problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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|---|---|---|---|---|---|---|---|---|---|
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 | 39,549 |
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... | 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... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,550 |
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 t... | 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 (tran... | 544000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,551 |
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 ... | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,552 |
64. One glass of coffee was poured into a cup, and then a quarter glass of milk from a glass filled with milk was added to this cup. After thoroughly mixing the contents of the cup, it was topped up from the glass of milk. How much milk and how much coffee is left in the cup? | 64. When a quarter of a cup of milk was added to the cup of coffee already in the mug, the mug then contained $\frac{5}{4}$ cups of coffee with milk. In this case, the milk constitutes 1 part, and the coffee - 4 such parts. To top up the cup with milk, one needs to pour out a quarter of a cup of the contents of the mug... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,554 |
65. One glass contains water, the other contains the same amount of milk. A spoonful of milk is taken from the second glass, poured into the first glass, and mixed. Then a spoonful of the mixture is taken from the first glass and poured into the second glass. Which is greater: the amount of milk in the water or the amo... | 65. Let there be $a$ spoons of liquid in each glass. Pouring a spoon of milk into the glass of water, we get $a+1$ spoons of the mixture, in which milk constitutes $\frac{1}{a+1}$ part. Taking a spoon of the mixture, in which water constitutes $\frac{a}{a+1}$ part, and pouring it into the glass of milk. In it, there wi... | \frac{1}{+1} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,555 |
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 $\f... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,556 |
68. When Vasya Verkhoglyadkin was given this problem: "Two tourists set out from $A$ to $B$ simultaneously. The first tourist walked half of the total time at a speed of $5 \mathrm{km} /$ h, and the remaining half of the time at a speed of 4 km/h. The second tourist walked the first half of the distance at a speed of 5... | 68. If the length of the path is $a$ km and the first tourist walked it in $x$ hours, then $5 \cdot \frac{x}{2} + 4 \cdot \frac{x}{2} = a$, from which $x = \frac{2a}{9}$. The second tourist walked this path in $\frac{a}{2 \cdot 5} + \frac{a}{2 \cdot 4} = \frac{9a}{40}$. Let's compare $\frac{2a}{9}$ and $\frac{9a}{40}, ... | \frac{2a}{9}<\frac{9a}{40} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,558 |
69. In the store, there are candies at 2 p. and 3 p. per 1 kg, each type for the same amount of money. At what price should the mixture of these candies be sold to keep the total cost of the candies the same? | 69. Let the cost of each type of candy be $a$ rubles. Then at a price of 2 rubles per 1 kg, there was $a: 2=\frac{a}{2}$ (kg), and at 3 rubles $-\frac{a}{3}$ kg. In total, there was $\frac{a}{2}+\frac{a}{3}=\frac{5 a}{6}$ (kg), so the average price per 1 kg of the mixture will be $2 a: \frac{5 a}{6}=\frac{12}{5}=2.4$, ... | 2.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,559 |
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
$$
2 a:\left(\frac{a}{2.4}+... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,560 |
71. The bus traveled three equal-length segments of the route: the first at an average speed of 50 km/h, the second at 30 km/h, and the third at 70 km/h. What is the average speed of the bus over the entire route? | 71. Let the length of one segment be $a$ km, then the bus spent $\left(\frac{a}{50}+\frac{a}{30}+\frac{a}{70}\right)$ hours to travel the entire path of length $3 a$ km. Therefore, the average speed of movement is
$$
3 a:\left(\frac{a}{50}+\frac{a}{30}+\frac{a}{70}\right)=\frac{3 \cdot 1050}{21+35+15} \approx 44(\text... | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,561 |
72. Find the product:
$$
\left(1-\frac{1}{4}\right) \cdot\left(1-\frac{1}{9}\right) \cdot\left(1-\frac{1}{16}\right) \ldots\left(1-\frac{1}{225}\right)
$$ | 72. $\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \ldots\left(1-\frac{1}{225}\right)=$ $=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \ldots \cdot \frac{224}{225}=\frac{1 \cdot 3}{2^{2}} \cdot \frac{2 \cdot 4}{3^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdot \ldots \cdot \frac{14... | \frac{8}{15} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,562 |
73. An electronic computer can quickly calculate the sum
$$
\frac{2}{1 \cdot 3}+\frac{2}{3 \cdot 5}+\frac{2}{5 \cdot 7}+\ldots+\frac{2}{99 \cdot 101}
$$
However, you can compete with it if you uncover the secret and rewrite the terms differently. How can this be done? | 73. Since $\frac{2}{1 \cdot 3}=1-\frac{1}{3} ; \frac{2}{3 \cdot 5}=\frac{1}{3}-\frac{1}{5} ; \cdots \frac{2}{99 \cdot 101}=\frac{1}{99}-\frac{1}{101}$, we will have $1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\ldots+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}$. | \frac{100}{101} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,563 |
74. Establish the rule according to which the table is composed, and exclude the "extra" number:
| $\frac{1}{3}$ | $\frac{1}{8}$ | $\frac{23}{7}$ |
| :---: | :---: | :---: |
| $3 \frac{2}{7}$ | $\frac{4}{11}$ | $\frac{1}{3}$ |
| 0.125 | $\frac{5}{13}$ | $\frac{4}{11}$ | | 74. By representing the numbers in the table in a different form, we get
| $\frac{1}{3}$ | $\frac{1}{8}$ | $\frac{23}{7}$ |
| :---: | :---: | :---: |
| $\frac{23}{7}$ | $\frac{4}{11}$ | $\frac{1}{3}$ |
| $\frac{1}{8}$ | $\frac{5}{13}$ | $\frac{4}{11}$ |
From this table, it is clear that the numbers written in it, exc... | \frac{5}{13} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,564 |
75. Find all fractions with single-digit denominators, each of which is greater than $\frac{7}{9}$ but less than $\frac{8}{9}$. | 75. We will replace the given fractions with equal ones by multiplying the numerator and denominator by $2,3, \ldots, 8$. As a result, we get:
1) $\frac{14}{18}$ and $\frac{16}{18}$; between these fractions, there is one fraction $\frac{15}{18}$, which, after simplification, gives $\frac{5}{6}$.
2) $\frac{21}{27}$ and ... | \frac{5}{6};\frac{4}{5};\frac{6}{7};\frac{7}{8} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,565 |
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 $\frac{100}{b}=\frac{90}{c}$. T... | 19 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,566 |
78. "Just you wait!"-the wolf growled, noticing the hare 30 m away, he rushed after it when it had 250 m left to reach its hiding place. Will the wolf catch the hare if he runs 600 m per minute, while the hare runs 550 m per minute? | 78. The hare will run 250 m in $\frac{5}{11}$ min $\left(250: 550=\frac{5}{11}\right)$, while the wolf needs to run 280 m $(250+30=280)$ to reach the hiding place, which will take $\frac{7}{15}$ min $\left(280: 600=\frac{7}{15}\right)$. Let's compare $\frac{5}{11}$ and $\frac{7}{15}$. Since $\frac{5}{11}=\frac{75}{165}... | \frac{77}{165}>\frac{75}{165} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,568 |
79. Three geologists had to reach a station located 60 km from their base in 3 hours. Can they do this if they have a motorcycle at their disposal, on which no more than two can ride with a maximum speed of $50 \mathrm{km} / h$, while a pedestrian walks at a speed of 5 km/h? | 79. Two people get on a motorcycle and drive for 1 hour, while the third walks this hour behind them. After driving 50 km in 1 hour, one of the geologists walks the remaining 10 km on foot and arrives at the station on time $(10: 5=2,1+2=3)$. The second person drives the motorcycle to meet the third when they are 45 km... | 2\frac{46}{55} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,569 |
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.... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,571 |
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 "hu... | 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... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,572 |
84. The numbers $a, b$, and $c$ satisfy the following condition:
$\frac{b+c-a}{a}, \frac{a+c-b}{b}$, and $\frac{a+b-c}{c}$ are natural and equal to each other. Can
the expression $\frac{(a+b)(b+c)(a+c)}{a b c}$ equal $12$? | 84. From the condition, by term-by-term division, we get
$$
\frac{b}{a}+\frac{c}{a}-1=\frac{a}{b}+\frac{c}{b}-1=\frac{a}{c}+\frac{b}{c}-1 \text { or } \frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}
$$
Let $\frac{a+b}{c}=n$, then $\frac{(b+c)(a+b)(a+c)}{a b c}=n^{3}$, but $n^{3} \neq 12$, since 12 is not a cube of a natura... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,574 |
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 | 39,575 |
86. Decode the numerical riddle
$$
\overline{\text { SLOV, O }}+\overline{\text { SLOV, O }}=\overline{\text { PESNYA }}
$$
(Different letters correspond to different digits, and the same letters correspond to the same digits.) | 86. Let's write the addition in a column
Since the result is a whole number, then $O=5$. The letter П can only represent the digit 1, then $\mathrm{H}=0$. Since $\mathrm{C}>5$, we find by ... | 9453.5+9453.5=18907 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,576 |
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 | 39,578 |
2. Which of the points $A(0; -7)$ or $B(-4; 0)$ is farther: a) from the origin; b) from the point $C(-4; -7)$? | 2. a) Point $A$ is farther from the origin than point $B$;
b) Point $B$ is farther from point $C$. | )A,b)B | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,579 |
4. The product of the numbers in each row of the table
$$
\begin{array}{lll}
a_{1} & a_{2} & a_{3} \\
a_{4} & a_{5} & a_{6} \\
a_{7} & a_{8} & a_{9}
\end{array}
$$
is negative. What signs can the products of the numbers in the columns have? | 4. Since the products of the numbers in each row are negative, the product of all the numbers in the table is also negative. This means that the product of all numbers can be negative either in only one column or in all three; the products of the numbers can be positive only in two columns. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,581 |
5. Vitya and Misha argued: the first of them claimed that if $a=|b|$, then the equality $a+b=0$ is always true; the second, however, said that this equality is always false. Who is right? | 5. If $b \leqslant 0$, then $a+b=|b|+b=0$; if $b>0$, then $a+b=2 b \neq 0$. Therefore, neither of the boys is correct. | neither\of\the\boys\is\correct | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,582 |
7. We have 9 numbers: $-6,-4,-2,-1,1,2,3,4,6$. It is known that the sum of some of these numbers is -8. Write down these numbers. | 7. These numbers can be:
1) $-6,-2 ; 2)-6,-2,-1,1 ; 3)-6,-2,-4,4$;
2) $-6,-2,-1,1,4,-4 ; 5)-6,-4,2$;
3) $-6,-4,2,-1,1 ; 7)-6,-4,-2,-1,2,3$. | -6,-2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,584 |
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.
Si... | 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 | 39,585 |
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 | 39,586 |
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 | 39,589 |
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 | 39,590 |
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 | 39,591 |
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 | 39,592 |
17. In a $4 \times 4$ square, place numbers, not equal to zero, such that the sum of all numbers in the given square and in each $2 \times 2$ sub-square is equal to zero. | 17. The numbers are obviously the same in modulus but different in sign. The problem has infinitely many solutions. Here is one of them:
| 1 | -1 | 1 | -1 |
| :---: | :---: | :---: | :---: |
| 1 | -1 | 1 | -1 |
| -1 | 1 | -1 | 1 |
| -1 | 1 | -1 | 1 | | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,594 |
18. Is it possible to arrange 25 numbers in a $5 \times 5$ table of cells so that the sum of four numbers in each $2 \times 2$ square of cells is negative, while the sum of all numbers is positive? | 18. Arrange nine twos and 16 times -1 as follows:
| 2 | -1 | 2 | -1 | 2 |
| :---: | :---: | :---: | :---: | :---: |
| -1 | -1 | -1 | -1 | -1 |
| 2 | -1 | 2 | -1 | 2 |
| -1 | -1 | -1 | -1 | -1 |
| 2 | -1 | 2 | -1 | 2 | | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,595 |
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 resort... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,596 |
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 | 39,597 |
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 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,598 |
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:
| $a_{1}$ | $a_{2}$ | $a_{3}$ | $a_{4}$ | $a_{5}$ | $a_{6}$ | $a_{7}$ | $a_{8}$ | $a_{9}$ | $a_{10}$ | $a_{11}$ | $a_{12}$ | $a_{13}$ | $a_{14}$ | $a_{15}$ | $a_{16}$ | $a_{17}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,599 |
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 notic... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,600 |
24. Given 1989 numbers, each of which is 1 or -1. By calculating the sum of all these numbers, Ira got 846 as a result. Show that she made a mistake. Could the sum be 845? How many of the numbers would be 1 and -1 in this case? | 24. If the total sum is 846, this means there were 846 more 1s than -1s, since each time $1+(-1)=0$. Therefore, among the given numbers, -1 should occur $(1989-846): 2$ times. However, this number is not an integer, so the sum of all numbers cannot be 846. If the sum is 845, then -1 occurs $(1989-845): 2=572$ times, an... | 845,1417,572 | Number Theory | proof | Yes | Yes | olympiads | false | 39,601 |
25. Find two numbers if their sum, product, and quotient are equal to each other. | 25. $0.5+(-1)=0.5 \cdot(-1)=0.5:(-1)=-0.5$. | -0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,602 |
26. Find two integers such that their product equals the sum and difference of these numbers. | 26. Let the required numbers be $a$ and $b$. Then, by the condition, $a+b=ab$, from which $a=b:(b-1)$. This number is an integer only when $b=a=2$ or $a=b=0$. According to the condition, $ab=a-b$ as well, from which $b=a:(a+1)$, which gives an integer when $a=b=0$ or $b=2$ and $a=-2$. But in the second case, $a+b=0$, s... | =b=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,603 |
28. One of the students wrote on the board $1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9=21$, but instead of the asterisks, he put "+" or "-". The second student corrected several of the signs to their opposites and as a result got 20. Can we assert that at least one of the students made a mistake in the calculation? | 28. The sum of all numbers from 1 to 9 is 45. Since $45-21=24$, to get 21, the minus signs must be placed before numbers that sum up to 12. Such numbers, for example, are 2, 4, and 6. Then $1-2+3-4+5-$ $-6+7+8+9=21$. Changing the sign to the opposite changes the sum by no less than 2. Since $21-20=1<2$, one of the stud... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,605 |
29. Let the ratios $a: b$ and $c: d$ be equal, i.e., $a: b=c: d$. If the dividend $a$ is greater than the divisor $b$, then the dividend $c$ is greater than $d$. Take an arbitrary positive number $a$. Since $a:(-a)=-1$ and $(-a): a=-1$, we have $a:(-a)=(-a): a$. But $a>-a$, so based on the previous conclusion, from the... | 29. A rule that is valid for natural numbers $a, b, c$ and $d$ is improperly extended to negative numbers. For example, if $6: 3=(-6):(-3)=2$, then from $6>3$ it does not follow that $-6>-3$. | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,606 |
30. Establish the pattern in the arrangement of numbers in the sequence
$$
2,3,0,-2,-7,-14, \ldots
$$
and add two more numbers. | 30. Each number, starting from the third, is equal to the sum of the two preceding ones minus five. Thus $2+3-5=0, 0-2-5=-7$ and so on. Therefore, after the number -14 will follow the number $-7-14-5=-26$, and after it $-14-26-5=-45$. | -26,-45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,607 |
31. In the first line, three different integers are written: $a, b, c$. In the second line under them are three differences: $a-b, b-c, c-a$. The numbers in the third line are formed by the same rule and so on. Show that the number 855 cannot appear in any of the lines below the seventh. | 31. 32) $a ; b ; c$;
2) $a-b ; b-c ; c-a$;
3) $a-2 b+c ; b-2 c+a ; c-2 a+b$;
4) $3 c-3 b ; 3 a-3 c ; 3 b-3 a$;
5) $6 c-3 b-3 a$; $6 a-3 c-3 b ; 6 b-3 a-3 c$;
6) $9 c-9 a ; 9 a-9 b ; 9 b-9 c$;
7) $9 c-18 a+9 b ; 9 a-18 b+9 c ; 9 b-18 c+9 a$;
8) $27 b-27 a ; 27 c-27 b ; 27 a-27 c$.
The numbers in the eighth and all subs... | proof | Algebra | proof | Yes | Yes | olympiads | false | 39,608 |
32. $1,2,3, \ldots, n$ are represented in some arbitrary order and denoted as $a_{1}, a_{2}, \ldots, a_{n}$.
a) What is the sum $\left(a_{1}-1\right)+\left(a_{2}-2\right)+\ldots+\left(a_{n}-n\right)$?
b) What will be the product $\left(a_{1}-1\right) \cdot\left(a_{2}-2\right) \cdot \ldots \cdot\left(a_{n}-n\right)$ -... | 32. a) $\left(a_{1}-1\right)+\left(a_{2}-2\right)+\ldots+\left(a_{n}-n\right)=\left(a_{1}+a_{2}+\ldots+a_{n}\right)-(1+2+\ldots+n)=0$;
б) the product will be even if at least one of the factors is even. From the given $n$ numbers, when $n$ is odd, there will be one more odd number than even numbers. Therefore, if we f... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,609 |
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 | 39,610 |
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 | 39,611 |
3. The sum of two numbers is 13,5927. If in the larger of these numbers, the decimal point is moved one digit to the left, the result is the smaller number. Find these numbers. | 3. One of the sought numbers is 10 times larger than the other. Let the smaller number be $x$, then the larger one is $10 x$. We have $10 x + x = 13.5927$, which means $11 x = 13.5927$, from which $x = 13.5927 : 11 = 1.2357$. Therefore, the smaller number is 1.2357, and the larger one is 12.357. | 1.235712.357 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,612 |
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 | 39,613 |
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 | 39,614 |
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 plan... | 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 | 39,615 |
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 | 39,618 |
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 | 39,619 |
11. Vera bought 6 fewer notebooks than Misha and Vasya together, and Vasya bought 10 fewer notebooks than Vera and Misha together. How many notebooks did Misha buy? | 11. Let Vera bought $a$ notebooks, Vasya - $b$, and Misha - $c$. Then, according to the condition, $a+6=b+c$ and $b+10=a+c$. Adding these equations term by term, we get $a+b+16=a+b+2c$ from which $2c=16$, hence $c=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,620 |
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 | 39,621 |
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 divide... | 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 | 39,623 |
15. How many people are in the team if the average age of all team members is 25 years, the team leader is 45 years old, and the average age of the team members without the team leader is 23 years? | 15. Let there be $x$ people in the brigade, then we have $25 x=23(x-1)+45$, from which $2 x=22$ and $x=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,624 |
16. On one plot of 14 hectares, unsorted potatoes were planted, and on another plot of 4 hectares, sorted potatoes were planted. The average yield per hectare on these two plots is 20 centners higher than the average yield of the first plot. By how many centners is the yield of sorted potatoes higher than the yield of ... | 16. Let the average yield of non-variety potatoes be $x$ centners per hectare, and that of variety potatoes be $-y$ centners per hectare. Then the average yield of the two plots is $(x+20)$ centners per hectare. We have
$$
(x+20) \cdot 18=x \cdot 14+y \cdot 4
$$
from which $4 y-4 x=360 ; y-x=90$, which means that the... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,625 |
17. Helping the collective farm with the harvest, 5th Grade A class collected 45715 kg of potatoes over 5 days. How many students were there in total, and how much did each student collect on average per day, if the average daily output of one student is a whole number of kilograms? | 17. Let there be $x$ students, and the average daily output of a student be $y$ kg. Then $y \cdot x \cdot 5=45715$, from which $x y=9143$. Since $9143=41 \cdot 223$ and there cannot be 223 students in a class, there were 41 students, and each of them collected on average 223 kg of potatoes daily. | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,626 |
19. In a $3 \times 3$ square, the cells were filled with numbers $1,2,3, \ldots, 9$. Then, in each circle (Fig. 2), the arithmetic mean of the four surrounding numbers was written. After that, the arithmetic mean of the four obtained numbers was calculated. What is the largest number that can result from this? | 19. The largest number will be obtained in the case where the circles contain the largest possible arithmetic means. When determining these arithmetic means, the numbers in the corner cells are taken once, the number in the central cell is taken four times, and the numbers in the middle cells adjacent to the sides of t... | 6.125 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,628 |
20. The surname of a Russian writer consists of six letters. It is known that the numbers indicating the positions of these letters in the Russian alphabet are in the following ratios: the first is equal to the third; the second is equal to the fourth; the fifth is 9 more than the first; the sixth is 2 less than the su... | 20. Let the first number be $x$, then the second is $(3 x+4)$, the third is $x$, the fourth is $(3 x+4)$, the fifth is $(x+9)$, and the sixth is $(6 x+6)$. According to the condition
$$
x+(3 x+4)+x+(3 x+4)+(x+9)+(6 x+6)=83
$$
from which $x=4$. The fourth letter of the alphabet is $\Gamma ;(3 x+4=16)$ the sixteenth le... | GOGOL | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,629 |
21. When the young fishermen were asked how many fish each of them had caught, the first one answered: "I caught half the number of fish that my friend caught, plus 10 fish." The second one said: "And I caught as many as my friend, plus 20 fish." How many fish did the fishermen catch? | 21. Let the second one catch $x$ fish, then the first one catches $\frac{1}{2} x+10$ fish. According to the second fisherman's statement, he caught $\left(\frac{1}{2} x+10\right)+20$ fish. We have the equation $x=\left(\frac{1}{2} x+10\right)+20$, from which $x=60$, i.e., the second one caught 60 fish. Then the first o... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,630 |
22. Yura left the house for school 5 minutes later than Lena, but walked twice as fast as she did. How many minutes after leaving will Yura catch up to Lena? | 22. If Lena walks $a$ m in 1 min, then Yura walks $2a$ m. Let him catch up with her after $x$ min from his start. Then we have $a \cdot 5 + a \cdot x = 2a \cdot x$, from which $x=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,631 |
23. 25 oranges cost as many rubles as can be bought for 1 p. How many oranges can be bought for 3 p.?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 23. Let 25 oranges cost $x$ r., then for 1 r. one can buy $x$ oranges. One orange costs $1: x=x: 25$, from which $x=5$. Therefore, for 3 r. one can buy 15 oranges. | Number Theory | proof | Yes | Yes | olympiads | false | 39,632 | |
24. Preparing his son for school, the father bought him a new uniform for 24 r. How did he pay for the purchase, if he only had 5 r bills, and the cashier only had 3 r bills? | 24. If the buyer gave the cashier $x$ bills and received $y$ bills as change, we have the equation $5 x-3 y=24$, from which $x=\frac{24+3 y}{5}$. Since $x$ must be an integer, $24+3 y$ must be divisible by 5. This is possible with the smallest value of $y=2$. Therefore, the buyer gave 6 bills (30 p.) and received 6 p. ... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,633 |
26. In a competition of fun and ingenuity, 9 points were awarded for each correctly completed task, and 5 points were deducted for each uncompleted or incorrectly completed task. It is known that the team was offered no more than 15 tasks and scored 57 points. How many tasks did the team complete correctly? | 26. Let $x$ be the number of tasks completed correctly, and $y$ be the number of tasks completed incorrectly or not at all. Then we have $9 x-5 y=57$, from which $x=\frac{57+5 y}{9}$. Since $x$ is a natural number, $(57+5 y)$ must be divisible by 9. The smallest value of $y$ for which this condition is met is 3. In thi... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,635 |
27. If the team had scored 97 points in the last sport, the average number of points scored by it in one sport would have been 90; if it had scored 73 points in this last sport, the average number of points per sport would have been 87. In how many sports did the team compete? | 27. Let the team compete in $x$ sports and score $n$ points in all sports except the last one. Then we have
$$
(97+n): x=90 ;(73+n): x=87
$$
which means $97+n=90 x$ and $73+n=87 x$, so $x=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,636 |
28. In response to a question about his age, the grandfather answered: “The number expressing my age in years is a two-digit number equal to the sum of the number of its tens and the square of its units.” How old is the grandfather? | 28. Let the grandfather be $(10 a+b)$ years old. Then $10 a+b=a+b^{2}$, hence
$$
9 a=b^{2}-b, \text{ or } a=\frac{b(b-1)}{9}
$$
Since $a$ and $b$ are single-digit numbers, the equality holds only when $b=9$. In this case, $a=8$. Therefore, the grandfather is 89 years old. | 89 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,637 |
29. In the parking lot, there were passenger cars and motorcycles. The number of motorcycles with sidecars was half the number of those without sidecars. What could be the maximum number of cars if the total number of wheels on these cars and motorcycles was $115?$ | 29. If the number of motorcycles with sidecars is $x$, then the number without sidecars is $2x$. Let the number of cars be $y$, then the total number of wheels is $3x + 2x \cdot 2 + 4y = 7x + 4y$. According to the condition, $7x + 4y = 115$, from which
$$
y=\frac{115-7x}{4}
$$
The value of $y$ is greatest when $x$ is... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,638 |
30. Several girls were picking mushrooms. They divided the collected mushrooms among themselves as follows: one of them was given 20 mushrooms and 0.04 of the remainder, another - 21 mushrooms and 0.04 of the new remainder, the third - 22 mushrooms and 0.04 of the remainder, and so on. It turned out that everyone recei... | 30. Let the girls collected $x$ mushrooms. The first one received $20+0.04(x-20)$, the second one received $21+0.04(x-21-20-0.04(x-20))$ mushrooms. According to the condition $20+0.04 x-$ $-0.8=21+0.04(x-21-20-0.04(x-20))$, from which
$$
0.0016 x=0.192 \text { and } x=120 .
$$
Thus, 120 mushrooms were collected, each... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,639 |
31. When Vitya becomes twice as old, Kolya will be 4 years younger than Vera. Last year, Kolya was half as old as Vera, and 3.5 times younger than Vitya. How old is each of them now? | 31. Let Kolya be $x$ years old last year, then Vera is $2 x$ and Vitya is $3.5 x$. Now Kolya is $(x+1)$ years old, Vera is $(2 x+1)$ years old, and Vitya is $(3.5 x+1)$ years old. Vitya will be twice as old in $(3.5 x+1)$ years, then Kolya will be $(4.5 x+2)$ years old $((x+1)+$ $+(3.5 x+1)=4.5 x+2)$, and Vera will be ... | Kolya:5,Vera:9,Vitya:15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,640 |
32. When shooting at a target, Misha hit the bullseye several times, scored 8 points as many times, and hit the five several times. In total, he scored 99 points. How many shots did Misha take if $6.25\%$ of his shots did not score any points? | 32. Let's say Mishka hit the ten and the eight $x$ times each, and the five $y$ times. Then $10 x + 8 x + 5 y = 99$, or $18 x + 5 y = 99$. The term $5 y$ can end in 5 or 0, so $18 x$ must end in 4 or 9. But $18 x$ cannot end in 9, so $18 x$ must end in 4. Since $18 x < 99$, then $x = 3$. We have $18 \cdot 3 + 5 y = 99$... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,641 |
33. In a constructed school boarding house, several rooms were double-occupancy, and the rest were triple-occupancy. In total, the boarding house is designed for more than 30 but fewer than 70 places. When $\frac{1}{5}$ of the available places were occupied, the number of unoccupied places turned out to be the same as ... | 33. Let there be $x$ two-person rooms and $y$ three-person rooms in the boarding school. Then the total number of places in the boarding school is $2 x + 3 y$. If $\frac{1}{5}$ of the places were occupied, then $\frac{4}{5}$ of the places remained unoccupied, i.e., $\frac{4}{5}(2 x + 3 y)$. According to the condition, ... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,642 |
34. Vova is as many times older than his sister Galya as he is younger than his grandmother Katya. How old is each of them, if Galya is not yet 6 years old, and Vova together with his grandmother is already 112 years old? | 34. Let Vova be $x$ times older than Gala. If Gala is 5 years old, then Vova is $5 x$, and Grandma Katya is $5 x \cdot x=5 x^{2}$. According to the condition, it should be $5 x+5 x^{2}=112$. But this is impossible, since 112 is not divisible by 5. Gala cannot be 3 years old either, since 112 is not divisible by 3. She ... | Gala:2,Vova:14,Katya:98 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,643 |
35. The fish soup prepared by the tourists on the seashore after a successful fishing trip was good, but due to lack of culinary experience, they put too little salt in it, so they had to add salt at the table. The next time they made the same amount of fish soup, they put in twice as much salt as the first time, but e... | 35. Let the amount of salt that should have been added the first time be $a$ g, and the amount actually added be $x$ g, resulting in a shortage of $(a-x)$ g. Another time, $2x$ g of salt was added, leading to a shortage of $a-2x$. Since the amount of salt added the second time was half of what was added the first time,... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,644 |
36. The front tires of a car wear out after 25,000 km, while the rear tires wear out after 15,000 km. When is it advisable to swap the tires to ensure they wear out evenly? (Assume that the tires are swapped once, although drivers do this more frequently in practice.) | 36. It is advisable to change the tires so that their wear occurs at the same mileage for all wheels. Let's say they should be changed at $x$ km of mileage, then the front wheels have $(25000-x)$ km left to wear. But if they are put in place of the rear wheels, they can only travel $\frac{15000}{25000}=\frac{3}{5}$ of ... | 9375 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,645 |
37. Two people, who have one bicycle, need to get from point $A$ to point $B$, which is 40 km away from $A$. The first person walks at a speed of 4 km/h and rides the bicycle at 30 km/h, while the second person walks at a speed of 6 km/h and rides the bicycle at 20 km/h. What is the minimum time they can take to reach ... | 37. Let the first person travel $x$ hours by vehicle and $y$ hours on foot, then the second person travels $y$ hours by vehicle and $x$ hours on foot. We have the equation $30 x + 4 y = 20 y + 6 x$, from which $x = \frac{2}{3} y$. Therefore, the first person walked $y$ hours and traveled $\frac{2}{3} y$ hours by vehicl... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,646 |
38. While doing his homework, Petya was in a hurry to get to football and, as always in such cases, made mistakes. Instead of squaring the number, he doubled it and got a two-digit number, written with the same digits as the required square, only in reverse order. What is the correct answer that Petya should have obtai... | 38. Let Pete need to square the number $x$. He should have obtained $x^{2}=10 a+b$, but in fact, he got $2 x=10 b+a$. This means $x=\frac{10 b+a}{2}$. From this, it follows that the number $x$ is a single-digit number (since $x^{2}$ is a two-digit number), and the number $10 b+a$ is an even two-digit number. The number... | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,647 |
39. The sum of four consecutive odd numbers is equal to the cube of a single-digit number. Find all such quartets of numbers. | 39. The sum of four consecutive odd numbers is an even number, so single-digit numbers should be sought among the numbers $2,4,6,8$.
1) $2^{3}=8 ;(2 x-3)+(2 x-1)+(2 x+1)+(2 x+3)=8$, from which $8 x=8 ; x=1-$ does not satisfy the condition, since $2 x-3<0$.
2) $4^{3}=64 ; 8 x=64$, from which $x=8 ; 13+15+17+19=4^{3}$;
3... | 13,15,17,19;51,53,55,57;125,127,129,131 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,648 |
40. The sum of three consecutive odd numbers is equal to the fourth power of a single-digit number. Find all such triples of numbers. | 40. The sum of three consecutive odd numbers is an odd number, and if the fourth power is an odd number, then the base is also odd. Therefore, single-digit numbers should be sought among the numbers $3,5,7,9$.
1) $(2 x-1)+(2 x+1)+(2 x+3)=3^{4}, 6 x+3=3^{4}$, from which $2 x+1=3^{3}$ and $x=$ $=13 ; 2 x-1=25 ; 2 x+1=27 ... | 25,27,292185,2187,2189 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,649 |
41. On one of the three cards, the number 18 was written, on another - the number 75, and on the third - some two-digit number. If you add up all the different six-digit numbers that can be formed by arranging these cards in a row, the result is 2606058. What number is written on the third card? | 41. Let the number on the third card be the one with $a$ tens and $b$ units, i.e., $\overline{a b}$. Then, by arranging the cards in a row, we can obtain the following numbers:
$\overline{1875 a b}, \overline{18 a b 75}, \overline{7518 a b}, \overline{75 a b 18}, \overline{a b 7518}, \overline{a b 1875 .}$
We can rep... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,650 |
42. Ten identical books cost no more than 11 p., while 11 such books cost more than $12 \mathrm{p}$. How much does one book cost? | 42. Let one book cost $x$ k. Then we have $x \cdot 10 \leqslant 1100$ and $x \cdot 11 > 1200$ or $x \leqslant 110$ and $x > 109 \frac{1}{11}$. Therefore, $x=110$ k., i.e., the book costs 1 p. 10 k. | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 39,651 |
44. At a mathematics evening, there were students from grades V and VI. Together, they were more than 70 but less than 90. The total number of rows of chairs in which the students sat in the hall was 3 more than the number of sixth-graders in each row, and the fifth-graders sat 3 in each row. How many sixth-graders and... | 44. If there were $x$ rows, then in each of them there were $(x-3)$ sixth graders. Since there were 3 fifth graders in each row, there were $x$ students in each row. Then the total number of students at the evening was $x \cdot x=x^{2}$. According to the condition $x^{2}>70$ and $x^{2}<90$. This condition is satisfied ... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 39,653 |
45. Four combines of brand "K" and three combines of brand "N" harvest as much in 5 days as three combines of brand "K" and five of brand "N" do in 4 days. Which brand of combine harvests more in a day? | 45. Let a combine harvester of brand "K" harvest $x$ hectares in one day, and a combine harvester of brand "N" harvest $y$ hectares. Then we can write the equation $5(4 x+3 y)=4(3 x+5 y)$, from which $8 x=5 y$. From the last equation, it is clear that $y>x$. Answer. A combine harvester of brand "N" harvests more in a d... | y>x | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,654 |
46. During the holidays for eight classes of the school, each with the same number of students, a trip to the museum was organized, and for the rest of the students in the school, who turned out to be 15% more, - a cultural trip to the puppet theater. How many students are there in the school, if it is known that there... | 46. Let in one of the eight classes that participated in the excursion, there were $x$ students, then the museum was visited by $8 x$ students. By the condition $8 x>230$, hence $x>28 \frac{3}{4}$. The puppet theater was visited by $8 x+x \cdot 0.15=9.2 x$. Therefore, the total number of students in the school is $8 x+... | 516 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 39,655 |
47. When summarizing the results of the scrap metal collection competition, it turned out that 6A class collected more scrap metal than 6B and 6V combined, but 6A and 6B together collected as much as 6V and 6G. Moreover, 6B and 6G collected more than 6A and 6V. How did the classes place in the competition? | 47. Let 6 A collect a kg, 6 B-b kg, 6 V-v kg, and 6 G-g kg. Then, according to the conditions:
1) $\mathrm{a}>6+\mathrm{B}$
2) $6+r>a+B$
3) $a+\sigma=B+r$.
Subtracting the left side of the third equation from the left side of the second inequality, and the right side from the right side, we get $r-a>a-r$, from which $... | 6G,6A,6B,6V | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 39,656 |
51. Prove that $\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \frac{7}{8} \cdot \ldots \cdot \frac{99}{100}<\frac{1}{10}$.
## Search for solutions
Consider the product \( P = \frac... | 51. Let's take the obvious equality $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdots \frac{98}{99} \cdot \frac{99}{100}=\frac{1}{100}$. Since $\frac{2}{3}>\frac{1}{2}$; $\frac{4}{5}>\frac{3}{4}$, and so on, by replacing $\frac{2}{3}$ with $\frac{1}{2}$; $\frac{4}{5}$ with $\frac{3}{4}$, and so ... | proof | Inequalities | proof | Yes | Yes | olympiads | false | 39,659 |
1. We have 6 sticks that are $1 \mathrm{~cm}$ long, 3 sticks that are $2 \mathrm{~cm}$ long, 6 sticks that are $3 \mathrm{~cm}$ long, and 5 sticks that are $4 \mathrm{~cm}$ long. Is it possible to form a square using all the sticks without breaking them and without overlapping them? | 1. The perimeter of the square should be equal to $1 \mathrm{~cm} \cdot 6+2 \mathrm{~cm} \cdot 3+3 \mathrm{~cm} \cdot 6+$ $+4 \mathrm{~cm} \cdot 5=50 \mathrm{~cm}$, so the side should be 50 cm:4. As can be seen, the length of the side is not expressed as an integer, so it is impossible to form a square. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,660 |
2. Find the length of the side of a square whose area is numerically equal to its perimeter. | 2. Let the side length of the square be $a$, then the area $S=a \cdot a$, and the perimeter $P=4 a$. According to the condition $a \cdot a=4 a$, from which $a=4$. | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,661 |
3. Does there exist a 1989-gon such that all its sides can be intersected by a straight line at points other than the vertices? | 3. If such polygons and a straight line existed, there would be an equal number of vertices on both sides of the line, which is impossible since 1989 is not divisible by 2. | proof | Geometry | proof | Yes | Yes | olympiads | false | 39,662 |
4. On one of the sides of a triangle, 60 points are taken, and on another - 50. (These points are distinct from the vertices of the triangle.) Each vertex at the base is connected by straight line segments to the points taken on the opposite side. How many intersection points of these segments are formed in this case? | 4. Each of the 60 segments drawn from one vertex intersects each of the 50 segments drawn from another vertex, so we get 3000 such points $(50 \cdot 60=$ $=3000)$. | 3000 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 39,663 |
10. Help the gardener find a point in the garden, which has the shape of a quadrilateral, such that the sum of the distances from this point to the vertices of the quadrilateral is the smallest. | 10. Let the garden be represented in Figure 18. Draw the diagonals of this quadrilateral, and let their intersection point be $M$. Take any point $K$ inside the quadrilateral, different from point $M$, and connect it to the vertices. Since $A C<A K+C K$ and $B D<B K+D K$, then $A C+B D<A K+D K+C K+B K$ or $A M+C M+B M+... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,669 |
11. A school plot of rectangular shape has a perimeter of $160 \mathrm{m}$. How will its area change if the length of each side is increased by 10 m? | 11. Let the length of the plot be $x$ m, then its width is $(80-x)$ m (Fig. 19) (the semi-perimeter is $160: 2=80$). If $A D=x$, then $C D=80-x ; C P=10$ m and $E K=$ $=10$ m. The area of the plot will increase by an amount equal to the sum of the areas of the rectangles $A D E M$ and $C E K P$, i.e., by $x \cdot 10+(8... | 900 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,670 |
12. A wooden cube, with an edge length of $5 \mathrm{~cm}$, is painted and then cut into cubes with edges of 1 cm. How many of the resulting cubes will have three, two, and one painted face? | 12. With three painted faces, we get 8 cubes (according to the number of vertices of the given cube), with two - \(3 \cdot 12=36\), with one - \(9 \cdot 6=54\) and none - \(3 \cdot 3 \cdot 3=27\)
:(2 \cdot 2 \cdot 2)=15625000$. Therefore, the lengt... | 31 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,672 |
14. A cube with an edge length of $n$ was cut into cubes, each with an edge length of 1. The total volume of the resulting cubes will obviously remain the same, but the surface area will undoubtedly increase. By what factor? | 14. As a result of cutting, we get $n \cdot n \cdot n$ cubes. The surface area of the original cube is $6 n \cdot n$, the surface area of each resulting cube is 6, and the total surface area of all such cubes is $6 n \cdot n \cdot n$. Therefore, the surface area increases by $n$ times. | n | Geometry | math-word-problem | Yes | Yes | olympiads | false | 39,673 |
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