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Example 78. A triangular pyramid is cut by a plane into two polyhedra. We will find the ratio of the volumes of these polyhedra, given that the cutting plane divides the edges converging at one vertex of the pyramid in the ratio $1: 2, 1: 2, 2: 1$, counting from this vertex.
Construction of the image. Let the quadrila... | Solution. Let $V$ be the volume of the pyramid $SABC$, and $V_{1}$ the volume of the pyramid $PAQR$. Then $V_{1}=V-V_{2}$.
Construct $[SO]$ and assume that $[SO]$ is the image of the height of the pyramid $SABC$ (thus, two parameters are used). Construct $(AO)$ and $[PM] \|[SO]$. Then $M \in (AO)$. To simplify the cal... | 25:2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,589 |
Example 79. A sphere touches the base of a regular triangular pyramid $S A B C$ at point $B$ and its lateral edge $S A$. Let's find the radius of the sphere, if $|A B|=a$ and $|S A|=b$.
Construction of the image. Let the quadrilateral $S A B C$ with its diagonals (Fig. 88) be the image of the given pyramid. This image... | S o l u t i o n. First of all, note that if we have the image of the center of the sphere, point $P$, then $|P B|$ will be the desired length of the radius of the sphere $\Omega$. Thus, having an image of the sphere itself is not necessary for calculating its radius. We will construct point $P$ based on the following c... | \frac{\sqrt{3}(2b-)}{2\sqrt{3b^{2}-^{2}}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,590 |
Example 80. Given a cube $A B C D A_{1} B_{1} C_{1} D_{1}$, with an edge length of $a$. A sphere $\Omega$ is drawn through the vertices $A$ and $C$ and through the midpoints of the edges $B_{1} C_{1}$ and $C_{1} D_{1}$. Let's find the radius of this sphere.
Construction of the image. Let the figure $A B C D A_{1} B_{1... | The solution is as follows. First of all, let's find the point $M$ - the center of the sphere $\Omega$. Since $A \in \Omega$ and $C \in \Omega$, the point $M$ belongs to the set of points equidistant from points $A$ and $C$. It is not difficult to guess that this set of points is the diagonal plane $B B_{1} D_{1} D$. S... | \frac{\sqrt{41}}{8} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,591 |
Example 81. A trihedral angle is formed by the planes $\alpha$, $\beta$, and $\gamma$, with $\alpha \perp \gamma, \beta \perp \gamma$, and $\widehat{\alpha}=2 \varphi$. The sphere $\Sigma$ touches the plane $\gamma$ at point $B$, and intersects the planes $\alpha$ and $\beta$ along congruent circles $\omega_{1}$ and $\... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,592 | |
1. In a box, there are 7 red and 4 blue pencils. Without looking into the box, a boy randomly draws pencils. What is the minimum number of pencils he needs to draw to ensure that there are both red and blue pencils among them? | 1. Answer: 8 pencils. | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,593 |
3. Can 54 notebooks be divided into three piles so that each pile has an odd number of notebooks? | 3. It is impossible, otherwise it would mean that the sum of three odd numbers equals the even number 54. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,595 |
8. Calculate: $(3 \cdot 5 \cdot 7+4 \cdot 6 \cdot 8)(2 \cdot 12 \cdot 5-20 \cdot 3 \cdot 2)$. | 8. The difference in the second parenthesis is zero, and therefore the entire expression is zero. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,600 |
9. Anya and Vanya are climbing the stairs of a multi-story building. Anya is going to the second floor, while Vanya is going to the sixth floor. How many times longer is Vanya's path?
Puc. 1
 | 9. Anya needs to climb the height of one floor; while Vanya needs to climb the height of five floors, i.e., Vanya's path will be five times longer. | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,601 |
10. Write the number 7 using five twos and arithmetic operation signs. Find several solutions. | 10. 11) $2 \cdot 2 \cdot 2-2: 2$; 2) $2+2+2+2: 2$; 3) $22: 2-2 \cdot 2$.
10. 11) $2 \cdot 2 \cdot 2-2 \div 2$; 2) $2+2+2+2 \div 2$; 3) $22 \div 2-2 \cdot 2$. | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,602 |
29. Calculate: $1111 \cdot 1111$. | 29. Mentally perform the multiplication of the given numbers "in a column". A n s w e r. 1234321. | 1234321 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,611 |
30. How many zeros does the product of all integers from 1 to 30 inclusive end with?
## Problems for the sixth grade | 30. Zeros will result from multiplying by $5, 10, 15, 20$ and 30.
Multiplying by 25 will add two zeros. Thus, the entire product will end with seven zeros. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,612 |
31. Name the number consisting of 11 hundreds, 11 tens, and 11 units. | 31. $1100+110+11=1221$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
31. $1100+110+11=1221$. | 1221 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,613 |
32. There are different numbers of nuts in two piles. Is it always possible to move some nuts from one pile to the other so that the number of nuts in both piles becomes equal? | 32. No, not always. It is necessary for the total number of nuts in the two piles to be even. | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,614 |
33. The difference between a three-digit number and an even two-digit number is 3. Find these numbers. | 33. The desired even two-digit number can only be 98. Consequently, the three-digit number is 101. | 101 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,615 |
34. Name three integers whose product is equal to their sum.
35. Find $12.5\%$ of the number 44. | 34. Students usually name the triplet of numbers $1,2,3$. Indeed, $1 \cdot 2 \cdot 3=1+2+3$. However, the condition of the problem is also satisfied by the triplets of numbers $-1,-2,-3$ and $-n, 0, n$. More points should be given for each of the last two triplets than for the first triplet. | -n,0,n | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,616 |
36. Name a fraction greater than $\frac{11}{23}$ but less than $\frac{12}{23}$. | 36. The fractions can be represented as $\frac{22}{46} \frac{24}{46}$. Then, the fraction $\frac{23}{46}=\frac{1}{2}$ satisfies the condition of the problem. | \frac{1}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,617 |
38. Find the smallest natural number that, when increased by 1, would be divisible by $4, 6, 10, 12$. | 38. This number will be the least common multiple of the given numbers, decreased by 1. Answer. 59. | 59 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,619 |
39. All odd numbers from 27 to 89 inclusive are multiplied together. What is the last digit of the resulting product? | 39. The product ends with the digit 5. Indeed, if in the multiplication of several odd numbers at least one factor ends with the digit 5, then the entire product ends with the digit 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,620 |
40. What is one and a half thirds of 100? | 40. One and a half thirds is a third plus half of a third, i.e. $\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$. Answer. 50. | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,621 |
41. Can 5 rubles be exchanged into coins of 1, 2, 3, and 5 kopecks in such a way that there are an equal number of each type of coin? | 41. No, it is not possible, since $1+2+3+5=11$, and 500 is not divisible by 11. | No | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,622 |
43. An adult needs 16 kg of salt for consumption over 2 years and 8 months. How much salt does an adult need for one year? | 43. 2 years 8 months make up 32 months. If 32 months require 16 kg of salt, then 12 months require 6 kg of salt. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,624 |
44. Is it possible for the sum and product of three natural numbers to both be prime numbers? | 44. It can be. For example: $3 \cdot 1 \cdot 1=3,3+1+1=5$. | 3\cdot1\cdot1=3,3+1+1=5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,625 |
46. Calculate: $79 \cdot 81+1$. | 46. $79 \cdot 81+1=(80-1)(80+1)+1=80^{2}-1+1=6400$.
46. $79 \cdot 81+1=(80-1)(80+1)+1=80^{2}-1+1=6400$.
(Note: The original text and the translation are identical as the content is a mathematical expression which is universal and does not change in translation.) | 6400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,627 |
47. Cut a square into two equal trapezoids. | 47. The line of the cut passes through the center of symmetry of the square and divides the square into two equal rectangular trapezoids (Fig. 5). | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,628 |
48. Can each of the mean terms in a proportion be less than each of the extremes? | 48. It can be. For example: $1:(-2)=(-2): 4$. | 1:(-2)=(-2):4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,629 |
51. Calculate: $\quad\left(5 \frac{1}{10}\right)^{2}$. | 51. $\left(5 \frac{1}{10}\right)^{2}=\left(5+\frac{1}{10}\right)^{2}=25+1+\frac{1}{100}=26.01$. | 26.01 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,632 |
52. Divide the number 10 into two parts, the difference between which is 5. | 52. A n s w e r. 7.5 and 2.5 . | 7.52.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,633 |
53. Prove that among 11 natural numbers, there will definitely be two such that their difference is divisible by 10. | 53. Since there are only 10 different digits, among 11 natural numbers there will be at least two that end in the same digit. The difference between such numbers is divisible by 10. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,634 |
54. A boy has 25 copper coins. Will there be at least 7 coins of the same denomination among them? | 54. Copper coins come in denominations of one, two, three, and five kopecks. If there were no more than six coins of each denomination, there would be no more than 24 coins in total. Therefore, there will be at least seven coins of the same denomination. | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 44,635 |
55. What type of triangle is it if the sum of any two of its angles is greater than $90^{\circ}$? | 55. If the sum of two angles of a triangle is greater than $90^{\circ}$, then the third angle is less than $90^{\circ}$, i.e., it will be acute. Since this is the case for each angle of the triangle, the given triangle is acute-angled. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,636 |
56. Prove that the sum of any two consecutive odd numbers is divisible by 4. | 56. The sum of two consecutive odd numbers can be represented as $(2 n+1)+(2 n+3)=4(n+1)$, which proves the statement of the problem. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,637 |
57. Prove that for any integer $n$ the number $n^{2}+n+1$ is odd. | 57. Of two consecutive integers, one is even. Therefore, the number $n(n+1)=n^{2}+n$ will be even for any integer $n$. But then the number $n^{2}+n+1$ will be odd. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,638 |
59. A number was increased by $25 \%$. By what percentage does the resulting value need to be decreased to return to the original number?
10 | 59. Answer. By $20 \%$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The note is for your understanding and should not be included in the output. Here is the correct format:
59. Answer. By $20 \%$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,640 |
60. Which numbers are more numerous among the first 100 natural numbers: those divisible by 3, or those divisible by at least one of the numbers 5 or 7?
## Problems for the eighth grade | 60. Among the first 100 natural numbers, there are 33 numbers divisible by 3, 20 numbers divisible by 5, and 14 numbers divisible by 7. Since the numbers 35 and 70 were counted both as divisible by 5 and by 7, the total number of numbers that are divisible by at least one of the numbers 5 or 7 will be $20+14-2=32$. The... | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,641 |
61. Insert arithmetic operation signs between the digits 12345 so that the result equals 1. | 61. There are several solutions. For example: $1+2-3-$ $-4+5=1 ; 1-2+3+4-5=1$. | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,642 |
64. My friend's name is Petr Taev. If you rearrange the letters in his first and last name to form one word, it can become the name of my friend's profession. What is this profession? | 64. Answer. Therapist. | Therapist | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,644 |
65. Can a quadrilateral have three acute angles? | 65. Yes. For example, a quadrilateral in which three angles are $80^{\circ}$ each and one is $120^{\circ}$. | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,645 |
66. Can a quadrilateral have three obtuse angles? | 66. Yes. For example, a quadrilateral in which three angles are $100^{\circ}$ each and one is $60^{\circ}$. | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,646 |
68. Find the number which, after being increased by one fifth of itself and then by one, gives 10. | 68. We have: $x+0.2x+1=10$, from which $x=7.5$. | 7.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,648 |
69. Factorize the number 899. | 69. We have: $899=900-1=(30-1)(30+1)=29 \cdot 31$. | 29\cdot31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,649 |
70. The product of two two-digit numbers is 777. Find these numbers. | 70. We have: $777=7 \cdot 111=7 \cdot 37=21 \cdot 37$. | 21\cdot37 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,650 |
71. Does the equation $2 x+6 y=91$ have solutions in integers? | 71. Does not have, since the left side of the equation will be an even number for any integers $x$ and $y$, while the right side is an odd number. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,651 |
72. Diagonals of two faces are drawn from the same vertex of a cube. Find the magnitude of the angle between these diagonals. | 72. By connecting the ends of the diagonals with a segment, we obtain an equilateral triangle (Fig. 7). Therefore, the angle between the specified diagonals contains \(60^{\circ}\).
## Fig. 7
?
Fig. 2
 | 73. Since the diameter of the large semicircle is equal to the sum of the diameters of the small semicircles, the length of the large semicircle is also equal to the sum of the lengths of the small ones. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,653 |
74. From a bag, $60 \%$ of the flour it contained was taken, and then $25 \%$ of the remainder. What percentage of the original amount of flour is left in the bag? | 74. After taking $60 \%$ of the flour from the bag, $40 \%$ remained. When $25 \%$ of the remainder, i.e., a quarter of it, was taken, $30 \%$ of the original amount of flour remained in the bag. | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,654 |
75. Can the sum of four consecutive integers be a prime number?
## Problems for the ninth grade | 75. The sum of four consecutive integers is an even number. There is only one even prime number. This number is 2. Since $-1+0+1+2=2$, the answer to the question of the problem is affirmative. | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,655 |
80. A perfectly smooth sphere, having the same size as the Earth, is encircled by a wire at the equator. This wire was extended by one meter and arranged so that an equal gap was formed between the wire and the surface of the sphere. Would a mouse be able to squeeze through the resulting gap? | 80. If the circumference of a circle increases by 1 m, then the radius of the circle will increase by $1: 2 \pi \approx 16 \text{ cm}$, i.e., the gap formed will be large enough for not only a mouse but also a cat to pass through. | 16 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,660 |
81. Come up with a statement that is true only for the numbers 2 and 5. | 81. For example: $(x-2)(x-5)=0$. | (x-2)(x-5)=0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,661 |
82. There is a water tap and two containers: a three-liter and a five-liter. How can you get 4 liters of water in the larger one? | 82. Let's fill a five-liter container and pour 3 liters from it into a three-liter container. Then we will empty the three-liter container and transfer the 2 liters of water remaining in the five-liter container into it. Finally, we will fill the five-liter container from the tap and pour 1 liter from it into the three... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,662 |
83. Compare the cube of the hypotenuse with the sum of the cubes of the legs. | 83. If \(a, b, c\) are the legs and hypotenuse of a right triangle, respectively, then \(c^{2}=a^{2}+b^{2}\). Then \(c^{3}=c a^{2} + c b^{2} > a^{3} + b^{3}\), since \(c > a\) and \(c > b\).
Puc. 8
^{2}+1=0$. It is clear that for any $x$ and $y$, the left side of the equation will be no less than one, i.e., it cannot be equal to zero. | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,670 |
91. Find the smallest natural number that, when divided by $4, 5, 6$, and 12, gives a remainder that is two units less than the divisor each time. | 91. From the condition of the problem, it follows that if 2 is added to the required number, then it will be divisible by each of the given numbers. This means that the required number is equal to the least common multiple of the given numbers, decreased by 2. Answer. 58. | 58 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,671 |
93. Each side of one triangle is less than any side of another triangle. Can the area of the first triangle be greater than the area of the second? | 93. Let's consider an isosceles triangle where the base is at least twice as large as the largest side of the given triangle. By decreasing the height of the isosceles triangle, we can make its area arbitrarily small, while each of its sides will remain larger than any side of the given triangle. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,673 |
95. How many roots does the equation $\cos x=2^{x}$ have? | 95. One of the roots of the given equation is $x=0$. In addition, the equation has infinitely many negative roots. To convince yourself of this, it is enough to mentally visualize the graphs of the functions $y=\cos x$ and $y=2^{x}$ (see Fig. 9). | notfound | Calculus | math-word-problem | Yes | Yes | olympiads | false | 44,675 |
96. Is the sum of nine consecutive natural numbers even or odd? | 96. If nine consecutive natural numbers start with an even number, then the sum of them will be even, since among the nine addends there will be 5 even and 4 odd. Otherwise, the sum will be odd. | notfound | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,676 |
97. For the first half of the journey, the motorcyclist traveled at a speed of 30 km/h, and for the second half - at a speed of 60 km/h. Find the average speed of the motorcyclist. | 97. Let the entire distance be denoted as $a$ km. The motorcyclist spent $0.5a : 30 + 0.5a : 60 = a : 40$ hours on the entire journey. Therefore, the motorcyclist traveled at an average speed of 40 km/h.
## Puc. 9
: x=100$ is true for any digit $x$. 102. Since $\operatorname{tg} 45^{\circ}=1$, then $\lg \operatorname{tg} 45^{\circ}=0$. Therefore, the given product is also zero. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,681 |
108. The purchase of a book cost 1 ruble and one third of the book's cost. What is the cost of the book? | 108. Answer. 1 ruble 50 kopecks. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,685 |
109. What is the last digit of the difference
$$
1 \cdot 2 \cdot 3 \cdot 4 \ldots 13-1 \cdot 3 \cdot 5 \cdot 7 \ldots 13 ?
$$ | 109. The minuend ends in 0, and the subtrahend ends in 5. Therefore, the difference ends in 5. | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,686 |
110. A figure made of 24 matches is shown in Figure 3. It is required to remove 4 matches so that a large square and 3 small squares remain. (Find two solutions.)
Fig. 3
 | 110. The two required solutions to the problem are shown in Figures 10 and 11.
Fig. 10
 | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,687 |
115. One hundred apples are lying in a row, one meter apart from each other. The gardener picked up the first apple, placed a basket in its place, and threw the apple into it. What distance will the gardener travel if he carries the remaining 99 apples to the basket one by one? | 115. To carry an apple to the basket, one needs to travel the path from the basket to the apple and back to the basket. Therefore, the entire required path will be equal to $2(1+2+3+\ldots+99)=$ $=99 \cdot 100=9$ km 900 m. | 9900 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 44,692 |
116. How many lines can divide a plane into 5 parts? | 116. Two lines can divide the plane only into 3 or 4 parts (see Fig. 15). Three lines can divide the plane only into 4, 6, and 7 parts (see Fig. 16). Four lines can divide the plane into 5 parts only if these lines are parallel (see Fig. 17). It is obvious that with any arrangement of five or more lines, the number of ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,693 |
117. How many negative roots does the equation
$$
x^{4}-5 x^{3}+3 x^{2}-7 x+1=0 ?
$$
have? | 117. For any negative $x$, all terms on the left side of the equation will be positive, i.e., the given equation does not have negative roots. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 44,694 |
118. A circle is inscribed in a square, and then a new square is inscribed in the circle. Find the ratio of the areas of these squares. | 118. It is clear that the second square can be placed so that its vertices fall on the points of tangency of the circle with the sides of the first square (see Fig. 18). After this, it is not difficult to verify that the desired ratio of the areas of the squares is 2.
. Connect point $M$ with the center $O$ of the circle. If point $M$ were the midpoint of both chords, then two perpendiculars to segment $OM$ would pass through point $M$, which is impossible. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,697 |
121. Can the difference of the squares of two integers be equal to 222? | 121. Let $a$ and $b$ be integers. Then the difference of their squares is $(a+b)(a-b)$. If $a$ and $b$ are of different parity, then the number $(a+b)(a-b)$ is odd. If $a$ and $b$ are of the same parity, then the number $(a+b)(a-b)$ is divisible by 4. But the number 222 is even and not divisible by 4.
Puc. 20
. In the right triangle $A B C$, the segment $K L$ is the midline, i.e., $|K L|=\frac{1}{2}|A C|=R$. Similarly, we prove that $|K P|=|P M|=|M L|=R$. Therefore, the quadrilateral $K P M L$ is a rhombus, and its perimeter is... | 4R | Geometry | math-word-problem | Yes | Yes | olympiads | false | 44,699 |
123. Prove that the expression
$$
(x-4)(x-6)+3
$$
is positive for any value of $x$. | 123. We have: $(x-4)(x-6)+3=x^{2}-10 x+27=(x-5)^{2}+2 \geqslant 2$ for any $x$. | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,700 |
124. The side of an equilateral triangle is twice the side of a regular hexagon. Find the ratio of the areas of these two figures. | 124. Let the side of a regular hexagon be $a$. Obviously, a regular hexagon consists of six equilateral triangles with side $a$, and a regular triangle - of four (see Fig. 21 and 22). Thus, the ratio of the area of a regular triangle to the area of a regular hexagon is $\frac{4}{6}=\frac{2}{3}$.
 Prove that if the sum of two integers is odd, then their product is even.
b) Prove that for any integer $n$, the number $(3n-1)(5n+2)$ is even.
c) Prove that for any integers $k$ and $n$, the number $(k-n)(k+n+1)$ is even.
d) Prove that the equation $(x-y)(x+y+1)=315$ has no solutions in integers. | 1. a) If the sum of two integers is odd, then these numbers have different parity, i.e., one is even and the other is odd. The product of an even number and an odd number is an even number.
b) For any integer $n$, the numbers $3n-1$ and $5n+2$ are integers, and their sum $3n-1+5n+2=8n+1$ is an odd number. Therefore, t... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,703 |
2. a) Prove that the square of an integer is either divisible by 4 without a remainder or leaves a remainder of 1.
b) Prove that the sum of two consecutive even numbers cannot be a perfect square.
c) Prove that the sum of the squares of two or three odd numbers cannot be a perfect square.
d) Prove that in an integer... | 2. a) The squares of even and odd numbers respectively have the form $(2 n)^{2}=4 n^{2}$ and $(2 n+1)^{2}=4 n(n+1)+1$, where $n$ is an integer. It is clear that the first square is divisible by 4, while the second, when divided by 4, leaves a remainder of 1.
b) The sum of two consecutive even numbers has the form $2 n... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,704 |
3. a) Prove that the product of two consecutive even numbers is divisible by 8.
b) Prove that the square of an odd number, decreased by one, is divisible by 8.
c) Prove that the difference of the squares of two odd numbers is divisible by 8.
d) Prove that the equation \( x^{2} = 4 y^{2} + 12 y + 5 \) has no solution... | 3. a) The product of two consecutive even numbers can be represented as $2 n(2 n+2)=4 n(n+1)$, where $n$ is an integer. Since one of the numbers $n$ or $n+1$ is even, the number $4 n(n+1)$ is divisible by 8 for any integer $n$.
b) We have: $(2 n+1)^{2}-1=4 n(n+1)$. Since the obtained number is divisible by 8 (see the ... | 1,81 | Number Theory | proof | Yes | Yes | olympiads | false | 44,705 |
4. a) Prove that the product of three consecutive integers is divisible by 6.
b) Prove that the product of three consecutive integers, the first of which is even, is divisible by 24.
c) Prove that the square of a prime number greater than three, when divided by 24, leaves a remainder of 1.
d) Prove that the differen... | 4. a) The product of three consecutive integers can be represented as $n(n+1)(n+2)$, where $n \in Z$. We will prove that one of the three factors is necessarily divisible by 3. Indeed, every integer has one of the following three forms: $3k, 3k+1, 3k+2$, where $k \in Z$. Then, if $n=3k$, $n$ is divisible by 3; if $n=3k... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,706 |
5. a) Prove that the square of an integer is either divisible by 9 or leaves a remainder of 1 when divided by 3.
b) Prove that for no natural number \( n \) the number \( 3n - 1 \) can be a perfect square.
c) Prove that for no integer \( n \) the number \( n^2 + 1 \) is divisible by 3.
d) Prove that none of the numb... | 5. a) Every integer has one of the following three forms: $3 k, 3 k+1, 3 k+2$, where $k \in Z$. Then for the square of an integer we get: $(3 k)^{2}=9 k^{2}, (3 k+1)^{2}=3(3 k^{2}+2 k)+1$ and $(3 k+2)^{2}=3(3 k^{2}+4 k+1)+1$. Thus, if a number is divisible by 3, then its square is divisible by 9; if the number is not d... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,707 |
6. a) Prove that the product of two consecutive integers is either divisible by 6 or leaves a remainder of 2 when divided by 18.
b) Prove that there does not exist an integer \( n \) such that the number \( 3n + 1 \) is the product of two consecutive integers.
c) Prove that for no integer \( n \) can the number \( n^... | 6. a) If one of two consecutive integers is divisible by 3, then the product of these numbers is divisible by 6, since one of them is necessarily even. If both numbers are not divisible by 3, then their product has the form $(3 k+1)(3 k+2) = 9 k(k+1)+2$. For any integer $k$, a number of this form, when divided by 18, g... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,708 |
7. a) Prove that the square of an integer cannot end in any of the digits $2, 3, 7, 8$.
b) Prove that for no natural number $n$ can the numbers $5n+2$ and $5n+3$ be perfect squares.
c) Prove that for no integer $n$ is the number $n^2 + 3$ divisible by 5.
d) Find all natural values of $n$ for which the number $123 \l... | 7. a) Any integer can be represented in the form: \( n = 10k + r \), where \( k \) is an integer and \( r = 0, 1, 2, \ldots, 9 \). Based on the equality \( n^2 = 10(10k^2 + 2kr) + r^2 \), we conclude that the numbers \( n^2 \) and \( r^2 \) end with the same digit. It remains to verify directly that \( r^2 \) does not ... | 4 | Number Theory | proof | Yes | Yes | olympiads | false | 44,709 |
8. a) Prove that the product of two consecutive integers can end only in one of the digits $2, 6, 0$.
b) Prove that no permutation of the digits in the number 13579148 can result in a number that is the product of two consecutive integers.
30
c) Prove that for no natural number $n$ can the number $5n + 3.5 + (-1)^n \... | 8. a) Taking into account that any integer can be represented in the form $10 k+r(0 \leqslant r \leqslant 9)$, consider products of the form $(10 k+r)(10 k+r+1)$, assigning $r$ values from 0 to 9 sequentially.
b) Note that among the digits of the given number, the digits $0,2,6$ are absent (see problem 8 a).
c) It is... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,710 |
9. a) Prove that two consecutive odd numbers are always coprime.
b) Prove that for any natural number $n$, the numbers $14n + 3$ and $21n + 4$ are coprime.
c) Prove that for any natural number $n$, the numbers $7 \cdot 10^n + 4$ and $7 \cdot 10^n + 11$ are coprime. | 9. a) If two consecutive odd numbers had a common divisor other than one, then this divisor would also be a divisor of their difference, i.e., the number 2.
b) If the numbers $14 n+3$ and $21 n+4$ had a common divisor other than one, then this divisor would also be a divisor of the numbers $3(14 n+3)=42 n+9$ and $2(21... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,711 |
10. a) Given an irreducible fraction. Prove that the fraction complementing it to one is also irreducible.
b) Prove that if the fraction $\frac{m-n}{m+n}$ is irreducible, then the fraction $\frac{m}{n}$ is also irreducible.
c) Prove that if for some natural number $n$ the fraction $\frac{5 n+2}{10 n+7}$ is reducible,... | 10. a) Let $\frac{a}{b}$ be a proper irreducible fraction. Then the fraction that complements it to one is $1-\frac{a}{b}=\frac{b-a}{b}$. If the latter fraction were reducible, i.e., if the numbers $b$ and $b-a$ had a common divisor other than one, then this same divisor would also be a divisor of their difference $b-(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,712 |
11. a) The difference of two integers is 2. Prove that the product of these numbers, increased by one, is a perfect square.
b) Prove that the product of four consecutive integers can be represented as the product of two consecutive even numbers.
c) Prove that the product of four consecutive integers, increased by one... | 11. a) If the smaller of the numbers is denoted by $n$, then the larger number will be $n+2$. Then $(n+2 n) n+1=(n+1)^{2}$.
b) We have: $\quad n(n+1)(n+2)(n+3)=n(n+3)(n+1)(n+$ $+2)=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)$. It is obvious that the numbers $n^{2}+3 n+2$ and $n^{2}+3 n$ are even and their differenc... | proof | Algebra | proof | Yes | Yes | olympiads | false | 44,713 |
12. a) Prove that if an integer is not divisible by 5, then its square has the form \(5k + 1\) or \(5k - 1\).
b) Prove that if an integer \(n\) is not divisible by 5, then the number \(n^4 - 1\) is divisible by 5.
c) Prove that the fifth power of an integer ends with the same digit as the integer itself.
d) Prove th... | 12. a) If a number is not divisible by 5, then the square of this number ends in one of the digits: $1, 4, 6, 9$. It is obvious that a number with such a last digit has one of two forms: $5k+1$ or $5k-1$.
b) If $n$ is not divisible by 5, then $n^2$ has the form $5k+1$ or $5k-1$ (see problem 12a). Then either $n^2-1$ o... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,714 |
13. a) Integers \(a, b, c\) are such that their sum is divisible by 6. Prove that the number \(a^{5} + b^{3} + c\) is also divisible by 6.
b) Prove that for any integer \(n\), the number \(n^{5} - 5 n^{3} + 4 n\) is divisible by 120.
c) Prove that for any odd \(n\), the number \(n^{3} + 3 n^{2} - n - 3\) is divisible... | 13. a) We have: \(a^{5}+b^{3}+c-(a+b+c)=\)
\[
=(a-1) a(a+1)\left(a^{2}+1\right)+(b-1) b(b+1)
\]
40
If the subtrahend and the difference are divisible by 6, then the minuend is also divisible by 6.
b) Let's transform the given expression as follows:
\[
\begin{aligned}
& n^{5}-5 n^{3}+4 n=n\left(n^{2}-1\right)\left(... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,715 |
14. Prove that the third digit from the end of the square of a number ending in 5 can only be 0, 2, or 6. | 14. An integer ending in 5 can always be represented in the form $10 k+5$, where $k \in Z$. Then $(10 k+5)^{2}=100 k^{2}+$ $+100 k+25=100 k(k+1)+25$. It is clear that the third digit from the end of the resulting number coincides with the last digit of the number $k(k+1)$, which can only be 0, 2, or 6 (see problem $8 \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,716 |
15. Prove that for no integer $n$ the numbers $\frac{n-5}{6}$ and $\frac{n-1}{21}$ can both be integers. | 15. The numbers $n-5$ and $n-1$ cannot both be divisible by 3 for any integer $n$ (since their difference is not divisible by 3), whereas both numbers 6 and 21 are divisible by 3. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,717 |
16. Prove that the difference between a natural number and the sum of its digits is divisible by 9. | 16. Let $x$ be the digit of a given natural number standing at the $k+1$ position from the end of this number. The statement of the problem follows from the fact that the difference $x \cdot 10^{k}-x=999 \ldots 9 x$ is divisible by 9. This thereby proves the divisibility rule for 9: a number is divisible by 9 if and on... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,718 |
17. Find all three-digit numbers starting with the digit 3 and having exactly 3 divisors. | 17. It is easy to notice that only the square of a prime number can have exactly three divisors. The only three-digit number starting with the digit 3 and having exactly three divisors is 361. | 361 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,719 |
18. Find all integers $n$ for which $n^{5}+3$ is divisible by $n^{2}+1$. | 18. Consider the expression:
$$
\frac{n^{5}+3}{n^{2}+1}=\frac{\left(n^{5}-n\right)+(n+3)}{n^{2}+1}=n\left(n^{2}-1\right)+\frac{n+3}{n^{2}+1}
$$
It is clear that we need to find all such integers $n$ for which the expression $\frac{n+3}{n^{2}+1}$ is an integer. It is easy to verify that there are only 5 required value... | -3,-1,0,1,2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,720 |
19. Find all prime numbers $p$ such that the numbers $p+4$ and $p+8$ are also prime. | 19. It is not hard to notice that the number 3 satisfies the requirements of the problem. Let's show that the problem has no other solutions. Indeed, every prime number not equal to 3 has one of two forms: $3k+1$ or $3k+2$, where $k$ is a non-negative integer. But in the case $p=3k+1$, the number $p+8$ is divisible by ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,721 |
20. Prove that for no integer $n$ the number $n^{2}+1$ is divisible by 7. | 20. Any integer can be represented in the form $n=7 k+r$, where $r$ is one of the numbers $0,1,2,3,4,5,6$. Then $n^{2}+1=$ $=(7 k+r)^{2}+1=7 k(7 k+2 r)+r^{2}+1$. By direct verification, we can see that $r^{2}+1$ is not divisible by 7 for any of the possible values of $r$. Therefore, $n^{2}+1$ is not divisible by 7 for ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,722 |
21. Prove that there are infinitely many composite numbers of the form $2^{n}-1$, where $n$ is an odd natural number. | 21. Let $n=3(2 k+1)$, where $k \in N$. Then $2^{3(2 k+1)}-1=$ $=\left(2^{2 k+1}\right)^{3}-1=\left(2^{2 k+1}-1\right)\left(2^{4 k+2}+2^{2 k+1}+1\right)$. Thus, for any natural $k$ we get a composite number. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 44,723 |
22. Find all natural $n$ for which the number $n^{2}+3 n$ is a perfect square. | 22. For $n=1$ the number $n^{2}+3 n$ equals 4, i.e., is a perfect square. Let now $n>1$. Obviously, in this case
$$
(n+1)^{2}<n^{2}+3 n<(n+2)^{2}
$$
Since a perfect square cannot exist between the squares of two consecutive natural numbers, the problem has a unique solution $n=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 44,724 |
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