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5-1. A square with a side of 100 was cut into two equal rectangles. They were placed next to each other as shown in the figure. Find the perimeter of the resulting figure.
|
Answer: 500.
Solution. The perimeter of the figure consists of 3 segments of length 100 and 4 segments of length 50. Therefore, the length of the perimeter is
$$
3 \cdot 100 + 4 \cdot 50 = 500
$$
|
500
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-2. In a sports tournament, a team of 10 people participates. The regulations stipulate that 8 players from the team are always on the field, changing from time to time. The duration of the match is 45 minutes, and all 10 participants on the team must play an equal number of minutes. How many minutes will each player be on the field during the game?
|
Answer: 36.
Solution. In total, the players will spend $8 \cdot 45=360$ minutes on the field. This time needs to be divided equally among 10 players, so each will be on the field $360 / 10=36$ minutes.
|
36
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-3. How many two-digit numbers exist where at least one of the digits is smaller than the corresponding digit in the number $35?$
For example, the numbers 17 and 21 are valid, while the numbers 36 and 48 are not.
|
Answer: 55.
Solution. First, let's find the number of two-digit numbers that do not meet the condition. In the units place, any digit from 5 to 9 can stand, and in the tens place, from 3 to 9. The total number of numbers that do not suit us will be exactly $7 \cdot 5=35$. Now we can count the number of two-digit numbers that meet the condition: we need to subtract the number of unsuitable numbers from their total number, which is 90: $90-35=55$.
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-5. Along a straight alley, 400 lamps are placed at equal intervals, numbered in order from 1 to 400. At the same time, from different ends of the alley, Alla and Boris started walking towards each other at different constant speeds (Alla from the first lamp, Boris from the four hundredth). When Alla was at the 55th lamp, Boris was at the 321st lamp. At which lamp will their meeting occur? If the meeting occurs between two lamps, indicate the smaller number of these two.
|
Answer. At the 163rd lamppost.
Solution. There are a total of 399 intervals between the lampposts. According to the condition, while Allа walks 54 intervals, Boris walks 79 intervals. Note that $54+79=133$, which is exactly three times less than the length of the alley. Therefore, Allа should walk three times more to the meeting point than to the 55th lamppost, i.e., $54 \cdot 3=162$ intervals. And she will be at the 163rd lamppost.
|
163
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-6. On a rectangular table of size $x$ cm $\times 80$ cm, identical sheets of paper of size 5 cm $\times 8$ cm are placed. The first sheet touches the bottom left corner, and each subsequent sheet is placed one centimeter higher and one centimeter to the right of the previous one. The last sheet touches the top right corner. What is the length $x$ in centimeters?
|
Answer: 77.
Solution I. Let's say we have placed another sheet of paper. Let's look at the height and width of the rectangle for which it will be in the upper right corner.
Let's call such a rectangle the current one. Notice that for each new current rectangle, both the width and the height are 1 cm larger than the previous one. Initially, when there was only one sheet of paper, the width of the large rectangle was 8 cm, and at the end, it was 80 cm. Thus, a total of $(80-8): 1=72$ sheets of paper were added. The height of the current rectangle also increased by $72 \cdot 1$ cm, initially it was 5, so $x=5+72=77$.
Solution II. As in the first solution, let's look at the length and width of the current rectangles. Again, notice that for each new current rectangle, both the length and the width are 1 cm larger than the previous one. However, we will draw a different conclusion: specifically, the difference between the width and the height of the current rectangle is always the same! (Such a value that does not change during a certain process is called an invariant.) Since initially the width was 3 cm greater than the height, i.e., $8-5=3$ cm, at the end it should also be 3 cm greater, so the answer is $x=80-3=77$ cm.
|
77
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5-8. In the "Young Photographer" club, 300 children attend. During the class, they divided into 100 groups of 3 people, and in each group, each person took one photo of the other two in their group. No one took any other photos. In total, there were 100 photos of "boy+boy" and 56 photos of "girl+girl". How many "mixed" groups were there, that is, groups that had both a boy and a girl?
|
Answer: 72.
Solution: There were a total of 300 photos, so the number of photos of "boy+girl" was $300-100-56=144$. Each mixed group provides two photos of "boy+girl", while non-mixed groups do not provide such photos. Therefore, there were exactly $144 / 2=72$ mixed groups.
## 6th grade
|
72
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6-1. The physical education teacher lined up the class so that everyone was facing him. To the right of Kolya, there are 12 people, to the left of Sasha - 20 people, and to the right of him - 8 people. How many people are standing to the left of Kolya?
|
Answer: 16.
Solution: Since there are 20 people to the left of Sasha and 8 people to the right of him, there are a total of 28 people in the row, not counting Sasha. Therefore, including Sasha, there are 29 people in the class. Then, to the left of Kolya, there are $29-12-1=16$ people (first subtracting the 12 people who are to the right of Kolya, and then Kolya himself).
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6-5. Grandma baked 21 batches of dumplings, with $N$ dumplings in each batch, $N>70$. Then she laid out all the dumplings on several trays, with 70 dumplings on each tray. What is the smallest possible value of $N$?
|
Answer: 80.
Solution: The total number of baked buns is $21 \cdot N$. This number must be divisible by 70 to be able to distribute them into several trays of 70 each. $70=2 \cdot 5 \cdot 7$, and 21 is already divisible by 7. Therefore, $N$ must be divisible by 10, and the smallest such $N$ is 80.
|
80
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6-7. In a confectionery store, the saleswoman laid out 91 candies of several varieties in a row on the counter. It turned out that between any two candies of the same variety, there was an even number of candies. What is the smallest number of varieties there could have been?
|
Answer: 46.
Solution. We will prove that there could not have been three or more candies of the same type. Indeed, let candies of the same type $A, B$, and $C$ lie in that exact order. Suppose there are $2x$ candies between $A$ and $B$, and $2y$ candies between $B$ and $C$, then there are $2x + 2y + 1$ candies between $A$ and $C$, which is an odd number of candies, contradicting the condition.
Since there are no more than 2 candies of each type, there must be at least 46 types. Let's provide an example where there could be 46 types. For instance, the candies can be arranged in pairs: two of the first type, two of the second type, and so on for 45 pairs, with one more candy of the 46th type at the end. There are many other examples.
|
46
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7-1. Petya has stickers. If he gives 5 stickers to each of his friends, he will have 8 stickers left. If he wants to give 6 stickers to each of his friends, he will be short of 11 stickers. How many friends does Petya have?
|
Answer: 19.
Solution I. Suppose Petya gave 5 stickers to each of his friends. Next, he wants to give each friend one more sticker. For this, he needs to spend $8+11=19$ stickers, and he gives one sticker to each friend, so he has 19 friends.
Solution II. Let Petya have $x$ friends. Then $5 x+8=6 x-11$, from which $x=19$.
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7-3. A secret object is a rectangle measuring $200 \times 300$ meters. Outside the object, there is a guard at each of the four corners. An intruder approached the perimeter of the secret object from the outside, and all the guards ran to him by the shortest paths along the external perimeter (the intruder remained in place). Three of the guards ran a total of 850 meters to reach the intruder. How many meters did the fourth guard run to reach the intruder?
|
Answer: 150.
Solution. Note that no matter where the violator is, two guards in opposite corners will run a distance equal to half the perimeter in total.
Therefore, all four guards will run a total distance equal to the perimeter (two in diagonally opposite corners will run a total of half the perimeter, and the other two will also run half the perimeter). Therefore, the fourth guard will run a distance that complements 850 to the perimeter, which is 1000, i.e., 150 meters.
|
150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7-4. In a giraffe beauty contest, two giraffes, Tall and Spotted, made it to the final. 135 voters are divided into 5 districts, each district is divided into 9 precincts, and each precinct has 3 voters. The majority of voters on each precinct choose the winner in their precinct; in the district, the giraffe that wins the majority of precincts in the district wins; finally, the giraffe that wins the majority of districts is declared the winner of the final. The giraffe Tall won. What is the minimum number of voters who could have voted for him?
|
Answer: 30.
Solution: For High to win the final, he must win in 3 districts. To win in a district, High must win in 5 precincts of that district. In total, he needs to win in at least $3 \cdot 5=15$ precincts. To win in a precinct, at least 2 voters must vote for him. Therefore, at least 30 voters are needed.
Comment: In problems with questions like "what is the greatest" and "what is the least," the solution usually consists of two parts: estimation and example. Estimation is the proof that a greater (or lesser, depending on the question) answer cannot be achieved, and the example is the proof that the given answer can be achieved.
In the solution above, we did not provide an example (i.e., did not prove) that 30 voters are sufficient. For completeness, the solution is missing the phrase: "From the above reasoning, it is clear that 30 voters for High can be sufficient: 2 voters per 5 precincts in any 3 districts."
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7-5. In a row, there are 1000 toy bears. The bears can be of three colors: white, brown, and black. Among any three consecutive bears, there is a toy of each color. Iskander is trying to guess the colors of the bears. He made five guesses:
- The 2nd bear from the left is white;
- The 20th bear from the left is brown;
- The 400th bear from the left is black;
- The 600th bear from the left is brown;
- The 800th bear from the left is white.
It turned out that exactly one of his guesses is incorrect. What could be the number of the bear whose color Iskander did NOT guess correctly? Select all possible answers.
|
Answer: 20.
Solution: Since among any three consecutive bears there is a bear of each color, the numbers of all bears of a certain color have the same remainder when divided by 3. Indeed, let's look at the bears with numbers $n, n+1$, and $n+2$, as well as with numbers $n+1, n+2$, and $n+3$. In both cases, there will be bears of all three colors only if the bears with numbers $n$ and $n+3$ are of the same color.
At least one of the guesses "the 20th bear is brown" and "the 600th bear is brown" is incorrect: since the numbers 200 and 600 give different remainders when divided by 3. Also, at least one of the guesses "the 2nd bear is white" and "the 20th bear is brown" is incorrect: since the numbers 2 and 20 give the same remainder when divided by 3, therefore, the bears with these numbers must be of the same color. Thus, the guess "the 20th bear is brown" is definitely incorrect.
Comment: As follows from the solution, the situation described in the problem is possible: all bears with numbers giving a remainder of 1 when divided by 3 must be black, a remainder of 2 - white, and a remainder of 0 - brown. This is the only possible situation.
|
20
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7-6. In an ornithological park, there are several species of birds, a total of 2021 individuals. The birds sat in a row, and it turned out that between any two birds of the same species, there was an even number of birds. What is the smallest number of bird species that could have been there?
|
Answer: 1011.
Solution. Estimation. We will prove that there could not have been three or more birds of the same species. Indeed, suppose birds of the same species $A, B$, and $C$ sit in that exact order. Let there be $2x$ birds between $A$ and $B$, and $2y$ birds between $B$ and $C$, then there are $2x + 2y + 1$ birds between $A$ and $C$, which is an odd number of birds, contradicting the condition. Since there are no more than 2 birds of each species, there must be at least 1011 species.
Example. We will provide an example where 1011 species could have been present. For instance, the birds could sit in pairs: two of the first species, two of the second species, and so on up to the 1010th species, with one more bird of the 1011th species at the end. There are many other examples.
|
1011
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7-7. The recruits stood in a row, facing the same direction. Among them were three brothers: Peter, Nikolai, and Denis. In front of Peter were 50 people, in front of Nikolai 100, and in front of Denis 170. On the command "About face!" everyone turned to face the opposite direction. As a result, it turned out that in front of one of the brothers there are now four times as many people as in front of another. How many recruits could there be, including the brothers? List all possible options.
|
Answer: 211.
Solution. Let there be $x$ people in front of Peter, $y$ people in front of Nicholas, and $z$ people in front of Denis. There are three possible cases.
- $x=4 y$. Then $4 y+50=y+100$, from which $y$ is not an integer, which is impossible.
- $y=4 z$. Then $4 z+100=z+170$, from which $z$ is not an integer, which is impossible.
- $x=4 z$. Then $4 z+50=z+170$, from which $z=40$. In total, there are 211 recruits.
|
211
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-1. Two rectangles $8 \times 10$ and $12 \times 9$ are overlaid as shown in the figure. The area of the black part is 37. What is the area of the gray part? If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: 65.
Solution. The area of the white part is $8 \cdot 10-37=43$, so the area of the gray part is $12 \cdot 9-43=65$
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-4. In a giraffe beauty contest, two giraffes, Tall and Spotted, made it to the final. 105 voters are divided into 5 districts, each district is divided into 7 precincts, and each precinct has 3 voters. The majority of voters on each precinct choose the winner of their precinct; in a district, the giraffe that wins the majority of precincts wins the district; finally, the giraffe that wins the majority of districts is declared the winner of the final. The giraffe Tall won. What is the minimum number of voters who could have voted for him?
|
Answer: 24.
Solution. For High to win the final, he must win in 3 districts. To win a district, High must win in 4 precincts of that district. In total, he needs to win in at least $3 \cdot 4=12$ precincts. To win in a precinct, at least 2 voters must vote for him. Therefore, at least 24 voters are needed.
Comment. In problems with questions like "what is the greatest" and "what is the least," the solution usually consists of two parts: estimation and example. Estimation is the proof that a greater (or lesser, depending on the question) answer cannot be achieved, and the example is the proof that the given answer can be achieved.
In the solution above, we did not provide an example (i.e., did not prove) that 24 voters are sufficient. For completeness, the solution is missing the phrase: "From the above reasoning, it is clear that 24 voters for High can be sufficient: 2 voters per 4 precincts in any 3 districts."
|
24
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8-8. A chess tournament is held according to the following system: each of the 15 students from the "White Rook" school must play one game with each of the 20 students from the "Black Bishop" school, i.e., a total of 300 games should be played. At any given time, no more than one game is played.
After $n$ games, a spectator named Sasha, who has watched all the games and knows all the participants, exclaims: "I can definitely name one of the participants of the next game!" What is the smallest $n$ for which this could happen?
|
Answer: 280.
Solution: Estimation. Suppose fewer than 280 games have passed, i.e., more than 20 games remain. Then, among the participants of the "White Rook" school, there are at least two students who have not yet played all their games. In this case, Sasha cannot accurately name the participant of the next game from the "White Rook" school. Similarly, Sasha cannot name the participant of the next game from the "Black Bishop" school. Therefore, $n$ must be at least 280.
Example. We have just proven that if $n<280$, Sasha cannot name either of the participants of the next game. But is there a situation when $n=280$ where Sasha can still name someone who will play in the next game? It turns out, yes! For this, the 14 students of the "White Rook" school must play all their games, which is exactly 280 games. Then, in the next game, the remaining student from the "White Rook" school will definitely participate - and this is the one Sasha will name.
## 9th Grade
|
280
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-1. Segment $P Q$ is divided into several smaller segments. On each of them, a square is constructed (see figure).
What is the length of the path along the arrows if the length of segment $P Q$ is 73? If necessary, round the answer to 0.01 or write the answer as a common fraction.
|
Answer: 219.
Solution. Note that in each square, instead of going along one side, we go along three sides. Therefore, the length of the path along the arrows is 3 times the length of the path along the segment, hence the answer $73 \cdot 3=219$.
|
219
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-3. Arina wrote down all the numbers from 71 to 81 in a row without spaces, forming a large number 717273...81. Sofia started appending the next numbers to it (i.e., she first appended 82, then 83, ...). She stopped when the large number became divisible by 12. The last number she appended was $N$. What is $N$?
|
Answer: 88.
Solution. A number is divisible by 12 if and only if it is divisible by 3 and by 4. For a number to be divisible by 4, the number formed by its last two digits must also be divisible by 4. Therefore, the last number that Sofia writes must be divisible by 4.
The nearest number that is divisible by 4 is 84, but the number 71727374...84 has a sum of digits equal to 158, which means it is not divisible by 3. The next number that is divisible by 4 is 88. The sum of the digits of the number 71727374...88 is 216, which means the entire number is divisible by 3.
|
88
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-5. A circle is divided into 100 equal arcs by 100 points. Next to the points, numbers from 1 to 100 are written, each exactly once. It turns out that for any number $k$, if a diameter is drawn through the point with the number $k$, then the number of numbers less than $k$ on either side of this diameter will be equal. What number can be written at the point diametrically opposite the point with the number $83?$
|
Answer: Only 84.
Solution: Consider the odd number $2 m+1$. Let's mentally discard it and the number diametrically opposite to it. According to the condition, among the remaining numbers, all numbers less than $2 m+1$ are divided into two groups of equal size. Therefore, among the remaining numbers, there is an even number of numbers less than $2 m+1$. In total, the number of numbers less than $2 m+1$ is also even - there are $2 m$ of them. From this, it follows that the number diametrically opposite to $2 m+1$ must be greater than it!
Then, opposite 99 can only be 100, opposite 97 - only 98 (since the numbers 99 and 100 are already opposite each other), opposite 95 - only 96 (since all larger numbers are already opposite each other), and so on. Therefore, opposite 83 stands 84.
Comment: Note that any variant where for all $m=1,2, \ldots, 50$ the numbers $2 m-1$ and $2 m$ are in diametrically opposite positions, is suitable.
|
84
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-6. Petya wants to place 99 coins in the cells of a $2 \times 100$ board so that no two coins are in cells that share a side, and no more than one coin is in any cell. How many ways are there to place the coins?
|
Answer: 396.
Solution. Note that there will be exactly 1 empty column. Then, to the left of it, there are exactly two ways to arrange the tiles, and to the right of it, there are also exactly two ways to arrange the tiles.
In total, there are $2 \cdot 2=4$ ways to arrange the tiles for a fixed empty column - each way on the left side of the picture must be combined with each way on the right side of the picture.
However, this reasoning does not work when the empty column is the farthest left or the farthest right - in those cases, there are not 4, but only 2 ways to arrange the tiles. Therefore, in total, there will be $98 \cdot 4 + 2 + 2 = 396$ ways.
|
396
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-8. All digits in the notation of 6-digit natural numbers $a$ and $b$ are even, and in the notation of any number between them, there is an odd digit. Find the largest possible value of the difference $b-a$.
---
Translation:
9-8. All digits in the notation of 6-digit natural numbers $a$ and $b$ are even, and in the notation of any number between them, there is an odd digit. Find the largest possible value of the difference $b-a$.
|
Answer: 111112.
Solution. Estimation. We will prove that to a 9-digit number $a$, less than 888 888, all digits of which are even, one can add a number not exceeding 111112 so that all its digits will again be even. If among the digits of the number $a$, except for the first one, there is a digit less than 8, then it can be increased by 2. Otherwise, the number has the form $A 88888$. Since $A<8$, one can add 111112 to it, and the result will be $(A+2) 00000$. If, however, $a=888$ 888, then all larger 6-digit numbers contain an odd digit, so there cannot be a suitable number $b$ among them.
Example. It remains to check that the difference 111112 is possible. The reasoning from the estimation suggests that examples could be the numbers $a=288888$ and $b=400$ 000. Indeed: any number between them has either the digit 3 or the digit 9.
|
111112
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-1. There are several bowls on the table, each containing several grapes. Different bowls may contain different numbers of grapes. If 12 bowls are each added 8 more grapes, the average number of grapes in all the bowls will increase by 6. How many bowls are on the table
|
Answer: 16.
Solution: Let the number of bowls be $n$. The total number of berries increased by $12 \cdot 8=96$. Since the average number of berries increased by 6, the total number of berries should have increased by $6n$. Therefore, $6n=96$, from which we find $n=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-2. The large rectangle in the figure consists of 20 identical smaller ones. The perimeter of figure $A$ is 56 cm, the perimeter of figure $B$ is 56 cm. What is the perimeter of figure $C$? Provide your answer in cm.
| | $A$ | | |
| ---: | :--- | :--- | :--- |
| | | | |
| | | | |
| $B$ | | | $C$ |
| | | | |
|
Answer: 40.
Solution: Let the horizontal size of the rectangle be $x$, and the vertical size be $y$. From the condition, we get the system of equations
$$
\left\{\begin{array}{l}
6 x+2 y=56 \\
4 x+6 y=56
\end{array}\right.
$$
We need to find what $2 x+6 y$ equals. Solving the system, we find $x=8, y=4$, from which $2 x+6 y=40$.
|
40
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-4. Initially, a natural number was displayed on the calculator screen. Each time, Olya added a natural number to the current number \( n \) on the calculator screen, which \( n \) did not divide. For example, if the screen showed the number 10, Olya could add 7 to get 17.
Olya repeated this operation five times, and the number 200 appeared on the screen. What is the largest initial number for which this could have happened?
|
Answer: 189.
Solution. Estimation. Note that Olya increased the number on the screen by at least 2 each time, because any number is divisible by 1. If Olya added two five times, the initial number would have been 190, and it would not have been possible to add two to it. Therefore, Olya must have added a number greater than two at least once. Consequently, she increased the number by at least 11 in total, from which $n \leqslant 189$.
Estimation. For $n=189$, this is possible: $200=189+2+2+2+2+3$.
|
189
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-5. For each natural number from 1 to 999, Damir subtracted the last digit from the first digit and wrote all 1000 differences on the board. For example, for the number 7, Damir wrote the number 0 on the board, for the number 105 he wrote $(-4)$, and for the number 61 he wrote 5.
What is the sum of all the numbers on the board
|
Answer: 495.
Solution: Note that for single-digit numbers, zeros are recorded on the board, which do not affect the sum. For numbers where the first and last digit are the same, zeros are also recorded on the board, which do not affect the sum.
Almost all other numbers can be paired: a number and the number obtained by swapping the first and last digit. For example, 17 and 71, 122 and 221, 103 and 301. If for one number in the pair, the number $x$ is recorded on the board, then for the other number in the pair, $-x$ is recorded, resulting in a zero contribution to the sum of all numbers on the board from $x$ and $-x$.
It remains to consider the numbers that do not have a pair. These are numbers ending in 0! For these, everything needs to be calculated directly. Numbers of the form $a 0$ (where $a=1,2, \ldots 9$) will contribute $a$ to the calculated sum, and there is one such number for each $a$. Numbers of the form $a * 0$ will also contribute $a$ to the calculated sum, and there are 10 such numbers. In total, there are 11 addends, each equal to $a$, in the calculated sum.
Therefore, the calculated sum is
$$
1 \cdot 11+2 \cdot 11+3 \cdot 11+\ldots+9 \cdot 11=45 \cdot 11=495
$$
|
495
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-7. Parabola $\Pi_{1}$ with branches directed upwards passes through points with coordinates $(10,0)$ and $(13,0)$. Parabola $\Pi_{2}$ with branches directed upwards also passes through the point with coordinates $(13,0)$. It is also known that the vertex of parabola $\Pi_{1}$ bisects the segment connecting the origin and the vertex of parabola $\Pi_{2}$. At what abscissa does parabola $\Pi_{2}$ intersect the $O x$ axis again?
|
Answer: 33.
Solution. We will use the following fact twice: if $x_{1}$ and $x_{2}$ are the x-coordinates of the points where the parabola intersects the x-axis, then the x-coordinate of the vertex is $\frac{x_{1}+x_{2}}{2}$ (the x-coordinate of the vertex is the midpoint of the segment with endpoints $x_{1}$ and $x_{2}$).
Applying this fact, we get that the x-coordinate of the vertex of the parabola $\Pi_{1}$ is $\frac{10+13}{2}=11.5$, and then the x-coordinate of the vertex of the parabola $\Pi_{2}$ is $2 \cdot 11.5$.
If we denote by $t$ the sought x-coordinate of the second point of intersection of the parabola $\Pi_{2}$ with the x-axis, then, applying the fact again, we get $\frac{13+t}{2}=2 \cdot 11.5$, from which $t=33$.
Comment. The condition that the branches of the parabolas are directed upwards was not used in the solution.
|
33
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10-8. In trapezoid $A B C D$, the bases $A D$ and $B C$ are 8 and 18, respectively. It is known that the circumcircle of triangle $A B D$ is tangent to the lines $B C$ and $C D$. Find the perimeter of the trapezoid.
|
Answer: 56.
Solution. Let's make the following observation. Through point $B$ on the circle, a line passes parallel to the chord $AD$. It is clear that then $B$ is the midpoint of the arc $AD$, that is, $BA = BD$ (indeed, $\angle 1 = \angle 2$ as alternate interior angles, $\angle 1 = \angle 3$ by the theorem on the angle between a tangent and a chord, therefore, $\angle 2 = \angle 3$).
Next, from point $C$, two tangents are drawn to the circle; hence, they are equal: $CD = CB = 18$.
It remains to find the length of side $AB$, which we will denote as $x$. For this, note that $\angle CBD = \angle BDA$ as alternate interior angles. Therefore, isosceles triangles $CBD$ and $BDA$ are similar as isosceles triangles with equal angles at the base. From the similarity, we get $\frac{x}{8} = \frac{18}{x}$, from which $x = 12$. Thus, the perimeter of the trapezoid is $12 + 18 + 18 + 8 = 56$.
## 11th Grade
|
56
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11-3. Point $O$ is the center of the circle. What is the value of angle $x$ in degrees?
|
Answer: 58.
Solution. Angle $ACD$ is a right angle since it subtends the diameter of the circle.
Therefore, $\angle CAD = 90^{\circ} - \angle CDA = 48^{\circ}$. Also, $AO = BO = CO$ as they are radii of the circle, so $\angle OCA = \angle ACO = 48^{\circ}$ and $x = \angle OBC = \angle OCB = 48^{\circ} + 10^{\circ} = 58^{\circ}$.
|
58
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11-4. At first, a natural number was displayed on the calculator screen. Each time, Tanya added to the current number \( n \) on the screen a natural number that \( n \) did not divide. For example, if the screen showed the number 10, Tanya could add 7 to get 17.
Tanya repeated this operation five times, and the number 100 appeared on the screen. What is the largest initial number for which this could have happened?
|
Answer: 89.
Solution. Estimation. Note that Tanya increased the number on the screen by at least 2 each time, because any number is divisible by 1. If Tanya added two five times, the initial number would have been 90, and it would not have been possible to add two to it. Therefore, Tanya must have added a number greater than two at least once. Consequently, she increased the number by at least 11 in total, from which $n \leqslant 89$.
Estimation. For $n=89$, this is possible: $100=89+2+2+2+2+3$.
|
89
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11-6. Petya uses all possible ways to place the signs + and - in the expression $1 * 2 * 3 * 4 * 5 * 6$ in the places of the asterisks. For each arrangement of the signs, he calculates the resulting value and writes it on the board. On the board, some numbers may appear multiple times. Petya adds up all the numbers on the board. What is the sum obtained by Petya?
|
Answer: 32.
Solution. Note that each of the digits $2,3,4,5,6$ will contribute zero to Petya's sum: each will equally often appear with a + sign and with a - sign. The digit 1 will appear in all sums with a + sign as many times as there are addends in total. Since each of the asterisks can take two values, there will be $2^{5}=32$ addends. Therefore, Petya's sum is 32.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11-7. Given a parallelogram $A B C D$, where $\angle B=111^{\circ}$ and $B C=B D$. On the segment $B C$, a point $H$ is marked such that $\angle B H D=90^{\circ}$. Point $M$ is the midpoint of side $A B$. Find the angle $A M H$. Give your answer in degrees.
|
Answer: $132^{\circ}$.
Solution. Note that $\angle D M B=90^{\circ}$, since $D A=D B$, and in the isosceles triangle $B D A$, the median $D M$ is also the altitude. Since angles $D H B$ and $D M B$ are right angles, points $M, B, H$, and $D$ lie on the same circle. It is clear that we need to find the angle $D M H$, as $\angle A M H=\angle D M H+90^{\circ}$.
Angles $D M H$ and $D B H$ are equal, as these angles subtend the same arc $D H$ in the circle. Therefore, we need to find the angle $D B H$. This can be easily found from the isosceles triangle $D B C$ if we know the angle $C$ at the base of this triangle: $\angle D B C=180^{\circ}-2 \angle C$. But the angle $C$ is equal to $180^{\circ}-111^{\circ}=69^{\circ}$. Thus, $\angle D B C=42^{\circ}$, and $\angle A M H=\angle D M H+90^{\circ}=42^{\circ}+90^{\circ}=132^{\circ}$.
|
132
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11-8. In a caravan, there are 100 camels, both one-humped and two-humped, and there are at least one of each. If you take any 62 camels, they will have at least half of the total number of humps in the caravan. Let $N$ be the number of two-humped camels. How many values (in the range from 1 to 99) can $N$ take?
|
Answer: 72.
Solution. If there were $N$ two-humped camels in total, then there were $100-N$ one-humped camels, and there were $100+N$ humps in total. Let's line up the camels: first the one-humped ones, and then the two-humped ones. It is clear that if the condition for 62 camels is met for the first 62 camels, then it is met for any 62 camels. There are two possible cases: the first 62 camels are all one-humped, or there is at least one two-humped camel among them.
1) Suppose the first 62 camels are one-humped. Then, by the condition, $62 \geqslant \frac{1}{2}(100+N)$, from which $N \leqslant 24$.
2) Among the first 62 camels, there are two-humped ones, let their number be $y$. Then, by the condition, $62+y \geqslant \frac{1}{2}(100+38+y)$, from which $y \geqslant 14$. Then $N \geqslant 14+38=52$.
Thus, the number of two-humped camels lies in the ranges from 1 to 24 inclusive or from 52 to 99 inclusive. This gives $24+48=72$ options.
|
72
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1.
When multiplying two two-digit numbers, a four-digit number $A$ is obtained, where the first digit matches the second, and the second-to-last digit matches the last. Find the smallest $A$, given that $A$ is divisible by 51.
|
Answer: 1122.
Solution.
Notice that $A=\overline{x x y y}=x \cdot 11 \cdot 100+y \cdot 11=11 \cdot(100 x+y)$. Since 51 and 11 are coprime, then $100 x+y$ is divisible by 51. The minimum $x=1$, so $y=2$ (the only number from 100 to 109 divisible by 51 is 102).
|
1122
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Variant 1.
Find the number of four-digit numbers for which the last three digits form an increasing arithmetic progression (numbers cannot start with zero).
|
Answer: 180.
Solution: The difference of the progression cannot be greater than 4.
For $d=1$ - there are eight suitable progressions of digits: from 012 to 789.
For $d=2$ - six: from 024 to 579.
For $d=3$ - four: from 036 to 369.
For $d=4$ - two: 048 and 159.
In total, $8+6+4+2=20$ options for the last three digits. By choosing the first digit in nine ways for each of them, we get the answer $20 \cdot 9=180$.
|
180
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Variant 1.
Find the ratio $\frac{16 b^{2}}{a c}$, given that one of the roots of the equation $a x^{2}+b x+c=0$ is 4 times the other.
|
Answer: 100.
Solution.
By Vieta's theorem, $-\frac{b}{a}=x_{1}+x_{2}=5 x_{2}$ and $\frac{c}{a}=x_{1} \cdot x_{2}=4 x_{2}^{2}$. Express $x_{2}=-\frac{b}{5 a}$ from the first equation and substitute it into the second: $\frac{c}{a}=\frac{4 b^{2}}{25 a^{2}}$. Then find $\frac{b^{2}}{a c}=\frac{25}{4}$.
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Variant 1.
It is known that
$$
\frac{1}{\cos (2022 x)}+\operatorname{tg}(2022 x)=\frac{1}{2022}
$$
Find $\frac{1}{\cos (2022 x)}-\operatorname{tg}(2022 x)$.
|
Answer: 2022.
Solution 1.
\[
\begin{aligned}
& \frac{1}{\cos 2A} + \tan 2A = \frac{1 + \sin 2A}{\cos 2A} = \frac{\cos^2 A + \sin^2 A + 2 \sin A \cdot \cos A}{\cos^2 A - \sin^2 A} = \frac{(\cos A + \sin A)^2}{(\cos A - \sin A)(\cos A + \sin A)} = \frac{\cos A + \sin A}{\cos A - \sin A} \\
& \frac{1}{\cos 2A} - \tan 2A = \frac{1 - \sin 2A}{\cos 2A} = \frac{\cos^2 A + \sin^2 A - 2 \sin A \cdot \cos A}{\cos^2 A - \sin^2 A} = \frac{(\cos A - \sin A)^2}{(\cos A - \sin A)(\cos A + \sin A)} = \frac{\cos A - \sin A}{\cos A + \sin A}
\end{aligned}
\]
Therefore, if \(\frac{1}{\cos (2022x)} + \tan (2022x) = \frac{1}{2022}\), then \(\frac{1}{\cos (2022x)} - \tan (2022x) = 2022\).
Solution 2.
Consider the product \(\left(\frac{1}{\cos \alpha} + \tan \alpha\right)\left(\frac{1}{\cos \alpha} - \tan \alpha\right) = \frac{1}{\cos^2 \alpha} - \tan^2 \alpha = 1\). Therefore, the desired expression is 2022.
|
2022
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 6. Variant 1.
In the district, there are three villages $A, B$, and $C$ connected by dirt roads, with any two villages being connected by several (more than one) roads. Traffic on the roads is two-way. We will call a path from one village to another either a road connecting them or a chain of two roads passing through a third village. It is known that villages $A$ and $B$ are connected by 34 paths, and villages $B$ and $C$ by 29 paths. What is the maximum number of paths that can connect villages $A$ and $C$?
|
Answer: 106.
Solution.
Let there be $k$ roads between cities $A$ and $B$, $m$ roads between cities $B$ and $C$, and $n$ roads between cities $A$ and $C$. Then the number of paths from $A$ to $B$ is $k + mn$, and the number of paths from $B$ to $C$ is $m + kn$. We have the system of equations $k + mn = 34$, $m + kn = 29$, where the unknowns are natural numbers greater than 1. Subtracting the second equation from the first, we get: $(m - k)(n - 1) = 5$. We need to check all divisors of 5: 1 and 5. Therefore, $n = 2$ or $n = 6$. For each of them, we find $m$ and $k$ by solving the corresponding system of linear equations. If $n = 2$, then $k = 8$ and $m = 13$. If $n = 6$, then $k = 4$ and $m = 5$. The number of paths connecting cities $A$ and $B$ is $n + km$. In the first case, $n + km = 2 + 8 \cdot 13 = 106$, and in the second case, $-n + km = 6 + 4 \cdot 5 = 26$. Therefore, the desired answer is 106.
|
106
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1.
103 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there are at least two even numbers. What is the minimum number of even numbers that can be in the entire circle?
|
Answer: 42.
Solution.
We will show that there will be 3 consecutive numbers, among which there are at least 2 even numbers. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them there are at least 6 even numbers. But these 15 numbers can also be divided into 5 sets of 3 consecutive numbers. Therefore, by the Pigeonhole Principle, in one of these triplets there are at least 2 even numbers.
Let's fix these 3 numbers. Among them, there are at least 2 even numbers. The remaining 100 numbers can be divided into 20 sets of 5 consecutive numbers. In each such set, there will be at least 2 even numbers. Thus, the total number of even numbers is at least $2 + 2 \cdot 20 = 42$. Such a situation is possible. We can number the numbers in a circle. The even numbers can be those with numbers $1, 2, 6, 7, \ldots, 96, 97, 101, 102$. (Other examples are also possible.)
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.4. In a table containing $A$ columns and 100 rows, natural numbers from 1 to $100 \cdot A$ were written in ascending order, starting from the first row. The number 31 is in the fifth row. In which row is the number 100?
|
Answer: in the 15th row.
Solution. From the condition, it follows that $A \leq 7$, since for $A \geq 8$ the number 31 would be located before the fifth row. Similarly, we get that $A \geq 7$, otherwise the number 31 would be located after the fifth row. Therefore, $A=7$. Since $100=7 \cdot 14+2$, the number 100 is located in the 15th row.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.5. In the village of Matitika, along a straight road, five friends live in the following order: Alya, Bella, Valya, Galina, and Dilya. Each of them found the sum of the distances (in meters) from her house to the houses of the others. Bella named the number 700, Valya - 600, Galina - 650. How many meters are there between Bella's and Galina's houses?
|
Answer: 150 meters.
Solution. Let's denote the houses of the friends with the letters A, B, V, G, D.
It is easy to see that the total distance from B to the other houses is AB + 3BV + 2VG + GD, and from V to the other houses is AB + 2BV + 2VG + GD. These values differ by BV, so the distance between the houses B and V is $700-600=100$ meters.
Similarly, the distance between the houses V and G is $650-600=50$ meters.
Therefore, the distance between the houses B and G is $100+50=150$ meters.
Comment. By similar reasoning, the following result can be obtained: if Bella named the number $x$, Vanya $-y$, Galya $-z$, then the distance between Bella's and Galya's houses is $x+z-2y$ meters.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.6. In a convex quadrilateral $A B C D$, side $B C$ is half the length of $A D$. Diagonal $A C$ is perpendicular to side $C D$, and diagonal $B D$ is perpendicular to side $A B$. Find the larger acute angle of this quadrilateral, given that the smaller one is $36^{\circ}$.
|
Answer: $84^{\circ}$.
Solution. Let point $M$ be the midpoint of side $A D$. Since angles $A B D$ and $A C D$ are right angles, angles $B$ and $C$ of quadrilateral $A B C D$ are obtuse, and angles $A$ and $D$ are acute, as angles of right triangles $A B D$ and $A C D$. Let $\angle A=\alpha$ be the given angle, and $\angle D=\beta$ be the angle to be found.
Triangles $A B D$ and $A C D$ are right triangles, so their medians to $A D$ are equal to half the hypotenuse, thus $A M=B M=C M=D M=B C$. From this, it follows that (1) $\angle A B M=\alpha$ (since $\triangle A M B$ is isosceles), (2) $\angle D C M=\beta$ (since $\triangle C M D$ is isosceles), (3) $\angle C B M=\angle B C M=60^{\circ}$ (since $\triangle B M C$ is equilateral).
The sum of the angles of quadrilateral $A B C D$ is $360^{\circ}$, so the equation $\alpha+\left(\alpha+60^{\circ}\right)+\left(\beta+60^{\circ}\right)+\beta=360^{\circ}$ holds, from which it follows that $\alpha+\beta=120^{\circ}$ or $\beta=120^{\circ}-\alpha$.
In this case, if $\alpha=36^{\circ}$, then $\beta=120^{\circ}-\alpha=120^{\circ}-36^{\circ}=84^{\circ}$.
|
84
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.7. In the city of Bukvinsk, people are acquainted only if their names have the same letters, otherwise - they are not. Several residents of Bukvinsk were asked how many acquaintances they have in the city. Martin said 20, Klim - 15, Inna - 12, Tamara - 12. What did Camilla answer?
|
Answer: 15 acquaintances
Solution. Note that all five students listed in the condition are acquainted with each other. Therefore, Martin has 16 acquaintances outside this group, while Inna and Tamara each have 8. However, all of Inna's acquaintances are acquainted with Martin, and all of Tamara's acquaintances are also, with no other acquaintances for Martin. Since $16=8+8$, there are no common acquaintances of Inna and Tamara outside our group. Their common acquaintances can only be people with the letter A in their name, so there are no such people outside the considered group. In this case, Camilla is acquainted with exactly the same people as Klim (and they are also acquainted with each other), meaning she has 15 acquaintances.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.8. In the cells of an $8 \times 8$ board, natural numbers from 1 to 64 (each appearing once) are placed such that numbers differing by 1 are in adjacent side-by-side cells. What is the smallest value that the sum of the numbers on the diagonal from the bottom left to the top right corner can take?
|
Answer: 88.
Solution. We will color the cells of the board in a checkerboard pattern. Let the considered diagonal be black. We will move through the cells according to the numbers placed. Consider the moment when we occupy the last cell on the diagonal. Before this, we must have visited all the cells on one side of it, so we have visited at least 19 black cells (7 on the diagonal and 12 on this side). Since white and black cells alternate during the traversal of the board, we have also visited at least 19 white cells, so the number of the current cell is no less than \(2 \cdot 19 + 1 = 39\). Since all cells on the diagonal have numbers of the same parity, the numbers in the other cells are no less than \(1, 3, \ldots, 13\). Thus, the sum of the numbers on the diagonal is no less than \(1 + 3 + \ldots + 13 + 39 = 88\). The diagram shows that this value is achievable.
| 58 | 57 | 48 | 47 | 42 | 41 | 40 | 39 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 59 | 56 | 49 | 46 | 43 | 12 | 13 | 38 |
| 60 | 55 | 50 | 45 | 44 | 11 | 14 | 37 |
| 61 | 54 | 51 | 8 | 9 | 10 | 15 | 36 |
| 62 | 53 | 52 | 7 | 18 | 17 | 16 | 35 |
| 63 | 4 | 5 | 6 | 19 | 20 | 21 | 34 |
| 64 | 3 | 26 | 25 | 24 | 23 | 22 | 33 |
| 1 | 2 | 27 | 28 | 29 | 30 | 31 | 32 |
|
88
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 2.1. Condition:
In the campus, rooms are numbered consecutively $1,2,3,4 \ldots, 10,11, \ldots$ For room numbering, stickers with digits were purchased, with the digits 1, 2, and 3 being purchased in equal quantities, and the digit 5 being purchased three more than the digit 6. How many rooms are there in the campus, given that there are fewer than 100?
|
# Answer: 66
## Solution.
In each decade up to the sixth, the digits "5" and "6" are equal, so there are at least 50 rooms. Since the digits "1", "2", and "3" are equal, they must appear in each decade, meaning the number of rooms will be at least 53. Then the digit "5" will be four more than the digit "6". Therefore, the minimum number of rooms will reach 60, and the digit "5" will be ten more than the digit "6", as there are 11 fives among the numbers from 50 to 59 (but note one digit "6" in the number 56). Among the numbers from 60 to 66, there will be seven digits "6" and one digit "5", so the difference will decrease by 6, and the number of digits "5" will exceed the number of digits "6" by exactly four. Note that the difference in the number of digits "5" and "6" will decrease further until they are equal at number 69. If we take numbers from the next decades, the difference in the number of digits "5" and "6" will not exceed one when reaching the numbers 75, 76, or 85, 86, or 95, 96. Therefore, there is only one answer.
|
66
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3.1. Condition:
Vanya bought balloons, red ones were 7 times more than blue ones. While Vanya was walking home, some of the balloons burst, and among the burst balloons, there were 3 times fewer red ones than blue ones. What is the smallest number of balloons Vanya could have bought?
|
# Answer: 24
## Solution.
Notice that at least one red balloon has burst, which means at least three blue balloons have burst. Therefore, there are at least three blue balloons, which means there are at least $7 \cdot 3=21$ red balloons, so the total number of balloons must be at least 24. For example, Vanya could have been carrying 21 red balloons and 3 blue ones, and 1 red balloon and all 3 blue balloons could have burst.
#
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 6.1. Condition:
Petya thought of a natural number and wrote down the sums of each pair of its digits on the board. After that, he erased some of the sums, and the numbers $2,0,2,2$ remained on the board. What is the smallest number Petya could have thought of?
|
Answer: 2000
## Solution.
Since among the sums there is a 0, the number must contain at least two digits 0. If the number has only three digits, there will be three pairwise sums, while the condition states there are at least four. Therefore, the number must have at least four digits. A sum of 2 can be obtained either as $1+1$ or as $2+0$. If the number contains the digit 2, then it will have four digits, three of which are $-2,0,0$, which means the number cannot be less than 2000. At the same time, 2000 meets the condition. If 2 is obtained as $1+1$, then at least four digits $-1,1,0,0$ are required. With these, it is impossible to get pairwise sums of $2,0,2$, 2, which means there must be at least 5 digits, i.e., the number will definitely be greater than 2000.
|
2000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 7.1. Condition:
A carpenter took a wooden square and cut out 4 smaller equal squares from it, the area of each of which was $9 \%$ of the area of the larger one. The remaining area of the original square was $256 \mathrm{~cm}^{2}$.
Find the side of the original square. Express your answer in centimeters.
|
Answer: 20
Solution.
Let the side of the larger square be 10x. The area of the smaller square is $9 \%$ of the area of the larger square, which is $9 / 100 \cdot 100 x^{2}=9 x^{2}$. After removing four smaller squares, the remaining area is $100 x^{2}-4 \cdot 9 x^{2}=64 x^{2}=256 \text{ cm}^{2}$. Therefore, $x^{2}=4$. Hence, $x=2$ cm, and the entire side is $10 x=20$ cm.
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 7.2 (7 points)
There are 22 kg of blueberries in a box. How can you measure out 17 kg of blueberries using a two-kilogram weight and a balance scale in two weighings.
|
# Solution:
Place the weight on one pan and balance the scales using all the blueberries $12=10+2$, then divide the 10 kg into equal parts of 5 kg each. And we get $12+5=17$ kg.
| Criteria | Points |
| :--- | :---: |
| Correct algorithm of actions | 7 |
| Incorrect solution | 0 |
Answer: 17 kg.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 7.3 (7 points)
A piece has fallen out of a dictionary, the first page of which is numbered 213, and the number of the last page is written with the same digits in some other order. How many pages are in the missing piece?
|
# Solution:
The number of the last page is 312 (it must be even). Then the number of pages is 312 - 212 = 100.
| Criteria | Points |
| :--- | :---: |
| Correct solution | 7 |
| Calculated 312 - 213 + 1 = 100, without explaining why 1 is added | 6 |
| Obtained the answer 99 | 4 |
| Noted the different parity of the first and last pages, without further progress | 2 |
| Incorrectly determined the number of the last page | 0 |
Answer: 100 pages (50 sheets)
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.6. On the board, 2011 numbers are written. It turns out that the sum of any three written numbers is also a written number. What is the smallest number of zeros that can be among these numbers?
|
Answer: 2009.
Solution. Let $n=2011$. Arrange the written numbers in non-decreasing order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}$. Since the number $a_{1}+a_{2}+a_{3}$ is written, then $a_{1}+a_{2}+a_{3} \geqslant a_{1}$, hence $a_{2}+a_{3} \geqslant 0$. Similarly, we get $a_{n-2}+a_{n-1}+a_{n} \leqslant a_{n}$, hence $a_{n-2}+a_{n-1} \leqslant 0$. Therefore, $0 \geqslant a_{n-2}+a_{n-1} \geqslant a_{2}+a_{3} \geqslant 0$; thus, $a_{2}+a_{3}=a_{n-2}+a_{n-3}=0$. Since $a_{2} \leqslant a_{3} \leqslant a_{n-2} \leqslant a_{n-1}$, it follows that $a_{2}=a_{3}=a_{n-2}=a_{n-1}$, and therefore, $a_{2}=a_{3}=\ldots=a_{n-1}=0$. Thus, among the written numbers, there are at least 2009 zeros.
An example with 2009 zeros and the numbers $1, -1$ shows that there can be exactly 2009 zeros.
Comment. Only the answer - 0 points.
Provided an example showing that there can be exactly $2009-1$ points.
Proved only that there must be at least 2009 zeros - 5 points.
Only shown that $a_{2}+a_{3} \geqslant 0$ (or $\left.a_{n-1}+a_{n-2} \leqslant 0\right)-$ 1 point.
Only shown that among the written numbers, there are no more than two positive and/or no more than two negative - 3 points.
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10.8. A straight rod 2 meters long was sawn into $N$ sticks, the length of each of which is expressed in whole centimeters.
For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle?
(A. Magazinov)
|
Answer. $N=102$.
Solution. First solution. Let $N \leqslant 101$. Cut the stick into $N-1$ sticks of 1 cm each and one stick of $201-N$ cm. It is impossible to form a rectangle from this set, as each side of the rectangle is less than half the perimeter, and thus the stick of length $201-N \geqslant 100$ cm cannot be part of any side. Therefore, $N \geqslant 102$.
We will show that for $N=102$, the desired rectangle can be found. To do this, note that among all the sticks, there will be two sticks of 1 cm each. Indeed, if this were not the case, the total length of the sticks would be at least $2 \cdot 101 + 1 = 203$ cm, which is incorrect.
Set aside these two sticks. Let the lengths of the remaining sticks be $a_{1}, a_{2}, \ldots, a_{100}$ cm, then we have $a_{1} + a_{2} + \ldots + a_{100} = 198$. Among the 100 numbers $A_{1} = a_{1}, A_{2} = a_{1} + a_{2}, A_{3} = a_{1} + a_{2} + a_{3}, \ldots, A_{100} = a_{1} + a_{2} + \ldots + a_{100}$, there will be two that give the same remainder when divided by 99. Let these be $A_{k}$ and $A_{\ell}, k < \ell$. The number $A_{\ell} - A_{k}$ is strictly greater than zero and strictly less than 198, and it is divisible by 99. Thus, $A_{\ell} - A_{k} = 99 = a_{k+1} + a_{k+2} + \ldots + a_{\ell}$.
Thus, we have found several sticks with a total length of 99 cm. Set these aside as well. The remaining sticks also have a total length of 99 cm. Therefore, we can form a rectangle of $1 \times 99$ cm.
Second solution. We will provide another proof that for $N=102$, it is possible to form a rectangle.
Let the lengths of the sticks in the set, expressed in centimeters, be $a_{1}, a_{2}, \ldots, a_{102}$. We have $a_{1} + a_{2} + \ldots + a_{102} = 200$. Consider a circle of length 200 and divide it into 102 red points, forming arcs of lengths $a_{1}, a_{2}, \ldots, a_{102}$. These points are some 102 vertices of a regular 200-gon $T$ inscribed in this circle. The vertices of $T$ are paired into opposite pairs. There are 100 such pairs, and 102 red points, so among the red points, there will be two pairs of opposite points.
These two pairs of points divide the circle into two pairs of equal arcs. Thus, we have divided all the sticks into four groups $A, B, C, D$, where the total lengths in groups $A$ and $C$, as well as in groups $B$ and $D$, are equal. Therefore, it is possible to form a rectangle using each group to form one of its sides.
Comment. Only the correct answer - 0 points.
An example is provided showing that $N \geqslant 102 - 1$ point. It is proven that $N=102$ works, but its minimality is not justified - 5 points.
|
102
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. How many natural numbers exist such that the product of all digits of such a number, multiplied by their quantity, equals 2014?
|
2. Answer: 1008.
Solution. The number 2014 is divisible only by the digits 1 and 2, so the number can only contain the digits 1 and 2. Moreover, the digit 2 can only appear once. If the number consists only of 1s, there are 2014 of them, and there is only one such number. If there is a 2, then there are 1007 digits of 1, and the 2 can be in any position, resulting in 1007 such numbers. Hence the answer.
Grading criteria:
Correct answer with proper justification: 7 points. Incorrect answer with correct idea of justification: 2 points. Only the correct answer provided: 1 point.
3) Answer: AC can be 6 cm, 8 cm, 12 cm, 16 cm, 18 cm.
4) C is the midpoint of AB. There are two cases: D is the midpoint of AC or D is the midpoint of BC, in each case AC = 12 cm.
$\xrightarrow[A C=12]{A} B$
2) D is the midpoint of AB. There are two cases: C is the midpoint of AD or C is the midpoint of BD; in the first case AC = 6 cm, and in the second case AC = 18 cm.
3) D is the midpoint of AC and C is the midpoint of BD. In this case AC = 16 cm.
4) D is the midpoint of BC and C is the midpoint of AD. In this case AC = 8 cm.
BC and C is the midpoint of AD. In this case AC = 8 cm.
Grading criteria:
Correct answer and all necessary diagrams provided: 7 points. Four answers with justification provided: 4 points. Three answers with justification provided: 2 points. Only the correct answer provided with no justification: 1 point.
|
1008
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Find the value of the fraction
$$
\frac{2 \cdot 2020}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+2020}}
$$
|
Solution. Let's denote the denominator of the fraction by $q$. By repeatedly applying the formula for the sum of an arithmetic progression, we get that
$$
q=\frac{2}{1 \cdot 2}+\frac{2}{2 \cdot 3}+\frac{2}{3 \cdot 4}+\ldots+\frac{2}{2020 \cdot 2021}
$$
Now, using the identity $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$, we get
$$
q=\frac{2}{1}-\frac{2}{2}+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+\frac{2}{4}-\ldots+\frac{2}{2020}-\frac{2}{2021}=\frac{2 \cdot 2020}{2021}
$$
Substituting this expression into the original fraction and simplifying it, we get the answer: 2021.
|
2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Papa, Masha, and Yasha are going to school. While Papa takes 3 steps, Masha takes 5 steps. While Masha takes 3 steps, Yasha takes 5 steps. Masha and Yasha counted that together they made 400 steps. How many steps did Papa take?
|
Answer: 90 steps.
Solution. 1st method. Let's call the distance equal to 3 steps of Masha and 5 steps of Yasha a Giant's step. While the Giant makes one step, Masha and Yasha together make 8 steps. Since they made 400 steps together, the Giant would have made 400:8=50 giant steps in this time. If the Giant made 50 steps, then Masha made 150 steps. Now let's count them in "fives." 150 is 30 times 5 steps. This means that the father made 30 times 3 steps, that is, 90 steps.
2nd method. While Masha makes 3$\cdot$5=15 steps, the father makes $3 \cdot 3=9$ steps, and Yasha makes $5 \cdot 5=25$ steps. Together, in this time, Masha and Yasha will make $15+25=40$ steps. And while they make 400 steps, the father will also make 10 times more steps, that is, 9$\cdot$10=90 steps.
## 5th grade. Recommendations for checking.
Each problem is scored out of 7 points. Each score is an integer from 0 to 7. Some guidelines for checking are provided below. Naturally, the creators cannot foresee all cases. When evaluating a solution, it is important to determine whether the provided solution is generally correct (although it may have flaws) - in which case the solution should be scored no less than 4 points. Or if it is incorrect (although it may have significant progress) - in which case the score should not exceed 3 points.
|
90
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Dad, Masha, and Yasha are going to school. While Dad takes 3 steps, Masha takes 5 steps. While Masha takes 3 steps, Yasha takes 5 steps. Masha and Yasha counted that together they made 400 steps. How many steps did Dad take?
|
Answer: 90 steps.
Solution. 1st method. Let's call the distance equal to 3 steps of Masha and 5 steps of Yasha a Giant's step. While the Giant makes one step, Masha and Yasha together make 8 steps. Since they made 400 steps together, the Giant would have made 400:8=50 giant steps in this time. If the Giant made 50 steps, then Masha made 150 steps. Now let's count them in "fives." 150 is 30 times 5 steps. This means that the father made 30 times 3 steps, which is 90 steps.
2nd method. While Masha makes 3*5=15 steps, the father makes 3*3=9 steps, and Yasha makes 5*5=25 steps. Together, Masha and Yasha will make 15+25=40 steps in this time. And while they make 400 steps, the father will also make 10 times more steps, i.e., 9*10=90 steps.
In the museum, there are 16 halls arranged as shown in the diagram. In half of them, paintings are exhibited, and in the other half, sculptures. From any hall, you can go to any adjacent one (having a common wall). During any tour of the museum, the halls alternate: a hall with paintings - a hall with sculptures - a hall with paintings, and so on. The tour starts in hall A, where paintings are displayed, and ends in hall B.
a) Mark with crosses all the halls where paintings are displayed. Solution. See the diagram.
b) A tourist wants to visit as many halls as possible (travel from hall A to hall B), but visit each hall no more than once. What is the maximum number of halls he can visit? Draw one of his longest routes and prove that he could not visit more halls.
Answer: 15.
Solution. One of the possible routes is shown in the diagram. Let's prove that if the tourist wants to visit each hall no more than once, he cannot visit more than 15 halls. Note that the route starts in a hall with paintings (A) and ends in a hall with paintings (B). This means that the number of halls with paintings that the tourist has passed is one more than the number of halls with sculptures. Since the number of halls with paintings that the tourist could pass is no more than 8, the number of halls with sculptures is no more than 7. Therefore, the route cannot pass through more than
15 halls.
## 6th Grade. Recommendations for Checking.
Each problem is scored out of 7 points. Each score is an integer from 0 to 7. Some guidelines for checking are provided below. Naturally, the creators cannot foresee all cases. When evaluating a solution, it should be determined whether the provided solution is generally correct (although it may have flaws) - in which case the solution should be scored no less than 4 points. Or if it is incorrect (although it may have significant progress) - in which case the score should be no higher than 3 points.
|
90
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The clock shows half past eight. What is the angle between the hour and minute hands?
|
Answer: $75^{\circ}$.
Solution. At the moment when the clock shows half past eight, the minute hand points to the number 6, and the hour hand points to the midpoint of the arc between the numbers 8 and 9 (see figure). If two rays are drawn from the center of the clock to the adjacent
6 numbers on the clock face, the angle between them is $360^{\circ} \div 12 = 30^{\circ}$. The angle between the clock hands when they show half past eight is two and a half times larger. Therefore, it is $75^{\circ}$.
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a number palindromic if it reads the same from left to right as it does from right to left. For example, the number 12321 is palindromic.
a) Write any palindromic five-digit number that is divisible by 5.
b) How many five-digit palindromic numbers are there that are divisible by 5?
a) Solution. Any palindromic number ending in 5. For example, 51715.
b) Answer. 100.
|
Solution. A number that is divisible by 5 must end in 5 or 0. A palindromic number cannot end in 0, as then it would have to start with 0. Therefore, the first and last digits are 5. The second and third digits can be anything - from the combination 00 to the combination 99 - a total of 100 options. Since the fourth digit repeats the second, there will be 100 different numbers in total.
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Sasha, Lёsha, and Kolya started a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.)
|
Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m.
|
19
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The museum has 16 halls, arranged as shown in the diagram. In half of them, paintings are exhibited, and in the other half, sculptures. From any hall, you can go to any adjacent one (sharing a common wall). During any tour of the museum, the halls alternate: a hall with paintings - a hall with sculptures - a hall with paintings, and so on. The tour starts in hall A, where paintings are displayed, and ends in hall B.
a) Mark with crosses all the halls where paintings are exhibited. Solution. See the diagram.
b) A tourist wants to visit as many halls as possible (travel from hall A to hall B), but visit each hall no more than once. What is the maximum number of halls he can visit? Draw one of his longest routes and prove that he could not have visited more halls.
|
Answer: 15.
Solution: One of the possible routes is shown in the diagram. Let's prove that if a tourist wants to visit each hall no more than once, they will not be able to see more than 15 halls. Note that the route starts in a hall with paintings (A) and ends in a hall with paintings (B). Therefore, the number of halls with paintings that the tourist has passed is one more than the number of halls with sculptures. Since the number of halls with paintings that the tourist could pass is no more than 8, the number of halls with sculptures is no more than 7. Thus, the route cannot pass through more than
15 halls.
## 7th Grade. Recommendations for Checking.
Each problem is scored out of 7 points. Each score is an integer from 0 to 7. Below are some guidelines for checking. Naturally, the problem setters cannot foresee all cases. When evaluating a solution, it should be determined whether the provided solution is generally correct (although it may have flaws) - in which case the solution should be scored no less than 4 points. Or if it is incorrect (although it may have significant progress) - in which case the score should be no more than 3 points.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Sasha, Lёsha, and Kolya start a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.)
|
Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m.
|
19
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. We will call a number palindromic if it reads the same from left to right as it does from right to left. For example, the number 12321 is palindromic. How many five-digit palindromic numbers are there that are divisible by 5?
|
Answer: 100.
Solution: A number that is divisible by 5 must end in 5 or 0. A mirrored number cannot end in 0, as then it would have to start with 0. Therefore, the first and last digits are 5. The second and third digits can be anything from the combination 00 to the combination 99 - a total of 100 options. Since the fourth digit repeats the second, there will be a total of 100 different numbers.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.4. 2011 Warehouses are connected by roads in such a way that from any warehouse you can drive to any other, possibly by driving along several roads. On the warehouses, there are $x_{1}, \ldots, x_{2011}$ kg of cement respectively. In one trip, you can transport an arbitrary amount of cement from any warehouse to another warehouse along the road connecting them. In the end, according to the plan, there should be $y_{1}, \ldots, y_{2011}$ kg of cement on the warehouses respectively, with
$$
x_{1}+x_{2}+\ldots+x_{2011}=y_{1}+y_{2}+\ldots+y_{2011}
$$
What is the minimum number of trips required to fulfill the plan for any values of the numbers $x_{i}$ and $y_{i}$ and any road network?
(P. Karasev)
|
Answer. In 2010 trips.
Solution. First, we show that the plan cannot always be completed in 2009 trips. Suppose (with any road scheme) that initially all the cement is located at one warehouse $S$, and it needs to be evenly distributed to all warehouses. Then, cement must be delivered to each warehouse, except $S$, in some trip; it is clear that these 2010 trips are distinct, so the total number of trips must be at least 2010.
We need to show that the plan can always be completed in 2010 trips. We will prove by induction on $n$ that with $n$ warehouses, it is always possible to manage with $n-1$ trips. The base case for $n=1$ is obvious.
Let $n>1$. Since from any warehouse you can reach any other, there exists a route that passes through all warehouses (possibly more than once). Consider any such route and the warehouse $A$ that appears on this route the latest for the first time. Then, if we remove warehouse $A$ and all roads leading from it, it is still possible to reach any warehouse from any other (using the previous roads of the route).
We can assume that $A$ is the warehouse with number $n$. If $y_{n} \leqslant x_{n}$, then we will transport $x_{n}-y_{n}$ kg of cement from $A$ to any connected warehouse, and then forget about it and all roads leading from it. By the induction hypothesis, for the remaining warehouses, the plan can be completed in $(n-1)-1$ trips. In total, through $(n-2)+1$ trips, the required distribution of cement will be achieved.
If $y_{n}>x_{n}$, then we have already shown that from the distribution where the $i$-th warehouse has $y_{i}$ kg, it is possible to achieve the distribution where the $i$-th warehouse has $x_{i}$ kg in $n-1$ trips. By performing all these transports in reverse order (and in the opposite direction), we will implement the required plan.
Comment. Only the answer - 0 points.
Only an example is provided, showing that the plan cannot always be completed in 2009 trips - 2 points.
Only proved that 2010 trips are always sufficient - 5 points.
|
2010
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Pentagon $A B C D E$ is inscribed in circle $\omega$. Diagonal $A C$ is the diameter of circle $\omega$. Find $\angle B E C$, if $\angle A D B=20^{\circ}$.
|
Answer: $70^{\circ}$.
Solution. Fig. 4. Since $\angle A D B=20^{\circ}$, the arc $A B$ is $40^{\circ}$. Since $A C$ is a diameter, the arc $A B C$ is $180^{\circ}$, so the arc $B C$ is $180^{\circ}-40^{\circ}=140^{\circ}$. The angle $B E C$ subtends the arc $B C$, which means it is equal to $140^{\circ} / 2=70^{\circ}$.
## Criteria
4 p. Correct solution.
0 p. Only the correct answer.
Fig. $4:$ to problem 2
|
70
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. Inside the magician's hat, there live 100 rabbits: white, blue, and green. It is known that if 81 rabbits are randomly pulled out of the hat, there will definitely be three of different colors among them. What is the minimum number of rabbits that need to be taken out of the hat to ensure that there are definitely two of different colors?
#
|
# Answer: 61.
Solution. We will prove that if 61 rabbits are randomly pulled out of the hat, then among them there will be two of different colors. Suppose the opposite: let there be $a \geqslant 61$ rabbits of some color (for example, white). Let the second color by number of rabbits be blue. Then there are at least $\frac{100-a}{2}$ blue rabbits in the hat. Therefore, the total number of white and blue rabbits is at least
$$
a+\frac{100-a}{2}=\frac{100+a}{2} \geqslant \frac{161}{2}=80.5
$$
Since the number of rabbits is an integer, the total number of white and blue rabbits is at least 81, which contradicts the condition.
We will show that 60 rabbits may not be enough. Suppose there are 60 white rabbits and 20 blue and 20 green rabbits in the hat. Then it is possible that all the pulled-out rabbits are white. On the other hand, if 81 rabbits are pulled out, then among them there will definitely be rabbits of all three colors.
## Criteria
4 points. Correct solution.
3 points. Proven that 61 rabbits are enough.
1 point. Shown that 60 rabbits may not be enough.
0 points. Only the correct answer.
|
61
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In an $n \times n$ square, there are 1014 dominoes (each covering two adjacent cells). No two dominoes share any points (even corner points). For what smallest $n$ is this possible?
|
Answer. For $n=77$.
Solution. Attach four cells to the right and below each domino so that they form a $2 \times 3$ rectangle (if the domino is vertical) or a $3 \times 2$ rectangle (if it is horizontal). If the rectangles of two dominoes have at least one common cell, then the dominoes have a common point. Therefore, if 1014 dominoes, which do not have common points, fit into an $n \times n$ square, then all the constructed rectangles of 6 cells each must fit into a $(n+1) \times (n+1)$ square, obtained by adding a row below and a column to the right of the $n \times n$ square. Hence,
$$
(n+1)^{2} \geqslant 6 \cdot 1014=6084=78^{2}
$$
Thus, $n \geqslant 77$.
It remains to provide an example where this is possible for $n=77$. For this, place the dominoes horizontally in the first row, starting from the first cell, with a gap of one cell; a total of 26 dominoes will fit. Do the same with all other odd-numbered rows. The total number of dominoes will then be $26 \cdot 39=1014$; they obviously do not have common points.
|
77
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. (7 points) Lёsha did not hesitate to calculate the sum
$$
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017}
$$
and wrote it on the board. How many times is the digit 1 written in the final result?
|
Answer: 2013.
Solution. Transform the expression:
$$
\begin{aligned}
9+99+999+\ldots+\underbrace{9 \ldots 9}_{2017} & =(10-1)+(100-1)+\ldots+\left(10^{2017}-1\right)= \\
& =\underbrace{1 \ldots 10}_{2017}-2017=\underbrace{1 \ldots 1}_{2013} 09093 .
\end{aligned}
$$
Criteria. Any correct solution: 7 points.
It is shown that the original sum is equal to $\underbrace{1 \ldots 10}_{2017}-2017$, but further solution is missing or contains an arithmetic error: 5 points. Only the correct answer is provided: 1 point.
|
2013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. (7 points) Several sages lined up in a column. All of them wore either black or white caps. It turned out that among any 10 consecutive sages, there were an equal number of sages with white and black caps, while among any 12 consecutive sages - not an equal number. What is the maximum number of sages that could be
Answer: 15 sages.
|
Solution. We will prove that there cannot be more than 15 sages. Suppose the opposite, that there are at least 16 sages. Sequentially number all the sages. Consider nine consecutive sages. If we add one of the two neighboring sages to them, then among them there will be an equal number of sages with white and black hats, so on any sages, between whom there are 9 sages, hats of the same color are worn.
Without loss of generality, let the first sage wear a black hat. Then the eleventh sage also wears a black hat. If the twelfth sage wears a white hat, then among the first twelve sages there will be an equal number of white and black hats. Therefore, the twelfth sage wears a black hat, from which it follows that the second sage also wears a black hat. Similarly, considering the sages from the second to the eleventh, we get that the third and thirteenth sages wear black hats. Considering the sages from the third to the twelfth, we get that the fourth and fourteenth sages wear black hats. Similarly, the fifth and fifteenth, and the sixth and sixteenth sages wear black hats. But then among the first ten sages, the first six wear black hats, so there will be more black hats. Contradiction.
15 sages can be: let the first 5 and the last 5 sages wear black hats, and the remaining 5 wear white hats. It is easy to see that the condition of the problem will be satisfied.
Criteria. Any correct solution: 7 points.
Proved that there cannot be more than 15 sages, but no example is given of how to put the hats on 15 sages: 6 points.
Proved that on two sages, between whom stand 9 sages, hats of the same color are worn, but further reasoning is missing or incorrect: 2 points.
An example of arranging 15 sages that satisfies the condition is given, but it is not proved that more sages cannot be placed: 1 point.
Only the correct answer is given: 0 points.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Variant 1.
Petya has four cards with digits $1,2,3,4$. Each digit appears exactly once. How many natural numbers greater than 2222 can Petya make from these cards?
|
Answer: 16.
Solution: Let's find out how many different numbers can be formed from these cards: the first digit can be chosen in 4 ways, the second can be appended in 3 ways, the third in 2 ways, and the last one is uniquely determined. That is, a total of 24 different numbers can be obtained (it is also possible to verify this explicitly by listing all suitable numbers). Let's find out how many of the listed numbers will not suit us. These are all numbers starting with 1, there are 6 of them. And 2 more numbers: 2134, 2143, because if the number starts with 2, the second digit can only be 1. Then the suitable numbers are $24-6-2=16$.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Variant 1.
Nine integers from 1 to 5 are written on the board. It is known that seven of them are not less than 2, six are greater than 2, three are not less than 4, and one is not less than 5. Find the sum of all the numbers.
|
Answer: 26.
Solution. A number not less than 5 is 5. There is exactly one number 5. Three numbers are not less than 4, so exactly two numbers are equal to 4. Six numbers are greater than 2, meaning all of them are not less than 3. Therefore, exactly three numbers are equal to 3. Seven numbers are not less than 2, so one number is equal to 2. In total, there are 9 numbers, hence two numbers are equal to 1. The sum of all numbers is $1+1+2+3+3+3+4+4+5=26$.
|
26
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 3. Option 1.
Café "Buratino" operates 6 days a week with a day off on Mondays. Kolya said that from April 1 to April 20, the café was open for 17 days, and from April 10 to April 30, it was open for 18 days. It is known that he made a mistake once. What was the date of the last Tuesday in April?
|
Answer: 29
Solution: Since there are exactly 21 days from April 10 to April 30, each day of the week occurred exactly 3 times during this period. Therefore, this statement cannot be false. This means the first statement is false, and there were only 2 Mondays from April 1 to April 20 (there could not have been four, as at least 22 days would be needed for that). This could only happen if April 1 was a Tuesday. Therefore, the last Tuesday in April was the 29th.
|
29
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Variant 1.
A rectangle was cut into three rectangles, two of which have dimensions 9 m $\times$ 12 m and 10 m $\times$ 15 m. What is the maximum area the original rectangle could have had? Express your answer in square meters.
|
Answer: 330
Solution. Since the sizes of the two rectangles are fixed, in order for the original rectangle to have the maximum area, the third rectangle must have the largest area. Since the two given rectangles do not have the same sides, the largest area will be obtained by attaching the smaller side of one rectangle to the larger side of the other. This results in a rectangle of size $12 \times(9+15)$ or a rectangle of size $15 \times(10+12)$. The area of the first is 288, and the second is 330, so the answer is 330.
Variant 2.
A rectangle was cut into three rectangles, two of which have dimensions 8 m $\times 12$ m and 10 m $\times 14$ m. What is the maximum area the original rectangle could have had? Express your answer in square meters.
Answer: 308
|
330
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Variant 1.
In the addition example, where the numbers were written on cards, two cards were swapped, resulting in the incorrect expression: $37541+43839=80280$. Find the error and write down the correct sum.
|
Answer: 80380
Solution. Let's start checking the example from right to left. There are no errors in the units and tens place, but an error appears in the hundreds place. This means that one of the digits in this place $-2, 8$ or $5-$ is transposed.
Let's consider the following cases:
|
80380
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 6. Option 1.
Nезнайка named four numbers, and Ponchik wrote down all their pairwise sums on six cards. Then he lost one card, and the numbers left on the remaining cards were $270, 360, 390, 500, 620$. What number did Ponchik write on the lost card?
|
Answer: 530.
Solution. Let the original numbers be $a \leq b \leq c \leq d$. Suppose the card with the maximum sum is lost. Then this sum is $c+d$. Therefore, $a+b=270$ and $a+b+c+d>270+620=890$. On the other hand, the sum of all numbers on the cards is $3a+3b+2c+2d=270+360+390+500+620=2140$. We get that $2140>1780+2c+2d$, hence $c+d<180$. This is a contradiction. Similarly, we can prove that the card with the smallest numbers was not lost. Therefore, the sum of all numbers on the cards is $270+620=890$. All pairwise sums can be divided into the following groups: 1) $a+b, c+d$, 2) $a+c, b+d$, 3) $a+d, b+c$. The sum of the numbers in each group is 890. This is only possible if the lost card has the number $890-360=530$, since 620 and 270, 390 and 500 form pairs.
|
530
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Variant 1.
101 natural numbers are written in a circle. It is known that among any 3 consecutive numbers, there is at least one even number. What is the smallest number of even numbers that can be among the written numbers?
|
Answer: 34.
Solution: Consider any 3 consecutive numbers. Among them, there is an even number. Fix this number and its neighbor, and divide the remaining 99 into 33 sets of 3 consecutive numbers. In each such set, there is at least one even number. Thus, the total number of even numbers is no less than $1+33=34$. Such a situation is possible. Number the numbers in a circle. The even numbers can be those with numbers $1,4,7, \ldots, 100$
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# 8. Variant 1.
Identical coins are laid out on a table in the shape of a hexagon. If they are laid out so that the side of the hexagon consists of 2 coins, then 7 coins are enough, and if the side consists of 3 coins, then a total of 19 coins are required. How many coins are needed to build a hexagon with a side consisting of 10 coins?
|
Answer: 271.
Solution.
Method 1.
We will divide the coins into layers (contours) from the center. The first layer contains 1 coin, the second layer contains 6, and so on. Notice that each new layer contains 6 more coins than the previous one (if we remove the coins at the vertices, we get exactly as many coins as there were in the previous layer). Therefore, the total number of coins can be calculated using the formula $1+6+12+18+24+30+36+42+48+54=271$.
Method 2.
Let the side of the hexagon contain $n$ coins. Consider two opposite sides. Each of them consists of $n$ coins. To each of these sides, there are 2 more sides attached. But we have already counted one coin from each of these sides, so there are $n-1$ coins left on each of these sides, and two coins are counted twice. In total, there will be $2n + 4(n-1) - 2 = 6n - 6$. It remains to notice that each new layer of coins is built around the existing one. Then the total will be: $19 + (6 \cdot 4 - 6) + (6 \cdot 5 - 6) + (6 \cdot 6 - 6) + (6 \cdot 7 - 6) + (6 \cdot 8 - 6) + (6 \cdot 9 - 6) + (6 \cdot 10 - 6) = 19 + 18 + 24 + 30 + 36 + 42 + 48 + 54 = 271$.
|
271
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There were 10 chatterboxes sitting in a circle. At first, one of them told one joke, the next one clockwise told two jokes, the next one told three, and so on around the circle, until one of them told 100 jokes at once. At this point, the chatterboxes got tired, and the next one clockwise told 99 jokes, the next one told 98, and so on around the circle, until one of them told just one joke, and everyone dispersed. How many jokes did each of these 10 chatterboxes tell in total?
|
Answer: 1000 jokes.
Solution. Let's number the chatterboxes from 1 to 10 clockwise, starting with the one who told the first joke. Then for any pair of chatterboxes with numbers $k$ and $k+1 (1 \leq k \leq 9)$, the $(k+1)$-th chatterbox initially tells one more joke per round than the $k$-th, for 10 rounds. After the chatterboxes get tired, the $(k+1)$-th chatterbox tells one less joke than the $k$-th, also for 10 rounds. Thus, the first and second told the same number of jokes, the second and third - the same number, and so on. Therefore, all of them told the same number of jokes - each one-tenth of the total. The total number of jokes told is $1+2+\cdots+99+100+99+\cdots+2+1=(1+99)+(2+98)+\cdots+(99+1)+100=100 \cdot 100=10000$. Each of the 10 chatterboxes told $\frac{10000}{10}=1000$ jokes.
Comment. It is proven that each chatterbox told one-tenth of the total number of jokes - 4 points. The total number of jokes told is found - 3 points. For potentially useful ideas and approaches in the absence of a solution - 2-3 points.
|
1000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In triangle $ABC$, the bisector $BD$ was drawn, and in triangles $ABD$ and $CBD$ - the bisectors $DE$ and $DF$ respectively. It turned out that $EF \| AC$. Find the angle $DEF$.
|
Answer: $45^{\circ}$.
Solution. Let segments $B D$ and $E F$ intersect at point $G$. From the condition, we have $\angle E D G = \angle E D A = \angle D E G$, hence $G E = G D$. Similarly, $G F = G D$. Therefore, $G E = G F$, which means $B G$ is the bisector and median, and thus also the altitude in triangle $B E F$. From this, $D G$ is the median and altitude, and thus also the bisector in triangle $E D F$, from which $\angle D E G = \angle E D G = \angle F D G = \angle G F D$. Since the sum of the four angles mentioned in the last equality is $180^{\circ}$ degrees, each of them is $45^{\circ}$ degrees.
Comment. A complete and justified solution - 7 points. A generally correct reasoning with minor gaps or inaccuracies - up to 5 points.
|
45
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How many different triangles can be formed from: a) 40 matches; b) 43 matches? Solution. We need to find the number of triples of natural numbers $\mathrm{x}, \mathrm{y}, \mathrm{z}$ such that $\mathrm{x} \leq \mathrm{y} \leq \mathrm{z}$, $\mathrm{x}+\mathrm{y}+\mathrm{z}=40$ and $\mathrm{x}+\mathrm{y}>\mathrm{z}$. From these inequalities, it follows that $\mathrm{z}$ can take values satisfying the inequalities $14 \leq z \leq 19$. If $z=19$, then $x+y=21$, and $x \leq y \leq 19$. Therefore, $11 \leq y \leq 19$, and we have 9 triangles with $z=19$. Similarly, we establish that the number of triangles for which $\mathrm{z}=18,17,16,15,14$ is respectively $8,6,5,3,2$, and in total we have 33 triangles. Similarly, we find that the number of triangles with a perimeter of 43 is 44.
|
Answer: a) $33 ;$ b) 44.
|
33
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8.6. First solution. Divide all the coins into two parts of 20 coins each and weigh them. Since the number of counterfeit coins is odd, one of the piles will weigh more. This means that there is no more than one counterfeit coin in it. Divide it into two piles of 10 coins and weigh them. If the scales are in balance, then all 20 weighed coins are genuine. If one of the pans outweighs, then there are 10 genuine coins on that pan, and among the other 10 coins, there is exactly one counterfeit. Divide these 10 coins into three piles of 4, 4, and 2 coins. In the third weighing, compare the two piles of 4 coins. If they balance, then all 8 coins are genuine and we have found 18 genuine coins. If one of the piles outweighs, then there are 4 genuine coins in it, the other pile contains the counterfeit coin, and the 2 set aside coins are genuine. In total, 16 genuine coins have been found.
|
The second solution. Divide all the coins into five equal piles, each containing 8 coins, and number them. Place the 1st and 2nd piles on one side of the scales, and the 3rd and 4th piles on the other.
Consider the first case - the scales balance. Then either there is one fake coin on each side, or all the coins being weighed are genuine. Then we will weigh the 1st and 2nd piles. If they balance, then all 16 coins are genuine. If one of the piles outweighs the other, then it contains 8 genuine coins. The third weighing will compare the 3rd and 4th piles to determine the next 8 genuine coins.
Now consider the second case - the scales do not balance. Let's assume for definiteness that the 1st and 2nd piles outweigh the others, then among them there is no more than one fake coin. The second weighing will compare the 1st and 2nd piles. If they balance, then all 16 coins are genuine. If one of the piles outweighs the other, then it contains 8 genuine coins, and the other has exactly one fake coin. Consequently, the 3rd and 4th piles contain exactly two fake coins, and the 5th pile contains 8 genuine coins. Thus, a total of 16 genuine coins have been found.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Postman Pechkin is riding a bicycle along a highway. He noticed that every 4.5 kilometers, a suburban bus overtakes him, and every 9 minutes, a suburban bus passes him in the opposite direction. The interval of bus movement in both directions is 12 minutes. At what speed is Pechkin riding?
|
Answer: 15 km/h
Solution: Let x (km/h) and y (km/h) be the speeds of the cyclist and the bus, respectively. Since the interval between bus movements is 12 minutes (1/5 hour), the distance between two consecutive buses is y/5 km. Therefore, at the moment the cyclist meets a bus, the distance between the cyclist and the next oncoming bus is y/5 km. Since their meeting will occur in 9 minutes (3/20 hour), and the closing speed is (x+y), we have $3(x+y)/20=y/5$. The time between two consecutive overtakes is 4.5/x, on the other hand, it is equal to $1/5 + 4.5/y$. Thus, we have the system of equations $3(x+y)=4y$ and $4.5/x=1/5+4.5/y$, from which $x=15$ km/h.
Grading criteria: Correct solution - 7 points, the solution process is correct but the answer is wrong due to an arithmetic error - **5** points, the system is set up but the subsequent solution is incomplete or incorrect - **3** points, in all other cases - 0 points.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Given a square $A B C D$. Point $N$ lies on side $A D$, such that $A N$ : $N D=2: 3$, point $F$ lies on side $C D$ and $D F: F C=1: 4$, point $K$ lies on side $A B$, such that $A K: K B=1: 4$. Find the angle $K N F$.
|
Solution. Mark point $E$ on side $B C$ such that $B E: E C=2: 3$. Let $M-$ be the midpoint of $K E$. Then $\triangle K M N-$ is an isosceles right triangle and $\triangle M N F-$ is an isosceles triangle. Therefore, $\angle K N F=$ $\angle K N M+\angle M N F=45^{\circ}+90^{\circ}=135^{\circ}$.
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.1 The number 100 was divided by some number less than 50, with a remainder of 6. By what number could the division have occurred?
|
Let $x$ be the number by which they divided, $m$ be the quotient: $100 = m \cdot x + 6$ (since 6 is the remainder, then $x > 6$, and $m \geq 2$, since $x < 50$) $m \cdot x = 94 = 2 \cdot 47$. From this, it is clear that the numbers $m$ and $x$ are exactly the numbers 2 and 47, i.e., $m = 2$, $x = 47$. (since the number 94 can only be represented as a product of two natural numbers in four ways: $1 \cdot 94, 94 \cdot 1, 2 \cdot 47, 47 \cdot 2$, then all ways except $2 \cdot 47$ are eliminated).
## Answer: 47.
|
47
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 9.1
All three-digit numbers are written in a row: $100101102 \ldots 998$ 999. How many times in this row does a zero follow a two?
## Number of points 7 Answer: 19
#
|
# Solution
Since a three-digit number cannot start with zero, the two followed by a zero cannot be in the units place of any three-digit number in the sequence. Let's assume the two is in the tens place of a three-digit number. Then the zero following it is in the units place of the same number, i.e., the number ends in 20. There are 9 such numbers: 120, 220, ..., 920. Finally, if the two followed by a zero is in the hundreds place, then the corresponding three-digit number starts with 20. There are 10 such numbers: 200, 201, ..., 209. Therefore, the zero will follow the two 19 times in total.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7.2. Harry, Ron, and Hermione wanted to buy identical waterproof cloaks. However, they lacked the money: Ron was short by a third of the cloak's price, Hermione by a quarter, and Harry by one fifth of the cloak's price. When the price of the cloak dropped by 9.4 sickles during a sale, the friends pooled their savings and bought three cloaks, spending all their money. How many sickles did one cloak cost before the price reduction?
|
Answer: 36 sickles.
Solution. Let the initial cost of the cloak be $x$ sickles, then Ron had $\frac{2}{3} x$ sickles, Hermione had $\frac{3}{4} x$ sickles, and Harry had $\frac{4}{5} x$ sickles. During the sale, the cloak cost $(x-9.4)$ sickles, and three cloaks cost $3(x-9.4)$ sickles. Since the friends bought three cloaks, spending all their money, we have $\frac{2}{3} x+\frac{3}{4} x+\frac{4}{5} x=3(x-9.4)$. Solving this equation, we get: $x=36$.
Grading criteria.
"+" A correct and justified solution and the correct answer are provided
“士” The equation is correctly and justifiedly formulated, but the solution is incomplete or contains a computational error
"耳" The correct answer is provided and it is shown that it satisfies the condition
“耳” Only the correct answer is provided
“-" An incorrect solution is provided or it is missing
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6.2. Lines parallel to the sides of a square form a smaller square, the center of which coincides with the center of the original square. It is known that the area of the cross formed by the smaller square (see the figure on the right) is 17 times the area of the smaller square. How many times is the area of the original square larger than the area of the smaller square?
|
Answer. 81 times.
Solution. Let the square have dimensions of 1 cm $\times 1$ cm, and the larger square have dimensions of $n$ cm $\times n$ cm. Then the area of the cross is $(2 n-1)$ cm $^{2}$ (the vertical column has dimensions $n \times 1$, the horizontal row has dimensions $1 \times n$, and the area of the square is counted twice). From the equation $2 n-1=17$, we get that $n=9$. Therefore, the area of the square is $9 \times 9=81$ cm $^{2}$.
Comment. The correct answer was obtained by trial and error - 5 points.
|
81
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.3. From the natural numbers $1,2, \ldots, 101$, a group of numbers is chosen such that the greatest common divisor of any two numbers in the group is greater than two. What is the maximum number of numbers that can be in such a group?
|
Answer: 33. Solution: Estimation. Let's divide the numbers $1,2, \ldots, 101$ into 34 sets: $A_{0}=\{1,2\}, A_{1}=\{3,4,5\}$, $A_{2}=\{6,7,8\}, \ldots, A_{33}=\{99,100,101\}$ (i.e., $A_{k}$ for $k \geq 1$ consists of three numbers $3 k, 3 k+1,3 k+2$). In the desired group of numbers, there cannot be a number from $A_{0}$ (otherwise, the GCD of two numbers from the group, one of which belongs to $A_{0}$, would be $\leq 2$), and from each $A_{k}$ for $k \geq 1$, no more than one number can be in the group, since otherwise, if the group includes adjacent numbers, their GCD $=1$, and if the group includes numbers $3 k$ and $3 k+2$, their GCD $\leq 2$ (since their difference is 2). Therefore, the total number of numbers in the desired group is no more than 33 (no more than one from each $A_{k}$ for $k \geq 1$). Example. Consider the group of numbers $\{3,6,9, \ldots, 99\}$: this group contains 33 numbers, and all of them are divisible by 3.
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.5. In a convex pentagon $P Q R S T$, angle $P R T$ is half the size of angle $Q R S$, and all sides are equal. Find angle $P R T$.
|
Answer: $30^{\circ}$.
Solution. From the condition of the problem, it follows that $\angle P R Q+\angle T R S=\angle P R T(*)$.
First method. We use the "folding" method. Reflect triangle $P Q R$ symmetrically relative to line $P R$, and triangle $T S R$ - relative to line $T R$ (see Fig. 11.5a). From the equality (*) and the equality $R Q=R S$, it follows that the images of points $Q$ and $S$ are the same point $O$.
Note that triangle $T O P$ is equilateral. Moreover, $O R=O P=O T$. Therefore, $O$ is the center of the circumscribed circle of triangle $P R T$. Then $\angle P R T = 0.5 \angle P O T = 30^{\circ}$.
Second method. We will prove that $Q P T S$ is a parallelogram (see Fig. 11.5b). Indeed, using the equality of the angles at the bases of the isosceles triangles $P Q R$ and $R S T$ and the equality (*), we get: $\angle Q P T + \angle P T S = \angle Q P R + \angle R P T + \angle R T P + \angle S T R = \angle P R Q + \angle T R S + (180^{\circ} - \angle P R T) = 180^{\circ}$.
Thus, $P Q \| S T$ and $P Q = S T$ (by condition), so $Q P T S$ is a parallelogram. Then $Q S = P T$, hence triangle $Q R S$ is equilateral. Therefore, $\angle P R T = 0.5 \angle Q R S = 30^{\circ}$.
Grading criteria.
"+" A complete and justified solution is provided.
"±" A generally correct solution is provided, containing minor gaps or inaccuracies (for example, the fact that the images of points $Q$ and $S$ coincide under the symmetries is used but not justified).
"Ғ" The correct answer is obtained based on the fact that $Q P T S$ is a parallelogram, but this is not proven.
"-" Only the answer is provided or the answer is obtained by considering a regular pentagon.
"-" An incorrect solution is provided or no solution is provided.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11.6. A stack consists of 300 cards: 100 white, 100 black, and 100 red. For each white card, the number of black cards lying below it is counted; for each black card, the number of red cards lying below it is counted; and for each red card, the number of white cards lying below it is counted. Find the maximum possible value of the sum of the three hundred resulting numbers.
|
Answer: 20000.
Solution. First method. The number of different permutations of cards is finite. Therefore, their arrangement with the largest indicated sum exists (possibly not unique).
Let the cards lie in such a way that this sum is maximal. Without loss of generality, we can assume that the top card is white. Then, in this arrangement, pairs of cards CB, BK, and WB cannot lie consecutively from top to bottom, otherwise, the sum can be increased by swapping them in such pairs (symmetric pairs do not increase the desired sum when swapped). Therefore, the cards must lie as follows (from top to bottom): WW...WBB...BBKK...KKWW...W...
The length of each subsequent series of cards of the same color cannot be less than the length of the previous series. Indeed, if, for example, in the arrangement with the largest sum, there is a fragment ...WWWBBK..., then the K card can be moved to the top: ...KWWWBB..., increasing the sum. Since the number of cards of each color is the same, the lengths of all series must be the same (otherwise, there will be more cards of the color that ends up at the bottom than cards of another color). Then the series of the same color can be rearranged "in a cycle," without changing the sum, that is, to get such an arrangement of cards: 100 white cards on top, 100 black cards below them, and 100 red cards at the bottom. Therefore, the desired sum is $100 \cdot 100 + 100 \cdot 100 = 20000$.
Second method. Let the number of cards of each of the three colors be $n$. Using the method of mathematical induction, we will prove that for the specified sum $S$, the inequality $S \leq 2 n^{2}$ holds.
Base of induction. For $n=1$, by enumeration, we verify that $S \leq 2$. Induction step: Suppose the inequality is true for $n$ cards of each color. We will prove that it is true if the number of cards of each color is $n+1$. Consider how the sum $S$ can increase if one card of each color is added. Without loss of generality, we can assume that the white card is added to the very top of the stack, and the added black and red cards are the topmost among the cards of their color. Let there be $b$ previously lying black cards above the first red card from the top, and $w$ previously lying white cards above the first black card from the top. Then the white card adds $n+1$ to the sum (considering all black cards lying below it), the black card adds $n+1$ (considering all red cards lying below it) and $w$, due to it lying below $w$ old white cards, and the red card adds no more than $n-w$ due to white cards lying below it and $b$ due to it lying below $b$ old black cards. In total, $S \leq 2 n^{2} + n + 1 + n + 1 + w + n - w + b = 2 n^{2} + 3 n + b + 2$. Considering that $b \leq n$, we get: $S \leq 2 n^{2} + 4 n + 2 = 2(n+1)^{2}$.
Thus, the statement is proved for all natural $n$. For $n=100$, we get that $S \leq 2 \cdot 100^{2} = 20000$. This value is achieved, for example, with the following arrangement: 100 white cards on top, 100 black cards below them, and 100 red cards at the bottom.
Grading criteria.
“+" A complete and well-reasoned solution is provided
“士” A generally correct solution is provided, containing minor gaps or inaccuracies
“Ғ” The correct answer is obtained based on the assumption that the lengths of all monochromatic series of cards are the same, but this is not proven
"Ғ" The solution contains correct ideas on how to maximize the sum by rearranging the cards, but the solution is not completed or contains errors
"-" Only the answer is provided or the answer is obtained by considering only specific cases
“-" An incorrect solution is provided or it is absent
|
20000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. In a right triangle $ABC$ with angle $\angle A=60^{\circ}$, a point $N$ is marked on the hypotenuse $AB$, and the midpoint of segment $CN$ is marked as point $K$. It turns out that $AK=AC$. The medians of triangle $BCN$ intersect at point $M$. Find the angle between the lines $AM$ and $CN$.
|
Solution. By the property of a right triangle with an angle of $30^{\circ}$, we get that $A B: A C=2: 1$. Therefore, $A B: A K=2: 1$. By the property of medians in a triangle, $\mathrm{BM}: \mathrm{MK}=2: 1$. Then, by the converse of the angle bisector theorem, we get that $A M$ is the bisector of angle $\angle \mathrm{BAK}$. Let $\angle \mathrm{B}=\bar{\alpha}=$. Then $\angle \mathrm{K} M=\alpha$,
$\angle \mathrm{KAC}=60^{\circ}-2 \alpha$. Since triangle KAC is isosceles, $\angle \mathrm{KCA}=60^{\circ}+\alpha$. By the theorem of the sum of angles in a triangle, we get that the angle between lines $A M$ and $C N$ is $180^{\circ}-\left(60^{\circ}-\alpha+60^{\circ}+\alpha\right)=60^{\circ}$.
Criteria. If the solution is incorrect - 0 points. If it is proven that triangle OXY is isosceles - 3 points. If the solution is correct - 7 points.
|
60
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Vitya Pereperepkin always calculates percentages incorrectly in surveys: he divides the number of people who answered in a certain way by the number of all the others. For example, in the survey "What is your name?", conducted among 7 Anyas, 9 Ols, 8 Yuls, Vitya counted $50 \%$ Yuls.
Vitya conducted a survey in his school: what kind of triangle is one with sides $3,4,5$? According to his calculations, $5 \%$ answered "acute-angled", $5 \%$ - "obtuse-angled", $5 \%$ - "such a triangle does not exist", $50 \%$ - "right-angled", and the remaining $a \%$ - "it depends on the geometry". What is $a$?
|
Answer: $a=110$ (in the $2-nd$ variant $a=104$). When Vitya calculates that some part of the students constitutes $5 \%$ of the entire class, in reality, it constitutes $5 / 100=1 / 20$ of the remaining students, $1 / 21$ of the entire class. Similarly, Vitya's $50 \%$ is one third of the class. Therefore, all those who answered in the first four ways constitute a share of $1 / 21+1 / 21+1 / 21+1 / 3=10 / 21$ of the entire class. Thus, the remaining students constitute $11 / 10$ of all except them, that is, according to Vitya's opinion, $110 \%$.
|
110
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In a checkers tournament, students from 10th and 11th grades participated. Each player played against every other player once. A player received 2 points for a win, 1 point for a draw, and 0 points for a loss. There were 10 times more 11th graders than 10th graders, and together they scored 4.5 times more points than all the 10th graders. How many points did the most successful 10th grader score?
|
Answer: 20.
Solution: 1. Let $a$ be the number of tenth graders who participated in the tournament, earning $b$ points. Then, $10a$ eleventh graders played, earning $4.5b$ points. In each match, 2 points are played for, and a total of $11a$ players play $\frac{11 a(11 a-1)}{2}$ matches. Therefore, from the condition of the problem, we have the equation $11 a(11 a-1)=5.5 b$, from which it follows that $b=2 a(11 a-1)$.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9.5. Each face of a cube $1000 \times$ $\times 1000 \times 1000$ is divided into $1000^{2}$ square cells with side 1. What is the maximum number of these cells that can be painted so that no two painted cells share a side?
$$
\text { (S. Dolgikh) }
$$
|
Answer. $3 \cdot 1000^{2}-2000=$ $=2998000$ cells.
Solution. Consider an arbitrary coloring that satisfies the condition. Divide all the cells on the surface into "edges" as shown in Fig. 2 - 500 edges around each of the eight vertices (one of the edges is marked in gray). Then, in the $k$-th edge, counting from the vertex, there will be $S_{k}=6 k-3$ cells. Since no two colored cells can be adjacent, in this edge there will be no more than $\left[\frac{S_{k}}{2}\right]=3 k-2=\frac{S_{k}-1}{2}$ colored cells. Summing over all 4000 edges and noting that their total area is $6 \cdot 1000^{2}$, we get that the total number of colored cells does not exceed $\frac{6 \cdot 1000^{2}-4000}{2}=$ $=3 \cdot 10^{6}-2000$.
It remains to provide an example showing that so many cells can be colored. Let's call two opposite faces of the cube the top and bottom, and the others - side faces. On each of the side faces, half of the cells can be marked in a checkerboard pattern. After that, on the top and bottom faces, half of the cells can also be colored in all rows except the two outermost, leaving them empty - see Fig. 3, where two side faces and the top face are visible. It is easy to see that with such a coloring, in each edge there will be the maximum possible number of colored cells. (Instead of checking each edge, one can note that the entire surface is divided into strips $1 \times 100$,
Fig. 2
Fig. 3
four of which are empty, and in each of the others, exactly half of the cells are colored.)
Remark. There are other optimal examples. In particular, in the given example, the coloring of the top face can be changed as follows: divide the top face into 4 triangles with diagonals, and in each of them, color the cells in a checkerboard pattern (so that the coloring of this triangle agrees with the coloring of the adjacent side face).
Comment. Only the answer - 0 points.
Only provided a correct example of coloring $N=$ $=3 \cdot 1000^{2}-2000$ cells - 1 point.
Only proved that in any coloring satisfying the conditions, there are no more than $N$ colored cells - 5 points.
|
2998000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9-4-1. In the figure, $O$ is the center of the circle, $A B \| C D$. Find the degree measure of the angle marked with a «?».
|
Answer: $54^{\circ}$.
Solution variant 1. Quadrilateral $A D C B$ is an inscribed trapezoid in a circle. As is known, such a trapezoid is isosceles, and in an isosceles trapezoid, the angles at the base are equal: $\angle B A D=\angle C B A=63^{\circ}$. Triangle $D O A$ is isosceles ($O A$ and $O D$ are equal as radii), so its base angles $D A O$ and $A D O$ are equal. Therefore,
$$
\angle D O A=180^{\circ}-\angle D A O-\angle A D O=180^{\circ}-63^{\circ}-63^{\circ}=54^{\circ}
$$
|
54
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.2. The least common multiple of four pairwise distinct numbers is 165. What is the maximum value that the sum of these numbers can take?
|
Answer: 268.
Solution. Since 165 is the least common multiple of four numbers, these numbers are divisors of 165. To maximize the sum of these numbers, it is sufficient to take the four largest divisors of 165. If one of them is the number 165 itself, then the LCM will definitely be equal to it.
Then the maximum sum will be
$$
165+\frac{165}{3}+\frac{165}{5}+\frac{165}{11}=165+55+33+15=268
$$
|
268
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.4. Given a cyclic quadrilateral $A B C D$. It is known that $\angle A D B=48^{\circ}, \angle B D C=$ $56^{\circ}$. Inside triangle $A B C$, a point $X$ is marked such that $\angle B C X=24^{\circ}$, and ray $A X$ is the angle bisector of $\angle B A C$. Find the angle $C B X$.
|
Answer: $38^{\circ}$.
Solution. Angles $B D C$ and $B A C$ are equal since they subtend the same arc. Similarly, angles $A D B$ and $A C B$ are equal. Then
$$
\angle A C X=\angle A C B-\angle X C B=\angle A D B-\angle X C B=48^{\circ}-24^{\circ}=24^{\circ}=\angle X C B .
$$
We obtain that $C X$ is the bisector of angle $A C B$, so $X$ is the point of intersection of the angle bisectors of triangle $A B C$.
Now it is easy to find the required angle:
$$
\angle C B X=\frac{\angle A B C}{2}=\frac{180^{\circ}-\angle B A C-\angle A C B}{2}=\frac{180^{\circ}-56^{\circ}-48^{\circ}}{2}=38^{\circ}
$$
|
38
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.5. On the board, the graph of the function $y=x^{2}+a x+b$ is drawn. Yulia drew two lines parallel to the $O x$ axis on the same drawing. The first line intersects the graph at points $A$ and $B$, and the second line intersects the graph at points $C$ and $D$. Find the distance between the lines if it is known that $A B=5, C D=11$.
|
Answer: 24.
Solution. Let the first line have the equation $y=s$, and the second line have the equation $y=t$. Then the distance between the lines is $(t-s)$.
The length of segment $A B$ is equal to the absolute value of the difference of the roots of the equation $x^{2}+a x+b=s$. We can express the difference of the roots using the formula for solving the quadratic equation $x^{2}+a x+(b-s)=0$:
$$
\frac{-a+\sqrt{a^{2}-4(b-s)}}{2}-\frac{-a-\sqrt{a^{2}-4(b-s)}}{2}=\sqrt{a^{2}-4(b-s)}=5
$$
from which we derive
$$
a^{2}-4(b-s)=25
$$
Similarly, we obtain
$$
a^{2}-4(b-t)=121
$$
Subtract the first equation from the second and we get
$$
121-25=\left(a^{2}-4(b-t)\right)-\left(a^{2}-4(b-s)\right)=4(t-s)
$$
We have $4(t-s)=96$, that is, $t-s=24$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9.6. On a line, two red points and several blue points are marked. It turned out that one of the red points is contained in exactly 56 segments with blue endpoints, and the other - in 50 segments with blue endpoints. How many blue points are marked?
|
Answer: 15.
Solution. Let there be $a$ blue points to the left of the first red point, and $b$ blue points to the right; $c$ blue points to the left of the second red point, and $d$ blue points to the right. Then $a b=56, c d=50$. Additionally, $a+b=c+d$ - the number of blue points.
Notice that among the numbers $c$ and $d$, exactly one is even, since the number 50 is divisible by 2 but not by 4. This means that among the numbers $a$ and $b$, exactly one is also even. There are exactly two ways to represent 56 as the product of an even number and an odd number: $56=8 \cdot 7$ and $56=56 \cdot 1$.
In the first case, $c=10, d=5$ work. As an example, you can place 5 blue points on the line, then a red one, then 3 more blue points, then another red one, and finally 7 more blue points.
In the second case, the total number of blue points is 57. Then it only remains to understand that the following system of equations has no integer solutions:
$$
\left\{\begin{array}{l}
c+d=57 \\
c d=50
\end{array}\right.
$$
One can check that the quadratic trinomial $t^{2}-57 t+50=0$, whose roots by Vieta's theorem should be the numbers $c$ and $d$, does not have a pair of natural solutions (for example, it is sufficient to verify that its values at $t=0$ and $t=1$ have different signs, meaning one of the roots lies in the interval $(0 ; 1))$.
Alternatively, one can simply enumerate the factorizations of the number 50 into $c$ and $d$.
|
15
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Each pair of numbers $A$ and $B$ is assigned a number $A * B$. Find $2021 * 1999$, if it is known that for any three numbers $A, B, C$ the identities are satisfied: $A * A=0$ and $A *(B * C)=(A * B)+C$.
|
Solution.
$A *(A * A)=A * 0=A *(B * B)=A * B+B$,
$A *(A * A)=(A * A)+A=0+A=A$, then $A * B+B=A$.
Therefore, $A * B=A-B$. Hence, $2021 * 1999=2021-1999=22$.
Answer: 22.
|
22
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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