problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4. The perimeter of a cardboard sheet in the shape of a square is 28 dm. How many square centimeters does its area contain? | 4. 5) $28: 4=7$ (dm) - the length of the side of the square;
2) $7 \times 7=49$ (sq. dm) - the area of the square;
3) $100 \times 49=4900$ (sq. cm).
Answer: the area of the square is 4900 sq. cm. | 4900 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,653 |
5. A part is machined from a metal blank. The chips obtained from machining 8 parts can be remelted into one blank. How many parts can be made from 64 blanks? | 5. 6) $64: 8=8$ (blanks) - can be manufactured from the waste when machining 64 parts;
2) $8: 8=1$ (blank) - can be manufactured from the waste when machining 8 parts;
3) $64+8+1=73$ (parts).
Answer: from 64 blanks, 73 parts can be manufactured. | 73 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,654 |
7. How will the area of a rectangle change if one of its sides is increased by 3 cm, and the other is decreased by 3 cm? | 7. Let $a$ and $b$ be the sides of a rectangle. Then its area is: $S=a \times b$. The sides of the modified rectangle will be
$$
(a-3) \text { and }(b+3)
$$
and the area will accordingly be:
$$
S_{1}=(a-3)(b+3)=a b-3 b+3 a-9
$$
Let's compute the difference between $S$ and $S_{1}$. It is equal to:
$$
\begin{aligned... | the\area\can\,\,\or\remain\the\same | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,656 |
1. How many strikes do the clocks make in a day if they strike once every half hour, and at each hour $1,2,3 \ldots 12$ times? | 1. $(1+2+3+4+5+6+7+8+9+10+11+12) \times 2+24=180$.
Answer: 180 hits. | 180 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,657 |
2. Three girlfriends - Vера, Olya, and Tanya went to the forest to pick berries. For berry picking, they had a basket, a small basket, and a bucket. It is known that Olya was not with the basket and not with the small basket, Vera - not with the small basket. What did each girl take with her for berry picking? Justify ... | 2. The task is about three friends (Vera, Olya, Tanya) and three items for picking berries (basket, basket, bucket). Let's create a table.
| | V | O | T |
| :---: | :---: | :---: | :---: |
| B | | | |
| L | | | |
| b | | | |
Olya was not with the basket and not with the basket. Therefore, Olya was with the ... | Verawaswiththebasket,Olya-withthebucket,Tanya-withthesmallbasket | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,658 |
3. There are 24 sparrows on three branches. When 4 sparrows flew from the first branch to the second, and 3 sparrows flew from the second to the third, the number of sparrows on all branches became equal. How many sparrows were originally on each branch? | 3. 4) $24: 3=8$ (sparrows) - was on each branch after the flight;
2) $8-3=5$ (sparrows) - was on the third branch initially;
3) $8+4=12$ (sparrows) - was on the first branch;
4) $5+12=17$ (sparrows) - was on the first and third branches before the flight;
5) $24-17=7$ (sparrows) - was on the second branch initially.
T... | 12,7,5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,659 |
4. The perimeter of a square is $20 \mathrm{~cm}$. By how many square centimeters will the area of the square increase if its perimeter is increased by 12 cm? | 4. Since all sides of a square are equal, then
1) $20: 4=5$ (cm) — the length of one side;
2) $5 \times 5=25\left(\mathrm{~cm}^{2}\right)$ - the area of the square;
3) $12: 4=3$ (cm) - by how much the side increased;
4) $5+3=8$ (cm) - the side of the new square;
5) $8 \times 8=64\left(\mathrm{~cm}^{2}\right)$ - the are... | 39\mathrm{~}^{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,660 |
5. What is the least number greater than 1992 that gives a remainder of $7$ when divided by $9$? | 5. Divide 1992 by 9 with a remainder
| $-1992\llcorner 9$ |
| :--- |
| $\frac{18}{-19}$ |
| $\frac{-18}{221}$ |
| $\frac{-12}{9}$ |
| 3 |
We obtained a remainder of 3. To make the remainder equal to 7, the divisor needs to be increased by four:
$$
1992+4=1996 \text {. }
$$
Answer: 1996. | 1996 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,661 |
6. Find the sum of all even numbers from 10 to 31. Calculate in different ways. | 6. Let's list all even numbers from 10 to 31:
$$
10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30
$$
a) Let's add the obtained numbers:
$$
10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 = 220.
$$
b) We can notice that the sum of numbers equally distant from the ends is 40:
$$
10 + 30 = 12 + 28 = 14 + 26 = \ldots =... | 220 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,662 |
1. What will be the result if we add the largest odd two-digit number and the smallest even three-digit number? | 1. The largest odd two-digit number is 99, and the smallest even three-digit number is 100.
$$
99+100=199
$$
Answer: The sum of the largest odd two-digit number and the smallest even three-digit number is 199. | 199 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,663 |
2. Write in digits the number equal to the sum of 22 million, 22 thousand, 22 hundred, and 22 units. | 2. Many believe that this number will be 22022222.
In fact, this number is 22024222, because if you add 22 thousand, which is 22000, and 22 hundred, which is 2200, and 22 units to 22 million, which is 22000000, you get 22024222.
Answer: 22024222. | 22024222 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,664 |
5. In a two-digit number, the number of tens is two times less than the number of units. If you subtract the sum of its digits from this two-digit number, you get 18. Find this number. | 5. According to the condition, in a two-digit number, the number of tens is twice less than the number of units, i.e., the number of units is twice the number of tens.
Two-digit numbers that satisfy this condition:
$$
12 ; \quad 24 ; \quad 36 ; \quad 48 .
$$
The sum of the digits of each of the obtained numbers:
$$... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,667 |
6. Three friends - Winnie-the-Pooh, Piglet, and Rabbit went for a walk in red, green, and blue shirts. Their shoes were of the same colors. Winnie-the-Pooh had the same color for his shirt and shoes, Piglet's shoes and shirt were not red, and Rabbit was wearing green shoes. How were the friends dressed? | 6. This task is about three friends (Winnie-the-Pooh, Piglet, and Rabbit) and the three colors of their shoes and shirts. Let's first find out the color of the shoes of each friend. We will create a table:
| | W | P | R |
| :---: | :---: | :---: | :---: |
| K | | | |
| 3 | | | |
| c | | | |
Based on the con... | Winnie-the-Poohinredshirtredshoes,Pigletingreenshirtblueshoes,Rabbitinblueshirtgreenshoes | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,668 |
2. What is the greatest number of Saturdays that can be in a year? | 2. Since a year can have 365 or 366 days, we will choose the larger of these numbers - 366. Saturday occurs once every seven days. Therefore, to find the number of Saturdays in a year, we need to divide 366 by 7 with a remainder. We get
, 20 numbers are divisible by 5 \((100: 5=20)\).
Of these, 4 nu... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,673 |
6. There are four children in the family. They are 5, 8, 13, and 15 years old. The children are named Anya, Borya, Vера, and Galya. How old is each child, if one girl attends a kindergarten, Anya is older than Borya, and the sum of Anya's and Vера's ages is divisible by 3? | 6. First solution method. The problem is about four children (Anya, Borya, Vera, and Galya) and four ages of children (5 years, 8 years, 13 years, and 15 years).
Let's represent the problem's condition in a table.
| Anya | Borya | Vera | Galya |
| :---: | :---: | :---: | :---: |
| $5,8,13,15$ | $5,8,13,15$ | $5,8,13,... | Verais5old,Boryais8old,Anyais13old,Galyais15old | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,675 |
7. Find the sum of all possible different two-digit numbers, all digits of which are odd. | 7. To find out how many such numbers there are, let's try to determine which digits can stand in the tens place. There are five such digits: $1,3,5,7,9$.
The second digit must also be odd, so it can also be chosen in five ways.
As can be seen from the diagram, the problem is about three ducks.
Answer: 3 ducks. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,679 |
4. There were plates on the shelf. First, from all the plates except two, $1 / 3$ part was taken, and then $1 / 2$ of the remaining plates. After this, 9 plates were left on the shelf. How many plates were on the shelf? | 4. Let's represent the condition of the problem in a drawing
1) $9 \times 2=18$ (plates) — remain after the first time plates were taken
2) $18-2=16$ (plates) - corresponds to $2 / 3$;
3) ... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,680 |
5. There were 5 times more strawberry bushes on the first bed than on the second. When 22 bushes were transplanted from the first bed to the second, the number of strawberry bushes on each bed became the same. How many bushes were there on each bed? | ## 5. First method of solving.
Let's build a graphical model of the problem's condition
1) $22: 2=11$ (bushes) - this is $1/5$ of all the bushes (that were on the second bed);
2) $11 \ti... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,681 |
6. A horse eats a stack of hay in 1 month, a goat - in 2 months, a sheep - in three months. How long will a horse, a goat, and a sheep together eat the same stack of hay
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。 | 6. 7) $6: 1=6$ (stacks) - a horse eats in half a year;
2) $6: 2=3$ (stacks) - a goat eats in half a year;
3) $6: 3=2$ (stacks) - a sheep eats in half a year;
4) $6+3+2=11$ (stacks) - all animals will eat in 6 months.
Therefore, the stack of hay they will eat in $6 / 11$ month.
Answer: one stack of hay the animals wil... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 42,682 | |
7. The teacher checked the work of three students - Alekseev, Vasiliev, and Sergeev, but did not take their papers with him. He told the students: “All of you have completed the work, and you received different grades (‘3’, ‘4’, ‘5’). Sergeev did not get a ‘5’, Vasiliev did not get a ‘4’, and I think Alekseev got a ‘4’... | 7. Since it is not clear which of the statements is true in this problem, we need to consider three cases.
Case 1. Let's assume the teacher told the truth to Sergeyev. Then, based on the problem's conditions, we will fill in the table. "Sergeyev did not get a '5'." We will put a '-' in the corresponding cell. "Vasilye... | Alexeev:5,Sergeyev:3,Vasilyev:4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,683 |
1. Between the numbers $1,2,3,4,5,6,7,8,9$ insert arithmetic operation signs and parentheses so that the resulting expression has a value of 100. | 1. a) $1+2+3+4+5+6+7+8 \times 9=100$;
b) $1 \times 2 \times 3+4+5+6+7+8 \times 9=100$;
c) $1+2 \times 3+4 \times 5-6+7+8 \times 9=100$;
d) $1 \times 2 \times 3 \times 4+(5+6-7)+8 \times 9=100$
e) $(1 \times 2+3) \times 4 \times 5+6-7-8+9=100$;
f) $(1+2+3) \times(4+5+6)-7+8+9=100$;
g) $((1+2): 3+4+5-6) \times 7+8 \... | 100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,685 |
2. How to cut half a meter from a piece of fabric that is $2 / 3$ of a meter without having a meter stick
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。
---
2. How to cut half a meter from a piece of fabric that is $2 / 3$ of a meter without having a meter stick | 2. Let's conditionally represent a segment 1 meter long and divide it into three equal parts.
To get half a meter, we need to cut off $1 / 4$ of the given piece. Therefore, we will proceed... | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,686 | |
3. A rectangular photograph with dimensions of 30 cm and 4 dm was enlarged several times to create a rectangular billboard. The area of the billboard is 48 m². What are its length and width? | 3. 4) $30 \times 40=1200\left(\mathrm{~cm}^{2}\right)$ - the area of the photograph;
2) $480000: 1200=400$ (times) - the area of the billboard is 400 times larger than the area of the photograph.
Since when each side of a rectangle is increased by $k$ times, its area increases by $k \times k$ times, in our case, each ... | thewidthofthebillboardis6,thelengthis8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,687 |
4. Let 2 cups and 2 pitchers weigh as much as 14 plates, 1 pitcher weighs as much as 1 cup and 1 plate. How many plates will balance a pitcher
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。 | 4. First method of solution (graphical).
Reduce the number of items on each scale by half.
 $22725: 9=2525$ (rubles) — the amount the students paid for one month.
To determine how much each student paid monthly, we need to know the number of students in the class. This is unknown in the problem. However, from the problem's condition, it follows that this is a natural number that is a divisor of 2525.
... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,689 |
1. Write the smallest four-digit number in which all digits are different. | 1. According to the problem, to record a four-digit number, four different digits should be used. They can be chosen from $0,1,2,3,4,5,6,7,8,9$.
The smallest digit in the thousands place can be $1: 1^{* * *}$.
The smallest digit in the hundreds place can be $0: 10^{* *}$.
Since we have already used 1 and 0, the digi... | 1023 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,692 |
2. Identify the counterfeit coin among eight coins with two weighings, given that the counterfeit coin is heavier than each of the others.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. Divide all the coins into three groups: two groups of three coins each and one of two. Place three coins from the first two groups on the scales. If the scales are in balance, the counterfeit coin is among the two remaining ones, and we can determine it with the second weighing.
If one of the scale pans tips during... | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 42,693 | |
3. Find the smallest number that gives a remainder of 1 when divided by 2, and a remainder of 2 when divided by 3. | 3. First, let's write down the numbers that give a remainder of 1 when divided by 2:
$$
3,5,7,9,11 \ldots
$$
Next, let's write down the numbers that give a remainder of 2 when divided by 3:
$$
5,8,11,14,17 \ldots
$$
We will choose from the obtained numbers the one that satisfies both conditions and is the smallest.... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,694 |
4. The perimeter of a rectangle is 48 cm, and its length is 2 cm more than its width. Find the area of this rectangle. | 4. This problem can be solved both analytically and arithmetically.
First method of solving.
According to the condition
$$
(a+b) \times 2=48
$$
from this we can conclude that $a+b=24$.
Using the second condition $a=b+2$, we get an equation to determine $b$:
$$
b+2+b=24, 2b=22, b=11(\mathrm{~cm})
$$
From the cond... | 143(\mathrm{~}^{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,695 |
5. Instead of the asterisks, place such digits so that the correct equation is obtained $5^{*}+* * 3=* * 01$.
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. $11 \times 13=143\left(\mathrm{~cm}^{2}\right)$ - area of the rectangle.
Third solution method.
Let's make another brief record of the problem statement:
1) $48-4=44$ (cm) - four wid... | 143\mathrm{~}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,696 |
6. A motorcyclist traveled 980 km over three days. In the first two days, he traveled 725 km, and on the second day, he traveled 123 km more than on the third day. How many kilometers did he travel each of these three days? | 6. Let's represent the condition of the problem in a drawing
980 km
1) $980-725=255$ (km) - traveled on the third day.
2) $255+123=378$ (km) - traveled on the second day.
3) $725-378=347$... | 347 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,697 |
7. In a cafe, three friends met: sculptor Belov, violinist Chernov, and artist Ryzhov. "It's great that one of us has white, one has black, and one has red hair, but none of us has hair of the color that our last name suggests," noted the black-haired man. "You're right," — said Belov. What color is the artist's hair? | 7. This problem can be solved through reasoning or using the tabular method.
The first solution method. Belov was talking to the black-haired person, which means his hair color is neither black nor white (due to the fact that the hair color should not indicate the surname). Therefore, Belov has red hair. Since Belov w... | TheartistRyzhovhasblackhair | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,698 |
1. How many tiles with dimensions $15 \times 15$ cm are needed to tile a wall that is 3 m 6 dm long and 27 dm wide $?$ | 1. The first method of solving.
1) $15 \times 15=225$ (cm$^2$ - the area of one tile;
2) $360 \times 270=97200\left(\mathrm{~cm}^{2}\right)$ - the area of the walls;
3) $97200: 225=432$ (tiles).
The second method of solving (Rukhov Arseny, student of School No. 25).
1) $360: 15=24$ (tiles) - will fit in one row along... | 432 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,699 |
3. The difference between two numbers is 157, and their quotient is 2. Find these numbers. | 3. From the condition that the quotient of the numbers is 2, it follows that the minuend is twice as large as the subtrahend. Then their difference equals the subtrahend, which means the subtrahend is the number 157, and the minuend is twice its size - 314.
Analytically, the solution can be presented as follows: from ... | 157314 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,701 |
4. Given the sequence of numbers $0,1,2,6,16,44,120, \ldots$ Extend it and write down the next two numbers. | 4. All numbers, starting from the third, in the given sequence are formed according to the rule:
$$
\begin{gathered}
a_{n}=\left(a_{n-1}+a_{n-2}\right) \times 2 \\
(0+1) \times 2=2 \\
(1+2) \times 2=6 \\
(2+6) \times 2=16 \\
(16+6) \times 2=44 \\
(44+16) \times 2=120 \\
(44+120) \times 2=328
\end{gathered}
$$
Thus, w... | 328 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,702 |
5. How many days have passed from March 19, 1990 to March 23, 1996 inclusive | 5. 6) \(365 \times 6 + 2 = 2192\) (days) - have passed from 19.03.1990 to 19.03.1996
2) \(23 - 19 = 4\) (days) - have passed from 19.03.1990 to 23.03.1996
3) \(2192 + 4 = 2196\) (days) - have passed from 19.03.1990 to 23.03.1996\).
Answer: a total of 2196 days have passed. | 2196 | Other | math-word-problem | Yes | Yes | olympiads | false | 42,703 |
6. It is known that the perimeter of one rectangle is greater than the perimeter of another rectangle. Compare the areas of these rectangles.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 6. The solution to this problem can be obtained by students as a result of conducting a computational experiment with various rectangles, for example:
a) $a=3 \text{~cm}, b=4 \text{~cm}$;
$$
\begin{aligned}
& a=4 \text{~cm}, b=2 \text{~cm} ; \\
& p_{2}=12 \text{~cm} ; \\
& S_{2}=8 \text{~cm}^{2}
\end{aligned}
$$
$$
\... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 42,704 |
7. Three soldiers need to reach the headquarters, located 60 km from the front line, within 3 hours. Can they do it, given that a pedestrian walks at a speed of 5 km/h and they have a motorcycle that can carry no more than two people at a speed of no more than 50 km/h? | 7. In one hour, two soldiers will travel 50 km on a motorcycle, while one will walk 5 km. Then, one of the two who were on the motorcycle can walk the remaining 10 km in 2 hours, meaning he will reach the headquarters in 3 hours. The second of the two who were on the motorcycle can return to pick up the pedestrian, tra... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,705 |
1. Which of the following time intervals is the largest? a) 1500 minutes b) 10 hours c) 1 day. | 1. To compare time intervals, it is necessary to express them in units of the same denomination, for example, in hours:
$1500: 60=25$ (hours);
1 day $=24$ hours.
Thus, the greatest time interval is 1500 minutes.
Answer: the greatest time interval is 1500 minutes. | 1500 | Other | MCQ | Yes | Yes | olympiads | false | 42,706 |
3. In the box, there are geometric shapes: triangles, squares, and circles. In total, there are 24 shapes. There are 7 times as many triangles as squares. How many of each shape could be in the box? | 3. The first method of solving. Let's represent the condition of the problem in the form of a drawing.
squares
triangles
, we c... | 22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,709 |
5. Vера, Nina, and Olya were playing with dolls. They dressed their dolls in one item each: either a coat, a jacket, or a dress. When their mother asked the girls what they dressed their dolls in, one of them joked and said: "Vера dressed the doll in a dress, and Olya in a coat." The second one replied: "Vера dressed t... | 5. Let's assume that in the first answer, the first part was correct (Vera put a dress on the doll). From this, we can conclude that Olya's doll was not in a coat. In the second answer, the first part will be incorrect, since we know that Vera put a dress on the doll, and therefore, she could not have put a coat on her... | Vera'\doll\is\in\\dress,\\Nina'\doll\is\in\\coat | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,710 |
6. By how much is the largest five-digit number greater than the smallest five-digit number | 6. The largest five-digit number is 99999, and the smallest is 10000.
$$
99999-10000=89999
$$
Answer: greater by 89999. | 89999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,711 |
7. A rectangular sheet of iron was divided into 2 parts such that the first part was 4 times larger than the second. What is the area of the entire sheet if the first part is $2208 \mathrm{~cm}^{2}$ larger than the second? | 7. Let's make a brief record of the problem:
1) $2208: 3=736\left(\mathrm{~cm}^{2}\right)$ - corresponds to one part;
2) $736 \times 4=2944\left(\mathrm{~cm}^{2}\right)$ - area of the fir... | 3680\mathrm{~}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,712 |
1. Find the sum of all three-digit numbers that can be written using the digits $1,2,3$ such that all digits in each number are different. | 1. Let's first determine the three-digit numbers that can be formed using the digits $1,2,3$ such that all digits in each number are different.
Thus, we have the following numbers that sati... | 1332 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,713 |
2. In the record 123456789, place the signs "plus" or "minus" between some of the digits so that the resulting expression equals 100. | 2. $123-45-67+89=100$ | 123-45-67+89=100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,714 |
3. Once in a train carriage, Tanya started to encrypt words by replacing letters with their numbers in the alphabet. When she encrypted the arrival and departure points of the train, she was surprised to find that they were written using only two digits: 211221 - 21221. Where is the train coming from and going to? | 3. Let's write out all the letters of the alphabet:
| $\mathrm{A}$ | B | V | G | D | E | Yo | Zh | Z | I | Y | K | L | M | N | O |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 1 | 1 | 1 | 1 | 1 | ... | BAKU—UFA | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,715 |
4. Anton was given a scale, and he started weighing his toys. A car was balanced by a ball and two cubes, and a car with a cube - two balls. How many cubes balance a car? (All of Anton's balls are the same, and the cubes are the same too.) | 4. First method of solving. Since the machine is balanced by the ball and two cubes, the machine with a cube will be balanced by the ball and three cubes. From the second condition, we have that 2 balls balance a ball and 3 cubes, meaning one ball by mass is equal to 3 cubes. Thus, the machine can be balanced by 5 cube... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,716 |
5. Vasya has to read an entertaining math book in three days. On the first day, he read half of the book, on the second day, he read a third of the remaining pages, and on the third day, he read a number of pages equal to half the number of pages he read in the first two days. Did Vasya manage to finish the book in thr... | 5. Let's represent the volume of the book as a segment of a certain length.
On the first day, Vasya read $1 / 2$ of the book, which constitutes half of the segment:
 $(72: 9-3) \times 2=10$
2) $72:(9-3) \times 2=24$
3) $72:((9-3) \times 2)=6$.
Answer: three different answers - $10, 24, 6$. | 10,24,6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,720 |
2. Two numbers were first multiplied, and then the larger number was divided by the smaller one, resulting in equal outcomes. What are these numbers? How many such pairs of numbers exist? | 2. Let $a$ and $b$ be the two numbers mentioned in the problem, with $a > b$. According to the problem,
$$
a \times b = a : b
$$
or
$$
a b^{2} - a = 0
$$
Factoring out the common factor, we get
$$
a \left(b^{2} - 1\right) = 0
$$
From this, we can conclude that $b = 1$, and $a$ can be any number.
Students might r... | 1, | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,721 |
3. The director of the St. Petersburg Academy of Sciences, Princess E.P. Vorontsova-Dashkova, a native of Kaluga Governorate, lived for 66 years. In the 18th century, she lived 46 years longer than in the 19th century. In what year was E.R. Vorontsova-Dashkova born and in what year did she die? | 3. First method of solving. Let's represent the condition of the problem in the form of a drawing
1) $66-46=20$ (years) - the doubled period of time she lived in the 19th century;
2) $20:... | 1746-1810 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,722 |
5. Three brothers - Ivan, Dmitry, and Sergey - teach different subjects in schools in Moscow, St. Petersburg, and Kaluga. Ivan does not work in Moscow, and Dmitry does not work in St. Petersburg. The Muscovite does not teach history. The one who works in St. Petersburg teaches chemistry. Dmitry teaches biology. What su... | 5. This task is about three brothers (Ivan, Dmitry, Sergey); the cities where they work (Moscow, Saint Petersburg, Kaluga), and their professions (historian, chemist, biologist).
First, let's determine the profession of each resident. We will create a table:
| | H | C | B |
| :--- | :--- | :--- | :--- |
| $\mathrm{M... | SergeyworksinKalugateacheshistoryorSergeyworksinSaintPetersburgteacheschemistry | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,724 |
1. Katya is 6 times younger than her great-grandfather; if a 0 is placed between the digits of her age, the age of her great-grandfather is obtained. How old is she | 1. The first way to solve the problem. From the problem statement, it follows that Katya's age is a two-digit number, and her great-grandfather's age is a three-digit number. The number of units in Katya's age, when multiplied by 6, gives a number in which the number of units will be the same as in Katya's age. Let's f... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,727 |
2. How many three-digit numbers do not contain the digit 8? | 2. First method of solving. There are a total of 900 three-digit numbers. To find out how many of them do not contain the digit 8, we need to subtract from 900 the number of three-digit numbers that contain the digit 8. Let's determine the number of these numbers. From 800 to 899, there are one hundred such numbers. Ot... | 648 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,728 |
3. Apples were divided into two unequal piles. When half of the apples from the first pile were moved to the second pile, and then half of the apples from the second pile were moved back to the first pile, there were 18 apples in the first pile and 8 in the second. How many apples were in each pile originally? | 3. The first method of solving.
1) $8 \times 2=16$ (beans) - was in the second pile after half was moved from the first pile;
2) $18-8=10$ (beans) - was in the first pile after half was moved from it;
3) $10 \times 2=20$ (beans) - was originally in the first pile;
4) $16-10=6$ (beans) - was originally in the second pil... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,729 |
4. How many four-digit numbers are there that are divisible by 45 and have the two middle digits as 9 and $7?$
untranslated part: $7?$
Note: The untranslated part is kept as is, to ensure the format and content match the original text as requested. | 4. The numbers satisfying the condition of the problem have the form $a 97 b$.
In the condition of the problem, it is stated that these numbers are divisible by 45, which means they are divisible by 5 and by 9. From the first statement, we can conclude that $b=0$ or 5: $a 970$ or $a 975$.
The most likely way to find ... | 6975,2970 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,730 |
5. In the second round of the Olympiad, 30 people participate. During the problem-solving, one of the students made 12 mistakes, and the others made fewer. Try to prove that at least three students made the same number of mistakes on the Olympiad. | 5. The first method of solving. If we exclude the student who made 12 mistakes, the remaining 29 people can be divided into groups based on the number of mistakes made: one group will include students who made one mistake, another will include those who made two mistakes, and so on, with the last group including those ... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 42,731 |
6. Given a triangle with side lengths of 7 cm, 12 cm, and 9 cm respectively. Explain how to construct a segment connecting a vertex and the opposite side with a length of 9 cm so that the perimeters of the two resulting triangles are equal. | 6. First method of solution. Since in the obtained triangles one side will be common ( $A D$ ), in order for the perimeters to be equal, it is necessary that the side of 9 cm be divided into parts, the difference in lengths of which would equal the difference of the other two sides ( $12-7=5 \text{ cm}$ ). Based on thi... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,732 |
1. How many times will a three-digit number increase if the same number is appended to it? | 1. Let's write down a random three-digit number (531) and append the same number to it, resulting in -531531. Let's find their quotient:
$$
531 \times 531: 531=1001
$$
The pattern found is also valid for other three-digit numbers.
Answer: The three-digit number will increase by 1001 times. | 1001 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,734 |
2. Carlson has a runny nose. He uses square handkerchiefs measuring 25 cm $\times 25 \mathrm{~cm}$. Over eight days, Carlson used $3 \mathrm{~m}^{2}$ of fabric. How many handkerchiefs did Carlson use per day? | 2. First method of solving.
1) $25 \times 25=625\left(\mathrm{~cm}^{2}\right)-$ area of one handkerchief;
2) $3 \times 10000=30000\left(\mathrm{~cm}^{2}\right)$ - contained in $3 \mathrm{~m}^{2}$;
3) $30000: 625=48$ (handkerchiefs) - Carlson used over 8 days
4) $48: 8=6$ (handkerchiefs) - Carlson used per day.
Second ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,735 |
3. If a schoolboy bought 11 notebooks, he would have 8 rubles left, but for 15 notebooks, he would be short of 12 rubles and 24 kopecks. How much money did the schoolboy have? | 3. Let's represent the condition of the problem in a drawing:
1) $15-11=4$ (notebooks) - the difference in the number of notebooks purchased;
2) $800+1224=2024$ (kop.) - the cost of 4 note... | 63 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,736 |
5. In a box, there are 7 blue and 5 red balls. What is the minimum number of balls that need to be taken out to ensure that among them there are at least 2 blue and 1 red? | 5. The worst-case scenario is if we draw 7 balls and they all turn out to be blue. To get another red ball, we need to draw one more ball. In this case, there will definitely be one red ball and 2 blue balls among the balls.
Answer: 8 balls. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,738 |
6. Each of the three children was bought candies by their parents. Vitya was given 5 candies, while Masha received fewer candies than Vitya, and Sasha received as many candies as Vitya and Masha combined. How many candies could the parents have bought for all the children? | 6. Based on the condition of the problem, Masha could have been bought $4, 3, 2$ or 1 candy, but then Sasha could have been bought $9, 8, 7$ or 6 candies. In total, the children could have been bought $18 (4+9+5=18), 16, 14$ or 12 candies.
Answer: $18, 16, 14$ or 12 candies. | 18,16,14,12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,739 |
7. The donkey had to share the feed (oats and hay) with both the horse and the cow.
1) If the donkey eats oats, then the horse eats the same as the cow.
2) If the horse eats oats, then the donkey eats what the cow does not eat.
3) If the cow eats hay, then the donkey eats the same as the horse.
Who always eats from th... | 7. The problem is about three animals - a donkey, a horse, and a cow, which ate oats and hay. Let's draw a table reflecting all possible combinations of the animals eating oats (O) and hay (S).
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Donkey | O | O | O | ... | The\donkey\always\eats\from\the\hay\feeder | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,740 |
1. From points $A$ and $B$, which are 70 km apart, a pedestrian and a cyclist set off simultaneously at speeds of 5 km/h and 15 km/h, respectively. What distance will be between them after 3 hours? | 1. In this problem, the direction of movement of the cyclist and the pedestrian is not specified, so there are four different cases.
Case 1. Moving towards each other:
1) $5+15=20($ km/h) ... | 10,40,100,130 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,741 |
2. The boy caught a fish. When asked how much the caught fish weighed, he said: “I think the tail weighs 1 kg, and the head weighs as much as the tail and half the body, and the body weighs as much as the head and the tail together.” How much does the fish weigh? | 2. From the problem statement, it is known that the tail ( $X$ ) weighs 1 kg. The weight of the head is equal to the weight of the tail plus an additional $1 / 2$ of the body weight (T):
$$
\Gamma=\mathrm{X}+\mathrm{T} / 2 \text { or } 2 \Gamma=2 \mathrm{X}+\mathrm{T} .
$$
Since the tail weighs 1 kg, then $2 \Gamma=2... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,742 |
3. This figure consists of 6 identical squares. Its perimeter is 84 cm. Find the area of this figure.
 | 3. 1 ) $84: 14=6$ (cm) - the length of the side of the square.
2) $6 \times 6=36$ (cm$^2$) - the area of one square.
3) $36 \times 6=216$ (cm$^2$) - the area of the entire figure.
Answer: 216 sq. cm. | 216 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,743 |
4. One of the digits of a four-digit number is zero. When the zero is erased, the number is reduced by nine times. In which position is the zero? Find such numbers. | 4. A four-digit number, one of the digits of which is zero, can have the form:
The last number can be immediately excluded, as in this case, the number $\boldsymbol{\text { a }}$ s will be ... | 2025,6075 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,744 |
5. How many numbers less than 96 are divisible by 2 and 3? | 5. We need to select numbers from 1 to 95 that are divisible by 2 and 3, which means they are divisible by 6. Such numbers are: $6,12,18$, $24,30,36,42,48,54,60,66,72,78,84,90$. There are 15 such numbers in total.
The number of such numbers could also be determined as follows:
$$
95: 6=15 \text { (remainder 5). }
$$
... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,745 |
6. How to cut a rectangle, which is 16 cm long and 9 cm wide, into two equal parts, from which a square can be formed?
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。 | 6. Let's first determine the area of the rectangle. It is equal to $144 \mathrm{~cm}^{2}(16 \times 9=144)$. Since $144=12 \times 12$, the length of the side of the square will be $12 \mathrm{cm.} \mathrm{The} \mathrm{area} \mathrm{of} \mathrm{one} \mathrm{part} \mathrm{should} \mathrm{be}$ $72 \mathrm{~cm}^{2}(144: 2=7... | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 42,746 | |
1. If from each of two numbers you subtract half of the smaller one, the remainder of the larger one is three times the remainder of the smaller one. How many times larger is the larger number than the smaller one? | 1. The first way to solve the problem is to represent the condition of the problem in the form of a drawing.
From the drawing, it is clear that the first number is twice as large as the sec... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,748 |
2. Three teams scored 285 points in the Olympiad. If the team from School No. 24 had scored 8 points less, the team from School No. 46 had scored 12 points less, and the team from School No. 12 had scored 7 points less, then all of them would have scored equally. How many points did the teams from Schools No. 24 and No... | 2. First method of solving. Let's make a brief record of the problem in the form of a drawing
1) $7+12+8=27$ (points) - this is how many fewer points all schools scored;
2) $285-27=258$ (p... | 187 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,749 |
3. There are three vessels with capacities of 6, 3, and 7 liters, respectively. The first vessel contains 4 liters, and the third vessel contains 6 liters of milk. Using these three vessels, it is necessary to divide the milk equally between two vessels. | 3. First, let's determine the total amount of milk:
$$
4+6=10 \text { (L). }
$$
Next, we need to find out how many liters should be in one container:
$$
\text { 10:2 = } 5 \text { (L). }
$$
Therefore, at the last pouring, the possible scenarios are $5-0-5, 5-5-0, 0-5-5$. The last two scenarios are impossible becaus... | 5-0-5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,750 |
4. Remove five out of twelve digits so that the remaining numbers add up to 1111.
$$
\begin{array}{r}
111 \\
333 \\
+\quad 777 \\
999 \\
\hline 1111
\end{array}
$$ | 4. The sum of the digits in the units place, 一 is a unit or a two-digit number ending in 1.
Case 1.
$$
\begin{array}{r}
111 \\
333 \\
+\begin{array}{l}
777 \\
799 \\
\hline 1111
\end{array}
\end{array}
$$
Since three digits are already crossed out, only one digit can be crossed out in the tens place - 9.
$$
\begin{... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,751 |
5. There were 2 minutes left before the electric train departed when the motorist was 2 km from the station. For the first minute, he drove at a speed of 30 km/h. At what speed should he drive during the second minute to catch the train? | 5. The first method of solving.
1) $2 \times 1000=2000$ (m) - need to travel;
2) $30 \times 1000=30000(\mathrm{m} / \mathrm{h})$ - speed of the car;
3) $30000: 60=500$ (m/s) - speed in the first minute;
4) $2000-500=1500$ (m) - left to travel in 1 minute;
5) $1500 \times 60=90000(\mathrm{m} / \mathbf{h})=90(\mathrm{km}... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,752 |
6. One side of the square was increased by 5 times, and the other side was reduced by 2 times, resulting in a rectangle with an area of $160 \mathrm{~cm}^{2}$. What is the side length of the square | 6. First method of solution.
1) $160: 5=32\left(\mathrm{~cm}^{2}\right)$ - area of half the square;
2) $32 \times 2=64\left(\mathrm{~cm}^{2}\right)$ - area of the original square;
3) $64: ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,753 |
7. Misha, Sergey, and Volodya participated in a math olympiad. When discussing who among them could be the winner, the following opinions were expressed: Misha and Sergey; Misha and Volodya; Sergey, but not Volodya.
It turned out that two of the boys received winner diplomas. Who among them became the winner of the ol... | 7. If the first assumption - Misha and Sergey turned out to be the winners - is true, then the second can be partially true, and the third assumption must then be false, but it contradicts the first.
If the second assumption - Misha and Volodya turned out to be the winners - is true, then the first assumption is parti... | MishaVolodyareceivedthewinner'diplomas | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,754 |
1. Calculate in the most rational way
$$
12 \times 171+29 \times 9+171 \times 13+29 \times 16
$$ | $$
\begin{aligned}
& 12 \times 171 + 29 \times 9 + 171 \times 13 + 29 \times 16 = \\
& = (171 \times 12 + 171 \times 13) + 29 \times 9 + 29 \times 16 = \\
& = 171 \times (12 + 13) + 29 \times (9 + 16) = \\
& = 171 \times 25 + 29 \times 25 = (171 + 29) \times 25 = \\
& = 200 \times 25 = 5000
\end{aligned}
$$
Answer: 50... | 5000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,755 |
3. From two points $A$ and $B$, the distance between which is 300 km, two cars set off simultaneously. The speed of the car that set off from $A$ is 40 km/h. Determine the speed of the second car, given that after two hours the distance between the cars was 100 km. | 3. Since the problem does not specify the direction they were traveling, we need to consider two cases: moving in the same direction and moving in different directions.
First case. Moving in the same direction. Let the first car travel from $A$ to $B$ at a speed of 40 km/h.
 $104: 10=10$ (remainder 4) (rubles) - the price of one book if there were 10 books;
2) $104: 60=1$ (remainder 44) (rubles) - the price of one book if there were 60 books.
Therefore, the price of one book is more than 1 ruble but less than 10 rubles. Thus, the price of one book can be 2, 3, 4, 5, 6, 7, 8, or 9 ru... | 2,4,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,759 |
6. In 1996, a farmer bought two sheep. This year he did not get any offspring. The first sheep gave birth to 1 sheep every three years, and the second sheep gave birth to 1 sheep every two years. All the born sheep gave birth to 1 sheep annually. How many sheep will the farmer have in the year 2000? | 6. Let's draw a diagram according to the problem statement.
Answer: 9 sheep. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,760 |
1. A sprinter runner covers a distance of 100 m in 10 seconds. Find his speed in m $/ \mathbf{s}$, m/min, km/h. | 1. 2) $100: 10=10(\mathrm{M} / \mathrm{s})$
2) $10 \times 60=600(\mathrm{m} / \mathrm{min})$
3) $600 \times 60=36000(\mathrm{m} / \mathrm{h})$
4) $36000: 1000=36$ (km/h).
Answer: $10 \mathrm{m} / \mathrm{s}, 600$ m/min, 36 km/h. | 10\mathrm{}/\mathrm{},600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,762 |
2. Arrange the order of operations in the expression
$$
1891-(1600: a+8040: a) \times c .
$$
and calculate its value when $a=40$ and $c=4$. Show how the expression can be modified without changing its numerical value. | $$
\begin{aligned}
& 1891-\left(1600: 40+8040^{5}: 40\right)^{4} \times 4= \\
& =1891-(40+201) \times 4= \\
& =1891-241 \times 4=1891-964=927 .
\end{aligned}
$$
Based on the properties of arithmetic operations, we can write:
$$
\begin{aligned}
& 1891-(1600: a+8040: a) \times c= \\
& =1891-(8040: a+1600: a) \times c= ... | 927 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,763 |
3. In this century, 200 years will be marked since the birth of the famous Russian mathematician, a native of Kaluga Governorate, Pafnuty Lvovich Chebyshev. Among the numbers that record his year of birth, the sum of the digits in the hundreds and thousands place is three times the sum of the digits in the units and te... | 3. Since in the 21st century the 200th anniversary will be marked, the first digit in the number will be -1, and the second will be 8, and the year of birth has the form $18 a c$.
The sum of the digits in the hundreds and thousands place is 9, and it is three times the sum of the digits in the units and tens place ($9... | 1821 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,764 |
4. Three boys participated in a "Russian Lotto" draw. Misha won 943 rubles more than Kolya, Vitya - 127 rubles more than Misha, and Misha and Kolya together - 479 rubles more than Vitya. How much money did each win? | 4. First method of solving. Let's make a brief record of the problem's condition in the form of a drawing:
From the drawing, it is clear that
1) $127+479=606$ (rubles) - Kolya won;
2) $60... | Kolyawon606rubles,Mishawon1549rubles,Vityawon1676rubles | Algebra | math-word-problem | Yes | Yes | olympiads | false | 42,765 |
5. Each of the three Greeks brought the same number of wreaths. Meeting nine Muses, they divided the wreaths in such a way that each Greek and each Muse had the same number of wreaths. How many wreaths did each Greek have at first? | 5. The least number of wreaths each Greek and Muse could receive after division is one. Then the Muses had 9 wreaths, and the Greeks brought a total of 12 wreaths $(3+9=12)$, with each Greek having 4 wreaths $(12: 3=4)$. If everyone received 2 wreaths, then the Muses would have a total of 18 wreaths $(2 \times 9=18)$. ... | 4m | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,766 |
1. The perimeter of quadrilateral $A B C D$ is 100 cm. The length of side $A B$ is 41 cm, side $B C$ is 18 cm shorter than side $A B$, but 6 cm longer than side $C D$. Find the length of side $A D$. | 1. The first method of solving.
1) $41-18=23$ cm - the length of side $B C$;
2) $23-6=17$ cm - the length of side $C D$;
3) $41+23+17=61$ cm - the sum of sides $A B, B C$, and $C D$;
4) $100-61=19$ cm - the length of side $A D$.
The second method of solving. Another approach can be used to solve this problem. First, d... | 19 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 42,769 |
2. "Well, Rabbit, just you wait!" the wolf shouted and rushed after the rabbit. Each step of the rabbit was half as long as the wolf's, but the rabbit took steps three times more frequently than the wolf. Will the wolf catch the rabbit? | 2. First method of solving. Let the hare's step have a certain unit of length, then the wolf's step is 2 such lengths. If the wolf, making one step, runs 2 units, then the hare in this time will make 3 steps and run 3 such units. Since 3 is greater than 2, the wolf will not catch the hare.
Second method of solving. Wh... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,770 |
3. Today is March 24, 2007. The sum of the digits of the year is 9. After how many years will the same sum of digits repeat? How many times will the same sum occur until the year 2107 inclusive? | 3. In the record of the year, the first digit does not change, so the sum of the second, third, and fourth digits should be equal to 7 $(9-2=7)$. Since the largest year is 2107, we need to consider numbers of the form $20 \mathrm{XX}$ and $210 \mathrm{X}$. Let's consider each of these cases separately.
Let's list all ... | 9,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 42,771 |
4. Roman, Fedia, Liza, Katya, and Andrey came to the club meeting. Roman came later than Liza, Fedia earlier than Roman and right after Katya. Katya came earlier than Liza, but was not the first. Who among the kids came to the meeting third? | 4. First solution method. Based on the condition that Roman came later than Liz, and Fedya came right after Katya, we have: LR, KF. Since Katya came earlier than Liz, and Fedya came earlier than Roman, we get the following situation: KFLR. Since Katya was not the first, it follows that the first was Andrey: AKFLR. Fedy... | Fedya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,772 |
5. Divide eleven cakes equally among six girls without cutting any cake into six equal pieces. | 5. Out of 11 cakes, we give each girl one, then we will have 5 cakes left $-11: 6=1$ (remainder 5). We will divide each of the 5 cakes into two equal halves. We will get a total of 10 pieces. They can be divided equally among 6 girls as follows: 10:6 = 1 (remainder 4). Thus, each girl will get one more half of a cake.
... | 1+1/2+1/4+1/12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 42,773 |
6. First, three, and then four people shook hands with each other. How many handshakes were there? Find the pattern in counting the number of handshakes and determine their number for 7 people. | 6. When solving this problem, we should use incomplete induction. Let's analyze the first situation and try to find a pattern in determining the number of handshakes based on the number of participants. In the first case, when there were three people, the following options are possible:
$$
\text { 1-2, 1-3, 2-3. }
$$
... | 21 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 42,774 |
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