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5. In a chess tournament (played in a round-robin system) with 30 participants, to achieve the 4th category rating, a player needs to score $60 \%$ of the points. What is the maximum number of chess players who can achieve the category rating by the end of the tournament? | 5. In total, $30 \cdot 29: 2=435$ games are played in the tournament, and thus 435 points are contested. The number of participants who become rated players cannot exceed $435: 17.5=24$ people. If 24 participants draw all their games against each other and win the rest of their games, they will accumulate the required ... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,689 |
6. In an international football tournament, the following results were recorded:
| | games | wins | draws | losses | points | goals |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: |
| Scotland | 3 | 3 | 0 | 0 | 6 | $7-1$ |
| Wales | 3 | 1 | 1 | 1 | 3 | $3-3$ |
| England | 3 | 1 | 1 | 1 | 1 | $2-3$ |
| Irelan... | 6. From the condition, it follows that England defeated Ireland and drew with Wales. The total score in these two matches is $2-0$, therefore, the results of the matches England-Wales and England-Ireland are $0-0$ and $2-0$, respectively. Further, Scotland conceded one goal from Ireland or Wales. Ireland scored 1 goal ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,690 |
7. In a chess tournament, 8 players participated. They scored $7, 6, 4, 4, 3, 2, 1, 5$ and 0.5 points respectively. How many points did the players who took the first four places lose in matches against the others? | 7. From the condition of the problem, it follows that the tournament winner and the participant who took second place did not lose a single point in their matches against the others. Further, the participants who shared third and fourth places could have scored 8 points against the rest, but scored 7 (they played one p... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,691 |
8. The hockey match "Dynamo" - "Spartak" ended with a score of $8-5$. Prove that there was a moment in the match when "Dynamo" still had to score as many goals as "Spartak" had already scored. | 8. Let $\mathrm{S}$ be the difference between the number of goals "Dynamo" has left to score and the number of goals already scored by "Spartak". Note that at the beginning of the match, this number was 8, and at the end of the match, it was -5, so at some point, this difference must have been equal to 0. | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 35,692 |
9. The Match of Giants. Kafelnikov won one set against Becker with a score of 6:3. In five games, the player who served won. Who served in the first game? | 9. The tennis player who served in the first game served five times. Let's say Kafelnikov won $x$ games out of these five and $y$ games out of the remaining four. Then, the total number of games in which the serving side lost is: $5-x+y=5$ (since according to the problem, in five games, the server won). Therefore, $x=y... | Kafelnikovservedfirst | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,693 |
10.Three chess players $A$, $B$ and $C$ played a match-tournament (each played the same number of games with each other). Could it have happened that in terms of points $\mathcal{A}$ took first place, $C$ - last, and in terms of wins, the opposite: $\mathcal{A}$ took last place, $C$ - first? | 10. Example of a tournament:
| № | Team | 1 | 2 | 3 | points | Place |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $\mathcal{A}$ | $* * * *$ | $\frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2} \frac{1}{2}$ | $101 \frac{1}{2} \frac{1}{2} \frac{1}{2}$ | $6 \frac{1}{2}$ | 1 |
| 2 | $B$ | $\... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,694 |
11. Homework. Fedya was supposed to divide a certain number by 4 and add 15 to it, but Fedya multiplied this number by 4 and subtracted 15, yet he still got the correct answer. What was this number? | 11. Solving the equation: $0.25 x + 15 = 4 x - 15$, we get the answer: 8. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,695 |
12. difficult task. The teacher gave a difficult task in class. As a result, the number of boys who solved the task turned out to be equal to the number of girls who did not solve it. Who is there more of in the class - those who solved the task or girls?
## 71 Tournaments | 12. Let the number of boys and girls who solved the problem be $A$ and $B$, respectively, then the number of girls who did not solve it is also $A$. Therefore, the number of those who solved the problem is equal to the number of girls. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,696 |
14. The sum of the digits of some natural number $A$ is $B$, the sum of the digits of number $B$ is $C$. It is known that the sum of the numbers $A$, $B$, and $C$ is 60. What is the number $A$? Are you sure you have found all solutions? | 14. If the number $A$ is a two-digit number, and $B$ and $C$ are one-digit numbers, then: $A=\overline{x y}=10 x+y ; B=C=x+y$. From this: $A+B+C=12 x+3 y=60$ and $4 x+y=20$. Considering that $x$ and $y$ are digits, we find that $x=y=4$, or $x=5, y=0$. If the numbers $A$ and $B$ are two-digit numbers, and the number $C$... | 44,47,50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,698 |
1. To dig or not to dig? And where to dig? Along a straight road, there are six houses. Where should a well be dug so that the sum of the distances from it to the houses is the smallest? | 1. Let points $A_{1}, A_{2} \ldots A_{6}$ be the locations of houses, and point $X$ be the location of the well. To minimize the sum $X A_{1} + X A_{6}$, point $X$ should lie on the segment $A_{1} A_{6}$. Similarly, the sums $X A_{2} + X A_{5}$ and $X A_{3} + X A_{4}$ will be minimized if $X$ lies on the segments $A_{2... | The\well\should\be\built\between\the\third\\fourth\house | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,700 |
3. From 22 matches, form the outline of a rectangle with the largest possible area. Do not break the matches.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | 3. The sum of the length and width of a rectangle that can be formed using 22 matches is 11. By enumeration, we confirm that the rectangle with the largest area is $6 \times 5$. | 6\times5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,702 |
4. Linear function. The distance between villages $\mathcal{A}$ and $B$ is 3 km. In village $\mathcal{A}-300$ students, and in village $B-$ 200 students. Where should a school be built to minimize the total distance traveled by students on their way to school? | 4. It is clear that the school should be built on the segment $A B$, but where exactly? Let the distance from village $A$ to the school be $x$, then the total distance traveled by all schoolchildren on the way
=2$. Answ... | 18750\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,705 |
7. $A$ if there are eleven houses? Eleven houses stand along a straight street. Where should a well be dug to minimize the sum of the distances from it to the houses? | 7. Reasoning similarly to problem 1, we get that the well should be built next to the sixth house, counting from the beginning of the street. | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,706 |
9. Satisfied Pikes. 40 pikes were released into the pond. A pike is satisfied if it has eaten three other pikes (satisfied or hungry). What is the maximum number of pikes that can be satisfied? | 9. We will feed the pikes in three stages. Initially, we will feed 9 pikes, after which there will be 9 full and 4 hungry pikes left. Then, with the nine full pikes, we will feed three hungry pikes, and with three "newly" full pikes, we will feed the last hungry pike. In total, 13 pikes have been fed. Note that as each... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,708 |
10. Transportation of cargo. Several boxes together weigh 10 tons, with each of them weighing no more than one ton. What is the minimum number of three-ton trucks needed to haul this cargo? | 10. Let's show that a) four three-ton trucks may not be enough, b) five three-ton trucks will always be enough. Note that at any moment during loading, there is a truck with less than 2 tons of cargo (otherwise, the total weight of the cargo would be more than 10 tons). Therefore, we will load the trucks with boxes in ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,709 |
11. Which polygons are more numerous? On a circle, there are 1997 white points and one red point. Polygons are considered with vertices at these points. Which are more numerous: polygons with a red vertex or without it? | 11. Each purely "white" polygon can be uniquely associated with a "red" polygon: a pentagon with a hexagon, a 56-gon with a 57-gon, and so on. However, in this case, the red triangles remain extra - there is nothing to associate them with. Therefore, there are more polygons with a red vertex. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,710 |
12.B In the forest, there is a rectangular highway (by the inner edge) measuring 7 km by 5 km and 200 m wide. Draw where a tourist can set up a tent so that the tent is no closer than 1 km from the highway. | 12. Solution (see figure): the tourist can set up the tent outside the figure, as well as inside the rectangle. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,711 |
13. An example of division. Kolya wrote an example of multiplying two-digit numbers on the board. Then he erased all the digits and replaced them with letters: the same digits with the same letters, and different ones with different letters. The resulting equation was: $\overline{c d} \cdot \overline{a b}=\overline{e f... | 13. The number $\overline{e f f e}=1000 e+100 f+10 f+e=$ $1001 e+110 f$ - is divisible by 11 (since $1001=11 \times 91$). However, the numbers $\overline{a b}$ and
$\overline{c d}$ are not ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,712 |
14. Who goes home by what? Alyosha, Borya, and Vitya study in the same class. One of them goes home from school by bus, another by tram, and the third by trolleybus. One day, after classes, Alyosha went to see his friend off to the bus stop. When a trolleybus passed by them, the third friend shouted from the window: "B... | 14. Alyosha cannot ride either the bus or the trolleybus, therefore, he rides the tram. Borya does not ride the trolleybus, so he rides the bus. It leaves only the trolleybus for Vitya. | Alyoshatram,Boryabus,Vityatrolleybus | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,713 |
1. From Alcuin's problems. How many jumps will it take for the hound to catch up with the hare if initially they are separated by a distance of 150 feet, the hare moves away from the dog by 7 feet with each jump, and the dog runs faster than the hare and gets 9 feet closer with each jump? | 1. For each jump, the greyhound gets 2 feet closer to the hare. Therefore, she will catch up to it in 75 jumps. | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,715 |
2. What is average speed? A car travels from point $A$ to point $B$ at a speed of 40 km/h, and returns at a speed of 60 km/h. What is the average speed of the car? | 2. If the distance from $A$ to $B$ is $n$, then the average speed on the path $2 A B$ is $V_{\text {avg }}=2 n:\left(\frac{n}{40}+\frac{n}{60}\right)=\frac{240 n}{5 n}=48 \mathrm{km} /$ h. | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,716 |
3. From Nizhny Novgorod to Astrakhan, a steamboat takes 5 days, and on the return trip - 7 days. How long will it take for rafts to float from Nizhny Novgorod to Astrakhan? | 3. When the steamboat goes from Nizhny Novgorod to Astrakhan (downstream), it covers $\frac{1}{5}$ of the distance in a day, and when it goes back - $\frac{1}{7}$ of the distance. Therefore, $\frac{1}{5}-\frac{1}{7}=\frac{2}{35}$ - two current speeds. From this, $\frac{1}{35}$ of the distance per day is the speed of th... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,717 |
4. The matter is settled. A rower, passing under a bridge, lost his hat. After 15 minutes, he noticed the loss and caught the hat a kilometer from the bridge. What is the speed of the river's current? | 4. The rower noticed the loss after 15 minutes, so he will catch up with it in 15 minutes; hence, the hat drifted 1 kilometer in 30 minutes, and the speed of the river current is $2 \kappa$ km/h. (Why will the rower catch up with the hat in 15 minutes? Imagine the situation taking place on a train, with the hat left on... | 2\kappa | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,718 |
6. Two pedestrians are walking towards each other along a straight road at a speed of 5 km/h. The initial distance between them is 10 kilometers. A fly, which flies at a speed of $14 \mathrm{km} / \mathrm{h}$, takes off from the first pedestrian, flies straight to the second, lands on him, and without losing a second, ... | 6. If the pedestrians walk at a speed of $5 \mathrm{km} / \mathrm{h}$, they will meet in one hour. In this time, the fly will fly $14 \times 1=14$ km. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,720 |
7. Train and Pole. A train 18 m long passes by a pole in 9 seconds. How much time will it need to pass a bridge 36 m long? | 7. Draw a picture! The lead car will pass the bridge in 18 seconds, and the last car will still be traveling on the bridge for another 9 seconds. Answer: 27 seconds. | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,721 |
8. Task by L. Carroll. Couriers from places $A$ and $B$ move towards each other uniformly, but at different speeds. After meeting, one needed another 16 hours, and the other needed another 9 hours to reach their destination. How much time does it take for each to travel the entire distance from $A$ to $B$? | 8. Let $t$ be the time it took for the couriers to reach the meeting point, and $v_1$ and $v_2$ be the speeds of the couriers. Then $\left(v_{1}+v_{2}\right) t=v_{1}(16+t)=v_{2}(9+t) ; 16 v_{1}=v_{2} t$ and $v_{1} t=9 v_{2} ; \frac{16 v_{1}}{v_{1} t}=\frac{v_{2} t}{9 v_{2}}$ or $\frac{16}{t}=\frac{t}{9}$ and $t^{2}=16 ... | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,722 |
9. Pete and the Hooligan. Pete was riding on an escalator. When he was in the middle of the escalator, a hooligan ran past him, snatched his hat, and threw it onto the opposite escalator. Pete wants to get his hat back as quickly as possible. Should he run down the escalator or up? | 9. Let's assume that Petya's own speed (v) is greater than the speed of the escalator. Then it all depends on whether Petya can get to the top faster than his hat can get there. 1. If yes, then it doesn't matter whether he runs down or up. Petya approaches the hat at a speed of $v$, regardless of whether he is moving "... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,723 |
10. Draw a diagram! Two ferries leave simultaneously from opposite banks of a river and cross it perpendicular to the banks. The speeds of the ferries are constant. The ferries meet each other 720 meters from the nearest bank. Upon reaching the bank, they immediately head back. On the return trip, they meet 400 meters ... | 10. The total distance they traveled by the time of their first meeting (720 m from one of the banks) is equal to the width of the river. When they meet for the second time, the total distance is three times the width of the river, which required three times the time. By the time of the first meeting, one of the ferrie... | 1760 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,724 |
11. A Trip for Milk. Anton went to a dairy store. He had no money, but he had empty bottles - six one-liter bottles (worth 20 kopecks each) and six half-liter bottles (worth 15 kopecks each). In the store, there was milk sold by the liter for 22 kopecks. What is the maximum amount of milk he could bring home? He had no... | 11. By returning six half-liter bottles and one liter bottle, Anton will receive 1 ruble 10 kopecks, which will be the cost of 5 liters of milk. The 5 liters of milk he buys can be carried home in the remaining liter bottles. Let's ensure that he won't be able to carry more than 5 liters. If he returns not one liter bo... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,725 |
12. How many parts should a worker produce to exceed the plan by 40 parts by no less than $47 \%$? | 12. Let's pay attention to the fact that $1 \%$ of the plan for 40 parts is 0.4 parts. Next, $47 \% - 18.8$ parts. But you can't make 0.8 of a part, so the worker must make no less than 59 parts. | 59 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,726 |
13. Pawn and Knight. Let's place a pawn on a chessboard. Can a knight, placed on one of the free squares, visit all the other free squares and return to the starting square, visiting each free square only once? | 13. Note that the knight changes the color of the square it stands on with each move. Since one square is occupied by a pawn, it needs to make 63 moves and end up on the initial square, which is impossible. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,727 |
1. A plane is painted in 2 colors. Prove that there are two points 1 meter apart that are painted the same color. | 1. Consider an equilateral triangle (make a drawing!) with a side of 1 meter. Two of its three vertices are painted the same color. The distance between them is exactly 1 meter. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,730 |
2. Balls. In a box, there are 100 black and 100 white balls. What is the smallest number of balls that need to be taken out, without looking into the box, to ensure that among them there are 2 balls of the same color? To ensure there were 2 white balls? | 2. Out of three balls, there will definitely be 2 of the same color. Therefore, three balls are sufficient. Note that two balls are not enough, as they can be of different colors. It may happen that we initially draw 100 black balls, and only then - 2 white ones. In total: 102 balls. | 102 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,731 |
3. Strange but true. Prove that in your class there will be 2 people who have the same number of friends among classmates. | 3. Let there be $n$ people in your class, then for each of your classmates (including, of course, yourself), the number of friends can be any number from 0 (when a person is not friends with anyone) to $n-1$ (friends with everyone), and note that these two "extremes" cannot be realized simultaneously. Thus, there are $... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,732 |
5. Pete's Pockets. Can Pete distribute 44 coins into 10 pockets so that the number of coins in each pocket is different? | 5. Let's "order" Petya's pockets in ascending order. Then, the smallest pocket may not contain any coins, the next one must contain at least 1 coin, and so on. Finally, in the last, tenth pocket, there must be at least 9 coins. Therefore, in all 10 pockets, there will be at least $1+2+\ldots+9=45$ coins. Answer: he wil... | hewillnotbeableto | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,734 |
6. Seven mushroom pickers collected 100 mushrooms, and each picker collected a different number of mushrooms. Prove that there are three pickers who collected no fewer than 50 mushrooms. | 6. Let seven mushroom pickers have collected $a_{1}, a_{2}, \ldots, a_{7}$ mushrooms, respectively, and $a_{1}<a_{2}<\ldots<a_{7}$. Suppose there are no three mushroom pickers who collected at least 50 mushrooms, then $a_{5} \leq 15$. Therefore, $a_{4} \leq 14 ; a_{3} \leq 13 ; a_{2} \leq 12 ; a_{1} \leq 11$. Hence,
$a... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 35,735 |
7. On planet Earth, the ocean occupies more than half of the surface area. Prove that in the world's ocean, there are two diametrically opposite points. | 7. Suppose that each point in the world's ocean corresponds to a diametrically opposite point on land, then the world's ocean and land are centrally symmetrical, and their areas are equal, which contradicts the condition of the problem. Therefore, in the world's ocean, two diametrically opposite points can be indicated... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,736 |
8. Triangles on a plane. On a plane, 20 points are chosen, no three of which lie on the same line. Does there exist a triangle with vertices at these points and at least one angle less than $10^{\circ}?$ | 8. Note that all points lie on one side of some line $A A_{1}$. The sum of the angles $\angle A_{1} A A_{2}, \angle A_{2} A A_{3}, \ldots, \angle A_{18} A A_{19}$ is less than $180^{\circ}$, from which it follows that each of these 18 angles is less than $10^{\circ}$.
. We will rotate the line around one of these points until it meets each of the other points (thus forming 33 lines or 32 angles), after which we will sequentially connect the remainin... | proof | Geometry | proof | Yes | Yes | olympiads | false | 35,739 |
11. A three-digit number is thought of, which with any of the numbers 543, 142, and 562, matches one digit in the same place, while the other two do not match. What is the number thought of? | 11. If the first digit of the desired number is 5, then either the second digit is 4 or the third is 2 (since a match with the second number is required). Both cases lead to a contradiction: a match with either the first or the third number will be in two digits, so the first digit cannot be 5. Reasoning similarly, we ... | 163 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,740 |
12. Given a 1998-digit number, every two adjacent digits of which form a two-digit number divisible by 17 or 23. The last digit of the number is 1. What is the first? | 12. Let's start recording this number from the end. At some point, we will notice that the number has the form ...92346...9234692346851. Further, we see that the digits 92346 are repeating, so we subtract from 1998 the number of digits that do not belong to this cycle - 3. We divide the resulting number by 5 (the numbe... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,741 |
13. The Forgotten Planet. On each of the planets in a certain solar system, there is an astronomer observing the nearest planet. The distances between the planets are all different. Prove that if the number of planets is odd, then some planet is not observed by anyone. | 13. Since the distances between the planets are different, there are two planets whose distance is the smallest of all. Astronomers on these planets are looking at "each other". If an astronomer from one of the remaining planets observes one of them, then there will not be enough astronomers to observe the remaining pl... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 35,742 |
1. Balls and Boxes. Two players take turns removing balls from two boxes. On their turn, each can take any number of balls from any one (only one) box. The player who takes the last ball wins. Who wins with correct play, and how should one play to win? | 1. If the number of balls in the boxes is the same, the second player wins: on their turn, they can always take the same number of balls as the first player, but from a different box. 2. If the number of balls in the boxes is different, the first player wins: by their move, they can equalize the number of balls in the ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,745 |
2. Strategic game. There is a pile of 31 stones. Two players take turns making moves. In one move, it is allowed to divide any of the existing piles into two piles. The player who cannot make a move loses. Who will win with correct play - the first or the second? | 2. The result of the game does not depend on what moves they will make, since the game will end in 30 moves. This means the second player will win.
## 91 Let's play a little | The\\player\will\win | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,746 |
3. Loves? Or doesn't love? Two girls are playing a game - tearing petals off a daisy. In one move, they can tear off either 1 petal or 2 adjacent petals. The girl who tears off the last petal wins. Who will win with correct play? | 3. With correct play, the second girl wins. With any first move, there is an opportunity to make such a second move that will divide all the remaining petals into two symmetrical parts (see figure). Then the second girl makes moves symmetrically to the first girl's moves. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,747 |
4. Coins on the table. Two players, taking turns, place identical coins on a rectangular table so that they do not touch each other. The player who cannot make a move loses. Who wins with correct play, and what should the winning strategy be? | 4. With correct play, the first player wins. The first move he makes is at the intersection of the diagonals of the rectangle, and subsequently, his moves are made symmetrically to the second's relative to this point.
 if in any of the equalities there is only one *, he places his number there, 2) if not, he places the numbers in any of the equalities where there is an odd number of *. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,751 |
8. Double Chess. Prove that in "double chess" (moves are made according to standard rules, but each player makes 2 moves in a row) white can, at the very least, not lose. | 8. To solve the problem, one must firmly remember how the knight moves in chess! Note that if an arbitrary knight move is made from the initial position and then the knight is returned to its original position, the initial position will be restored, and the turn will pass to the black side. Now, suppose that the whites... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 35,752 |
9. Minus to Plus. The line contains several minuses. Two players, taking turns, change one or two adjacent minuses to pluses. The player who changes the last minus wins. Who will win with correct play? | 9. The first player must, on their turn, split the minuses into two equal groups. Then we play similarly to problem 3. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,753 |
10. Push the checker. In the extreme cells of row 1 to 20, there are white and black checkers. Two players, taking turns, move their checker one or two cells forward or backward, if possible (jumping over the checker is not allowed). The player who cannot move their checker loses. Who wins with correct play - the first... | 10. The second player wins. No matter what move the first player makes, the second can always move in such a way that the number of cells between them is a multiple of 3, while also reducing the distance between the pieces. At some point, the distance between them will become 0, after which the first player will be for... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,754 |
11. Divisibility by 1998. A certain natural number was multiplied by 2, and then 1 was added to it, then it was again multiplied by 2 and 1 was added, and so on many times. Could the result be a number divisible by 1998? | 11. Since $2 n+1$ is an odd number, starting from the second operation, only odd numbers can be obtained. Therefore, a number divisible by 1998 cannot be obtained. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,755 |
12. Chocolate Coupons. A chocolate candy factory places a coupon in each box of candies. For every ten coupons collected, a customer receives a free box of candies. What fraction of the cost of a box does one coupon represent? | 12. Common answer: the coupon costs as much as $\frac{1}{10}$ of the chocolate box. This answer is incorrect because the "prize" chocolate box also contains a coupon. Therefore, the customer gets the prize for 9 coupons, not 10. Answer: the cost of the coupon corresponds to $\frac{1}{9}$ of the cost of the chocolate bo... | \frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,756 |
13. Lucky Tickets. Bus tickets have numbers from 000001 to 999999. A ticket is called lucky if the sum of its first three digits equals the sum of the last three digits. Prove that the sum of the numbers of all lucky tickets is divisible by $9, 13, 37,$ and 1001. | 13. A happy ticket with the number $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}$ corresponds to a unique happy ticket with the number $b_{1} b_{2} b_{3} b_{4} b_{5} b_{6}$, such that $a_{1}+b_{1}=9 ; a_{2}+b_{2}=9 ; \ldots ; a_{6}+b_{6}=9$. Therefore, the sum of all the numbers of happy tickets is divisible by 999999, and thus... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,757 |
14. In the kingdom of King Pea, a rumor spread that the Serpent Gorynych had finally been killed. King Pea knew that this could have been done by one of the Russian heroes: Ilya Muromets, Dobrynya Nikitich, or Alyosha Popovich. Summoning them to his court, the king began to question them. Each hero spoke three times, a... | 14.Let's consider the cases: 1) if the Snake was killed by Ilya Muromets, then he did not lie once but three times, which contradicts the condition, 2) if the Snake was killed by Alyosha Popovich, then all three statements of Ilya will be true, which also contradicts the condition. Only the assumption that the Snake wa... | DobrynyaNikitich | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,758 |
1. Which clocks more often show the exact time: those that are one minute behind or those that have stopped? | 1. A clock that is 1 minute behind every 24 hours shows the correct time once every 2 years. A clock that has stopped shows the correct time twice a day. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,760 |
3. What word is encoded in the number 222122111121, if each letter has been replaced by its number in the alphabet? | 3. In the encrypted word, letters with the following alphabet numbers may appear: $2,22,21,1,11,12$, that is, $\delta, \phi, y, a, \check{u}, \kappa$. By brute force, we find that this word is "fufajka". | fufaika | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,762 |
5. Fifteen numbers are written in a row on the board. Can it happen that the sum of any three consecutive numbers is positive, while the sum of any four consecutive numbers is negative?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 5. No. If the sum of any three consecutive numbers is positive, then the sum of twelve consecutive numbers is also positive (four "triples"). On the other hand,
## 101 More Problems!
if the sum of any four consecutive numbers is negative, then the sum of twelve consecutive numbers should be negative. These two condit... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,764 |
6. It is known that the numbers $3 x+y$ and $4 x+y$ are positive. Can the number $6 x+5 y$ be negative? | 6. Yes, if $x=1$, and $y=-1.4$, for example. | Yes,if1,-1.4,forexample | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,765 |
7. It is known that the numbers $x+y$ and $4 x+y$ are positive. Can the number $8 x+5 y$ be negative? | 7. The number $x+y>0$, Therefore, $4 x+4 y>0$. Adding $4 x+y>0$, we get that the number $8 x+5 y>0$, hence, it cannot be negative. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,766 |
8. Is the statement correct: "if you add the square of a negative number to the same negative number, the result will always be a positive number"? | 8. The statement is incorrect, for example: $-0.5 + (-0.5)^2 < 0$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,767 |
9. A $5 \times 5$ square is filled with numbers such that the product of the numbers in each row is negative. Prove that the product of the numbers in some column is also negative. | 9. Note that the product of all numbers in the table is negative (the product in each row is negative). If the product of numbers in each column were positive, then the product of all numbers in the table would be positive. Contradiction. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,768 |
10. Represent the number 987654321 as the sum of two addends so that each of them consists of the same nine digits, but written in a different order. | 10. For example: $123456789+864197532=987654321$ or $123456798+864197523=987654321$. | 123456789+864197532=987654321 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,769 |
11. Find 1000 natural numbers whose sum equals their product. | 11. The sought numbers can be: 998 units, 2 and 1000. | 998,21000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,770 |
12. From the numbers $21,19,30,25,3,12,9,15,6,27$, select three numbers such that their sum equals 50. | 12. The sum of the numbers 19, 25, and 6 is 50. | 19+25+6=50 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,771 |
13. The warehouse has nails in boxes weighing $24, 23, 17$ and 16 kg. Can the warehouse keeper issue 100 kg of nails from the warehouse without opening the boxes? | 13. For example: 4 boxes - at 17 kg each and 2 boxes - at 16 kg each. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,772 |
14.Can there be two lucky tickets among ten consecutive tickets? A ticket is considered lucky if the sums of its first three and last three digits are equal. | 14. Yes, for example: 199883 and 199892. | Yes | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,773 |
15. Find the sum: $1+2+3+\ldots+1999$. | 15. Let $S=1+2+3+\ldots+1999$. Then: $S=1999+1998+\ldots+3+2+1$ and $2 S=(1+1999)+(2+1998)+\ldots+(1999+1)=2000 \cdot 1999$. Therefore: $S=1000 \cdot 1999$. Answer: 1999000. | 1999000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,774 |
16. Find the sum: $1+3+5+\ldots+1999$. | 16. Let $S=1+3+5+\ldots+1999$. Rewrite it as: $S=1999+1997+\ldots+1$. Then $2 S=2000 \cdot 1000 . S=1000000$. | 1000000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,775 |
17. What is the value of the sum: $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{1024}$ ? | 17. Since $S=1+\frac{1}{2}+\ldots+\frac{1}{1024}=1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{512}-\frac{1}{1024}\right)$, we get that $S=2-\frac{1}{1024}$. Answer: $1 \frac{1023}{1024}$. | 1\frac{1023}{1024} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,776 |
18.Using the digit 4 four times, parentheses, and operation signs, represent each number from 0 to 10. | 18. For example, like this:
$$
\begin{array}{lll}
4+4-4-4=0 & (4-4) \times 4+4=4 & 4+4+4-4=8 \\
4: 4+4-4=1 & (4+4 \times 4): 4=5 & 4+4+4: 4=9 \\
4: 4+4: 4=2 & 4+(4+4): 4=6 & (44-4): 4=10 \\
(4+4+4): 4=3 & 4+4-4: 4=7 &
\end{array}
$$ | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,777 |
19. Will the quotient and remainder change if the dividend and divisor are increased by three times
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a repetition of the instruction and should not be part of the translation. Here is the corrected output:
19. Will the quotient and remainder change if the... | 19. Let $a=b \cdot c+d$, where $a$ is the dividend, $b$ is the divisor, $c$ is the quotient, and $d$ is the remainder. Then $3 a=3 b \cdot c+3 d$, therefore, the quotient will not change, but the remainder will triple? | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,778 |
22. Can the numbers $1,2,3, \ldots, 11,12$ be arranged in a table with 3 rows and 4 columns so that the sum of the numbers in each of the four columns is the same? | 22. It is impossible, since the sum $1+2+\ldots+11+12=(13 \times 12): 2=13 \times 6$ is not divisible by 4. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,780 |
23. Can the numbers $1,2,3, \ldots, 11,12$ be arranged in a table with 3 rows and 4 columns so that the sum of the numbers in each of the three rows is the same? | 23. Possible. Example:
The above text has been translated into English, preserving the original text's line breaks and formatting. | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,781 |
24. Does there exist an integer that when divided by 9 gives a remainder of 2, and when divided by 6 gives a remainder of 1? | 24. Let's translate the problem into "algebraic language." The phrase "the number $m$ when divided by 9 gives a remainder of 2" can be written as: $m=9 n+2$, where $n$ is a natural number, and the phrase "the number $m$ when divided by 6 gives a remainder of 1" can be written as: $m=6 k+1$, where $k$ is a natural numbe... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,782 |
25. Find the number that when divided by 2 gives a remainder of 1, when divided by 3 gives a remainder of 2, when divided by 4 gives a remainder of 3, and when divided by 5 gives a remainder of 4. | 25. A number greater than the desired one by 1 will be divisible by $2,3,4,5$, that is, by 60. The smallest suitable number is 59. | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,783 |
26. For a three-digit number $x$, it is known that if you subtract 7 from it, the result will be divisible by 7, if you subtract 8, the result will be divisible by 8, and if you subtract 9, the result will be divisible by 9. What is this number? | 26. The desired number is divisible by 7, 8, and 9. Therefore, this number is 504. There are no other such three-digit numbers. | 504 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,784 |
27. Does there exist a number that decreases by 58 times when the first digit is crossed out? | 27. Let $x$ be the crossed-out digit, then $x \cdot 10^{n}+y=58 y$, or $x \cdot 10^{n}=57 y$. The right-hand side of this equation is divisible by 19, but the left-hand side is not. Contradiction. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,785 |
28. Baron Münchhausen claimed that he managed to find such a natural number, the product of the digits of which is equal to 6552. Prove that, as always, he was lying. | 28. To check Münchhausen's statement, let's factorize the number 6552 into prime factors. We get: $6552=2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 13=2^{3} \times 3^{2} \times 7 \times 13$. Since the number $13$ is prime, meaning it cannot be represented as a product of single-digit factors, and it itself is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,786 |
30. What digit does the number $3^{100}$ end with? | 30. The number $3^{100}=81^{25}$, and therefore, ends in 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,788 |
31. What is the remainder when the number $2^{99}$ is divided by 7?
95 More problems! | 31. Note that $2^{99}=8^{33}=(7+1)^{33}=(7+1) \ldots(7+1)$. Expand the brackets. The resulting terms will be divisible by 7, except for 1. Thus, $2^{99}$ can be written in the form: $7 x$ +1 . Answer: 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,789 |
32. Is the number $10^{1998}+8$ divisible by 9? | 32. Note that $10^{1998}+8=100 \ldots 008$ (with a total of 1997 zeros). The sum of the digits of this number is divisible by 9, hence the number itself is also divisible by 9. | 10^{1998}+8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,790 |
35. How many diagonals does a thirty-four-sided polygon have? | 35. Each vertex of the polygon is connected by diagonals to all other vertices, except for two, which it is connected to by a side. Thus, each vertex of the thirty-four-sided polygon is connected by diagonals to 31 vertices. Note that in this counting system, each diagonal is counted twice. In total: 527 diagonals. | 527 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,793 |
36. The ball is tightly wrapped in a rope net, in which exactly three ropes come out of each node. Can the net contain exactly 2001 nodes? | 36. Suppose this is possible. If we count the number of strings coming out of each knot, we get 2001.3. This is impossible, because in such a counting system, we count each string twice. Therefore, the network cannot consist of exactly 2001 knots. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,794 |
37. In two boxes, there are 70 coins. It is known that in the first box, $\frac{5}{9}$ of the total number of coins are gold, and the rest are silver, in the second box, $\frac{7}{17}$ of the number of coins are silver, and the rest are gold. How many coins are in each box | 37.Let $9 x$ be the number of coins in the first box, and $17 y$ be the number in the second, then $9 x + 17 y = 70$, where $x$ and $y$ are natural numbers, and $1 \leq y \leq 4$. By enumeration, we find that $y=2$, and $x=4$. Answer: in the first -36 coins, in the second -34. | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,795 |
38. On the board, the numbers are written: $1 ; 2 ; \ldots 1996 ; 1997$. It is allowed to erase any two numbers and write down the absolute value of their difference instead. Prove that it is impossible to achieve that only 0 remains on the board. | 38.When we erase numbers $a$ and $b(a>b)$, the sum of the numbers on the board decreases by $(a+b)-(a-b)=2b$ - an even number. Since the sum $1+2+\ldots+1997=\frac{(1+1997) \cdot 1997}{2}=999 \cdot 1997-$ is odd, an even number 0 cannot be obtained from it. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,796 |
39. Calculate: $2379 \cdot 23782378-2378 \cdot 23792379$. | 39. Let $a=2378$, then the desired expression is: $(a+1)(10000 a+a)-a(10000(a+1)+(a+1))=$ $=(a+1) \cdot 10001 a-a \cdot 10001 \cdot(a+1)=0$ | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,797 |
41.On the farmyard, geese and piglets were wandering around. A boy counted the number of heads, there were 30, then he counted the total number of legs, there were 84. Can you find out how many geese and how many piglets were on the farmyard? | 41.If only geese were wandering around the farmyard, there would be 60 legs in total, the "extra" legs, which number 24, belong to the piglets - two for each. Therefore, there were 12 piglets, and 18 geese. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,799 |
42.A brick weighs 2 kg and half a brick. How much does the brick weigh? | 42.It follows from the condition that half a brick weighs 2 kg. Therefore, a whole brick weighs 4 kg. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,800 |
43. A cup and a saucer cost 2500 rubles, while 4 cups and 3 saucers cost 8870 rubles. Find the price of a cup and the price of a saucer. | 43.4 cups and 4 saucers cost 10000 rubles, while 4 cups and 3 saucers cost 8870 rubles, therefore, the price of one saucer: $10000-8870=1130$ rubles, the price of one cup: $2500-1130=1370$ rubles. | 1370 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,801 |
45. A peasant bought a cow, a goat, a sheep, and a pig, paying 1325 rubles. The goat, pig, and sheep together cost 425 rubles. The cow, pig, and sheep together cost 1225 rubles, and the goat and pig together cost 275 rubles. Find the price of each animal. | 45.All the animals together cost 1325 rubles, and all except the cow cost 425 rubles, so the cow costs: $1325-425=900$ rubles. The goat, pig, and sheep cost 425 rubles, and the same without the sheep cost 275 rubles, so the sheep costs $425-275=150$ rubles. All, except the goat, cost 1225 rubles, so the goat costs: $13... | 900,150,100,175 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,803 |
46.Cargo was delivered to three warehouses. 400 tons were delivered to the first and second warehouses, 300 tons were delivered to the second and third warehouses together, and 440 tons were delivered to the first and third warehouses. How many tons of cargo were delivered to each warehouse separately? | 46. Let's write the condition as: $I$ warehouse $+I I$ warehouse $=400$ t.; $I I$ warehouse $+I I I$ warehouse $=300$ t.; $I$ warehouse $+I I I$ warehouse $=440$ t.
Then, "adding" these three conditions, we get that the doubled sum of the loads on the three warehouses is 1140 t., and the sum is 570 t. From this: on th... | 270,130,170 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,804 |
47. 6 carps are heavier than 10 perches, but lighter than 5 pikes; 10 carps are heavier than 8 pikes. What is heavier: 2 carps or 3 perches? | 47. Since 6 carp are heavier than 10 perch, it is clear that 6 carp are even heavier than 9 perch. Therefore, 2 carp are heavier than 3 perch. This means that two of the three conditions in the problem are redundant. | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,805 |
48. How old is each of us, if I am now twice as old as you were when I was as old as you are now, and we are both 63 years old in total. | 48. Translating the condition of the problem into the language of algebra, we get the system:
| | Me | You |
| :--- | :--- | :--- |
| Now | $2 x$ | $y$ |
| Then | $y$ | $x$ |
$\quad\left\{\begin{array}{c}2 x+y=63 \\ 2 x-y=y-x\end{array}\right.$. From which $x=18, y=27$. Answer: I am 36 years old, You are 27 years ol... | I36old,You27old | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,806 |
50.What is the 1997th digit in the decimal expansion of the fraction $\frac{1}{7}=0.142857 \ldots ?$ | 50.If we divide 1 by 7 using long division, we get that $\frac{1}{7}=0.(142857)$. The remainder of 1997 divided by 6 is 5, Therefore, the digit at the 1997th place is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,808 |
51. Little One eats a jar of jam in six minutes, while Karlson is twice as fast. How long will it take them to eat the jam together | 51. The question of the problem can also be formulated as follows: "How long would it take for three Little Ones to eat the jam?" (According to the condition of the problem, Carlsson can be equated to two Little Ones). It is clear that three Little Ones would finish the jam three times faster than one. Answer: in 2 min... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,809 |
52. One fountain fills the pool in 2.5 hours, the other in 3.75 hours. How long will it take for both fountains to fill the pool? | 52. If both fountains together fill the pool in $x$ hours, then, considering that the first fountain fills $\frac{2}{5}$ of the pool per hour, and the second fills $\frac{4}{15}$ of the pool per hour, we get the equation: $\frac{2}{5}+\frac{4}{15}=\frac{1}{x}$, from which $x=1.5$. Answer: in 1.5 hours. | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,810 |
54. Two workers had the same salary. The salary of the first worker was increased by $100 \%$, while that of the second worker was increased by only $50 \%$. By what percentage is the new salary of the first worker greater than that of the second? | 54. Let the original salary be $x$, then after the raise, the first person's salary will be $2x$, and the second person's salary will be $1.5x$. But $0.5x$ is one third of $1.5x$. Answer: $33 \frac{1}{3} \%$. | 33\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,812 |
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