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 |
|---|---|---|---|---|---|---|---|---|---|
9. A store needed to receive 185 kg of candies in closed boxes from the warehouse. The warehouse has boxes of candies weighing 16 kg, 17 kg, and 21 kg. What boxes and how many could the store receive? | 9. We notice that $185=37 \cdot 5$ and $16+21=37$. Therefore, the store could have received 5 boxes weighing 21 kg each and 5 boxes weighing 16 kg each. Since the number 185 has been factored into prime factors, the problem has a unique solution. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,318 |
10. A kilogram of one type of candy is 80 k more expensive than another. Andrey and Yura bought 150 g of candies, which included both types of candies, with Andrey having twice as much of the first type as the second, and Yura having an equal amount of each. Who paid more for their purchase and by how much?
14 | 10. Andrei has 100 g of one type of candy and 50 g of another, while Yura has 75 g of each type. Andrei's candies are more expensive than Yura's by the same amount that 25 g of one type is more expensive than 25 g of the other. Since 1 kg is more expensive by 80 k., then 25 g is more expensive by 2 k. $(80: 40=2)$. The... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,319 |
11. How many strikes will a clock make in a day if it strikes the whole number of hours and also marks the midpoint of each hour with one strike? | 11. $(1+2+3+\ldots+12) \cdot 2+24=13 \cdot 6 \cdot 2+24=180$.
Translating the above text into English, while preserving the original text's line breaks and format:
11. $(1+2+3+\ldots+12) \cdot 2+24=13 \cdot 6 \cdot 2+24=180$. | 180 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,320 |
12. The Zmey Gorynych has 2000 heads. A legendary hero can cut off 33, 21, 17, or 1 head with one strike of his sword, but in return, the Zmey grows 48, 0, 14, or 349 heads respectively. If all heads are cut off, no new ones grow. Can the hero defeat the Zmey? How should he act? | 12. $2000=94 \cdot 21+26$. This means that after cutting off 21 heads 94 times, 26 heads will remain. We notice that if 17 heads are cut off, 14 will grow back, i.e., the number of heads will decrease by 3. Then, we perform this three times, and $26-3 \cdot 3=$ $=17$ will remain, which can be cut off with the final blo... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,321 |
13. In a new nine-story building, where the first floor is allocated for stores, Sergei's family received apartment 211. On which floor and in which entrance is this apartment located, if on the third floor of one of the entrances of this building are apartments from 55 to 60? (All entrances and floors are the same.) | 13. Since there are 6 apartments on one floor in one entrance (60-54=6), there are 48 apartments on eight floors of one entrance. Since $211=48 \cdot 4 + 19$, apartment 211 is in the fifth entrance, and since $19=6 \cdot 3+1$, this apartment is on the fifth floor. | 第五入口，第五层 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,322 |
14. Decode the record ${ }_{* *}+{ }_{* * *}={ }_{* * * *}$, if it is known that both addends and the sum will not change if read from right to left. | 14. In the first addend, the tens and units digits are the same. Taking the smallest of such numbers - the number 11, then the second addend should be 989 or 999 (to get a four-digit number as the sum), but $11+989=$ $=1000$ and $11+999=1010$ do not satisfy the condition. Let the first addend be 22, then $22+999=1021$.... | 22+979=1001 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,323 |
16. The store released two batches of notebooks. If the second batch had 5 more notebooks than it actually did, the product of the numbers expressing the quantities of notebooks in each batch would have increased by 1250, and if the first batch had 7 more notebooks, then this product would have increased by 3150. How m... | 16. Since increasing the second factor by 5 increases the product by 1250, the first factor is $1250: 5=250$, i.e., in the first batch there were 250 notebooks. When the first factor is increased by 7, the product increases by 3150, so the second factor is: $3150: 7=450$, i.e., in the second batch there were 450 notebo... | 700 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,325 |
17. Find the largest number that, when divided by 31, gives a quotient of 30. | 17. The number will be the largest if when divided by 31 it yields the largest possible remainder. This remainder will be 30. Therefore, the desired number is: $30 \cdot 31 + 30 = 32 \cdot 30 = 960$. | 960 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,326 |
18. In a bank document, it was recorded that an equal amount of money was issued daily, and the initial sum of money in rubles, the number of days, the daily issued amount, and the remaining sum after the last issuance ended in 7, 5, 3, and 1, respectively. Upon seeing this record, the auditor immediately noticed that ... | 18. In this case, there is a dividend ending in the digit 7, a divisor being the digit 5, a quotient being the digit 3, and a remainder being the digit 1. It is known that the dividend equals the product of the divisor and the quotient, plus the remainder. However, the product of the divisor and the quotient ends in th... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,327 |
20. A magazine consists of 16 nested double sheets. On which double sheet will the sum of the numbers indicating the page numbers be the greatest? | 20. The page numbers range from 1 to 64. Pages on double sheets are arranged as follows:
$$
\begin{aligned}
& 1-\bar{h}-1,2,63,64 ; \\
& 2-\bar{и}-3,4,61,62 ;
\end{aligned}
$$
16th-31, 32, 33, 34. Since in the sequence the sums of numbers equidistant from the ends are the same, i.e., $1+64=2+63=\ldots=32+33$, the sum... | 130 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,329 |
21. Five participants of the competition became its prize winners, scoring 20, 19, and 18 points and taking first, second, and third places, respectively. How many participants won each prize place if together they scored 94 points? | 21. Three prize-winning participants scored 57 points ( $20+19+18=57$ ), so the other two scored 37 points ( $94-57=37$ ). Since $37=19+18$, one of them scored 19, and the other scored 18 points. Therefore, the first place was taken by one participant, the second place by two participants, and the third place by two pa... | 1,2,2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,330 |
22. Members of the "Handy Hands" club had to cut a piece of wire 102 cm long into parts 15 cm and 12 cm long, and so that there were no scraps. The children found all possible solutions. Find them as well. | 22. Since the total length of the wire is an even number (102 cm) and the sum of the lengths of the segments that are 12 cm each is even, the sum of the lengths of the segments that are 15 cm each must also be even. Therefore, the number of pieces that are 15 cm each must be even. Let's consider all possible cases:
1) ... | )2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,331 |
23. Represent the number 231 as the sum of several natural numbers so that the product of these addends also equals 231. | 23. Since $231=3 \cdot 7 \cdot 11$, but $3+7+11=21$, this number can be represented as a sum of addends, the product of which is 231, as follows:
$$
231=3+7+11+\underbrace{1+1+\ldots+1}_{210 \text { ones }}
$$ | 231=3+7+11+\underbrace{1+1+\ldots+1}_{210\text{ones}} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,332 |
24. L. N. Tolstoy's Problem. Five brothers divided their father's inheritance equally. The inheritance included three houses. Since the three houses could not be divided into 5 parts, the three older brothers took them, and in return, the younger brothers were given money. Each of the three brothers paid 800 rubles, an... | 24. Two younger brothers received 2400 rubles, and each received 1200 rubles. $(2400: 2=$ $=1200)$. Therefore, the total value of the inheritance, i.e., three houses, is 6000 rubles. $(1200 \times$ $\times 5=6000)$. Then one house costs 2000 rubles. $(6000: 3=2000)$. | 2000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,333 |
27. Using the digit 7 four times, operation signs, and parentheses, represent all numbers from 1 to 10 inclusive. | 27. $7-7+7: 7=1 \quad(7 \cdot 7-7): 7=6$
$7: 7+7: 7=2 \quad(7-7) \cdot 7+7=7$
$(7+7+7): 7=3 \quad(7 \cdot 7+7): 7=8$
$77: 7-7=4 \quad(7+7): 7+7=9$
$7-(7+7): 7=5 \quad(77-7): 7=10$ | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,336 |
28. How many standard sheets of paper will be required to print all numbers from 1 to 1000000 inclusive, if printing on one side of the sheet? Each sheet has 30 lines with 60 characters each, and there are intervals between numbers, each equivalent to two characters. | 28. For printing single-digit numbers, 9 characters are required; for two-digit numbers - $180(2 \cdot 90=180)$; for three-digit numbers $-2700(900 \cdot 3=2700)$; for four-digit numbers $36000(4 \cdot 9000=36000)$; for five-digit numbers $-450000(5 \cdot 90000=450000)$; for six-digit numbers $-5400000(6 \cdot 900000=5... | 4383 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,337 |
30. Think of a single-digit number: double it, add 3, multiply by 5, add 7, using the last digit of the resulting number, write down a single-digit number, add 18 to it, and divide the result by 5. What number did you get? No matter what number you thought of, you will always get the same number in the final result. Ex... | 30. Let the number thought of be $x$. Then we sequentially obtain:
1) $2 x ; 2) 2 x+3 ; 3) 10 x+15=10(x+1)+5$
2) $10(x+1)+12 ; 5) 2 ; 6) 2+18=20$; 7) $20: 5=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,339 |
31. The quotient of dividing one number by another is an integer that is two times smaller than one of them and six times larger than the other. Find this quotient. | 31. Let the dividend be $a$, and the divisor be $b$, then according to the condition, the quotient is $a: 2$ or $6 b$, i.e., $a: 2 = 6 b$, from which $a = 12 b$ and $a: b = 12$. Therefore, the quotient $a: b$ is 12. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,340 |
32. If mom gives each of her children 13 notebooks, she will have 8 notebooks left; if she gives them 15 notebooks each, all the notebooks will be distributed. How many notebooks did mom have? | 32. If the number of children was $a$, then the total number of notebooks was $15a$. When dividing $15a$ by 13, the quotient is $a$ and the remainder is 8. Therefore, $15a = 13a + 8$, from which $2a = 8$ and $a = 4$. Thus, the total number of notebooks was $60 (15 \cdot 4 = 60)$. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,341 |
33. By dividing a certain integer by 15, Sergei got a remainder of 8, and by dividing it by 20, he got a remainder of 17. Show that Sergei made a mistake. | 33. Let Sergey divide the number $a$, and the quotients he obtained are $b$ and $c$ respectively. Then $a=15b+8$ and $a=20c+17$, so $15b+8=20c+17$, or $15b=20c+9$. The left side of the obtained equation is divisible by 5, but the right side is not, since 9 is not divisible by 5. Therefore, the equation cannot be true f... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,342 |
34. In the book of records, a whole number expressing the number of meters of fabric sold was blotted out with ink. The total revenue could not be read either, but the end of this entry was visible: 7 r. 28 k. - and it is known that this amount does not exceed 500 r. The price of 1 m of fabric is 4 r. 36 k. Help the in... | 34. Since one factor ends with the digit 6 (436), and the product ends with the digit 8 (...728), the second factor must end with the digit 8 or 3. If it ends with the digit 8, then \(436 \cdot 8 = 3488\). We have
 in the row or column ... | 35. The sum of the numbers in each row or column is 110. Therefore, if, for example, the number 30 is chosen, then the sum of the numbers in the row without it: $14+32+8+$ $+26=80$. If you are given this sum, then you find $110-80=30$. | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,344 |
36. Multiply your shoe size by 2, add 39 to the product, multiply the resulting sum by 50, add 40 to the product, and subtract your birth year from the sum. You will get a four-digit number, the first two digits of which are your shoe size, and the last two digits are your age at the end of the calendar year 1990. Expl... | 36. Let the size of your shoe be $\overline{a b}$ (a two-digit number). Perform the following steps sequentially:
1) $\bar{a} \bar{b} \cdot 2 ; 2) 2 \cdot \overline{a b}+39$
2) $(2 \overline{a b}+39) \cdot 50=100 \cdot \overline{a b}+1950$;
3) $100 \cdot \bar{a} \bar{b}+1950+40=100 \overline{a b}+1990$.
If the birth y... | 100\overline{}+15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,345 |
37. Think of a three-digit number: double its first digit (the hundreds place), add 3 to the resulting product, multiply the obtained sum by 5, and add the second digit of the thought number (the tens place) to the resulting product. Finally, append the last digit of the thought number to the right of the obtained sum.... | 37. Let the number thought of be $100 a+10 b+c$. Then, performing the operations in sequence, we get
$$
((2 a+3) \cdot 5+b) \cdot 10+c, \text { or }(100 a+10 b+c)+150
$$
from which it is clear that to obtain the number thought of, it is sufficient to subtract 150 from the result obtained. If, for example, we obtained... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,346 |
38. At a math club meeting, the students were offered the following math trick:
Write down any number,
I'll take a look at it
And then on a piece of paper
I'll put it in an envelope in front of you.
Under your number, any
Write down a number again,
And I will allow myself
To write down one number.
We will draw... | 38. Let's say, for example, the numbers 58431 and 61388 were recorded. The guesser added such a third addend that, when added to the second number recorded by the children, it would result in a number consisting only of nines. For example, $61388 + 38611 = 99999$. The sum of the three addends can be easily obtained usi... | 158430 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,347 |
40. In a football tournament, each of the participating teams played against each other once. The teams scored $16,14,10,10,8,6,5,3$ points. How many teams participated in the tournament and how many points did the teams that played and finished in the top 4 positions lose? (A team gets 2 points for a win and 1 point f... | 40. If the number of teams is $n$, then the total number of games is $\frac{n(n-1)}{2}$, and the total number of points scored by them is $(n-1) n$. In our example, we have $16+14+10+10+8+$ $+6+5+3=72$. Therefore, $(n-1) \cdot n=72$, or $(n-1) \cdot n=8 \cdot 9$, so $n=9$, i.e., there were 9 teams. The maximum number o... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,349 |
41. There are 67 weights with masses of 1 g, 2 g, 3 g, ..., 67 g. Can they be divided into three groups of equal mass? | 41. Let's find the total (sum) mass of all the weights: $1+2+\ldots+67=\frac{(1+67) \cdot 67}{2}=$ $=\frac{68 \cdot 67}{2}=34 \cdot 67$ (see solutions to problems 11, 20). Since 34 and 67 are not divisible by 3 without a remainder, it is impossible to divide these weights into groups of equal mass. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,350 |
42. In a regular set of dominoes, there are 28 tiles. How many tiles would a domino set have if the values indicated on the tiles ranged not from 0 to 6, but from 0 to $12?$ | 42. If the values on the dominoes changed from 0 to 12, then the dominoes would be:
This means there would be 13 dominoes with zero, 12 dominoes with one (but without zero), 11 dominoes wit... | 91 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,351 |
43. Can several digits be erased from the beginning and the end of the four-hundred-digit number 86198619 ... 8619 so that the sum of the remaining digits equals 1986? | 43. The sum of the digits of the number 1986 is 24, and since $1986=82 \cdot 24+18$, it is sufficient to erase the first 2 digits from the left (8 and 6), leaving 1, 9, and 8 ($1+9+8=18$) and 82 groups of digits $6,1,9,8(6+1+9+8=24)$, and then erase all the remaining 67 digits $(400-5-4\cdot82=67)$. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,352 |
44. Find the sum of all possible different three-digit numbers, all digits of which are odd. | 44. Let's write the sum in ascending order of the addends: $111+113+\ldots+$ $+119+131+\ldots+311+313+\ldots+319+\ldots+911+913+\ldots+993+995+$ $+997+999$. The number of addends starting with the digits $1,3,5,7,9$ will be 25 each, so the total number of addends is $25 \cdot 5=125$. The sum of two addends equally dist... | 69375 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,353 |
45. Tanya and Olya tried for a long time to arrange nine numbers from 1 to 9 in a circle so that the sum of any three consecutive numbers would be divisible by 3 and greater than 12, but without success. Is it possible to arrange the numbers this way? | 45. Let the numbers arranged in a circle be $a_{1}, a_{2}, a_{3}, \ldots, a_{9}$. Then the sums of the numbers taken three at a time, in sequence, are $a_{1}+a_{2}+a_{3}, a_{2}+a_{3}+a_{4}, \ldots, a_{8}+a_{9}+a_{1}, a_{9}+a_{1}+a_{2}$. There are 9 such sums in total. The total sum of these sums contains each of the nu... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,354 |
46. Find 4 triples of non-negative integers such that each number from 1 to 81 inclusive can be represented as the sum of four numbers - one from each triple. | 46. Since the sum of four non-negative numbers must be 1, three of the four addends must be zeros. Therefore, the first triplet can be $1, 2, 3$, and each of the remaining triplets can have one zero. To get a sum of 4, one of the triplets, other than the first, must have the number 3. Let it be the second number in the... | \begin{pmatrix}1&2&3\\0&3&6\\0&9&18\\0&27&54\end{pmatrix} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,355 |
47. Over the past five years, 27 collective farmers have been awarded. Moreover, in each subsequent year, more were awarded than in the previous year. In the last year, three times as many were awarded as in the first year. How many collective farmers were awarded in the third year? | 47. If in the first year $x$ collective farm workers are awarded, then in the last year it is $3x$, and for these two years it is $4x$. The values $x=1$ and $x=2$ do not satisfy the condition, as in the second, third, and fourth years, 23 (27-4=23) and 19 (27-8=19) would be awarded, which does not meet the condition th... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,356 |
48. Having made 5 shots each at the target, Kolya and Petya scored $10,9,9,8,8,5$, $4,4,3,2$ points. With their first three shots, they scored the same number of points, and with their last three shots, Kolya scored three times more than Petya. How many points did each of them score with their third shot? | 48. With three shots, one can score no less than 9 points (2+3+4=9) and no more than 28 (10+9+9=28). Since 9 * 3 = 27, it is clear that Kolya scored 10, 9, and 8 points with his last three shots, while Petya scored 4, 3, and 2. The remaining points 9, 8, 5, and 4 were scored with the first two shots. If Kolya scored 5 ... | 10,2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,357 |
49. Sergei wrote down a certain five-digit number and multiplied it by 9. To his surprise, he got a number consisting of the same digits but in reverse order. What number did Sergei write down? | 49. Since multiplying a five-digit number by 9 results in a five-digit number, the leftmost digit of the original number is 1.
Since the number obtained after multiplying this number by 9 ends with the digit 1, the original number ends with the digit $9(9 \cdot 9=81)$
 | 52. Taking a number as an addend 99999 times means multiplying it by 99999 or by $(100000-1)$, i.e., appending 5 zeros to the given number and subtracting the given number from the result. In this case, we have
$$
\overline{\text { ПОТОП }} 00000-\overline{\text { ПОТОП }}=\ldots 285 .
$$
From this, we get $\Pi=5, \m... | 51715 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,361 |
53. The Merry Clown Nibumbum
Today is gloomy and sullen.
What is troubling Nibumbum?
He solved an example eight times,
And each time a different sum!
A sad case! (And you?)
When solving, don't forget
(That's the subtlety of the matter!)
Identical letters represent identical digits!
$$
\begin{array}{r}
\text { ... | 53. The sum of three $A$ ends in $A$, so $A=0$ or $A=5$. But, if $A=5$, then $(K+K+K+1)$ cannot end in $K$. Therefore, $\mathbf{A}=0, \mathrm{~K}=5$. Since (Ш + Ш + ІІ +1 ) ends in $\mathbf{A}=0$, then $Ш=3$. Since $\mathrm{K}+\mathrm{K}+\mathrm{K}=15$, then $\mathrm{C}=1$. We have
$$
\begin{aligned}
& 5 * 350 \quad 5... | 172050 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,362 |
54. The task is very difficult-
Not everyone can find:
What does the star,
Bicycle and hedgehog equal?
| BICYCLE | HEDGEHOG | 7 |
| :---: | :---: | :---: |
| + STAR | HEDGEHOG | 4 |
| 6 | BICYCLE HEDGEHOG | |
| 1 BICYCLE | 0 | STAR |
Decode the rebuses: | 54. (HEDGEHOG + HEDGEHOG + BICYCLE + 1) ends with the digit 0. Therefore, (HEDGEHOG + HEDGEHOG + BICYCLE) = 9 (or 19). The equation HEDGEHOG + HEDGEHOG + BICYCLE = 19 is impossible. Therefore, the sum 9 is possible, and from the cases 1+1+7=9, 2+2+5=9, 3+3+3=9, and 4+4+1=9, only 2+2+5=9 fits. As a result, HEDGEHOG = 2,... | 1503 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,363 |
59. Petya and Vasya calculated that if they walk to the neighboring village, which is four kilometers away from them, at a speed of 4 km per hour, they will be 10 minutes late for the football match being held there for the district championship. How should they act to arrive at the match simultaneously and gain the ma... | 59. To arrive simultaneously and gain the most time, Petya and Vasya should do the following: one of them rides the bicycle, while the other walks. After covering half the distance, the first one leaves the bicycle and continues on foot. The second one, upon reaching the bicycle, picks it up and rides it to the end of ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,368 |
60. Winnie-the-Pooh and Piglet set off to visit each other at the same time. But since Winnie-the-Pooh was composing another "hum" all the way, and Piglet was counting the crows flying by, they did not notice each other when they met. After the meeting, Piglet reached Winnie-the-Pooh's house in 4 min, and Winnie-the-Po... | 60. Let's illustrate schematically:
After meeting, Winnie-the-Pooh traveled the distance $C B$, which is the same distance Piglet traveled before the meeting, 4 times faster than Piglet tr... | 36 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,369 |
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 ... | 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 $5,1... | 73 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,370 |
62. Fill in the free circles with natural numbers so that they contain all numbers from 1 to 19 and in all five rows of each of the three directions (vertical and two diagonal) the sum of the numbers in one row is the same. | 62. The sum of the numbers in each row is $38(3+17+18=38)$, so between the numbers 3 and 16 there should be 19, between 16 and 10 - the number 12, and so on, until we get
Let's find the nu... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,371 |
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... | 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. Am... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,373 |
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)+\ld... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,374 |
1. The difference of two natural numbers, which are perfect squares, ends with the digit 2. What are the last digits of the minuend and subtrahend if the last digit of the minuend is greater than the last digit of the subtrahend? | 1. The squares of natural numbers end in the digits $0,1,4,9,6,5$. Among them, only $6-4=2$, so the minuend ends in the digit 6, and the subtrahend in the digit 4. | 6-4=2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,375 |
2. Write down the first 15 consecutive odd numbers in a row. Divide the numbers in this row into groups so that the first group contains one number, the second group contains the next two numbers, the third group contains three numbers, and so on. Find the sum of the numbers in each group and indicate the common patter... | 2. We have
$$
\begin{aligned}
& 1=1^{3} \\
& 3+5=8=2^{3} \\
& 7+9+11=27=3^{3} \\
& 13+15+17+19=64=4^{3} \\
& 21+23+25+27+29=125=5^{3}
\end{aligned}
$$
We can observe the following pattern: if we raise the number of elements in a group to the power of three, we get the sum of the numbers (elements) in that group. | n^3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,376 |
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 | 39,377 |
4. When raising the number 4 to different powers, Sasha obtained three numbers, all of which had three different digits. Without calculating the results, Andrey noticed that Sasha had made a mistake. What was his basis for this? | 4. Powers of the number 4 can end either in 6 or 4, therefore three powers cannot have three different unit digits. | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,378 |
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 touris... | 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.... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,379 |
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 | 39,380 |
7. Show that the number 425102348541 is not a perfect square of a natural number. | 7. If the given number is divisible by 3, then $425102348541 = 3 \cdot n$, but $n$ is not divisible by 3, since the given number is not divisible by 9. The number 3 is not a perfect square, so the given number is not a perfect square. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,381 |
8. In the record of a certain number, there are 123 ones, and the rest of the digits are zeros. Can this number be a perfect square? | 8. This number is divisible by 3, since the sum of its digits 123 is divisible by 3, but it is not divisible by 9, so this number cannot be a perfect square. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,382 |
9. In the example $(* *)^{3}=* * * 9$, replace the asterisks with digits to obtain a correct equation. | 9. A two-digit number raised to the cube should not exceed 22, since $22^{3}=10648$ is a five-digit number. The cube obtained ends in 9, so the number being cubed also ends in 9. As a result, we have $19^{3}=6859$. | 19^{3}=6859 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,383 |
10. In the example $(\Delta \square)^{\triangle}=\square \bigcirc \square$ the digits are encoded with geometric shapes. (Identical digits correspond to identical shapes, different ones to different shapes.) Decode these digits. | 10. From the condition, it is clear that raising a two-digit number to a power results in a three-digit number. Considering that even $10^{3}=1000$, the exponent can only be 2, i.e., $\Delta=2$. Since the base of the power ends with the same digit as the power, $\square$ can be $0,1,5$ or 6. The condition is only satis... | 26^{2}=676 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,384 |
11. In the example $(\Delta \triangle) \square=\Delta \square \square \triangle$ the digits are encoded with geometric shapes. Decode them. | 11. $11^{3}=1331$.
The translation is as follows:
11. $11^{3}=1331$. | 11^3=1331 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,385 |
12. Decode the rebus $(\mathrm{AP})^{M}=$ МИР, if the same letters encode the same digits, and different letters encode different digits. | 12. Since raising a two-digit number to a certain power resulted in a three-digit number, the exponent can only be 2. This means $\mathrm{M}=2$, then $\mathrm{A}=1$; P can be 5 or 6, but $15^{2}=225$ does not satisfy the condition, so $\mathrm{P}=6$. We have $16^{2}=256$. | 16^{2}=256 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,386 |
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 ... | 3025 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,387 |
14. While erasing the board in class, Ira accidentally erased three digits from the calculations of the cubes of two numbers, leaving only the unit digits - 8 and 9. Help her restore all the digits and find the numbers that were cubed. | 14. Since the cubes are four-digit numbers by condition, their bases are two-digit numbers. If a cube ends in the digit 8, then the base ends in the digit $2\left(2^{3}=8\right)$. In this case, the base can only be 12, since $22^{3}=10648$ is a five-digit number. Therefore, the only possibility is $12^{3}=1728$. If a c... | 12^3=1728,19^3=6859 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,388 |
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 | 39,389 |
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 perfe... | 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 ... | 7744 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,390 |
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... | 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 an... | 62512 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,391 |
18. Replace the letters with digits so that in each row (column) there are three-digit numbers that are perfect squares. (Identical letters must correspond to identical digits, different letters to different digits.)
| $\mathbf{T}$ | $\mathbf{i}$ | $\mathbf{P}$ |
| :---: | :---: | :---: |
| $\mathbf{U}$ | $\mathbf{B}$... | 18. There are 13 three-digit numbers that are perfect squares and have no repeated digits in their representation: $169,196,256,289,324,361,529,576,625$, $729,784,841,961$. Among them, the ones that satisfy the condition are $729,256,961$ :
| 7 | 2 | 9 |
| :--- | :--- | :--- |
| 2 | 5 | 6 |
| 9 | 6 | 1 | | \begin{pmatrix}\hline7&2&9\\\hline2&5&6\\\hline9&6&1\\\hline\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,392 |
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 pe... | 367 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,397 |
24. Returning from the collective farm garden, the boys were enthusiastically discussing the results of their work harvesting carrots. "What are the dimensions of the plot we worked on?" they asked their team leader Vasya. "I lost the measurement records somewhere," he replied, "but I remember that the plot is square a... | 24. The problem can be solved simply using a table of squares of numbers. But what if you don't have it at hand? Then we proceed as follows. The smallest four-digit number that is a perfect square is \(1024 (32^2 = 1024)\), and the largest is \(9801 (99^2 = 9801)\). Therefore, if the side length of the plot is \(x\) me... | 78,6084^2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,398 |
1. Prove that from any three natural numbers, it is always possible to choose two such that their sum is divisible by 2. | 1. Since natural numbers can be either even or odd, among three numbers there will be at least two numbers of the same parity (either even or odd). The sum of two numbers of the same parity is always an even number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,399 |
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 | 39,400 |
3. Prove or disprove the statement: "The difference between a three-digit number and the sum of its digits is always divisible by 9."
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. $100a + 10b + c - (a + b + c) = 99a + 9b = 9(11a + b)$. Since $9(11a + b)$ is divisible by 9 for any $a$ and $b$, the statement is true. | proof | Combinatorics | MCQ | Yes | Yes | olympiads | false | 39,401 |
4. Prove that the word XAXAXA is divisible by 7, if in it the letters X and A represent any digits. (Identical letters represent identical digits, different letters represent different digits.) | 4. Method I. HAHAHA $=10^{5} \mathrm{X}+10^{4} \mathrm{~A}+10^{3} \mathrm{X}+10^{2} \mathrm{~A}+10 \mathrm{X}+\mathrm{A}=$ $=10 \mathrm{X}\left(10^{4}+10^{2}+1\right)+\mathrm{A}\left(10^{4}+10^{2}+1\right)=\left(10^{4}+10^{2}+1\right) \cdot(10 \mathrm{X}+\mathrm{A})=$ $=10101 \cdot(10 \mathrm{X}+\mathrm{A})$. Since $10... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,402 |
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 | 39,403 |
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 s... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,404 |
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 | 39,405 |
8. Prove that if any five-digit number is appended to itself on the right (or left), the resulting ten-digit number is divisible by 11. | 8. Let's say we have a five-digit number $a$. After appending the same number to it, we get $100001 a$. Since the number 100001 is divisible by $11 (100001: 11=9091)$, then $100001 a$ is also divisible by 11. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,406 |
9. Prove that if to any three-digit number you append a three-digit number written with the same digits but in reverse order, the resulting number will be divisible by 11. | 9. Method I. Let us have the number $\overline{a b c}$, then according to the condition we get $\overline{a b c c b a}$ $\overline{a b c ~ c b a}=10^{5} a+10^{4} b+10^{3} c+10^{2} c+10 b+a=a\left(10^{5}+1\right)+10 b\left(10^{3}+1\right)+$ 136
$+10^{2} c(10+1)=100001 a+10 b \cdot 1001+10^{2} c \cdot 11$. Since each ter... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,407 |
10. Find the greatest and the least three-digit numbers, each of which is divisible by 6 and has the digit 7 in its representation. | 10. Each of the sought numbers must end in an even digit (since it is divisible by 6), so the digit 7 cannot be at the far right. To make the number the largest (smallest), the hundreds digit should be the largest (smallest) of the possible ones. Taking into account the condition of divisibility by 3, we get the sought... | 978174 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,408 |
11. If the sum of the first and second digits of a three-digit number, where the hundreds and units digits are the same, is divisible by 7, then the number is also divisible by 7. Prove it. | 11. We have the number $100a + 10b + a = 10(a + b) + 91a$. Since $(a + b)$ is divisible by 7 and 91 is divisible by 7 (91: 7 = 13), the number $100a + 10b + a$ is also divisible by 7. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,409 |
12. Prove that among eight natural numbers, there will be at least two numbers whose difference is divisible by 7. | 12. When dividing by 7, all possible different remainders are 7, namely $0,1,2,3,4,5,6$. Since the condition states that there are 8 numbers, there will be at least two numbers that give the same remainder when divided by 7. Therefore, their difference will be divisible by 7. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,410 |
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 | 39,411 |
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: ... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,412 |
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 | 39,413 |
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... | 547 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,414 |
17. Find the digits $a$ and $b$ in the number $\overline{42 a 4 b}$, given that this number is divisible by 72. | 17. If a number is divisible by 72, then it is divisible by 9 and 8. Since $4+2+4=10$, then $a+b=8$ or $a+b=17$. If $a+b=17$, then either $a=9$ and $b=8$, or $a=8$ and $b=9$. But in both cases, the resulting numbers are not divisible by 8. Therefore, $a+b=8$. Based on divisibility by 8, only $a=8$ and $b=0$, or $a=0$ a... | =8,b=0or=0,b=8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,415 |
18. Find the hundreds and units digits of the number $72 * 3 *$, if this number is divisible by 45. | 18. The desired number is divisible by 45, which means it is divisible by 9 and 5. The unit digit can be 0 or 5. If $72 * 30$, then the hundreds digit is 6, since $7+2+3+0=12$ and 18 is divisible by 9. If $72 * 35$, then $7+2+3+5=17$, so the hundreds digit is 1. Therefore, 72630 or 72135. | 72630or72135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,416 |
19. Andrey found the product of all numbers from 1 to 11 inclusive and wrote the result on the board. During the break, someone accidentally erased three digits, and the remaining number on the board is $399 * 68 * *$. Help restore the digits without recalculating the product. | 19. The two rightmost digits are zeros, since among the factors there is the number 10 and also $2 \cdot 5=10$. The product of numbers from 1 to 11 is divisible by 9,
therefore the sum of the digits of the result must be divisible by 9. The sum of the known digits is $3+9+9+6+8=35$. The number greater than 35 and clos... | 39916800 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,417 |
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 small... | 123457968 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,418 |
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 can... | 1023457896 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,419 |
22. When shopping at the "Sport Goods" store, Yura and his dad were supposed to receive 31 p in change. When the cashier gave them 1 p., 3 p., and 5 p. notes—altogether 10 notes—Yura immediately noticed that the cashier had made a mistake without counting the total amount. Thanking the boy, the cashier immediately corr... | 22. Since 31 is an odd number, the sums obtained from 1 r., 3 r., and 5 r. must either all be odd or two even and one odd. This means that the number of bills of 1 r., 3 r., and 5 r. must either all be odd or only one is odd. In both of these cases, the total number of all bills must be an odd number, but 10 is an even... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,420 |
23. When the cashier rang up a bill for 22 p. 85 k. for an album costing 6 r., a book costing 12 r., 9 boxes of colored pencils, and 15 rulers, the customer, although not knowing the cost of the pencils and rulers, immediately noticed that the cashier had made a mistake. What was the basis for his suspicion? | 23. If a box of pencils costs $x$ k., and a ruler costs $-y$ k., then pencils and rulers cost $(9 x+15 y)$ k. On the other hand, they cost 22 p. 85 k. $-(6+12)$ p. $=4$ p. 85 k. Therefore, $9 x+15 y=485$. The obtained equation cannot be true for any natural values of $x$ and $y$, since the left side is divisible by 3, ... | 9x+15485 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,421 |
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 eac... | 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 broug... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,422 |
26. To be able to use vending machines for buying water and postcards during a city tour, tourists approached the cashier with a request to exchange their money only for 3 k. and 5 k. coins. Each tourist had more than 8 k. Could the cashier always satisfy their request, having a sufficient amount of the required coins? | 26. Any number either divides by 3, then the remainder is 0, or does not divide by 3, then the remainder is 1 or 2. Therefore, it can be represented in one of three forms: $3n, 3n+1, 3n+2$. If we have a number of the form $3n$, then it is represented by an integer number of threes. If $3n+1>8$, then $n \geqslant 3$, an... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,424 |
27. Take any two three-digit numbers that are not divisible by 37, but such that their sum is divisible by 37. By appending one of these numbers to the other, you get a six-digit number. Check if it is divisible by 37. Formulate a statement using the considered example and prove it. | 27. If each of two three-digit numbers is not divisible by 37 (not all digits in their representation are the same), and the sum of them is divisible by 37, then, by appending one to the other, we will get a six-digit number that is divisible by 37. Let's prove this.
Let $a$ and $b$ be three-digit numbers satisfying t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,425 |
28. A field was divided into 9 plots, some of the resulting plots were again divided into 9 plots, some of the resulting plots were again divided into 9 plots, and so on. Could it result in 1986 plots? | 28. Let $n$ of the 9 plots obtained after the first division be divided, then there will be a total of $n \cdot 9 + (9 - n) = 8n + 9$ plots. And so after each division. Suppose after some division, 1986 plots were obtained. Then $8n + 9 = 1986$, from which $8n = 1977$. Since the number 1977 is not divisible by 8, this ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,426 |
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 b... | 205 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,427 |
30. Preparing for the club lesson, the children found such 2 natural consecutive numbers, the smallest of the possible ones, that the sum of the digits of each of them is divisible by 17. What numbers did the children find? | 30. The smallest number, different from 0, divisible by 17 is 17, the next one is 34. The sum of the digits of one of the desired numbers should equal 17, and the other should equal 34. To make the number as small as possible, its digits should be as large as possible. For example, if we take 7999, then $9 \cdot 3+7=34... | 88998900 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,428 |
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)... | 134 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,429 |
32. A number and its last digit are not divisible by 3 without a remainder. Prove that by appending its last digit to the number several times, you can obtain a number that is divisible by 3. What is the smallest number of times you need to append this digit? 2* | 32. Since this number is not divisible by 3, the remainders obtained when dividing by 3 its last digit and the number represented by the remaining digits (let's call it the "shortened" number) are either the same, or the "shortened" number is divisible by 3 without a remainder. If the "shortened" number is divisible by... | 1or2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,430 |
33. When a three-digit number, the two outer digits (left) of which are the same, and the right one is 5, is divided by a single-digit number, the remainder is 8. Find the dividend, divisor, and quotient. | 33. Since the remainder is 8 and the divisor is a single-digit number, it is equal to 9. The remainder when dividing the number $\overline{a a 5}$ by 9 is the same as the remainder when dividing the sum of the digits of this number by 9, so $2a=3$ or $2a=12$, from which $a=6$. Therefore, the dividend is 665, the diviso... | 665=73\cdot9+8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,431 |
34. After finishing reading the book, Vasya calculated that 301 digits were required to number all its pages. Show that he was wrong. | 34. For numbering pages with single-digit numbers, 9 digits are required, for double-digit numbers $-2 \cdot 90=180$ digits, and in total 189 digits. The remaining $301-189=112$ digits should be used for numbering pages with three-digit numbers, but 112 is not divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 39,432 |
35. By writing down 6 different numbers, none of which is 1, in ascending order and multiplying them, Olya got the result 135135. Write down the numbers that Olya multiplied. | 35. Since $135135=135 \cdot 1001$ and $1001=7 \cdot 11 \cdot 13$, and $135=5 \cdot 3 \cdot 9$, we have $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11 \cdot 13=135135$ | 3\cdot5\cdot7\cdot9\cdot11\cdot13=135135 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,433 |
36. The number $n \cdot(n+2)$ ends with the digit 4. Name the second to last digit of this number. | 36. The numbers $n$ and $(n+2)$ have the same parity, but since their product is even (ends in the digit 4), these numbers are even. Therefore, the product $n(n+2)$ is divisible by 4. However, numbers divisible by 4 are those where the last two digits form a number divisible by 4. Such numbers are $24, 44, 64, 84$. The... | 2,4,6,8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,434 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.