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9-6-1. Six pirates - a captain and five members of his crew - are sitting around a campfire facing the center. They need to divide a treasure: 180 gold coins. The captain proposes a way to divide the treasure (i.e., how many coins each pirate should receive: each pirate will receive a non-negative integer number of coi... | Answer: 59.
Solution variant 1. Let's number the pirates clockwise, starting from the captain: First, Second, ..., Fifth. The main observation in this problem is:
two pirates sitting next to each other cannot both vote "for": one of them will not vote "for" because the captain offers no more to him than to his neighb... | 59 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9-8-1. In Midcity, houses stand along one side of a street, each house can have $1,2,3, \ldots, 9$ floors. According to an ancient law of Midcity, if two houses on one side of the street have the same number of floors, then no matter how far apart they are from each other, there must be a house with more floors between... | Answer: 511.
Solution Variant 1. Estimation. Let $a_{k}$ be the maximum number of houses if all of them have no more than $k$ floors. Clearly, $a_{1}=1$. Let's find $a_{k}$. Consider the tallest house with $k$ floors, which is only one by the condition. Both to the left and to the right of it stand houses with $1,2,3,... | 511 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a finite set of cards. On each of them, either the number 1 or the number -1 is written (exactly one number on each card), and there are 100 more cards with -1 than cards with 1. If for each pair of different cards, the product of the numbers on them is found, and all these products are summed, the result is 1... | Answer: 3950.
Solution: Let the number of cards with 1 be $m$, and the number of cards with -1 be $k$. Then, among all pairs, there are $\frac{m(m-1)}{2}$ pairs of two 1s, $\frac{k(k-1)}{2}$ pairs of two -1s, and $m k$ pairs of 1 and -1. Therefore, the sum in the condition is $\frac{m(m-1)}{2}+\frac{k(k-1)}{2}-m k$, f... | 3950 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9.3. Consider natural numbers $a, b$, and $c$ such that the fraction
$$
k=\frac{a b+c^{2}}{a+b}
$$
is a natural number less than $a$ and $b$. What is the smallest number of natural divisors that the number $a+b$ can have?
(P. Kozlov) | Answer. Three divisors.
First solution. Since the number $a+b$ is greater than one, it has at least two distinct divisors. We will prove that there cannot be exactly two, i.e., that the number $a+b$ cannot be prime. Multiplying the equality from the condition by the denominator, we get $a b+c^{2}=k a+k b$ or, equivale... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.1. Natural numbers starting from 1 are written in a row. This results in a sequence of digits: 1234567891011121314... What digit is in the 2021st position? | Answer. 1.
Solution. Note that the sum of the digits of all single-digit and two-digit numbers is $1892021$. Therefore, the digit in the 2021st position belongs to the recording of some three-digit number.
Let $x$ be some three-digit number, then the sum of the digits in the sequence from 1 to $x$ is $n=189+3(x-99)$.... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The bathtub fills with cold water in 6 minutes 40 seconds, with hot water - in 8 minutes. In addition, if the plug is removed from a full bathtub, the water will drain in 13 minutes 20 seconds. How long will it take to fill the bathtub completely, provided that both taps are open, but the bathtub is not plugged? | Solution: First, we will convert the time in seconds to minutes: 6 minutes 40 seconds will be replaced by $6+2 / 3$, or $20 / 3$, and 13 minutes 20 seconds will be replaced by $13+1 / 3$, or $40 / 3$. Then, in one minute, the cold water will fill $3 / 20$ of the bathtub, the hot water will fill $1 / 8$ of the bathtub, ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The sides of a rectangle were reduced: the length - by $10 \%$, the width - by $20 \%$. As a result, the perimeter of the rectangle decreased by $12 \%$. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $20 \%$ and its width is reduced by $10 \%$? | Answer: by $18 \%$.
Solution. Let $a$ and $b$ be the length and width of the rectangle. After decreasing the length by $10 \%$ and the width by $20 \%$, we get a rectangle with sides $0.9 a$ and $0.8 b$, the perimeter of which is $0.88$ of the perimeter of the original rectangle. Therefore, $2(0.9 a + 0.8 b) = 0.88 \c... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In a six-digit number, one digit was crossed out to obtain a five-digit number. The five-digit number was subtracted from the original number, and the result was 654321. Find the original number. | Answer: 727023.
Solution. Note that the last digit was crossed out, as otherwise the last digit of the number after subtraction would have been zero. Let $y$ be the last digit of the original number, and $x$ be the five-digit number after crossing out. Then the resulting number is $10 x + y - x = 9 x + y = 654321$. Di... | 727023 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Each of the three friends either always tells the truth or always lies. They were asked the question: "Is there at least one liar among the other two?" The first answered: "No," the second answered: "Yes." What did the third answer? | Solution Since the first and second friends gave different answers, one of them is a liar, and the other is a knight. Moreover, the knight could not have answered “No” to the question posed, as in that case, he would have lied (there is definitely a liar among the two remaining). Therefore, the first one is a liar. He ... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Tourists arrived at the campsite. For lunch, each of them ate half a can of soup, a third of a can of stew, and a quarter of a can of beans. In total, they ate 39 cans of food. How many tourists were there? Answer: 36. | Solution. Note that 12 tourists ate 6 cans of soup, 4 cans of stew, and 3 cans of beans - that is, a total of 13 cans of food.
39 cans is 3 sets of 13 cans. One set is eaten by 12 people. Therefore, the total number of tourists is 36. | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Six people are standing in a circle, each of whom is either a knight - who always tells the truth, or a liar - who always lies. Each of them said one of two phrases: "There is a liar next to me" or "There is a liar opposite me." What is the minimum number of liars among them? Provide an example and prove that there ... | Answer: 2.
Solution. Let's number all the people standing clockwise (this way, people with numbers 1 and 4, 2 and 5, 3 and 6 will stand opposite each other).
Zero liars is obviously impossible (then there would be only knights and no one could say any of the phrases).
If there is one liar, let's say his number is 1,... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.2. In Solar City, 6 dwarfs eat donuts daily, 8 dwarfs eat donuts every other day, and the rest do not eat donuts at all. Yesterday, 11 dwarfs ate donuts. How many dwarfs will eat donuts today? | Answer: 9.
Solution: Of the 11 dwarfs who ate donuts yesterday, 6 dwarfs eat them daily, so the remaining $11-6=5$ eat them every other day. Therefore, these five will not eat donuts today, while the other $8-5=3$ from those who eat every other day will. So today, these three will eat donuts, as well as the six who al... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.4. Given nine cards with the numbers $5,5,6,6,6,7,8,8,9$ written on them. From these cards, three three-digit numbers $A, B, C$ were formed, each with all three digits being different. What is the smallest value that the expression $A+B-C$ can have? | # Answer: 149.
Solution. By forming the smallest sum of numbers $A$ and $B$, and the largest number $C$, we get the smallest value of the expression $A+B-C$. This is $566+567-988=145$. However, this partition is not valid: two numbers have the same digits. By swapping the digits 6 and 8 in the units place, we get the ... | 149 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5. A round table was sat by 10 people - liars and knights. Liars always lie, while knights always tell the truth. Each of them was given a coin. Then each of those sitting passed their coin to one of their two neighbors. After that, 5 people said: “I have one coin,” while the other 5 said: “I have no coins.” What is ... | # Answer: 7.
Solution. After passing the coins, each person sitting at the table can have 0, 1, or 2 coins. The total number of coins will be 10. Note that if a person lies, they will state a number of coins that differs from the actual number by 1 or 2. Since the total number of coins based on the answers differs fro... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.2. Little kids were eating candies. Each one ate 7 candies less than all the others together, but still more than one candy. How many candies were eaten in total? | Answer: 21 candies.
Solution: Let $S$ denote the total number of candies eaten by the children. If one of the children ate $a$ candies, then according to the condition, all the others ate $a+7$ candies, and thus together they ate $S=a+(a+7)=2a+7$ candies. This reasoning is valid for each child, so all the children ate... | 21 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.3 In triangle $A B C$, the median $A M$ is drawn (point $M$ lies on side $\mathrm{BC}$). It is known that angle $C A M$ is $30^{\circ}$, and side $A C$ is 2. Find the distance from point $B$ to the line $A C$.
Omвem: 1. | Solution: See fig.
Triangle СKM is equal to triangle ВHM (these are right triangles, the hypotenuses СM and ВM of which are equal, and the angles are the same).
Therefore, $\mathrm{BH}=$ СK. But in triangle СKA, the leg СK lies opposite the angle $30^{\circ}$ and is equal to half the hypotenuse АC: СK=1.
 | Answer: 120.
Solution: If we add the perimeters of the two squares, we get $100+40=140$ cm. This is more than the perimeter of the resulting figure by twice the side of the smaller square. The side of the smaller square is $40: 4=10$ cm. Therefore, the answer is $140-20=120$ cm. | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4-5. Given a figure consisting of 33 circles. You need to choose three circles that are consecutive in one of the directions. In how many ways can this be done? The image shows three of the desired ways.
![](https://cdn.mathpix.com/cropped/2024_05_06_1d29d5cb430346a58eb5g-02.jpg?height=369&width=366&top_left_y=1409&to... | Answer: 57.
Solution. The number of options along the long side is $1+2+3+4+5+6=21$. Along each of the other two directions, it is $-4+4+4+3+2+1=18$. The total number of options is $21+18 \cdot 2=57$ | 57 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4-8. In a chess club, 90 children attend. During the session, they divided into 30 groups of 3 people, and in each group, everyone played one game with each other. No other games were played. In total, there were 30 games of "boy+boy" and 14 games of "girl+girl". How many "mixed" groups were there, that is, groups wher... | Answer: 23.
Solution: There were a total of 90 games, so the number of games "boy+girl" was $90-30-14=46$. In each mixed group, two "boy+girl" games are played, while in non-mixed groups, there are no such games. In total, there were exactly $46 / 2=23$ mixed groups.
## Grade 5 | 23 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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 ... | 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 numbe... | 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 l... | 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 ... | 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 ... | 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 ... | 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"... | 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 ... | 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... | 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 | 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 ... | 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 ru... | 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 ... | 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... | 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;
... | 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 ... | 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 $2x + 2y + 1$ bird... | 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 fr... | 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+5... | 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.
Since the sum of the angles in triangle $D F E$ is $180^{\circ}$, we have $7 \alpha... | 9 | 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 th... | 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. I... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8-5. In a tournament, 6 teams $P, Q, R, S, T$ and $U$ participate, and each team must play against every other team exactly once. Each day, they are divided into 3 pairs, and all three matches are played simultaneously. The "Sports" channel has chosen which match it will broadcast each day:
$$
\begin{array}{c|c|c|c|c}... | Answer. Only in the 1st.
Solution. Let's look at team $P$: on the 1st, 3rd, and 5th days, it will play against teams $Q, T$, and $R$. Therefore, in the remaining two days, it must play against teams $S$ and $U$. Since on the 2nd day $S$ plays against $R$, $P$ has no choice but to play against $U$ on the 2nd day, and a... | 1 | Combinatorics | 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 spe... | 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 fro... | 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 ... | 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,... | 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... | 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 n... | 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.
 | Answer: 9.
Solution. Since $O B=O C$, then $\angle B C O=32^{\circ}$. Therefore, to find angle $x$, it is sufficient to find angle $A C O: x=32^{\circ}-\angle A C O$.
Since $O A=O C$, then ... | 9 | 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, an... | 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 greate... | 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 o... | 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 ... | 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 origi... | 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... | 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 a... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11-1. Twins Paolo and Sevilla are celebrating their birthday at a cafe with friends. If the final bill amount is divided equally among everyone, then each person should pay 12 euros. But if the bill is divided equally among everyone except Paolo and Sevilla, then each person should pay 16 euros. How many friends came t... | Answer: 6.
Solution. Let $n$ be the number of friends who arrived. Then we get the equation $12(n+2)=16n$, from which $n=6$. | 6 | Algebra | 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... | 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 numbe... | 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 great... | 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 ... | 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 ... | 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 $D M H$, a... | 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 i... | 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 digi... | 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... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Variant 1.
It is known that
$$
\left(x^{2}-x+3\right)\left(y^{2}-6 y+41\right)\left(2 z^{2}-z+1\right)=77
$$
Find $\frac{x y}{z}$. | # Answer: 6.
Solution.
$$
\begin{aligned}
& x^{2}-x+3=(x-0.5)^{2}+2.75 \geq 2.75 \\
& y^{2}-6 y+41=(y-3)^{2}+32 \geq 32 \\
& 2 z^{2}-z+1=2(z-0.25)^{2}+0.875 \geq 0.875
\end{aligned}
$$
Therefore, $\left(x^{2}-x+3\right)\left(y^{2}-6 y+41\right)\left(2 z^{2}-z+1\right) \geq 77$ and if at least one of the three inequa... | 6 | 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 throu... | 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 = ... | 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 1... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Variant 1.
Given a parallelogram $A B C D$. Let $B P$ and $C Q$ be the perpendiculars dropped from vertices $B$ and $C$ to diagonals $A C$ and $B D$ respectively (point $P$ lies on segment $A C$, and point $Q$ lies on segment $B D$). Find the ratio $\frac{10 B D}{A C}$, if $\frac{A P}{A C}=\frac{4}{9}$ and $\frac{D... | Answer: 6.
Solution: Let $O$ be the point of intersection of the diagonals. Note that points $B, C, Q, P$ lie on the same circle (segment $B C$ is seen from points $P$ and $Q$ at a right angle). Therefore, triangles $B O P$ and $C O Q$ are similar. Let $A C=2 a, B D=2 b$. Then $P O=a-\frac{8 a}{9}=\frac{a}{9}, Q O=$ $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.1. Draw a row of 11 circles, each of which is either red, blue, or green. Moreover, among any three consecutive circles, there should be at least one red, among any four consecutive circles, there should be at least one blue, and there should be more than half green. How many red circles did you get? | Answer: 3 red circles
Hint. The circles are arranged only as follows: ZZKSKZKSKZZ.
Solution. (1) Three non-overlapping triplets of circles can be identified, each containing at least one red circle. Therefore, there are no fewer than three red circles. (2) Two non-overlapping quartets of circles can be identified, ea... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7.2. Represent the number 32 as the product of three integer factors, the sum of which is 3. What is the smallest of the factors? | Answer: -4.
Example: $32=(-4) \cdot(-1) \cdot 8$.
Solution. The given factorization is unique. This can be proven.
If all three factors are positive, then the largest of them is not less than 4, and the sum is greater than 3, which contradicts the condition. Therefore, two of the factors are negative, and the third ... | -4 | Number Theory | 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 loc... | 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 be... | 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... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.7. Petya told Misha that in his class exactly two thirds of all the girls are blondes, exactly one seventh of the boys are blonds, and in total, a third of the class has light hair. Misha said: "You once told me that there are no more than 40 people in your class. 0 ! I know how many girls are in your class!" How man... | Answer: 12 girls
Solution. Let there be $x$ girls and $y$ boys in the class. From the problem statement, we have the following relationship:
$\frac{2}{3} x+\frac{1}{7} y=\frac{1}{3}(x+y)$,
which, after transformation, becomes $7 x=4 y$.
From the condition and the derived relationship, it follows that the number $x$... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.1. Represent the number 36 as the product of three integer factors, the sum of which is 4. What is the smallest of the factors? | Answer: -4.
Example: $36=(-4) \cdot(-1) \cdot 9$.
Solution. The given factorization is unique. This can be proven.
If all three factors are positive, then the largest of them is not less than 4 (since $3^{3}<36$), and the sum is greater than 4, which contradicts the condition. Therefore, two of the factors are negat... | -4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.3. Given a parallelogram $A B C D, \angle D=100^{\circ}, B C=12$. On side $A D$ there is a point $L$ such that $\angle A B L=50^{\circ}, L D=4$. Find the length of $C D$. | Answer: 8.
Solution. By the property of a parallelogram, $\angle A B C=\angle D=100^{\circ}, A D=B C=12$ and $C D=A B$. Therefore, $\angle C B L=\angle A B C-\angle A B L=100^{\circ}-50^{\circ}=50^{\circ}$ and $A L=A D-L D=12-4=8$. Since $\angle A L B=\angle C B L$ (as alternate interior angles when $A D$ and $B C$ ar... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8.4. Four boys and three girls went to the forest to pick mushrooms. Each found several mushrooms, in total they collected 70. No two girls collected the same amount, and any three boys together brought no fewer than 43 mushrooms. The number of mushrooms collected by any two children differed by no more than 5 times. M... | Answer: 5 mushrooms.
Solution. Any three boys collected at least 43 mushrooms together, so there is a boy who collected no less than 15 mushrooms (since $14 \cdot 3 < 43$). Therefore, this boy and the other three collected no less than $15 + 43 = 58$ pieces.
If there is a boy who collected no less than 15 pieces, the... | 5 | Logic and Puzzles | 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 ... | 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 als... | 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... | 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... | # 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... | 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 ha... | 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 eithe... | 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 $256 \mathrm{~cm}^{2}$.
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 fi... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Problem 7.5 (7 points)
On a plane, 6 lines are drawn and several points are marked. It turned out that on each line exactly 3 points are marked. What is the minimum number of points that could have been marked? | # Solution:
The vertices of the triangle, the midpoints of its sides, and the point of intersection of the medians - 7 points lying in threes on 6 lines (3 sides and 3 medians).
P.S. It doesn't have to be medians specifically.
Proof of the estimate: If we have a point through which at least 4 lines pass, then we wil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Misha has a $7 \times 7$ square of paper, all cells of which are white. Misha wants to color $N$ cells black. What is the smallest $N$ for which Misha can color the cells so that after coloring, no completely white rectangle with at least ten cells can be cut out from the square? | Answer: 4.
Solution. Divide the $7 \times 7$ square into 5 rectangles: four $3 \times 4$ rectangles (each corner of such a rectangle coincides with one of the corners of the $7 \times 7$ square) and a $1 \times 1$ square. If only three cells are colored, there will be a white rectangle consisting of 12 cells. Example ... | 4 | Combinatorics | 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} \leqs... | 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 ... | 102 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5. On the side $A D$ of the square $A B C D$, point $K$ is marked, and on the extension of ray $A B$ beyond point $B$ - point $L$. It is known that $\angle L K C=45^{\circ}, A K=1, K D=2$. Find $L B$. | Answer: $L B=2$.
Fig. 1: to the solution of problem 5
Solution. Note that $\angle L A C=45^{\circ}=\angle L K C$, which implies that quadrilateral $L A K C$ is cyclic. Then $\angle K C L=90^... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. An excursion group of 6 tourists is visiting attractions. At each attraction, three people take photos, while the others photograph them. After what minimum number of attractions will each tourist have photos of all other participants in the excursion?
Answer: after 4 attractions. | Solution. Evaluation. A total of $6 \cdot 5=30$ photographs need to be taken (considering only photographs between two people $A$ and $B$, that is, if person $A$ photographs 3 other participants $B, C, D$ in one photograph - this counts as 3 photographings $A \rightarrow B, A \rightarrow C, A \rightarrow D$).
At one l... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Find the sum of the numbers $1-2+3-4+5-6+\ldots+2013-2014$ and $1+2-3+4-5+6-\ldots-2013+2014$. | 1. Answer: 2.
Notice that for each term of the first sum, except for 1, there is an opposite term in the second sum. The sum of opposite numbers is 0. Therefore, the total sum is $1+1=2$.
Grading criteria:
Correct answer with proper justification: 7 points.
Incorrect answer with the right idea in the justification:... | 2 | Algebra | 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... | 1008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Misha painted all integers in several colors such that numbers whose difference is a prime number are painted in different colors. What is the smallest number of colors that Misha could have used? Justify your answer. | 5. Answer: 4 colors
Evaluation. Consider the numbers $1,3,6,8$. The difference between any two of them is a prime number, which means that all of them must be of different colors, and at least four colors are needed.
Example. Paint numbers of the form $4 \mathrm{k}$ in the first color, numbers of the form $4 \mathrm{... | 4 | 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+... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How to measure 8 liters of water when you are near a river and have two buckets with a capacity of 10 liters and 6 liters? (8 liters of water should end up in one bucket).
| Solution. Let's write the sequence of filling the buckets in the form of a table:
| | Bucket with a capacity of 10 liters | Bucket with a capacity of 6 liters | Comment |
| :--- | :--- | :--- | :--- |
| Initially | 0 liters | 0 liters | |
| Step 1 | 10 liters | 0 liters | Filled the first bucket from the river |
| S... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Snow White entered a room where there were 30 chairs around a round table. Some of the chairs were occupied by dwarfs. It turned out that Snow White could not sit down without having someone next to her. What is the minimum number of dwarfs that could have been at the table? (Explain how the dwarfs should have been ... | Answer: 10.
Solution: If there were three consecutive empty chairs at the table in some place, Snow White could sit down in such a way that no one would sit next to her. Therefore, in any set of three consecutive chairs, at least one must be occupied by a dwarf. Since there are 30 chairs in total, there cannot be fewe... | 10 | Logic and Puzzles | 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 ste... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How to measure 2 liters of water when you are near a river and have two buckets with a capacity of 10 liters and 6 liters? (2 liters of water should end up in one bucket).
| Solution. Let's write the sequence of filling the buckets in the form of a table:
| | Bucket with a capacity of 10 liters | Bucket with a capacity of 6 liters | Comment |
| :---: | :---: | :---: | :---: |
| Initially | 0 liters | 0 liters | |
| Step 1 | 10 liters | 0 liters | The first bucket is filled from the rive... | 2 | Logic and Puzzles | 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 ste... | 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
 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 p... | 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 di... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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