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11-6. Petya uses all possible ways to place the signs + and - in the expression $1 * 2 * 3 * 4 * 5 * 6$ in the places of the asterisks. For each arrangement of the signs, he calculates the resulting value and writes it on the board. On the board, some numbers may appear multiple times. Petya adds up all the numbers on ... | Answer: 32.
Solution. Note that each of the digits $2,3,4,5,6$ will contribute zero to Petya's sum: each will equally often appear with a + sign and with a - sign. The digit 1 will appear in all sums with a + sign as many times as there are addends in total. Since each of the asterisks can take two values, there will ... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,397 |
11-7. Given a parallelogram $A B C D$, where $\angle B=111^{\circ}$ and $B C=B D$. On the segment $B C$, a point $H$ is marked such that $\angle B H D=90^{\circ}$. Point $M$ is the midpoint of side $A B$. Find the angle $A M H$. Give your answer in degrees. | Answer: $132^{\circ}$.
Solution. Note that $\angle D M B=90^{\circ}$, since $D A=D B$, and in the isosceles triangle $B D A$, the median $D M$ is also the altitude. Since angles $D H B$ and $D M B$ are right angles, points $M, B, H$, and $D$ lie on the same circle. It is clear that we need to find the angle $D M H$, a... | 132 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,398 |
11-8. In a caravan, there are 100 camels, both one-humped and two-humped, and there are at least one of each. If you take any 62 camels, they will have at least half of the total number of humps in the caravan. Let $N$ be the number of two-humped camels. How many values (in the range from 1 to 99) can $N$ take? | Answer: 72.
Solution. If there were $N$ two-humped camels in total, then there were $100-N$ one-humped camels, and there were $100+N$ humps in total. Let's line up the camels: first the one-humped ones, and then the two-humped ones. It is clear that if the condition for 62 camels is met for the first 62 camels, then i... | 72 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,399 |
1. Variant 1.
When multiplying two two-digit numbers, a four-digit number $A$ is obtained, where the first digit matches the second, and the second-to-last digit matches the last. Find the smallest $A$, given that $A$ is divisible by 51. | Answer: 1122.
Solution.
Notice that $A=\overline{x x y y}=x \cdot 11 \cdot 100+y \cdot 11=11 \cdot(100 x+y)$. Since 51 and 11 are coprime, then $100 x+y$ is divisible by 51. The minimum $x=1$, so $y=2$ (the only number from 100 to 109 divisible by 51 is 102). | 1122 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,400 |
2. Variant 1.
Find the number of four-digit numbers for which the last three digits form an increasing arithmetic progression (numbers cannot start with zero). | Answer: 180.
Solution: The difference of the progression cannot be greater than 4.
For $d=1$ - there are eight suitable progressions of digits: from 012 to 789.
For $d=2$ - six: from 024 to 579.
For $d=3$ - four: from 036 to 369.
For $d=4$ - two: 048 and 159.
In total, $8+6+4+2=20$ options for the last three digi... | 180 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,401 |
3. Variant 1.
Find the ratio $\frac{16 b^{2}}{a c}$, given that one of the roots of the equation $a x^{2}+b x+c=0$ is 4 times the other. | Answer: 100.
Solution.
By Vieta's theorem, $-\frac{b}{a}=x_{1}+x_{2}=5 x_{2}$ and $\frac{c}{a}=x_{1} \cdot x_{2}=4 x_{2}^{2}$. Express $x_{2}=-\frac{b}{5 a}$ from the first equation and substitute it into the second: $\frac{c}{a}=\frac{4 b^{2}}{25 a^{2}}$. Then find $\frac{b^{2}}{a c}=\frac{25}{4}$. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,402 |
4. Variant 1.
It is known that
$$
\frac{1}{\cos (2022 x)}+\operatorname{tg}(2022 x)=\frac{1}{2022}
$$
Find $\frac{1}{\cos (2022 x)}-\operatorname{tg}(2022 x)$. | Answer: 2022.
Solution 1.
\[
\begin{aligned}
& \frac{1}{\cos 2A} + \tan 2A = \frac{1 + \sin 2A}{\cos 2A} = \frac{\cos^2 A + \sin^2 A + 2 \sin A \cdot \cos A}{\cos^2 A - \sin^2 A} = \frac{(\cos A + \sin A)^2}{(\cos A - \sin A)(\cos A + \sin A)} = \frac{\cos A + \sin A}{\cos A - \sin A} \\
& \frac{1}{\cos 2A} - \tan 2A... | 2022 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,403 |
5. Variant 1.
It is known that
$$
\left(x^{2}-x+3\right)\left(y^{2}-6 y+41\right)\left(2 z^{2}-z+1\right)=77
$$
Find $\frac{x y}{z}$. | # Answer: 6.
Solution.
$$
\begin{aligned}
& x^{2}-x+3=(x-0.5)^{2}+2.75 \geq 2.75 \\
& y^{2}-6 y+41=(y-3)^{2}+32 \geq 32 \\
& 2 z^{2}-z+1=2(z-0.25)^{2}+0.875 \geq 0.875
\end{aligned}
$$
Therefore, $\left(x^{2}-x+3\right)\left(y^{2}-6 y+41\right)\left(2 z^{2}-z+1\right) \geq 77$ and if at least one of the three inequa... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,404 |
# 6. Variant 1.
In the district, there are three villages $A, B$, and $C$ connected by dirt roads, with any two villages being connected by several (more than one) roads. Traffic on the roads is two-way. We will call a path from one village to another either a road connecting them or a chain of two roads passing throu... | Answer: 106.
Solution.
Let there be $k$ roads between cities $A$ and $B$, $m$ roads between cities $B$ and $C$, and $n$ roads between cities $A$ and $C$. Then the number of paths from $A$ to $B$ is $k + mn$, and the number of paths from $B$ to $C$ is $m + kn$. We have the system of equations $k + mn = 34$, $m + kn = ... | 106 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,405 |
7. Variant 1.
103 natural numbers are written in a circle. It is known that among any 5 consecutive numbers, there are at least two even numbers. What is the minimum number of even numbers that can be in the entire circle? | Answer: 42.
Solution.
We will show that there will be 3 consecutive numbers, among which there are at least 2 even numbers. This can be done, for example, as follows. Consider 15 consecutive numbers. They can be divided into 3 sets of 5 consecutive numbers, so among them there are at least 6 even numbers. But these 1... | 42 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,406 |
8. Variant 1.
Given a parallelogram $A B C D$. Let $B P$ and $C Q$ be the perpendiculars dropped from vertices $B$ and $C$ to diagonals $A C$ and $B D$ respectively (point $P$ lies on segment $A C$, and point $Q$ lies on segment $B D$). Find the ratio $\frac{10 B D}{A C}$, if $\frac{A P}{A C}=\frac{4}{9}$ and $\frac{D... | Answer: 6.
Solution: Let $O$ be the point of intersection of the diagonals. Note that points $B, C, Q, P$ lie on the same circle (segment $B C$ is seen from points $P$ and $Q$ at a right angle). Therefore, triangles $B O P$ and $C O Q$ are similar. Let $A C=2 a, B D=2 b$. Then $P O=a-\frac{8 a}{9}=\frac{a}{9}, Q O=$ $... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,407 |
1. Place a comma in each of the six numbers so that the equation becomes true: $2016+2016+$ $2016+2016+2016=46368$ | Answer: $20.16+20.16+20.16+201.6+201.6=463.68$ or $2.016+2.016+2.016+20.16+20.16=46.368$. Criteria for checking.
“+” - any correct answer is provided (example)
“士” - several examples are provided, among which there are both correct and incorrect ones
“-" - the correct example is not provided | 20.16+20.16+20.16+201.6+201.6=463.68 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,408 |
2. Yesterday, Nikita bought several pens: black ones - at 9 rubles each and blue ones - at 4 rubles each. Today, when he went to the same store, he found that the prices for the pens had changed: black pens now cost 4 rubles each, and blue pens - 9 rubles each. Seeing this, Nikita said with regret: “If I had bought the... | Answer: Nikita is wrong.
Solution. First method. Let Nikita buy $x$ black pens and $y$ blue pens, then he paid $9x + 4y$ rubles. After the price change, the cost of the pens became $4x + 9y$ rubles. Suppose Nikita is not wrong, then $(9x + 4y) - (4x + 9y) = 49$. Transforming the left side of this equation, we get: $5(... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,409 |
3. On a coordinate line, several points (more than two) are marked. Each point, except for the two extreme ones, is exactly in the middle of some two marked points. Can all segments, inside which there are no marked points, have different lengths? | Answer: they can.
Solution.
Mark on the coordinate
line the points corresponding to the numbers $0, 6, 8, 9, 12, 16$ (see Fig. 7.3). Then 6 is the midpoint of the segment [0;12], 8 is the ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,410 |
4. In three cells of a $3 \times 3$ table, numbers are placed (see the figure). It is required to fill in the remaining cells with numbers so that the sums of the numbers in all rows, columns, and main diagonals are equal. Prove that this can be done in a unique way, and fill in the table. | Solution. First method. Let the number in the empty cell of the top row be $x$, then the sum of the numbers in this row is $6+x$. For the sum of the numbers in the first column to be the same, the number in the empty cell of this column should be $x+2$. Considering the diagonal from the bottom-left cell to the top-righ... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,411 |
5. Along a straight section of the border, 15 posts have been set up. Around each post, several nearsighted spies were caught. Each of them honestly said how many other spies they saw. However, any spy could only see those who were near their post and the nearest neighboring posts. Can the number of spies caught around... | Answer: It is possible.
Solution. Number the poles from 1 to 15 from left to right. From the interrogation of all spies caught at the second pole, we learn the total number of spies at the first three poles, and from the interrogation of spies caught at the first pole, we learn the number of spies caught at the first ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,412 |
7.1. Draw a row of 11 circles, each of which is either red, blue, or green. Moreover, among any three consecutive circles, there should be at least one red, among any four consecutive circles, there should be at least one blue, and there should be more than half green. How many red circles did you get? | Answer: 3 red circles
Hint. The circles are arranged only as follows: ZZKSKZKSKZZ.
Solution. (1) Three non-overlapping triplets of circles can be identified, each containing at least one red circle. Therefore, there are no fewer than three red circles. (2) Two non-overlapping quartets of circles can be identified, ea... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,413 |
7.2. Represent the number 32 as the product of three integer factors, the sum of which is 3. What is the smallest of the factors? | Answer: -4.
Example: $32=(-4) \cdot(-1) \cdot 8$.
Solution. The given factorization is unique. This can be proven.
If all three factors are positive, then the largest of them is not less than 4, and the sum is greater than 3, which contradicts the condition. Therefore, two of the factors are negative, and the third ... | -4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,414 |
7.3. In a hat, there are three cards: blue, green, and red. Pete, Vasya, and Tolya were each given one of them and asked to name the colors. Pete said "blue," Vasya - "blue," and Tolya - "green." After this, the cards were put back into the hat and dealt again. This time, Pete said "blue," Vasya - "green," and Tolya - ... | Answer: Petya - green, Vasya - red, Tolya - blue.
Solution. Both Petya and Tolya named the same color both times, but they got different cards. This means that they each lied once. Therefore, Vasya told the truth: he had a blue card first,
| Petya | Vasya | Tolya |
| :---: | :---: | :---: |
| blue | blue | green |
| ... | Petya-green,Vasya-red,Tolya-blue | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,415 |
7.4. In a table containing $A$ columns and 100 rows, natural numbers from 1 to $100 \cdot A$ were written in ascending order, starting from the first row. The number 31 is in the fifth row. In which row is the number 100? | Answer: in the 15th row.
Solution. From the condition, it follows that $A \leq 7$, since for $A \geq 8$ the number 31 would be located before the fifth row. Similarly, we get that $A \geq 7$, otherwise the number 31 would be located after the fifth row. Therefore, $A=7$. Since $100=7 \cdot 14+2$, the number 100 is loc... | 15 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,416 |
7.5. In the village of Matitika, along a straight road, five friends live in the following order: Alya, Bella, Valya, Galina, and Dilya. Each of them found the sum of the distances (in meters) from her house to the houses of the others. Bella named the number 700, Valya - 600, Galina - 650. How many meters are there be... | Answer: 150 meters.
Solution. Let's denote the houses of the friends with the letters A, B, V, G, D.
It is easy to see that the total distance from B to the other houses is AB + 3BV + 2VG + GD, and from V to the other houses is AB + 2BV + 2VG + GD. These values differ by BV, so the distance between the houses B and V... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,417 |
7.7. Petya told Misha that in his class exactly two thirds of all the girls are blondes, exactly one seventh of the boys are blonds, and in total, a third of the class has light hair. Misha said: "You once told me that there are no more than 40 people in your class. 0 ! I know how many girls are in your class!" How man... | Answer: 12 girls
Solution. Let there be $x$ girls and $y$ boys in the class. From the problem statement, we have the following relationship:
$\frac{2}{3} x+\frac{1}{7} y=\frac{1}{3}(x+y)$,
which, after transformation, becomes $7 x=4 y$.
From the condition and the derived relationship, it follows that the number $x$... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,419 |
7.8. A) On a plane, three lines and $n$ points are drawn such that on either side of each line there are exactly two points (points lying on the line itself do not belong to either side). For which values of $n$ is this possible? | Answer: for $n$ equal to $4, 5, 6$ or 7.
Solution. Let's count how many points lie on the specified lines. Since for each line, four points do not lie on it, there should be $n-4$ points on the line. In this case, either all three lines intersect at one point, and if this is one of the marked points, we have counted i... | 0,1,3,4,6,7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,420 |
8.1. Represent the number 36 as the product of three integer factors, the sum of which is 4. What is the smallest of the factors? | Answer: -4.
Example: $36=(-4) \cdot(-1) \cdot 9$.
Solution. The given factorization is unique. This can be proven.
If all three factors are positive, then the largest of them is not less than 4 (since $3^{3}<36$), and the sum is greater than 4, which contradicts the condition. Therefore, two of the factors are negat... | -4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,421 |
8.2. Vasya replaced identical digits in two numbers with identical letters, and different digits with different letters. It turned out that the number ZARAZA is divisible by 4, and ALMAZ is divisible by 28. Find the last two digits of the sum ZARAZA + ALMAZ. | Answer: 32.
Solution. From the condition, it follows that the numbers ZARAZA and ALMAZ are divisible by 4. By the property of divisibility by 4, the numbers ZA and AZ are also multiples of 4. In particular, it follows from this that both digits 3 and A are even. But a number divisible by 4, in which the second-to-last... | 32 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,422 |
8.3. Given a parallelogram $A B C D, \angle D=100^{\circ}, B C=12$. On side $A D$ there is a point $L$ such that $\angle A B L=50^{\circ}, L D=4$. Find the length of $C D$. | Answer: 8.
Solution. By the property of a parallelogram, $\angle A B C=\angle D=100^{\circ}, A D=B C=12$ and $C D=A B$. Therefore, $\angle C B L=\angle A B C-\angle A B L=100^{\circ}-50^{\circ}=50^{\circ}$ and $A L=A D-L D=12-4=8$. Since $\angle A L B=\angle C B L$ (as alternate interior angles when $A D$ and $B C$ ar... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,423 |
8.4. Four boys and three girls went to the forest to pick mushrooms. Each found several mushrooms, in total they collected 70. No two girls collected the same amount, and any three boys together brought no fewer than 43 mushrooms. The number of mushrooms collected by any two children differed by no more than 5 times. M... | Answer: 5 mushrooms.
Solution. Any three boys collected at least 43 mushrooms together, so there is a boy who collected no less than 15 mushrooms (since $14 \cdot 3 < 43$). Therefore, this boy and the other three collected no less than $15 + 43 = 58$ pieces.
If there is a boy who collected no less than 15 pieces, the... | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,424 |
8.5. Two graphs of linear functions intersect at $x=2$. At $x=8$, the values differ by 8. At $x=20$, the value of one of the functions is 100. What could be the value of the other function | Answer: 76 or 124.
Solution 1. Let the value of the first function change by $d$ when the argument increases by $8-2=6$, then the second function changes by $d \pm 8$. From 2 to 20, the argument increases by 18, which is 3 times more than 6. Therefore, the value of the first function changes by $3d$, and the second fu... | 76or124 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,425 |
8.6. In a convex quadrilateral $A B C D$, side $B C$ is half the length of $A D$. Diagonal $A C$ is perpendicular to side $C D$, and diagonal $B D$ is perpendicular to side $A B$. Find the larger acute angle of this quadrilateral, given that the smaller one is $36^{\circ}$. | Answer: $84^{\circ}$.
Solution. Let point $M$ be the midpoint of side $A D$. Since angles $A B D$ and $A C D$ are right angles, angles $B$ and $C$ of quadrilateral $A B C D$ are obtuse, and ... | 84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,426 |
8.7. In the city of Bukvinsk, people are acquainted only if their names have the same letters, otherwise - they are not. Several residents of Bukvinsk were asked how many acquaintances they have in the city. Martin said 20, Klim - 15, Inna - 12, Tamara - 12. What did Camilla answer? | Answer: 15 acquaintances
Solution. Note that all five students listed in the condition are acquainted with each other. Therefore, Martin has 16 acquaintances outside this group, while Inna and Tamara each have 8. However, all of Inna's acquaintances are acquainted with Martin, and all of Tamara's acquaintances are als... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,427 |
8.8. In the cells of an $8 \times 8$ board, natural numbers from 1 to 64 (each appearing once) are placed such that numbers differing by 1 are in adjacent side-by-side cells. What is the smallest value that the sum of the numbers on the diagonal from the bottom left to the top right corner can take? | Answer: 88.
Solution. We will color the cells of the board in a checkerboard pattern. Let the considered diagonal be black. We will move through the cells according to the numbers placed. Consider the moment when we occupy the last cell on the diagonal. Before this, we must have visited all the cells on one side of it... | 88 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,428 |
# 2.1. Condition:
In the campus, rooms are numbered consecutively $1,2,3,4 \ldots, 10,11, \ldots$ For room numbering, stickers with digits were purchased, with the digits 1, 2, and 3 being purchased in equal quantities, and the digit 5 being purchased three more than the digit 6. How many rooms are there in the campus... | # Answer: 66
## Solution.
In each decade up to the sixth, the digits "5" and "6" are equal, so there are at least 50 rooms. Since the digits "1", "2", and "3" are equal, they must appear in each decade, meaning the number of rooms will be at least 53. Then the digit "5" will be four more than the digit "6". Therefore... | 66 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,433 |
# 3.1. Condition:
Vanya bought balloons, red ones were 7 times more than blue ones. While Vanya was walking home, some of the balloons burst, and among the burst balloons, there were 3 times fewer red ones than blue ones. What is the smallest number of balloons Vanya could have bought? | # Answer: 24
## Solution.
Notice that at least one red balloon has burst, which means at least three blue balloons have burst. Therefore, there are at least three blue balloons, which means there are at least $7 \cdot 3=21$ red balloons, so the total number of balloons must be at least 24. For example, Vanya could ha... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,434 |
# 4.1. Condition:
A swallow, a sparrow, and a canary perched on three poles along the highway. The swallow landed on a pole located 380 meters from the bus stop; the sparrow - on a pole that is 450 meters from the stop, right in the middle between the swallow and the canary. At what distance from the stop is the pole ... | Answer: 520, 1280
## Solution.
Since the sparrow is exactly in the middle, the distance between it and the swallow is equal to the distance between it and the canary. The distance between the sparrow and the swallow is $450-380=70$ meters (if the birds are on the same side of the bus stop) or $450+380=830$ meters (if... | 520,1280 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,435 |
# 5.1. Condition:
On Monday, Olya started training according to the following program: 6 days of training in a row, then 2 days of rest, then again 6 days of training, 2 days of rest, and so on. On which day of the week will the hundredth training session fall?
## Answer options:
$\square$ Monday
$\square$ Tuesday
... | # Solution.
Let's try to understand which cycle of the program (i.e., an 8-day segment consisting of 6 training days and 2 rest days) the hundredth training session falls on (not the hundredth day of the program). By dividing 100 by 6, we can determine how many cycles must pass before the hundredth training session—16... | Saturday | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false | 12,436 |
# 6.1. Condition:
Petya thought of a natural number and wrote down the sums of each pair of its digits on the board. After that, he erased some of the sums, and the numbers $2,0,2,2$ remained on the board. What is the smallest number Petya could have thought of? | Answer: 2000
## Solution.
Since among the sums there is a 0, the number must contain at least two digits 0. If the number has only three digits, there will be three pairwise sums, while the condition states there are at least four. Therefore, the number must have at least four digits. A sum of 2 can be obtained eithe... | 2000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,437 |
# 7.1. Condition:
A carpenter took a wooden square and cut out 4 smaller equal squares from it, the area of each of which was $9 \%$ of the area of the larger one. The remaining area of the original square was $256 \mathrm{~cm}^{2}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_c8e4d847c3bda2de5ad8g-13.jpg?height=6... | Answer: 20
Solution.
Let the side of the larger square be 10x. The area of the smaller square is $9 \%$ of the area of the larger square, which is $9 / 100 \cdot 100 x^{2}=9 x^{2}$. After removing four smaller squares, the remaining area is $100 x^{2}-4 \cdot 9 x^{2}=64 x^{2}=256 \text{ cm}^{2}$. Therefore, $x^{2}=4$... | 20 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,438 |
# 8.1. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Six islanders lined up in a row, each 1 meter apart from each other.
- The leftmost in the row said: "My nearest fellow tribesman in this row stands 3 meters away from me."
- The fourth said: "My neare... | # Answer:
## Solution.
Let's number all the islanders from left to right. Suppose the fourth is a liar. Then the third and fifth must be knights, otherwise the fourth would have told the ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,439 |
# 8.3. Condition:
On an island, there are two tribes: knights, who always tell the truth, and liars, who always lie. Six islanders lined up in a row, each 1 meter apart from each other.
- The second from the left in the row said: "My nearest fellow tribesman in this row stands 2 meters away from me."
- The third said... | Answer:
## Solution.
Let's number all the islanders from left to right. Suppose the third is a liar. Then the second and fourth must be knights, otherwise the third would have told the trut... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,441 |
# Problem 7.1 (7 points)
Find such natural numbers that their product is 2020 and their sum is 2020. # | # Solution:
Factorize 2020 and supplement with ones so that the sum equals 2020.
| Criteria | Points |
| :--- | :---: |
| Any correct example provided | 7 |
| Arithmetic error with correct idea | 4 |
| Incorrect solution | 0 |
Answer: 1010, 2, and 1008 | 1010,2,1008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,443 |
# Problem 7.2 (7 points)
There are 22 kg of blueberries in a box. How can you measure out 17 kg of blueberries using a two-kilogram weight and a balance scale in two weighings. | # Solution:
Place the weight on one pan and balance the scales using all the blueberries $12=10+2$, then divide the 10 kg into equal parts of 5 kg each. And we get $12+5=17$ kg.
| Criteria | Points |
| :--- | :---: |
| Correct algorithm of actions | 7 |
| Incorrect solution | 0 |
Answer: 17 kg. | 17 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,444 |
# Problem 7.3 (7 points)
A piece has fallen out of a dictionary, the first page of which is numbered 213, and the number of the last page is written with the same digits in some other order. How many pages are in the missing piece? | # Solution:
The number of the last page is 312 (it must be even). Then the number of pages is 312 - 212 = 100.
| Criteria | Points |
| :--- | :---: |
| Correct solution | 7 |
| Calculated 312 - 213 + 1 = 100, without explaining why 1 is added | 6 |
| Obtained the answer 99 | 4 |
| Noted the different parity of the fi... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,445 |
# Problem 7.4 (7 points)
Find all three-digit numbers that decrease by 6 times after the first digit is erased. # | # Solution:
Let $\overline{a b c}$ be the desired number. According to the condition, $100 \mathrm{a}+10 \mathrm{~b}+\mathrm{c}=6(10 \mathrm{~b}+\mathrm{c})$, from which we get $100a = 50 \mathrm{~b}+5 \mathrm{c}$, or $20 \mathrm{a}=10 \mathrm{~b}+\mathrm{c}$. Therefore, $\mathrm{c}=10(2 \mathrm{a}-\mathrm{b})$. Since... | 120,240,360,480 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,446 |
# Problem 7.5 (7 points)
On a plane, 6 lines are drawn and several points are marked. It turned out that on each line exactly 3 points are marked. What is the minimum number of points that could have been marked? | # Solution:
The vertices of the triangle, the midpoints of its sides, and the point of intersection of the medians - 7 points lying in threes on 6 lines (3 sides and 3 medians).
P.S. It doesn't have to be medians specifically.
Proof of the estimate: If we have a point through which at least 4 lines pass, then we wil... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,447 |
1. You are allowed to cut the pizza into equal parts, with no more than 6 parts. How can you evenly distribute 8 identical pizzas among 15 people? (There should be no "extra" slices of pizza left over) | Solution: For example, let's cut the first three pizzas into 5 equal parts each - we will get 15 equal pieces, each being $1 / 5$ of a pizza. Then, cut the remaining 5 pizzas into 3 parts each, resulting in 15 equal pieces, each being $1 / 3$ of a pizza. By giving each of the 15 people one piece of each type, we achiev... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,448 |
2. Given three not necessarily integer numbers. If each of the given numbers is increased by 1, then the product of the numbers also increases by 1. If each of the numbers is increased by 2, then their product also increases by 2. Find these numbers. | Solution. Let the required numbers be $a, b, c$. According to the condition, $(a+1)(b+1)(c+1)=a b c+1$ and $(a+2)(b+$ $2)(c+2)=a b c+2$. Expanding the brackets, we get $a b+b c+c a+a+b+c=0$ and $2(a b+b c+c a)+4(a+b+c)=-6$. From this, $a+b+c=-3$ and $a b+b c+c a=3$. Since $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=9... | =b==-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,449 |
3. Polycarp instead of the usual multiplication of three-digit numbers decided to simply "glue the numbers", appending one number to the other. The result turned out to be 7 times larger than the usual. What numbers was Polycarp multiplying? | Answer: 143 and 143.
Solution. Let the original numbers be denoted by $a$ and $b$. The result of the usual multiplication is $a b$, and the result of the "concatenation" is $-1000 a + b$. According to the condition, they are related by the equation $1000 a + b = 7 a b$, from which we get $b = \frac{1000 a}{7 a - 1}$. ... | 143143 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,450 |
4. On the diagonal $A C$ of a trapezoid with base $A D$, a point $E$ is chosen. Indicate a way to choose point $E$ so that triangles $A B C$ and $E C D$ have equal areas. | Answer. Point $E$ should be the intersection point of diagonal $A C$ and line $B K$, which is parallel to side $C D$.
Solution. The area of triangle $E C D$ is equal to the area of triangle $C... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,451 |
5. Misha has a $7 \times 7$ square of paper, all cells of which are white. Misha wants to color $N$ cells black. What is the smallest $N$ for which Misha can color the cells so that after coloring, no completely white rectangle with at least ten cells can be cut out from the square? | Answer: 4.
Solution. Divide the $7 \times 7$ square into 5 rectangles: four $3 \times 4$ rectangles (each corner of such a rectangle coincides with one of the corners of the $7 \times 7$ square) and a $1 \times 1$ square. If only three cells are colored, there will be a white rectangle consisting of 12 cells. Example ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,452 |
10.5. Find all numbers $a$ such that for any natural $n$ the number $a n(n+2)(n+3)(n+4)$ is an integer.
(O. Podlipsky) | Answer. $a=\frac{k}{6}$, where $k$ is any integer.
Solution. Substituting $n=1, n=3$ and $n=4$, we get that the numbers $60a, 630a$, and $24 \cdot 56a$ are integers. Therefore, $a$ is a rational number, and the denominator $q$ in its irreducible form is a divisor of the numbers 60, 630, and $24 \cdot 56$. Consequently... | \frac{k}{6} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,453 |
10.6. On the board, 2011 numbers are written. It turns out that the sum of any three written numbers is also a written number. What is the smallest number of zeros that can be among these numbers? | Answer: 2009.
Solution. Let $n=2011$. Arrange the written numbers in non-decreasing order: $a_{1} \leqslant a_{2} \leqslant \ldots \leqslant a_{n}$. Since the number $a_{1}+a_{2}+a_{3}$ is written, then $a_{1}+a_{2}+a_{3} \geqslant a_{1}$, hence $a_{2}+a_{3} \geqslant 0$. Similarly, we get $a_{n-2}+a_{n-1}+a_{n} \leqs... | 2009 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,454 |
10.7. In an acute scalene triangle $ABC$, points $C_{0}$ and $B_{0}$ are the midpoints of sides $AB$ and $AC$ respectively, $O$ is the center of the circumscribed circle, and $H$ is the orthocenter. The lines $BH$ and $OC_{0}$ intersect at point $P$, and the lines $CH$ and $OB_{0}$ intersect at point $Q$. It turns out ... | Solution. Let $a=BC, b=CA, c=AB, \alpha=\angle BAC$; then $AB'=c \cos \alpha, AC'=b \cos \alpha$. Suppose $BB'$ and $CC'$ are the altitudes of the triangle. Since $OB_0$ and $OC_0$ are the perpendicular bisectors of sides $AC$ and $AB$, the segments $B'B_0$ and $C'C_0$ are equal to the heights of the rhombus $OPHQ$, he... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,455 |
10.8. A straight rod 2 meters long was sawn into $N$ sticks, the length of each of which is expressed in whole centimeters.
For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle?
(A. Magazinov) | Answer. $N=102$.
Solution. First solution. Let $N \leqslant 101$. Cut the stick into $N-1$ sticks of 1 cm each and one stick of $201-N$ cm. It is impossible to form a rectangle from this set, as each side of the rectangle is less than half the perimeter, and thus the stick of length $201-N \geqslant 100$ cm cannot be ... | 102 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,456 |
Problem 1. A four-digit number is called wonderful if it is divisible by 25, the sum of its digits is divisible by 25, and the product of its digits is divisible by 25. Find all wonderful numbers. | Answer: 5875 and 8575.
Solution. The sum of the digits of a four-digit number does not exceed 36, so for a wonderful number, it must be equal to 25.
Since the wonderful number is divisible by 25, it ends in either 00, 50, 25, or 75.
If a four-digit number ends in 00 or 50, then the sum of its digits does not exceed ... | 58758575 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,458 |
Problem 2. Masha's classes at school end at 13:00, and her mother picks her up by car, and they drive home. One day, the classes ended at 12:00, and Masha started walking home. On the way, she met her mother, who, as usual, drove to pick up her daughter at 13:00 at the school. Then Masha and her mother drove home in th... | Answer: $12: 54$
Solution. Let Masha walk a distance of $l$. Then, on the way to school and on the way back, her mother drives $l$ less than usual. This means that the mother covers a distance of $2l$ in 12 minutes. Therefore, the distance $l$ she covers in 6 minutes. From this, it follows that the mother met Masha 6 m... | 12:54 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,459 |
Problem 3. The equations $x^{2}+2019 a x+b=0$ and $x^{2}+2019 b x+a=0$ have one common root. What can this root be, given that $a \neq b$? | Answer: $\frac{1}{2019}$.
Solution. Let the common root of the given equations be $r$. Then
$$
r^{2}+2019 a r+b=0=r^{2}+2019 b r+a
$$
From this, we obtain that $2019 r(a-b)=(a-b)$. Since $a \neq b$, it follows that $r=\frac{1}{2019}$
Criteria
4 p. A complete and justified solution is provided.
In the absence of a... | \frac{1}{2019} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,460 |
Task 4. Irina wrote down some integers from 0 to 999 in a row on the board. As a result, a long number was formed. Polina wrote down all the remaining integers from the same range on her part of the board, resulting in a second long number. Could these two long numbers have been the same?
Answer: they could not. | Solution. Suppose these two long numbers coincide. Then one of them will contain the number 100. In the other number, such a sequence of digits can only be obtained by appending the number 0 to a number that is a multiple of ten (for example, to the number 210, 0 was appended on the right).
But one of the long numbers... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,461 |
Problem 5. On the side $A D$ of the square $A B C D$, point $K$ is marked, and on the extension of ray $A B$ beyond point $B$ - point $L$. It is known that $\angle L K C=45^{\circ}, A K=1, K D=2$. Find $L B$. | Answer: $L B=2$.
Fig. 1: to the solution of problem 5
Solution. Note that $\angle L A C=45^{\circ}=\angle L K C$, which implies that quadrilateral $L A K C$ is cyclic. Then $\angle K C L=90^... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,462 |
Problem 6. An excursion group of 6 tourists is visiting attractions. At each attraction, three people take photos, while the others photograph them. After what minimum number of attractions will each tourist have photos of all other participants in the excursion?
Answer: after 4 attractions. | Solution. Evaluation. A total of $6 \cdot 5=30$ photographs need to be taken (considering only photographs between two people $A$ and $B$, that is, if person $A$ photographs 3 other participants $B, C, D$ in one photograph - this counts as 3 photographings $A \rightarrow B, A \rightarrow C, A \rightarrow D$).
At one l... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,463 |
1. Can the digits $2,0,1,9$ be placed in a $3 \times 3$ square (one digit per cell) such that each row, each column, and both diagonals contain three different digits? Provide an example or explain why it is not possible. | Solution. Answer: no. Consider the corner and central cells of a $3 \times 3$ square (a total of 5 cells). Any two of these cells lie either in the same column, or in the same row, or on the same diagonal. Since there are 5 cells in total, and there are 4 different digits to fill them, by the pigeonhole principle, ther... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,464 |
2. Find the largest positive number which, after erasing one of the digits in its decimal representation, can double. | Solution. Answer: 0.375. It is obvious that when a digit in the integer part of a positive number is erased, the number decreases. Therefore, the digit that was erased must be from the fractional part. Let the first digit $x$ after the decimal point in the original number not be 0. Then, if a digit other than the first... | 0.375 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,465 |
3. Chuk and Gek need to divide a triangular cake between themselves. Gek sets the condition that he will cut his portion with a straight cut, and Chuk agrees to this on the condition that he can mark a point $P$ in advance, through which the cut must pass. The cake has the same thickness everywhere and is uniform in ta... | Solution. Chuk's strategy is to choose point $P$ at the centroid of the triangle, while Gek's strategy is to cut parallel to one of the sides of the triangle. If both adhere to these strategies, the cake will be divided between Chuk and Gek in the ratio $4: 5$. | \frac{4}{5} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 12,466 |
4. Can 9 dominoes of size $1 \times 2$ be placed on a $10 \times 10$ board such that in each row and each column they occupy an odd number of cells? | Solution. Answer: no. Suppose it can be done. Each domino lies either vertically or horizontally. Since there are 9, the number of dominoes of one type (for example, horizontal) is no more than four. Then they can lie in no more than eight columns. In the remaining two columns, there will be only vertical dominoes, mea... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,467 |
5. Ten people are sitting at a round table. Each of them is either a knight (who always tells the truth) or a liar (who always lies). Each person named four other people as liars from among their non-neighbors at the table (each person has two neighbors at the table). Prove that someone named a liar the person sitting ... | Solution. Obviously, under our conditions, there is at least one knight and at least one liar. Therefore, there will be a pair of adjacent people where one is a knight and the other is a liar. A knight could accuse only liars of lying, and did not accuse his neighbor. Therefore, there are at least 5 liars. Similarly, t... | proof | Logic and Puzzles | proof | Yes | Yes | olympiads | false | 12,468 |
1. Find the sum of the numbers $1-2+3-4+5-6+\ldots+2013-2014$ and $1+2-3+4-5+6-\ldots-2013+2014$. | 1. Answer: 2.
Notice that for each term of the first sum, except for 1, there is an opposite term in the second sum. The sum of opposite numbers is 0. Therefore, the total sum is $1+1=2$.
Grading criteria:
Correct answer with proper justification: 7 points.
Incorrect answer with the right idea in the justification:... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,469 |
2. How many natural numbers exist such that the product of all digits of such a number, multiplied by their quantity, equals 2014? | 2. Answer: 1008.
Solution. The number 2014 is divisible only by the digits 1 and 2, so the number can only contain the digits 1 and 2. Moreover, the digit 2 can only appear once. If the number consists only of 1s, there are 2014 of them, and there is only one such number. If there is a 2, then there are 1007 digits of... | 1008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,470 |
4. In the warehouse, there are 25 white glass cups and 35 black porcelain cups. Each glass cup breaks into 17 pieces when dropped, and each porcelain cup breaks into 18 pieces. The guard repainted some of the glass cups black and some of the porcelain cups white. After that, he accidentally broke all the cups by droppi... | 4. Answer: it could not.
Suppose the number of black and white fragments turned out to be equal. Let \( x \) be the number of glass cups painted black, and \( y \) be the number of porcelain cups painted white. Then the total number of black fragments is \( 17x + 18(35 - y) \), and the total number of white fragments ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,471 |
5. Misha painted all integers in several colors such that numbers whose difference is a prime number are painted in different colors. What is the smallest number of colors that Misha could have used? Justify your answer. | 5. Answer: 4 colors
Evaluation. Consider the numbers $1,3,6,8$. The difference between any two of them is a prime number, which means that all of them must be of different colors, and at least four colors are needed.
Example. Paint numbers of the form $4 \mathrm{k}$ in the first color, numbers of the form $4 \mathrm{... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,472 |
1. There is a light bulb and 10 numbered buttons. It is known that exactly 5 of them are active. In one attempt, you are allowed to press any three buttons simultaneously. If at least one of them is active, the light bulb will turn on. How can you determine in no more than 9 attempts whether the button numbered 1 is ac... | Solution. Let's divide the buttons numbered from the second to the ninth into three groups of three buttons, for example, $(2,3,4),(5,6,7),(8,9,10)$. We will spend three attempts on each group, each attempt consisting of pressing button 1 and two buttons from one group (each attempt has a different pair). If the light ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,473 |
2. Find the value of the fraction
$$
\frac{2 \cdot 2020}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots+\frac{1}{1+2+3+\ldots+2020}}
$$ | Solution. Let's denote the denominator of the fraction by $q$. By repeatedly applying the formula for the sum of an arithmetic progression, we get that
$$
q=\frac{2}{1 \cdot 2}+\frac{2}{2 \cdot 3}+\frac{2}{3 \cdot 4}+\ldots+\frac{2}{2020 \cdot 2021}
$$
Now, using the identity $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+... | 2021 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,474 |
3. The mole's dwelling is a line of 100 mounds (surface exits), with adjacent mounds connected by underground passages ($1-2-3-\ldots-99-100$). The fox wants to catch the mole. At the initial moment, the mole is at some exit. Every minute, the mole moves to one of the adjacent mounds, and the fox approaches any arbitra... | Solution. We will show that the fox can catch the mole. We will provide an example of a strategy that will require no more than 200 attempts. First, assume that at the initial moment, the mole is near a pile with an odd number. Then, after two approaches to pile 1, the mole will either be caught or will be at one of th... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,475 |
4. A square $O M K N$ is inscribed in a circle such that vertex $O$ coincides with the center of the circle, and vertex $K$ lies on the circle. Chord $A B$ of the circle passes through vertex $M$, and chord $\mathrm{CD}$ passes through vertex $\mathrm{N}$. Prove that $A M \cdot M B=C N \cdot N D$. | Solution. Let the side of the square be $a$. Extend the side $K M$ of the square to intersect the circle at point $F$, and draw the radius $O F$. We obtain an isosceles triangle $K O M$, in which $O M$ is the height, so $F M = M K = a$. Chords $A B$ and $K F$ intersect at one point, then $A M \cdot M B = F M \cdot M K ... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,476 |
5. All natural numbers have been colored in two colors. Prove that among them, one can select numbers $A, B$ and $C$ of the same color such that $A: C=C: B$. | Solution. Consider only powers of two $2^{n}, n \in \mathbb{N}$. Let $A=2^{a}, B=2^{b}$ and $C=2^{c}$. Then the required equality reduces to the condition $a-c=c-b$. If the numbers of the considered form are colored such that there are no two consecutive numbers of the same color, then all numbers of the form $2^{2 n}$... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,477 |
1. Vasya can get the number 100 using ten twos, parentheses, and arithmetic operation signs: $100=(22: 2-2: 2) \cdot(22: 2-2: 2)$. Improve his result: use fewer twos and get the number 100. (It is sufficient to provide one example). | Solution. For example: 1) $100=222: 2-22: 2, \quad$ 2) $100=(2 \cdot 2 \cdot 2+2) \cdot(2 \cdot 2 \cdot 2+2)$. There are $u$ other solutions. | 100=(2\cdot2\cdot2+2)\cdot(2\cdot2\cdot2+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,478 |
3. How to measure 8 liters of water when you are near a river and have two buckets with a capacity of 10 liters and 6 liters? (8 liters of water should end up in one bucket).
| Solution. Let's write the sequence of filling the buckets in the form of a table:
| | Bucket with a capacity of 10 liters | Bucket with a capacity of 6 liters | Comment |
| :--- | :--- | :--- | :--- |
| Initially | 0 liters | 0 liters | |
| Step 1 | 10 liters | 0 liters | Filled the first bucket from the river |
| S... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,480 |
4. Snow White entered a room where there were 30 chairs around a round table. Some of the chairs were occupied by dwarfs. It turned out that Snow White could not sit down without having someone next to her. What is the minimum number of dwarfs that could have been at the table? (Explain how the dwarfs should have been ... | Answer: 10.
Solution: If there were three consecutive empty chairs at the table in some place, Snow White could sit down in such a way that no one would sit next to her. Therefore, in any set of three consecutive chairs, at least one must be occupied by a dwarf. Since there are 30 chairs in total, there cannot be fewe... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,481 |
5. Papa, Masha, and Yasha are going to school. While Papa takes 3 steps, Masha takes 5 steps. While Masha takes 3 steps, Yasha takes 5 steps. Masha and Yasha counted that together they made 400 steps. How many steps did Papa take? | Answer: 90 steps.
Solution. 1st method. Let's call the distance equal to 3 steps of Masha and 5 steps of Yasha a Giant's step. While the Giant makes one step, Masha and Yasha together make 8 steps. Since they made 400 steps together, the Giant would have made 400:8=50 giant steps in this time. If the Giant made 50 ste... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,482 |
1. Vasya can get the number 100 using ten threes, parentheses, and arithmetic operation signs: $100=(33: 3-3: 3) \cdot(33: 3-3: 3)$. Improve his result: use fewer threes and get the number 100. (It is sufficient to provide one example). | Solution. For example: 1) $100=333: 3-33: 3, \quad 2) 100=33 \cdot 3+3: 3$. There are other solutions as well. | 100=33\cdot3+3:3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,484 |
3. How to measure 2 liters of water when you are near a river and have two buckets with a capacity of 10 liters and 6 liters? (2 liters of water should end up in one bucket).
| Solution. Let's write the sequence of filling the buckets in the form of a table:
| | Bucket with a capacity of 10 liters | Bucket with a capacity of 6 liters | Comment |
| :---: | :---: | :---: | :---: |
| Initially | 0 liters | 0 liters | |
| Step 1 | 10 liters | 0 liters | The first bucket is filled from the rive... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,486 |
4. Dad, Masha, and Yasha are going to school. While Dad takes 3 steps, Masha takes 5 steps. While Masha takes 3 steps, Yasha takes 5 steps. Masha and Yasha counted that together they made 400 steps. How many steps did Dad take? | Answer: 90 steps.
Solution. 1st method. Let's call the distance equal to 3 steps of Masha and 5 steps of Yasha a Giant's step. While the Giant makes one step, Masha and Yasha together make 8 steps. Since they made 400 steps together, the Giant would have made 400:8=50 giant steps in this time. If the Giant made 50 ste... | 90 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,487 |
1. Vasya can get the number 100 using ten sevens, parentheses, and arithmetic operation signs: $100=(77: 7-7: 7) \cdot(77: 7-7: 7)$. Improve his result: use fewer sevens and get the number 100. (It is sufficient to provide one example). | Solution. For example: 1) $100=777: 7-77: 7$, 2) $100=7 \cdot 7+7 \cdot 7+7: 7+7: 7$. There are $u$ other solutions. | 100=7\cdot7+7\cdot7+7:7+7:7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,489 |
2. The clock shows half past eight. What is the angle between the hour and minute hands? | Answer: $75^{\circ}$.
Solution. At the moment when the clock shows half past eight, the minute hand points to the number 6, and the hour hand points to the midpoint of the arc between the numbers 8 and 9 (see figure). If two rays are drawn from the center of the clock to the adjacent
 Write any palindromic five-digit number that is divisible by 5.
b) How many five-digit palindromic numbers are there that are divisible by 5?
a) Solution. Any p... | Solution. A number that is divisible by 5 must end in 5 or 0. A palindromic number cannot end in 0, as then it would have to start with 0. Therefore, the first and last digits are 5. The second and third digits can be anything - from the combination 00 to the combination 99 - a total of 100 options. Since the fourth di... | 100 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,491 |
4. Sasha, Lёsha, and Kolya started a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.) | Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,492 |
5. The museum has 16 halls, arranged as shown in the diagram. In half of them, paintings are exhibited, and in the other half, sculptures. From any hall, you can go to any adjacent one (sharing a common wall). During any tour of the museum, the halls alternate: a hall with paintings - a hall with sculptures - a hall wi... | Answer: 15.
Solution: One of the possible routes is shown in the diagram. Let's prove that if a tourist wants to visit each hall no more than once, they will not be able to see more than 15 halls. Note that the route starts in a hall with paintings (A) and ends in a hall with paintings (B). Therefore, the number of ha... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,493 |
1. Replace the asterisk $(*)$ in the expression $\left(x^{3}-2\right)^{2}+\left(x^{2}+*\right)^{2}$ with a monomial so that after squaring and combining like terms, there are four terms. | Solution. Replace the asterisk (*) with $2 x: \quad\left(x^{3}-2\right)^{2}+\left(x^{2}+2 x\right)^{2}=$ $x^{6}-4 x^{3}+4+x^{4}+4 x^{3}+4 x^{2}=x^{6}+x^{4}+4 x^{2}+4$ | x^{6}+x^{4}+4x^{2}+4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,494 |
4. Sasha, Lёsha, and Kolya start a 100 m race at the same time. When Sasha finished, Lёsha was ten meters behind him, and when Lёsha finished, Kolya was ten meters behind him. How far apart were Sasha and Kolya when Sasha finished? (It is assumed that all the boys run at constant, but of course, not equal speeds.) | Answer: 19 m.
Solution: Kolya's speed is 0.9 of Lesha's speed. At the moment when Sasha finished, Lesha had run 90 m, and Kolya had run $0.9 \cdot 90=81$ m. Therefore, the distance between Sasha and Kolya was 19 m. | 19 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,496 |
6. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - ... | # Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of rai... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,498 |
1. Replace the asterisk (*) in the expression $\left(x^{4}-3\right)^{2}+\left(x^{3}+*\right)^{2}$ with a monomial so that after squaring and combining like terms, there are four terms. | Solution. $\quad$ Replace the asterisk (*) with $3 x: \quad\left(x^{4}-3\right)^{2}+\left(x^{3}+3 x\right)^{2}=$ $x^{8}-6 x^{4}+9+x^{6}+6 x^{4}+9 x^{2}=x^{8}+x^{6}+9 x^{2}+9$ | x^{8}+x^{6}+9x^{2}+9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,499 |
3. We will call a number palindromic if it reads the same from left to right as it does from right to left. For example, the number 12321 is palindromic. How many five-digit palindromic numbers are there that are divisible by 5? | Answer: 100.
Solution: A number that is divisible by 5 must end in 5 or 0. A mirrored number cannot end in 0, as then it would have to start with 0. Therefore, the first and last digits are 5. The second and third digits can be anything from the combination 00 to the combination 99 - a total of 100 options. Since the ... | 100 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 12,501 |
4. Vasya thought of two numbers. Their sum equals their product and equals their quotient. What numbers did Vasya think of? | Answer: $\frac{1}{2}, -1$
Solution. Let the numbers be $x$ and $y$. Then, according to the problem, we have: $x+y=xy=\frac{x}{y}$.
From the equation $xy=\frac{x}{y}$, it follows that either $x=0$ and $y \neq 0$, or $y^{2}=1$, and $x$ can be any value. If $x=0$, from the equation $x+y=xy$ we get $y=0$, which is a cont... | \frac{1}{2},-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,502 |
1. If the number $100^{10}$ is written as a sum of tens $(10+10+10+\ldots)$, how many addends will there be | Answer: $10^{19}$.
Solution. $100^{10}=10^{20}=10 \cdot 10^{19}$. Therefore, there will be $10^{19}$ addends. | 10^{19} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,505 |
3. Vasya thought of two numbers. Their sum equals their product and equals their quotient. What numbers did Vasya think of? | Answer: $\frac{1}{2},-1$
Solution. Let the numbers be $x$ and $y$. Then, according to the problem, we have: $x+y=xy=\frac{x}{y}$.
From the equation $xy=\frac{x}{y}$, it follows that either $x=0$ and $y \neq 0$, or $y^{2}=1$, and $x$ can be any value. If $x=0$, from the equation $x+y=xy$ we get $y=0$, which is a contr... | \frac{1}{2},-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,507 |
4. Each of the 10 dwarfs either always tells the truth or always lies. It is known that each of them loves exactly one type of ice cream: butter, chocolate, or fruit. First, Snow White asked those who love butter ice cream to raise their hands, and everyone raised their hands, then those who love chocolate ice cream - ... | # Answer. 4.
Solution. The gnomes who always tell the truth raised their hands once, while the gnomes who always lie raised their hands twice. In total, 16 hands were raised (10+5+1). If all the gnomes had told the truth, 10 hands would have been raised. If one truthful gnome is replaced by one liar, the number of rai... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,508 |
5. Plot the graph of the function $y=(\sqrt{x})^{2}+\sqrt{(x-1)^{2}}$. | Solution. The function $y=(\sqrt{x})^{2}+\sqrt{(x-1)^{2}}$ is defined for $x \geq 0$. Transform it into $y=x+|x-1|$. For $x \geq 1 \quad y=2 x-1$, for $0 \leq x<1 \quad y=1$. The graph is shown in the figure: | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,509 | |
2. On the lines containing the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ of a convex quadrilateral $\mathrm{ABCD}$, points $\mathrm{K}$ and $\mathrm{M}$ are taken respectively such that line $\mathrm{BK}$ is parallel to line $\mathrm{AD}$, and line $\mathrm{AM}$ is parallel to line $\mathrm{BC}$. Prove that line $\math... | Solution. Let the lines AC and BD intersect at point O, then triangles OKB and OAD are similar (by two pairs of corresponding angles when the lines BK and AD are parallel). Similarly, triangles OAM and OCB are similar. From the similarities, it follows that OK/OB = OA/OD and OA/OM = OC/OB. Multiplying these equalities,... | proof | Geometry | proof | Yes | Yes | olympiads | false | 12,513 |
3. The weights of four weights form an increasing geometric progression consisting of positive numbers. How can the heaviest weight be determined using no more than two weighings? | Solution. Note that the heaviest weight in a pair with any other is heavier than the two remaining ones. It is enough to make sure that the heaviest and the lightest together are heavier than the two middle ones. Let the weights of the weights be $m, m q, m q^{2}, m q^{3}$. Then $m+m q^{3}=m \cdot\left(1+q^{3}\right)=m... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 12,514 |
4. Can the polynomial $x^{2019} y^{2019}+x^{2018} y^{2018}+\ldots+x^{2} y^{2}+x y+1$ be represented as a product of two polynomials $f(x)$ and $g(y)$? | Solution. Suppose that for some $f(x)$ and $g(y)$, the polynomial $x^{2019} y^{2019} + x^{2018} y^{2018} + \ldots + x^{2} y^{2} + x y + 1$ is identically equal to $f(x) \cdot g(y)$. Then, for $x=0$, we get $x^{2019} y^{2019} + x^{2018} y^{2018} + \ldots + x^{2} y^{2} + x y + 1 = 1 = f(0) \cdot g(y)$, from which $g(y) =... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 12,515 |
5. On a coordinate plane, a grasshopper jumps from points with integer coordinates. The lengths of its jumps form a sequence of natural numbers: $1,2,3, \ldots$ Can the grasshopper return to the same point it started from, after making exactly 2222 jumps? | Solution. Let's see how the sum of the grasshopper's coordinates changes during a jump. By the Pythagorean theorem $x_{n}^{2} + y_{n}^{2} = n^{2}$, where $x_{n}$ is the difference in the abscissas before and after the jump, $y_{n}$ is the difference in the ordinates, and $n$ is the length of the jump. Note that if $n$ ... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 12,516 |
11.1. Does there exist a real $\alpha$ such that the number $\cos \alpha$ is irrational, while all the numbers $\cos 2 \alpha, \cos 3 \alpha, \cos 4 \alpha, \cos 5 \alpha$ are rational?
(V. Senderov) | Answer. Does not exist.
Solution. Suppose the opposite. Then the number $A = \cos \alpha + \cos 5 \alpha$ is irrational as the sum of a rational and an irrational number; on the other hand, $A = 2 \cos 2 \alpha \cos 3 \alpha$ is rational as the product of three rational numbers. Contradiction.
Remark. If we remove $\... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 12,517 |
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