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1. The Sleeping Passenger. When the passenger had traveled half of the entire journey, he went to sleep and slept until only half of the distance he had traveled while sleeping remained to be traveled. What part of the journey did he travel while sleeping? | 1. The passenger slept for two thirds of the second half of the journey, which is one third of the entire journey. | \frac{1}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,551 |
2. Morning in the store. At what price should a mixture of two types of candies be sold if the price of the first type is 10 rubles per kilogram, the second type is 15 rubles per kilogram, and the weight of the first type of candies is three times greater than that of the second? | 2. Let $3 x$ kg be the weight of candies of the first type, then their total cost is $45 x$ rubles, and the weight is $4 x$ kg. They should be sold at a price of $\frac{45 x}{4 x}$ rubles, that is, at 11 rubles and 25 kopecks per 1 kg. | 11.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,552 |
3. Roman Law. A man, dying, bequeathed: “If my wife gives birth to a son, let him be given $\frac{2}{3}$ of the estate, and the wife the remaining part. If a daughter is born, let her have $\frac{1}{3}$, and the wife $\frac{2}{3}$.” Twins, a son and a daughter, were born. How should the estate be divided? | 3. A wise legal decision is as follows: since the will states that the son should receive twice as much as the mother, and the mother should receive twice as much as the daughter, it follows that the son should receive $\frac{4}{7}$, the mother - $\frac{2}{7}$, and the daughter - $\frac{1}{7}$ of the inheritance.
Frac... | \frac{4}{7},\frac{2}{7},\frac{1}{7} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,553 |
4. Barrels of Honey. Three people want to divide among themselves seven full barrels of honey, seven barrels filled with honey halfway, and seven empty ones, but in such a way that both the honey and the containers are divided equally. How can this division be carried out without transferring honey from one barrel to a... | 4. First solution: the first person (as well as the second) should take three full barrels, one half-empty, and three empty; the third - one full, five half-empty, and one empty. Second solution: the first person (as well as the second) - two full barrels, three half-empty, and two empty; the third - three full, one ha... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,554 |
5. Which of the fractions is greater: $\frac{199719973}{199719977}$ or $\frac{199819983}{199819987}$? | 5. We will compare not the numbers themselves, but their complements to 1. $\frac{4}{199819987}<\frac{4}{199719977}$, from which: $\frac{199719973}{199719977}<\frac{199819983}{199819987}$. | \frac{199719973}{199719977}<\frac{199819983}{199819987} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,555 |
6. What can you buy for a ruble? Nine boxes of matches cost 9 rubles and some kopecks, while ten such boxes cost 11 rubles and some kopecks. How much does one box cost? | 6. Note that a box of matches costs more than $\frac{11}{10}$ rubles, but less than $\frac{10}{9}$ rubles. That is, more than 1.10 rubles, but less than 1.111 rubles. Answer: 1 ruble 11 kopecks. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,556 |
7. Physical problem. The volume of water increases by $10 \%$ when it freezes. By what percentage does the volume of ice decrease when it melts? | 7. If the volume of water increases by $\frac{1}{10}$ (i.e., becomes $\frac{11}{10}$) when it freezes, then when the ice melts, the volume decreases by $\frac{1}{11}$ (from $\frac{11}{10}$!), which is $9 \frac{1}{11} \%$. | 9\frac{1}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,557 |
8. How much does the dress cost? The worker's monthly pay, that is, for thirty days, is ten dinars and a dress. He worked for three days and earned the dress. What is the cost of the dress? | 8. Since for 30 days he will receive 10 dinars and a dress, then for 3 days he will receive a dinar and $\frac{1}{10}$ of a dress, therefore, the dress costs $1 \frac{1}{9}$ dinars. One can reason differently: since he earned a dress in three days, then in 27 days he should have received 10 dinars, and in 3 days accord... | 1\frac{1}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,558 |
9. Coffee with milk. First, $\frac{1}{6}$ of a cup of black coffee was sipped and then it was topped up with milk. Then $\frac{1}{3}$ of a cup was drunk and it was topped up with milk again. Then another half cup was drunk and it was topped up with milk once more. Finally, a full cup was drunk. What was drunk more: cof... | 9. If one cup of coffee was drunk, and milk was added first $\frac{1}{6}$ cup, then $\frac{1}{3}$ cup, and finally $\frac{1}{2}$ cup, that is, also one cup, which means that the coffee and milk were drunk in equal amounts. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,559 |
10.Fair Division. According to the will of a deceased father, three sons were to divide 7 horses among themselves so that the eldest would receive half, the middle one a quarter, and the youngest an eighth. The will greatly puzzled the heirs - for its implementation would have required cutting the horses into pieces. A... | 10. Notice that $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{7}{8}$, not 1. If the will is followed literally, the eldest son will receive 3.5 horses, the middle son 1.75 horses, and the youngest 0.875 horses. In this case, 0.875 horses will not go to anyone, and the neighbor, in fact, helped them divide the horses. He p... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,560 | |
11. From the "Greek Anthology". Four springs rise from the ground. The first fills the basin in one day, the second in 2 days, the third in 3 days, and the fourth in 4 days. How long will it take for all four springs to fill the basin? | 11. If all four sources were open, then in twelve days they would fill $12+6+4+3=25$ pools, therefore they would fill one pool in $\frac{12}{25}$ of a day. | \frac{12}{25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,561 |
12. The Big Laundry. After seven hours of washing, the length, width, and height of the soap piece were halved. For how many washes will the remaining soap last? | 12. Since the length, width, and height of the soap piece have been halved, its volume has decreased by 8 times, meaning that in 7 hours, the soap piece has decreased by $\frac{7}{8}$ of its volume (by $\frac{1}{8}$ of its volume per hour). Therefore, the soap will last for one more hour of heavy washing. | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,562 |
13. A traveler returned to the hotel. This time he had a chain with 7 links. He had to pay one link per day for his stay, but the hotel owner warned that he would accept no more than one cut link. Which links should be cut so that the traveler can stay in the hotel for 7 days and pay the owner daily? | 13. The third link should be cut. This will cause the chain to break into pieces consisting of one, two, and four links. On the first day, he should give one link, on the second day, two links, receiving one back in change, on the third day, one link again, on the fourth day, four links, and receive back the pieces of ... | Thethirdlinkshouldbecut | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,563 |
1. Compare the numbers. It is known that $2 \%$ of a positive number $A$ is greater than $3 \%$ of a positive number $B$. Is it true that $5 \%$ of number $A$ is greater than $7 \%$ of number $B ?$ | 1. Since $2 \%$ of number $A$ is greater than $3 \%$ of number $B$, then $4 \%$ of number $A$ is greater than $6 \%$ of number $B$, and moreover, $1 \%$ of number $A$ is greater than $1 \%$ of number $B$. "Adding" the last two statements, we get that $5 \%$ of number $A$ is greater than $7 \%$ of number $B$. Or: $0.02 ... | 0.05A>0.07B | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,566 |
2. Petya bought two books. The first one was $50 \%$ more expensive than the second. By what percentage is the second book cheaper than the first? | 2. The second book is one third cheaper than the first, that is, by $33 \frac{1}{3} \%$. | 33\frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,567 |
3. There was an equal amount of water in two barrels. The amount of water in the first barrel initially decreased by $10 \%$, and then increased by $10 \%$. The amount of water in the second barrel, on the contrary, initially increased by $10 \%$, and then decreased by $10 \%$. In which barrel did the amount of water b... | 3. Let there initially be $x$ liters of water in each barrel, then in the first barrel, after all changes, there became $x \cdot 0.9 \cdot 1.1 = 0.99 x$ liters of water, and in the second $x \cdot 1.1 \cdot 0.9$ - which is also $0.99 x$ liters of water. Answer: equally. | 0.99x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,568 |
4. Where is it cheaper? In one store, milk became cheaper by $40 \%$, and in another, it first became cheaper by $20 \%$, and then by another $25 \%$. Where did the milk become cheaper? The initial price of milk in each store was the same. | 4. Let the initial price of milk be $x$ rubles. In the first store, it became $40 \%$ cheaper, that is, $0,6 x$ rubles. In the second store, after the first reduction, the milk cost $0,8 x$, and after the second reduction, $0,8 x \cdot 0,75=0,6 x$ rubles. Thus, the milk in each of the stores is once again priced the sa... | 0.6x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,569 |
5. How has the purchasing power of the population changed? The product has become cheaper by $20 \%$. By what percentage can more of the product be bought for the same amount of money? | 5. The product became cheaper by $20 \%$. Therefore, all the previously purchased goods could be bought by spending $80 \%$ of the money, and the remaining $20 \%$ could buy an additional $\frac{1}{4}$ of the goods, which constitutes $25 \%$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,570 |
6. Dried mushrooms. The moisture content of fresh mushrooms is $99\%$, and that of dried mushrooms is $98\%$. How has the weight of the mushrooms changed after drying? | 6. Let the weight of fresh mushrooms be $100 x$ kg, then the weight of the dry matter in them is $x$ kg. After drying, the weight of the dry matter did not change and became $2 \%$ (one fiftieth) of the weight of the mushrooms. The weight of the dried mushrooms is $50 x$ kg, which means it has decreased by half. | 50x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,571 |
7. At the conference. $85 \%$ of the delegates at the conference know English, and $75 \%$ know Spanish. What fraction of the delegates know both languages? | 7. Note that $85 \%+75 \%=160 \%$, which exceeds the total number of conference delegates by $60 \%$. The excess is due to those people who know both languages - we counted them twice. Thus, not less than $60 \%$ of the conference delegates know both languages. | 60 | Other | math-word-problem | Yes | Yes | olympiads | false | 35,572 |
9. Seawater contains $5 \%$ salt. How many kilograms of fresh water need to be added to 40 kg of seawater to make the salt content $2 \%$? | 9. In 40 kg of seawater, there is $40 \cdot 0.05=2$ (kg) of salt, which in the new solution constitutes $2\%$, therefore, the solution should be $2: 0.02=100$ (kg). Answer: 60 kg of fresh water should be added. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,574 |
10. Vера and Аnya attend a math club, in which there are more than $91 \%$ boys. Find the smallest possible number of club participants. | 10. Let $x$ be the number of participants in the club, and $y$ be the number of girls. Then, according to the conditions of the problem, $0.09 x > y$ or $9 x > 100 y$, where $x$ and $y$ are natural numbers. Solving the problem by enumeration, we will verify that the smallest possible solution for $y=2$ is achieved at $... | 23 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,575 |
11. Who is more? Every tenth mathematician is a philosopher. Every hundredth philosopher is a mathematician. Who is more, philosophers or mathematicians?
## 33 Percentages | 11. Let $x$ be the number of mathematicians who are also philosophers, then the number of mathematicians is $10 x$, and the number of philosophers is $100 x$. Therefore, there are more philosophers. | there\\\philosophers | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,576 |
12. Mysterious Inscription. During excavations of an ancient city, a stone was found on which the following was engraved:
"In this inscription, the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are used, and: digit 0 is used ... times; digit 1 is used ... times; digit 2 is used ... times; digit 3 is used ... times; digit 4 is u... | 12. The inscription should look like this:
«In this inscription, the digits $0,1,2,3,4,5$, 6, 7, 8, 9 are used, and: the digit 0 is used 2 times; the digit 1 is used 2 times; the digit 2 is used 8 times; the digit 3 is used 4 times; the digit 4 is used 3 times; the digit 5 is used 2 times; the digit 6 is used 2 times;... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,577 |
13. Neat Pete numbered all the pages of a notebook from 1 to 192. Vasya tore out 25 sheets from this notebook and added up all 50 numbers written on them. Could he have obtained the number $1998$? | 13. Since the first page in the torn block has an odd number and the last page has an even number, the total sum must be odd. Therefore, the number 1998 could not have been obtained. | 1998couldnothavebeenobtained | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,578 |
14. A painted cube with an edge of 10 cm was sawn into cubes with an edge of $1 \mathrm{~cm}$. How many of them will have one painted face? How many will have two painted faces? | 14. $K$ each face of the cube adjoins $8 \times 8=64$ cubes, painted only on one side (make a drawing!). There are 6 faces, so 384 cubes are painted on one side. Reasoning similarly, we get that 96 cubes are painted on two sides. | 384 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,579 |
15. Battleship. On the game board for "Battleship," there is a four-cell "ship" $\square \square \square \square$. What is the minimum number of "shots" needed to definitely destroy it? | 15. Note that on a $10 \times 10$ board, 24 ships can fit simultaneously, so it is impossible to detect a ship with fewer shots. By shooting as shown in the first figure, we will definitely hit a ship. After that, it will take 4 shots to determine in which "direction" it is located, and another 2 shots to destroy it.
... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,580 | |
3. Visiting Mumbo-Jumbo. In the language of the Mumbo-Jumbo tribe, there are only two sounds: "ы" and "y". Two words mean the same thing if one can be obtained from the other by applying a certain number of the following operations: skipping consecutive sounds "ыy" or "yyыы" and adding the sounds "yы" at any position. ... | 3. Note that with any of the allowed operations, the number of letters "ы" and "y" in the word changes by the same amount. Therefore, from the word "уыу", where the number of letters "y" is greater, it is impossible to obtain the word «ыуы», where the number of letters «у» is less. Answer: no. | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,583 |
4. Another circus act. On the table, there are 6 glasses. Five of them are standing upright, while one is upside down. You are allowed to flip any 4 glasses at the same time. Is it possible to get all the glasses standing upright? | 4. First solution: Notice that the number of glasses standing incorrectly always remains odd, therefore, it cannot become equal to 0.
Second solution: let's associate the number 1 with each correctly standing glass, and the number -1 with each incorrectly standing glass. We will calculate the product of the numbers co... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,584 |
6. A grasshopper jumps along a straight line: the first jump is 1 cm, the second is $2 \mathrm{~cm}$, the third is $3 \mathrm{~cm}$, and so on. Can it return to the point where it started after the twenty-fifth jump? | 6. Let the grasshopper jump along the number line and start from the point with coordinate 0. Note that after 25 jumps, it will end up at a point with an odd coordinate (among the numbers from 1 to 25, the number of odd numbers is odd). Since 0 is an even number, it cannot return. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,586 |
11. Guess the number. Sixth-graders Petya, Vasya, Kolya, and Anton made the following statements. Petya: “The number is prime.” Vasya: “The number is 9.” Kolya: “The number is even.” Anton: “The number is 15.” It is known that only one of Petya's and Vasya's statements is true, just as only one of Kolya's and Anton's s... | 11. Note that if Vasya told the truth, then both Kolya and Anton lied. This means that Petya told the truth. From this, Anton lied. Therefore, the number thought of is 2. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,591 |
12. To number the pages of an encyclopedia, 6869 digits were required. How many pages are there in it? | 12.Note that there are 9 single-digit numbers, 99-9=90 two-digit numbers, digits - 90 × 2=180; 999-99=900 three-digit numbers; digits - 900 × 3=2700. Let's calculate the number of four-digit numbers in the page numbering of the book: 6869-2700-180-9=3980, 3980: 4=995. Adding up the obtained results, we get: 9+90+900+99... | 1094 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,592 |
13. About the cat. A ribbon was stretched around the Earth along the equator. Then the length of the ribbon was increased by 1 meter and again evenly positioned around the equator. Will a cat be able to crawl through the resulting gap? | 13. Let the radius of the Earth be $R$, and the formed gap be $h$, then $2 \pi(R+h)-2 \pi R=1$ and $2 \pi h=1$. Therefore, the size of the gap $h=\frac{1}{2 \pi} \approx 15$ cm - a gap quite sufficient for a cat to squeeze through. | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,593 |
14. Simplify the fraction: $\frac{37373737}{81818181}$ | 14. Solution: $\frac{37373737}{81818181}=\frac{37 \cdot 1010101}{81 \cdot 1010101}=\frac{37}{81}$. | \frac{37}{81} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,594 |
1. Incorrect entry. Petya wrote an example of multiplying two-digit numbers on the board. Then he erased all the digits and replaced them with letters. It turned out to be the equation: $\overline{a b} \cdot \overline{c d}=\overline{m n p k t}$. Prove that he made a mistake. | 1. The equality $\overline{a b} \cdot \overline{c d}=\overline{m n p k t}$ cannot hold, since the largest possible product of two-digit numbers $99 \cdot 99<100 \cdot 100<10000$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 35,596 |
2. In a certain kingdom, the share of blondes among blue-eyed people is greater than the share of blondes among all its inhabitants. Is the share of blue-eyed people among blondes greater or the share of blue-eyed people among the inhabitants of the kingdom? | 2. Let $Г Б$ be the number of blue-eyed blondes, $\Gamma$ be the number of blue-eyed people, Б be the number of blondes, and $B$ be the total number of people. Translating the condition of the problem into the "language of algebra": $\frac{\Gamma \zeta}{\Gamma}>\frac{E}{B}$. Multiplying both sides of the inequality by ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,597 |
3. $о$ about the fisherman and the fish. When asked how big the fish he caught was, the fisherman said: “I think that its tail is 1 kg, the head is as much as the tail and half of the body, and the body is as much as the head and the tail together.” How big is the fish? | 3. Let $2 x$ kg be the weight of the torso, then the head will weigh $x+1$ kg. From the condition that the torso weighs as much as the head and tail together, we get the equation: $2 x=x+1+1$. From this, $x=2$, and the whole fish weighs - 8 kg. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,598 |
4. How old is each of us? I am now twice as old as you were when I was as old as you are now. Together, we are 35 years old. How old is each of us? | 4. Let $2x$ be my current age, and $y$ be your current age, then $y$ was my age back then, and $x$ was your age back then. How many years have passed? We can calculate it in two ways: $2x - y = y - x$. Moreover, it is known that $2x + y = 35$. From the first equation, we get: $2y = 3x$. On the other hand, $y = 35 - 2x$... | I20old,you15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,599 |
5. The Metallurgist's Task. The first alloy consists of zinc and copper in a ratio of $1: 2$, while the second alloy contains the same metals in a ratio of $2: 3$. From how many parts of both alloys can a third alloy be obtained, containing the same metals in a ratio of $17: 27$? | 5. Let's write the condition of the problem in the form of a table:
| alloys: | weight of zinc (kg) | weight of copper (kg) | take for the new alloy (kg) |
| :---: | :---: | :---: | :---: |
| 1st alloy | $x$ | $2 x$ | $A$ |
| 2nd alloy | $2 y$ | $3 y$ | $B$ |
| new | $x+2 y$ | $2 x+3 y$ | |
Note that: $\frac{x+2 y}{... | 9:35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,600 |
6. A Herd of Elephants. Springs are bubbling at the bottom of the lake. A herd of 183 elephants could drink it dry in one day, while a herd of 37 elephants would take 5 days. How many days would it take for 1 elephant to drink the lake? | 6. Let's translate the problem into the language of algebra. Let the volume of the lake be $V$ liters, the elephant drinks $C$ liters of water per day, and $K$ liters of water flow into the lake from the springs per day. Then the following two equations hold: $183 C = V + K$ and $37 \cdot 5 C = V + 5 K$; from which we ... | 365 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,601 |
11. The Clock Hands Problem. At what time after 12:00 will the minute hand first catch up with the hour hand? | 11. At 13:00, the minute hand will lag behind the hour hand by 5 minute divisions. Before the "meeting," the hour hand will travel $x$ divisions, and the minute hand will travel $12x$ divisions. From the equation $x + 5 = 12x$, we get that $x = \frac{5}{11}$ minute divisions on the clock face. Answer: 1 hour $5 \frac{5... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,606 |
12.Cunning manipulations. With the number written on the board, it is allowed to perform the following operations: replace it with its double, or erase its last digit. How can you use these operations to get from the number 458 to the number 14? | 12.Let A be the doubling of a number, and B be the erasing of the last digit. Then: $458-\mathrm{B} \Rightarrow 45-\mathrm{A} \Rightarrow 90-\mathrm{B} \Rightarrow 9-\mathrm{A} \Rightarrow 18-\mathrm{A} \Rightarrow 36-\mathrm{A} \Rightarrow 72-\mathrm{B} \Rightarrow 7-\mathrm{A} \Rightarrow 14$. | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,607 |
13. Task $\Gamma$. Dudeney. If one and a half chickens lay one and a half eggs in one and a half days, then how many chickens plus another half chicken, laying eggs one and a half times faster, will lay ten eggs and a half in one and a half weeks? | 13.If one and a half chickens lay one and a half eggs in one and a half days, then one chicken will lay one egg in one and a half days. A chicken that "works" one and a half times faster will lay one and a half eggs in one and a half days (one egg per day). Therefore, this chicken will lay ten and a half eggs in ten an... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,608 |
14.Workdays. Three workers are digging a pit. They work in turns, and each works as long as the other two need to dig half the pit. Working this way, they dug the pit. How many times faster would they have finished the work if they had worked simultaneously
## 43 Languages of Mathematics | 14. Let each of them work for $a, b$, and $c$ hours respectively. Consequently, the work was completed in $a+b+c$ hours. Imagine that all this time they worked together. How many holes will they dig in total? In the first $a$ hours, the first worker will dig his share of the "total" hole, while the other two will dig h... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,609 |
1. $O$ about the wolf, the goat, and the cabbage. A wolf, a goat, and a cabbage need to be transported across a river. The boat can carry only one of the three, in addition to the ferryman. How can they be transported so that the goat does not eat the cabbage, and the wolf does not eat the goat? | 1. First: transport the goat. Second: leave it on the other bank and return to transport the wolf. $\boldsymbol{B}$-third: since the wolf cannot be left with the goat, return with the goat, leave it, and transport the cabbage to the other bank. In fourth: and, finally, transport the goat. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,611 |
2. Fake Coins. Out of nine coins, one is fake: it is lighter than the others. How can you determine which one it is with just two weighings on a balance scale without weights? | 2. Divide the coins into 3 piles of 3 coins each. First weighing: place 3 coins on each pan of the balance. There are two possible cases: 1. Balance, then the coins on the scales are genuine, and the fake coin is among those that were not weighed; 2. One of the piles is lighter, then the fake coin is in that pile. Seco... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,612 |
3. Other scales. Science has progressed. Now experts have high-precision electronic scales at their disposal. But the tasks are now more difficult. Here is one of them: there are 10 bags of coins, in nine bags the coins are genuine (the weight of one coin is 10 g), in one - counterfeit (the weight of one coin is 11 g).... | 3. Let's write a number from 1 to 10 on each bag and take a number of coins from each bag corresponding to its number, a total of 55 coins. If all of them were genuine, their weight would be 550 grams, but there are counterfeit coins among them, and the weight will be greater. If the counterfeit coins are in the first ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,613 |
4. Divide equally. There is milk in an eight-liter bucket. How can you measure out 4 liters of milk using a five-liter bucket and a three-liter jar? | 4. It is convenient to write the solution in the form of a table:
| 8-liter can | 8 | 3 | 3 | 6 | 6 | 1 | 1 | 4 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 5-liter can | 0 | 5 | 2 | 2 | 0 | 5 | 4 | 4 |
| 3-liter jar | 0 | 0 | 3 | 0 | 2 | 2 | 3 | 0 | | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,614 |
5. The Zmey Gorynych has 2000 heads. A legendary hero can cut off $1, 17, 21$ or 33 heads with one blow, but in return, $10, 14, 0$ or 48 heads grow back, respectively. If all heads are cut off, no new ones grow. Can the hero defeat the Zmey? | 5. We can propose the following tactic for cutting off the heads of the Snake: 1. Initially, we will cut off 21 heads (94 times); no new heads will grow, and the Snake will be left with 26 heads. 2. Next, we will cut off 17 heads three times (remember, 14 new heads will grow back each time) - after which 17 heads will ... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,615 |
9. Is this possible? Can three people cover a distance of 60 km in 3 hours if they have a two-person motorcycle at their disposal? The motorcycle's speed is 50 km/h, and the pedestrian's speed is 5 km/h. | 9. Yes. In the first hour, two of them ($\mathcal{A}$ and $B$) travel 50 km by motorcycle, while the third ($C$) walks 5 km. Then $B$, after getting off the motorcycle, walks the remaining distance to the destination in 2 hours. In the second hour, $C$ walks another 5 km, while $\mathcal{A}$ returns 40 km and waits for... | Yes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,618 |
10. Two million points. On the plane, two million points are marked. Can a line be drawn such that there would be exactly one million of these points on each side of it? | 10.Let's take a point lying to the right of the given set of points, so that it does not lie on any line that can be drawn through any two points belonging to the given set. Draw a line through it so that all points are to the left of it. Start rotating this line clockwise. It will sequentially pass through all the poi... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,619 |
12. Restore the digits. On the board, there was the product of three consecutive even numbers. During the break, Vasya erased some digits. As a result, on the board remained $87 * * * * * 8$. Help Petya find the missing digits in the product. | 12. These numbers $442 \cdot 444 \cdot 446=87526608$. | 87526608 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,621 |
13. From empty to empty. In one vessel, there are $2a$ liters of water, and the other is empty. From the first vessel, half of the water in it is poured into the second, then from the second, $\frac{1}{3}$ of the water in it is poured back into the first, then from the first, $\frac{1}{4}$ of the water in it is poured ... | 13. We will record the amount of water in the vessels after each transfer in the table.
| steps | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $2 a$ | $a$ | $\frac{4}{3} a$ | $a$ | $\frac{6}{5} a$ | $a$ | $\frac{8}{7} a$ | $a$ |
| 2 | 0 | $a$ | $\frac{... | a | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,622 |
14. Garden plot. The length of a rectangular plot of land was increased by $35 \%$, and the width was decreased by $14 \%$. By what percentage did the area of the plot change? | 14. Let $a$ be the length of the plot, and $b$ be its width. The length increased by 1.35 times, and the width increased by 0.86 times. This means the area increased by 1.161 times. Express 116.1% in percentages. | 16.1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 35,623 |
2. What is the day of the week today, if it is known that "when the day after tomorrow becomes yesterday, today will be as far from Sunday as the day that was today when yesterday was tomorrow"? | 2. Answer: Wednesday. Then "the day after tomorrow will become yesterday" - Friday, and "the day that was today when yesterday was tomorrow" - Monday. | Wednesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,626 |
4. In a family, there are six children. Five of them are respectively 2, 6, 8, 12, and 14 years older than the youngest, and the age of each child is a prime number. How old is the youngest? | 4. Suppose the child's age does not exceed 35 years. Let's list all the prime numbers: $2,3,5,7,11,13,17,19,23$, 29,31. It is clear that the age of the younger child is an odd number. The numbers 29 and 31 also do not fit. Let's find the age of the younger child. He cannot be 1 year old, because $1+8=9$. His age cannot... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,628 |
6. In which year does the thirteenth day never fall on a Monday? | 6. In any year, at least one Monday will be the 13th of the month - verify this yourself: if March 13 is a Tuesday, then August 13 is a Monday; if March 13 is a Wednesday, then May 13 is a Monday; if March 13 is a Thursday, then October 13 is a Monday; if March 13 is a Friday, then April 13 is a Monday; if March 13 is ... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,630 |
7. In one month, three Wednesdays fell on even dates. What date will the second Sunday of this month be?
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。 | 7. If in one month three Wednesdays fall on even dates, then the first Wednesday is the 2nd, the third is the 16th, and the fifth is the 30th (for example, if the first Wednesday is the 4th, then the fifth would be the 32nd), then the second Sunday will be the 13th. | Calculus | math-word-problem | Yes | Yes | olympiads | false | 35,631 | |
8. Happy New Year! What day of the week was January 1st if this January had four Fridays and four Mondays? | 8. Answer: January 1st was a Tuesday. | Tuesday | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,632 |
10. What does the year more often start with: Saturday or Sunday?
I've maintained the original text's line breaks and formatting as requested. | 10. Note that if among 28 consecutive years every fourth year is a leap year (i.e., there is no year whose number is divisible by 100 but not by 400), then these 28 years contain a whole number of weeks. Since January 1, 1901, was a Tuesday, January 1, 1929, will also be a Tuesday. It can be calculated that during thes... | NewYear'DaystartsonSundayoftenthanonSaturday | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,634 |
12. Cat and Mouse. If $n$ cats eat $n$ mice in $n$ hours, then how many mice will $p$ cats eat in $p$ hours? | 12.Let's assume that cats eat mice, not the other way around. If $n$ cats eat $n$ mice in $n$ hours, then one cat will eat 1 mouse in $n$ hours, and in 1 hour - $\frac{1}{n}$ of a mouse. Consequently, $p$ cats will eat $\frac{p^{2}}{n}$ mice in $p$ hours. | \frac{p^2}{n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,636 |
13. In a line, there are 20 numbers. It is known that the sum of any three consecutive numbers is greater than zero. Can we assert that the sum of all 20 numbers is greater than zero? | 13. No. For example:
| -5 | -5 | 11 | -5 | -5 | 11 | -5 | -5 | 11 | -5 | -5 | 11 | -5 | -5 | 11 | -5 | -5 | 11 | -5 | -5 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | No | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 35,637 |
14. To eat or to sleep? Let's assume that if a person does not eat for 7 days or does not sleep for 7 days, they will die. Suppose a person has not eaten or slept for a week. What should they do first by the end of the seventh day: eat or sleep, to stay alive? | 14. A person cannot eat and sleep at the same time, so the 7-day period after sleeping and after eating occurs at different times: either they have not slept for less than 7 days, or they have not eaten for less than 7 days. Otherwise, they would have already died. In the first case, they should eat, and in the second ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,638 |
1. The Knight's Journey. Can a knight move around all the squares of a chessboard, starting from square $a 1$, ending at square $h 8$, and visiting each square of the board exactly once? | 1. To traverse all 64 squares of a chessboard, visiting each square exactly once, a knight must make 63 moves. Since with each move the knight transitions from a white square to a black one or from a black square to a white one, after even-numbered moves the knight will land on squares of the same color as the starting... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,640 |
5. Beetles. Imagine that you managed to catch 25 beetles and placed each one on a cell of a $5 \times 5$ piece of a chessboard. Suppose now that each beetle crawls to an adjacent cell, either horizontally or vertically, on this piece of the board. Will there be any empty cells left? | 5. Since the number of cells on the $5 \times 5$ board is odd, the number of black and white cells cannot be equal. Suppose there are more black cells. Then one of the black cells will definitely remain empty, as only the bugs sitting on white cells move to black cells. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,644 |
6. Mice love cheese. A mouse is gnawing at a cube of cheese with an edge of 3, divided into 27 unit cubes. When the mouse eats any cube, it moves to the next cube that shares a face with the previous one. Can the mouse eat the entire cube except for the central cube? | 6. Let's paint all the unit cubes,
except the central one, in a "chessboard" pattern. Since the mouse eats white and black cubes alternately, the number of white cubes cannot exceed the num... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,645 |
7. The Eight Queens Problem. Can eight queens be placed on a chessboard so that they do not threaten each other? | 7. There are 92 solutions to this problem. The figure shows one of them. | 92 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,646 |
8. Rook Placement. On a chessboard, eight rooks are placed in such a way that they do not attack each other (there is one rook on each row and each column). Prove that an even number of rooks stand on the black squares of the chessboard. | 8. Let's introduce a coordinate system on a chessboard. The left-bottom field will have the coordinate $(1 ; 1)$, and the right-top field will have $(8,8)$. The sum of the coordinates of any white square is an odd number, while the sum of the coordinates of any
![](https://cdn.mathpix.com/cropped/2024_05_21_b6bdef2bf9... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,647 |
9. Tetromino. Can a "chessboard" $10 \times 10$ be covered with 25 rectangular tiles of size $4 \times 1$? | 9. We will color the cells of the board as shown in the figure. Notice that a rectangle cannot cover more than one colored cell entirely. There are a total of 26 colored cells, while there are 25 rectangles. Therefore, it is impossible to cover the $10 \times 10$ board with $1 \times 4$ rectangles. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,648 |
11. Another incorrect equation. In the equation $101-102=1$, move one digit so that it becomes correct. | 11. Answer: $101-10^{2}=1$. | 101-10^{2}=1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,650 |
13. In the senate of the country of knights and liars - there are 100 senators. Each of them is either a knight or a liar. It is known that: 1. At least one of the senators is a knight. 2. Out of any two arbitrarily chosen senators, at least one is a liar. Determine how many knights and how many liars are in the senate... | 13. Only one of the senators is a knight. The fact that there is at least one knight among the senators follows from the first statement, the existence of a second knight contradicts the second statement. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,652 |
14. Diophantine equation. Vasya said that he can solve the equation $19 x^{2}+97 x=1997$ in natural numbers. Prove that Vasya is wrong. | 14. The expression $19 x^{2}+97 x$ is even for any $x$, so it cannot equal 1997. Therefore, Vasya was wrong. | proof | Algebra | proof | Yes | Yes | olympiads | false | 35,653 |
1. Scientific organization of labor. There are logs of two types: 6 meters and 7 meters long. They need to be sawn into 1-meter logs. Which logs are more profitable to saw? | 1. To saw 42 one-meter logs from six-meter logs requires 35 cuts, while from seven-meter logs - 36. It can be considered that sawing six-meter logs is more profitable. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,655 |
2. The final test. And Koshchei said to Ivan the Tsarevich: "You, Vanya, shall live until tomorrow morning. In the morning, you shall appear before me. I will think of three digits $a, b, c$, and you will tell me three numbers $x, y, z$. I will listen to you and tell you the value of the expression $a x + b y + c z$. T... | 2. We can choose $x=100, y=10, z=1$. Since $a, b, c$ are digits, the sum $S=a x+b y+c z=100 a+10 b+c=\overline{a b c}$, and Koschei the Deathless will name the required number himself. | 100a+10b+ | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,656 |
3. Many numbers. Can 1998 consecutive numbers be composite? | 3. Yes. The numbers $1999!+2 ; 1999!+3 ; \ldots ; 1999!+1999$ will be composite, since $1999!+k$ is divisible by $k$ for $2 \leq k \leq 1999$. Note that $n$ consecutive numbers can be composite for any natural number $n$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,657 |
5. To the Mechanics and Mathematics Faculty! Several identical books and identical albums were bought. The books cost 10 rubles 56 kopecks. How many books were bought if the price of one book is more than one ruble higher than the price of an album, and 6 more books were bought than albums? | 5. Since each book is more expensive than a ruble, no more than 10 books were bought. Moreover, it is clear that no fewer than 7 books were bought (since at least one album was bought). The number 1056 is divisible by 8 and not divisible by $7,9,10$. Therefore, 8 books were bought. (MSU, Mechanics and Mathematics, 1968... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,659 |
6. 175 shaltev cost more than 125 boltayev, but less than 126 boltayev. Prove that 80 kopecks will not be enough to buy 3 shaltev and 1 boltayev. | 6. Let's prove that 80 kopecks are not enough to buy 3 shaltays and 1 boltay. Let one shaltay cost $x$ kopecks, and one boltay $y$ kopecks, then $126 y > 175 x > 125 y$. From this, it follows that $y > 25(7 x - 5 y) > 0$ and $y \geq 26$. Further, $7 x \geq 5 \cdot 26 = 130$. Hence:
63 Natural numbers
$x \geq 19$. The... | 83 | Inequalities | proof | Yes | Yes | olympiads | false | 35,660 |
7. Is the theorem correct: "The number $n^{2}+n+41$ is prime for any integer $n$"? | 7. The theorem is false because $f(41)=41^{2}+41+41=41 \cdot 43$ is not a prime. Generalization: note that the theorem: the number $n^{2}+n+a$ is prime for any integer $n,$ will not be true for any integer $a$. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,661 |
8. One or two? Let's take all natural numbers from 1 to 1000000 and for each of them, calculate the sum of its digits. For all the resulting numbers, we will again find the sum of their digits. We will continue this process until all the resulting numbers are single-digit. Among the million resulting numbers, 1 and 2 w... | 8. Let's use the statement: if a number when divided by 9 has a remainder of \( d \), then the sum of its digits will have the same remainder. Which numbers from 1 to 1000000 are more: those that have a remainder of 1 when divided by 9, or those that have a remainder of 2? In the range from 1 to 999999, there are an eq... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,662 |
9. For a two-digit number $x$, 6 statements are made:
a) $x$ is divisible by 3; b) $x$ is divisible by 5;
c) $x$ is divisible by 9; d) $x$ is divisible by 15;
e) $x$ is divisible by 25; f) $x$ is divisible by 45.
Find all such $x$ for which exactly three of these statements are true. | 9. Note that $x$ cannot be divisible only by a power of 3, or only by a power of 5 - in each of these cases, only two statements would be true. Therefore, $x$ is divisible by 15. Further, $x$ is not divisible by $5^{2}$ (or by $3^{2}$) - otherwise, at least 4 statements would be true. Therefore, the condition of the pr... | 15,30,60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,663 |
10. Centipedes and three-headed dragons. In a certain herd of centipedes and three-headed dragons, there are a total of 26 heads and 298 legs. Each centipede has one head. How many legs does a three-headed dragon have? | 10.Note: 1. the number of heads of dragons is divisible by 3, and the number of centipedes cannot exceed 7 (otherwise, there would be more than 298 legs), so the number of centipedes is either 2 or 5. 2. If there are 2 centipedes, then there are 8 dragons, and each dragon would have $\frac{218}{8}$ legs, which is impos... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,664 |
11. More about electronic scales. There are 11 bags of coins. In ten bags, the coins are genuine (each coin weighs 10 grams), and in one bag, they are counterfeit (each coin weighs 11 grams). How can you determine which bag contains the counterfeit coins with just one weighing? | 11. Let's assign the number 0 to one of the bags, the number 1 to another, and so on, and proceed as in problem 10.3. In this case, if the counterfeit coins are in bag 0, the total weight of all the taken coins will be 550 g., and the solutions to the other problems will coincide. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,665 |
12. The Sages' Test. A certain ruler, wishing to test three of his sages, said to them: “Before you are five caps: three black and two white. You will each be given a cap to wear. The one who first guesses the color of the cap on his head will receive a reward.” Then the sages' eyes were blindfolded and a cap was place... | 12. In this task, there are three options for distributing the caps: black, white, white; black, black, white; black, black, black. The first option is uninteresting because the sage wearing the black cap reasons as follows: "I see two white caps, and there are only two of them. Therefore, I must be wearing a black cap... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,666 |
13.One hundred different chips are placed in a row. Any two chips standing one apart can be swapped. Is it possible to rearrange all the chips in reverse order? | 13. Let's call the number of a chip its place "in line." Then, we notice that chips with odd numbers only swap places with chips with odd numbers. Therefore, the first chip can never become the hundredth. Hence, it is impossible to rearrange all the chips in reverse order. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,667 |
1. For training. Among the four statements: "number $a$ is divisible by 2", "number $a$ is divisible by 4", "number $a$ is divisible by 12", "number $a$ is divisible by 24" - three are true, and one is false. Which one? | 1. Note that: "number $a$ is divisible by 24" $\Rightarrow$ "number $a$ is divisible by 12" $\Rightarrow$ "number $a$ is divisible by 4" $\Rightarrow$ "number $a$ is divisible by 2". Therefore, only the statement "number $a$ is divisible by 24" can be false. | isdivisible24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,670 |
2. Who should be questioned first? The statements of three suspects in the case contradict each other, and Smith accuses Brown of lying, Brown accuses Jones, and Jones says that neither Brown nor Smith should be believed. Who would you, as an investigator, question first?
65 Basics of the deductive method | 2. Which of the suspects could have told the truth? It couldn't have been Jones, because then Smith's statement that Brown is lying would be false, which contradicts Jones' statement. It couldn't have been Brown either, because then Jones' statement that Smith is lying... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,671 |
3. Who stole the suitcase? One of the four gangsters stole a suitcase full of money. During the interrogation, Alex said that Louis stole the suitcase, Louis claimed that the culprit was Tom, Tom assured the investigator that Louis was lying. George insisted only that he was not guilty. During the investigation, it was... | 3. Since one of Louis's and Tom's statements is true, George lied and, consequently, he is the criminal. Answer: Tom told the truth. | George | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,672 |
4. Who participated in the robbery? It is known that out of six gangsters, exactly two participated in the robbery. In response to the question of who participated in the robbery, they gave the following answers: Harry: Charlie and George. James: Donald and Tom. Donald: Tom and Charlie. George: Harry and Charlie. Charl... | 4. Each segment on the diagram represents the testimony of one of the gangsters. For example: the segment Charlie-Harry is the testimony of George. For all segments except one, one vertex corresponds to a gangster who participated in the robbery, and the other to one who did not. At the two points corresponding to the ... | CharlieJames | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,673 |
5. The Case of Brown, Jones, and Smith. One of them committed a crime. During the investigation, each of them made two statements:
Brown: 1. I am not the criminal. 2. Neither is Jones.
Jones: 1. Brown is not the criminal. 2. Smith is the criminal. Smith: 1. Brown is the criminal. 2. I am not the criminal.
During the... | 5. The culprit cannot be Jones, because in this case none of the defendants would have lied twice, which contradicts the condition of the problem. It cannot be Smith either, because in this case there is no defendant who lied exactly once. The only possible option is that the crime was committed by Brown. | Brown | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,674 |
7. Two detectives met. Here is their dialogue:
- Do you have two sons?
- Yes, they are young, they don't go to school.
- By the way, the product of their ages equals the number of pigeons near us.
- That's not enough information.
- And I named the older one after you.
- Now I know how old they are.
How old are the so... | 7. The ages of the sons could have been equal, hence the product of their ages - a perfect square. Upon receiving information about the product of the ages, the second detective could not answer the question, therefore, the product can be factored in two ways. Answer: 1 and 4 years. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,676 |
8. A theft has been committed at the warehouse. The investigation has established: 1) The criminals transported the stolen goods by car; 2) The crime was committed by one of three people: $\mathcal{A}$, $B$, or $C$ (it could be all three); 3) $C$ never goes on a job without $\mathcal{A}$; 4) $B$ knows how to drive. Is ... | 8. If $B$ is guilty, then he must have an accomplice, since he does not know how to drive a car. If $B$ is not guilty, then someone must have committed this crime. But, since $C$ does not go on a job without $\mathcal{A}$, then $\mathcal{A}$ is guilty in any case. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,677 |
10. Scattered Witnesses. Brown, Jones, and Smith are witnesses to a bank robbery. Brown stated that the criminals fled in a blue "Buick." Jones claimed it was a black "Chrysler," and Smith said it was a "Ford," but not blue. Due to their scatterbrained nature, each of them correctly identified either only the make or o... | 10. The car could not have been blue, because otherwise both Smith's and Jones's statements about the make of the car would have been correct, which is impossible at the same time. Therefore, Brown correctly identified the make of the car, while Jones and Smith correctly identified the color of the car. Consequently, t... | black\"Buick" | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,679 |
11. Is it possible to lay out all 28 dominoes in a chain according to the rules of the game so that a "six" ends up on one end and a "five" on the other? | 11. According to the rules of the domino game, all "sixes," except for the double, are placed in pairs in the chain, unless they are at the end. If a "six" is at the end, one "six" will remain without a pair, which means it will also be at the end. Answer: no.
69 Fundamentals of the Deductive Method | no | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,680 |
13. Stack of Triangles. In the vertices of an equilateral triangle, the numbers $1,2,3$ are written. Is it possible to stack several such triangles so that the sum of the numbers in each vertex equals $1997 ?$
67 Basics of the Deductive Method | 13.If it were possible, the sum of all numbers would be an odd number, but this is incorrect, as the sum of the numbers on each card is 6. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 35,682 |
14.Write the equation: $x_{-}\left(x_{-}\left(x_{-} \ldots-(x-1) \ldots\right)\right)=1$ (the expression contains 1999 pairs of parentheses). | 14. After expanding the brackets, the signs before all $x$ standing next to each other will be different. Therefore, after expanding the brackets, the equation will take the form: $1=1$. Answer: $x$ - any number. | x- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 35,683 |
15. In each cell of a $3 \times 3$ table, numbers are placed. Can the sum of the numbers in any $2 \times 2$ square be negative, while the sum of the numbers in the entire table is positive? | 15. See the figure:
| 1 | 1 | 1 |
| :---: | :---: | :---: |
| 1 | -4 | 1 |
| 1 | 1 | 1 | | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 35,684 |
1. Boxing Championship. In the Olympic boxing tournament, 247 people participated. How many matches were needed to determine the winner? | 1. In the Olympic tournament, when after each match one boxer is eliminated, and only the winner continues to compete, 246 matches will be required. | 246 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,685 |
2. In a round-robin chess tournament, 7 players participated. How many games did they play? | 2. Let's calculate the total number of games played in the tournament. Each player played 6 games - in total: $6 \times 7=42$. Notice that in this way, we counted each game twice (since two players participate in each game!), so the total number of games played in the tournament is $42: 2=21$. We can also solve this us... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,686 |
3. How many chess players participated in the round-robin tournament if a total of 190 games were played in this competition? | 3. Since $19 \times 20: 2=190$, there were 20 chess players in the tournament. | 20 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 35,687 |
4. Odd or even? During a chess tournament, the number of players who played an odd number of games was counted. Prove that the number of such players is even. | 4. If you add up the number of games played by all chess players, you get an even number, since two people participate in each game. Therefore, the number of chess players who played an odd number of games is even. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 35,688 |
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