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8.1. Of the three boys named Anton, Vanya, and Sasha, only one always tells the truth. Anton said: "Vanya does not always tell the truth," Vanya said: "I do not always tell the truth," and Sasha said: "Anton does not always tell the truth." Who among them always tells the truth, given that at least one of them lied?
# | # Solution
1) It is easy to see that Vanya is telling the truth (if we assume that he is lying and the statement "I do not always tell the truth" is not true, then the truth would be: "I always tell the truth," which would lead to a contradiction).
2) Since Anton's statement has the same meaning, Anton is also telling... | Anton | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,736 |
8.2. At noon, a "Moskvich" left point A for point B. At the same time, "Zhiguli" left point B for point A on the same road. An hour later, the "Moskvich" was halfway between A and the "Zhiguli". When will it be halfway between the "Zhiguli" and B? (The speeds of the cars are constant and differ by less than a factor of... | Solution. Let the speeds of the "Moskvich" and "Zhiguli" be u and v, respectively. From the problem statement, it follows that if the speed of the "Moskvich" were $2 \mathrm{u}$, then its meeting with the "Zhiguli" (traveling at speed v) would occur one hour after the start of the journey. From this, it follows that if... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,737 |
8.3. Let $\mathrm{O}$ be an interior point of the square $\mathrm{ABCD}$ with side $\mathrm{AB}=1$, for which the equality
$$
A O^{2}+B O^{2}+C O^{2}+D O^{2}=2
$$
holds. Prove that O is the center of the square. | Solution. Let $x$ and $y$ be the distances from point $O$ to sides $AD$ and $AB$ respectively. Using the Pythagorean theorem, we can express the desired expression in terms of these numbers and transform it into the form:
$$
AO^{2}+BO^{2}+CO^{2}+DO^{2}=4\left(x-\frac{1}{2}\right)^{2}+4\left(y-\frac{1}{2}\right)^{2}+2
... | proof | Geometry | proof | Yes | Yes | olympiads | false | 5,738 |
8.4. A cargo weighing 13.5 tons is packed into a certain number of "weightless" boxes. The weight of each box with the cargo does not exceed 350 kg. Prove that this cargo can be transported using 11 one-and-a-half-ton trucks. | Solution. We will prove that with boxes not exceeding 350 kg (if their total weight is more than 1.2 tons), we can gather a weight from 1.2 to 1.5 tons. Arrange them in order, starting with the heaviest. If the first four together weigh more than 1.2 tons - they are already sufficient (the weight will not exceed 1.4 to... | proof | Other | proof | Yes | Yes | olympiads | false | 5,739 |
8.5. The faces of a cube are labeled with six different numbers from 6 to 11. The cube was rolled twice. The first time, the sum of the numbers on the four side faces was 36, the second time - 33. What number is written on the face opposite the one where the digit $10$ is written? | Solution: The sum of the numbers on all faces is
$$
6+7+8+9+10+11=51
$$
On the first roll, the sum of the numbers on the top and bottom faces is 51 $-36=15$, on the second roll $-51-33=18$. Therefore, the sum on the third pair of opposite faces is $51-15-18=18$. The sum of 18 can be obtained in two ways: $11+7$ or $1... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,740 |
8.6. Ostap Bender put new tires on the car "Gnu Antelope". It is known that the front tires of the car wear out after 25,000 km, while the rear tires wear out after 15,000 km (the tires are the same both in the front and in the rear, but the rear ones wear out more). After how many kilometers should Ostap Bender swap t... | Solution. Let Ostap Bender swap the tires after x kilometers. Then the rear tires have used up [x/15000] of their resource, and the front tires [x/25000]. After the swap, they can work for another
$$
25000 \cdot\left(1-\frac{x}{15000}\right) \text { and } 15000 \cdot\left(1-\frac{x}{25000}\right)
$$
kilometers, respe... | 9375 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,741 |
5.1. Construct a rectangle, where each side is greater than 1, using six rectangles $7 \times 1, 6 \times 1, 5 \times 1, 4 \times 1, 3 \times 1, 2 \times 1$ and a square $1 \times 1$. | Solution. From a rectangle $6 \times 1$ and a square $1 \times 1$, we form a rectangle $7 \times 1$. Similarly, we form rectangles $7 \times 1$ from pairs of rectangles $5 \times 1, 2 \times 1$ and $4 \times 1, 3 \times 1$. From the four resulting rectangles $7 \times 1$, a rectangle $7 \times 4$ is formed. | 7\times4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,742 |
5.3. To number the pages of a book, a total of 1392 digits were used. How many pages are in this book? | Solution. The first nine pages will require 9 digits, and for the next 90 pages, 2 digits are needed for each page, which means 2 * 90 digits are required. Let the book have x pages, then the pages with three digits will be x - 99, and the digits on them will be 3 * (x - 99). We get the equation: $9 + 2 \cdot 90 + 3 \c... | 500 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,744 |
5.4. At the crossroads, four travelers met: inhabitants of the city of liars (who always lie) and the city of knights (who always tell the truth) (not all were from the same city). The first said: “Besides me, there is exactly one inhabitant of my city here.” The second added: “And I am the only one from my city.” The ... | Solution. From the third statement, it follows that he is from the same city as the second, but then the second cannot be telling the truth. Therefore, there are at least two liars (the second and the third). Now, regardless of the first statement, it turns out that either he is right and, consequently, the fourth $\sq... | From\the\city\of\knights | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,745 |
5.5. The distance between two cars driving on a highway is 200 km. The speeds of the cars are 60 km/h and 80 km/h. What will be the distance between them after 1 hour? | Solution. There are four possible cases (a diagram should be made):
1) The cars are driving towards each other: $200-(60+80)=60$ km;
2) The cars are driving in opposite directions: $200+(60+80)=340$ km;
3) The cars are driving in the same direction, with the second car catching up to the first: $200+(60-80)=180$ km;
4... | 60,180,220,340 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,746 |
5.6. A student wrote a multiplication problem with two-digit numbers on the board. Then he erased all the digits and replaced them with letters. The result was the equation:
$$
A B \cdot C D=M L N K T
$$
Prove that the student made a mistake. | Solution. The equality $A B \cdot C D=M L N K T$ cannot hold, since the largest possible product of two-digit numbers
$$
99 * 99<100 * 100=10000
$$ | proof | Number Theory | proof | Yes | Yes | olympiads | false | 5,747 |
7.1. Prove that any triangle can be cut into four isosceles triangles. | Solution. Cut the triangle into two right triangles by the height dropped from the vertex with the largest angle. Then each of them is divided into two isosceles triangles by the median connecting the vertex of the right angle and the midpoint of the hypotenuse. | proof | Geometry | proof | Yes | Yes | olympiads | false | 5,748 |
7.2. There are 30 logs with lengths of 3 and 4 meters, the total length of which is 100 meters. How many cuts can be made to saw the logs into logs of 1 meter length? (Each cut saws exactly one log.)
# | # Solution.
First solution. Glue all the logs into one 100-meter log.
To divide it into 100 parts, 99 cuts are needed, 29 of which have already been made.
Second solution. If there were $m$ three-meter logs and $n$ four-meter logs,
then $m+n=30, 3m+4n=100$, from which $m=20, n=10$. Therefore, $20 \cdot 2 + 10 \cdot... | 70 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,749 |
7.4. Ivan Ivanovich bought a dog. Sasha thinks that this dog is a black poodle, Pasha considers her a white bichon, and Masha - a black bull terrier. It is known that each of the kids correctly guessed either the breed or the color of the dog's fur. Name the breed of the dog and the color of its fur. | Solution. The dog is black, i.e., the color was guessed by Sasha and Masha, because otherwise, they should have guessed the breed, and the dog cannot be both a poodle and a bull terrier at the same time. Therefore, Pasha guessed the breed, i.e., the dog is a bichon. | The\dog\is\\black\bichon | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,751 |
7.5. Comparing the fractions $\frac{111110}{111111}, \frac{222221}{222223}, \frac{333331}{333334}$, arrange them in ascending order. | # Solution
Consider the numbers
$$
\begin{aligned}
& 1-x=\frac{1}{111111}, 1-y=\frac{2}{222223}, \quad 1-z=\frac{3}{333334}, \text { and their reciprocals } \\
& \frac{1}{1-x}=111111, \frac{1}{1-y}=111111+\frac{1}{2}, \frac{1}{1-z}=111111+\frac{1}{3} \\
& \text { We see that } \frac{1}{1-x}1-z>1-y. \text{ Therefore, ... | x<z<y | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,752 |
1.1. On the Island of Knights and Liars, knights always tell the truth, while liars always lie. One day, a traveler interviewed seven residents of the island.
- I am a knight, - said the first.
- Yes, he is a knight, - said the second.
- Among the first two, there are no less than 50% liars, - said the third.
- Among ... | Answer: 5.
Solution: Suppose the first inhabitant is a knight. Then the second is also a knight, while the third and fourth are liars. If, on the other hand, the first is a liar, then the second is also a liar, while the third and fourth are knights. In either case, among the first four, there are exactly two knights ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,753 |
2.1. Vasya loves to collect mushrooms. He calculated that over the autumn he collected a number of mushrooms that is expressed by a three-digit number, the sum of the digits of which is 14. Then Vasya calculated that 8% of the collected mushrooms were white, and 14% were boletus. How many mushrooms did Vasya collect? | Answer: 950.
Solution. For $14 \%$ of the number of mushrooms to be an integer, it is necessary for the total number of mushrooms to be a multiple of 50. Then the last two digits are 00 or 50. But if a three-digit number ends with two zeros, the sum of the digits cannot be greater than 9, so the last two digits must b... | 950 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,754 |
3.1. Find all pairs of two-digit natural numbers for which the arithmetic mean is $25 / 24$ times greater than the geometric mean. In your answer, specify the largest of the arithmetic means for all such pairs. | Answer: 75.
Solution: Let $a, b$ be the required numbers (without loss of generality, we can assume that $a > b$) and let $\frac{a+b}{2}=25x$ and $\sqrt{ab}=24x$. Then $(\sqrt{a}+\sqrt{b})^2=a+b+2\sqrt{ab}=98x$ and $(\sqrt{a}-\sqrt{b})^2=$ $a+b-2\sqrt{ab}=2x$. From this, it follows that $\sqrt{a}+\sqrt{b}=7(\sqrt{a}-\... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,755 |
4.1. All natural numbers from 1 to 2017 inclusive were written in a row. How many times was the digit 7 written? | Answer: 602.
Solution. First, consider the numbers from 1 to 2000. Then the digit 7 can be in the third position from the end: numbers of the form $7 * *$ or $17 * *$ - there are 200 such numbers. It can be in the second position: $* 7 *$ or $1 * 7 *$ - there are also 200 such numbers; or the last position: $* * 7$ or... | 602 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,756 |
5.1. At the vertices of a cube, numbers $\pm 1$ are placed, and on its faces - numbers equal to the product of the numbers at the vertices of that face. Find all possible values that the sum of these 14 numbers can take. In your answer, specify their product. | Answer: -20160.
Solution. It is obvious that the maximum value of the sum is 14. Note that if we change the sign of one of the vertices, the sum of the numbers in the vertices will increase or decrease by 2. On the other hand, the signs of three faces will change. If their sum was $1, -1, 3, -3$, it will become $-1, 1... | -20160 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,757 |
6.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers | Answer: 250.
Solution. Note that any odd number $2n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of the f... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,758 |
7.1. The sequence is defined by the relations $a_{1}=1$,
$$
a_{2 n}=\left\{\begin{array}{ll}
a_{n}, & \text { if } n \text { is even, } \\
2 a_{n}, & \text { if } n \text { is odd; }
\end{array} \quad a_{2 n+1}= \begin{cases}2 a_{n}+1, & \text { if } n \text { is even, } \\
a_{n}, & \text { if } n \text { is odd. }\en... | Answer: 5.
Solution: The given rules are easily interpreted in terms of the binary system: if $n$ ends in 0 and 1 is appended to the right, then 1 is appended to the right of $a_{n}$. If $n$ ends in 1 and 0 is appended, then 0 is appended to the right of $a_{n}$. In all other cases, $a_{n}$ does not change (when 0 is ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,759 |
1.1. On the Island of Knights and Liars, knights always tell the truth, while liars always lie. One day, a traveler interviewed seven residents of the island.
- I am a knight, - said the first.
- Yes, he is a knight, - said the second.
- Among the first two, there are no less than 50% liars, - said the third.
- Among ... | Answer: 5.
Solution: Suppose the first inhabitant is a knight. Then the second is also a knight, while the third and fourth are liars. If, on the other hand, the first is a liar, then the second is also a liar, while the third and fourth are knights. In either case, among the first four, there are exactly two knights ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,760 |
4.1. Vasya loves to collect mushrooms. He calculated that over the autumn he collected a number of mushrooms that is expressed by a three-digit number, the sum of the digits of which is 14. Then Vasya calculated that 8% of the collected mushrooms were white, and 14% were boletus. How many mushrooms did Vasya collect? | Answer: 950.
Solution. For $14 \%$ of the number of mushrooms to be an integer, it is necessary for the total number of mushrooms to be a multiple of 50. Then the last two digits are 00 or 50. But if a three-digit number ends with two zeros, the sum of the digits cannot be greater than 9, therefore, the last two digit... | 950 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,763 |
5.1. How many four-digit numbers exist that contain the digit 9 in their notation, immediately followed by the digit 5? | Answer: 279.
Solution. For numbers of the form $95 * *$, the last two digits can be anything - there are $10 \cdot 10=100$ such numbers, and for numbers of the form $* 95 *$ and $* * 95$, the first digit cannot be 0, so there are $10 \cdot 9=90$ of each. The number 9595 was counted twice, so we get 279 numbers. | 279 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,764 |
6.1. All natural numbers from 1 to 2017 inclusive were written in a row. How many times was the digit 7 written? | Answer: 602.
Solution. First, consider the numbers from 1 to 2000. Then the digit 7 can be in the third position from the end: numbers of the form $7 * *$ or $17 * *$ - there are 200 such numbers. It can be in the second position: $* 7 *$ or $1 * 7 *$ - there are also 200 such numbers; or the last position: $* * 7$ or... | 602 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,765 |
7.1. How many numbers from 1 to 1000 (inclusive) cannot be represented as the difference of two squares of integers | Answer: 250.
Solution. Note that any odd number $2 n+1$ can be represented as $(n+1)^{2}-n^{2}$. Moreover, an even number that is a multiple of 4 can be represented as $4 n=(n+1)^{2}-(n-1)^{2}$. The numbers of the form $4 n+2$ remain. Note that a square can give remainders of 0 or 1 when divided by 4, so numbers of th... | 250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,766 |
8.5. From the right angle $C$ of triangle $ABC$, the median $CM$ is drawn. The circle inscribed in triangle $ACM$ touches side $CM$ at its midpoint. Find the angles of triangle $ABC$. | Answer: $30^{\circ}, 60^{\circ}, 90^{\circ}$.
Solution 1. Angle $\angle C$ is a right angle, i.e., $\angle C=90^{\circ}$. Let's find angles $\angle A$ and $\angle B$. Let $K, L$, and $N$ be the points of tangency of the circle inscribed in triangle $A C M$ with sides $C M, A M$, and $A C$ of this triangle, respectivel... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,770 |
9.6. Find all three-digit numbers that are five times the product of their digits. Answer: 175. | Solution. Let $\overline{a b c}$ be the desired three-digit number. Then, by the condition, $100 a+10 b+c=5 a b c$. From this, we get $c=5(a b c-2 b-20 a)$, so $c$ is divisible by 5. But $c$ cannot be zero, since otherwise the product of the digits is also zero. Therefore, $c=5$. Thus, we have $100 a+10 b+5=25 a b \Lef... | 175 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,771 |
9.8. In an isosceles triangle $A B C$, the bisector $B P$ is drawn. Prove that if $\angle B A C=100^{\circ}$, then $A P+P B=B C$. | Solution. Since triangle $ABC$ is isosceles, the angles at its base are equal:
$$
\angle ABC = \angle ACB = \frac{1}{2}\left(180^{\circ} - \angle BAC\right) = 40^{\circ}
$$
By the condition, $BP$ is the bisector of angle $\angle ABC$, so $\angle ABP = \angle PBC = 20^{\circ}$. Let $Q$ be a point on the base $BC$ such... | proof | Geometry | proof | Yes | Yes | olympiads | false | 5,772 |
3.1. Among all integer solutions of the equation $20 x+19 y=2019$, find the one for which the value of $|x-y|$ is minimal. In the answer, write the product $x y$. | Answer: 2623.
Solution. One of the solutions to the equation is the pair $x=100, y=1$. Therefore, the set of all integer solutions is $x=100-19 n, y=1+20 n, n \in \mathbb{Z}$. The absolute difference $|x-y|=$ $|100-19 n-1-20 n|=|99-39 n|$ is minimized when $n=3$, and the corresponding solution is $(x, y)=(43,61)$. We ... | 2623 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,775 |
4.1. A teacher at a summer math camp took with him for the whole summer several shirts, several pairs of pants, several pairs of shoes, and two jackets. At each lesson, he wore pants, a shirt, and shoes, and he wore a jacket on some lessons. On any two lessons, at least one of the items of his clothing or shoes was dif... | Answer: 126.
Solution: Let the teacher bring $x$ shirts, $y$ pairs of trousers, $z$ pairs of shoes, and 2 jackets. Then he can conduct $3 x y z$ lessons (the number 3 means: 2 lessons in each of the jackets and 1 lesson without a jacket). If he has one more shirt, the number of lessons will increase by $3 y z$. If he ... | 126 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,776 |
4.4. A teacher at a summer math camp took with him for the whole summer several shirts, several pairs of pants, several pairs of shoes, and two jackets. At each lesson, he wore pants, a shirt, and shoes, and he wore a jacket on some lessons. On any two lessons, at least one of the items of his clothing or shoes was dif... | Answer: 216.
Note. In versions $4.2,4.3,4.4$, answers $252,360,420$ respectively are also counted as correct. | 216 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,777 |
5.1. Find $\frac{S_{1}}{S_{2}}$, where
$S_{1}=\frac{1}{2^{2019}}+\frac{1}{2^{2018}}-\frac{1}{2^{2017}}+\ldots+\frac{1}{2^{3}}+\frac{1}{2^{2}}-\frac{1}{2}, \quad S_{2}=\frac{1}{2}+\frac{1}{2^{2}}-\frac{1}{2^{3}}+\ldots+\frac{1}{2^{2017}}+\frac{1}{2^{2018}}-\frac{1}{2^{2019}}$
(in both sums, the signs of the terms alte... | Answer: -0.2.
Solution. For the sum $S_{1}$ we have
$$
\frac{1}{2}\left(\frac{1}{2}\left(\frac{1}{2} S_{1}-\frac{1}{2^{2020}}\right)-\frac{1}{2^{2020}}\right)+\frac{1}{2^{2020}}=S_{1}-\frac{1}{2^{3}}-\frac{1}{2^{2}}+\frac{1}{2}
$$
Solving this equation for $S_{1}$, we get
$$
S_{1}=\frac{1-2^{2019}}{7 \cdot 2^{2019}... | -0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,778 |
5.3. Find $\frac{S_{1}}{S_{2}}$, where
$S_{1}=\frac{1}{3^{2019}}+\frac{1}{3^{2018}}-\frac{1}{3^{2017}}+\ldots+\frac{1}{3^{3}}+\frac{1}{3^{2}}-\frac{1}{3}, \quad S_{2}=\frac{1}{3}+\frac{1}{3^{2}}-\frac{1}{3^{3}}+\ldots+\frac{1}{3^{2017}}+\frac{1}{3^{2018}}-\frac{1}{3^{2019}}$
(in both sums, the signs of the terms alte... | Answer. $-\frac{5}{11} \approx-0.45$. | -\frac{5}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,779 |
6.1. On the sides $A C$ and $B C$ of triangle $A B C$, points $D$ and $E$ are taken, respectively. Segments $A E$ and $B D$ intersect at point $F$. Find the area of triangle $C D E$, if the areas of triangles $A B F, A D F$, and $B E F$ are $1, \frac{1}{2}$, and $\frac{1}{4}$, respectively. | Answer. $\frac{15}{56} \approx 0.27$.
Solution. Let $S_{1}=S_{A F D}=\frac{1}{2}, S_{2}=S_{A B F}=1, S_{3}=S_{B E F}=\frac{1}{4}, S_{4}=S_{D E F}, S_{5}=S_{C D E}$. Since the ratio of the areas of triangles with equal heights is equal to the ratio of the lengths of the bases, then
$$
\frac{S_{4}}{S_{3}}=\frac{F D}{F ... | \frac{15}{56} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,780 |
7.1. Solve the equation
$$
3 \cos \frac{4 \pi x}{5}+\cos \frac{12 \pi x}{5}=2 \cos \frac{4 \pi x}{5}\left(3+\operatorname{tg}^{2} \frac{\pi x}{5}-2 \operatorname{tg} \frac{\pi x}{5}\right)
$$
In the answer, write the sum of its roots on the interval $[-11 ; 19]$. | Answer: 112.5.
Solution. Let $\frac{\pi x}{5}=z$. Then, using the formula for the cosine of a triple angle, we get
$$
\begin{gathered}
3 \cos 4 z+4 \cos ^{3} 4 z-3 \cos 4 z=2 \cos 4 z \cdot\left(3+\operatorname{tg}^{2} z-2 \operatorname{tg} z\right) \\
\left(2 \cos ^{2} 4 z-3-\operatorname{tg}^{2} z+2 \operatorname{t... | 112.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,785 |
8.1. It is known that $P(x)$ is a polynomial of degree 9 and $P(k)=2^{k}$ for all $k=1,2,3, \ldots, 10$. Find $P(12)$. | Answer: 4072.
Solution. Let $P(x)$ be a polynomial of degree $n$ and $P(k)=2^{k}$ for all $k=1,2,3, \ldots, n+1$. We will find $P(n+m+2), m=0,1, \ldots$. By the binomial theorem for any $k \in \mathbb{N}$ we have
$$
2^{k}=2 \cdot(1+1)^{k-1}=2 \sum_{i=0}^{k-1} C_{k-1}^{i}=2 \sum_{i=0}^{k-1} \frac{(k-1)(k-2) \ldots(k-i... | 4072 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,786 |
9.1. Two spheres touch the plane of triangle $A B C$ at points $A$ and $B$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 7, and the distance between their centers is 13. The center of a third sphere with radius 5 is at point $C$, and it touches each of the first two spheres e... | Answer. $\sqrt{30} \approx 5.48$.
Solution. Let $O_{1}$ and $O_{2}$ be the centers, and $r_{1}$ and $r_{2}$ be the radii of the spheres that touch the plane $A B C$ at points $B$ and $A$ respectively, $r_{3}=5$ be the radius of the third sphere. By the problem's condition, $r_{1}+r_{2}=r=7$, $O_{1} O_{2}=d=13$. Let al... | \sqrt{30} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,787 |
9.2. Two spheres touch the plane of triangle $A B C$ at points $B$ and $C$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 7, and the distance between their centers is 17. The center of a third sphere with radius 8 is located at point $A$, and it touches each of the first two s... | Answer. $2 \sqrt{15} \approx 7.75$. | 2\sqrt{15} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,788 |
9.4. Two spheres touch the plane of triangle $A B C$ at points $B$ and $C$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 11, and the distance between their centers is $5 \sqrt{17}$. The center of a third sphere with radius 8 is located at point $A$, and it touches each of the... | Answer. $2 \sqrt{19} \approx 8.72$. | 2\sqrt{19} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,790 |
9.5. Two spheres touch the plane of triangle $A B C$ at points $A$ and $B$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 11, and the distance between their centers is $\sqrt{481}$. The center of a third sphere with radius 9 is at point $C$, and it touches each of the two firs... | Answer. $3 \sqrt{10} \approx 9.49$. | 3\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,791 |
9.6. Two spheres touch the plane of triangle $A B C$ at points $B$ and $C$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 12, and the distance between their centers is $4 \sqrt{29}$. The center of a third sphere with radius 8 is located at point $A$, and it touches each of the... | Answer. $4 \sqrt{5} \approx 8.94$. | 4\sqrt{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,792 |
9.7. Two spheres touch the plane of triangle $A B C$ at points $A$ and $B$ and are located on opposite sides of this plane. The sum of the radii of these spheres is 13, and the distance between their centers is $\sqrt{505}$. The center of a third sphere with radius 8 is located at point $C$, and it touches the first tw... | Answer. $2 \sqrt{21} \approx 9.17$ | 2\sqrt{21} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,793 |
10.1. The sequences $\left\{x_{n}\right\},\left\{y_{n}\right\}$ are defined by the conditions $x_{1}=11, y_{1}=7, x_{n+1}=3 x_{n}+2 y_{n}$, $y_{n+1}=4 x_{n}+3 y_{n}, n \in \mathbb{N}$. Find the remainder of the division of the number $y_{1855}^{2018}-2 x_{1855}^{2018}$ by 2018. | Answer: 1825.
Solution. For all $n \in \mathbb{N}$, the numbers $x_{n}$ and $y_{n}$ are odd. Notice that
$$
2 x_{n+1}^{2}-y_{n+1}^{2}=2\left(3 x_{n}+2 y_{n}\right)^{2}-\left(4 x_{n}+3 y_{n}\right)^{2}=2 x_{n}^{2}-y_{n}^{2}
$$
Therefore, $2 x_{n}^{2}-y_{n}^{2}=\ldots=2 x_{1}^{2}-y_{1}^{2}=242-49=193$.
Let $p=1009$. ... | 1825 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,794 |
1. How many weeks can a year overlap? Assume that a year overlaps with a week if at least 6 and one day of this week falls within the given year. | Answer: On the 53rd 54th week. Solution. If there are 365 days in a year (a non-leap year), then since $365=52 \cdot 7+1$, there are no fewer than 53 weeks in a year. If it is a leap year, meaning there are 366 days in a year ( $366=52 \cdot 7+2$ ), there is a situation where the year starts with the last day of the we... | 54 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,795 |
4. How many numbers divisible by 4 and less than 1000 do not contain any of the digits $6,7,8,9$ or 0. | Answer: 31. Solution. According to the condition, these numbers consist only of the digits 1, 2, 3, 4, and 5. Out of such numbers, only one single-digit number is divisible by 4: 4. Among the two-digit numbers, the following are divisible by 4: $12, 24, 32, 44, 52$. If we prepend 1, 2, 3, 4, or 5 to all these two-digit... | 31 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,796 |
# Task 2.
At the time when a lion cub, located 6 minutes away, set off for a waterhole, the second, having already quenched his thirst, headed back along the same path 1.5 times faster than the first. At the same time, a turtle, located 32 minutes away, set off for the waterhole along the same path. After some time, t... | Answer: 9.57
## B-9
A crocodile swam across a river 42 m wide in a straight line in 6 seconds. During this time, it was carried downstream by 28 m. After resting and regaining strength, it returned to the same point along the same straight line, but this time fighting the current, in 14 seconds. How many seconds woul... | 9.57 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,797 |
Task 1. Misha has a set of nine cards with the letters of the word "LOMONOSOV". On the reverse side of each card, Misha wrote a digit such that the digits on cards with the same letters are the same, and the digits on cards with different letters are different. It turned out that the following equality is true:
$$
\ma... | Answer: $8+\frac{2}{3}+2+9+\frac{2}{6}=20$ and variants where the pairs of digits 8 and 9, 3 and 6 are swapped.
Solution. The left side of the correct equality is less than the value of $10+9+10=29$, which means the digit corresponding to the letter "O" can be either one or two. In this case, the equality $\frac{\math... | 8+\frac{2}{3}+2+9+\frac{2}{6}=20 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,799 |
Problem 2. Verify that $1009=15^{2}+28^{2}$, and represent the number 2018 as the sum of two squares of natural numbers. | Answer: $2018^{2}=13^{2}+43^{2}$.
Solution. The given equality can be verified directly. Let $a=15, b=28$. Then $1009=a^{2}+b^{2}, 2018=2 a^{2}+2 b^{2}=(a+b)^{2}+(a-b)^{2}=13^{2}+43^{2}$.
Answer to variant 2: $3866^{2}=29^{2}+55^{2}$. | 2018=13^{2}+43^{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,800 |
Problem 3. In a right triangle $ABC$ with a right angle at $C$, the bisector $BD$ and the altitude $CH$ are drawn. A perpendicular $CK$ is dropped from vertex $C$ to the bisector $BD$. Find the angle $HCK$, if $BK: KD=3: 1$. | Answer: $30^{\circ}$.
Solution. Let $M$ be the midpoint of $B D$. Then $C M$ is the median of the right triangle $C B D$ and $C M=M B=M D$. In addition, $C K$ is the height and median of triangle $M C D$, so $M C=C D$ and triangle $C M D$ is equilateral. Then $\angle C D M=60^{\circ}$ and $\angle C B M=$ $\angle C B D... | 30 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,801 |
Problem 4. A clock shows exactly one o'clock. A mosquito and a fly are sitting at the same distance from the center on the hour and minute hands, respectively. When the hands coincide, the insects swap places. How many times greater is the distance that the mosquito has traveled in half a day compared to the distance t... | Answer: $83 / 73$.
Solution. The mosquito and the fly move in a circle. In the first hour, the mosquito will cover 11/12 of this circle (at the beginning, it was on the hour hand, pointing to 1, at the end - on the minute hand, pointing to 12). In the second hour, the mosquito will cover $3 / 12$ of the circle (it was... | \frac{83}{73} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,802 |
Problem 5. Each cell of a $3 \times 3$ table is painted in one of three colors such that cells sharing a side have different colors. Among all possible such colorings, find the proportion of those in which exactly two colors are used. | Answer: $1 / 41$.
Solution. The central cell can be colored in any of the three colors, let's call this color $a$. Each of the four cells that share a side with the central cell can be colored in any of the two remaining colors. Let the cell above the central one be colored in color $b$. The third color will be called... | \frac{1}{41} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,803 |
Problem 6. From 24 identical wooden cubes, a "pipe" was glued - a $3 \times 3 \times 3$ cube with the "core" of three cubes removed (see figure). Is it possible to draw a diagonal in each square on the surface of the "pipe" so that a closed path is formed, which does not pass through any vertex more than once?
# | # Answer: No.
Solution. The number of diagonals on the surface of the "tube" is the same as the number of faces of the cubes on this surface: \(4 \cdot 9 + 2 \cdot 8 + 12 = 64\). A non-self-i... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,804 |
Problem 7. On the table lie cards with numbers from 1 to 8: one card with the number 1, two with the number 2, three with the number 3, and so on. Petya and Vasya take one card each in turn and place them in one deck (Petya starts). After Vasya's turn, Petya can say "stop," and then all the unchosen cards are removed f... | Answer: Yes, he can (Vasya).
Solution. First, note that any odd number can be represented as the difference of squares of integers, since $2 k+1=(2 k+1)^{2}-k^{2}$, and any number divisible by 4, since $4 k=(k+1)^{2}-(k-1)^{2}$. Secondly, according to the conditions of the problem, the last card from the stack will be... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,805 |
Problem 1. Two cars have covered the same distance. The speed of the first was constant and three times less than the initial speed of the second. The second car traveled the first half of the distance without changing its speed, then it suddenly reduced its speed by half, traveled another quarter of the distance at a ... | Answer: $\frac{5}{3}$.
Solution. Let $V_{1}$ be the speed of the first car, $V_{0}$ the initial speed of the second, and the distance covered by each car be $S$. Then $V_{0}=3 V_{1}$. The first car spent the entire journey time $t_{1}=\frac{S}{V_{1}}$, while the second car spent time
$$
t_{2}=\frac{\frac{S}{2}}{V_{0}... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,806 |
Problem 2. Vanya thought of a two-digit number, then swapped its digits and multiplied the resulting number by itself. The result turned out to be four times larger than the number he thought of. What number did Vanya think of? | Answer: 81.
Solution. Let the intended number be $\overline{m n}=10 m+n$. Then $4 \overline{m n}=\overline{n m}^{2}$. Therefore, $\overline{n m}^{2}$ is divisible by 4, and $\overline{n m}$ is divisible by 2, so the digit $m$ is even (and not zero). Moreover, $\overline{m n}=\overline{n m}^{2}: 4=(\overline{n m}: 2)^{... | 81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,807 |
Problem 3. We will call a composite natural number $n$ "interesting" if all its natural divisors can be written in ascending order, and each subsequent divisor is divisible by the previous one. Find all "interesting" natural numbers from 20 to 90 (inclusive). | Answer: $25,27,32,49,64,81$.
Solution. "Interesting" can only be numbers of the form $n=p^{k}, k=2,3,4, \ldots$, where $p-$ is a prime number.
Indeed, if we consider the number $n=a \cdot p^{k}$, GCD $(a, p)=1$, then in the sequence of divisors $1, p, \ldots, p^{k}$, the divisor $a$ cannot stand. The situation $\ldot... | 25,27,32,49,64,81 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,808 |
Problem 4. Solve the equation:
$$
(x+1)^{2}+(x+3)^{2}+(x+5)^{2}+\ldots+(x+2021)^{2}=x^{2}+(x-2)^{2}+(x-4)^{2}+\ldots+(x-2020)^{2}
$$ | Answer: $-0.5$.
Solution. The left and right sides of the equation are respectively $1011 x^{2}+2(1+3+5+\ldots+2021) x+(1^{2}+3^{3}+5^{2}+\ldots+2021^{2})$ and $1011 x^{2}-2(2+4+6+\ldots+2020) x+2^{2}+4^{2}+6^{2}+\ldots+2020^{2}$. The equation is reduced to $2(1+2+3+4+\ldots+2021) x=-1^{2}+2^{2}-3^{2}+4^{2}+\ldots-201... | -0.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,809 |
Problem 5. On the lateral sides $AB$ and $BC$ of an isosceles triangle $ABC$, points $M$ and $N$ are marked such that $AM = MN = NC$. On the side $AC$, points $P$ and $Q$ are chosen such that $MQ \parallel BC$ and $NP \parallel AB$. It is known that $PQ = BM$. Find the angle $MQB$. | Answer: $36^{\circ}$. The answer can also be given in the form $\arccos \left(\frac{\sqrt{5}+1}{4}\right)$, etc.
Solution. Let $\angle A=\angle C=\alpha$, then by the property of parallel lines $\angle N P C=$ $\angle A Q M=\alpha$. It is not difficult to prove that $M N \| A C$ (through the similarity of triangles, o... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,810 |
1.1. If the 200th day of some year is Sunday and the 100th day of the following year is also Sunday, then what day of the week was the 300th day of the previous year? Enter the number of this day of the week (if Monday, then 1, if Tuesday, then 2, etc.). | Answer. 1.
Solution. Between the $200-\mathrm{th}$ day of the year and the $100-\mathrm{th}$ day of the next year, there are either $165+100=265$ or $166+100=266$ days. The number 266 is divisible by 7, while the number $265-$ is not. Therefore, the current year has 366 days, meaning it is a leap year. Thus, the previ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,813 |
2.1. How many terms will there be if we expand the expression $\left(4 x^{3}+x^{-3}+2\right)^{2016}$ and combine like terms? | Answer: 4033.
Solution. In the resulting sum, there will be monomials of the form $k_{n} x^{3 n}$ for all integers $n \in[-2016 ; 2016]$ with positive coefficients $k_{n}$, i.e., a total of $2 \cdot 2016+1=4033$ terms. | 4033 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,814 |
3.1. Znayka cut out a semicircle from paper. Neznaika marked a point $D$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A D$ and $D B$. Find the area of the remaining figure if the length of the part of the chord lying inside it, passing through point $D$ p... | Answer: 28.27 (exact value: $9 \pi$).
Solution. The area $S$ of the resulting figure is
$$
S=\frac{\pi}{2}\left(\left(\frac{A B}{2}\right)^{2}-\left(\frac{A D}{2}\right)^{2}-\left(\frac{D B}{2}\right)^{2}\right)=\frac{\pi}{8}\left(A B^{2}-A D^{2}-B D^{2}\right)
$$
Since $A B=A D+D B$, we get
$$
S=\frac{\pi}{4} A D ... | 28.27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,815 |
3.2. Znayka cut out a semicircle from paper. Neznaika marked a point $C$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A C$ and $C B$. Find the area of the remaining figure if the length of the part of the chord lying inside it, passing through point $C$ p... | Answer: 50.27 (exact value: $16 \pi$). | 50.27 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,816 |
3.3. Znayka cut out a semicircle from paper. Neznaika marked a point $M$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A M$ and $M B$. Find the area of the remaining figure if the length of the part of the chord lying inside it, passing through point $M$ p... | Answer: 21.99 (exact value: $7 \pi$). | 7\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,817 |
3.4. Znayka cut out a semicircle from paper. Neznaika marked a point $D$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A D$ and $D B$. The area of the remaining figure turned out to be $8 \pi$. Find the length of the part of the chord lying inside it, pass... | Answer: 5.66 (exact value: $4 \sqrt{2}$ ). | 4\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,818 |
3.5. Znayka cut out a semicircle from paper. Neznaika marked a point $C$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A C$ and $C B$. The area of the remaining figure turned out to be $10 \pi$. Find the length of the part of the chord lying inside it, pas... | Answer: 6.32 (exact value: $2 \sqrt{10}$ ). | 2\sqrt{10} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,819 |
3.6. Znayka cut out a semicircle from paper. Neznaika marked a point $M$ on the diameter $A B$ of this semicircle and cut off two semicircles from Znayka's semicircle with diameters $A M$ and $M B$. The area of the remaining figure turned out to be $16 \pi^{3}$. Find the length of the part of the chord lying inside it,... | Answer: 25.13 (exact value: $8 \pi$). | 8\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,820 |
4.1. The function $f$ satisfies the equation $(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$ for each value of $x$, not equal to 0 and 1. Find $f\left(\frac{2016}{2017}\right)$. | Answer: 2017.
Solution. Substitute $\frac{1}{x}$ for $x$ in the equation. Together with the original equation, we get a system of two linear equations in terms of $f(x)$ and $f\left(\frac{1}{x}\right)$:
$$
\left\{\begin{array}{l}
(x-1) f(x)+f\left(\frac{1}{x}\right)=\frac{1}{x-1} \\
\left(\frac{1}{x}-1\right) f\left(... | 2017 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,821 |
5.1. From point $A$ to point $B$, which are 10 km apart, a car departed at 7:00. After traveling $2 / 3$ of the distance, the car passed point $C$, from which a cyclist immediately set off for point $A$. As soon as the car arrived at $B$, a bus immediately departed from $B$ in the opposite direction and arrived at $A$ ... | Answer: 6.
Solution. Let's plot the graphs of the car's movement (segment $K L$), the bus's movement (segment $L M$), and the cyclist's movement (segment $N P$) on the axes $(t ; s)$, where $t$ is time (in hours) and $s$ is the distance (in kilometers) from point $A$. Let $Q$ be the intersection point of $L M$ and $N ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,822 |
6.1. Find the sum of all integers \( x \in [-3 ; 13] \) that satisfy the inequality
$$
\left(1-\operatorname{ctg}^{2} \frac{\pi x}{12}\right)\left(1-3 \operatorname{ctg}^{2} \frac{\pi x}{12}\right)\left(1-\operatorname{tg} \frac{\pi x}{6} \cdot \operatorname{ctg} \frac{\pi x}{4}\right) \leqslant 16
$$ | Answer: 28.
Solution. Let $t=\frac{\pi x}{12}$, then the inequality takes the form
$$
\left(1-\operatorname{ctg}^{2} t\right)\left(1-3 \operatorname{ctg}^{2} t\right)(1-\operatorname{tg} 2 t \cdot \operatorname{ctg} 3 t) \leqslant 16
$$
Since
$$
\begin{gathered}
1-\operatorname{ctg}^{2} t=-\frac{\cos 2 t}{\sin ^{2}... | 28 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,823 |
7.1. Two spheres are inscribed in a dihedral angle, touching each other. The radius of one sphere is twice that of the other, and the line connecting the centers of the spheres forms an angle of $45^{\circ}$ with the edge of the dihedral angle. Find the measure of the dihedral angle. Write the cosine of this angle in y... | Answer: 0.56 (exact value: $\frac{5}{9}$).
Solution. Let the sought dihedral angle be denoted by $\alpha$. Let $R$ and $r$ be the radii, and $O_{R}$ and $O_{r}$ be the centers of the larger and smaller spheres, respectively. Let $A$ and $B$ be the feet of the perpendiculars dropped from $O_{R}$ and $O_{r}$ to the edge... | \frac{5}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,824 |
7.2. Two spheres are inscribed in a dihedral angle, touching each other. The radius of one sphere is three times that of the other, and the line connecting the centers of the spheres forms an angle of $60^{\circ}$ with the edge of the dihedral angle. Find the measure of the dihedral angle. Write the cosine of this angl... | Answer: 0.33 (exact value: $\frac{1}{3}$ ). | 0.33 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,825 |
8.1. Calculate $\frac{1}{2 \sqrt{1}+\sqrt{2}}+\frac{1}{3 \sqrt{2}+2 \sqrt{3}}+\frac{1}{4 \sqrt{3}+3 \sqrt{4}}+\ldots+\frac{1}{100 \sqrt{99}+99 \sqrt{100}}$. If necessary, round the answer to two decimal places. | Answer: 0.9.
Solution. Since
$$
\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}=\frac{(n+1) \sqrt{n}-n \sqrt{n+1}}{(n+1) n(n+1-n)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}},
$$
we get that the number from the problem condition is equal to
$$
1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{99}}-... | 0.9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,830 |
9.1. Find the smallest natural number $n$ for which the decimal representation of $n$ together with $n^{2}$ uses all the digits from 1 to 9 exactly once. | Answer: 567.
Solution: If $n \geqslant 1000$, then $n^{2} \geqslant 10^{6}$, so the decimal representations of $n$ and $n^{2}$ together contain at least 11 digits. If, however, $n \leqslant 316$, then $n^{2} \leqslant 99856$, and the decimal representations of $n$ and $n^{2}$ together contain no more than 8 digits. Th... | 567 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,831 |
3. Find a two-digit number, the digits of which are different and the square of which is equal to the cube of the sum of its digits. | Answer: 27.
Solution: Let's write the condition as $\overline{A B}^{2}=(A+B)^{3}$. Note that the sum of the digits $(A+B)$ does not exceed 17, i.e., $\overline{A B}^{2} \leq 17^{3}$. Moreover, this number $\overline{A B}^{2}=n^{6}$, where $n$ is some natural number that does not exceed 4. However, 1 and 2 do not work ... | 27 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,834 |
4. On a circle, 100 points are marked, painted either red or blue. Some of the points are connected by segments, with each segment having one blue end and one red end. It is known that no two red points belong to the same number of segments. What is the maximum possible number of red points? | Answer: 50.
Solution: Take 50 red and 50 blue points. The first red point is not connected to any other, the second is connected to one blue, ..., the 50th is connected to 49 blues. Obviously, there cannot be more than 50 red points, because if there are 51 or more, then there are no more than 49 blues, hence the numb... | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 5,835 |
5. How many diagonals in a regular 32-sided polygon are not parallel to any of its sides | Answer: 240.
Solution: In a 32-sided polygon, there are $32*(32-3)/2=464$ diagonals in total. We can divide the sides into 16 pairs of parallel sides. It is not hard to notice that if we fix a pair, i.e., 4 vertices, the remaining vertices can be connected in pairs by diagonals parallel to this pair. There will be a t... | 240 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,836 |
6. For natural numbers $m$ and $n$, it is known that $3 n^{3}=5 m^{2}$. Find the smallest possible value of $m+n$. | Answer: 60.
Solution: Obviously, if $m, n$ contain prime factors other than 3 or 5, then they can be canceled out (and reduce $\mathrm{m}+\mathrm{n}$).
Let $n=3^{a} \cdot 5^{b}, m=3^{c} \cdot 5^{d}$. Then the condition implies that $3 a+1=2 c, 3 b=2 d+1$. The smallest possible values are: $a=1, b=1, c=2, d=1$, from w... | 60 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,837 |
7. Pete and Vasya are playing a game. On the board, there is a number: 11223334445555666677777. In one move, it is allowed to erase any number of identical digits. The player who erases the last digit wins. Pete goes first. Can he play in such a way as to guarantee a win? | Answer: Yes, he can.
Solution: On his first move, Petya can, for example, erase all the digits 7. The remaining digits can be written as:
113335555 | 666644422.
After this, Petya symmetrically (relative to the middle line) repeats the moves of Bacu. | Yes,hecan | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,838 |
1.1. On the Island of Misfortune, there live knights who always tell the truth, and liars who always lie. One day, a tourist met five inhabitants of the island and asked them: "How many liars are there among you?" The first answered: "One," the second answered: "Two," the third answered: "Three," the fourth answered: "... | Answer: 4.
Solution: Among the islanders surveyed, exactly one is a knight, since they all gave different, and all possible answers. Then the remaining four are liars. | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,841 |
2.1. On a grid paper, a square made up of several cells is shaded, with its sides lying on the grid lines. It is known that to get a larger square under the same condition, 47 more cells need to be shaded. Find the side length of the original square. | Answer: 23.
Solution. If the side of the original square was $n$, and the side of the obtained square became larger by $k$, then to obtain it, one needs to color $(n+k)^{2}-n^{2}=2 n k+k^{2}$ cells, i.e., $2 n k+k^{2}=47$. Therefore, $k$ is odd, and $k^{2}<47$, so $k \leqslant 5$. If $k=5$, then $10 n+25=47$, and $n$ ... | 23 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,842 |
3.1. At a sumo wrestling tournament, 20 sumo-tori (sumo wrestlers) arrived. After weighing, it was found that the average weight of the sumo-tori is 125 kg. What is the maximum possible number of wrestlers who weigh more than 131 kg, given that according to sumo rules, people weighing less than 90 kg cannot participate... | Answer: 17.
Solution: Let $n$ be the number of sumo-tori that weigh more than 131 kg. Their total weight is more than $131 n$ kg, and the total weight of the remaining ones is no less than $90(20-n)$. Therefore, $\frac{131 n+90(20-n)}{20}<125$, from which $41 n<35 \cdot 20$, i.e., $n<\frac{700}{41}=17 \frac{3}{41}$. T... | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,843 |
4.1. Once, in a company, the following conversation took place:
- We must call Misha immediately! - exclaimed Vanya.
However, no one remembered Misha's phone number.
- I remember for sure that the last three digits of the phone number are consecutive natural numbers, - said Nastya.
- And I recall that the first five... | Answer: 7111765.
Solution. Note that three consecutive ones cannot stand at the beginning, as 111 is not divisible by 9. Therefore, among the first three digits, there is one that is different from one. If this is the second or third digit, then since the first five form a palindrome and the last three are consecutive... | 7111765 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,844 |
5.1. From one point on a circular track, a pedestrian and a cyclist started simultaneously in the same direction. The cyclist's speed is $55\%$ greater than the pedestrian's speed, and therefore the cyclist overtakes the pedestrian from time to time. At how many different points on the track will the overtakes occur? | Answer: 11.
Solution: Let's assume the length of the path is 55 (in some units), and the speeds of the pedestrian and the cyclist are $100 x$ and $155 x$. Then, the overtakes will occur every $1 / x$ units of time. During this time, the pedestrian covers 100 units, which means he will be 10 units away in the opposite ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,845 |
6.1. We will call a natural number interesting if all its digits, except the first and last, are less than the arithmetic mean of the two adjacent digits. Find the largest interesting number. | Answer: 96433469.
Solution. Let $a_{n}$ be the n-th digit of the desired number. By the condition, $a_{n}9$. Similarly, there cannot be four negative differences. Therefore, the maximum number of such differences can be 7, and the desired number is an eight-digit number. To make it the largest, the first digit should ... | 96433469 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,846 |
7.1. Find the smallest natural solution of the inequality $2^{x}+2^{x+1}+\ldots+2^{x+2000}>2^{2017}$. Answer. 17. | Solution. The value $x=17$ is clearly a solution. If $x \leqslant 16$, then we have
$$
2^{x}+2^{x+1}+\ldots+2^{x+2000} \leqslant 2^{16}+2^{17}+\ldots+2^{2015}+2^{2016}<2^{2017}
$$
since this inequality reduces sequentially to the following: $2^{16}+2^{17}+\ldots+2^{2015}<$ $2^{2017}-2^{2016}=2^{2016}, 2^{16}+2^{17}+\... | 17 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 5,847 |
1. On the Island of Knights and Liars, there live knights who always tell the truth and liars who always lie. One day, five residents of this island were asked in turn how many knights were among them.
- One, - answered the first.
- Two, - answered the second.
- Three, - answered the third.
- Don't believe them, they ... | Solution. Among the first three, there is no more than one knight, since they all give different answers. If we consider the 4th and 5th, one of them is a knight and the other is a liar. Therefore, there is one or two knights in total. In any case, one of the first three told the truth. This means there are two knights... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,848 |
2. First-grader Petya was laying out a contour of an equilateral triangle with the tokens he had, so that each of its sides, including the vertices, contained the same number of tokens. Then, with the same tokens, he managed to lay out the contour of a square in the same way. How many tokens does Petya have, if each si... | Answer: 24.
Solution. Let the side of the triangle contain $x$ chips, and the side of the square - $y$ chips. The total number of chips, counted in two ways, is $3 x-3=4 y-4$ (we account for the corner chips being counted twice). From the problem statement, it follows that $y=x-2$. Therefore, we get the equation $3(x-... | 24 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,849 |
3. Petrov and Vasechkin were solving the same arithmetic problem. A certain number had to be divided by 2, multiplied by 7, and 1001 subtracted. Petrov performed all the operations correctly, while Vasechkin got everything mixed up: he divided by 8, squared the result, and also subtracted 1001. It is known that Petrov ... | Answer: 295.
Solution. Note that the number 1001 is divisible by 7 without a remainder. This means that the number Petrov obtained must be divisible by 7. But it is a prime number, so Petrov got 7. Let's reverse the operations Petrov performed and find the original number: $\frac{7+1001}{7} \cdot 2=288$. Now, let's re... | 295 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,850 |
4. Let's call a natural number $n$ squareable if the numbers from 1 to $n$ can be arranged in such an order that each member of the sequence, when added to its position, results in a perfect square. For example, the number 5 is squareable, as the numbers can be arranged as: 32154, in which $3+1=2+2=1+3=4$ and $5+4=4+5=... | Answer: 9 and 15.
Solution. The number 7 cannot be squareable, since both numbers 1 and 6 must be in the third position, which is impossible.
The number 9 is squareable, as the numbers from 1 to 9 can be arranged in the following order: $8,2,6$, $5,4,3,9,1,7$, thus the required condition is satisfied.
The number 11 ... | 915 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,851 |
5. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face? | Answer: 16.
Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).
Consider a face containing the vertex where the number 6 is placed. ... | 16 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,852 |
6. On graph paper (the side of a cell is 1 cm) a rectangle is drawn, the sides of which lie on the grid lines, with one side being 5 cm shorter than the other. It turned out that it can be cut into several pieces along the grid lines and form a square from them. What can the side of this square be? Find all possible va... | # Answer: 6 cm.
Solution. Let the larger side of the rectangle be $k$ cm, $k>5$, and the side of the obtained square be $n$ cm. Then, from the equality of areas, we get $k(k-5)=n^{2}$. Note that $k^{2}-5k=k^{2}-6k+9=(k-3)^{2}$ for $k>9$. Therefore, $6 \leqslant k \leqslant 9$. For $k=6,7,8,9$, we get that $k(k-5)$ equ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,853 |
7. All natural numbers, the sum of the digits of each of which is equal to 5, were arranged in ascending order. What number is in the 125th place?
Otvet: 41000. | Solution. Let's calculate the number of $n$-digit numbers, the sum of the digits of each of which is equal to 5, for each natural $n$. Subtract 1 from the leading digit, we get a number (which can now start with zero), the sum of the digits of which is equal to 4. Represent the digits of this number as cells, in each o... | 41000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 5,854 |
1.1. From the village to the station along one road, a dachnik A set off on foot and a motorcyclist with a passenger - dachnik B - at the same time. Not reaching the station, the motorcyclist dropped off the passenger and immediately headed back to the village, while dachnik B continued on foot to the station. Meeting ... | Answer: $\frac{5}{6}$.
Solution. Let the distance from the village to the station be 1. Since both cottagers spent the same amount of time traveling to the station and used the same motorcycle, the distance they walked is the same (this can also be seen on the graphs of their movement). Let this distance be $x$. Then ... | \frac{5}{6} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 5,855 |
2.1. Find the integer part of the number $a+\frac{9}{b}$, where $a$ and $b-$ are respectively the integer and fractional part of the number $\sqrt{76-42 \sqrt{3}}$. | Answer: 12.
Solution. The given number is $\sqrt{76-42 \sqrt{3}}=\sqrt{(7-3 \sqrt{3})^{2}}=7-3 \sqrt{3}=1+(6-3 \sqrt{3})$, where $6-3 \sqrt{3} \in(0 ; 1)$. Therefore, $a=1, b=6-3 \sqrt{3}$. Thus, $a+\frac{9}{b}=1+\frac{9}{6-3 \sqrt{3}}=1+\frac{3}{2-\sqrt{3}}=1+3(2+\sqrt{3})=7+3 \sqrt{3}$. Since $12<7+3 \sqrt{3}<13$, t... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,856 |
3.1. Solve the system of equations:
$$
\left\{\begin{array}{l}
2^{x+2 y}+2^{x}=3 \cdot 2^{y} \\
2^{2 x+y}+2 \cdot 2^{y}=4 \cdot 2^{x}
\end{array}\right.
$$ | Answer: $x=y=\frac{1}{2}$.
Solution.
$$
\begin{gathered}
\left\{\begin{array}{l}
2^{x} \cdot\left(2^{y}\right)^{2}+2^{x}=3 \cdot 2^{y}, \\
\left(2^{x}\right)^{2} \cdot 2^{y}+2 \cdot 2^{y}=4 \cdot 2^{x}
\end{array}\right. \\
2^{x}=u(>0), \quad 2^{y}=v(>0) \\
\left\{\begin{array}{l}
u v^{2}+u-3 v=0, \\
u^{2} v+2 v-4 u=... | (\frac{1}{2},\frac{1}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,857 |
# 3.2. Solve the system of equations:
$$
\left\{\begin{array}{l}
2^{x+y}-2^{x-y}=4 \\
2^{x+y}-8 \cdot 2^{y-x}=6
\end{array}\right.
$$ | Answer: $x=\frac{5}{2}, y=\frac{1}{2}$.
Solution.
$$
\begin{aligned}
& 2^{x+y}=u(>0), \quad 2^{y-x}=v(>0) \\
& \left\{\begin{array}{l}
u-v=4 \\
u-\frac{8}{v}=6
\end{array}\right. \\
& \left\{\begin{array}{l}
u=v+4 \\
v+4-\frac{8}{v}=6
\end{array}\right. \\
& v^{2}+4 v-6 v-8=0 \\
& v^{2}-2 v-8=0 \\
& {\left[\begin{arr... | \frac{5}{2},\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,858 |
3.4. Solve the system of equations:
$$
\left\{\begin{array}{l}
3^{x+y}-3^{x-y}=18 \\
3^{x+y}-27 \cdot 3^{y-x}=24
\end{array}\right.
$$ | Answer: $x=\frac{5}{2}, y=\frac{1}{2}$. | \frac{5}{2},\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 5,859 |
4.1. On the side $A C$ of triangle $A B C$, points $E$ and $K$ are taken, with point $E$ lying between points $A$ and $K$ and $A E: E K: K C=3: 5: 4$. The median $A D$ intersects segments $B E$ and $B K$ at points $L$ and $M$ respectively. Find the ratio of the areas of triangles $B L M$ and $A B C$. Answer: $\frac{1}{... | Solution. Since the median of a triangle divides its area in half, $S(ABD) = S(ADC)$ and $S(BMD) = S(MDC)$, which means $S(ABM) = S(AMC) = \frac{3}{2} S(AMK)$. At the same time, $\frac{S(ABM)}{S(AMK)} = \frac{BM}{MK}$, so $\frac{BM}{MK} = \frac{3}{2}$. Therefore, $\frac{S(BLM)}{S(LMK)} = \frac{3}{2}$, which means $S(AB... | \frac{1}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 5,860 |
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