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7.2. The road from point A to point B first goes uphill and then downhill. A cat takes 2 hours and 12 minutes to travel from A to B, and the return trip takes 6 minutes longer. The cat's speed going uphill is 4 km/h, and downhill is 5 km/h. How many kilometers is the distance from A to B? (Provide a complete solution, not just the answer.)
|
Solution: When a cat goes uphill, it takes 15 minutes for 1 km, and when it goes downhill, it takes 12 minutes. That is, when the direction changes, the time spent on 1 km changes by 3 minutes. Since the cat spent 6 minutes more on the return trip, the uphill section on the return trip is 2 km longer.
Let the length of the uphill section on the way from A to B be $x$. Then we can set up the equation $15 x + 12(x + 2) = 132$, from which $x = 4$. Therefore, the total distance is $x + (x + 2) = 4 + 6 = 10$ km.
Comment: Other solutions are possible. For example, a system of equations could have been set up and solved immediately. Or it could have been proven that the difference between the two sections of the path is 2 km, then the answer could be guessed, and then its uniqueness could be proven.
Criteria: Only the answer - 0 points.
Answer with verification - 1 point.
Correct system of equations - 2 points.
Proven that the difference between the two sections of the path is 2 km - 1 point.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. On a certain island, there live 100 people, each of whom is either a knight, who always tells the truth, or a liar, who always lies. One day, all the inhabitants of this island lined up, and the first one said: "The number of knights on this island is a divisor of the number 1." Then the second said: "The number of knights on this island is a divisor of the number 2," and so on until the hundredth, who said: "The number of knights on this island is a divisor of the number 100." Determine how many knights can live on this island. (Find all answers and prove that there are no others.)
|
Solution: If there are no knights, then all the speakers are lying, since 0 is not a divisor of any natural number.
If there are knights, let there be $a$ of them. Then only the people with numbers $a k$ for $k=1,2, \ldots$ are telling the truth. On the other hand, since exactly $a$ people are telling the truth, $k$ changes precisely from 1 to $a$.
Thus, $a$ is such a number that the person with number $a \cdot a$ is still in the row $\left(a^{2} \leqslant 100\right)$, while the person with number $a \cdot(a+1)$ is not $(a(a+1)>100)$. Clearly, only $a=10$ fits, as for $a10$ - the first.
In total, there are either 0 or 10 knights.
Criteria: Only the answer - 0 points.
The case where there are no knights - 1 point.
Noted that the people with numbers $a, 2 a, \ldots, k a$ should be telling the truth - 2 points.
Proven that $k=a$ - another 2 points.
Proven that then $a=10$ - another 2 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. On a cubic planet, there live cubic mice, and they live only on the faces of the cube, not on the edges or vertices. It is known that different numbers of mice live on different faces, and the number on any two adjacent faces differs by at least 2. What is the minimum number of cubic mice that can live on this planet, given that there is at least one mouse on each face?
|
Solution: We will prove that no three consecutive numbers can be the number of mice on the faces. Indeed, if there were $x, x+1$, and $x+2$ mice on some three faces, then $x$ and $x+1$ would have to be on opposite faces. But then $x+2$ mice could not be anywhere.
Consider the first 8 natural numbers. Among the first three, there is at least one missing, and among the next three, there is at least one missing. Therefore, the minimum sum is $1+2+4+5+7+8=27$. It is possible to distribute 27 mice. For this, we will divide all the faces into pairs of opposite faces. On the first pair, there will be 1 and 2 mice, on the second pair - 4 and 5, on the third pair - 7 and 8. Thus, quantities differing by 1 are on opposite faces, and on adjacent faces, the difference is at least 2.
Criteria: Only the answer - 1 point.
Only an example (with or without justification) - 2 points.
Only the estimate - 5 points
Proved that no three consecutive numbers can be the number of mice on the faces - 2 points.
|
27
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. The perimeter of triangle $ABC$ is 24 cm, and the segment connecting the point of intersection of its medians with the point of intersection of its angle bisectors is parallel to side $AC$. Find the length of $AC$.
|
Answer: 8 cm.
Solution. Let AK be the median from vertex A, M - the point of intersection of the medians ABC, and I - the point of intersection of its angle bisectors AA1, BB1, CC1. Draw a line through K parallel to AC, intersecting the angle bisector BB1 at point P - its midpoint. By Thales' theorem, $PI: IB1 = KM: MA = 2$, so $BI: IB1 = 1$. By the property of angle bisectors AI and CI in triangles $ABB1$ and $CBB1$, we have
$AB: AB1 = BI: IB1 = CB: CB1 = 1$. Therefore, $AC = \frac{1}{2}(AB + BC) = \frac{1}{3}(AB + BC + AC) = 8$.
Evaluation. Just the answer: 0 points. Finding the ratio $BI: IB1 = 1: 3$: 1 point. Finding the ratios $AB: AB1 = BI: IB1 = CB: CB1 = 1: 2$: 2 points.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. What is the smallest number of colors needed to color all the cells of a 6 by 6 square so that in each row, column, and diagonal of the square, all cells have different colors? Explanation: a diagonal of the square is understood to mean all rows of at least two cells running diagonally from one edge of the square to the other at an angle of $45^{\circ}$ or $135^{\circ}$ to the horizontal.
|
Answer: In 7 colors.
Solution. Let's provide an example of coloring in 7 colors that satisfies the condition of the problem. Consider a 7 by 7 square, and color it in the required way using 7 colors with a known technique: the coloring of each subsequent row is obtained from the coloring of the previous row by a cyclic shift of 2 cells. Then, we select the left lower corner 6 by 6 square from it, which will be the required example.
Let's prove the minimality of 7 colors. Each row of the square contains 6 cells, so at least 6 colors are required for a correct coloring. Suppose we managed to color the square in 6 colors in the required way, then there will be exactly 6 cells of each color, and they will all be in different verticals and horizontals of the square. Let's call the color of the left lower corner cell A of the square black, and show that the remaining 5 black cells cannot be located in the upper right 5 by 5 square in different verticals and horizontals and at the same time not lie on its main diagonal, already controlled by cell A. Otherwise, we can assume that at least three of them would be located below the main diagonal in a "staircase" figure consisting of 10 cells, made up of 4 strips of 4, 3, 2, and 1 cells respectively. However, it is directly verified that for any cell of the "staircase," cells not in the same row, column, or diagonal can correctly contain no more than one black cell. The possible positions of this second black cell are shown in the figure. Cells controlled by it are colored light gray.
Therefore, it is impossible to color the 6 by 6 square in the required way in 6 colors.
Grading criteria. Any explicit and correct example with 7 colors: 3 points. Its correctness does not need to be proven.
Noted that there should be no fewer than 6 colors: 1 point. Proven that there should be no fewer than 7 colors: 4 points.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. In the city, there are 9 bus stops and several buses. Any two buses have no more than one common stop. Each bus has exactly three stops. What is the maximum number of buses that can be in the city
|
Answer: 12.
Solution: if some stop is common for 5 routes, then no two of them have any more common stops, which means there are at least $1+5*2=11$ stops, which contradicts the condition. Therefore, each stop is the intersection of no more than 4 routes, then the total number of stops made by buses is no more than 94, and since there are 3 buses on each route, the total number of buses is no more than $94 / 3=12$.
Buses in the city can run, for example, as follows (stops are marked with numbers): $(1,2,3),(4,5,6),(7,8,9)$ $(1,4,7),(2,5,8),(3,6,9)$ $(1,5,9),(2,6,7),(3,4,8)$ $(3,5,7),(2,4,9),(1,6,8)$.
Criteria: only the example - 3 points; only the estimation that it cannot be more than 12 - 3 points; only the remark that one stop cannot be in more than 4 routes - 1 point; only the answer - 0 points.
## 8th grade
## Each problem is worth 7 points
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. In the city, there are 9 bus stops and several buses. Any two buses have no more than one common stop. Each bus has exactly three stops. What is the maximum number of buses that can be in the city?
|
Answer: 12.
Solution: if some stop is common for 5 routes, then no two of them have any more common stops, which means there are at least $1+5*2=11$ stops, which contradicts the condition. Therefore, each stop is the intersection of no more than 4 routes, then the total number of stops made by buses is no more than 94, and since there are 3 buses on each route, the total number of buses is no more than $94 / 3=12$.
Buses in the city can operate, for example, as follows (stops are denoted by numbers):
$(1,2,3),(4,5,6),(7,8,9)$
$(1,4,7),(2,5,8),(3,6,9)$
$(1,5,9),(2,6,7),(3,4,8)$
$(3,5,7),(2,4,9),(1,6,8)$.
Criteria: only the example - 3 points; only the estimation that it cannot be more than 12 - 3 points; only the remark that one stop cannot be in more than 4 routes - 1 point; only the answer - 0 points.
## 9th grade
Each problem is worth 7 points
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. The electronic clock on the building of the station shows the current hours and minutes in the format HH:MM from 00:00 to 23:59. How much time in one day will the clock display four different digits?
|
Answer: 10 hours 44 minutes.
Solution. Each possible combination of four digits burns on the clock for one minute. Consider separately the time of day from 00:00 to 19:59 and from 20:00 to 23:59. In the first case, the number of valid combinations according to the problem's conditions will be: 2 (the tens digit of the hour in the first position), multiplied by 5 (the tens digit of the minute in the third position, except for the one already burning in the first position), multiplied by 8 (the units digit of the hour, not equal to the digits in the 1st and 3rd positions) and multiplied by 7 (the units digit of the minute in the fourth position, not equal to the digits in the 1st, 2nd, and 3rd positions). This results in 560 combinations, burning for 560 minutes, which is 9 hours and 20 minutes per day.
In the second case, the number of valid combinations according to the problem's conditions will be: 1 (the tens digit of the hour in the first position, equal to 2), multiplied by 3 (the units digit of the hour in the second position, not equal to 2), multiplied by 4 (the tens digit of the minute in the third position, except for those already burning in the 1st and 2nd positions) and multiplied by 7 (the units digit of the minute in the fourth position, not equal to the digits in the 1st, 2nd, and 3rd positions). This results in 84 combinations, burning for 84 minutes, which is 1 hour and 24 minutes per day. Together, we get 10 hours and 44 minutes.
## Grading Criteria.
The idea of dividing the consideration into cases for the time of day from 00:00 to 19:59 and from 20:00 to 23:59: 2 points.
The idea of counting the number of options by digits in a convenient order of position: 3 points.
The correct calculation after this: 2 points.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. Find the number of different ways to arrange all natural numbers from 1 to 9 inclusive in the cells of a 3 by 3 table, one number per cell, such that the sums of the numbers in each row and each column are equal. The table cannot be rotated or reflected.
|
Answer: 72 ways
Solution. Among the numbers from 1 to 9, there are 5 odd numbers. Since the sums in all rows and columns are $\frac{1}{3}(1+2+\ldots+9)=15$ - which are odd, there must be an odd number of odd numbers in each row and each column. This is only possible if one row contains three odd numbers, and the other rows contain one each, and one column contains three odd numbers, and the other columns contain one each. Therefore, all odd numbers must be in the union of a certain row and a certain column. At the intersection of these row and column, the number $15+15-(1+3+5+7+9)=5$ is written. Moreover, 7 and 9 cannot be in the same row or column. Therefore, we can write the number 5 in one of the nine cells, then write the number 9 in one of the four cells of the same row or column as 5, and then write the number 7 in two cells of the same row or column, not adjacent to 9. This gives us 9.4$\cdot$2=72 ways. Then we verify that at the intersection of the column and row containing 7 and 9, only 2 can be placed: 6 and 8 do not fit the sum, and 4 would give another 4 with 7. After this, it is clear that the remaining even numbers can be placed in the remaining cells of the table, and there is only one way to do this.
Grading Criteria. Noted that all odd numbers are in the union of a certain row and a certain column: 2 points.
Proved that the number 5 is written at the intersection of these row and column: 1 point.
Noted that 7 and 9 cannot be in the same row or column: 1 point.
Arrangements of the remaining odd digits in 72 ways: 2 points.
Lack of precise justification that the sums in rows and columns are 15: deduct 1 point. Arrangements differing by rotation or reflection are considered the same: deduct 2 points.
Loss of part of the solutions: deduct 2 points or more.
All required arrangements are constructed, but it is not proved that there are no others: no more than 3 points.
|
72
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. In a fairy-tale country, every piglet either always lies or always tells the truth, and each piglet reliably knows whether every other piglet is a liar. One day, Nif-Nif, Naf-Naf, and Nuf-Nuf met for a cup of tea, and two of them made statements, but it is unknown who exactly said what. One of the three piglets said: “Nif-Nif and Naf-Naf both always lie.” Another said: “Nif-Nif and Nuf-Nuf both always lie.” Determine how many liars are among the three piglets.
|
# Answer: Two
Solution: If at least one of the statements is true, then the piglets mentioned in it are liars, which means there are at least two liars. At the same time, the one making this true statement must be telling the truth. Therefore, there are no more than two liars. In total, if at least one of the spoken phrases is true, then there are two liars. Note that this is possible, for example, if the first phrase is said by Nuf-Nuf.
If both statements are false, then those who say them are liars. This means there are at least two liars. At the same time, for each phrase to be false, at least one honest piglet must be mentioned. This means there is one, so there are no more than two liars. Thus, in this case, there are exactly two liars. Note that this situation is possible if the only honest piglet is Nif-Nif and he is silent.
Criteria: Only the answer - 0 points.
Example that there can be 2 liars, no other progress - 1 point.
Enumerative solution, missing cases - no more than 4 points.
For the absence of verification that there can indeed be two liars, do not deduct points.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Provide an example of a natural number that is a multiple of 2020, such that the sum of its digits is also a multiple of 2020.
|
Solution: For example, the number 20202020...2020, where the fragment 2020 repeats 505 times, fits. Such a number is obviously divisible by 2020, and the sum of its digits is $505 \times 4=2020$.
Criteria: Any correct answer with or without verification - 7 points.
Correctly conceived example, but with a calculation error (for example, 2020 repeats 500 times) - 3 points.
|
2020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. On a certain island, only knights, who always tell the truth, and liars, who always lie, live. One day, 99 inhabitants of this island stood in a circle, and each of them said: "All ten people following me in a clockwise direction are liars." How many knights could there be among those standing in the circle?
|
Answer: 9 knights.
Solution. Note that all people cannot be liars, because then it would mean that each of them is telling the truth. Therefore, among these people, there is at least one knight. Let's number all the people so that the knight is 99th in the sequence. Then, the 10 people with numbers from 1 to 10 are liars (this follows from the knight's statement). Moreover, the 10 people with numbers 89-98 are also liars, since they made an incorrect statement (there is a knight among the ten people standing after them). Therefore, since these 10 people stand in a row along our circle after person number 88, he told the truth and, consequently, is also a knight. Repeating the reasoning for him, we get that people with numbers 78-87 are liars, and so on. In the end, we get that the knights are people with numbers 99, 88, 77, ..., 11, that is, there are 9 of them in total.
For completeness, note that we have proven that if the arrangement is possible, it must look exactly like this. It is not difficult to check that there are no contradictions, and that the people could indeed stand this way.
Criteria. Only the answer - 0 points.
Only the answer with an example of the arrangement - 2 points.
The case where there are no knights is not considered - no points deducted.
The obtained arrangement is not verified - no points deducted.
|
9
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Vera Aleksandrovna urgently needed to cut out three 20-sided polygons (not necessarily identical) from one rectangular sheet of paper. She can take this sheet and cut it along a straight line into two parts. Then take one of the resulting parts and cut it along a straight line. Then take one of the available pieces, cut it, and so on. What is the minimum number of cuts Vera Aleksandrovna will have to make to ensure that among the resulting pieces there are the three 20-sided polygons she needs?
|
Answer: 50 cuts.
Solution: With each cut, the total number of paper pieces increases by 1 (one piece turns into two new pieces), so after $n$ cuts, there will be $(n+1)$ pieces of paper. Let's calculate how many vertices all the pieces together can have after $n$ cuts. With each cut, the total number of vertices increases by 2 (if cutting through two vertices), 3 (if cutting through a vertex and a side), or 4 (if cutting through sides). Since there were initially 4 vertices, after $n$ cuts, the total number of vertices in all pieces together will not exceed $4 n+4$.
Suppose that after $n$ cuts, we have found three 20-sided polygons. Since the total number of pieces obtained will be $n+1$, there will be $n+1-3$ pieces in addition to these 20-sided polygons. Each of these pieces has at least three vertices, so the total number of vertices is at least $20 \cdot 3+3(n-2)=3 n+54$.
Thus, $4 n+4 \geqslant 3 n+54$, from which $n \geqslant 50$.
Now let's show how to get three 20-sided polygons with 50 cuts. Here is one way: cut the original sheet into three rectangles (2 cuts) and convert each rectangle into a 20-sided polygon with 16 cuts by cutting off triangles from the corners $(3 \times 16=48$ cuts $)$. In total, exactly 50 cuts.
Criteria: Only the answer - 0 points.
Example of 50 cuts - 2 points.
Proof of the minimality of 50 - 5 points. These points are combined from proving the following statements:
- After $n$ cuts, the number of vertices is no more than $4 n+4$ - 2 points;
- After $n$ cuts, the number of vertices is no less than $3 n+54$ - 2 points;
- From the two previous estimates, $n \geqslant 50$ - 1 point.
|
50
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. On a 10 by 10 cell board, some 10 cells are marked. For what largest $n$ is it always possible, regardless of which cells are marked, to find a rectangle consisting of several cells, the perimeter of which will be at least $n$? The length or width of the rectangle can be equal to one cell.
|
Answer: $n=20$.
Solution. First, we prove that when $n=20$, it is always possible to find such a rectangle. Suppose 10 cells are colored. If there is a column or row without colored cells, then a rectangle of $1 \times 9$ with a perimeter of 20 (or even $1 \times 10$ with a perimeter of 22) can be cut from it. Now, let's assume that there is a colored cell in each column and each row. If the colored cell in the top row is in the column numbered $k$, then a rectangle of $1 \times 9$ with a perimeter of 20 can be cut from this column from the bottom.
When $n=22$, the required result is not always possible. For this, we color the cells along the diagonal. In this case, a rectangle can only be cut from the lower (or only from the upper part), but then the sum of the length and width cannot exceed 10, and the entire perimeter cannot exceed 20.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Let's call a four-digit number $\overline{a b c d}$ curious if the sum of the two-digit numbers $\overline{a b}$ and $\overline{c d}$ equals the two-digit number $\overline{b c}$. For example, the number 1978 is curious because 19+78=97. Find the number of curious numbers. Answer. 36.
|
Solution. Let's form the equation $\overline{a b}+\overline{c d}=10(a+c)+b+d=\overline{b c}=10 b+c$, from which we get $10 a+9 c+d=9 b$. The difference $9(b-c)$ is divisible by 9, so the sum $10 a+d=9(b-c)$ is also divisible by 9, which is equivalent to the divisibility by 9 of the sum $a+d=9(b-c-a)$. The sum of two different numbers from the interval from 0 to 9 is no less than 1 and no more than 17, so the latter is only possible when $a+d=9$. Substituting this into the previous equation, we get $b=a+c+1$, so the quadruple $(a, b, c, d)$ can be written in the form $(a, a+c+1, c, 9-a)$, where $a, c$ are any pair of digits satisfying the relations $a \geq 1, a+c \leq 8$. For each fixed $a=1,2, \ldots, 8$, the value of $c$ can be any from 0 to $8-a$, a total of $9-a$ options. Therefore, the total number of curious numbers is the sum $8+7+\ldots+1=36$.
Evaluation criteria. . ($\cdot$) It is proven that $a+d=9(b-c-a)$ is divisible by 9: 2 points. . ($\cdot$) It is proven that from this follows $a+d=9$: another 2 points. . ($\cdot$) The form of the quadruple of digits is written as $(a, a+c+1, c, 9-a)$: another 1 point. . ($\cdot$) The number of curious quadruples is found: 2 points.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. We consider all possible tilings of an 8 by 8 chessboard with dominoes, each consisting of two adjacent squares. Determine the maximum natural number \( n \) such that for any tiling of the 8 by 8 board with dominoes, one can find some rectangle composed of \( n \) squares of the board that does not contain any domino in its entirety. The lengths of the sides of the rectangle in squares can be any natural numbers, starting from one.
|
Answer. $n=4$.
Solution. 1) We will prove that $n \leq 4$. Consider the following tiling of an 8 by 8 chessboard with dominoes. Divide the board into 2 by 2 squares, color each of them in red and blue in a checkerboard (relative to the 4 by 4 board) pattern, and divide the red squares into pairs of horizontal dominoes, and the blue squares into pairs of vertical dominoes. Any rectangle with an area greater than 4 either contains a 1 by 5 rectangle or a 2 by 3 rectangle. It is easy to verify that for the considered tiling, a 1 by 5 rectangle or a 2 by 3 rectangle in any position on the board contains at least one entire domino.
2) We will prove that for any tiling of an 8 by 8 board with dominoes, one can find some rectangle of area 4, composed of cells, that does not contain any entire domino. Consider the domino in the tiling that contains cell d4; we can, by rotating the board if necessary, assume it is horizontal. We will consider in detail the case where it consists of cells c4, d4; the case where it consists of cells d4, e4 is analogous. Consider the cells c3, d3 and c5, d5. If at least one of these cells, say c3 (other cases are analogous), lies in a horizontal domino of the tiling, then the vertical rectangle c2, c3, c4, c5 does not contain any entire domino of the tiling. If all these cells lie in vertical dominoes of the tiling, then the dominoes containing c3 and d3 are vertical, so the rectangle b3, c3, d3, e3 does not contain any entire domino of the tiling. Note that by choosing cell d4 to start the reasoning, we ensure that the selected 1 by 4 strip cannot extend beyond the board.
Grading criteria. ($\cdot$) If the point 1) that $n \leq 4$ is proven: 2 points. ($\cdot$) If the point 2) that one can always find some suitable rectangle of area 4: 5 points. ($\cdot$) If an example of a tiling is provided where any rectangle with more than 4 cells always contains an entire domino, and it is correct, the score is not reduced for lack of meticulous verification. ($\cdot$) In a generally correct reasoning about selecting a 1 by 4 strip of cells that does not contain any entire domino, the possibility of extending beyond the edge of the board is not considered: minus 2 points. ($\cdot$) In the reasoning, the possibility of the sought rectangle extending beyond the edge of the board is allowed: the solution is given 0 points.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. What is the minimum sum of digits in the decimal representation of the number $f(n)=17 n^{2}-11 n+1$, where $n$ runs through all natural numbers?
# Answer. 2.
|
Solution. When $n=8$, the number $f(n)$ is 1001, so the sum of its digits is 2. If $f(n)$ for some $n$ had a sum of digits equal to 1, it would have the form $100, \ldots 00$ and would either be equal to 1 or divisible by 10. The function of a real variable $f(x)$ reaches its minimum at $x=\frac{11}{34}1$, and $f(n)$ cannot take the value 1. Furthermore, it is easy to notice that $f(n)$ is always an odd number, so it cannot be divisible by 10. Therefore, the minimum sum of the digits of the number $f(n)=17 n^{2}-11 n+1$ is 2, and it is achieved when $n=8$.
Grading Criteria. Correct answer with verification for $n=8$: 2 points. Proven that $f(n)>1$ and that $f(n)$ cannot take the value 1: 2 points. Proven that $f(n)$ is always an odd number, so it cannot be divisible by 10: 3 points.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Inside an isosceles triangle $\mathrm{ABC}$ with equal sides $\mathrm{AB}=\mathrm{BC}$ and an angle of 80 degrees at vertex $\mathrm{B}$, a point $\mathrm{M}$ is taken such that the angle
$\mathrm{MAC}$ is 10 degrees, and the angle $\mathrm{MCA}$ is 30 degrees. Find the measure of angle $\mathrm{AMB}$.
|
Answer: 70 degrees.
Solution. Draw a perpendicular from vertex B to side AC, and denote the points of its intersection with lines AC and CM as P and T, respectively. Since angle MAC is less than angle MCA, side CM of triangle MAC is shorter than side AM, so point M is closer to C than to A, and therefore T lies on the extension of segment CM. Point T lies on the perpendicular bisector AP of segment AC, so triangle ATC is isosceles, hence angle TAC is 30 degrees, and therefore angles BAT and MAT are 20 degrees each. The measures of angles $\mathrm{ABP}=\mathrm{ABC} / 2=80 / 2=40$ and $\mathrm{AMT}=\mathrm{MAC}+\mathrm{MCA}=10+30=40$ are equal, so triangles ABT and AMT are congruent by the common side AM and the adjacent angles. Therefore, the corresponding sides $\mathrm{AB}$ and $\mathrm{AM}$ are equal, and triangle $\mathrm{AMB}$ is isosceles. Consequently, its angle AMB at the base MB is $(180- BAM) / 2=(180-(50-10)) / 2=(180-40) / 2=70$ degrees.
Grading criteria. Proof that T lies on the extension of segment CM: 1 point. Proof of the isosceles nature of triangle ATC: 1 point. Proof of the isosceles nature of triangle AMB: 4 points. Finding the angle AMB: 1 point.
|
70
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2. It is known that $70 \%$ of mathematicians who have moved to IT regret their change of activity. At the same time, only $7 \%$ of all people who have moved to IT regret the change. What percentage of those who have moved to IT are mathematicians, if only they regret the change of activity?
|
Solution. Let a total of $x$ people went into IT, and $y$ of them are mathematicians. According to the condition of the change in activity, on the one hand, $0.07 x$ people regret, and on the other - $0.7 y$. From this, we get that $0.07 x=0.7 y$, from which $y / x=0.1$, that is, $10 \%$.
Criteria. Only the answer - 1 point.
Solution in a particular case (let there be 100 people in total...) without reference to the general - no more than 5 points.
|
10
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. It is known that all krakozyabrs have horns or wings (possibly both). According to the results of the world census of krakozyabrs, it turned out that $20 \%$ of the krakozyabrs with horns also have wings, and $25 \%$ of the krakozyabrs with wings also have horns. How many krakozyabrs are left in the world, if it is known that their number is more than 25 but less than $35 ?$
|
Answer: 32.
Solution: Let $n$ be the number of krakozyabrs with both wings and horns. Then the number of horned krakozyabrs is $-5 n$, and the number of winged krakozyabrs is $-4 n$. Using the principle of inclusion-exclusion, the total number of krakozyabrs is $5 n + 4 n - n = 8 n$. There is only one integer between 25 and 35 that is divisible by 8.
Criterion: Answer, answer with verification - 1 point.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. What is the maximum number of rooks that can be placed on an 8x8 chessboard so that each rook attacks no more than one other? A rook attacks all squares on the same row and column it occupies.
|
Answer: 10.
Solution: It is clear that in each column and row there are no more than two rooks. Let $k$ rooks be placed while satisfying the condition. On each square where a rook is placed, write the number 0. In each of the 8 columns, perform the following operation: if there are two numbers in the column, add 1 to both; if there is one number, add 2 to it (do nothing in an empty column). Then perform the same operation for each row. Clearly, on the square of each of the $k$ rooks, the result will be either 3 or 4 (if 2 is written, then this rook is under attack by two others), so the sum $S$ of all the written numbers is at least $3k$. On the other hand, since we added no more than 2 in each of the 8 columns and then in each of the 8 rows, $S \leq 32$. Therefore, $3k \leq S \leq 32$, from which $k \leq 10$. An example is easily constructed.
Criterion: Only the answer - 0 points. An example is worth 2 points, the estimation of the number of rooks is worth 5 points (points are cumulative).
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. At the school for slackers, a competition on cheating and giving hints was organized. It is known that $75 \%$ of the students did not show up for the competition at all, and all the rest participated in at least one of the competitions. When the results were announced, it turned out that $10 \%$ of all those who showed up participated in both competitions, and that those who participated in the hinting competition were one and a half times more than those who participated in the cheating competition. Find the smallest possible number of students in the school for slackers.
|
Answer: 200.
Solution. Let the number of students in our school be $n$ people. $\frac{n}{4}$ people attended the competitions. Taking into account $\frac{n}{40}$ people who participated in both competitions, the number of participants in the competition with hints was $\frac{3}{5}\left(\frac{n}{4}+\frac{n}{40}\right)=\frac{33 n}{200}$. For this number to be an integer, given the mutual simplicity of the numbers 33 and 200, $n$ must be divisible by 200. In the competition involving copying, $\frac{2}{5}\left(\frac{n}{4}+\frac{n}{40}\right)=\frac{11 n}{100}$ people participated, and it is sufficient for $n$ to be divisible by 100. Considering everything, it is necessary and sufficient for $n$ to be divisible by 200, in particular, to be no less than 200.
Example: 200 students: 50 people participated in the competitions, 33 people in the competition with hints, and 22 people in the copying competition, with 5 people participating in both.
Criterion. Answer with verification: 2 points.
Correct formulas: 3 points.
Correct conclusions about the divisibility of $n$ by 200: 2 points.
Lack of an example with 200: minus 1 point.
|
200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Ellie and Toto painted daisies in the field. On the first day, Ellie painted one fourteenth of the entire field. On the second day, she painted twice as much as on the first day, and on the third day, she painted twice as much as on the second day. Toto, in total, painted 7000 daisies. How many daisies were there in total on the field, given that all of them were painted? (Find all possible answers and prove that there are no others)
|
Solution. Let there be $n$ daisies in the field. Then Ellie painted $n / 14 + 2 n / 14 + 4 n / 14 = 7 n / 14 = n / 2$ daisies in total. This means Toto also painted half of the field, from which it follows that half of the field is 7000 daisies, and the entire field is 14000.
Criteria. Only the answer - 1 point.
The answer with a check that it is correct - 2 points.
|
14000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Anton from the village was given several zucchini, and he decided to give them to his friends. He gave half of the received zucchini to Arina, and a third (also from the received amount) to Vera. It turned out that after this, the number of zucchini Arina had became a square of some natural number, and the number of zucchini Vera had became a cube (they had no zucchini before). Find the smallest possible number of zucchini that Anton could have received from the village. (Find the answer and prove that it is minimal).
|
Solution. Let Anton receive $n$ zucchinis. Since both half and a third of $n$ are integers, $n$ is divisible by 6, that is, $n=6k$ for some natural number $k$. Then it is known that $3k$ is the square of a natural number, and $2k$ is a cube. Let $k=2^p 3^q m$. In other words, let $p$ and $q$ be the highest powers of two and three, respectively, by which $k$ is divisible. Then $3k=2^p 3^{q+1} m$, and since this is a square, $p$ and $q+1$ are even numbers. On the other hand, $2k=2^{p+1} 3^q m$ is a cube, so $p+1$ and $q$ are divisible by 3. Note that then $p$ is at least 2 (because $p+1$ is divisible by 3), and $q$ is at least 3 (because $q$ is divisible by 3 and $q \neq 0$ from the condition of the evenness of $q+1$). Then $k$ is divisible by at least 4 and 27, so $k$ is at least 108, and $n$ is at least 648. It is clear that $648=6 \times 4 \times 27$ fits the construction.
Criteria. Only the answer - 1 point.
Verification that the answer fits - 1 point (in this solution, this verification implicitly follows from the fact that we are looking for exactly the suitable numbers).
Proven that $n$ is divisible by 6 - 1 point.
Proof that $k$ must be divisible by 4 and 27 (or that $n$ is divisible by 8 and 81, which is the same) - 2 points for each fact.
All the above points are cumulative.
The answer can be left in the form of a product; no points are deducted for the lack of an explicit notation.
|
648
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. What two digits need to be appended to the right of the number 2013 so that the resulting six-digit number is divisible by 101? Find all possible solutions.
|
Answer: 94, the obtained number will be equal to 201394.
Solution: The remainder of dividing the number $\overline{2013 x y}$ by 101 is $\overline{x y}+7$ and this must be divisible by 101. This is greater than 0 but less than 202, so $\overline{x y}+7=101, \overline{x y}=94, x=9, y=4$.
Grading: Just the answer with verification: 1 point. Failure to justify $\overline{x y}+7=101$ minus points.
|
94
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.4. On the cells of an 8 by 8 board, chips are placed such that for each chip, the row or column of the board in which it lies contains only one chip. What is the maximum possible number of chips on the board?
|
Answer: 14.
Solution: Let's match each chip to the row or column of the board in which it is the only one. If it is the only one in both, we match it to the row. From the condition, it follows that different chips are matched to different rows and columns. If not all rows and columns are matched, then their total number does not exceed 14, in each of them there is no more than one chip, so the total number of chips does not exceed 14. If, however, all rows, say, are matched, then there is a chip in each row - a total of 8.
Example of placing 14 chips: all cells of the left column and the bottom row are filled, except for the bottom left corner cell.
Evaluation: 5 points. Example: 2 points. Any incorrect estimate with any reasoning - 0 points.
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. At a diplomatic reception, there are 99 persons, each of whom has heard of no fewer than $n$ other attendees. If A has heard of B, it does not automatically mean that B has heard of A. For what minimum $n$ is it guaranteed that there will be two attendees who have heard of each other?
|
Answer. When $n=50$.
Solution. Let's call a situation where one of the guests has heard about another a half-acquaintance. If each guest has heard about at least 50 other participants at the reception, then there are at least $99 \cdot 50$ half-acquaintances, which is more than the total number of pairs of guests at the reception, equal to $\frac{\mathbf{9} \boldsymbol{9}}{\mathbf{2}} \underline{\underline{\mathbf{8}}} \boldsymbol{9 4}$: Therefore, in some pair, there are at least two half-acquaintances, and the guests in this pair have heard about each other.
Let's provide an example where each has heard about exactly 49 other guests, but there are no two guests who have heard about each other. Arranging the guests in a circle, we assume that each has heard only about the next 49 guests in the clockwise direction. In this case, each has been heard about only by the 49 guests located in front of them in the clockwise direction. These two sets do not intersect, so for each guest, there is no other guest who has heard about them and whom they have heard about.
Criteria. Proof that $n$ is not greater than $50: 4$ points. Example that $n$ is not less than 50: 3 points.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. Find all natural $n$, for which on a square grid of size $n$ by $n$ cells, it is possible to mark $n$ cells, each in different rows and different columns, that can be sequentially visited by a knight's move in chess, starting from some cell, without landing on the same cell twice, and returning to the starting cell. The knight can only land on the marked cells.
|
Answer. $n=4$.
Solution. An example for $n=4$ is not difficult:
We will prove that for $n \neq 4$, the required set of cells does not exist.
Suppose that for a given $n$, it is possible to mark $n$ cells in different rows and different columns of an $n \times n$ board, which can be sequentially traversed by a knight's move and return to the initial marked cell. Consider the horizontal displacements of the knight during its sequential moves to the marked cells in the columns of the board. The knight either moves to a cell in column a (the first from the left) from a cell in column b (the second from the left), and then moves to a cell in column c (the third from the left), or vice versa. In the first case (the second case is symmetric to the first), to avoid repeating columns, the knight can only move to column b from column d, and can only move from column c to column e. All preceding moves to column d and subsequent moves from column e are made with a horizontal displacement of 2, except for one with a displacement of 1, which will close the chain of the knight's movements and can only occur at the right edge of the board. In total, the knight will move horizontally 2 times by one cell and $n-2$ times by two cells. The same can be said about its vertical movements along the rows of the board. Each move of the knight involves a displacement of one cell in one direction and two cells in the other, so in the total set of all $2n$ displacements in both directions, the number of displacements by one cell must equal the number of displacements by two cells, from which $2n-4=4$ and $n=4$.
Grading criteria. The idea of considering moves along the three outermost columns (rows) of the board: 2 points.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Two athletes with constant speeds run on an oval track of a sports ground, the first of them runs the track completely 5 seconds faster than the second. If they run on the track from the same starting point in the same direction, they will meet again for the first time after 30 seconds. How many seconds will it take for them to meet again for the first time if they run on the track from the same starting point in opposite directions
|
Answer. In 6 seconds.
Solution. Let the length of the track be $S$ meters, and the speeds of the first and second runners be $x$ and $y$ meters per second, respectively. From the first condition: $\frac{S}{x} + 5 = \frac{S}{y}$, and from the second condition $\frac{S}{x-y} = 30$, since in this case the first runner catches up with the second runner who is initially $S$ meters behind. From the second equation, we express $S = 30(x - y)$, substitute it into the first equation, make the substitution $t = \frac{x}{y} > 1$, and obtain the equation $t^2 - \frac{13}{6} t + 1 = 0$, from which $t = \frac{3}{2} = \frac{x}{y}$. Express $y = \frac{2}{3} x$, substitute it into the second equation, we have $\frac{S}{x} = 10$, that is, $S = 10 x$.
If the runners run in opposite directions, their speeds will add up, and they will meet after covering the track in $\frac{S}{x+y} = \frac{S}{\frac{5}{3} x} = \frac{3}{5} \frac{S}{x} = 6$ seconds.
Grading criteria. Correct formulation of the system of equations: 3 points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. In a row from left to right, all natural numbers from 1 to 37 are written in such an order that each number, starting from the second to the 37th, divides the sum of all numbers to its left: the second divides the first, the third divides the sum of the first and second, and so on, the last divides the sum of the first thirty-six. The number 37 is on the first place from the left, what number is on the third place?
|
Answer: 2.
Solution. If the first number is the prime number 37, then the second must be 1, and the third must be a divisor of the number $37+1=38$, that is, 2 or 19. However, 19 must be in the last position, since the number 37 minus an even number must divide the sum of all the other numbers and itself, that is, divide the sum of all the numbers, which is $\frac{(1+37) \cdot 37}{2}=19 \cdot 37$, and must equal 19, as 37 is already occupied. Therefore, the number 2 is in the third position.
Grading Criteria. The solution does not require providing an example of the arrangement of numbers that satisfies the condition. If a complete correct example with the number 2 in the third position is provided without proving its uniqueness: 3 points. For constructing an example of the arrangement, if everything else is correctly proven, no additional points are added.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. What is the maximum number of 2 by 2 squares that can be placed on a 7 by 7 grid of squares so that any two placed squares share no more than one common cell? The 2 by 2 squares are placed along the grid lines such that each covers exactly 4 cells. The squares do not extend beyond the boundaries of the board.
|
Answer: 18 squares.
Solution: First, let's provide an example of laying out 18 squares of 2 by 2: 9 of them cover the left bottom square of size 6 by 6 of the board, and the other 9 cover the right top square of size 6 by 6 of the board.
We will prove that it is impossible to lay out 19 squares correctly. Note that if a cell of the board, adjacent to the edge, is covered by two squares, then they intersect in at least two cells, and if a cell of the board, not adjacent to the edge, is covered by three squares, then two of them intersect in at least two cells. Consequently, the edge cells of the board can be covered by squares no more than once, and the internal cells no more than twice. Therefore, the squares can contain no more than \(24 \cdot 1 + 25 \cdot 2 = 74\) cells in total, and the total number of squares cannot exceed \(74 / 4 = 18.5\) pieces.
Grading criteria: Any correct example: 2 points. Proof that the number of squares is no more than 18: 5 points.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. A bus with programmers left Novosibirsk for Pavlodar. When it had traveled 70 km, Pavel Viktorovich set off from Novosibirsk in a car along the same route, and caught up with the programmers in Karasuk. After that, Pavel drove another 40 km, while the bus traveled only 20 km in the same time. Find the distance from Novosibirsk to Karasuk, if both the car and the bus traveled at constant speeds. (Provide a complete solution, not just the answer.)
|
Solution. Since by the time the car has traveled 40 km, the bus has traveled half that distance, its speed is exactly half the speed of the car. However, when the bus has traveled 70 km after the car's departure, the car will have traveled 140 km and will just catch up with the bus. According to the problem, this happened in Karasuk, so 140 km is the answer.
Criteria. Only the answer -1 point.
Answer with verification (for example, for specific speeds) - 2 points. Proven that the car's speed is twice the speed of the bus - 3 points.
|
140
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Come up with at least one three-digit PAU number (all digits are different) such that $(П+\mathrm{A}+\mathrm{У}) \times \Pi \times \mathrm{A} \times \mathrm{Y}=300$ (it is sufficient to provide one example)
|
Solution: For example, PAU = 235 is suitable. There are other examples as well. Criteria: Any correct example - 7 points.
|
235
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Students in the seventh grade send each other New Year's stickers on Telegram. It is known that exactly 26 people received at least one sticker, exactly 25 - at least two stickers, ..., exactly 1 - at least 26 stickers. How many stickers did the students in this class receive in total, if it is known that no one received more than 26 stickers? (find all possible answers and prove that there are no others)
|
# Answer: 351
Solution: Note that exactly 1 sticker was received by one person, as it is precisely in this case that the difference between those who received at least 1 and those who received at least 2. Similarly, one person received exactly $2, 3, \ldots, 26$ stickers, so the total number of stickers is $1+\ldots+26=351$.
Criteria: Only the answer - 0 points.
Answer with verification that it is possible - 1 point.
Answer written in an open form (ellipsis left) or arithmetic error - minus 1 point.
|
351
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. Each cell of a $5 \times 5$ table is painted in one of several colors. Lada shuffled the rows of this table so that no row remained in its original position. Then Lera shuffled the columns so that no column remained in its original position. To their surprise, the girls noticed that the resulting table was the same as the original one. What is the maximum number of different colors this table can be painted with?
|
# Answer: 7.
Solution: Let's renumber the colors and reason about numbers instead. Both columns and rows could have been cyclically permuted or divided into a pair and a triplet. If a cyclic permutation of columns was used, then all columns consist of the same set of numbers, i.e., no more than five different numbers. The same situation applies to a cyclic permutation of rows.
If, however, both columns and rows were divided into a pair and a triplet, the table can be conditionally divided into squares $2 \times 2$ and $3 \times 3$, as well as rectangles $2 \times 3$ and $3 \times 2$, in each of which both columns and rows are cyclically permuted. In the squares, there can be no more than two and three different numbers, respectively (because the set of numbers in each column is the same), and in the rectangles, no more than one (since both in columns and rows, the set of numbers must be the same).
Example for seven numbers see in the figure.
Criteria: Only the estimate - 3 points
Only the example - 3 points.
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. In each of the four volleyball teams, there are six players, including a captain and a setter, and these are different people. In how many ways can a team of six players be formed from these four teams, such that there is at least one player from each team and there must be a pair of captain and setter from at least one team?
|
Answer: 9720.
Solution. Case 1. Three players, including the captain and the setter, are chosen from one of the teams, and one player is chosen from each of the remaining three teams. The team is chosen in four ways, the third player from it in another four ways, and the three players from the remaining teams in $6 \cdot 6 \cdot 6=216$ ways, resulting in $4 \cdot 4 \cdot 6 \cdot 6 \cdot 6=16 \cdot 216=3456$ possibilities for this case.
Case 2. Two players are chosen from two teams, with at least one team including the captain and the setter, and one player is chosen from each of the remaining two teams. The pair of teams can be chosen in $C_{4}^{2}=6$ ways, and choosing a pair of players from each can be done in $C_{6}^{2} \cdot C_{6}^{2}$ ways. However, we need to exclude $\left(C_{6}^{2}-1\right) \cdot\left(C_{6}^{2}-1\right)$ selections where neither of the two teams includes both the captain and the setter simultaneously, leaving us with $2 \cdot C_{6}^{2}-1=29$ ways. There are $6 \cdot 6=36$ ways to choose one player from each of the remaining two teams, resulting in a total of $6 \cdot 29 \cdot 36=6264$ ways in this case. In total, $6264+3456=9720-$ is the answer to the problem.
Grading criteria. Reducing the solution to cases 1 and 2: 1 point. Considering each of cases 1 and $2: 3$ points.
|
9720
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. For what minimum natural $n$ can $n$ distinct natural numbers $s_{1}, s_{2}, \ldots, s_{n}$ be found such that $\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right)=\frac{7}{66} ?$
|
Answer. $n=9$.
Solution. We can assume that $1<s_{1}<s_{2}<\ldots<s_{n}$, then for any $k=1, \ldots, n$ the inequality $s_{k} \geq k+1$ holds. Therefore, $\frac{7}{66}=\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \ldots\left(1-\frac{1}{s_{n}}\right) \geq\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right) \ldots\left(1-\frac{1}{n+1}\right)=\frac{1}{2} \cdot \frac{2}{3} \cdot \ldots \cdot \frac{n}{n+1}=\frac{1}{n+1}$, from which $n \geq 9$.
Example of nine numbers satisfying the condition: $2,3,4,5,6,8,9,10,11$.
Grading criteria. Estimation $n \geq 9: 5$ points. Example of nine numbers satisfying the condition: 2 points.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. A finite set of distinct real numbers $X$ is called good if each number in $X$ can be represented as the sum of two other distinct numbers in $X$. What is the minimum number of elements that a good set $X$ can contain?
#
|
# Answer: 6.
Solution. From the condition, it follows that $X$ contains no less than three numbers, which means there are non-zero numbers in it. By multiplying all numbers by minus one if necessary, we can assume that $X$ contains positive numbers. Let's choose the largest number $M$ from them. According to the condition, it is equal to the sum of two other distinct numbers from $X$, each of which is smaller than it. From the maximality of $M$, it follows that both of these numbers are positive. Therefore, $X$ contains no less than three distinct positive numbers. Consider the smallest positive number $\boldsymbol{m}$, it is equal to the sum of two other distinct numbers from $X$, one of which must be negative due to the minimality of $\boldsymbol{m}$. Now let's choose the smallest negative number $\mathrm{N}$ from $X$, it is also equal to the sum of two other distinct numbers from $X$, each of which is greater than it. From the minimality of $\mathrm{N}$, it follows that both of these numbers are negative. Therefore, $X$ contains no less than three distinct negative numbers. Thus, a good set $X$ must contain no fewer than six numbers.
An example of a good set with six elements is the set $X=\{-3,-2,-1,1,2,3\}$.
Grading Criteria. ($\cdot$) An example of a good set with six numbers is provided: 2 points. ($\cdot$) It is proven that $X$ contains no less than 3 positive or negative numbers: 3 points. ($\cdot$) It is proven that $X$ must contain numbers of different signs: 2 points.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. What is the maximum number of colors in which all cells of a 4 by 4 square can be painted so that any 2 by 2 square of cells necessarily contains at least two cells of the same color?
|
Answer: In 11 colors.
Solution. We will prove that the maximum number of colors under the conditions of the problem does not exceed 11. Consider in a 4x4 square five 2x2 squares: four corner ones and the central one. The corner 2x2 squares do not intersect, and the central one shares one common cell with each of the corner ones. According to the condition, each corner 2x2 square contains cells of no more than three different colors, totaling no more than 12. But if there are no repeating colors among these four sets of three colors in the corner 2x2 squares, then the central 2x2 square must contain cells of four different colors - one from each set of three, which contradicts the condition. Therefore, the total number of different colors in which the cells of the 4x4 square are painted does not exceed 12-1=11. If, however, in at least one of the corner 2x2 squares no more than two different colors are used, then the total number of different colors is immediately no more than 11.
An example of coloring a 4x4 square in 11 colors, satisfying the condition of the problem, looks like this: the bottom horizontal row of the square is painted in different colors $1,2,3$ and 4, the second horizontal row - entirely in color 5, the third horizontal row in different colors $6,7,8$ and 9, and the fourth horizontal row - in colors $10,7,11,9$.
Grading criteria. ($\cdot$) Proof of the maximum of 11 colors: 4 points, ($\cdot$) If the case where in one of the corner 2x2 squares no more than two different colors are used is not considered at all in the proof of the maximum of 11 colors: minus 1 point. ($\cdot$) Example of a correct coloring in 11 colors: 3 points.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. Find the number of five-digit numbers containing two digits, one of which is divisible by the other.
|
Answer. $89760=9 \cdot 10^{4}-2 \cdot 5!$.
Solution. We will count the number of five-digit numbers in which no digit is divisible by any other digit, and then subtract this number from the total number of five-digit numbers, which is 90000, to get the answer to the problem.
Note that a number, none of whose digits are divisible by any other, does not contain the digit 0 (0 is divisible by any digit), 1 (any digit is divisible by 1), and does not contain identical digits (they always divide each other). Therefore, the smallest digit of such a five-digit number can be $2, 3, 4$ or 5. If it is 2, then the other four digits must be odd, specifically $3, 5, 7$ and 9, but 9 is divisible by 3. If it is 3, the other four digits can only be $4, 5, 7, 8$, but 8 is divisible by 4. The only numbers left are those composed of the digits $4, 5, 6, 7, 9$ or $5, 6, 7, 8, 9$. In each case, there are $5!=120$ possible permutations of these digits, totaling 240 numbers. Therefore, the number of numbers sought in the problem is 90000 $- 240 = 89760$.
Grading criteria. The idea of applying the principle of subtraction: 3 points. Correct calculation of the number of five-digit numbers: 1 point. Correct calculation of the number of five-digit numbers in which no digit is divisible by any other digit: 3 points. If the solution does not account for the divisibility of 0 by any digit or the divisibility of any digit by 1 or equal digits by each other: minus 3 points.
|
89760
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. There are 100 coins, 99 of which are genuine and weigh the same, and 1 is counterfeit and lighter than the others. Dmitry has a pair of balance scales without weights, which always show incorrect results (for example, if the left pan is heavier, they will show either balance or the right pan being heavier, but it is unknown which). Help Dmitry find 98 genuine coins.
|
# Solution:
Let's number the coins from 1 to 100. Weigh the first coin against the second. If the scales show equality, then one of them is fake, and all the other coins are genuine, and we have achieved the desired result. If the scales do not balance, assume the coin numbered 1 is heavier. Then the second coin is definitely genuine (otherwise, the scales would have shown the correct result). Set aside the second coin in the bag for genuine coins.
Re-number the remaining coins from 1 to 99, and repeat the process.
Either we will get one new genuine coin with each weighing, and after 98 weighings, we will have found 98 genuine coins, or at some point, the scales will show equality, and we will achieve the desired result.
Criteria: Algorithm only without justification - 1 point. Case of equality not considered in a similar solution, otherwise justified correctly - 5 points.
|
98
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. How much of a $5 \%$ and a $20 \%$ salt solution in water should be taken to obtain 90 kg of a $7 \%$ solution
|
Answer: 78 kg of 5% solution and 12 kg of 20% solution.
Solution. Let the mass of the 5% solution be $x$ kg, the mass of the 20% solution will be $90-x$ kg, and the total mass of salt in the 5% and 20% solutions is equal to the mass of salt in 90 kg of 7% solution: $\frac{5}{100} x+\frac{20}{100}(90-x)=\frac{7}{100} 90$, from which $1800-15 x=630$ and $x=78$.
Grading criteria. Answer found by guessing with verification: 1 point. Equations written but not solved: 2 points.
|
78
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. In a kindergarten, each child was given three cards, each of which had either "MA" or "NYA" written on it. It turned out that 20 children could form the word "MAMA" from their cards, 30 children could form the word "NYANYA," and 40 children could form the word "MANYA." How many children had all three cards the same?
|
Answer: 10 children.
Solution. Let's denote the number of children who received three "MA" cards as $x$, two "MA" cards and one "NA" card as $y$, two "NA" cards and one "MA" card as $z$, and three "NA" cards as $t$. Then, the word "MAMA" can be formed by all children from the first and second groups and only them, the word "NANA" - by all children from the third and fourth groups and only them, and the word "MANA" - by all children from the second and third groups and only them. Therefore, $x+y=20, z+t=30, y+z=40$, so the desired number is $x+t=(x+y)+(z+t)-(y+z)=20+30-40=10$ children.
Grading criteria. A guessed answer with some examples is not to be evaluated. Formulating but not solving the system of equations: 3 points. Misinterpreting the problem: 0 points.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. For seven natural numbers $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ it is known that all of them are different prime numbers. Find all values that the smallest of these seven numbers can take.
|
Answer: 3.
Solution. From the condition, it follows that $a, b, c$ are also primes. If the smallest of the seven numbers were equal to two, the last four numbers would be different even numbers, which means they could not all be prime. If all seven numbers are greater than three, due to their primality, they are not divisible by 3, and their remainders when divided by 3 are 1 or -1. Consider the remainders of the numbers $a, b, c$ when divided by 3. If all their remainders are equal, then the number $a+b+c$ is divisible by 3. If two of them are equal and the third has the opposite sign, then one of the numbers $a+b-c, a+c-b, b+c-a$ is divisible by 3. In both cases, one of the seven numbers is divisible by 3 and, by assumption, is greater than 3, which contradicts its primality. Therefore, the smallest of the seven numbers in the condition can only be equal to 3. An example of such numbers: $a=7, b=13, c=17, a+b-c=3, a+c-b=11, b+c-a=23, a+b+c=37$.
Grading criteria. Correct answer with an example: 2 points. Proof of the minimality of the number 3: 5 points. If the case with two is not considered - deduct 2 points.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Paramon set out from point A to point B. At $12^{00}$, when he had walked half the way to B, Agafon ran out from A to B, and at the same time, Solomon set out from B to A. At $13^{20}$, Agafon met Solomon, and at $14^{00}$, he caught up with Paramon. At what time did Paramon and Solomon meet?
|
Answer: At 13 o'clock.
Solution: Let the distance between A and B be $\mathrm{S}$ km, and the speeds of Paramon, Solomon, and Agafon be $x, y, z$ km per hour, respectively. Then from the condition, we get: $\frac{S / 2}{z-x}=2, \frac{S}{y+z}=\frac{4}{3}$, from which $x+y=\frac{1}{2} S$. Therefore, Paramon and Solomon will meet after $\frac{S / 2}{S / 2}=1$ hour past noon, that is, at 13 o'clock.
Instructions: Answer with verification: 1 point. Formulation of equations: 3 points.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.2. The median $A M$ of triangle $A B C$ divides the segment $P R$, parallel to side $A C$, with endpoints on sides $\mathrm{AB}$ and $\mathrm{BC}$, into segments of lengths 5 cm and 3 cm, starting from side $\mathrm{AB}$. What is the length of side AC?
|
Answer: 13 cm.
Solution. Let the ends of the segment be denoted as $\mathrm{P}$ and $\mathrm{R}$, and the point of intersection with the median $\mathrm{AM}$ as $\mathrm{Q}$, with $\mathrm{P}$ lying on side $\mathrm{AB}$ and $\mathrm{R}$ on side $\mathrm{BC}$. Draw the midline $\mathrm{MN}$ of the triangle, its length is half the length of $\mathrm{AC}$. Using the similarity of triangles $\mathrm{ANM}$ and $\mathrm{APQ}$, and $\mathrm{MAC}$ and $\mathrm{MRQ}$. The similarity ratio of the first pair is $\mathrm{k}=\mathrm{AM} / \mathrm{AQ}$, and of the second pair is $\mathrm{AM} / \mathrm{MQ}$, their sum equals one. Then $k \cdot N M=k \cdot \frac{A C}{2}=5,(1-k) \cdot A C=3$, from which $A C=13$.
Instructions. The midline is drawn and both similarities are noted: 2 points.
|
13
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Daniil has 6 cards with letters, from which he managed to form the word WNMWNM shown in the picture. Note that this word has a remarkable property: if you rotate it 180 degrees, you get the same word. How many words with such a property can Daniil form using all 6 cards at once?
|
Answer: 12 words.
## Solution:
(1) According to the problem, Danil has 2 cards with the letter $\mathrm{N}$, which remains the same when flipped, and 4 cards with the letter M, which turns into the letter W when flipped. Clearly, to get a word with the desired properties, we need to arrange 2 letters M and one $\mathrm{N}$ in the first half, and the second half of the word will be uniquely determined.
(2) The letter $\mathrm{N}$ can be placed in three ways, after which the letter M can be placed on the first of the remaining positions in two ways (flip or not), and on the last position also in two ways. In total, there are $3 * 2 * 2=12$ options.
## Criteria:
Only the answer - 1 point.
Only the idea that the first three letters uniquely determine the word (part (1) of the solution) - 2 points (can be combined with the previous one).
After noting (1), listing all options with at least one missed case - no more than three points.
All 12 options are listed, but it is not explained why there are no other options - 1 point.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2 The banker leaves home, and at that exact moment, a car from the bank arrives to take him to the bank. The banker and the car always leave at the same time, and the car always travels at a constant speed. One day, the banker left home 55 minutes earlier than usual and, for fun, started walking in the direction opposite to the bank. The car caught up with him and took him to the bank 10 minutes later than usual. Find the ratio of the car's speed to the banker's speed.
|
Answer. The speed of the car is 12 times greater than the speed of the banker.
Solution. Indeed, the car delivered the banker to the bank 10 minutes later than usual, which means it caught up with him 5 minutes later than usual, i.e., from the moment he usually leaves his house. Therefore, the car traveled from the banker's house to the meeting point (or capture point) for 5 minutes. The banker, having left 55 minutes earlier than usual and being caught (captured) by the car, had walked from his house for $55+5=60$ minutes by the time of the meeting. Thus, the car travels 12 times faster than the banker walks.
Grading criteria. ($\cdot$) Proven that the car traveled from the banker's house to the meeting point for 5 minutes: 3 points. ($\cdot$) Proven that the banker had walked for 60 minutes by the time of the meeting: 2 points. ($\cdot$) Proven that the car travels 12 times faster than the banker walks: 2 points.
9.3.. Let $\sqrt{x}-\sqrt{y}=10$. Prove that $x-2 y \leq 200$.
Proof. Rewrite the condition as $\sqrt{x}=\sqrt{y}+10$. Both sides of the equation are non-negative, so we square both sides: $x=100+20 \sqrt{y}+y$. Then $x-2 y=100+20 \sqrt{y}-y=200-(10-\sqrt{y})^{2} \leq 200$, which is what we needed to prove.
Grading criteria. Not mentioning the non-negativity of both sides of the equation before squaring: minus 1 point.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. In a row from left to right, there are $n$ coins. It is known that two of them are counterfeit, they lie next to each other, the left one weighs 9 grams, the right one 11 grams, and all the remaining ones are genuine and each weighs 10 grams. The coins are weighed on balance scales, which either show which of the two pans is heavier, or are in equilibrium, indicating that the weights on both pans are the same. For what maximum $n$ can the coin weighing 11 grams be found in three weighings?
|
Answer. $n=28$.
Solution. Let $n=28$. Divide all 28 coins into three piles: the first pile contains coins numbered $11,13,15,17,19,21,23,25,27$, the second pile contains coins numbered $12,14,16,18,20,22,24,26,28$, and the third pile contains coins numbered from 1 to 10. The first weighing compares the first and second piles. Note that if the scales are not in balance, the heavy coin is in the heavier pile. Indeed, if the numbers of both fake coins are greater than 10, then the light one will fall into one of the two weighed piles, and the heavy one will fall into the other, causing the heavier pile to tip. If the numbers of both fake coins are not greater than 10, then both will fall into the third pile, and the scales will remain in balance. If the numbers of the fake coins are 10 and 11, the heavy one will fall into the first pile, and the second will consist entirely of genuine coins, causing the first pile to tip. It is important to note that if a similar division is made from the beginning of the numbers rather than the end, it is incorrect when the fake coins are in the 18th and 19th positions. If the scales are in balance, then the heavy and light coins are in the third pile of 10 coins. In this "worst" case, we divide the third pile similarly into three: the first pile contains numbers $6,8,10$, the second pile contains numbers $5,7,9$, and the third pile contains numbers $1,2,3,4$. The second weighing compares the first and second piles to determine which of the three piles contains the heavy coin. If balance is again achieved and it is together with the light one in the third pile of 4 coins, we compare coins 3 and 4. If they are equal, the heavy one is the 2nd; if one is heavier, that one is the heavy one. The same applies if the heavy one is in the pile of 3 coins - we compare the second and third, the heavier one is the heavy one.
If the heavy coin is found, for example, in the first pile during the first weighing, we divide it into three groups of three: the first pile contains coins $19,23,27$, the second pile contains coins $17,21,25$, and the third pile contains coins $11,13,15$. We weigh the first and second piles to find the pile of three coins containing the heavy one. The final weighing of the second and third coins in this pile identifies the desired one. Similarly, if the heavy coin is found in the second pile during the first weighing. Note that the light coin participates in the third weighing if and only if the first two weighings result in balance.
If there are more than 28 coins, the number of possible positions for the heavy coin will be more than $27=3^3$, which is greater than the number of possible outcomes of three weighings, equal to 3.3.3. Therefore, it is impossible to uniquely associate the number of the heavy coin with a specific outcome of three weighings and find it.
Grading criteria. Algorithm for finding the heavy coin when $n=28$: 5 points ($\cdot$) If similar divisions as indicated in the author's solution are made but from the beginning of the numbers, and the rest is the same: deduct 2 points. ($\cdot$) If the correct division is made but there is no clear explanation that if the scales are not in balance during the first weighing, the heavy coin is in the heavier pile: deduct 1 point. ( $($ )
Proof that the heavy coin cannot be found with certainty in three weighings when $n>28$: 2 points.
|
28
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Find the value of the expression $\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{2}{1+x y}$, given that $x \neq y$ and the sum of the first two terms of the expression is equal to the third.
|
Answer: 2.
Solution. Let's write the condition of the equality of the sum of the first two terms to the third one as: $\frac{1}{1+x^{2}}-\frac{1}{1+x y}=\frac{1}{1+x y}-\frac{1}{1+y^{2}} \quad$ and bring it to a common denominator:
$\frac{x(y-x)}{(1+x^{2})(1+x y)}=\frac{y(y-x)}{(1+y^{2})(1+x y)}$. Given $x \neq y$, we cancel $x-y$ and $1+x y$: $x(1+y^{2})=y(1+x^{2})$. The latter is equivalent to $(x-y)(x y-1)=0$, and again we cancel $x-y$, obtaining $x y=1$ and the desired expression equals $\frac{2}{1+x y} \cdot 2=2$.
Grading criteria. Any suitable $x \neq y$ guessed and the answer: 1 point.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. On the table, 28 coins of the same size but possibly different masses are arranged in a triangular shape (see figure). It is known that the total mass of any triplet of coins that touch each other pairwise is 10 g. Find the total mass of all 18 coins on the boundary of the triangle.
|
Answer: 60 g.
Solution 1: Take a rhombus made of 4 coins. As can be seen from the diagram, the masses of two non-touching coins in it are equal. Considering such rhombi, we get that if we color the coins in 3 colors, as shown in the diagram, then the coins of the same color will have the same mass.
Now it is easy to find the sum of the masses of the coins on the boundary: there are 6 coins of each color there, and the sum of the masses of three differently colored coins is 10 g; therefore, the total mass of the coins on the boundary is $6 \cdot 10=60$ g.
Solution 2: All coins except the central one can be divided into 9 triplets, and all internal coins except the central one can be divided into 3 triplets. This means that the coins on the boundary weigh as much as 9-3=6 triplets, i.e., 60 g.
Criteria: Only the answer - 0 points.
Rhombi are considered and it is shown that two coins in them have the same mass - 2 points.
In addition to this, it is proven that the weights in the entire triangle have the form shown in the diagram (coloring is done, but there is no further progress) - another 2 points.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. A set for playing lotto contains 90 barrels numbered with natural numbers from 1 to 90. The barrels are somehow distributed among several bags (each bag contains more than one barrel). We will call a bag good if the number of one of the barrels in it equals the product of the numbers of the other barrels in the same bag (for example, the bag "2, 3, 6" is good, while "4, 5, 10" is not). What is the maximum possible number of good bags?
|
Answer: 8.
Solution: In each good bag, there are no fewer than three barrels. The smallest number in each good bag must be unique; otherwise, the largest number in this bag would not be less than $10 \times 11=110$, which is impossible. For the same reason, if a good bag contains a barrel with the number 1, it must also contain another barrel with a single-digit number. Therefore, the number of good bags does not exceed 8.
On the other hand, an example can be provided where there are exactly 8 good bags. We can gather 8 good bags $(2,17,34),(3,16,48),(4,15,60),(5,14,70),(6,13,78),(7,12,84),(8$, $11,88)$, and $(9,10,90)$. All other numbers can be placed in a separate bag that is not good.
Criteria: Example only - 3 points.
Estimate only - 3 points.
Noting that each bag must contain a single-digit number - 1 point, can be combined with the example.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. Represent the number 1000 as the sum of the maximum possible number of natural numbers, the sums of the digits of which are pairwise distinct.
|
Answer: 19.
Solution: Note that the smallest natural number with the sum of digits A is a99..99, where the first digit is the remainder, and the number of nines in the record is the incomplete quotient of the division of A by 9. From this, it follows that if A is less than B, then the smallest number with the sum of digits A is also less than the smallest number with the sum of digits B. Consider the smallest natural numbers with the sums of digits $1,2,3, \ldots, 19,20$, they are respectively $a_{1}=1, a_{2}=2, \ldots a_{9}=9, a_{10}=19, a_{11}=29, \ldots a_{18}=99, a_{19}=199, a_{20}=299$. The sum of the first 19 of them is 775, and the remainder of its division by 9 is 1, which is the same as the remainder of the division of 1000 by 9. Increase this sum by 225, taking 234 instead of 9, with the same sum of digits, and we get a representation of 1000 as the sum of 19 numbers $1,2,3,4,5,6,7,8,19,29,39,49,59,69,79,89,99,199,234$ with different sums of digits from 1 to 19.
Suppose that 1000 can be represented as the sum of $n \geq 20$ natural numbers $b_{1}, b_{2}, \ldots b_{n}$ with different sums of digits $s\left(b_{1}\right)<s\left(b_{2}\right)<\ldots<s\left(b_{n}\right)$. In this case, $s\left(b_{k}\right) \geq k, k=1,2, \ldots n,$ and, consequently, $b_{k} \geq a_{k}, k=1,2, \ldots n$. Therefore, $1000=b_{1}+b_{2}+\ldots+b_{n} \geq a_{1}+a_{2}+\ldots+a_{n} \geq a_{1}+a_{2}+\ldots+a_{20}=107$ - a contradiction.
Grading criteria: Finding a representation of 1000 as the sum of 19 numbers with different sums of digits: 3 points. Proof of the maximality of 19: 4 points.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Experienced sawyer Garik knows how to make cuts. In one day of continuous work, he can saw 600 nine-meter logs into identical three-meter logs (they differ from the original only in length). How much time will the experienced sawyer Garik need to saw 400 twelve-meter logs (they differ from the nine-meter logs only in length) into the same three-meter logs?
|
Answer: one day
Solution: to turn a 9-meter log into 3 three-meter pieces, 2 cuts are needed. Therefore, Garik makes $2 * 600=1200$ cuts per day. To turn a 12-meter log into three-meter pieces, 3 cuts are needed, and for 400 logs, $400 * 3=1200$ cuts are needed, which means it will take the same amount of time.
Criteria: only answer -1 point.
|
1
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. In Nastya's room, 16 people gathered, each pair of whom either are friends or enemies. Upon entering the room, each of them wrote down the number of friends who had already arrived, and upon leaving - the number of enemies remaining in the room. What can the sum of all the numbers written down be, after everyone has first arrived and then left?
|
Answer: 120
Solution: Consider any pair of friends. Their "friendship" was counted exactly once, as it is included in the sum by the person who arrived later than their friend. Therefore, after everyone has arrived, the sum of the numbers on the door will be equal to the total number of friendships between people. Similarly, each "enmity" will be counted exactly once by the person who left earlier. Therefore, after everyone has left, the sum of the numbers on the door will be increased by the sum of all "enmities." In total, the overall sum of the numbers on the door will be equal to the sum of the total number of friendships and enmities, which is precisely the number of pairs of people who arrived, i.e., $16 * 15 / 2 = 120$.
Criteria: answer only - 0 points.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Losharik is going to visit Sovunya along the river at a speed of 4 km/h. Every half hour, he launches paper boats that float to Sovunya at a speed of 10 km/h. With what time interval do the boats arrive at Sovunya? (Provide a complete solution, not just the answer.)
|
Solution: If Losyash launched the boats from one place, they would arrive every half hour. But since he is walking, the next boat has to travel a shorter distance than the previous one. In half an hour, the distance between Losyash and the last boat will be $(10-4) \cdot 0.5=3$. This means that the distance between adjacent boats is 3 km, and the next boat will have to travel exactly this distance when the previous one has already arrived at Sovunya. The speed of the boat is 10 km/h, which means 1 km in 6 minutes, so 3 km will be covered by the boat in 18 minutes, which is the answer.
Criteria: Correctly found the distance of 3 km - 3 points.
Only the answer or the answer with verification - 1 point.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. Katya wrote a four-digit number on the board that was divisible by each of its digits without a remainder (there were no zeros in the number). Then she erased the first and last digits, and the number 89 remained on the board. What could have been written on the board initially? (Find all options and show that there are no others.)
|
Solution: The original number was divisible by 8. Therefore, the number formed by its last three digits is also divisible by 8. From the condition, this number has the form $\overline{89 x}$. Clearly, $x=6$ fits, and other numbers divisible by 8 do not fit into this decade. The entire number is divisible by 9, which means the sum of its digits is divisible by 9. Currently, it is $8+9+6=23$. It is 4 short of the nearest number divisible by 9, and 13 short of the next one. Therefore, this number is 4896. Since it is divisible by 8 and 9 by construction, it is also divisible by 4 (since 4 is a divisor of 8), and by 6 (due to divisibility by 8 and 9).
Criteria: Only the answer, answer with verification - 1 point.
One of the digits is found unambiguously - 3 points.
It is proven that it can only be 4896, but not verified that it fits - 7 points.
|
4896
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. Vikentiy has two jars, a red one and a blue one, and a pile of 20 pebbles. Initially, both jars are empty. A move in Vikentiy's game consists of transferring a pebble from the pile to one of the jars or returning a pebble from one of the jars to the pile. The number of pebbles in the jars determines the game position. After each move, the number of pebbles in the red jar is always not less than the number of pebbles in the blue jar; and during the game, no position can be repeated. What is the maximum number of moves Vikentiy can make?
|
Answer: 110.
Solution: The position in Vikenty's game is uniquely defined by a pair of non-negative integers $(x, y)$, and $x+y \leq 20$, where $0 \leq y \leq x \leq 20$ are the numbers of pebbles in the blue and red jars, respectively. In total, there are $21+19+17 \ldots+1=121$ positions in Vikenty's game. We will call a position even if the sum of the numbers of pebbles in the jars is even, and odd otherwise. In total, there are $11+10+\ldots+1=66$ even positions and $121-66=55$ odd positions in the game. Each move in the game changes the parity of the position, so the game cannot contain more than 55 even and 56 odd positions (taking into account that it starts with an even position $(0,0)$), a total of 111 positions, and consist of no more than $111-1=110$ moves.
We will provide an example of a game where Vikenty can make 110 moves. In this case, he makes the only possible first move from $(0,0)$ to $(1,0)$, then he sequentially passes through all positions $(x, y)$ with odd $x=1,3,5, . ., 9$ from $(x, 0)$ to $(x, x)$, after which he moves to $(x+1, x)$, and all positions with even $x=2,4, . ., 10$ from $(x, x-1)$ to $(x, 0)$, after which he moves to $(x+1,0)$. After all of the above, he ends up in the position $(11,0)$, after which his strategy changes slightly. He sequentially passes through all positions $(x, y)$ with odd $x=11,13,15, .17$ from $(x, 0)$ to $(x, 19-x)$, after which he moves to $(x+1,19-x)$, and all positions with even $x=12,14, . ., 18$ from $(x, 20-x)$ to $(x, 0)$, after which he moves to $(x+1,0)$. In the end, he ends up in the position $(19,0)$ and makes the last move to $(20,0)$. The unvisited positions are $(2,2),(4,4), \ldots,(10,10),(11,9),(13,7), \ldots,(19,1)$ - exactly 10 of them, that is, exactly $121-10=111$ positions were visited. Exactly 110 moves were made.
Grading criteria. ($\cdot$) Proved that the number of moves is no more than 110: 3 points. ($\cdot$) Construction of an example with 110 moves with precise justification: 4 points. ($\cdot$) Lack of precise justification of the example: minus 2 points.
|
110
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. The ore contains $21 \%$ copper, enriched - $45 \%$ copper. It is known that during the enrichment process, $60 \%$ of the mined ore goes to waste. Determine the percentage content of ore in the waste.
|
Answer: $5 \%$.
Solution. In the extracted 100 kg of ore, there are 21 kg of copper. From these 100 kg of enriched ore, 40 kg will be obtained, containing 18 kg of copper. Therefore, the 60 kg of waste that went to the dump contain 3 kg of copper, that is, $5 \%$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. Find the number of different arrangements in a row of all natural numbers from 1 to 10 such that the sum of any three consecutive numbers is divisible by 3.
|
Answer: $4! \cdot 2 \cdot 3! \cdot 3! = 1728$
Solution. From the condition, it follows that the remainders of the numbers standing two apart when divided by 3 are equal. Therefore, the numbers standing at positions 1, 4, 7, and 10, as well as those at positions 2, 5, and 8, and at positions 3, 6, and 9, have equal remainders when divided by 3. Among the numbers from 1 to 10, four identical remainders of 1 are given only by 1, 4, 7, and 10, so any permutation of these numbers can be placed at positions 1, 4, 7, and 10, giving a total of 4! options. Similarly, at positions 2, 5, and 8, any arrangement of the numbers 2, 5, 8, or the numbers 3, 6, and 9 can be placed, and at positions 3, 6, and 9, the opposite. In total, we get $4! \cdot 3! \cdot 3! \cdot 2$ options.
|
1728
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. All natural numbers from 1 to 100 are written in some order in a circle. For each pair of adjacent numbers, the sum is calculated. Out of the hundred resulting numbers, what is the maximum number that can be divisible by 7?
|
Answer: 96.
Solution. For the sum of a pair of adjacent numbers to be divisible by 7, their remainders when divided by 7 must sum to 7: 0+0, 1+6, 2+5, and 3+4. Let's call such pairs of remainders suitable pairs. Among the numbers from 1 to 100, there are 14, 15, 15, 14, 14, 14, and 14 numbers with remainders 0, 1, 2, 3, 4, 5, and 6, respectively. A sequence of numbers with alternating suitable pairs of remainders will be called a correct series. If a correct series consists of \( x \) numbers, it will yield \( x-1 \) sums that are divisible by 7. If there are \( n \) such series, they will yield \( 15+14-n=29-n \) or \( 14+14-n=28-n \) such sums. Similarly, for other groups of remainders, we see that the total number of sums divisible by 7 is 100 minus the number of correct series into which all numbers from 1 to 100 are divided. For each pair of suitable remainders, there is at least one correct series, so the total number of sums divisible by 7 does not exceed \( 100-4=96 \).
An example for 96 can be constructed easily: first, all numbers divisible by 7 go in a row, then numbers with remainders 1 and 6 alternate, followed by numbers with remainders 2 and 5 alternating, and finally, all numbers with remainders 3 and 4 alternate. Since the quantities of numbers in groups with different remainders differ by no more than 1, this is possible.
|
96
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. Lesha wrote down a number, and then replaced the same digits with the same letters, and different digits with different letters. He ended up with the word NOVOSIBIRSK. Could the original number be divisible by 3?
|
Answer: Yes, it could.
Solution: For example, 10203454638.
Criteria: Any correct example without verification - 7 points.
|
10203454638
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. The steamship "Raritet" after leaving the city moves at a constant speed for three hours, then drifts for an hour, moving with the current, then moves for three hours at the same speed, and so on. If the steamship starts its journey from city A and heads to city B, it takes 10 hours. If it starts from city B and heads to city A, it takes 15 hours. How long would it take to travel from city A to city B on a raft?
|
Answer: 60 hours.
Solution: Let the speed of the steamboat be $U$, and the speed of the river be $V$. When the steamboat travels from B to A, it arrives at the destination just before the fourth engine stop, meaning it travels 12 hours at a speed of $U-V$ towards A, and then 3 hours back at a speed of $V$ towards B. Therefore, the distance between the cities is $S=12(U-V)-3 V=12 U-15 V(1)$.
When the steamboat travels from A to B, it manages to break down twice and then continue for another two hours on its own. So, it travels 8 hours at a speed of $U+V$ towards B and 2 hours at a speed of $V$ with the current towards B. Thus, the distance is $S=8(U+V)+2 V=8 U+10 V(2)$. Equating the two, we get $12 U-15 V=8 U+10 V$ or $4 U=25 V$ or $(U / V)=(25 / 4)$. The time it will take to travel by raft is $S / V=12(U / V)-15(V / V)=12 *(25 / 4)$ - 15 $=60$ hours.
Criteria: Only the answer - 0 points. Correctly formulated equation (1) or (2) - 1 point (these points are cumulative).
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. A square box 3 by 3 is divided into 9 cells. It is allowed to place balls in some cells (possibly a different number in different cells). What is the minimum number of balls that need to be placed in the box so that each row and each column of the box contains a different number of balls?
|
Answer: 8.
Solution: Add up the number of balls in all rows and columns. Since these are 6 different non-negative numbers, this sum is at least $0+1+\ldots+5=15$.
Now notice that the sum of the numbers in the rows is equal to the sum of the numbers in the columns, since these sums are equal to the total number of balls in the box. Therefore, the sum of the six numbers is twice the sum of the numbers in the rows, which means it is even. Thus, twice the number of balls in the box is greater than 15, so the number of balls is greater than 7. An example with 8 is shown alongside.
Criteria: only the answer - 0 points, only the example - 2 points, only noted that the sum of 6 numbers is not less than $15-1$ point. Only the correct estimate - 3 points.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find all natural numbers $n$ such that $\frac{1}{n}=\frac{1}{p}+\frac{1}{q}+\frac{1}{p q}$ for some primes $p$ and $q$
|
Answer. $n=1$.
Solution. Bringing the expression in the condition to a common denominator, we get: $n(p+q+1)=pq$. From the simplicity of $p$ and $q$, it follows that the divisors of the right-hand side can only be the numbers $1, p, q$, and $pq$, one of which must equal $p+q+1$. Since $1, p$, and $q$ are less than $p+q+1$, we obtain $pq=p+q+1$ and $n=1$. Rewriting the last equation as $(p-1)(q-1)=2$, we get $p=2, q=3$ or $p=3, q=2$.
Grading Criteria. Guessed $n=1$ and $p=2, q=3$ or $p=3, q=2$: 1 point. Proved that $n=1$ is the only number that can satisfy the condition: 5 points. Shown that it satisfies the condition for $p=2, q=3$ or $p=3, q=2$: 2 points. (It is sufficient to simply provide the example $p=2, q=3$ or $p=3, q=2$)
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1. A kilogram of meat with bones costs 165 rubles, a kilogram of meat without bones costs 240 rubles, and a kilogram of bones costs 40 rubles. How many grams of bones are there in a kilogram of meat with bones?
|
Answer: 375 grams.
Solution: Let $x$ kilos of bones be in a kilogram of meat with bones. Then $40 x + 240(1-x) = 165$, from which $x=0.375$.
|
375
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. From the highway, four roads sequentially lead to four villages A, B, C, D. It is known that the route by road/highway/road from A to B is 9 km, from A to C - 13 km, from B to C - 8 km, from B to D - 14 km. Find the length of the route by road/highway/road from A to D. Explain your answer.
|
Answer: 19 km.
Solution. Let's add the distances from $A$ to $C$ and from $B$ to $D$. Then the highway segment from the turn to $B$ to the turn to $C$ will be counted twice, while the highway segments from the turn to $A$ to the turn to $B$ and from the turn to $C$ to the turn to $D$, as well as the four roads from the highway to the villages, will be counted once. Therefore, if we now subtract the distance from $B$ to $C$ from the resulting sum, the difference will include each of the three highway segments and the access roads to $A$ and $D$ exactly once, which is precisely the path from $A$ to $D$. Thus, the length of this path is $13+14-8=19$ km.
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. What is the maximum number of different rectangles that an 8 by 8 chessboard can be cut into? All cuts must follow the grid lines. Rectangles are considered different if they are not equal as geometric figures.
|
Answer: 12.
Solution. Let's list the possible sizes of different integer rectangles of minimal areas that can fit along the grid lines on an 8 by 8 board in ascending order of these areas: 1 by 1, 1 by 2, 1 by 3, 1 by 4, 2 by 2, 1 by 5, 1 by 6, 2 by 3, 1 by 7, 1 by 8, 2 by 4, 3 by 3, 2 by 5. There are already 13 rectangles, and the sum of their areas is 73, which is greater than the area of the board. Therefore, it is impossible to cut the board into more than 12 rectangles in the required manner.
On the other hand, the sum of the areas of all of them except the 3 by 3 is exactly 64, and it is possible to provide an example of such a partition into all 12 rectangles except the 3 by 3: the first four verticals of the board can be cut into strips of width 1 and lengths 1 and 7, 2 and 6, 3 and 5, and 8, respectively. The remaining 4 verticals can be divided into two vertical rectangles 2 by 7, composed of rectangles 2 by 5 and 2 by 2, and 2 by 4 and 2 by 3. Above them, add a horizontal strip 1 by 4.
Other examples of such a partition are also possible.
Grading criteria. It is proven that the number of rectangles in the partition does not exceed 12: 4 points. A correct example for 12 rectangles is provided: 3 points.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. It is known that the values of the quadratic trinomial $a x^{2}+b x+c$ on the interval $[-1,1]$ do not exceed 1 in absolute value. Find the maximum possible value of the sum $|a|+|b|+|c|$. Answer. 3.
|
Solution. Substituting the values $x=0,1,-1$ from the interval $[-1,1]$ into the polynomial $a x^{2}+b x+c$, we obtain three inequalities: $-1 \leq c \leq 1$, $-1 \leq a+b+c \leq 1$, and $-1 \leq a-b+c \leq 1$. Adding the second and third inequalities, we also get $-1 \leq a+c \leq 1$. Subtracting the second from the third (they are double and symmetric!), we have $-1 \leq b \leq 1$. Subtracting $-1 \leq c \leq 1$ from $-1 \leq a+c \leq 1$, we get $-2 \leq a \leq 2$. Due to the symmetry of the problem's conditions with respect to multiplication by -1, we can assume the coefficient $a$ is positive. If $b, c \geq 0$, then $|a|+|b|+|c|=$ $a+b+c \leq 1$ as proven. If $b<0, c \geq 0$, then $|a|+|b|+|c|=a-b+c \leq 1$, as proven. If $b \geq 0, c<0$, then $|a|+|b|+|c|=a+b-c=(a+b+c)-2 c \leq 1+2=3$. If $b, c<0$, then $|a|+|b|+|c|=a-b-c=(a-b+c)-2 c \leq 1+2=3$. Thus, the sum $|a|+|b|+|c|$ under the conditions of the problem does not exceed 3.
The value 3 is achieved, for example, on the polynomial $f(x)=2 x^{2}-1$: its minimum value is reached inside the interval at the vertex of the parabola when $x=0$, and its maximum values are reached at the ends of the interval when $x=1,-1$.
Grading criteria. Only the boundaries of the coefficients of the equation and their sums from statements 1-3 are found: 2-3 points. The estimate $|a|+|b|+|c| \leq 3$ is proven: 5 points. An example is provided and justified where this boundary is achieved: 2 points. If the example is provided without justification: minus 1 point. Any incorrect boundary: 0 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. What is the maximum number of integers that can be written in a row so that the sum of any five consecutive ones is greater than zero, and the sum of any seven consecutive ones is less than zero?
|
Answer. Ten.
Solution. The sum of any seven written numbers is negative, while the sum of the five outermost numbers from this seven is positive, which means the sum of the two leftmost and the sum of the two rightmost numbers from this seven are negative. Therefore, the sum of any two adjacent numbers, with at least five more numbers to the right or left, is negative.
This means that if we assume more than ten numbers are written, the sum of any two adjacent numbers is negative.
In every sequence of five consecutive numbers, the sum of all is positive, while the sum of two pairs of adjacent numbers is negative, so the outermost and middle numbers of each five are positive. If there are at least nine (or eleven) numbers, each of them will be the outermost in some five, meaning all written numbers will be positive, which contradicts the condition of the negativity of the sums of sevens. Therefore, the number of written numbers is no more than ten.
Let's construct an example for ten numbers. From the previous reasoning, it follows that the numbers with indices $1,3,5,6,8,10$ should be positive, while the others should be negative. Suppose all positive numbers are equal to ${ }^{x}$, and all negative numbers are equal to ${ }^{-y}$. The sum of any five consecutive numbers will be $3 x-2 y>0$, and the sum of any seven consecutive numbers will be $4 x-3 y<0$, from which $\frac{4}{3} x<y<\frac{3}{2} x$. We can take, for example, $x=5, y=7$, then the desired example will be: 5,-7,5,-7,5,5,-7,5,-7,5.
Grading criteria. Noted that the sum of the two leftmost and the sum of the two rightmost numbers from this seven are negative: 1 point. Noted that if there are more than ten numbers, the sum of any two adjacent numbers is negative: 2 points. Noted that the outermost and middle numbers of each five are positive: 1 point. Noted that if there are at least eleven numbers, all numbers will be positive: 1 point. Any correct example for ten numbers: 2 points.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. There are 100 boxes numbered from 1 to 100. One of the boxes contains a prize, and the host knows where it is. The audience can send the host a batch of notes with questions that require a "yes" or "no" answer. The host shuffles the notes in the batch and, without reading the questions aloud, honestly answers all of them. What is the minimum number of notes that need to be sent to definitely find out where the prize is?
|
Answer: 99.
Solution: To be able to definitively determine which of the 100 boxes contains the prize, it is necessary to have the possibility of receiving at least 100 different answers to one set of questions. Since the host's answers for different prize positions can only differ by the number of "yes" responses, it is necessary to have the possibility of receiving at least 100 different counts of "yes." Therefore, at least 99 questions are required (from 0 to 99 "yes").
Example with 99 questions. Let the $k$-th question be: "Is the number of the box containing the prize less than or equal to $k$?" Then, if the number of "yes" answers is zero, the prize is in the hundredth box; if one, it is in the 99th box, and so on.
Criteria: Only the evaluation -3 points, only the example -3 points. Only the answer -0 points.
|
99
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.2. Find the number of all five-digit numbers $\overline{a b c d e}$, all digits of which are distinct and $ad>e$.
|
Answer: 1134.
Solution. For the correct notation of a number satisfying the condition of the problem, one needs to arbitrarily select a quintet of different digits from 10 possible ones, and then arrange two of them to the left of the maximum in ascending order and the two remaining to the right of the maximum in descending order. There are two different cases possible.
1) The selected quintet does not contain zero. Then it can be chosen in $C_{9}^{5}$ ways, then choose two non-maximum and non-zero digits from them in $C_{4}^{2}$ ways, and arrange them to the left of the largest in ascending order in a unique way, and write the two remaining to the right of the maximum in descending order. In total, we have $C_{9}^{5} \cdot C_{4}^{2}=126 \cdot 6=756$ such numbers.
2) The selected quintet contains zero. Then its non-zero digits can be chosen in $C_{9}^{4}$ ways, then choose two non-maximum digits from them in $C_{4}^{2}$ ways, and then arrange them to the left of the largest in ascending order in a unique way, and write one remaining and zero to the right of the maximum in descending order. In total, we have $C_{9}^{4} \cdot C_{3}^{2}=126 \cdot 3=378$ such numbers.
In total, we get $756+378=1134$ numbers.
Grading criteria. The idea of separately considering the cases when there is a 0 among the digits of the number and when there is not: 1 point. Correct consideration of each case: 3 points.
|
1134
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.3. Prove that for any $0 \leq x, y \leq 1$ the inequality $\frac{x}{1+y}+\frac{y}{1+x} \leq 1$ holds.
Proof 1. Replace the ones in the denominators of the fractions on the left side of the inequality with $0 \leq x \leq 1$ and $0 \leq y \leq 1$ respectively. In this case, the denominators of the fractions will not increase and will remain positive, and the values of the fractions will not decrease: $\frac{x}{1+y}+\frac{y}{1+x} \leq \frac{x}{x+y}+\frac{y}{y+x}=\frac{x+y}{x+y}=1$ - which is what we needed to prove.
Proof 2. Eliminate the denominators and combine like terms, we get $x^{2}+y^{2} \leq 1+x y$. If $x \leq y$, then the required inequality is obtained by adding the inequalities $y^{2} \leq 1$ and $x^{2} \leq x y$. If $y \leq x$, then the required inequality is obtained by adding the inequalities $x^{2} \leq 1$ and $y^{2} \leq x y$.
10.4. Point $M$ is the midpoint of side $A B$ of triangle $A B C$. Points $P$ and $Q$ are chosen on segment $C M$ such that $P$ is closer to $M$, $Q$ is closer to $C$, and $C Q=2 P M$. It turns out that $B Q=A C$. Find the measure of angle $A P M$.
|
Answer. $A P M=90^{\circ}$.
Solution. Mark a point T on the extension of CM beyond M such that MT=MP. In this case, segments $\mathrm{AB}$ and TP are bisected by their intersection point M, so quadrilateral ATBP is a parallelogram. In particular, segments AP and BT are equal and parallel. Consider triangles $\mathrm{BQT}$ and $\mathrm{APC}$. In them, sides $\mathrm{BQ}$ and $\mathrm{AC}$ are equal by condition, QT and PC are equal by construction, and sides AP and BT are equal as proven above. Therefore, triangles BQT and APC are equal, meaning their corresponding angles BTQ and APC are equal. From the parallelism of AP and BT, it follows that the sum of these angles is $180^{\circ}$, so each of them is $90^{\circ}$. Then the angle APM, supplementary to APC, is also $90^{\circ}$.
Remark. Another solution can be obtained if a point E is marked on the extension of CM beyond M such that $\mathrm{ME}=\mathrm{MQ}$ and it is noticed that triangle $\mathrm{ACE}$ is isosceles, with $\mathrm{P}$ being the midpoint of its base.
Evaluation Criteria. An attempt is made to set $\mathrm{MT}=\mathrm{MP}$ or $\mathrm{ME}=\mathrm{MQ}$ without further progress: 1 point. It is noted in this case that triangle ACE is isosceles: another 1 point.
|
90
|
Inequalities
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
9.1. Petya wrote 10 integers on the board (not necessarily distinct).
Then he calculated the pairwise products (that is, he multiplied each of the written numbers by each other). Among them, there were exactly 15 negative products. How many zeros were written on the board?
|
Answer: 2.
Solution. Let there be $A$ positive numbers and $B$ negative numbers on the board. Then $A+B \leq 10$ and $A \cdot B=15$. Since a negative product is obtained when we multiply a negative and a positive number. From this, it is easy to understand that the numbers $A$ and $B$ are 3 and 5 (1). Therefore, $A+B=8$ and there are exactly two zeros on the board.
Criteria. Correct answer without justification - 1 point. Correctly found intermediate relationship (1) - 4 points. Correct answer with justification - 7 points.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. In a trapezoid, one lateral side is twice as large as the other, and the sum of the angles at the larger base is 120 degrees. Find the angles of the trapezoid.
|
Answer: 90 and 30 degrees.
Solution: Let the vertices of the trapezoid be A, B, C, D, with the larger base AD, and assume CD is twice as long as AB. Choose a point E on AD such that BE is parallel (and equal) to CD, and let M be the midpoint of segment BE. Then triangle ABM is isosceles with a 60-degree angle at vertex B, so it is equilateral, and point M is equidistant from A, B, and E. Therefore, M is the center of the circle containing points A, B, and E, and BE is its diameter. Angle BAE is an inscribed angle subtending this diameter, so its measure is 90 degrees, which means the measure of angle BAD is also 90 degrees, and the measure of angle CDA is 30 degrees.
Criteria: The idea of constructing segment BE: 2 points.
|
90
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.4. $N$ different natural numbers, not exceeding 1000, are written in a circle such that the sum of any two of them, standing one apart, is divisible by 3. Find the maximum possible value of $N$.
#
|
# Answer: 664.
Solution. Consider the remainders of the numbers when divided by 3. Divisibility by 3 means that in each pair of numbers standing one apart, either both numbers are divisible by 3, or one has a remainder of 1 and the other has a remainder of 2 when divided by 3. Among the numbers from 1 to 1000, 333 are divisible by 3, 334 give a remainder of 1, and 333 give a remainder of 2.
1) Suppose \( \mathrm{N} \) is odd. Then the remainders 1 and 2 cannot alternate throughout the circle, so all numbers in the circle will be divisible by 3, and there will be no more than 333 of them.
2) Suppose \( \mathrm{N} \) is even. The numbers are divided into two cycles of equal length \( \mathrm{N}/2 \): those in even positions and those in odd positions. In each of them, either all numbers are divisible by 3, or they alternate with remainders of 1 and 2.
A cycle of numbers divisible by 3 has a length of no more than 333. A cycle with alternating remainders has an even length, not exceeding 666.
If there are numbers divisible by 3 among the listed ones, then there is a cycle of such numbers with a length of no more than 333. The second cycle would then consist of no more than 333 even numbers with remainders of 1 and 2, i.e., no more than 332 numbers. Then the total number of numbers would be no more than 664. This number is achievable: one cycle contains numbers divisible by 3 from 3 to 996, and the second: all numbers from 1 to 498 that are not divisible by 3.
If there are no numbers divisible by 3, then the lengths of the cycles are equal and even, so the total number of numbers written should be no more than 667 and divisible by 4, i.e., no more than 664.
Criteria. Consider the case 1) \( \mathrm{N} \) is odd: 1 point. Consider the case 2) \( \mathrm{N} \) is even - the idea of two cycles: 1 point, estimating the length of each cycle: 1 and 2 points respectively. Constructing an example: 2 points.
|
664
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.3. For what smallest $n$ is the following condition satisfied: if $n$ crosses are placed in some cells of a $6 \times 6$ table in any order (no more than one per cell), then there will definitely be three cells forming a strip of length 3, either vertical or horizontal, each containing a cross?
|
Answer. $n=25$.
Solution. If there are no fewer than 25 crosses, then one of the rows of the table contains no fewer than 5 crosses, and no more than one empty cell. Then either the three left cells of this row, or the three right cells of it, all contain crosses and form the desired strip.
If there are fewer than 25 crosses, they can be arranged so that there are no three crosses forming a strip. For this, all cells of one main diagonal, and the cells of two parallel diagonals of length 3, should be left empty.
Grading criteria. Estimation (if more than 24, then a strip will always be found): 3 points. Example of arranging 24 or fewer crosses without three crosses in a row: 4 points.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. Find all natural numbers $x$ such that the product of all digits in the decimal representation of $x$ equals $x^{2}-10 x-22$
|
Answer: $x=12$.
Solution: First, the product of all digits of a natural number is non-negative, so $x^{2}-10 x-22$, from which $x \geq \frac{10+\sqrt{188}}{2}$, that is, $x \geq 12$. Second, if in the product of all digits of a natural number, all digits except the first are replaced by tens, the product will not decrease, but will not exceed the number itself, so the product of all digits of the number does not exceed the number itself, hence $x^{2}-11 x-22 \leq 0$, from which $x \leq \frac{11+\sqrt{209}}{2}$, that is, $x \leq 12$. For $x=12$, the condition is obviously satisfied, therefore, this is the only answer.
Grading criteria. Only the answer with verification: 1 point. Proof that the product of all digits of the number does not exceed the number itself: 3 points.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. Lesha wrote down a number, and then replaced the same digits with the same letters, and different digits with different letters. He ended up with the word NOVOSIBIRSK. Could the original number be divisible by 9?
|
Answer: Yes, it could.
Solution: For example, 10203454638.
Criteria: Any correct example without verification - 7 points.
|
10203454638
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.4. The steamship "Raritet" after leaving the city moves at a constant speed for three hours, then drifts for an hour, moving with the current, then moves for three hours at the same speed, and so on. If the steamship starts its journey from city A and heads to city B, it takes 10 hours. If it starts from city B and heads to city A, it takes 15 hours. How long would it take to travel from city A to city B on a raft?
|
Answer: 60 hours.
Solution: Let the speed of the steamboat be $U$, and the speed of the river be $V$. When the steamboat travels from B to A, it arrives at the destination just before the fourth engine stop, meaning it travels 12 hours at a speed of $U-V$ towards A, and then 3 hours back at a speed of $V$ towards B. Therefore, the distance between the cities is $S=12(U-V)-3 V=12 U-15 V(1)$.
When the steamboat travels from A to B, it manages to break down twice and then continue for another two hours on its own. So, it travels 8 hours at a speed of $U+V$ towards B and 2 hours at a speed of $V$ with the current towards B. Thus, the distance is $S=8(U+V)+2 V=8 U+10 V$ (2). Equating the two, we get $12 U-15 V=8 U+10 V$ or $4 U=25 V$ or $(U / V)=(25 / 4)$. The time it will take to travel by raft is $S / V=12(U / V)-15(V / V)=12 *(25 / 4)$ - 15 $=60$ hours.
Criteria: Only the answer - 0 points. Correctly formulated equation (1) or (2) - 1 point (these points are cumulative).
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.1. Find all four-digit numbers $\overline{x y z t}$, where all digits $x, y, z, t$ are distinct and not equal to 0, such that the sum of all four-digit numbers obtained from $\overline{x y z t}$ by all possible permutations of the digits is 10 times the number $\overline{x x x x}$.
|
Answer. The number 9123 and all numbers obtained from it by permuting the last three digits, a total of 6 answers.
Solution. The number of four-digit numbers obtained from $\overline{x y z t}$ by all possible permutations of the digits is 24, in each of which each of the digits $x, y, z, t$ appears exactly 6 times in each place. Therefore, the sum of all such numbers is $6(x+y+z+t)(1000+100+10+1)=6666(x+y+z+t)$, which, by condition, equals $11110 x$. From this, $3(y+z+t)=2 x$. According to the condition, all digits $x, y, z, t$ are distinct and not equal to 0, so the left side of the equation is no less than $3(1+2+3)=18$, from which we get the only possibility $x=9$ and $y, z, t$ are any permutation of the digits $1,2,3$.
Evaluation Criteria. If a complete correct answer with verification is provided: 1 point. Correct finding of the sum on the left side of the equation: 3 points.
|
9123
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.5. Around a circle, 32 numbers $a_{1}, a_{2}, \ldots, a_{32}$ are written, each of which is either -1 or 1. In one operation, each number $a_{n}, n=1,2, \ldots, 32$ is replaced by the product $a_{n} a_{n+1}$ of it and the next number in the cycle, with indices considered cyclically, $a_{33}=a_{1}, a_{34}=a_{2}$, and so on. Prove that for any initial set of numbers $a_{1}, a_{2}, \ldots a_{32}$, after some finite number of operations, a set of 32 ones will always be obtained. Find the smallest number $\mathrm{N}$ of operations such that after applying $\mathrm{N}$ operations, a set of 32 ones will always be obtained from any initial set of numbers.
|
Answer: 32.
Solution: We will prove by induction on $n$ that if $2^n$ numbers are written in a circle, the answer to the problem is $2^n$. The base case $n=1$ is straightforward: either $\{1,-1\} \rightarrow\{-1,-1\} \rightarrow\{1,1\}$, or $\{-1,-1\} \rightarrow\{1,1\}$, in any case, two operations are always sufficient, but fewer are not always enough.
Inductive step. Suppose $2^{n+1}$ numbers are written in a circle and the statement is true for any $2^n$ plus-minus ones in a circle. Consider separately the set A, consisting of $2^n$ numbers in the positions with odd indices, and set B - consisting of $2^n$ numbers in the positions with even indices. Note that after performing two operations, each number $a_k, k=1,2, \ldots, 32$ will be replaced by the product $\left(a_k a_{k+1}\right)\left(a_{k+1} a_{k+2}\right)=a_k a_{k+2}$, since $a_{k+1}^2=1$. This means that, as a result of performing two operations on the initial set of $2^{n+1}$ numbers, sets A and B change as if one operation had been performed on each of them. By the inductive hypothesis, to get sets of all ones from sets A and B, $2^n$ operations are sufficient, so to get all ones from the initial set, twice as many operations are needed, i.e., $2 \cdot 2^n = 2^{n+1}$ operations. An example of a set of numbers for which exactly $2^{n+1}$ operations are needed is a set of 31 ones and one minus one. After two operations, it will give similar sets of numbers A and B, for which, by induction, exactly $2^n$ operations are needed, so for the initial set, exactly $2 \cdot 2^n = 2^{n+1}$ operations are needed.
For the number of numbers $32=2^5$, the number of necessary operations is 32.
Evaluation criteria. It is proven that 32 operations are sufficient: 4 points. It is proven that fewer than 32 operations may not be enough: 3 points.
|
32
|
Combinatorics
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
11.4. Find all natural numbers $n$ that can be represented as $n=\frac{x+\frac{1}{x}}{y+\frac{1}{y}}$, for some natural numbers $x$ and $y$.
|
Answer. $n=1$
Solution. Transform the equality in the condition to the form $n=\frac{\left(x^{2}+1\right) y}{\left(y^{2}+1\right) x}$. Note that the numbers $x^{2}+1$ and $x$, as well as the numbers $y^{2}+1$ and $y$ are coprime, so $x$ in the denominator can only cancel out with $y$ in the numerator, meaning $y$ is divisible by $x$ and, in particular, $y$ is not less than $x$. Similarly, $y^{2}+1$ in the denominator can only cancel out with $x^{2}+1$ in the numerator, so $y^{2}+1$ divides $x^{2}+1$, in particular, does not exceed it, from which it follows that $y$ does not exceed $x$. Therefore, $y$ equals $x$ and $n=1$. The number $n=1$ is obtained for any pair of equal $x$ and $y$, for example, when they are both equal to 1.
Grading criteria. A statement in the solution like “the numbers $x^{2}+1$ and $x$, as well as the numbers $x^{2}+1$ and $y$ are coprime” can be considered correct without detailed explanations. If there are no references to coprimality: the score will not exceed 2 points. If it is not explicitly stated that the number $n=1$ is still obtained for any pair of equal $x$ and $y$, for example, when they are equal to something specific: 1 point will be deducted.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.2 Sergey arranged several (more than two) pairwise distinct real numbers in a circle so that each number turned out to be equal to the product of its neighbors. How many numbers could Sergey have arranged?
|
Answer: 6.
Solution: Let's denote the two adjacent numbers as $a$ and $b$. Then, next to them stands $b / a$, followed by $1 / a, 1 / b, a / b$, and again $a$. Thus, it is impossible to arrange more than 6 numbers.
If 3 numbers can be arranged, then $a=1 / a$, which means $a$ is 1 or -1. In the first case, $b$ and $b / a$ coincide. In the second case, we get the triplet: $-1, b, -b$, where $-1=b * (-b)$, meaning this triplet is $-1, 1, -1$.
If 4 numbers can be arranged, then $a=1 / b$, resulting in the quartet of numbers $a, 1 / a, 1 / a^2, 1 / a$. There are repeating numbers.
If 5 numbers can be arranged, then $a=a / b$, which means $b=1$. This results in the quintet of numbers $a, 1, 1 / a, 1 / a, 1$. There are repeating numbers.
Finally, 6 numbers can be arranged: $2, 3, 3 / 2, 1 / 2, 1 / 3, 2 / 3$.
Criteria: Proving that there are no more than 6 numbers - 3 points.
For considering each case - 1 point each.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.1. A number is called good if any two adjacent digits in its notation differ by at least 4. Vera wrote some good number, and then replaced identical digits with identical letters, and different ones with different letters. Could she have ended up with the word NOVOSIBIRSK?
|
Answer: For example, the number 82729161593 could work ( $\mathrm{H}=8, \mathrm{O}=2, \mathrm{~B}=7, \mathrm{C}=9$, I = 1, B = 6, $\mathrm{P}=5, \mathrm{~K}=3$).
Criterion: any valid example without verification - 7 points.
|
82729161593
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. Find the angle $D A C$, given that $A B=B C$ and $A C=C D$, and the lines on which points $A, B, C, D$ lie are parallel, with the distances between adjacent lines being equal. Point $A$ is to the left of $B$, $C$ is to the left of $B$, and $D$ is to the right of $C$ (see figure).
|
Answer: 30 degrees.
Solution: Let the line on which point $B$ lies be $b$, the line on which point $C$ lies be $c$, and the line on which point $D$ lies be $d$.
Suppose line $A C$ intersects line $b$ at point $M$. By Thales' theorem, $A M = M C$, since the distances between parallel lines are equal. Therefore, $B M$ is the median in the isosceles triangle $A B C$. Thus, $B M$ is also the altitude, meaning that angle $A M B$ is a right angle, as lines $b$ and $c$ are parallel, so $A C$ is perpendicular to line $c$. Let $A C = 2 x$, then the distance between the parallel lines is $x$. Extend $A C$ to intersect line $d$ at point $E$. Then $C E = x$, as the distance between the parallel lines, and $C D = 2 x$, since $A C = 2 x$. Additionally, in triangle $C E D$, angle $E$ is a right angle. Therefore, in the right triangle $C E D$, the hypotenuse is twice the length of one of the legs. This means that angle $E C D$ is 60 degrees, and the adjacent angle $A C D$ is 120 degrees. Then, in the isosceles triangle $A C D$, the two acute angles are each 30 degrees.
Criteria: It is proven that $A C$ is perpendicular to line $c-2$ points.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. Find the maximum odd natural number that cannot be represented as the sum of three distinct composite numbers.
|
Answer: 17.
Solution: An odd number gives a remainder of 1 or 3 when divided by 4. In the first case, the desired representation has the form \( n = 4k + 1 = 4(k-4) + 8 + 9, k \geq 5, n \geq 21 \), in the second case - \( n = 4k + 3 = 4(k-3) + 6 + 9, k \geq 4, n \geq 19 \). On the other hand, the three smallest composite numbers are 4, 6, 8, the sum of which is 18, so the number 17 cannot be represented in the required form. It is the answer to the problem.
Grading: Simply providing the answer with verification: 0 points. Failure to justify the non-representability of 17 in the desired form - minus 2 points.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. In the cells of an 8 by 8 board, tokens are placed such that for each token, the row or column of the board in which it lies contains no more than three tokens. What is the maximum possible number of tokens on the board?
|
Answer: 30.
Solution: By swapping the verticals and horizontals, we assume that only the left $x$ verticals and the bottom $y$ horizontals contain more than 3 chips. From the condition, it follows that there are no chips at all in the lower left rectangle at the intersection of these verticals and horizontals. Each vertical passing through the chips in these horizontals contains no more than three chips, and each horizontal passing through the chips in these verticals contains no more than three chips. Then the total number of chips on the board does not exceed $3(8-x)+3(8-y)=48-3(x+y)$, which does not exceed 30 when $x+y \geq 6$.
Let $x+y \leq 5$ and $x \leq y$, then $x \leq 2$. We estimate the total number of chips on the board from above as $3(8-x)+x(8-y)=24+5x-xy$. For $x=2$, this is no more than 30, for $x=1, x=0$ - no more than 29 and 24, respectively.
Example of placing 30 chips: all cells of the three left verticals and the three bottom horizontals are filled, except for the lower left 3 by 3 square. Evaluation: 5 points for the estimate. 2 points for the example. Any incorrect estimate with any reasoning - 0 points.
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.3. Given a triangle $A B C$, side $A B$ is divided into 4 equal segments $A B_{1}=B_{1} B_{2}=B_{2} B_{3}=B_{3} B$, and side $A C$ into 5 equal segments $A C_{1}=C_{1} C_{2}=C_{2} C_{3}=C_{3} C_{4}=C_{4} C$. How many times larger is the area of triangle $A B C$ compared to the sum of the areas of triangles $C_{1} B_{1} C_{2}, C_{2} B_{2} C_{3}, C_{3} B_{3} C_{4}, C_{4} B C$?
|
Answer: 2 times.
Solution: Let the area of $A B_{1} C_{1}$ be $S$. Then the area of $B_{1} C_{1} C_{2}$ is also $S$, since $B_{1} C_{1}$ is the median in $A B_{1} C_{2}$. Similarly, the area of $B_{1} B_{2} C_{2}$ is $2S$, as $C_{2} B_{1}$ is the median in $A B_{2} C_{2}$. The area of $B_{2} C_{2} C_{3}$ is half the area of $A B_{2} C_{2}$, because these triangles share the same height, and the base $C_{2} C_{3}$ is half the length of the base $A C_{2}$. Therefore, the area of $B_{2} C_{2} C_{3}$ is $2S$. From similar reasoning, the area of $B_{2} C_{3} B_{3}$ is half the area of $A B_{2} C_{3}$, which is $3S$. Since $C_{3} C_{4}$ is one-third of $A C_{3}$, the area of $B_{3} C_{3} C_{4}$ is one-third of $A B_{3} C_{3}$ and equals $3S$. Similarly, the areas of $B_{3} C_{4} B$ and $C_{4} B C$ are each $4S$. Therefore, the total area of the shaded part is $S + 2S + 3S + 4S = 10S$, and the unshaded part is $S + 2S + 3S + 4S = 10S$, meaning the shaded area is half of the entire triangle.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. Masha and Misha set out to meet each other simultaneously, each from their own house, and met one kilometer from Masha's house. Another time, they again set out to meet each other simultaneously, each from their own house, but Masha walked twice as fast, and Misha walked twice as slow as the previous time. This time, they met one kilometer from Misha's house. What is the distance between Masha's and Misha's houses?
|
# Answer: 3 km.
Solution: We will prove that they spent the same amount of time on the first and second occasions. Suppose this is not the case, and they spent less time on the second occasion. Then, on the first occasion, Masha walked 1 km, and on the second occasion, less than 2 km (her speed was twice as high, but the time was less). Misha walked 1 km on the second occasion, so on the first occasion, he walked more than 2 km (for similar reasons). Therefore, the total distance is more than 3 km (1 km from Masha and more than 2 km from Misha on the first occasion), but at the same time, it is less than 3 km (1 km from Misha and less than 2 km from Masha on the second occasion). This is impossible. If they walked longer on the second occasion, similar reasoning will lead to a contradiction (due to the symmetry of the situations).
Thus, they moved for the same amount of time both times. Therefore, on the second occasion, Masha walked 2 km, and Misha walked 1 km. Consequently, their houses are 3 kilometers apart.
Criteria: answer, answer with verification - 0 points, not proven that the time was the same in the first and second cases, when using this - no more than 2 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. Given the number 1836549, you can take two adjacent non-zero digits and swap their places, after which you subtract 1 from each of them. What is the smallest number that can result from these operations?
|
Answer: 1010101
Solution: The digits in the number alternate in parity: odd, even, etc. Note that with the described operation, even and odd numbers swap places, and then 1 is subtracted from them, thereby not disrupting the order of parity. Thus, it is impossible to obtain a number less than 1010101 (in each place, the smallest odd or even digit is placed, respectively).
Let's show how this result can be achieved (we perform the operation on the underlined digits, sometimes on two pairs simultaneously): $\underline{\underline{83}} 65 \underline{49}>12 \underline{7658} 3>125 \underline{6743}>12 \underline{56543}>12 \underline{54343}>1 \underline{23} 4343>1 \underline{214343}>$ 1014343
Next, each pair 43 is transformed into 23, then into 21, and then into 01, after which we get 1010101.
## Criteria for Determining Winners and Prize Winners of the All-Siberian Open Mathematics Olympiad for Schoolchildren (2015-2016 academic year)
According to the Regulations, the winners and prize winners of the Olympiad were determined based on the results of the Final Stage of the Olympiad. The total number of winners and prize winners was 380 out of 1578 participants, which is $24.08 \%$. The number of winners was 85, which is $5.38 \%$.
Based on the overall ranking of participants and taking into account the noticeable gaps in the scores of the groups of participants at the top of the ranking, the Olympiad jury developed the following criteria for determining winners and prize winners: The maximum possible score is 35 points.
## 11th Grade:
Winners:
Participants who scored more than $77 \%$ of the maximum possible points, i.e., from 27 to 35 points;
Prize winners:
2nd degree - more than $62 \%$ of the maximum possible points, i.e., from 22 to 26 points
3rd degree - more than $51 \%$ of the maximum possible points, i.e., from 18 to 21 points
10th Grade:
Winners:
Participants who scored more than $85 \%$ of the maximum possible points, i.e., from 30 to 35 points;
Prize winners:
2nd degree - more than $62 \%$ of the maximum possible points, i.e., from 22 to 29 points
3rd degree - more than $51 \%$ of the maximum possible points, i.e., from 18 to 21 points
9th Grade:
Winners:
Participants who scored more than $85 \%$ of the maximum possible points, i.e., from 30 to 35 points;
Prize winners:
2nd degree - more than $65 \%$ of the maximum possible points, i.e., from 24 to 29 points
3rd degree - more than $51 \%$ of the maximum possible points, i.e., from 18 to 23 points
8th Grade:
Winners:
Participants who scored more than $82 \%$ of the maximum possible points, i.e., from 29 to 35 points;
Prize winners:
2nd degree - more than $62 \%$ of the maximum possible points, i.e., from 22 to 28 points
3rd degree - more than $48 \%$ of the maximum possible points, i.e., from 17 to 21 points
7th Grade:
Winners:
Participants who scored more than $85 \%$ of the maximum possible points, i.e., from 30 to 35 points;
Prize winners:
2nd degree - more than $71 \%$ of the maximum possible points, i.e., from 24 to 29 points
3rd degree - more than $42 \%$ of the maximum possible points, i.e., from 15 to 21 points
Co-Chair of the Mathematics Jury
A.Yu. Avdyushenko
|
1010101
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.1. Find all positive integer solutions of the equation $(n+2)!-(n+1)!-(n)!=n^{2}+n^{4}$. Answer. $n=3$.
|
Solution. Rewrite the equation as $n!=\left(n^{*}\left(n^{2}+1\right)\right) /(n+2)$. Transforming the right side, we get $n!=n^{2}-2 n+5-10:(n+2)$. The last fraction will be an integer for $n=3$ and $n=8$, but the latter number is not a solution (substitute and check!)
Grading criteria. Acquiring extraneous solutions: minus 3 points. Guessed and verified answer: 1 point.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2 To buy an apartment, you need to take either exactly 9 small, 6 medium, and one large loan, or exactly 3 small, 2 medium, and 3 large loans. How many only large loans would be required to buy an apartment?
|
Answer: 4 large loans.
Solution: Let's denote small, medium, and large loans by the letters m, c, and b, respectively. Rewrite the condition using these notations:
$$
9 \mathrm{M}+6 \mathrm{c}+\sigma=3 \mathrm{~m}+2 \mathrm{c}+3 b
$$
Simplifying, we get $6 \mathrm{~m}+4 \mathrm{c}=2$ b or $3 \mathrm{~m}+2 \mathrm{c}=$ b, which means $3 \mathrm{~m}+2 \mathrm{c}+3 b=4 b$.
Criteria: Only the answer - 0 points.
Correctly formulated equation - 1 point.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.1. It is known that the sum of the digits of number A is 59, and the sum of the digits of number B is 77. What is the minimum sum of the digits that the number A+B can have?
|
Answer. 1.
Solution. It is sufficient to consider A=9999995, B=999999990000005, then $\mathrm{A}+\mathrm{B}=1000000000000000$, the sum of the digits is 1 - the minimum possible.
Grading Criteria. Correct answer and example: 7 points. Presence of arithmetic errors: minus 1-2 points. Any other answer: 0 points.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.2. On an island, 20 people live, some of whom are knights who always tell the truth, and the rest are liars who always lie. Each islander knows for sure who among the others is a knight and who is a liar. When asked by a visitor how many knights live on the island, the first islander answered: "None," the second: "No more than one," the third: "No more than two," the fourth: "No more than three," and so on, the twentieth stated: "No more than nineteen." So how many knights live on the island?
|
Answer: 10.
Solution: If the first islander were a knight, he would have lied in his answer, which cannot be the case. Therefore, the first is a liar, and there are no more than 19 knights on the island. This means the twentieth islander told the truth, so he is a knight, and there is at least one knight on the island. Then, if the second islander were a knight, there would already be two knights including the twentieth, and he would have lied, so the second is a liar, and there are no more than 18 knights. Therefore, the nineteenth told the truth and is a knight. Continuing this way, it is easy to see that the first to tenth islanders are all liars, and the eleventh to twentieth are all knights.
Grading Criteria: Established that the first is a liar: 1 point. Established that the first is a liar and the second is a knight: 2 points. Given the correct answer and constructed an example with verification: 3 points. Any incorrect answer: 0 points.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. Find the value of the expression $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}$, given that $\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}=5$ and $x+y+z=2$.
|
Answer: 7.
Solution. Transform: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x+y+z}{y+z}+\frac{y+x+z}{x+z}+\frac{z+x+y}{x+y}-3=$ $=(x+y+z)\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3=2 \cdot 5-3=7$.
Grading Criteria. Presence of arithmetic errors: minus 1-2 points. Correct answer calculated on some example: 1 point.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.3. There is a steamship route between the cities of Dzerzhinsk and Lviv. Every midnight, a steamship departs from Dzerzhinsk, arriving exactly eight days later in Lviv. How many steamships will the steamship "Raritet" meet on its way to Dzerzhinsk if it departs from Lviv exactly at midnight and spends the same eight days on the journey?
|
Answer: 17.
Solution: As "Raritet" departs from Lviv, a steamer arrives there, which left Dzerzhinsk 8 days ago. By the time "Raritet" arrives at its final destination, 8 days have passed since the initial moment, and at this moment, the last steamer departs from Dzerzhinsk, which "Raritet" meets on its way. Thus, "Raritet" will meet all the steamers that departed from Dzerzhinsk, starting 8 days before the initial time and ending 8 days after the initial time - a total of 17 steamers (if it is considered that the ships meet when one departs and the other arrives at the same moment).
Criteria: only the answer, answer with verification - 0 points.
If it is considered that "Raritet" does not meet the first and/or last steamers - do not deduct points.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.5. Egor, Nikita, and Innokentiy took turns playing chess with each other (two play, one watches). After each game, the loser gave up their place at the board to the spectator (there were no draws). In the end, it turned out that Egor participated in 13 games, and Nikita in 27. How many games did Innokentiy play?
|
Answer: 14.
Solution: On the one hand, there were no fewer than 27 games. On the other hand, a player cannot skip two games in a row, meaning each player participates in at least every other game. Therefore, if there were at least 28 games, Egor would have participated in at least 14, which contradicts the condition. Thus, exactly 27 games were played, and Nikita participated in all of them. In 13 of these, his opponent was Egor, so in the remaining 14, it was Innokentiy, and this is the answer.
Criteria: Only the answer, answer with verification - 0 points.
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The bathtub fills up in 23 minutes from the hot water tap, and in 17 minutes from the cold water tap. Pete first opened the hot water tap. After how many minutes should he open the cold water tap so that by the time the bathtub is full, one and a half times more hot water has been added than cold water?
|
# 9.2. Answer. In 7 minutes.
In one minute, the hot water tap fills $\frac{1}{23}$ of the bathtub, and the cold water tap fills $\frac{1}{17}$ of the bathtub. After filling the bathtub, the hot water should make up $\frac{3}{5}$ of the bathtub, and the cold water should make up $\frac{2}{5}$ of the bathtub. Therefore, the hot water tap should be open for $\frac{3}{5} / \frac{1}{23} = \frac{69}{5}$ minutes, and the cold water tap should be open for $\frac{2}{5} / \frac{1}{17} = \frac{34}{5}$ minutes. Thus, the cold water tap should be opened $\frac{69}{5} - \frac{34}{5} = 7$ minutes after the hot water tap is opened.
Grading Remarks. If the answer is only guessed, give 0 points. If the answer is guessed and its correctness is neatly verified, give 3 points. If the equation or system is correctly set up but solved with errors, give 4 to 6 points depending on the type of error.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Among all natural numbers from 1 to 20 inclusive, some 10 numbers were painted blue, and the other 10 - red, then all possible sums of pairs of numbers, one of which is blue and the other is red, were counted. What is the maximum number of different sums that can be among the hundred obtained numbers?
|
9.6. Answer. A maximum of 35 different numbers.
The sum of a blue and a red number can be a natural number from $1+2=3$ to $19+20=39$ inclusive, so there cannot be more than 37 different numbers. Moreover, note that one of the numbers $3,4, . ., 13$ must not be the sum of a blue and a red number. Otherwise, if the number 1 is blue, then 3 can only be represented as $1+2$, so 2 is red. The number 4 can only be represented as $1+3$, so 3 is also red. Next, $5=1+4=2+3$, but 2 and 3 are red, so only the first representation is valid, hence 4 is also red, and so on, we get that all 11 numbers $2,3, . ., 12$ are red, which is impossible. Similarly, we show that if all the numbers $29,30, \ldots, 39$ are sums of a blue and a red number, then 11 numbers $9, . ., 19$ are of the same color. Thus, there cannot be more than $37-2=35$ different numbers. One possible example where there are exactly 35 is as follows: $1,11,12, \ldots, 19$ are blue, and $2,3, \ldots, 10,20$ are red.
Grading remarks. Only the correct answer - 0 points. The correct answer and an example where it is achieved - 3 points. A correct proof of the maximum number 35 is worth 4 points. Attempts to solve where other estimates, different from 35, are proven, are graded 0 points.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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