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3. (7 points) A seller bought a batch of pens and sold them. Some customers bought one pen for 10 rubles, while others bought 3 pens for 20 rubles. It turned out that the seller made the same profit from each sale. Find the price at which the seller bought the pens. | Answer: 5 rubles.
Solution. Let the purchase price of a pen be $x$. Then the profit from one pen is $10-x$, and from 3 pens is $20-3x$. Solving the equation $10-x=20-3x$, we get $x=5$.
Criteria. Correct solution by any method: 7 points.
If it is not justified that the purchase price of the pen should be 5 rubles, bu... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11-5. Santa Claus is preparing gifts. He has distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest - 50. How many bags are the candies distributed into? How many candies are in the smallest gift? | Answer: 10 packages, 5 candies.
Solution. Let's number the gifts from the smallest to the largest, from 1 to n. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, the ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. The numbers from 1 to 10 were written in some order and resulted in the numbers \(a_{1}, a_{2}, a_{3}, \ldots, a_{10}\), and then the sums \(S_{1}=a_{1}\), \(S_{2}=a_{1}+a_{2}\), \(S_{3}=a_{1}+a_{2}+a_{3}\), \ldots, \(S_{10}=a_{1}+a_{2}+a_{3}+\ldots+a_{10}\) were calculated. What is the maximum number of prime num... | Answer: 7.
Among the numbers from 1 to 10, there are five odd numbers. Adding an odd number changes the parity of the sum. Let $y_{1}^{\prime}, y_{2}, y_{3}, y_{4}, y_{5}$ be the odd numbers, in the order they appear on the board. After adding $y_{2}$, the sum will become even and greater than 2, as it will after addi... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. In the figure, two equal triangles are depicted: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# 4. CONDITION
The sequence of numbers $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots, \mathrm{a}_{\mathrm{n}}, \ldots$ satisfies the relations $\mathrm{a}_{\mathrm{n}}=\mathrm{a}_{\mathrm{n}-1} \cdot \mathrm{a}_{\mathrm{n}-3}$ for $\mathrm{n}=4,5,6, \ldots$ Find $\mathrm{a}_{2019}$, given that $\mathrm{a}_{1... | Solution. It is clear that all members of this sequence are equal to $\pm 1$. We find:
$$
\begin{aligned}
& a_{n}=\left(a_{n-1}\right) \cdot a_{n-3}=\left(a_{n-2} \cdot a_{n-4}\right) \cdot a_{n-3}=\left(a_{n-2}\right) \cdot a_{n-4} \cdot a_{n-3}= \\
& =\left(a_{n-3} \cdot a_{n-4}\right) \cdot a_{n-4} \cdot a_{n-3}=a_... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A 2019-digit number written on the board is such that any number formed by any two adjacent digits (in the order they follow) is divisible by 13. Find the last digit of this number, given that the first digit is 6. | 1. Answer: 2.
Two-digit numbers divisible by 13: $13,26,39,52,65,78,91$. If the first digit of the number is 6, then the second digit must be 5 (forming the number 65, which is divisible by 13), the third digit is 2 (forming 52), and the fourth digit is again 6 (forming 26). Thus, the digits will be arranged in triple... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. If $x=5 m+4$, then $(5 m+4)^{2}+1=25 m^{2}+40 m+17$ is not divisible by 5. | Answer: $\left\{\begin{array}{c}x=5 m+2, \\ y=5 m^{2}+4 m+1 .\end{array}\right.$ or $\left\{\begin{array}{c}x=5 m+3, \\ y=5 m^{2}+6 m+2 .\end{array}\right.$ where $m \in \mathbb{Z}$.
5 points - the solution is correct, complete, and contains no errors.
4 points - if the answer is incorrect, the solution method is cor... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In a school chess tournament, boys and girls competed, with the number of boys being five times the number of girls. According to the tournament rules, each chess player played against every other player twice. How many players in total participated if it is known that the boys scored exactly twice as many points in... | Answer: 6 players.
Solution. Let $d$ girls and $5d$ boys participate in the tournament. Then the total number of players was $d + 5d = 6d$; playing two matches each with every other, they played a total of $2 \cdot \frac{1}{2} \cdot 6d(6d-1) = 6d(6d-1)$ matches. Since each match awards one point, the total number of p... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
10.7. For natural numbers $a>b>1$, define the sequence $x_{1}, x_{2}, \ldots$ by the formula $x_{n}=\frac{a^{n}-1}{b^{n}-1}$. Find the smallest $d$ such that this sequence does not contain $d$ consecutive terms that are prime numbers, for any $a$ and $b$.
(V. Senderov) | 10.7. Answer. 2.
For $a=4, b=2$ we have $\frac{a^{1}-1}{b^{1}-1}=3, \frac{a^{2}-1}{b^{2}-1}=5$. It remains to show that more than two consecutive prime numbers will not occur.
We will prove a stronger statement than required: for $n \geqslant 2$ at least one of the numbers $\frac{a^{n}-1}{b^{n}-1}, \frac{a^{n+1}-1}{b... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. There are 25 coins, 12 of which are counterfeit and differ in weight by exactly 1 g from the genuine ones. All coins weigh an integer number of grams. Some may be lighter than the genuine ones, while others may be heavier. There are balance scales without weights, with a needle that shows the difference in weight. W... | Solution. We will prove that one weighing is sufficient. Set aside the coin under investigation, and place the rest on the scales, 12 coins on each side. If the scales show a difference in an even number of grams, then the coin is genuine; if in an odd number, then it is counterfeit. Indeed, if an odd number of counter... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Find the hypotenuse of a right triangle if the height drawn to it is 1 cm, and one of the angles of the triangle is $15^{\circ}$. If the answer is not an integer, round it to the tenths. | Solution. Let's call the original triangle $ABC$. Let $CH$ be the height drawn to the hypotenuse; $\angle C=90^{\circ}$, and $\angle A=15^{\circ}$. Draw the median $CM$. It is clear that $CM=MA=MB$, so triangle $CMA$ is isosceles $\left(CM=MA \text{ and } \angle MCA=\angle MAC=15^{\circ}\right)$. Note that $\angle BMC=... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.2 In the fishing, 11 experienced fishermen and $n$ children participated. Together they caught $n^{2}+$ $5 n+22$ fish, with all experienced fishermen catching the same amount, and all children catching the same amount, but each 10 less than an experienced fisherman. Who was there more of at the fishing - experienced ... | Solution: Let each child catch $m$ fish. Then $n m+11(m+10)=n^{2}+5 n+22$. From this, $(n+11) m=n^{2}+3 n-88$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+5 n-88=$ $(n+11)(n-6)-22$, so 22 is divisible by $n+11$. The only divisor of 22 greater than 11 is 22 itself, so $n+11=22, n=11$. Therefore, the... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4 Is there a rectangular box, all three dimensions of which (height, width, and depth) are expressed as irrational numbers, while the surface area and volume are integers? | Solution 1: Consider a box with sides $\sqrt{2}-1, \sqrt{2}-1,3+2 \sqrt{2}$. Its volume is $(\sqrt{2}-1)^{2}(3+2 \sqrt{2})=(3-2 \sqrt{2})(3+2 \sqrt{2})=1$, and the surface area is $2(\sqrt{2}-1)^{2}+4(\sqrt{2}-1)(3+2 \sqrt{2})=2(3-2 \sqrt{2})+4(\sqrt{2}+1)=10$.
Solution 2: Consider the polynomial $f(x)=x^{3}-30 x^{2}+... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9.5 What is the largest number of non-overlapping groups into which all integers from 1 to 20 can be divided so that the sum of the numbers in each group is a perfect square? | Solution: A group consisting of a single number can only be formed by 4 squares. The remaining 16 numbers must be divided into groups of at least two. Therefore, there will be no more than 12 groups in total. Let's check that exactly 12 groups are not possible. Indeed, in such a case, the numbers 1, 4, 9, 16 would form... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the numerical value of the expression
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{2}{a b+1}
$$
if it is known that $a$ is not equal to $b$ and the sum of the first two terms is equal to the third. | Answer: 2.
Solution. Let's bring the condition to a common denominator
$$
\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}=\frac{2}{a b+1}
$$
we get
$$
\frac{\left(a^{2}+b^{2}+2\right)(a b+1)-2\left(a^{2}+1\right)\left(b^{2}+1\right)}{\left(a^{2}+1\right)\left(b^{2}+1\right)(a b+1)}=0
$$
expand all brackets in the numerator, c... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Find the number of integer points $(x, y)$ satisfying the equation $\frac{1}{|x|}+\frac{1}{|y|}=\frac{1}{2017}$. | Solution. Transform the original equation $\frac{1}{|x|}+\frac{1}{|y|}=\frac{1}{2017} \Rightarrow$
$2017(|x|+|y|)-|x||y|=0 \Rightarrow 2017 \cdot|x|+2017 \cdot|y|-|x||y|-2017^{2}=-2017^{2}$, from which it follows that $(|x|-2017)(|y|-2017)=2017^{2}$. Since 2017 is a prime number and $|x|$ and $|y|$ are natural numbers... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. The brother left the house 6 minutes after his sister, following her, and caught up with her after 12 minutes. How many minutes would it have taken him to catch up if he had walked twice as fast? Both the brother and the sister walk at a constant speed. | # Answer: 3 minutes.
Solution. Since the brother walked for 12 minutes before meeting his sister, and the sister walked for 18 minutes, the brother's speed was $3 / 2$ times the sister's speed. If the brother's speed is 3 times the sister's speed, which is 2 times faster than before, then the difference of 6 minutes w... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In the neighboring houses on Happy Street in Sunny Village, two families of 10 people each lived, with the average weight of the members of one family being 1 kg more than the average weight of the members of the other family. After the eldest sons from both families left to study in the city, it turned out that the... | Answer: by 1 kg or by 19 kg.
Solution. The total weight of the members of one of the families before the departure of the elder sons was 10 kg more than the total weight of the members of the other family. If the heavier family remained heavier, then its total weight became 9 kg more than the total weight of the other... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5 The numbers $x$, $y$, and $z$ satisfy the equations
$$
x y + y z + z x = x y z, \quad x + y + z = 1
$$
What values can the sum $x^{3} + y^{3} + z^{3}$ take? | Solution 1: Let $x y z=p$. Then, from the condition $x y+y z+z x$ is also equal to $p$. Therefore, by Vieta's theorem, the numbers $x, y$, and $z$ are the roots of the polynomial $t^{3}-t^{2}+p t-p$. However, the number 1 is a root of such a polynomial, so one of the numbers is equal to 1. Then the other two numbers ar... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.5 On an island, there live 25 people: knights, liars, and tricksters. Knights always tell the truth, liars always lie, and tricksters answer the questions posed to them in turn, alternating between truth and lies. All the islanders were asked three questions: "Are you a knight?", "Are you a trickster?", "Are you a li... | Solution. Each knight will answer "yes" to the first question and "no" to the other two. Each liar will answer "yes" to the first two questions and "no" to the last one. The tricksters can be divided into those who answered the first question truthfully (tricksters of the first type) and those who answered the first qu... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.1. What is the maximum number of L-shaped pieces
| |
| :--- |
consisting of three $1 x 1$ squares, that can be placed in a 5x7 rectangle? (The L-shaped pieces can be rotated and flipped, but they cannot overlap). | Solution: The area of the corner is 3, and the area of the rectangle is 35, so 12 corners cannot fit into the rectangle. The image below shows one way to place 11 corners in the rectangle.
A... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. In triangle $A B C$, the lengths of the sides are known: $A B=4, B C=5, C A=6$. Point $M$ is the midpoint of segment $B C$, and point $H$ is the foot of the perpendicular dropped from $B$ to the angle bisector of angle $A$. Find the length of segment $H M$. If necessary, round your answer to the hundredths.
# | # Answer. 1.
Solution. Let $D$ be the intersection point of line $B H$ with line $A C$. Triangle $A B D$ is isosceles because in it the bisector and the altitude from vertex $A$ coincide. Therefore, $H$ is the midpoint of segment $B D$. Then $H M$ is the midline of triangle $B C D$. Note that $C D = A C - A D = A C - ... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. The school stage of the Magic and Wizardry Olympiad consists of 5 spells. Out of 100 young wizards who participated in the competition,
- 95 correctly performed the 1st spell
- 75 correctly performed the 2nd spell
- 97 correctly performed the 3rd spell
- 95 correctly performed the 4th spell
- 96 correctly performed... | # Answer: 8.
Solution. The number of students who correctly performed all spells is no more than 75, since only 75 students correctly performed the second spell. The number of students who made mistakes in the 1st, 3rd, 4th, or 5th spells is no more than $(100-95)+(100-97)+(100-95)+(100-96)=$ $5+3+5+4=17$. If a studen... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Rational numbers a, b, and c are such that $(a+b+c)(a+b-c)=4 c^{2}$. Prove that $\mathrm{a}+\mathrm{b}=0$.
---
The translation maintains the original text's line breaks and formatting. | Solution. The initial equality is equivalent to the following $(a+b)^{2}-c^{2}=4 c^{2}$, or $(a+b)^{2}=5 c^{2}$. If $c \neq 0$, we get $((a+b) / c)^{2}=5 .|(a+b) / c|={ }^{-}$. On the left, we have a rational number, since the sum, quotient, and absolute value of rational numbers are rational, while on the right, we ha... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find $AB$, if $CA=9, CB=4$. | Answer: 6.
Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. A road 28 kilometers long was divided into three unequal parts. The distance between the midpoints of the extreme parts is 16 km. Find the length of the middle part. | Answer: 4 km.
Solution. The distance between the midpoints of the outermost sections consists of half of the outer sections and the entire middle section, i.e., twice this number equals the length of the road plus the length of the middle section. Thus, the length of the middle section $=16 * 2-28=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.5. The older brother took identical uncolored cubes from Misha and used them to build a large cube. After that, he completely painted some (not all) faces of the large cube red. When the paint dried, Misha disassembled the large cube and found that exactly 343 small cubes had no red faces. How many faces of the large... | Solution: We will call a small cube that has a red face painted. The size of the large cube is greater than 7 (since only the unpainted cubes amount to $343=7^{3}$, and there are also painted ones), but less than 9 (since all "internal" cubes are unpainted - no more than $7^{3}$). Therefore, it is equal to 8. Out of $8... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.1. A cube lies on a plane. Each face is marked with 1 to 6 points such that the sum of points on opposite faces is always 7. The cube is rolled over the plane as shown in the picture. How many points will be on the top face of the cube when it lands on the last cell?
![](https://cdn.mathpix.com/cropped/2024_05_06_ff... | Answer: 5
Solution. To find out the answer, you need to imagine rolling the die in the opposite direction and each time keep track of where the face that will end up on top will be. It turns out that this face will be at the bottom in the initial position of the die. Since the top face has 2 dots, the bottom face now ... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. On the board, there are five "equations" of the form $x^{2}+\ldots x+\ldots=0$. Two players take turns filling in the dots with natural numbers from 1 to 10, with each number being used only once. The game ends when all numbers are filled in. The player who makes the first move wants to have as many equation... | Answer. 3. Solution. To obtain three equations, each having two distinct roots, the first player needs to use the three initial moves to write the largest of the numbers not yet written in the place of the coefficient of $x$ in the "equation" where no number has yet been written. This is possible because if no more tha... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 10.5
The altitudes AD and $\mathrm{BE}$ of an acute-angled triangle $\mathrm{ABC}$ intersect at point $\mathrm{H}$.
The circumcircle of triangle $\mathrm{ABH}$ intersects sides $\mathrm{AC}$ and $\mathrm{BC}$ at points $\mathrm{F}$ and $\mathrm{G}$, respectively. Find FG, if $\mathrm{DE}=5$ cm.
## Number of p... | # Answer:
$\mathrm{FG}=10 \mathrm{~cm}$
## Solution
We have only one numerical given.
So, the length of FG will either be equal to ED or a multiple of it. It doesn't seem to be equal. It should be larger (judging by the diagram), by some factor. What do we know about such relationships? There is the midline of a tr... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Cut a $3 \times 9$ rectangle into 8 squares.
7 points are awarded for a complete solution to each problem
The maximum total score is 35 | 5. First, cut the rectangle into three squares of size $3 \times 3$. Leave two of them, and from the third, cut out a square of size $2 \times 2$. Cut the remaining part into 5 squares of size $1 \times 1$. In total, you will have 8 squares. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. The cells of the pyramid are filled according to the following rule: above every two adjacent numbers, their arithmetic mean is written. Some numbers were erased, and the structure shown in the figure was obtained. What number was in the bottom right cell? (The arithmetic mean of two numbers is their sum d... | Answer: 6.
Solution. Let's restore the numbers in the table by going through it from top to bottom. For example, if the numbers 21 and $x$ are in the second row, then from $18=\frac{1}{2}(21+x)$ we get $x=15$. Similarly, in the third row, we get that next to the number 14 is 16, and next to it is -26; in the last row,... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.1. Find $x^{2}+y^{2}+z^{2}$, if $x+y+z=2, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$. | # 9.1. Answer: 4.
## 9th grade
By getting rid of the denominators in the second equation, we get $x y + y z + z x = 0$. Squaring the first equation, we get $x^{2} + y^{2} + z^{2} + 2(x y + y z + z x) = 4$. From this, we obtain the answer. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.4. For four consecutive natural numbers, it is known that the largest of them is a divisor of the product of the other three. Find all values that the largest of these numbers can take. | Answer: 6.
First solution. Let our numbers be $n-3, n-2, n-1, n$, where $n-3 \geqslant 1, n \geqslant 4$. Since the numbers $n-1$ and $n$ are coprime, $(n-3)(n-2)$ is divisible by $n$. Note that $(n-3)(n-2) = n^2 - 5n + 6 = n(n-5) + 6$, and since both numbers $(n-3)(n-2)$ and $n(n-5)$ are divisible by $n$, 6 is also d... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.4 The Oddball marked the centers of 17 cells in an $N \times N$ grid such that the distance between any two marked points is greater than 2. What is the smallest value that $N$ can take? | Solution: We will show that in an $8 \times 8$ square (and then in any smaller size), it is impossible to mark the cells in this way. Indeed, let's divide the square into $2 \times 2$ squares. In each of them, the pairwise distances between the centers of the cells do not exceed $\sqrt{2}$, so no more than one cell is ... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Dima wrote a sequence of zeros and ones in his notebook. Then he noticed that a one follows a zero 16 times, a one follows 01 seven times, and a zero follows 01 eight times. What could be the last two digits on the board? (In your answer, indicate all options and prove that there are no others). | Answer. 01.
Solution. The combination 01 occurs 16 times in the tetrad. After it, $7+8=15$ times there is a 0 or 1, and one time there is not. Therefore, one of the combinations 01 stands at the end of the line.
Criteria. Full solution - 7 points. Partial examples of sequences with the correct answer - 1. Only answer... | 1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 5. CONDITION
A tourist goes on a hike from $A$ to $B$ and back, and completes the entire journey in 3 hours and 41 minutes. The route from $A$ to $B$ first goes uphill, then on flat ground, and finally downhill. Over what distance does the road pass on flat ground, if the tourist's speed is 4 km/h when climbing uphi... | Solution. Let $x$ km of the path be on flat ground, then $9-x$ km of the path (uphill and downhill) the tourist travels twice, once (each of the ascent or descent) at a speed of 4 km/h, the other at a speed of 6 km/h, and spends $(9-x) / 4+(9-x) / 6$ hours on this part. Since the tourist walks $2 x / 5$ hours on flat g... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Three circles with radii 1, 2, 3 touch each other externally at three points. Find the radius of the circle passing through these three points. | Answer: 1.
Solution: Let $\mathrm{O}_{1}, \mathrm{O}_{2}$ and $\mathrm{O}_{3}$ be the centers of the given circles, K, M, N the points of tangency, such that $\mathrm{O}_{1} \mathrm{~K}=\mathrm{O}_{1} \mathrm{~N}=1, \mathrm{O}_{2} \mathrm{~K}=\mathrm{O}_{2} \mathrm{M}=2$ and $\mathrm{O}_{3} \mathrm{~N}=$ $\mathrm{O}_{... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.5. What is the minimum number of cells that need to be marked on a 5 by 5 board so that among the marked cells there are no adjacent ones (having a common side or a common vertex), and adding any one cell to these would violate the first condition? | Answer: 4 cells.
Solution: Estimation. Divide the board into four parts (see fig.). In each of them, a cell must be marked, otherwise the black cell contained in it can be added.
Example. The four black cells in the figure satisfy both conditions.
Criteria: Answer only - 0 points. Estimation only - 5 points. Example... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. In a white $10 \times 10$ square, on the first move, a $1 \times 1$ cell rectangle is painted, on the second move - a $1 \times 2$ cell rectangle, on the third - $1 \times 3$ and so on, as long as it is possible to do so. After what minimum number of moves could this process end? (Cells cannot be painted over again.... | # Answer: after 6 moves.
Solution. Evaluation. On the board, 16 rectangles of size $1 \times 6$ can be highlighted, of which a maximum of $1+2+3+4+5=15$ will contain colored cells, meaning that a sixth move is always possible. Example. The process is illustrated in the diagram, showing a scenario where a seventh move ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.1
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
Now it is easy to count the resulting pieces. For convenience, they are highl... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.3
## Condition:
A sheet of paper was folded like an accordion as shown in the figure, and then folded in half along the dotted line. After that, the entire resulting square stack was cut along the diagonal.
 balls. Each is painted in some color. If you take out any three balls from the box, there will definitely be at least one red and at least one blue among them. How many balls can be in the box | Answer: 4. Solution. In the box, there are no more than two red and blue balls (otherwise, it would be possible to draw three red or three blue balls) and no more than one ball of other colors (otherwise, it would be possible to draw one blue or red ball and two balls of other colors). Therefore, there are no more than... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# 3. Option 1.
In a box, there are chips. Tolya and Kolya were asked how many chips are in the box. Tolya answered: “Less than 7”, and Kolya answered: “Less than 5”. How many chips can be in the box if it is known that one of the answers is correct? Find all the options. In the answer, write their sum. | Answer: 11.
Solution: If there are 7 or more chips in the box, then both boys are lying. If there are 4 or fewer chips in the box, then both boys are telling the truth. If there are 5 or 6 chips in the box, then Tolya is telling the truth, and Kolya is lying. | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
6. Variant 1. In the kindergarten, 5 children eat porridge every day, 7 children eat porridge every other day, and the rest never eat porridge. Yesterday, 9 children ate porridge. How many children will eat porridge today? | Answer: 8.
Solution. Of the 9 boys who ate porridge yesterday, 5 boys eat it every day, so the remaining $9-5=4$ eat it every other day. Therefore, these four will not eat porridge today, while the other $7-4=3$ of those who eat every other day will. So today, these three, as well as the five who always eat porridge, ... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7. Variant 1. On a rectangular plot measuring $5 \times 6$ meters, there is a triangular flower bed (see figure). Find the area of this flower bed (in square meters).
 | Answer: 10.
Solution. The area of the flower bed can be calculated as the difference between the area of the plot and the sum of the areas of the two triangular parts not occupied by the flower bed:
: four $3 \times 4$ and a $1 \times 1$ square. If only three cells are shaded, there will be a white rectangle of 12 cells. How to shade 4 cells is shown in the following figure:
 | # 3. Answer. 4.
Let the cells of a $5 \times 5$ square be painted in such a way that no more corners can be painted. Consider the 4 corners marked on the diagram. Since none of these corners can be painted, at least one cell in each of them must be painted. Note that one corner cannot paint cells of two marked corners... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. We will call a number greater than 25 semi-prime if it is the sum of some two distinct prime numbers. What is the maximum number of consecutive natural numbers that can be semi-prime? | 4. Answer. 5.
Note that an odd semiprime number can only be the sum of two and an odd prime number.
Let's show that three consecutive odd numbers \(2n+1, 2n+3\), and \(2n+5\), greater than 25, cannot all be semiprimes simultaneously. Assuming the contrary, we get that the numbers \(2n-1, 2n+1\), and \(2n+3\) are prim... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) The function $f$ is such that for any $x>0, y>0$ the equality $f(x y)=f(x)+f(y)$ holds. Find $f(2019)$, if $f\left(\frac{1}{2019}\right)=1$. | Solution. When $y=1 \quad f(x)=f(x)+f(1), f(1)=0$.
When $x=2019 \quad y=\frac{1}{2019} f(1)=f(2019)+f\left(\frac{1}{2019}\right)$,
$f(2019)=f(1)-f\left(\frac{1}{2019}\right)$
$f(2019)=-f\left(\frac{1}{2019}\right)=-1$.
Answer. $f(2019)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2. Find the GCD of all numbers obtained by all possible permutations of the digits of the number 202120222023
Solution. By the divisibility rule, all these numbers are divisible by 9 (the sum of the digits is 18). A sufficient condition to prove that there are no other numbers is that the difference between any two ... | Answer: 9.
## Criteria
7 points - complete solution;
3 points - GCD found, but not proven that there can be no other divisors. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4.1. Among fifty consecutive natural numbers, exactly 8 are divisible by 7 without a remainder. What is the remainder when the eleventh number in the sequence is divided by 7?
# | # Answer: 3
Solution. The closest numbers that give the same remainder when divided by 7 differ by 7. Therefore, among fifty consecutive numbers, 6 groups of numbers with the same remainder are formed, each containing 7 numbers, and only one group containing 8 numbers, with the first and last numbers among them. Thus,... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. We will build a sequence of numbers in the following way. We will place the number 7 in the first position, and after each number, we will place the sum of the digits of its square, increased by one. For example, the second position will hold the number 14, since $7^{2}=49$, and $4+9+1=14$. The third position will h... | Solution Let's continue finding the first few terms of the sequence: $7 ; 14 ; 17 ; 20 ; 5 ; 8 ; 11 ; 5 ; \ldots$ - the number 5 is repeated. This means the sequence has a period of length 3: the numbers 5; 8; 11 will repeat subsequently. The number 8 is in the sixth position, so for any $\mathrm{k}>0$, the number 8 wi... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.2 Points A, B, C, D are consecutive vertices of a regular n-gon. What is n if $\angle \mathrm{ACD}=120^{\circ}$? | Solution. We consider the given n-gon as inscribed.
Each small arc between adjacent vertices is $\frac{1}{\mathrm{n}} \cdot 360^{\circ}$. The angle $\mathrm{ACD}$ is an inscribed angle and s... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9.3 Let a, b, c be non-negative integers such that $28a + 30b + 31c = 365$. Prove that $a + b + c = 12$. | Solution. If it were $\mathrm{a}+\mathrm{b}+\mathrm{c} \leq 11$, then we would have $28 \mathrm{a}+30 \mathrm{~b}+31 \mathrm{c} \leq 31(\mathrm{a}+\mathrm{b}+\mathrm{c}) \leq 31 \cdot 11 < 365$. Therefore, $\mathrm{a}+\mathrm{b}+\mathrm{c} \geq 12$. We will show that the inequality $\mathrm{a}+\mathrm{b}+\mathrm{c} \ge... | 12 | Number Theory | proof | Yes | Yes | olympiads | false |
Problem 7.8. In a chess tournament, 30 chess players participated, each playing against each other exactly once. A win was awarded 1 point, a draw - $1 / 2$, and a loss - 0. What is the maximum number of chess players who could have ended up with exactly 5 points by the end of the tournament? | Answer: 11.
Solution. Let $N-$ be the number of chess players with 5 points.
Assume $N \geqslant 12$. In each game between two playing chess players, 1 point is played, so the sum of points of all chess players who scored 5 points at the end of the tournament is no less than $\frac{N \cdot(N-1)}{2}$. Then, by the pig... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6.4. The frame for three square photographs has the same width everywhere (see figure). The perimeter of one opening is 60 cm, the perimeter of the entire frame is 180 cm. What is the width of the frame? | Answer: 5 cm.
Solution. From the condition of the problem, it follows that the length of one hole's side is 15 cm. Let $d$ cm be the width of the frame, then the perimeter of the rectangle is $8 \cdot 15 + 12 d = 120 + 12 d$ (cm). According to the condition, $120 + 12 d = 180$, which means $d = 5$.
Similar reasoning ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6.5. The arithmetic mean of four numbers is 10. If one of these numbers is erased, the arithmetic mean of the remaining three increases by 1; if instead another number is erased, the arithmetic mean of the remaining numbers increases by 2; and if only the third number is erased, the arithmetic mean of the remaining inc... | Answer: will decrease by 6.
Solution. From the fact that the arithmetic mean of four numbers is 10, it follows that the sum of these numbers is 40. Similarly, the sum of three numbers (excluding the first) is 33, the sum of three numbers (excluding the second) is 36, and the sum of three numbers (excluding the third) ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9.8. On the board, there are $N \geqslant 9$ different non-negative numbers, each less than one. It turns out that for any eight different numbers on the board, there is a ninth, different from them, such that the sum of these nine numbers is an integer. For which $N$ is this possible?
(F. Nilov) | Answer. Only for $N=9$.
Solution. It is clear that for $N=9$ the required is possible - it is sufficient to write 9 different positive numbers on the board with a unit sum. We will show that for $N>9$ the required is impossible. Suppose the opposite; let $S$ be the sum of all numbers on the board.
Choose arbitrary nu... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8.2. There is a balance scale without weights and 11 visually identical coins, among which one may be counterfeit, and it is unknown whether it is lighter or heavier than the genuine coins (genuine coins have the same weight). How can you find at least 8 genuine coins in two weighings? | Solution. Let's divide the coins into three piles of three coins each. Compare pile 1 and pile 2, and then compare pile 2 and pile 3. If all three piles weigh the same, then all the coins in them are genuine, and we have found 9 genuine coins. Otherwise, one of the piles differs in weight from the others, and the count... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8.5. Find all numbers of the form $22 \ldots 2$ that can be represented as the sum of two perfect squares. | Answer: 2.
Solution. Let $22 \ldots 2=a^{2}+b^{2}$ for some integers $a$ and $b$.
If the numbers $a$ and $b$ are even, then the sum of their squares is divisible by 4, but the number $22 \ldots 2$ is not.
Thus, the numbers $a$ and $b$ can only be odd:
$$
a=2 k+1, b=2 l+1(k, l \in Z)
$$
Therefore, the sum
$$
a^{2}... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Task 10.1. Find any natural $x$ such that the value of the expression $2^{x}+2^{8}+2^{11}$ is a square of a natural number. | Answer: 12.
Solution. Note that $x=12$ works: $2^{12}+2^{8}+2^{11}=\left(2^{6}\right)^{2}+\left(2^{4}\right)^{2}+2 \cdot 2^{6} \cdot 2^{4}=\left(2^{6}+2^{4}\right)^{2}$. Remark. The number $x=12$ is the only one that fits the condition of the problem. | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. There are 36 balls lying in a circle, each of which is either red or blue (balls of each of these colors are present). It is known that:
- for any red ball, there is exactly one red ball such that there is exactly one ball between them;
- for any red ball, there is exactly one red ball such that there ar... | Answer:
(a) (2 points) 12.
(b) (2 points) 24.
Solution. By the condition, there will be two red balls, between which lies exactly one ball. Number the balls clockwise with numbers \(1, 2, \ldots, 36\) such that the 1st and 3rd balls are red. From the condition, it follows that the 35th ball is blue (otherwise, the f... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. The school principal, the caretaker, and the parent committee, failing to agree with each other, each bought a carpet for the school auditorium, which is $10 \times 10$. After thinking about what to do, they decided to place all three carpets as shown in the picture: the first carpet $6 \times 8$ - in one ... | Answer: 6.
Solution. We will measure all dimensions in meters and the area in square meters.
Let's look at the overlap of the second and third carpets. This will be a rectangle $5 \times 3$ (5 along the horizontal, 3 along the vertical), adjacent to the right side of the square room, 4 units from the top side, and 3 ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. The figure shows two equal triangles: $A B C$ and $E B D$. It turns out that $\angle D A E = \angle D E A = 37^{\circ}$. Find the angle $B A C$.
 | Answer: 7.
Fig. 4: to the solution of problem 8.7
Solution. Draw segments $A D$ and $A E$ (Fig. 4). Since $\angle D A E=\angle D E A=37^{\circ}$, triangle $A D E$ is isosceles, $A D=D E$.
... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. The graph of the function $f(x)=\frac{1}{12} x^{2}+a x+b$ intersects the $O x$ axis at points $A$ and $C$, and the $O y$ axis at point $B$, as shown in the figure. It turned out that for the point $T$ with coordinates $(3 ; 3)$, the condition $T A=T B=T C$ is satisfied. Find $b$.
Fig. 11: to the solution of problem 10.7
Solution. Let point $A$ have coordinates $\left(x_{1} ; 0\right)$, and point $C$ have coordinates $\left(x_{2} ; 0\right)$. From the con... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a room, 15 chairs are arranged in a circle. Three jewelers, when no one is watching, sit on three adjacent chairs, and the one sitting in the middle chair hides a diamond in the chair he is sitting on. The inspector has several detectors that show whether someone has sat on a chair or not. What is the minimum num... | Answer: 9 detectors.
Instructions. Evaluation. Consider five chairs such that between any two nearest ones there are two other chairs. If there is no detector in two nearest ones, then by adding to the two chairs between them any of them, we get two possible seating arrangements of jewelers, in which the detector read... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. A jeweler has six boxes: two contain diamonds, two contain emeralds, and two contain rubies. On each box, it is written how many precious stones are inside.
It is known that the total number of rubies is 15 more than the total number of diamonds. What is the total number of emeralds in the boxes?
Fig. 1: to the solution of problem 8.4
Notice that $\angle ABK = \angle CBL$, since they both complement $\angle ABL$ to ... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. Petya runs down from the fourth floor to the first floor 2 seconds faster than his mother rides the elevator. Mother rides the elevator from the fourth floor to the first floor 2 seconds faster than Petya runs down from the fifth floor to the first floor. How many seconds does it take for Petya to run down from the ... | Answer: 12 seconds.
Solution. Between the first and fourth floors, there are 3 flights, and between the fifth and first floors, there are 4. According to the problem, Petya runs 4 flights 2 seconds longer than it takes his mother to ride the elevator, and 3 flights 2 seconds faster than his mother. Therefore, it takes... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On the number line, points with integer coordinates are painted red and blue according to the following rules: a) points whose coordinate difference is 7 must be painted the same color; b) points with coordinates 20 and 14 should be painted red, and points with coordinates 71 and 143 - blue. In how many ways can all... | Answer. In eight ways.
Solution. From part a), it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers $0,1,2,3,4,5$, and 6. The point $0=14-2 \cdot 7$ must be colored the same as 14, i.e., red. Similarly, the point $1=71-10 ... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. Given a rectangle $A B C D$. Point $M$ is the midpoint of side $A B$, point $K$ is the midpoint of side $B C$. Segments $A K$ and $C M$ intersect at point $E$. How many times smaller is the area of quadrilateral $M B K E$ compared to the area of quadrilateral $A E C D$? | Answer: 4 times.
Solution. Draw segments $MK$ and $AC$. Quadrilateral $MBKE$ consists of triangles $MBK$ and $MKE$, while quadrilateral $AEC D$ consists of triangles $AEC$ and $ACD$. We can reason in different ways.
1st method. Triangles $MBK$ and $ACD$ are right-angled, and the legs of the first are half the length ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.1. There are 40 pencils of four colors - 10 pencils of each color. They were distributed among 10 children so that each received 4 pencils. What is the smallest number of children that can always be selected to ensure that pencils of all colors are found among them, regardless of the distribution of pencils?
(I. Bo... | 10.1. Answer. 3 boys.
We will show that it is always possible to choose three boys such that they have pencils of all colors. Since there are 10 pencils of each color and each boy received 4 pencils, at least one boy must have received pencils of at least two different colors. It remains to add to him two boys who hav... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11.1. Plot the graph of the function $\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$. | Answer. The graph of the function will be the line $y = 4$.
## Solution.
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2 \cos 2 x+3}+\sqrt{4 \cos ^{4} x+2 \cos 2 x+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x-2+4 \sin ^{2} x+3}+\sqrt{4 \cos ^{4} x+4 \cos ^{2} x-2+3}$
$\mathrm{y}=\sqrt{4 \sin ^{4} x+4 \sin ^{2} x+1}+\sqrt{4 \cos ^{4} x+4 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.7. In three of the six circles of the diagram, the numbers 4, 14, and 6 are recorded. In how many ways can natural numbers be placed in the remaining three circles so that the products of the triples of numbers along each of the three sides of the triangular diagram are the same?
$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that
$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$
Find the length of segment $AE$, if $DC = 9$.
Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common s... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.3. What is the maximum number of digits that a natural number can have, where all digits are different, and it is divisible by each of its digits? | Answer: 7 digits.
Evaluation. There are 10 digits in total. The number cannot contain the digit 0, so there are no more than 9. If all 9, then the digit 5 must be at the end of the number (divisibility rule for 5), but in this case, the number must also be divisible by 2. Therefore, the digit 5 is not in this number. ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Dima was supposed to arrive at the station at 18:00. By this time, his father was supposed to pick him up in a car. However, Dima managed to catch an earlier train and arrived at the station at 17:05. He didn't wait for his father and started walking towards him. On the way, they met, Dima got into the car, and they... | Answer: 6 km/h
Solution. Dima arrived home 10 minutes earlier, during which time the car would have traveled the distance Dima walked twice. Therefore, on the way to the station, the father saved 5 minutes and met Dima at 17:55. This means Dima walked the distance from the station to the meeting point in 50 minutes, s... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In triangle $ABC$, the median from vertex $A$ is perpendicular to the bisector of angle $B$, and the median from vertex $B$ is perpendicular to the bisector of angle $A$. It is known that side $AB=1$. Find the perimeter of triangle $ABC$. | Answer: 5.
Solution. Let $A M$ be the median drawn from vertex $A$. Then, in triangle $A B M$, the bisector of angle $B$ is perpendicular to side $A M$, i.e., the bisector is also an altitude. Therefore, this triangle is isosceles, $A B = B M = 1$. Hence, $B C = 2 B M = 2$. Similarly, from the second condition, we get... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. There are three vessels with volumes of 3 liters, 4 liters, and 5 liters, without any markings, a water tap, a sink, and 3 liters of syrup in the smallest vessel. Can you, using pourings, obtain 6 liters of a water-syrup mixture such that the amount of water is equal to the amount of syrup in each vessel? | Solution.
For example, as follows (see the table below, c - syrup, w - water, f - final mixture).
| |  | 4-liter container | 5-liter container |
| :---: | :---: | :---: | :---: |
| Pour the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. After a football match, the coach lined up the team as shown in the figure, and commanded: "Run to the locker room, those whose number is less than that of any of their neighbors." After several people ran away, he repeated his command. The coach continued until only one player was left. What is Igor's num... | Answer: 5.
Solution. It is clear that after the first command, the players left are $9,11,10,6,8,5,4,1$. After the second command, the players left are $11,10,8,5,4$. After the third - $11,10,8,5$. After the fourth - $11,10,8$. Therefore, Igor had the number 5. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.7. In triangle $ABC$, the bisector $AL$ is drawn. Points $E$ and $D$ are marked on segments $AB$ and $BL$ respectively such that $DL = LC$, $ED \parallel AC$. Find the length of segment $ED$, given that $AE = 15$, $AC = 12$.
Fig. 5: to the solution of problem 9.7
Solution. On the ray $AL$ beyond point $L$, mark a point $X$ such that $XL = LA$ (Fig. 5). Since in the quadrilateral $ACXD$ the diagonals ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Task № 2.3
## Condition:
Dmitry has socks in his wardrobe: 14 pairs of blue socks, 24 pairs of black socks, and 10 pairs of white socks. Dmitry bought some more pairs of black socks and found that now the black socks make up 3/5 of the total number of socks. How many pairs of black socks did Dmitry buy? | Answer: 12
Exact match of the answer -1 point
Solution by analogy with task №2.1.
# | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 6.4
## Condition:
On the faces of a cube, 6 letters are drawn: A, B, V, G, D, E. The picture shows three images of the cube from different angles. Which letter is drawn on the face opposite the face with the letter $\mathrm{A}$?
![](https://cdn.mathpix.com/cropped/2024_05_06_4cfa7cdc7a8a51b9e752g-25.jpg?he... | Solution by analogy with problem №6.1.
## Condition:
Anya, Borya, and Vasya took the same test consisting of 6 questions, each of which can be answered with "yes" or "no." The answers are presented in the table:
| Question No. | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| Anya | no |... | 3 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
10.6. On the board, the expression $\cos x$ is written. It is allowed to add or multiply several expressions written on the board (an expression can be used multiple times) and write the new obtained expression on the board. Is it possible to get an expression that takes the value 0 when $x=\pi$ after several actions?
... | Answer. Yes, it is possible.
Solution. The first action is to append $\cos ^{2} x$, the second is to append $\cos ^{2} x+\cos x$. Since $\cos \pi=-1$, the value of the last expression at $x=\pi$ is 0.
Comment. An answer without presenting the required expression - 0 points.
Only presenting any correct required expre... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. Three lines intersect to form 12 angles, and $n$ of them turn out to be equal. What is the maximum possible value of $n$? | Answer: 6.
Solution: Three lines limit a certain triangle. If this triangle is equilateral, then out of twelve angles, six are $60^{\circ}$, and the other six are $120^{\circ}$.
Can any external angle of the triangle be equal to its internal angle? It is equal to the sum of the non-adjacent internal angles, so it is ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# 8. Variant 1
In trapezoid $A B C D(A D \| B C)$, the bisectors of angles $D A B$ and $A B C$ intersect on side $C D$. Find $A B$, if $A D=5, B C=2$. | Answer: 7.
Solution.
Mark point $L$ on side $AB$ such that $LB = BC$. Let $K$ be the intersection point of the angle bisectors of $\angle DAB$ and $\angle ABC$. Then triangles $LBK$ and $BC... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.5. The hedgehogs collected 65 mushrooms and divided them so that each hedgehog got at least one mushroom, but no two hedgehogs had the same number of mushrooms. What is the maximum number of hedgehogs that could be | Answer: 10
Solution: If there were 11 hedgehogs, then together they would have collected no less than $1+2+3+\ldots+10+11=66$ mushrooms, which exceeds the total number of mushrooms collected. Therefore, there were fewer than 11 hedgehogs. We will show that there could have been 10 hedgehogs. Suppose the first found 1 ... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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