of Eifyadl, runic stones are sold. It is known that the first merchant offers a fixed discount of $\mathrm{n} \%$ for every 5th stone purchased, while the second merchant increases the discount by $1 \%$ for each subsequent stone purchased (0% for the 1st stone, 3% for the 4th stone, and so on), but not exceeding $20 \%$. An order for 100 runic stones has been placed. Assume that all stones must be purchased from only one merchant.
a) Find the minimum discount rate for the first merchant so that it is more advantageous for the buyer to purchase all the stones from him.
b) What is the maximum number of runic stones that it would be more advantageous to buy from the first merchant with the discount found in part a)?
c) Suppose now that the buyer has increased the order to 175 stones. Upon learning this, the second merchant increased the discount limit to $30 \%$ and the discount increment to $2 \%$. Can the first merchant compete with the second?
|
# Solution and Evaluation Criteria:
a) Let's assume the cost of one rune stone without a discount is 1 unit of currency. We find the average cost of a rune stone from the first merchant:
$(20(1-n)+80) / 100=(100-20 n) / 100$
For the second merchant:
$(1+0.99+0.98+\ldots+0.81+0.8 * 80) / 100=82.1 / 100$
Then:
$100-20 n>82.1$
$17.9 / 20 \quad n>0.895$
Thus, $n=90\%$
Answer: $90\%$.
## Evaluation Criteria:
Logically justified solution - 5 points, logically justified solution with arithmetic error - 0 points
b) We find the average cost of a rune stone from the second merchant:
$(1+0.98+\ldots+0.72+0.7 * 159) / 175=125.06 / 175$
And from the first merchant:
$(35 *(1-n)+140) / 175=(175-35 n) / 175$
Then:
$125.06>175-35 n$
$n>(175-125.06) / 35$
$n>49.94 / 35$
$n>1.427$, which is impossible, as the maximum discount is $100\%$, and paying extra for the purchase of a stone significantly affects the merchant's costs. Answer - no, $n>100\%$
## Evaluation Criteria:
Logically justified solution - 7 points,
justified solution with arithmetic error - 3.5 points, incorrect answer - 0 points
c) Let the number of stones be X. We write the general equation for the average cost of a stone from the first and second merchants:
$1: (x / 5 * (1-0.9) + (x - x / 5)) / x$
$2: (18.1 + 0.8 * (x - 20)) / x$
We get:
$0.1 x / 5 + x - x / 5 < 18.1 + 0.8 * (x - 20)$
$x - 0.18 x - 0.8 x < 2.1$
$0.02 x < 2.1$
$x < 105$
Then: $x=104$ stones.
Answer: 104.
Evaluation Criteria:
Logically justified solution - 8 points,
solution with arithmetic error - 0 points, unsolved part a) - 0 points.
|
104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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