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61. One hundred students, each with a number on their T-shirt, stood in a circle in the following sequence of their numbers: $1,2,3, \ldots, 100$. On command, they started to leave: the one with number 1 on the T-shirt stayed, the one with number 2 left, the one with number 3 stayed, the one with number 4 left, and so on, every other one around the circle until only one student remained. What number is on the T-shirt of this student?
|
61. When those with even numbers leave, 50 students with odd numbers remain: $-1,3,5, \ldots, 99$. Next, those with numbers $3,7,11, \ldots$ leave, after which 25 students remain. The last remaining number is 97, since $1+4 \cdot(25-1)=1+96=97$, i.e., the remaining numbers are $1,5,9,13, \ldots, 97$. Next, numbers $5,13, \ldots$ leave. In this process, 12 students leave, the last of whom has the number 93, since $5+8 \cdot(12-1)=93$, and 13 remain. Their numbers are:
$$
1,9,17,25,33,41,49,57,65,73,81,89,97
$$
Since the last one to leave had the number 93, number 97 remains, and then $1,17,33,49,65,81,97$ leave. After this, $25,57,89$ leave, followed by the number 41 and, finally, 9. The number 73 remains.
|
73
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
64. First, two numbers 2 and 3 are written, then they are multiplied and their product, the number 6, is recorded, then the product of the two last single-digit numbers $3 \cdot 6=18$, then $1 \cdot 8=8$, then $8 \cdot 8=64$ and so on. As a result, we get the following sequence of digits:
## $2361886424 \ldots$
Which digits cannot appear in such a sequence? Which digit stands at the thousandth place?
|
64. As a result of analyzing the sequence of digits $236188642483261224832612248 \ldots$, we notice that starting from the tenth digit, the group of digits 48326122 repeats (as in the previous problem, let's call it the period). Before the first period, there are 9 digits, and the period itself consists of 8 digits. Among these digits, there are no $5, 7, 9$, and 0. Therefore, these digits will be absent in the sequence, no matter how long it is extended. To find the digit in the thousandth place, since there are 9 digits before the period, there will be 991 digits in the first and subsequent periods up to the desired digit $(1000-9=991)$. Since the period consists of 8 digits, there will be 123 complete periods plus 7 more digits of the 124th period $(991=8 \cdot 123+7)$. The seventh digit of the period is 2. Therefore, the digit in the thousandth place in the sequence is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
65. How many terms of the sum
$$
1+2+3+\ldots
$$
are needed to obtain a three-digit number, all of whose digits are the same?
20
## Search for solutions
|
65. Let's calculate the sum $1+2+3+\ldots+n$. We can write it in another way, namely $n+(n-1)+\ldots+3+2+1$. By adding the corresponding terms of these sums (the first with the first, the second with the second, etc.), we get $n+1, 2+(n-1)=n+1$, and so on. Therefore, the doubled sum $1+2+3+\ldots+n$ is $(n+1)+(n+1)+\ldots+(n+1)=(n+1) n$, which means
$$
1+2+3+\ldots+n=\frac{(n+1) n}{2}
$$
According to the problem, $\frac{(n+1) n}{2}=\overline{a a a}$. Since $\overline{a a a}=111 \cdot a$ and $111=37 \cdot 3$, then $(n+1) n$ must be divisible by 37. This means either $n=37$, or $n+1=37$. If $n=37$, then $n+1=38$, and thus $\frac{38 \cdot 37}{2}=19 \cdot 37=703$, which does not satisfy the condition. If $n+1=37$, then $n=36$, and thus $\frac{37 \cdot 36}{2}=37 \cdot 18=666 ; 1+2+3+\ldots+36=666$. Therefore, $n=36$, i.e., there are 36 terms.
## Square and Cube of a Number
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Establish the rule according to which the table is constructed, and fill in the missing numbers:
| 9 | 81 | 2 |
| :---: | :---: | :---: |
| 16 | 256 | 2 |
| 11 | 11 | |
| 6 | 216 | 3 |
| 5 | | 3 |
|
3. Since $9^{2}=81,16^{2}=256,6^{3}=216$, the numbers in the middle column are powers, the left column contains their bases, and the numbers in the right column are the exponents. Therefore, the number 1 is missing in the right column, and in the middle column, $125\left(5^{3}=125\right)$.
|
125
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Tourists from one of the groups bought various souvenirs, with each of them taking a set of souvenirs of the same cost and as many as the cost of one souvenir in rubles. All tourists paid with ten-ruble bills (one or several), and each received change that did not match the change received by any of the other tourists. What change could each tourist have received, and what is the maximum number of tourists that could have been in the group?
|
5. The amount in rubles that each tourist paid for a set of souvenirs is expressed by a number of the form $a^{2}$, where $a$ is the number of souvenirs or the price of one souvenir in rubles. Since the squares of natural numbers can only end in the digits $1,4,5,6,9$, there can only be 5 different changes, namely 9 p., 6 p., 5 p., 4 p., and 1 p. The maximum possible number of tourists is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Vitya found the smallest possible natural number, which when multiplied by 2 gives a perfect square, and when multiplied by 3 gives a perfect cube. What number did Vitya find?
|
6. The desired number contains the factors $2^{3}=8$ and $3^{2}=9$. Thus, the smallest number is $8 \cdot 9=72$. Then $72 \cdot 2=144=12^{2}$ and $72 \cdot 3=216=6^{3}$. The desired number is 72.
|
72
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. The total corn harvest in centners collected from a certain plot of land is expressed by a four-digit number using the digits 0, 2, 3, and 5. When the average yield per hectare was calculated, it turned out to be the same number of centners as the area of the plot in hectares. Determine the total corn harvest.
|
13. Since the total harvest is equal to the product of the yield and the area of the plot, the four-digit number representing the total harvest is the square of some two-digit number. This square cannot end in the digits 0, 2, or 3. Therefore, the units digit of this four-digit number can only be 5. The tens digit can only be 2, then the hundreds digit is 0, and the thousands digit is 3. Thus, the total harvest is 3025 tons, and the yield per hectare is 55 tons, since $55^{2}=3025$.
|
3025
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15. Find a three-digit number that is equal to the cube of the sum of its digits.
|
15. Since $4^{3}=64$ and $10^{3}=1000$, the sum of the digits can be $5,6,7,8$ or 9. From these five numbers, we will select by trial those that satisfy the condition of the problem:
$$
5^{3}=125,6^{3}=216,7^{3}=343,8^{3}=512,9^{3}=729
$$
The condition is satisfied by $512=(5+1+2)^{3}$.
|
512
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. The pedestrian named to the traffic inspector the number of the car whose driver had grossly violated the traffic rules. This number is expressed as a four-digit number, the unit digit of which is the same as the tens digit, and the hundreds digit is the same as the thousands digit. Moreover, this number is a perfect square. What is this number?
|
16. Since the square of a number can end in the digits $0,1,4,9,5$ or 6, the last two digits of the desired number can be $00,11,44,99,55$ or 66. This four-digit number is the square of a two-digit number, the digits of which are the same. Such a two-digit number cannot be 11 or 22, since their squares are three-digit numbers. It is not difficult to find among the seven other numbers that the specified requirements are met by $88\left(88^{2}=7744\right)$.
|
7744
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
17. The amount in rubles, transferred by one of the enterprises to the World Fund, is expressed as the largest even five-digit number such that the first three digits form a number that is a perfect square, and the last three digits form a number that is a perfect cube. What amount has the enterprise transferred to the World Fund?
|
17. The last three digits of the desired number represent a number that is the cube of an even number. There are two possible cases: $8^{3}=512$ and $6^{3}=216$. However, since there is no square of a natural number that ends in 2, the second case is ruled out. The largest three-digit number that is a perfect square and ends in 5 is 625. Therefore, the desired number is 62512.
|
62512
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
23. Find a three-digit number, the square of which is a six-digit number such that each subsequent digit, counting from left to right, is greater than the previous one.
|
23. Since a square can end with the digits $0,1,4,5,6$ or 9, and each subsequent digit of a six-digit number is greater than the previous one, a six-digit number that is the square of a three-digit number can end with either the digit 6 or 9. If it ends with 6, it would be written as 123456. But this number is not a perfect square. Therefore, it must end with the digit 9, which means the sought three-digit number ends with the digit 3 or 7. The leftmost digit of the six-digit number cannot be 4, as the number 456789 is not a perfect square. Therefore, it is less than 4. Let's consider $1 * * * * 9$, then the three-digit number is either $4 * 7$ or $3 * 7$. But 407, $417,427,437,447$ do not satisfy the condition, and $457^{2}=208849$. (We get a number with the leftmost digit not 1, but 2.) We will look for numbers of the form $3 * 7$. Checking shows that such a number is $367\left(367^{2}=134689\right)$.
|
367
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. How many numbers from 1 to 100 are there, each of which is divisible by 3 but does not contain the digit 3 in its representation?
|
2. There are a total of 33 numbers up to 100 that are divisible by 3 (100: 3=33). Among these numbers, those that have the digit 3 in their representation are: $3,30,33,36,39,63,93$, i.e., 7 numbers. Therefore, the numbers divisible by 3 but not containing the digit 3 are 26 $(33-7=26)$.
|
26
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. If 7 is subtracted from the thought three-digit number, the resulting difference will be divisible by 7; if 8 is subtracted, the resulting difference will be divisible by 8; if 9 is subtracted, the resulting difference will be divisible by 9. What is the smallest possible number thought of?
|
5. The intended number is divisible by $7, 8, 9$. The smallest number divisible by 7, 8, and 9 is $7 \cdot 8 \cdot 9=504$. This number meets the requirement of the problem.
|
504
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. To the question: "How many two-digit numbers are there such that the sum of their digits is 9?" - Stepa Verkhoglyadkin started listing all two-digit numbers in a row, selecting the ones he needed. Show him a shorter way to solve the problem.
|
6. For two-digit numbers, except for 99, the statement is true: “If a number is divisible by 9, then the sum of its digits is 9, and vice versa.” Up to 99 inclusive, there are 11 numbers divisible by 9 (99: 9=11). However, among them, two numbers (9 and 99) are not considered in this problem, so the number of numbers satisfying the condition of the problem is 9 (11-2=9).
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Find the smallest six-digit number that is divisible by 3, 7, and 13 without a remainder.
|
7. The desired number must be divisible by $3 \cdot 7 \cdot 13=273$, and the smallest six-digit number $100000=366 \cdot 273+82$. If we add 191 to it, we get $100191=367 \cdot 273$. This is the desired number.
|
100191
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. For the school library, 17 identical books were purchased. How much do they cost if for 9 such books they paid more than 11 p. 30 k., but less than 11 p. 40 k.?
|
13. It is known that 9 books were paid $\overline{113 x}$ k. Since this number must be divisible by 9, then $1+1+3+x=9$, from which $x=4$. Therefore, one book costs 126 k. $=1$ r. 26 k., and 17 books $-126 \cdot 17=2142$ k. $=21$ r. 42 k.
|
21
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. How many natural numbers less than 100 are there that: a) are divisible by 2 but not by 3; b) are divisible by 2 or by 3; c) are not divisible by 2 or by 3?
|
14. a) There are $49(98: 2=49)$ even numbers from 1 to 99. Among them, there are numbers that are divisible by 3 (such numbers are divisible by 6). There are a total of 96:6 $=16$. Therefore, the numbers divisible by 2 but not by $3,49-16=33$.
b) There are 49 numbers divisible by 2, and 33 numbers divisible by 3 (99: 3=33), and 16 numbers divisible by 6. Therefore, the numbers divisible by 2 or 3 are $49+33-16=66$.
c) The numbers not divisible by 2 or 3 are $99-66=33$.
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15. The distance in kilometers that the airplane has flown is expressed as a four-digit number divisible by 45, and the two middle digits are 39. Find this distance if it does not exceed 5000 km.
|
15. The desired number is divisible by 5, so it ends in the digit 5 or 0. It is also divisible by 9, so the sum of its digits is divisible by 9. Such numbers are 6390 and 1395. Since $6390>5000$, and $1395<5000$, the desired distance is 1395 km.
|
1395
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
16. A store received fewer than 600 but more than 500 plates. When they started arranging them in tens, they were short of three plates to make a complete number of tens, and when they started arranging them in dozens (12 plates), there were 7 plates left. How many plates were there?
|
16. If there were three plates short of a full number of tens, this means that, as with counting by dozens, there were 7 plates left. This means that the number of plates minus seven is divisible by both 10 and 12, i.e., by 60. Among the numbers less than 600 and greater than 500, only the number 540 is divisible by 60, so the number of plates was $540+7=547$
|
547
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20. Find the smallest natural number divisible by 72, in the representation of which all digits from 1 to 9 appear.
|
20. The desired number must be divisible by 72, which means it must be divisible by 9 and 8; $1+2+\ldots+9=45$, which is divisible by 9. For the number to be divisible by 8, it must end with three digits that form a three-digit number divisible by 8. Considering all this, the digits should be arranged so that the smallest ones are to the left. Such a number is 123457968.
|
123457968
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21. Find the smallest natural number that is a multiple of 36 and in whose representation all 10 digits appear exactly once.
|
21. Since the desired number is divisible by 36, it must be divisible by 4 and 9. In this number, the last two digits should represent the largest two-digit number divisible by 4. This number is 96. The remaining digits should be written from the digit 9 to the left in descending order, but since the leftmost digit cannot be 0, the leftmost digit is 1. Since the sum of all single-digit numbers is 45, the desired number is divisible by 9. The desired number is 1023457896.
|
1023457896
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
24. Mowgli asked his monkey friends to bring him some nuts. The monkeys gathered an equal number of nuts and carried them to Mowgli. But on the way, they quarreled, and each monkey threw a nut at every other monkey. As a result, Mowgli received only 35 nuts. How many nuts did the monkeys gather, if it is known that each of them brought more than one nut?
|
24. Since the monkeys collected nuts equally and threw them away equally, they brought them equally. The number 35 is divisible by 5, so we have $35=5 \cdot 7$. There can be 2 cases:
1) There were 5 monkeys, each brought 7 nuts, and each threw away 4 nuts, so each collected $7+4=11$.
2) There were 7 monkeys, each brought 5 nuts, and each threw away 6 nuts, so each collected $5+6=11$.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
29. When from the numbers from 1 to 333 Tanya excluded all numbers divisible by 3 but not divisible by 7, and all numbers divisible by 7 but not divisible by 3, she ended up with 215 numbers. Did she solve the problem correctly?
|
29. There are 333 numbers in total, of which 111 numbers are divisible by 3 ( $333: 3=111$ ); among these numbers, 15 are divisible by 7 ( $333:(3 \cdot 7)=333: 21=15$ (remainder 18)). Therefore, the numbers divisible by 3 but not by 7 total 96 numbers ( $111-15=96$ ). There are 47 numbers in total that are divisible by 7 ( $333: 7=47$ (remainder 4)). Among these 47 numbers, 15 are divisible by 3, so the numbers divisible by 7 but not by 3 total 32 numbers ( $47-15=32$ ). Thus, 128 numbers need to be crossed out (excluded) ( $96+32=128$ ), leaving $333-128=205$ numbers. By obtaining 215 numbers, Tanya made a mistake.
|
205
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
31. How many natural numbers not exceeding 500 and not divisible by 2, 3, or 5 are there?
|
31. Let's calculate the number of numbers divisible by 2, or by 3, or by 5. 250 numbers are divisible by 2 (500: 2 = 250). 166 numbers are divisible by 3 (500: 3 = 166 (remainder 2)). At the same time, numbers divisible by both 2 and 3, i.e., by 6, are counted twice. There are 83 such numbers (500: 6 = 83 (remainder 2)). Therefore, the number of numbers divisible by 2 or 3 is 250 + 166 - 83 = 333. 100 numbers are divisible by 5 (500: 5 = 100). However, numbers divisible by both 5 and 2, i.e., by 10, are included here. There are 50 such numbers (300: 10 = 50). Since they are already included in the numbers divisible by 2, they need to be excluded here: 100 - 50 = 50. Among these numbers, those divisible by both 5 and 3, i.e., by 15, are included. There are 33 such numbers. They are already included in the numbers divisible by 3, so they need to be excluded here: 50 - 33 = 17. However, by excluding numbers divisible by both 2 and 5, as well as by 3 and 5, we have excluded numbers divisible by 2, 3, and 5, i.e., by 30, twice. There are 16 such numbers. Thus, the total number of numbers up to 500 that are divisible by 2, or by 3, or by 5, is 333 + 17 + 16 = 366, and therefore, the number of numbers not divisible by 2, 3, or 5 is 500 - 366 = 134.
|
134
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 number, move 63 to the beginning of the number and you get 63999999.
## Prime and Composite Numbers.
|
63999999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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
|
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 numbers can be obtained
|
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) $1+2+3-4+5+6+7+8-9=19$
5) $1-2+3+4+5+6+7+8-9=23$
6) $1+2+3+4+5+6+7-8+9=29$;
7) $1+2+3+4+5+6-7+8+9=31$
8) $1+2+3-4+5+6+7+8+9=37$
9) $1-2+3+4+5+6+7+8+9=41$
10) $2-1+3+4+5+6+7+8+9=43$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 $p+14=17$. If $p=3 k+1$, then $p+14=3 k+15=3(k+5)$-composite. If $p=3 k+2$, then $p+10=3 k+12=3(k+4)$-composite. We have the solution: $p=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
If $p=3 k$, then only when $k=1$ is the number $p$ prime. This number is $p=3$. Then $2 p+1=7$ and $4 p+1=13$ are also prime numbers. Therefore, when $p=3$, all three numbers are prime.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 numbers $p, p_{1}$, and $p_{2}$ are consecutive odd numbers, and the only such numbers are $3, 5, 7; p=5; 5=3+2$ and $5=7-2$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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
|
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
|
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 so on. Thus, we have a repeating group of digits $8,4,2$ and 6, which are the last digits of the powers of 8. Since 1980 is divisible by 4, $8^{1980}$ ends in the digit 6, and therefore, the given difference ends in the digit 5. Hence, it is divisible by 5, but not by 10.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 more twos. Since $5 \cdot 2=10$, then $(5 \cdot 2)^{12}=10^{12}$, which means that the given product ends with twelve zeros.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 digits) is divisible by 11, then the number is divisible by 11.
|
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 divisible by 9, and an alternating sum of digits divisible by 11. Since $1+2+3+4+5+6+7+8+9=45$ and $45: 9$, such a nine-digit number is indeed divisible by 9. The digits cannot be arranged in such a way that their alternating sum equals zero, as 45 is not divisible by 2. Therefore, the digits must be arranged so that their alternating sum equals 11, $45-11=34$, $34: 2=17$. One such number is 238471596. Therefore, the number 2384715960 is divisible by 1980 without a remainder.
|
2384715960
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 ( $4752862-4=$ $=4752858 ; 4752858=2 \cdot 3 \cdot 11 \cdot 23 \cdot 31 \cdot 101$ ). From this, it follows that the most plausible answer could be: the ship has 2 propellers, 3 funnels, 101 people on board, and it set sail on November 23, 1931. Checking shows that the captain being 27 years old does not satisfy the condition.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 total, not just in the rows?
Do your calculations with the right aim.
Note. Remember that there are fewer than 10 rows in the tadpole column.
|
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, there were 25 tadpoles (5 rows).
|
25
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 satisfies the condition of the problem. Therefore, 441 books were placed on each shelf.
|
441
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 it be?
|
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
|
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 numbers is 3.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 $42 \cdot \frac{1}{7}=6 ; 42 \cdot \frac{1}{3}=14 ; 42 \cdot \frac{1}{2}=21$; $42-(6+21+14)=1$. Therefore, 1 student of this class lives on Shipbuilding Street.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 could have been?
|
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
|
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 number of cuts. For the numbers 5 and 7, the least common multiple is 35. Therefore, taking 7 six-meter logs and making 35 cuts, we will get $6 \cdot 7=42$ pieces, while from five eight-meter logs with the same number of cuts, we will get $8 \cdot 5=40$ pieces. Therefore, it is more profitable to take six-meter logs.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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
|
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
|
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 is it to travel by bus than to walk?
|
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
|
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
He ate half of what was left
And plus half a pastry.
With an empty wallet—alas!
He entered the house sadly...
Tell me, how many pastries were there?
|
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. So, $1+0.5=1.5$ is half of the cookies he had at that point, and there were a total of $1.5 \cdot 2=3$ cookies. The first time he ate half the cookies and another half cookie, so 3.5 is half of the cookies he had. This means there were initially $3.5 \cdot 2=7$ cookies.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 and arrived at the tourist base 15 minutes earlier than the scheduled time. At what speed were the tourists walking before they met the bus, if the bus's speed was 60 km/h?
|
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 foot. They covered this distance in 1.5 hours $(5-3.25=$ $=1.75 ; 1.75-0.25=1.5)$. Therefore, their speed was 5 km/h $(7.5: 1.5=5)$.
150
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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
|
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
|
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, which is 15 kg. Therefore, $\frac{3}{4} x=15$, from which $x=\frac{15 \cdot 4}{3}=20$ (kg).
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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 of the cut, we can find the length of the path $2000: 4=500$ (m).
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 number of parts that can be obtained from 20 identical blanks, and what will be the waste in this case?
|
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 material results in swarf equivalent to $6: 3=2$ blanks. Using this material, 2 blanks are obtained, and 2 parts are machined from them, leaving swarf equivalent to $2: 3=\frac{2}{3}$ of a blank. Thus, in total, there is material for $\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$ blanks. Using this material, one more blank is obtained, from which one part is machined. As a result, 29 parts are obtained, and there is swarf equivalent to $\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$ of a blank.
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
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 to Sasha, if each of them should contribute the same amount?
|
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
|
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
|
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
|
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
$$
\begin{aligned}
a_{1} & =\frac{a_{2}+a_{3}+\ldots a_{n}}{2} ; \\
a_{2} & =\frac{a_{1}+a_{3}+\ldots+a_{n}}{2} ; \\
\ldots & \ldots \ldots \ldots+a_{n-1} \\
a_{n} & =\frac{a_{1}+a_{2}+\ldots+a_{n}}{2} ; \\
2\left(a_{1}\right. & \left.+a_{2}+\ldots+a_{n}\right)=(n-1) \cdot\left(a_{1}+a_{2}+\ldots+a_{n}\right), \text { hence } 2=n-1 ; n=3 .
\end{aligned}
$$
Thus, there were 3 numbers written.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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
|
59. Find the non-negative integer values of $n$ for which $\frac{30 n+2}{12 n+1}$ is an integer.
|
59. $\frac{30 n+2}{12 n+1}=\frac{(24 n+2)+6 n}{12 n+1}=2+\frac{6 n}{12 n+1}; \quad \frac{6 n}{12 n+1}$
will be an integer only when $n=0$, for other values of $n$ it is a proper fraction. Therefore, only when $n=0$ is $\frac{30 n+2}{12 n+1}$ an integer (2).
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
60. A barrel contains exactly 30 liters of linseed oil. For three construction brigades, 3 barrels were filled from it, each of which holds a whole number of liters, and the capacity of the first is $\frac{2}{3}$ of the capacity of the second or $\frac{3}{5}$ of the capacity of the third. How many liters of linseed oil are left in the barrel?
|
60. Let's express the capacities of the second and third buckets in terms of the capacity of the first one and schematically write it as: $\mathrm{II}=\frac{3}{2} \mathrm{I} ; \mathrm{III}=\frac{5}{3} \mathrm{I}$. Since each bucket contains a whole number of liters and the smallest number divisible by 2 and 3 is 6, the capacity of the first bucket is 6 liters, the second is 9 liters, and the third is 10 liters. Therefore, 5 liters remain in the barrel $(30-(6+9+10)=5)$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
61. The number of books issued to readers from the library is $\frac{1}{16}$ of the number of books on the shelves. After transferring 2000 books from the library to the reading room, the number of books absent from the shelves became $\frac{1}{15}$ of the number of books remaining on the shelves. How many books does the library have?
|
61. Before transferring books to the reading room, the number of books present in the library was 16 times greater than the number of books issued, which means that the number of issued books (absent from the library) constitutes $\frac{1}{17}$ of the total number of books in the library. After issuing 2000 books (transferring to the reading room), the number of absent books became $\frac{1}{16}$ of the total number of books. Therefore, 2000 books constitute $\frac{1}{16}-\frac{1}{17}=\frac{1}{272}$ of the total number of books. Thus, the library has $2000 \cdot 272=544000$ books.
|
544000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
62. There were books on three shelves. On the bottom shelf, there were 2 times fewer books than on the other two, on the middle shelf - 3 times fewer than on the other two, and on the top shelf - 30 books. How many books are there in total on the three shelves?
|
62. Let the number of books on the lower shelf be 1, then the number of books on the two other shelves will be 2 units, and on the three shelves - 3 units. This means the number of books on the lower shelf is $\frac{1}{3}$ of all the books. Similarly, the number of books on the middle shelf is $\frac{1}{4}$ of all the books. Therefore, the number of books on the upper shelf will be $1-\left(\frac{1}{3}+\frac{1}{4}\right)=\frac{5}{12}$, or 30 books. Thus, the total number of books is $30: \frac{5}{12}=72$.
|
72
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
66. After the tourists have walked 1 km and half of the remaining distance, they still have to walk a third of the entire distance and 1 km. What is the total distance?
|
66. After the tourists have walked 1 km, half of the remaining distance is one third of the entire distance plus 1 km, which means the entire remaining distance is $\frac{2}{3}$ of the entire distance plus 2 km. Thus, the entire distance is $\frac{2}{3}$ of the entire distance plus $2+1=3$ (km), which means 3 km is $\frac{1}{3}$ of the entire distance $\left(1-\frac{2}{3}=\frac{1}{3}\right)$. Therefore, the entire distance is 9 km $(3 \cdot 3=9)$.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
70. A certain amount was spent on strawberries at 2 r. 40 k. per 1 kg of one variety and the same amount on another variety at 1 r. 60 k. Find the average price of 1 kg of the strawberries purchased.
|
70. Let the amount spent on each type of strawberry be $a$ r. Then the first type purchased is $a: 2.4=\frac{a}{2.4}$ (kg), and the second type is $-\frac{a}{1.6}$ kg. In total, $\left(\frac{a}{2.4}+\frac{a}{1.6}\right)$ kg were purchased for a total of $2a$ r. The average price per 1 kg is
$$
2 a:\left(\frac{a}{2.4}+\frac{a}{1.6}\right)=\frac{2 \cdot 2.4 \cdot 1.6}{2.4+1.6}=1.92, \text{ i.e., } 1 \text{ r. } 92 \text{ k. }
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
71. The bus traveled three equal-length segments of the route: the first at an average speed of 50 km/h, the second at 30 km/h, and the third at 70 km/h. What is the average speed of the bus over the entire route?
|
71. Let the length of one segment be $a$ km, then the bus spent $\left(\frac{a}{50}+\frac{a}{30}+\frac{a}{70}\right)$ hours to travel the entire path of length $3 a$ km. Therefore, the average speed of movement is
$$
3 a:\left(\frac{a}{50}+\frac{a}{30}+\frac{a}{70}\right)=\frac{3 \cdot 1050}{21+35+15} \approx 44(\text { km } / \mathrm{h}) .
$$
|
44
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
76. Andrei, Boris, and Sergey were competing in a 100-meter race and started simultaneously. When Andrei finished, Boris was 10 meters behind him, and when Boris finished, Sergey was 10.62 meters behind him. How far was Sergey from Andrei when Andrei finished? (It is assumed that each runs at a constant speed.)
|
76. Let Andrey run $a$ m per second, Boris $b$ m, and Sergey $c$ m. Andrey ran the 100 m distance in $\frac{100}{a}$ seconds, and during this time, Boris ran 90 m, i.e., he spent $\frac{90}{b}$ seconds. When Boris ran 100 m in $\frac{100}{b}$ seconds, Sergey ran 90 m during this time, so $\frac{100}{b}=\frac{90}{c}$. Therefore, we have $\frac{100}{a}=\frac{90}{b}$ and $\frac{100}{b}=\frac{90}{c}$, from which $\frac{90}{b}=\frac{81}{c}$, so $\frac{100}{a}=\frac{81}{c}$. The last 160
means that if Andrey finished, then Sergey was 19 m behind him $(100-81=19)$
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
81. At exactly 24:00, the battery received an order to begin artillery preparation at the earliest time that will pass until the hour and minute hands overlap again. At what time should the artillery preparation begin?
|
81. In 1 hour, the minute hand makes a full revolution and will point to 12, while the hour hand will have traveled $\frac{1}{12}$ of the circumference and will point to 1. This means that after 1 hour, the minute hand, following the hour hand, will be at a distance equal to $\frac{1}{12}$ of the circumference from it. Since in 1 hour the minute hand will travel $\frac{11}{12}$ of the circumference more than the hour hand ($1-\frac{1}{12}=\frac{11}{12}$), it will "catch up" with the hour hand after 1 hour plus an additional $\frac{1}{11}$ hour $\left(\frac{1}{12} : \frac{11}{12} = \frac{1}{11}\right)$. Therefore, the hour and minute hands will coincide again, starting from 12, after $1 \frac{1}{11}$ hours, i.e., at 1 hour 5 minutes 27 seconds. This is the time (to the nearest second) when the artillery preparation should begin.
|
1
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
82. Andrey and Grandpa Grisha went mushroom picking sometime between six and seven in the morning, at the moment when the clock hands were aligned. They returned home between twelve and one in the afternoon, at the moment when the clock hands were pointing in exactly opposite directions. How long did their mushroom "hunt" last?
|
82. In 1 hour, the minute hand makes a full revolution (its end describes a circle), while the hour hand travels $\frac{1}{12}$ of the circumference. Therefore, in one hour, the minute hand will travel more than the hour hand by $\frac{11}{12}$ of the circumference. By the time the minute hand travels from 12 to 6, i.e., $\frac{1}{2}$ of the circumference, the hour hand will travel $\frac{1}{24}$ of the circumference, i.e., at 6:30, the minute hand will lag behind the hour hand by $\frac{1}{24}$ of the circumference. It will catch up with the hour hand in $\frac{1}{22}$ of an hour $\left(\frac{1}{24} : \frac{11}{12} = \frac{1}{22}\right)$. The clock will show $6 \frac{6}{11}$ hours $\left(6 \frac{1}{2} + \frac{1}{22} = 6 \frac{6}{11}\right)$. Similar reasoning leads to the conclusion that when the hour hand is between 12 and 1, and the minute hand is directly opposite to it, the clock will show $12 \frac{6}{11}$ hours. Therefore, the mushroom "hunt" lasted 6 hours $\left(12 \frac{6}{11} - 6 \frac{6}{11} = 6\right)$.
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
85. To which number can the expression
$$
\frac{(a+b)(b+c)(a+c)}{a b c}
$$
be equal if the numbers $a, b$ and $c$ satisfy the condition of problem 84?
|
85. From the solution of the previous problem, we have $a+b=c n ; b+c=a n ; a+c=b n$. After subtracting, we get $a-c=n(c-a)$. The equality is possible when $a-c=0$, i.e., $a=c$. Similarly, $a=b$, so $a=b=c$, and from this, $\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}=2$. Therefore, $\frac{(a+b)(b+c)(a+c)}{a b c}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. On a horizontal coordinate line, several integers are marked as points, the sum of which is equal to 25. If each point is moved 5 unit segments to the left, then the sum of the numbers corresponding to these points will become equal to -35. How many numbers were marked on the line?
|
1. The sum of the numbers decreased by $25-(-35)=60$. Therefore, $60: 5=12$ numbers were marked.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Please think in silence,
Consider, a rare case:
A squirrel sat on a pine,
On the very middle branch.
Then jumped up by five,
Then descended by seven.
(You all should remember,
As in a class lesson.)
74
## Then the agile squirrel once again
Jumped up by four,
Then by nine more
And sat at the very top.
Sitting and looking from the height
At stumps, birches, and bushes.
And how many branches the pine has,
We should calculate together.
|
8. From the middle branch to the upper one, there will be $+5-7+4+9=11$ branches. Therefore, the total number of branches is $11 \cdot 2+1=23$.
|
23
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. In one line, 19 numbers were written. The sum of any three consecutive numbers is positive. Can the sum of all 19 numbers be negative?
|
9. Consider, for example, the sequence of numbers $\underbrace{-7,4,4, \ldots,-7,4,4,-7}_{19 \text { numbers }}$. The sum of any 3 consecutive numbers in this sequence is 1, i.e., positive, while the sum of all 19 numbers is -1, i.e., negative. Answer: It can. Find another similar example.
|
-1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. Based on the first three lines, establish the rule by which the number $c$ is obtained from the numbers $a$ and $b$, and fill in the missing numbers in the empty cells:
| $a$ | $b$ | $c$ |
| :---: | :---: | :---: |
| 5 | -11 | 6 |
| $-2.7$ | $-2.3$ | 5 |
| 32 | -18 | -14 |
| -17 | 5 | |
| 14 | -14 | |
|
12. Obviously, $c=-(a+b)$, so the missing numbers are $-(-17+5)=12$; $-(14-14)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
13. Find the sum:
$$
-100-99-98-\ldots-1+1+2+\ldots+101+102
$$
|
13. Since $-100-99-98-\ldots-1+1+2+\ldots+99+100=0$, and $101+$ $+102=203$, the desired sum is 203.
|
203
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. Find the sum:
$$
-101-99-97-\ldots+95+97
$$
|
14. Since $-97-95-\ldots+95+97=0$, and $-101-99=-200$, the desired sum is -200.
|
-200
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
15. Find the sum:
$$
1+2-3-4+5+6-7-8+\ldots+301+302
$$
|
15. Let's write the sum as follows: $1+(2-3-4+5)+(6-7-8+9)+\ldots+(298-$ $-299-300+301)+302$. In each parenthesis, we get 0, so the final result is $1+302=303$.
|
303
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
19. On the number line, points $A, B, C, D$ were marked. It is known that the first three points correspond to the numbers 5, 8, and -10. What number corresponds to point $D$, if when the direction of the number line is reversed, the sum of the four numbers corresponding to these points does not change?
|
19. Let the point $D$ correspond to the number $x$. The sum of the numbers before changing the direction of the line is $5+8-10+x$, and after changing the direction it is $-5-8+10-x$. According to the condition, $5+8-10+x=5-8+10-x$, from which $2x = -6$ and $x = -3$. The problem can be solved more simply without resorting to an equation. Since the sum did not change when the signs were reversed, it is equal to zero. And if we have the terms 8, 5, and -10, then the fourth term is -3, because in this case $8+5-10-3=0$. Therefore, the point $D$ corresponds to the number -3.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
20. The variable $a$ can take the values: $-32,-30,-28, \ldots, 2$, $4,6,8$, and the variable $b: -17,-15,-13, \ldots, 11,13$. How many different values can the expression $a+b$ take? Find the product of the largest and smallest values of the expression $a+b$.
76
|
20. Since the first variable can take 21 values, and the second can take 16, the expression $a+b$ can take a total of $21 \cdot 16=336$ different values. The maximum value of the expression $a+b$ is 21 $(8+13=21)$, and the minimum value is $-49(-32+(-17)=-49)$. Their product $(-49) \cdot 21=-1029$.
|
-1029
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
21. Simplify the expression
$-1-(-1-(-1-(-1-(\ldots$, if the expression contains:
a) 1989 ones; b) 1990 ones.
|
21. Observing
$$
\begin{aligned}
& -1-(-1)=-1+1=0 \\
& -1-(-1-(-1))=-1 ; \\
& -1-(-1-(-1-(-1)))=0 ; \\
& -1-(-1-(-1-(-1-(-1))))=-1
\end{aligned}
$$
we notice that with an even number of ones, the result is 0, and with an odd number of ones, the result is -1. Therefore, in case a) we have -1, and in case b) we have 0. [^0]
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
22. Given 5 numbers: $a_{1}=-1, a_{2}=0, a_{3}=1, a_{4}=2, a_{5}=3$. The sixth is the product of the first and the second, the seventh is the product of the second and the third, the eighth is the product of the third and the fourth, and so on. What is the last non-zero number in this sequence?
|
22. Let's construct a table:
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| -1 | 0 | 1 | 2 | 3 | 0 | 0 | 2 | 6 | 0 | 0 | 0 | 12 | 0 | 0 | 0 | 0 |
Starting from $a_{14}$, all numbers will be zeros, so the last non-zero number will be $a_{13}=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
23. Given 5 numbers: $a_{1}=1, a_{2}=-1, a_{3}=-1, a_{4}=1, a_{5}=-1$. The following numbers are determined as follows: $a_{6}=a_{1} \cdot a_{2}, a_{7}=a_{2} \cdot a_{3}, a_{8}=a_{3} \cdot a_{4}$ and so on. What is $a_{1988}$?
|
23. Let's construct a table:
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | -1 | -1 | 1 | -1 | -1 | 1 | -1 | -1 | 1 | -1 | -1 |
We notice that in the obtained sequence, the group of numbers $1, -1, -1$ repeats periodically. If there are 1988 numbers, then the number of complete periods will be $662 (1988: 3=$
This means that the number at the 1988th position will be -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A game participant was offered 30 questions. For each correct answer, they were awarded 7 points, and for an incorrect answer (or no answer), 12 points were deducted. How many correct answers did the participant give if they scored 77 points?
|
1. Let the number of correct answers be $x$, then the number of incorrect answers is $30-x$. We have $7 x-12 \cdot(30-x)=77 ; 19 x=437 ; x=23$.
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. If from the number of students in class $5 \mathrm{~A}$ we subtract 23 and multiply the resulting difference by 23, we get the same as if we subtracted 13 from this number and multiplied the difference by 13. How many students are there in class 5 A?
|
2. Let there be $x$ students in class 5A, then $(x-23) \cdot 23=(x-13) \cdot 13$, from which $10 x=360$ and $x=36$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The lion is older than the porcupine
By two and a half times.
According to the hoopoe's information
Three years ago
The lion was seven times older,
Than the porcupine.
Take all into account and weigh:
How old are they together? -
Allow me to ask you.
|
4. Let the hedgehog be $x$ years old, then the lion is $2.5 x$. Three years ago, the hedgehog was $x-3$, and the lion was $2.5 x-3$. According to the problem,
$7(x-3)=2.5 x-3$
from which $x=4$. Therefore, the hedgehog and the lion together are 14 years old.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. - I am two years older than the lion, said the wise owl.
- And I am half your age, the porcupine replied to the owl. The lion looked at him and said proudly, slightly wrinkling his nose: - I am four years older, Than you, honorable quill-nose. And how old are they all together? Check your answer twice.
|
5. Let the hedgehog be $x$ years old, then the owl is $2x$, and the lion is $(2x-2)$. We have the equation $(2x-2)-x=4$, from which $x=6$. Therefore, the hedgehog is 6 years old, the owl is 12, the lion is 10, and together they are 28.
|
28
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. The number of trees planted by students during the clean-up day is a three-digit number, in which the number of tens is 3 more than the number of hundreds, and the number of units is 4 less than the number of tens, and the half-sum of all digits is equal to the digit of the tens. How many trees did the students plant?
|
6. Let tens be $x$, then hundreds are $x-3$, and units are $x-4$. According to the problem,
$$
\frac{x+(x-3)+(x-4)}{2}=x
$$
from which $x=7$. Therefore, the desired number is 473.
|
473
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
9. Petya has as many sisters as brothers, while his sister Ira has twice as few sisters as brothers. How many boys and how many girls are there in this family?
|
9. If there are $x$ girls, then there are $x+1$ boys. Without Irina, there will be $x-1$ girls, then from the condition we have
$$
2(x-1)=x+1
$$
from which $x=3$. Therefore, there are 3 girls and 4 boys in the family.
|
3
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
10. Aunt Masha is three years younger than Sasha and his same-aged friend Pasha combined. How old is Sasha when Aunt Masha was as old as Pasha is now?
|
10. Let Sasha and Pasha be $a$ years old, then Aunt Masha is $(2 a-3)$ years old. Aunt Masha was $a$ years old $(a-3)$ years ago $(2 a-3-a=a-3)$. But so many years ago, Sasha was 3 years old $(a-(a-3)=3)$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
12. If to a certain three-digit number, first the digit 7 is appended on the left, and then on the right, the first of the resulting four-digit numbers will be 3555 more than the second. Find this three-digit number.
|
12. Let the desired three-digit number be $x$. If we prepend the digit 7, we get the number $7000+x$; if we append the digit 7, we get the number $10x+7$. According to the problem,
$$
(7000+x)-(10x+7)=3555
$$
from which $9x=3438$ and $x=382$. The problem can also be solved by trial and error.
|
382
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
14. In the workshop, five clocks were checked. It turned out that clock No. 3 is 3 minutes behind clock No. 2 per day. Clock No. 1 is 1 minute behind clock No. 2 per day, clock No. 4 is behind clock No. 3, and clock No. 5 is behind clock No. 1. If the readings of all five clocks are added together and the sum is divided by 5, the exact time is obtained. It is also known that one of these clocks is accurate. Which one exactly?
|
14. Let clock No. 1 show $x$ min, then clock No. $2-(x+1)$, clock No. $3-(x-2)$, clock No. $4-(x-3)$ and clock No. $5-(x-1)$. The sum of the readings of all clocks $x+(x+1)+(x-2)+(x-3)+(x-1)=5 x-5$. Since $(5 x-5): 5=$ $=x-1$, and this is the reading of clock No. 5, then these clocks show the exact time.
|
5
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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