problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
38. Find the smallest natural number divisible by 63, the sum of whose digits is 63.
## Search for solutions | 38. The smallest natural number whose sum of digits is 63 is the number written with seven nines $(9 \cdot 7=63)$. But the number 9999999 is not divisible by 7, since 9999990 is divisible by 7, and 9 is not divisible by 7. Replace 0 with the number 63. But the number 99999963 is not the smallest. To get the desired num... | 63999999 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,436 |
2. By multiplying four numbers, Nina got a result where the units digit is 0. What numbers did she multiply and what result did she get, if the multipliers are consecutive prime numbers? | 2. Since the product of prime numbers ends in 0, then among the factors there are 2 and 5. Since the factors are consecutive prime numbers, we have \(2 \cdot 3 \cdot 5 \cdot 7 = 210\).
140 | 210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,437 |
3. As soon as Dima named the number 17 - the sum of four prime numbers, Andrey immediately found their product, even though Dima did not name the addends. How did Andrey reason? What is the product he found? | 3. Since $17: 4=4$ (remainder 1), the prime numbers should be sought among numbers close to four. These are $2, 3, 5$ and 7. Indeed, they satisfy the condition $2+3+5+7=17$, and then the desired product $2 \cdot 3 \cdot 5 \cdot 7=210$. | 210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,438 |
4. What digit can the product of two non-single-digit prime numbers end with?
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 4. Simple ambiguous numbers are odd and do not end in 5, i.e., they can end in the digits $1,3,7$ or 9. The product of two such numbers ends in one of these four digits, since $1 \cdot 1=1,3 \cdot 3=9,7 \cdot 7=49$, $9 \cdot 9=81,3 \cdot 1=3,7 \cdot 1=7,9 \cdot 1=9,3 \cdot 7=21,3 \cdot 9=27,7 \cdot 9=63$. | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,439 | |
5. Is the number $1986^{1986}+1990^{1990}$ prime or composite? | 5. The number $1986^{1986}$ ends in the digit 6, and the number $1990^{1990}$ ends in the digit 0, so their sum ends in the digit 6, and such a number is even and different from 2, meaning it is composite. | composite | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,440 |
6. Is each of the following numbers prime or composite:
$$
1991^{1991}+1, 1991^{1991}-1 ?
$$ | 6. $1991^{1991}$ ends with the digit 1, so $1991^{1991}+1$ ends with the digit 2, and the number $1991^{1991}-1$ ends with the digit 0. Each of these numbers is composite. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,441 |
7. By writing down the digits from 1 to 9 in a row, Oleg obtained a nine-digit number. Will this number be prime or composite? And if the order of the digits in this number is changed arbitrarily, will the answer to this question change? | 7. Oleg received the number 123456789, the sum of whose digits is
$$
(1+9)+(2+8)+(3+7)+(4+6)+5=45
$$
but $45: 9$, so the given number is also divisible by 9, and therefore it is composite. No matter how the order of the digits in the written number is changed, the sum of its digits does not change, so all resulting n... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,442 |
8. On New Year's Eve, Father Frost gave the children the following task: using all nine digits from 1 to 9 once each, insert "+" or "-" between every two adjacent digits so that after performing the operations, all possible two-digit prime numbers can be obtained. Help the children complete this task. How many such num... | 8. If we place only the "+" sign between the digits, we will get 45 after performing the operations. There are only 10 prime two-digit numbers less than 45. These numbers are $11,13,17,19,23,29,31,37,41,43$. They can be obtained as follows:
1) $1+2+3+4+5+6-8-9=11$
2) $1+2+3+4+5+6-7+8-9=13$
3) $1+2+3+4-5+6+7+8-9=17$
4) ... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,443 |
10. To enter the castle of Arifmos, you need to enter a code: write down in ascending order the first 10 prime numbers of the natural series, each once. In the resulting multi-digit number, without rearranging the digits, cross out half of the digits so that the remaining ones express: a) the smallest possible number; ... | 10. Writing down the specified prime numbers, we get 2357111317192329. Since this number has 16 digits, we need to leave 8.
a) To get the smallest number, we need to keep the smallest single-digit numbers on the left. These are the ones. Therefore, we need to remove $2,3,5,7,3,7,9$, and also the eighth digit 3. As a r... | 1111122977192329 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,445 |
11. Can we obtain a prime number by adding the first two, three, four, or five consecutive odd numbers of the natural series? | 11. We will obtain the specified sums:
$1+3=4=2 \cdot 2$
$1+3+5=9=3 \cdot 3$
$1+3+5+7=16=4 \cdot 4$
$1+3+5+7+9=25=5 \cdot 5$
Thus, it is impossible to obtain a prime number. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,446 |
12. After completing her math homework, Lena prepared a problem for the math club session. She found two two-digit prime numbers that can be obtained from each other by swapping digits, and their difference is a perfect square. What numbers did Lena find? | 12. The numbers found by Lena cannot contain even digits and the digit 5. They can only be written using the digits $1,3,7,9$. We have the numbers:
1) 31 and 13, but $31-13=18$ is not a square;
2) 71 and 17, but $71-17=54$ is not a square;
3) 97 and 79, but $97-79=18$ is not a square;
4) 73 and 37, and $73-37=36=6^{2}$... | 7337 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,447 |
13. "My phone number," said the trip leader to the kids, "is a five-digit number. The first digit represents a prime number, and the last two digits are obtained from the previous pair, which represents a prime number, by rearrangement and form a perfect square. The number, reversed to this phone number, is even." What... | 13. Since the reversed number is even, the first digit of the number is 2. (This is the only even prime number.) Of the exact squares 16, 25, 36, 49, 64, 81, only the number 16 gives a reversed prime number 61. Therefore, the phone number of the leader is $2-61-16$. | 26116 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,448 |
14. The sum of the squares of two certain prime numbers ends with the digit 9. Find all such prime numbers. | 14. Since the sum of two addends is an odd number (ending in the digit 9), one addend is even, and the other is odd. Since the parity of a square and its base is the same, one of the two prime numbers is even, and the other is odd. But there is only one even prime number - 2, so its square is 4, which means the square ... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,449 |
15. When classmates asked Kolya about the number of the apartment his family received in the new building, Kolya, a math enthusiast, replied: “Our apartment number is a three-digit number that is a perfect cube. If you swap the hundreds and units digits, you get a prime number.” What is Kolya's apartment number? | 15. Among three-digit numbers, only 5 are perfect cubes: $5^{3}=125$, $6^{3}=216$, $7^{3}=343$, $8^{3}=512$, $9^{3}=729$. Only the number 125 satisfies the condition, as only 521 is prime. | 125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,450 |
16. Help Pinocchio arrange all 28 domino bones into four piles so that the sums of the points in the piles are four consecutive prime numbers. | 16. Let's find the sum of all the points on the dominoes. The domino tiles look like this:
```
\(\begin{array}{lllllll}00 & 11 & 22 & 33 & 44 & 55 & \underline{66}\end{array}\)
\(\begin{array}{lllllll}01 & 12 & 23 & 34 & 45 & 56 & 12\end{array}\)
\(\begin{array}{llllll}02 & 13 & 24 & 35 & 46 & 21\end{array}\)
\(\begin{... | 37,41,43,47 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,451 |
17. What odd digit can the sum of two prime numbers end with, if it is not a single-digit number? | 17. If the sum of two numbers ends in an odd digit, then one addend is even, and the other is odd. Since among even numbers, the only prime number is 2, one of the addends must be 2. If the sum ends in the digit 1, then the second addend must end in the digit 9. Such a case is possible, for example, \(19+2=21\). If the... | 1,3,5,9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,452 |
18. Math enthusiasts Kolya and Tolya have been trying for a long time, but in vain, to find three different digits such that all three-digit numbers formed from them without repetition are prime. Help them with this. | 18. Among the digits, there should not be any even number or five. Therefore, for writing three-digit numbers, one can use $1,3,7,9$.
a) Among the numbers $137,173,317,371,713$, and 731, the last three numbers are composite. For example, $371=53 \cdot 7$.
b) Among the numbers $139,193,319, \ldots$ the number 319 is c... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,453 |
19. Preparing for a math evening, the kids came up with this problem: "Insert 12 consecutive prime numbers (from 3 to 41) into the circles (Fig. 1) so that the sums of the three numbers at the vertices of the triangles are equal. Additionally, the sums of the numbers on the inner and outer circles should also be the sa... | 19. The sum of all prime numbers from 3 to 41 inclusive is 236. Therefore, the sum of the numbers on the inner (and outer) circle is: \(236: 2=118\), and the sum of the numbers at the vertices of each triangle is: \(236: 4=59\). The larger of the given numbers should obviously be written in the circles of the inner cir... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,454 |
20. Find $p$, if each of the numbers $p, p+10$, and $p+14$ is prime.
 | 20. All natural numbers can be divided into 3 groups:
1) divisible by 3, i.e., of the form $3 k$;
2) giving a remainder of 1 when divided by 3, i.e., of the form $3 k+1$;
3) giving a remainder of 2 when divided by 3, i.e., of the form $3 k+2$.
If $p=3 k$, then it is prime only when $k=1$; then $p+10=13$ and $p+14=17$.... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,455 |
21. Find $p$, if each of the numbers $p, 2p+1$, and $4p+1$ is prime. | 21. The number $p$ can be represented in one of three forms: $3 k, 3 k+1$ or $3 k+2$. For $p=3 k+1$, we have:
$2 p+1=2(3 k+1)+1=6 k+3=3(2 k+1)$ - a composite number. For $p=3 k+2$, we have:
$4 p+1=12 k+9=3(4 k+3)$ - a composite number.
Fig. 12
$, then the next one is $2 n+1$, and the largest is $2 n+3$. There are three possible cases:
a) $n=3 k$, then $2 n+3=6 k+3=3(2 k+1)$ is not a prime number;
b) $n=3 k+1$, then $2 n+1=6 k+3$, i.e., composite;
c) $n=3 k+2$, then $2 n-1=6 k+3$, i.e.,... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,457 |
23. Find all such prime numbers, so that each of them can be represented as the sum and difference of two prime numbers. | 23. Since the desired number $p$ must be the sum of two prime numbers, $p>2$. All such prime numbers are odd. To get odd numbers when adding and subtracting two numbers, one of the components must be odd and the other even. Since the only even prime number is 2, then $p=p_{1}+2$ and $p=p_{2}-2$. Therefore, the prime nu... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,458 |
24. Misha tried for a long time in vain to find two numbers such that their product and sum would be divisible by some prime number, but at least one of them would not be divisible by this prime. Help Misha get out of this situation. | 24. We need to find out if such numbers exist. If $a \cdot b$ is divisible by a prime number $p$, then at least one of the numbers $a$ or $b$ must be divisible by $p$. But if, for example, $a$ is divisible by $p$, then the divisibility of the sum implies that $b$ is also divisible by $p$. Similarly, if $b$ is divisible... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,459 |
25. Find three prime numbers, the product of which is three times their sum. | 25. Let the required numbers be $a, b$ and $c$, then $a b c=3(a+b+c)$, from which $a+b+c=a b c: 3$. The equality is possible under the condition that one of the numbers is equal to 3. Let $a=3$, then $3+b+c=b c$. Of the two prime numbers $b$ and $c$, at least one is odd. Let $b=2 n+1$, then $3+2 n+1+c=2 c n+c$, from wh... | 2\cdot3\cdot5=3\cdot(2+3+5) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,460 |
26. Prove that if the product $1 \cdot 2 \cdot 3 \ldots n$ for $n>3$ is not divisible by $n+1$, then $4n+1$ is a prime number. | 26. Suppose that $n+1$ is a composite number, then $n+1=a \cdot b$ and $a<n+1$, $b<n+1$. Since $1 \cdot 2 \cdot 3 \ldots n$ is the product of all numbers less than $n+1$, there will be numbers among the factors equal to $a$ and $b$. Therefore, this product will be divisible by $n+1$. However, by the condition, it is no... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,461 |
27. The boys from two sixth-grade classes made 123 little chairs for kindergartens in the school workshops. How many boys worked and how many chairs did each of them make, if they made them equally? | 27. Since $123=41 \cdot 3$ and the numbers 41 and 3 are prime, it is most plausible that there were 41 boys and each of them made 3 stools. | 41 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,462 |
28. In all the carriages of a passenger train, 737 tourists were evenly distributed. How many carriages were there and how many tourists were in each carriage? | 28. Since $737=67 \cdot 11$ and the numbers 67 and 11 are prime, it is clear that there were 11 cars, and in each car there were 67 tourists. | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,463 |
29. The product of some prime numbers is 30030. Is their sum a prime or a composite number? | 29. $30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$.
$2+3+5+7+11+13=41$. The number 41 is prime. | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,464 |
30. Baron Munchausen claimed that he managed to find such a natural number, the product of all the digits of which is 6552. Show that he was lying. | 30. The number 6552 can be represented by the following factorization: $6552=2^{3} \cdot 3^{2} \cdot 7 \cdot 13$, or $6552=8 \cdot 9 \cdot 7 \cdot 13$. Since there is no digit 13 in the decimal system, Münchhausen's statement is incorrect. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,465 |
31. Prove that a three-digit number, in which all digits are the same, in its prime factorization contains the number 37. | 31. Let's consider the number $\overline{a a a}$. It can be written as $\overline{a a a}=100 a+10 a+a=$ $=111 \cdot a=37 \cdot 3 \cdot a$. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,466 |
32. In each cell of a $4 \times 4$ square table, a prime number must be written such that the product of the numbers in each column and each row equals 1989. If this cannot be done, prove it; if it can, write the numbers.
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。 | 32. Since $1989=3 \cdot 3 \cdot 13 \cdot 17$, i.e., it can be represented as the product of four prime numbers, the problem has a solution:
| 3 | 3 | 13 | 17 |
| :---: | :---: | :---: | :---: |
| 3 | 3 | 17 | 13 |
| 13 | 17 | 3 | 3 |
| 17 | 13 | 3 | 3 | | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,467 | |
33. To answer the question: "Is the difference $9^{1986}-8^{1980}$ divisible by 5 and by 10?" - one could call upon a computer. But perhaps you would like to compete with it? Give it a try. | 33. Since $9^{1}=9, 9^{2}=81, 9^{3}=729$, it is clear that the odd powers of nine end in the digit 9, and the even powers end in the digit 1. Let's consider the powers of the number 8: $8^{1}=8 ; 8^{2}=64$-ends in the digit $4 ; 8^{3}$ ends in the digit $2 ; 8^{4}$ ends in the digit $6 ; 8^{5}$ ends in the digit 8, and... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,468 |
34. To calculate the product of all natural numbers from 1 to 50 inclusive, of course, it's better to use a computer. However, you can easily state the number of zeros at the end of this product without resorting to a computer. How many are there? | 34. Among the factors of the number $1 \cdot 2 \cdot 3 \cdot \ldots \cdot 50$, 10 numbers are divisible by 5 $(50: 5=10)$, and out of these, 2 numbers are divisible by $5^{2}=25$. Therefore, the product of the numbers from 1 to 50, when factored into prime factors, will contain 12 fives, and there will undoubtedly be m... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,469 |
35. Prove that if in a three-digit number the last two digits are the same and the sum of its digits is divisible by 7, then in the prime factorization of this number there is the number 7. | 35. Let's have $\overline{a b b}=100 a+11 b$. Transform this sum $100 a+11 b=$ $=(98 a+7 b)+(2 a+4 b)=7(14 a+b)+2(a+2 b)$. By the condition $a+2 b=7 k$ (since it is divisible by 7). As a result, we get $\overline{a b b}=7 \cdot(14 a+b+2 k$ ). | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,470 |
36. Let's say we have two three-digit numbers that give the same remainder when divided by 7. By appending one number to the other, we get a six-digit number. Show that the resulting number in its prime factorization contains the number 7. | 36. Let one number be $(7 a+k)$, and the other $(7 b+k)$. Then the six-digit number will be
$$
1000(7 a+k)+(7 b+k) \text { or } 1000(7 b+k)+(7 a+k)
$$
Consider one of them:
$$
1000(7 a+k)+(7 b+k)=7(1000 a+b)+1001 k
$$
Since $1001=7 \cdot 11 \cdot 13$, the six-digit number can be represented as: $7(1000 a+b+11 \cdot... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,471 |
37. By calculating the sum of all distinct prime divisors of a certain six-digit number, in which all digits are the same, Stepan Rasteryaykin got 70, and Petya Ugadaykin got 80. Show that they were both wrong. | 37. Since $\overline{\text { aaaaaa }}=1001 \cdot 111 a=7 \cdot 11 \cdot 13 \cdot 3 \cdot 37 a$, the desired sum is: $7+11+13+3+37+m=71+m$, where $m$ is the sum of the distinct prime divisors of the single-digit number $a$, different from 3 and 7. For $a=1$ we have $m=0$, since the number 1 has no prime divisors. For $... | 71,73,76 | Number Theory | proof | Yes | Yes | olympiads | false | 39,472 |
38. Members of the "Eureka" math club claim that if in a six-digit number the sums of the digits in the first and fourth, second and fifth, and third and sixth positions (counting from left to right) are equal to each other, then this number in its prime factorization contains 37. What do you think? | 38. Let's consider $\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}$. We can write this number as a sum of place values and transform it as follows:
$100000 a_{1}+10000 a_{2}+1000 a_{3}+100 a_{4}+10 a_{5}+a_{6}=$ $=\left(100 a_{1}+100 a_{4}\right)+\left(10 a_{2}+10 a_{5}\right)+\left(a_{3}+a_{6}\right)+\left(99900 a_{1... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,473 |
39. Is it possible to form a ten-digit number using each digit exactly once that is divisible by 1980?
Note. To solve this problem, additional information about the divisibility of a number by 11 is required. If the difference between the sums of the digits in the even and odd positions (the alternating sum of the dig... | 39. The desired number must end with the digit 0, then when dividing it by 1980, we can reduce it by 10. The resulting nine-digit number must be divisible by 198. Since $198=2 \cdot 9 \cdot 11$, this nine-digit number must be divisible by 2, 9, and 11. Therefore, it must end with an even digit, have a sum of digits div... | 2384715960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,474 |
40. The steamship has $a$ funnels, $b$ screws, and $c$ people on board. It set off on the $n$-th day, $k$-th month, of the year $(1900+p)$. The product of the six numbers $a$, $b$, $c$, $n$, $k$, $p$, increased by the number whose cube is the captain's age in years, equals 4752862. Determine the unknowns. | 40. Since $2^{3}=8,3^{3}=27,4^{3}=64,5^{3}=125$, it is clear that the captain is either 27 years old or 64 years old. If the captain is 64 years old, then from the number 4752862, 4 should be subtracted and the resulting number should be factored into 6 factors ( $4752862-4=$ $=4752858 ; 4752858=2 \cdot 3 \cdot 11 \cdo... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,475 |
41. After persistent and prolonged searches, Andrey found a two-digit number that does not end in 0 and has the property that, by inserting a zero between the tens and units digits, a three-digit number is obtained that is divisible without a remainder by the original two-digit number. When he showed his result to the ... | 41. Let the desired number be $\overline{a b}$, then the three-digit number is $\overline{a o b}$. According to the condition, $\overline{a b} \cdot n = \overline{a o b}$ 146
or $(10 a + b) \cdot n = 100 a + b$, from which $10 a(10 - n) = b(n - 1)$. Therefore, $110 - n$, from which $n > 5$. Since the left side of the e... | 15,18,45 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,476 |
1. When Vitya, determining the greatest common divisor of three numbers, named 15 as the answer, Vera, not knowing these numbers but knowing that their product is 3000, noticed that he was wrong. Is she right? | 1. Since $3000=2^{3} \cdot 3 \cdot 5^{3}$, the greatest common divisor of such three numbers can only take the following values: $1,2,5$ or 10. Therefore, Vera is correct. | Vera | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,477 |
2. What can the least common multiple of three consecutive natural numbers be equal to? | 2. Let's consider 3 numbers: $n, n+1, n+2$. If $n+1$ is even, then the LCM of these numbers is equal to their product. For example, LCM $(3,4,5)=60$. If $n+1$ is odd, then $n$ and $n+2$ are even, and in this case, the LCM of such numbers is equal to their half-product, i.e., $\frac{n(n+1) \cdot(n+2)}{2}$.
For example,... | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,478 |
3. By the puddle, the squad In twos, in threes, in fours
Of long-legged tadpoles, No matter how the commander arranged,
Formed a very neat column: There was always someone left over
Exactly five in every row. In any group of five.
Now everyone is happy, everyone is delighted!
But how many tadpoles are there in tot... | 3. The desired number, when divided by 2, 3, and 4, each time gives a remainder of 1, but is divisible by 5. LCM $(2,3,4)=3 \cdot 4=12$, but the number $12+1=13$ is not divisible by 5. However, the number $12 \cdot 2+1=25$ is the smallest number that satisfies the condition. Since the number of rows is less than 10, th... | 25 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,479 |
4. While working on fruit harvesting, class 5A collected 560 kg of apples, 5B - 595 kg, and 5V - 735 kg. All the collected apples were packed into boxes, with each box containing the same maximum possible number of kilograms. How many such boxes were required for each class? | 4. The number of kilograms of apples in each box is the greatest common divisor of the numbers 560, 595, and 735. Since $560=2^{4} \cdot 5 \cdot 7, 595=5 \cdot 7 \cdot 17$ and $735=3 \cdot 5 \cdot 7^{2}$, then GCD $(560,595,735)=35$. This means that 35 kg of apples were placed in each box. Then the first squad needed 1... | 16,17,21 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,480 |
5. Three military trains passed through the railway station. The first had 462 soldiers, the second had 546, and the third had 630. How many cars were in each train if it is known that each car had the same number of soldiers and that this number was the largest possible? | 5. The number of soldiers traveling in each carriage is the greatest common divisor of the numbers 462, 546, and 630. GCD $(462,546,630)=42$. Therefore, in the first train there were 11 carriages $(462: 42=11)$, in the second $13(546: 42=13)$, in the third $-15(630: 42=15)$. | 11,13,15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,481 |
6. The numbers 100 and 90 were divided by the same number. In the first case, the remainder was 4, and in the other case, it was 18. By what number were they divided? | 6. The sought divisor divides the numbers $96(100-4=96)$ and $72(90-18=72)$. This divisor will be common for the numbers 96 and 72. Since the remainders upon division are 4 and 18, the divisor is greater than 18. Such a number is 24, since the GCD $(96,72)=24$ | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,482 |
7. In one school library, there are 24850 books, and in another - 55300. When these books were being arranged on shelves equally, 154 books were left in the first library, and 175 in the other. How many books were placed on each shelf? | 7. The desired number is a common divisor of the numbers $24696(24850-154=24696)$ and $55125(55300-175=55125)$. Since $24696=2^{3} \cdot 3^{2} \cdot 7^{3}$ and $55125=3^{2} \cdot 5^{3} \cdot 7^{2}$, the common divisors are $3,7,9,21,49,63,147,441$. Since the divisor must be greater than the remainder, only 441 satisfie... | 441 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,483 |
8. On a circular track that is 660 m long, a relay race is being held, with each stage being $150 \mathrm{~m}$. The start and finish are in the same place. What is the minimum number of stages that can be in this relay? | 8. Since 660 is not divisible by 150 without a remainder, we will find the least common multiple of these numbers: $660=2^{2} \cdot 3 \cdot 5 \cdot 11, 150=2 \cdot 3 \cdot 5^{2}$, LCM $(660,150)=$ $=660 \cdot 5=3300$. Therefore, the smallest number of stages in this relay is $3300: 150=22$. | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,484 |
9. The cruise ship "Happy" arrives at the seaport once every 3 days, the cruise ship "Lucky" - once every 4 days, and the cruise ship "Reliable" - once every 5 days. Last Monday, all three cruise ships were in this port. After how many minimum days will they all arrive at this port again, and what day of the week will ... | 9. All three steamships will be in the port after a number of days that is the least common multiple of the numbers 3, 4, and 5, i.e., after $3 \cdot 4 \cdot 5=60$ days. This will be on the fourth day of the week, counting from Tuesday, i.e., on Friday, since $60: 7=8$ (remainder 4). | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,485 |
10. Find the greatest common divisor of all five-digit numbers written using the digits
$$
1,2,3,4,5
$$
without repetition. | 10. Each of such five-digit numbers is divisible by 3, since $1+2+$ $+3+4+5=15$, and $15: 3$. These numbers do not have any common divisors greater than 3, as, for example, $12345=3 \cdot 5 \cdot 823$, but the number 12354 does not divide evenly by either 5 or 823. Therefore, the greatest common divisor of these number... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,486 |
11. In one class, there are fewer than 50 students. On Gorky Street, $\frac{1}{7}$ of the students from this class live, on Pushkin Street - $\frac{1}{3}$, on Gaidar Street - $\frac{1}{2}$, and on Shipbuilding Street - the remaining students. How many students live on Shipbuilding Street? | 11. The number of students in the class must be a multiple of the numbers 7, 3, and 2. The least common multiple of these numbers is $42(7 \cdot 3 \cdot 2=42)$. It is less than 50, so it satisfies the condition. Then $42 \cdot \frac{1}{7}=6 ; 42 \cdot \frac{1}{3}=14 ; 42 \cdot \frac{1}{2}=21$; $42-(6+21+14)=1$. Therefo... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,487 |
12. Following the order of King Pea, General Mushtralkin tried to arrange all the soldiers in rows first by 2, and then by $3,4,5,6,7,8,9,10$, but to his surprise, each time the last row was incomplete, as there were respectively
$$
1,2,3,4,5,6,7,8,9
$$
soldiers left. What is the smallest number of soldiers there cou... | 12. If we add 1 to the desired number, we get a number that is divisible by each of the numbers from 2 to 10 inclusive. Therefore, the desired number is the least common multiple of the numbers from 2 to 10, decreased by 1, i.e., \(5 \cdot 7 \cdot 8 \cdot 9-1=2520-1=2519\). | 2519 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,488 |
13. At the lumber yard, there were logs 6 m and 8 m long of the same thickness. Which logs are more advantageous to take in order to get more pieces with fewer cuts when cutting them into 1-meter pieces? | 13. It is more profitable to take logs from which more pieces can be obtained with fewer or the same number of cuts. From a six-meter log, 6 pieces can be obtained with 5 cuts, while from an eight-meter log, 8 pieces can be obtained with 7 cuts. Let's compare the number of pieces obtained from these logs with the same ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,489 |
2. Fishermen caught 80 fish in a pond with a net, marked them, and released them back into the pond. The next day, they caught 150 fish, 5 of which were marked. How many fish are there in the pond in total?
Note. The marked fish are evenly distributed among the others. | 2. Obviously, the marked fish mixed uniformly with all the fish in the pond. Among the caught fish, the marked ones constitute $5: 150=\frac{1}{30}$. Therefore, the total number of fish in the pond is $80 \cdot 30=2400$ (fish). | 2400 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,493 |
3. When asked how many sheep were in the flock, the shepherd replied: "60 sheep are drinking water, and the rest, 0.6 of all the sheep, are grazing." How many sheep are in the flock? | 3. 60 sheep constitute 0.4 of all the sheep $(1-0.6=0.4)$. Therefore, the total number of sheep in the flock is $60: 0.4=150$. | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,494 |
6. A grasshopper can jump exactly 0.5 m in any direction. Can it move exactly 7.3 m in several jumps? | 6. In 14 jumps, a grasshopper can move 7 meters $(0.5 \cdot 14=7)$. To move 0.3 meters $(7.3-7=0.3)$, it can do so in two jumps, "tracing" the sides of a triangle with legs of 0.5 meters and a base of 0.3 meters. | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,497 |
9. Half of the way the tourists walked, and the other half they traveled by bus, spending a total of 5.5 hours on the entire journey. If they had traveled the entire distance by bus, it would have taken them 1 hour. How much time in total will the tourists spend if they walk the entire distance? How many times faster i... | 9. Half of the way the tourists traveled by bus in 0.5 hours (1:2=0.5), then on foot they walked the other half in 5 hours (5.5-0.5). Therefore, the entire journey will take the tourists 10 hours (5 * 2=10). Traveling by bus is 10 times faster than walking. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,500 |
10. The numbers 2.75 and 8 have the property that the product of these numbers equals the sum of their digits: $2.75 \cdot 8=2+7+5+8=22$. Find at least one more such pair of unequal numbers. | 10. 11) $0.275 \cdot 80=22 ; 2+7+5+8+0=22$
2) $2.6 \cdot 5=13 ; 2+6+5=13$
3) $26 \cdot 0.5=13 ; 2+6+0+5=13$
4) $6.5 \cdot 2=13 ; 6+5+2=13$.
(You can provide as many solutions as you like.) | 2.6\cdot5=13;2+6+5=13 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,501 |
11. The bear was carrying pastries from the bazaar,
But on the forest edge
He ate half the pastries
And plus half a pastry.
He walked, walked, sat down to rest
And to the "coo-coo" of the cuckoo
He ate half the pastries again
And plus half a pastry.
It got dark, he quickened his pace,
But on the porch of the cottage
... | 11. On the porch of the little house, the bear ate all the remaining cookies, which were half a cookie and half a cookie. This means that half a cookie is half of the remainder, i.e., the third time he ate one cookie. Therefore, when he ate half the cookies and another half cookie the second time, one cookie was left. ... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,502 |
12. A wristwatch is 5 minutes slow per hour; 5.5 hours ago they were set to the correct time. Now, on the clock showing the correct time, it is 1 PM. In how many minutes will the wristwatch show 1 PM? | 12. In 60 minutes, the hand of the watch "counts" 55 minutes. They need to "count" 5.5 hours. They will do this in 6 hours $\left(\frac{5.5 \cdot 60}{55}=6\right)$. Therefore, 1 hour of the day they will show after 0.5 hours or 30 minutes $(6-5.5=0.5)$. | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,503 |
13. Minutkin usually wound the clock to the limit twice a day: in the morning at 8:30 and at night when going to bed. In the morning, he had to make 9 full turns of the clock's crown, and at night - 11. At what time did Minutkin go to bed? | 13. Over the night, the clock spring unwound by 9 turns of the head, during the day - by 11, and over 24 hours - a total of 20 turns $(9+11=20)$. One turn of the clock head allows the spring to work for 1.2 hours $(24: 20=1.2)$. Thus, from the time Minutkin went to bed until 8:30 AM, 10 hours and 48 minutes passed $(1.... | 9:42 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,504 |
14. While conducting observations on a geographical site, students recorded the temperature daily from April 15 to April 19 inclusive. The average temperature over these days was $17.5^{\circ}$. What were the temperatures on April 19 and April 15, if the temperature increased by $1.5^{\circ}$ daily? | 14. If the temperature increased daily by $1.5^{\circ}$ on the 15th, 16th, 17th, 18th, and 19th, then the average temperature was on the 17th. If the temperature on April 17th was $17.5^{\circ}$, then on April 19th it was $17.5^{\circ}+3^{\circ}=20.5^{\circ}$, and on April 15th it was $17.5^{\circ}-3^{\circ}=14.5^{\cir... | 14.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,505 |
15. There are 35 students in the class. Together, they are 280 years old. Is there a group of 25 students among them whose combined age is at least 225 years? | 15. The average age of a student in this class is 8 years $(280: 35=8)$. If it were possible to find 25 students in the class whose total age is at least 225 years, then the average age of this group would be at least 9 years $(225: 25=9)$. Then the remaining 10 students would have a total age of no more than 55 years ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,506 |
16. From a school plot of rectangular shape measuring $70 \times 35$ m, 14.7 tons of corn were harvested. Prove that among the squares of this plot, each of which has a size of $0.7 \times 0.7$ m, there will certainly be some that yielded the same harvest. | 16. The entire plot consists of 5000 squares of the specified size $\left(\frac{70 \cdot 35}{0.7 \cdot 0.7}=\right.$ $=5000)$. On average, each square yields 294 g $(14.7$ c: $5000=294$ g $)$. If we assume that no squares have the same yield, then the total yield from the entire plot would be no less than $(0+1+2+\ldot... | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 39,507 |
17. When dividing one number by another, we obtained a decimal fraction, the integer part of which is equal to the divisor, and after the decimal point, the dividend is written. What are these numbers? | 17. Let the required numbers be $a$ and $b$, then $a: b=b, a$ or after multiplying by 10 we get $10 a: b=10 b+a$. Since $a$ does not divide evenly by $b$, then the number 10 must divide by $b$, which means $b=2$ or $b=5(b \neq 10$, otherwise we would get $a=100+a$, which is impossible). If $b=5$, then $2 a=50+a$, from ... | =5,b=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,508 |
18. A group of tourists was supposed to arrive at the station at 5 o'clock. By this time, a bus was supposed to arrive from the tourist base to pick them up. However, arriving at the station at 3:15 PM, the tourists, not waiting for the bus, started walking to the tourist base. Meeting the bus on the way, they got in a... | 18. Since the bus arrived at the tourist base 15 minutes earlier, it should have traveled from the meeting point with the tourists to the station for $\frac{1}{8}$ hours (15 minutes / 2 = $=7.5$ minutes $=\frac{1}{8}$ hours). In this time, it would travel 7.5 km $(60: 8=7.5)$, which means the tourists walked 7.5 km on ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,509 |
20. In the drama club, the number of boys is $80 \%$ of the number of girls. What percent of the number of boys does the number of girls represent in this drama club? | 20. Let the number of girls be $a$, then the number of boys is $0.8 a$. The number of girls constitutes $\frac{a}{0.8 a} \cdot 100 \% = 125 \%$ of the number of boys. | 125 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,511 |
21. A plot of land in the shape of a rectangle was allocated for corn. After some time, the length of this plot was increased by $35\%$, and the width was decreased by $14\%$. By what percentage did the area of the plot change? | 21. Let the length of the plot be $a$, and the width be $b$, then its area is $a b$. After increasing the length by $35\%$, it became $1.35 a$, and after decreasing the width, it became $0.86 b$. The area of the plot became: $1.35 a \times 0.86 b = 1.161 a b$. Therefore, the area increased by $1.161 a b - a b = 0.161 a... | 16.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,512 |
22. Of the houses built in one district of the city this year, more than $94 \%$ have more than five floors. What is the smallest possible number of houses in this case? | 22. The smallest number of houses will be in the case where the largest possible percentage of houses having no more than five floors constitutes the smallest number of houses. This is possible if $5 \%$, i.e., $\frac{1}{20}$, part, constitutes one house. Then the total number of houses will be $1 \cdot 20=20$. | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,513 |
23. In the workshop, there are no more than 100 workers, a third of them are women, $8 \%$ of the workers have a reduced working day. How many workers are there in the workshop? How many of them are women and how many people have a reduced working day? | 23. Since $8 \%=\frac{2}{25}$, the number of workers in the workshop must be a multiple of 25. In addition, since a third of the workers are women, the number of workers is a multiple of 3. Such a number, not exceeding 100, is 75. Therefore, the number of workers is 75. The number of women in the workshop is $75: 3=25$... | 75,25,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,514 |
24. The moisture content of freshly cut grass is $70 \%$, and the moisture content of hay is $16 \%$. How much grass needs to be cut to obtain 1 ton of hay? | 24. In 1 t of dry hay, there will be $840 \mathrm{kr}(100 \%-16 \%=84 \%, 1000 \cdot 0,84=840)$. In fresh-cut grass, the dry mass constitutes $30 \%(100 \%-70 \%=30 \%)$. Therefore, the amount of grass needed is $840: 0,3=2800$ (kg). Thus, to obtain 1 t of hay, 2.8 t of grass must be cut. | 2800 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,515 |
25. Fresh mushrooms contain $90 \%$ water, and when they were dried, they became lighter by 15 kg with a moisture content of $60 \%$. How much did the fresh mushrooms weigh? | 25. If there were $x$ kg of fresh mushrooms, then the dry mass in them is $0.1 x$ kg. After drying, the dry mass became $40\%$, and the water $-60\%$. This means that at a moisture content of $60\%$, the mushrooms became $\frac{0.1 x \cdot 100\%}{40\%}=\frac{1}{4} x$ (kg), and $ \frac{3}{4} x$ kg of water evaporated, w... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,516 |
26. The price of a product was first reduced by $10 \%$, and then by $15 \%$. What is the overall percentage reduction? Would it be the same if the price was first reduced by $15 \%$, and then by $10 \%$? | 26. Let the original price be $a$. After a $10\%$ reduction, it became $0.9a$, and after a $15\%$ reduction, it became $0.765(0.9a \cdot 0.85 = 0.765a)$, which is $23.5\%$ lower than the original $(100\% - 76.5\% = 23.5\%)$. If the price had first been reduced by $15\%$, it would have become $0.85a$; if it had then bee... | 23.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,517 |
28. Determine the length of the path over which the amount of grain in the seeder's box will decrease by $14 \%$, if the box has a working width of 4 m, it was filled with 250 kg of grain, and the sowing rate is 175 kg per 1 ha. | 28. If the amount of grain in the seeder's box decreases by $14 \%$, then 35 kg of grain will be sown $(250 \cdot 0.14=35)$. This constitutes $\frac{35}{175}=\frac{1}{5}$ of the sowing norm per 1 ha, which means that an area of $\frac{1}{5}$ ha, or $2000 \mathrm{~m}^{2}(10000: 5=2000)$, has been sown. Knowing the width... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,519 |
29. In a consignment store, apples were sold for 60 cents per 1 kg on the first day. On the second day, after the price reduction, they sold 50% more apples than on the first day, and revenue increased by 12.5%. What was the price of apples after the reduction? | 29. Let $a$ kg of apples be sold on the first day, then the revenue on that day was $60 a$ k. After the price reduction on the second day, the revenue was $60 a \cdot 1.125$ k. and they sold $1.5 a$ kg of apples. The price per kg of apples was $60 a \times 1.125 : 1.5 a = 45($ k. $)$. | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,520 |
30. In one of the semi-cottage productions, when casting blanks in molds and machining parts from them, only $66 \frac{2}{3} \%$ of the material from each blank was used for each part, and the rest went to chips. To save on the scarce material, the chips were remelted and cast into blanks again. What is the maximum num... | 30. Having obtained 20 parts from 20 blanks, one-third of the total material will end up as swarf, i.e., the equivalent of $6 \frac{2}{3}$ blanks $\left(20: 3=6 \frac{2}{3}\right)$. From this material, 6 blanks are obtained, and there is still material left for $\frac{2}{3}$ of a blank. Manufacturing 6 parts from this ... | 29 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,521 |
31. Compare the fractions:
$$
\text { a) } \frac{373737}{777777} \text { and } \frac{37}{77} ; \text { b) } \frac{41}{61} \text { and } \frac{411}{611}
$$ | 31. a) Since $\frac{373737}{777777}=\frac{37 \cdot 10101}{77 \cdot 10101}=\frac{37}{77}$, the fractions are equal.
b) Let's find the fractions that complement the given ones to 1, and compare them.
$1-\frac{41}{61}=\frac{20}{61} ; \quad 1-\frac{411}{611}=\frac{200}{611} ; \quad \frac{20}{61}=\frac{200}{610}$.
Since ... | )\frac{373737}{777777}=\frac{37}{77};\quadb)\frac{41}{61}<\frac{411}{611} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,522 |
32. Which fraction is larger:
$$
\frac{200200201}{200200203} \text { or } \frac{300300301}{300300304} ?
$$ | 32. $\frac{200200201}{200200203}=1-\frac{2}{200200203} ; \frac{300300301}{300300304}=1-\frac{3}{300300304}$; $\frac{2}{200200203}=\frac{6}{600600609} ; \frac{3}{300300304}=\frac{6}{600600608}$.
Since $\frac{6}{600600609}<\frac{6}{600600608}$, it follows that $\frac{200200201}{200200203}>\frac{300300301}{300300304}$. | \frac{200200201}{200200203}>\frac{300300301}{300300304} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,523 |
35. It is said that when asked how many students he had, the ancient Greek mathematician Pythagoras answered: “Half of my students study mathematics, a quarter study nature, a seventh part spends time in silent contemplation, and the remaining part consists of 3 maidens.” How many students did Pythagoras have? | 35. 36) $\frac{1}{2}+\frac{1}{4}+\frac{1}{7}=\frac{25}{28}$; 2) $1-\frac{25}{28}=\frac{3}{28}$; 3) $3: \frac{3}{28}=28$. | 28 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,526 |
37. Find the smallest number that, when divided by $\frac{35}{66}$, $\frac{28}{165}$, and $\frac{25}{231}$, results in natural numbers. | 37. The numerator of the fraction of the desired number should be the least common multiple of the numbers 35, 28, and 25, and the denominator should be the greatest common divisor of the numbers 66, 165, and 231. The first of these numbers is $4 \cdot 5^{2} \cdot 7=700$, and the second is $3 \cdot 11=33$. Therefore, t... | \frac{700}{33} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,527 |
38. Find the largest number by which each of the fractions $\frac{154}{195}$, $\frac{385}{156}$, and $\frac{231}{130}$ can be divided to yield natural numbers. | 38. The numerator of the fraction of the desired number should be the greatest common divisor of the numerators of the given fractions, and the denominator should be the least common multiple of the denominators of these fractions. Since $154=2 \cdot 7 \cdot 11, 385=5 \cdot 7 \cdot 11$ and $231=3 \cdot 7 \cdot 11$, the... | \frac{77}{780} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,528 |
39. Yura borrowed a book from Lena for 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 Yura manage to finish the book in three days? | 39. On the second day, Yura read $\frac{1}{2} \cdot \frac{1}{3}=\frac{1}{6}$ of the book, so in 2 days $\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$. On the third day, he read $\frac{2}{3}: 2=\frac{1}{3}$. This is exactly what he needed to read, as $1-\frac{2}{3}=\frac{1}{3}$. | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,529 |
40. Misha paid at the cafeteria cash register for 3 dishes, and Sasha - for 2 (all dishes at the same price). At the table, Grisha joined them, and the three of them ate 5 dishes. During the settlement, it turned out that Grisha should pay his friends 50 k. How much of this amount should he give to Misha and how much t... | 40. Grisha paid $\frac{1}{3}$ of the cost of all dishes, which is 50 k. Therefore, the cost of all dishes is $50 \cdot 3=150$ (k.), and one dish costs 30 k. ($150: 5=30$). Misha paid $30 \cdot 3=90$ (k.), and Sasha - 60 k. Therefore, Grisha should give Misha 40 k. $(90-50=40)$, and Sasha 10 k. $(60-50=10)$. | 40 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,530 |
41. The road from home to school takes Seryozha 30 minutes. Once on the way, he remembered that he had forgotten a pen at home. Seryozha knew that if he continued on to school at the same speed, he would arrive there 9 minutes before the bell, but if he returned home for the pen, he would, walking at the same speed, be... | 41. Let the distance from home to school be $a$, then the speed is $\frac{a}{30}$. To return home and come back to the place from which he returned, Sergei had to spend 20 minutes $(9+11=20)$. In this time, he would have traveled the distance $\frac{a}{30} \cdot 20=\frac{2}{3} a$. Therefore, Sergei walked from home to ... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,531 |
42. How to cut off half a meter from a piece of fabric that is $\frac{2}{3}$ meters, if you have nothing to measure with?
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。 | 42. Since $\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$ and $\frac{2}{3}: 4=\frac{1}{6}$, the piece should be folded into four and a quarter should be cut off. The remaining part will be half a meter. | Logic and Puzzles | other | Yes | Yes | olympiads | false | 39,532 | |
43. Without using any unit of length or other auxiliary means, how can you cut exactly half a meter from a piece of fabric that is $\frac{8}{15}$ m? | 43. Fold the piece in half (we get $\frac{4}{15}$ m), fold the resulting piece in half again (we get $\frac{2}{15}$ m), then fold it in half once more (we get $\frac{1}{15}$ m), and finally, fold the resulting piece in half again (we get $\frac{1}{30}$ m). If we cut off this part from the piece, i.e., $\frac{8}{15}-\fr... | \frac{1}{2} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,533 |
44. The fabric shrinks by $\frac{1}{18}$ in length and by $\frac{1}{14}$ in width during washing. What length of fabric should be taken to have $221 \mathrm{~m}^{2}$ after washing, if the width before washing was 875 mm? | 44. The width of the material after washing $0.875 \cdot \frac{13}{14}=\frac{7}{8} \cdot \frac{13}{14}=\frac{13}{16}$ (m), length $-221: \frac{13}{16}=272$ (m), which is $\frac{17}{18}$ of the original. Therefore, we need to take $272: \frac{17}{18}=288(\mathrm{M})$. | 288 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,534 |
45. Calculate in the most rational way:
$$
x=\frac{225+375 \cdot 138}{375 \cdot 139-150}
$$ | 45. Since $375 \cdot 139-150=375 \cdot 138+375-150=375 \cdot 138+225$, then $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,535 |
46. Calculate in the most rational way:
$$
333\left(\frac{71}{111111}+\frac{573}{222222}-\frac{2}{7 \cdot 37 \cdot 3}\right)
$$ | 46. $\frac{333 \cdot 71}{111 \cdot 1001}+\frac{333 \cdot 573}{222 \cdot 1001}-\frac{2 \cdot 333}{7 \cdot 111}=\frac{3 \cdot 71}{1001}+\frac{3 \cdot 573}{2 \cdot 1001}-\frac{2 \cdot 3}{7}=\frac{3 \cdot 71-3 \cdot 286}{1001}+$ $+\frac{3 \cdot 573}{2 \cdot 1001}=\frac{3 \cdot 573}{2 \cdot 1001}-\frac{3 \cdot 215}{1001}=\f... | \frac{3}{14} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,536 |
47. In one vessel, there are $a$ 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 into the second, and s... | 47. After pouring from the second vessel into the first, the first will contain $\frac{1}{2} a+\frac{1}{3} \cdot \frac{1}{2} a=\frac{4}{6} a=\frac{2}{3} a$, and the second will contain $\frac{1}{2} a-\frac{1}{3} \cdot \frac{1}{2} a=\frac{2}{6} a=\frac{1}{3} a$. Reasoning this way, we get the table
| pouring | 1 | 2 | ... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,537 |
48. On the sheet, several non-zero numbers are written, each of which is equal to half the sum of the others. How many numbers are written? | 48. Let there be $n$ numbers, then
$$
\begin{aligned}
a_{1} & =\frac{a_{2}+a_{3}+\ldots a_{n}}{2} ; \\
a_{2} & =\frac{a_{1}+a_{3}+\ldots+a_{n}}{2} ; \\
\ldots & \ldots \ldots \ldots+a_{n-1} \\
a_{n} & =\frac{a_{1}+a_{2}+\ldots+a_{n}}{2} ; \\
2\left(a_{1}\right. & \left.+a_{2}+\ldots+a_{n}\right)=(n-1) \cdot\left(a_{1}... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,538 |
49. The sum of several numbers is 10. Can the sum of their squares be less than 0.2? | 49. It is possible. For example, if we take 0.01 as an addend 1000 times, we get 10, the sum of the squares of these numbers is 0.1, i.e.
$$
\underbrace{0.01^{2}+0.01^{2}+\ldots+0.01^{2}}_{1000 \text { times }}=0.0001 \cdot 1000=0.1<0.2
$$ | 0.1<0.2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 39,539 |
50. In the numerator of the fraction, there is a number whose leftmost digit is 1, followed by 1989 sixes. In the denominator, there is a number whose rightmost digit is 4, and before it, to the left, there are 1989 sixes. Vasya simplified this fraction by crossing out all the sixes in the numerator and the denominator... | 50. If the numerator $166 \ldots 6$ is multiplied by 4, we get $66 \ldots 64$. Therefore, $\frac{166 \ldots 6}{66 \ldots 64}=\frac{1}{4}$. The answer is indeed correct with this set of digits, but this does not mean that such cancellation is valid. For example, if you replace the sixes with fives or any other digits, t... | \frac{1}{4} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,540 |
52. By adding the same number to the numerator and the denominator of the fraction $\frac{11}{41}$, Ira obtained $\frac{3}{8}$ after simplification, while Olya, by subtracting a certain number from the numerator of the fraction $\frac{37}{63}$ and adding it to the denominator, obtained $\frac{3}{17}$. What numbers did ... | 52. Let Ira add the number $x$, then $\frac{11+x}{41+x}=\frac{3}{8}$, from which $88+8 x=$ $=123+3 x, 5 x=35$ and $x=7$. Indeed $\frac{11+7}{41+7}=\frac{18}{48}=\frac{3}{8}$. Let Olya add and subtract the number $y$, then $\frac{37-y}{63+y}=\frac{3}{17}$, from which $629-17 y=189+3 y$; $y=22$. Indeed $\frac{37-22}{63+2... | 7,22 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,542 |
53. By subtracting a certain number from the numerator of the fraction $\frac{537}{463}$ and adding it to the denominator, Tanya obtained $\frac{1}{9}$ after simplification. What number did she subtract from the numerator and add to the denominator? | 53. $\frac{537-x}{463+x}=\frac{1}{9}$, hence $x=437$. | 437 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,543 |
54. Continuing her research with fractions, Tanya added 4 to the numerator of a certain fraction and 10 to the denominator. After simplifying, to her surprise, she got the original fraction. She tried the same with another fraction, but the result did not repeat. What fraction did Tanya originally have? | 54. Let the original fraction be $\frac{a}{b}$, then $\frac{a+4}{b+10}=\frac{a}{b}$, from which $a b+4 b=$ $=a b+10 a, 2 b=5 a$ and $\frac{a}{b}=\frac{2}{5}$. Indeed $\frac{2+4}{5+10}=\frac{6}{15}=\frac{2}{5}$. For this case, such a fraction is unique. | \frac{2}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,544 |
55. When Tanya introduced Svetlana to her problem (see problem 54), Svetlana decided to find another interesting property of the fraction $\frac{2}{5}$. She discovered that if 2 is added to the numerator of this fraction, and the denominator is multiplied by 2, then after simplification, the original fraction is obtain... | 55. No, not correct, there are as many such fractions as you like. Let's determine their form. Let the original fraction be $\frac{a}{b}$, then $\frac{a+2}{2 b}=\frac{a}{b}$, from which $a=2$. With the numerator equal to 2 and any odd denominator, the fraction will be irreducible. Therefore, the sought fractions have t... | \frac{2}{2n+1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,545 |
56. Generalizing problem 55, the members of the math club established that to the numerator of an irreducible fraction, one can add not only the number 2, multiplying the denominator by the same number at the same time. They found the general form of such a number. Find and write it down. | 56. Let the desired number be $x$, then $\frac{a+x}{b \cdot x}=\frac{a}{b}$, from which $a+x=a x$ and $x=\frac{a}{a-1}$. This number will be an integer only when $a=2$, and then we have the case considered in problem 55. If, for example, we have $a=5$, then $x=\frac{5}{4}$, and then
$$
\frac{5+\frac{5}{4}}{b \cdot \fr... | \frac{}{-1} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,546 |
57. Is the fraction that complements a given irreducible fraction to 1 also reducible? Will the pattern you have noticed always be true? Prove it. | 57. Let the fraction $\frac{a}{b}$ be irreducible. The fraction that complements the given one to 1 is $1-\frac{a}{b}=\frac{b-a}{b}$. Since by condition $a$ and $b$ have no common divisors other than 1, the numerator $b-a$ of the obtained fraction does not divide any divisor of the number $b$, which means the fraction ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,547 |
58. Find natural numbers $a$ and $b$, such that the number, which is the reciprocal of their difference, is three times greater than the number, which is the reciprocal of their product. | 58. Let $a>b$. By the condition $\frac{1}{a-b}=3 \cdot \frac{1}{a b}$, hence $a b=3 a-3 b$ and $a=\frac{3 b}{3-b}$. The number $a$ will be a natural number if $b=2$. In this case, $a=6$. Therefore, $a=6, b=2$. | =6,b=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,548 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.