problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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4. Given a cube. $A, B$ and $C$ are the midpoints of its edges (see figure). What is the angle $ABC$?
Answer. $120^{\circ}$. | Solution. 1st method. Draw diagonals $D E \| B C$ and $E F \| A B$ and let $K$ be the point on the extension of diagonal $D E$ beyond point $E$ (see figure). Then $\angle A B C = \angle F E K$. But triangle $D E F$ is equilateral, so
^{2}$ is also equal to $4^{n}-15$; therefore, we will assume that $4^{n}-15=x^{2}$, where $x$ is a natural number. From the equation $2^{2 n}-15=x^{2}$, we get: $2^{2 n}-x^{2}=15$, and usi... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.1
## Condition:
Grandma is embroidering her grandchildren's names on their towels. She embroidered the name "ANNA" in 20 minutes, and the name "LINA" in 16 minutes. She spends the same amount of time on the same letters, and possibly different times on different letters. How long will it take her to embroi... | Express the answer in minutes.
Answer: 12
Exact match of the answer - 1 point
Solution.
Since the grandmother spends the same amount of time on identical letters, she will spend $20: 2=10$ minutes on the syllable "NA". Then she will spend
$16-10=6$ minutes on the syllable "LI". Therefore, she will spend $6+6=12$ m... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.3
## Condition:
Petya is teaching his cat to read and for this, he makes him cards with words, burning them. He burned the word "MAMA" in 20 minutes, and the word "MISKA" in 35 minutes. He spends the same amount of time on identical letters, and possibly different times on different letters. How long will ... | Express the answer in minutes.
Answer: 50
Exact match of the answer - 1 point
Solution by analogy with problem № 3.1.
## Condition:
Grandma is embroidering her grandchildren's names on their towels. She embroidered the name "LANA" in 15 minutes, and the name "ALLA" in 12 minutes. She spends the same amount of time... | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 4.1
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a five-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars above me than knights below me!"
How many liars can live in this building? | # Answer: 3
## Exact match of the answer -1 point
## Solution.
Notice that the person living on the 5th floor is definitely lying, as there is no one living above them, including liars. Therefore, there are 0 liars above them, and 0 or more knights below. Now consider the resident on the 1st floor. They definitely t... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 4.3
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a six-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars above me than knights below me!" How many liars can live in this building? | Answer: 3
Exact match of the answer - 1 point
Solution by analogy with task No. 4.1.
## Condition:
On an island of knights, who always tell the truth, and liars, who always lie, a five-story building was built. One resident lived on each floor. One day, each of them made the same statement:
"There are more liars b... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.1
## Condition:
In a correct equation, identical digits were replaced with the same letters, and different digits with different letters. The result is
$$
P+\mathrm{P}+\mathrm{A}+3+Д+\mathrm{H}+\mathrm{U}+\mathrm{K}=\mathrm{U} \mathrm{U}
$$
What can U be equal to? | # Answer:
$\circ 1$
$\circ 2$
$\checkmark 3$
$\circ 4$
० 5
० 6
○ 7
○ 8
$\circ 9$
$\circ 0$
Exact match of the answer - 1 point
## Solution.
Notice that the expression involves 9 different letters, meaning all digits except one are used. On the left, we have the sum of eight different single-digit numbers. ... | 3 | Logic and Puzzles | 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. 7.1
## Condition:
On the Misty Planet, santiks, kubriks, and tugriks are in circulation. One santik can be exchanged for 1 kubrik or 1 tugrik, 1 kubrik can be exchanged for 3 santiks, and 1 tugrik can be exchanged for 4 santiks. No other exchanges are allowed. Jolly U, initially having 1 santik, made 20 ex... | # Answer: 6
## Exact match of the answer -1 point
## Solution.
To increase the number of santiks, we need to exchange them for kubriks or tugriks. In essence, we need to perform a double exchange, that is, first convert a santik into a tugrik or kubrik at a 1:1 ratio, and then increase the number of coins. Since the... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 7.2
Condition:
In the city of Abracodabra, funtics, tubrics, and santics are in circulation. One funtic can be exchanged for 1 tubric or 1 santic, 1 tubric for 5 funtics, and 1 santic for 2 funtics. No other exchanges are allowed. Lunatic, initially having 1 funtic, made 24 exchanges and now has 40 funtics... | Answer: 9
Exact match of the answer -1 point
Solution by analogy with task No. 7.1.
## Condition:
On the planet Mon Calamari, dataries, flans, and pegats are in circulation. One datary can be exchanged for 1 flan or 1 pegat, 1 flan - for 2 dataries, and 1 pegat - for 4 dataries. No other exchanges are allowed. Mera... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. $\left(7\right.$ points) Calculate $\frac{(2009 \cdot 2029+100) \cdot(1999 \cdot 2039+400)}{2019^{4}}$. | # Solution.
$2009 \cdot 2029+100=(2019-10) \cdot(2019+10)+100=2019^{2}-10^{2}+100=2019^{2}$.
$1999 \cdot 2039+400=(2019-20) \cdot(2019+20)+400=2019^{2}-20^{2}+400=2019^{2}$.
Then $\frac{2019^{2} \cdot 2019^{2}}{2019^{4}}=1$
Answer. 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (7 points) Find all natural solutions to the equation $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}$.
# | # Solution.
1 method. $2 n-\frac{1}{n^{5}}=3-\frac{2}{n}, 2 n-3=\frac{1}{n^{5}}-\frac{2}{n}, 2 n-3=\frac{1-2 n^{4}}{n^{5}}$.
For $n=1$ the equality is true, for $n>1 \quad 2 n-3>0, \frac{1-2 n^{4}}{n^{5}}<0$.
Answer. $n=1$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (7 points) A traveler was riding in a bus and saw a two-digit number on a kilometer post. He fell asleep and woke up an hour later to see a three-digit number on the kilometer post, where the first digit was the same as the second digit an hour ago, the second digit was zero, and the third digit was the same as the ... | Solution. Let the first number the traveler saw be $\overline{x y}=10 x+y$. After an hour, the number became $\overline{y o x}=100 y+x$. After 2 hours, it became $\overline{y z x}=100 x+10 z+x$. Since the bus's speed is constant, $\overline{y o x}-\overline{x y}=\frac{1}{2} \cdot(\overline{y z x}-\overline{y o x})$, th... | 45 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. For natural numbers $a$ and $b$, it is known that $5a-1$ is divisible by $b$, $a-10$ is divisible by $b$, but $3a+5$ is not divisible by $b$. What values can the number $b$ take? | Answer: 49.
Solution. The number $(5 a-1)-5 \cdot(a-10)=49$ is divisible by $b$, so either $b=1$, or $b=7$, or $b=49$. If the number $(3 a+5)-3 \cdot(a-10)=35$ were divisible by $b$, then $3 a+5$ would be divisible by $b$, which is not the case. Therefore, the options $b=1$ and $b=7$ are impossible. The numbers $a=10$... | 49 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. There is an infinite grid plane, none of the points of which are painted blue. For one ruble, you can choose a cell and paint all its sides blue, even if some side was already painted. What is the minimum amount of money you need to pay to get a grid square $1001 \times 1001$, all grid lines inside and on the bounda... | Answer: 503000 rubles.
Solution: We will call a segment in the solution any segment connecting grid nodes at a distance equal to the side of a cell. Let $s$ be our $1001 \times 1001$ square. Any cell either has two common segments with the perimeter of $s$ (in which case it falls into a corner of $s$), or has no more ... | 503000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. The numbers from 1 to 8 are arranged in a circle. A number is called large if it is greater than its neighbors, and small if it is less than its neighbors. Each number in the arrangement is either large or small. What is the greatest possible sum of the small numbers? | Answer: 13.
Instructions. Adjacent numbers cannot be of the same type, so large and small numbers alternate, and there are four of each. 8 is large. 7 is also large, since a small number must be less than two numbers, and seven is less than only one. 1 and 2 are small. 6 and 5 cannot both be small, as they are less th... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9.2 In Pokémon hunting, 11 adults and $n$ children participated. Together, they caught $n^{2}+3 n-2$ Pokémon, with all adults catching the same number, and all children catching the same number, but each child catching 6 fewer than an adult. How many children participated in the game? | Solution: Let each child catch $m$ pokemons. Then $n m+11(m+6)=n^{2}+3 n-2$. From this, $(n+11) m=n^{2}+3 n-68$. Therefore, the right side is divisible by $n+11$. We have $n^{2}+3 n-68=$ $n(n+11)-8(n+11)+20$, so 20 is divisible by $n+11$. The only divisor of 20 greater than 10 is 20 itself, so $n+11=20, n=9$.
## Crite... | 9 | Algebra | 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 25 can be divided so that the sum of the numbers in each group is a perfect square? | Solution 1: A group consisting of a single number can only be formed by 5 squares. The remaining 20 numbers must be divided into groups of at least two. Therefore, there will be no more than 15 groups in total. Let's check that exactly 15 groups are not possible. Indeed, in such a case, the numbers 1, 4, 9, 16, 25 form... | 14 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5.1. The numbers $415, 43, 7, 8, 74, 3$ are written on cards (see figure). Arrange the cards in a row so that the resulting ten-digit number is the smallest possible. | Answer: $\square 3415 \square 43 \square 74 \square 7 \square 8$
The correct answer can simply be recorded as the number 3415437478.
+ correct answer provided
- incorrect answer or no answer provided | 3415437478 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5.3. One side of a rectangle was increased by 3 times, and the other was reduced by 2 times, resulting in a square. What is the side of the square if the area of the rectangle is $54 \mathrm{m}^{2} ?$ | Answer: 9 m.
Let's reduce the side of the given rectangle by half (see Fig. 5.3a). Then the area of the resulting rectangle will be 27 m $^{2}$ (see Fig. 5.3b). Next, we will increase the other side by three times, that is, "add" two more rectangles (see Fig. 5.3c). The area of the resulting figure will become $27 \cd... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5.4. The number 61 is written on the board. Every minute, the number is erased from the board and replaced with the product of its digits, increased by 13. That is, after one minute, the number on the board will be $19 (6 \cdot 1 + 13 = 19)$. What number will be on the board after an hour? | Answer: 16.
Let's consider the numbers that will be written on the board over the first few minutes:
| After one minute | $6 \cdot 1+13=\mathbf{1 9}$ |
| :---: | :---: |
| After two minutes | $1 \cdot 9+13=\mathbf{2 2}$ |
| After three minutes | $2 \cdot 2+13=\mathbf{1 7}$ |
| After four minutes | $1 \cdot 7+13=\math... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# 6. CONDITION
At the factory, there are exactly 217 women, among whom 17 are brunettes, and the remaining 200 are blondes. Before New Year's, all of them dyed their hair, and each of these women wrote on "VKontakte" the surnames of exactly 200 women from the factory, whom they believed to be definitely blondes. Each ... | Solution. According to the problem, the correct list of all 200 blondes will be on "Vkontakte" exactly for 17 female workers at the plant: brunettes will write exactly this list, and a blonde will never write it, as otherwise she would have to include herself in it. Therefore, if a certain list appears not 17 times, bu... | 13 | Combinatorics | proof | Yes | Yes | olympiads | false |
7.4 Vanya wrote the numbers $1,2,3, \ldots, 13$ in his notebook. He multiplied five of them by 3, and the rest by 7, then added all the products. Could the result have been 433? | Solution. Let's try to figure out the situation. Let A be the sum of the five numbers that need to be multiplied by 3, and B be the sum of the other eight numbers (which need to be multiplied by 7). Consider the sum after multiplication: 3A + 7B. We can write it as (taking into account that the sum of all thirteen numb... | 433 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7.5 In triangle ABC, the median BM is drawn. It is known that $\angle \mathrm{BAC}=30^{\circ}, \angle \mathrm{BMC}=45^{\circ}$. Find angle BAC. | Solution. See fig.
C
$\triangle \mathrm{AMB}: \angle \mathrm{AMB}=135^{\circ} \Rightarrow \angle \mathrm{ABM}=15^{\circ}$
Let $\mathrm{AC}=2 \mathrm{~b}$. Then $\mathrm{AM}=\mathrm{MC}=\mat... | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem №3
The base and the lateral side of an isosceles triangle are 34 and 49, respectively.
a) Prove that the midline of the triangle, parallel to the base, intersects the inscribed circle of the triangle.
b) Find the length of the segment of this midline that is contained within the circle. | # Answer: 8.
## Solution
a) Let $\mathrm{O}$ be the center of the inscribed circle in triangle $ABC$ with sides $AB = AC = 49$, $BC = 34$, and $AH$ be the height of the triangle. Points $M$ and $N$ are the midpoints of sides $AB$ and $AC$, respectively, and $K$ is the intersection point of $AH$ and $MN$. Since $MN$ i... | 8 | Geometry | proof | Yes | Yes | olympiads | false |
1. An army of mice attacked a grain warehouse. To fight the mice, one cat was released into the warehouse on the first day, a second cat on the next day, and each subsequent day, one more cat was released. A mouse eats 100 grams of grain per day, and a cat eats 10 mice per day. After several days, there were no mice le... | Solution. Let the entire process last $n$ days, then by its end, there were $n$ cats on the warehouse, with the first cat eating $10 n$ mice, the second cat eating 10( $n-1)$ mice, $\ldots, n$-th cat - 10 mice, meaning there were a total of $10(1+2+\ldots+n)$ mice. Let's calculate how much grain the mice ate.
Ten mice... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task № 3.4
## Condition:
Cupcakes are sold in boxes of 2, eclairs - in boxes of 6, and gingerbread - in boxes of 15. You can only buy whole boxes, you cannot open them. Alice bought an equal number of cupcakes, eclairs, and gingerbread. What is the smallest number of boxes she could have taken | Answer: 22
Exact match of the answer - 1 point
Solution by analogy with task №3.1
# | 22 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 5.1
## Condition:
A Dog, a Cat, and a Mouse are running around a circular lake. They all started simultaneously in the same direction from the same point and finished at the same time, each running at a constant speed. The Dog ran 12 laps, the Cat ran 6 laps, and the Mouse ran 4 laps. How many total overta... | # Answer: 13
## Exact match of the answer -1 point
## Solution.
At the moment when the faster runner catches up with the slower one, he is ahead by one lap. The Dog has lapped the Cat by 6 laps, meaning it has caught up with her 6 times, with the last catch-up at the finish line not counting as a lap, so the Dog has... | 13 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task No. 5.3
## Condition:
A Dog, a Cat, and a Mouse are running around a circular lake. They all started in the same direction from the same point and finished at the same time, each running at a constant speed.
The Dog ran 12 laps, the Cat ran 7 laps, and the Mouse ran 3 laps. How many total overtakes were made ... | # Answer: 15
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 15 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 5.4
## Condition:
A Dog, a Cat, and a Mouse are running around a circular lake. They started simultaneously in the same direction from the same point and finished simultaneously, all running at constant speeds.
The Dog ran 12 laps, the Cat ran 5 laps, and the Mouse ran 2 laps. How many total overtakes were ... | Answer: 17
Exact match of the answer -1 point
Solution by analogy with task №5.1
# | 17 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
# Task № 6.1
## Condition:
Anya, Tanya, and Vanya had identical cardboard squares with a side of 16 cm. Each of them cut off two rectangles from their square, as shown in the figure, and all 6 rectangles are the same. The perimeter of Anya's figure is 88 cm, and the perimeter of Vanya's figure is 82 cm.
(b+d)
$$
Since Katya increased each number by 1, her sum is
$$
(a+c+2)(b+d+2)=(a+c)(b+d)+2(a+b+c+d)+4
$$
greater than Masha's by $2(... | 4046 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10.1. Every second, the computer outputs a number equal to the sum of the digits of the previous number, multiplied by 31. On the first second, the number 2020 was displayed. What number will be displayed on the 2020th second | Answer: 310.
Solution: Let's calculate the first few numbers displayed on the screen.
On the first second, the number $a_{1}=2020$ is displayed, then
on the second second, the number $a_{2}=(2+0+2+0) \cdot 31=4 \cdot 31=124$ is displayed;
on the third second, the number $a_{3}=(1+2+4) \cdot 31=7 \cdot 31=217$;
on ... | 310 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11.1. Two runners, starting simultaneously at constant speeds, run on a circular track in opposite directions. One of them runs the loop in 5 minutes, while the other takes 8 minutes. Find the number of different meeting points of the runners on the track, if they ran for at least an hour.
# | # Solution.
Let the length of the track be $\mathrm{S}$ meters. Then the speeds of the runners are $\mathrm{S} / 5$ and $\mathrm{S} / 8$ meters per minute, respectively.
Assume the runners start from the same point. They will meet again after $\mathrm{S} /(\mathrm{S} / 5+\mathrm{S} / 8)=40 / 13$ minutes.
Now, let's ... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
11.2. How much greater is one of two positive numbers than the other if their arithmetic mean is $2 \sqrt{3}$ and the geometric mean is $\sqrt{3}$?
(Hint: the geometric mean of two numbers $m$ and $n$ is the number $p=\sqrt{\mathrm{mn}}$). | # Solution.
Let the unknown numbers be denoted by $x$ and $y$. Then, from the problem statement, we get:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ \frac { x + y } { 2 } = 2 \sqrt { 3 } , } \\
{ \sqrt { x y } = \sqrt { 3 } , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3 .
\end... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the record of three two-digit numbers, there are no zeros, and in each of them, both digits are different. Their sum is 40. What could be their sum if the digits in them are swapped? | Answer: 103.
Solution. In all numbers, the tens digit is 1. Otherwise, the larger number is at least 21, and the other two are at least 12. Their sum is no less than $12+12+21=44$, which is not equal to 40. The sum of the units digits is 10 (zeros are not allowed). Therefore, the sum of the numbers with swapped digits... | 103 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. Find the largest natural number $\mathrm{n}$, for which the system of inequalities
$1<x<2, 2<x^{2}<3, \ldots, n<x^{n}<n+1$
has a solution. (6 points) | Solution. From the condition $: n=4$.
Rewrite the inequalities as:
$\left\{\begin{aligned} & 16^{3} \text{, then for } n=5 \text{ the given system is already inconsistent: the intervals } [\sqrt[3]{3}, \sqrt[4]{4}] \text{ and } [\sqrt[5]{2}, \sqrt[5]{6}] \text{ do not intersect.} \\ & \text{For } n=4 \text{, it is not... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. There is a bag with 16 letters: А, А, А, А, В, В, Д, И, И, М, М, Н, Н, Н, Я, Я. Anna, Vanya, Danya, and Dima each took 4 letters from it, after which the bag was empty. How many of them could have successfully spelled out their names? Explain your answer. | 2. Answer: 3.
Solution. All of them could not have formed their names, as there were not enough letters D. Anna, Vanya, and Dima could have drawn cards from which they could form their names, while Danya would have been left with the cards: V, I, M, Ya. | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. Alexei drew 7 lines on a plane, which divided it into several parts. Then he chose two adjacent parts (adjacent parts are those that share a common side), counted how many sides each of them contains, and added these two numbers. What is the largest number he could have obtained? Explain your answer.
. Return the line we removed. It will be a side for both small parts (+... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11.3. The numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ form a geometric progression. Among them, there are both rational and irrational numbers. What is the maximum number of terms in this progression that can be rational numbers? | Answer: 3.
Example: let $a_{1}=1, q=\sqrt{2}$, we get the geometric progression $1, \sqrt{2}, 2, 2\sqrt{2}, 4$. Evaluation. If there are 4 rational numbers among them, then there will be two consecutive rational members of the geometric progression. This means that the common ratio of the progression (the ratio of the... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11.5. A cube of size $n \times n \times n$, where $n$ is a natural number, was cut into 99 smaller cubes, of which exactly one has an edge length different from 1 (each of the others has an edge length of 1). Find the volume of the original cube. | Answer: 125.
Let $m$ be the length of the edge of the cube, different from the unit cube. We get the equation $n^{3}-m^{3}=98$ (in natural numbers). Further, $(n-m)\left(n^{2}+n m+m^{2}\right)=98$. The numbers $m$ and $n$ are of the same parity, otherwise $n^{3}-m^{3}$ would be an odd number. Moreover, if $n$ and $m$ ... | 125 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.1. In a box, there are oranges, pears, and apples, a total of 60 fruits. It is known that there are 3 times more apples than non-apples, and there are 5 times fewer pears than non-pears. How many oranges are in the box? | Answer: 5.
Solution. Since there are 3 times more apples than non-apples, apples make up $\frac{3}{4}$ of the total number of fruits, i.e., there are $\frac{3}{4} \cdot 60=45$ apples. Since there are 5 times fewer pears than non-pears, pears make up $\frac{1}{6}$ of the total number of fruits, i.e., there are $\frac{1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. Oleg bought a chocolate bar for $n$ rubles, and after some time, he sold it for 96 rubles. It turned out that he sold the chocolate bar for exactly $n \%$ more than he bought it for. For how many rubles did Oleg buy the chocolate bar? | Answer: 60.
Solution. From the condition of the problem, it follows that $96=n \cdot\left(1+\frac{n}{100}\right)$. Transforming this equation, we get
$$
0=n^{2}+100 n-9600 \quad \Leftrightarrow \quad 0=(n+160)(n-60)
$$
Thus, $n=60$, since the chocolate bar cannot cost a negative number of rubles. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. Masha has three identical dice, on the faces of each of which six different prime numbers are written, with a total sum of 87.
Masha rolled all three dice twice. The first time, the sum of the numbers that came up was 10, and the second time, the sum was 62.
Exactly one of the six numbers did not come up... | # Answer: 17.
Solution. Note that the number 10 can be uniquely represented as the sum of three prime numbers: $10=2+3+5$. This means that the numbers $2,3,5$ are on the dice and they came up the first time.
Note that if the even number 62 can be represented as the sum of three prime numbers, then one of them is even... | 17 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. In the cells of a $2 \times 35$ table (2 rows, 35 columns), non-zero real numbers are placed, and all numbers in the top row are distinct. For any two numbers in the same column, the following condition is satisfied: one number is the square of the other.
(a) (1 point) What is the maximum number of negati... | Answer: (a) 35. (b) 12.
Solution. (a) In any column, there can be no more than one negative number, so the total number of negative numbers is no more than 35. There can be exactly 35 if, for example, the top numbers are $-1, -2, -3, \ldots, -35$, and below them are the numbers $1^{2}, 2^{2}, 3^{2}, \ldots, 35^{2}$ re... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. In a convex quadrilateral $A B C D$, the bisectors of angles $A$ and $C$ are parallel, and the bisectors of angles $B$ and $D$ intersect at an angle of $46^{\circ}$, as shown in the figure. How many degrees does the acute angle between the bisectors of angles $A$ and $B$ measure?
. Additionally, let $\angle A=2 \alpha, \angle B=2 \beta, \angle C=2 \gamma, \angle D=2 \delta$. Since the sum of the angles of quadrilateral
 (2 points) $n=27$
(b) (2 points) $n=26$ ? | Answer: (a) 14. (b) 105.
Solution. Let's number the zeros from left to right with numbers from 1 to 13. Denote the number of crosses before the first zero by $a_{1}$, between the first and second zero by $a_{2}, \ldots$, after the thirteenth zero by $a_{14}$. We get that $a_{1}+a_{2}+\ldots+a_{14}=n$. The condition th... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8. Given a triangle $A B C$, in which $\angle A=42^{\circ}$ and $A B<A C$. Point $K$ on side $A C$ is such that $A B=C K$. Points $P$ and $Q$ are the midpoints of segments $A K$ and $B C$ respectively. How many degrees does angle $A C B$ measure, if it is known that $\angle P Q C=110^{\circ} ?$
Fig. 4: to the solution of problem 8.8
Solution. Mark a point $L$ on the extension of side $C A$ beyond point $A$ such that $A L = A B$ (Fig. 4).
Notice that line $P Q$ is para... | 49 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.2. What is the greatest number of consecutive natural numbers, each of which has exactly four natural divisors (including 1 and the number itself)? | Answer: three numbers.
Solution. Suppose there are four consecutive numbers that satisfy the condition. Note that among four consecutive numbers, one is divisible by 4. Then, in the prime factorization of this number, there are at least two twos. If there is another prime divisor $p$ different from two, then the numbe... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Does there exist a natural number such that after increasing it by 18, the sum of its digits decreases by 18? | Answer: can
Solution. For example, the number 982 works. $982+18=1000$.
| | |
| :--- | :--- |
| | |
| | |
| ${ }_{15-x}$ | $\times$ |
## Grading Criteria.
Only answer: 0 points.
Correct example provided: 7 points. | 982 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. On the Island of Logic, there live 40 knights (who always tell the truth), 25 liars (who always lie), and several sophists. A sophist can only utter phrases that a knight or a liar could not say in their place. For example, standing next to a liar, a sophist can say, "We are both liars" (because if he were a knight,... | Answer. 1) There are exactly 26 liars on the island; 2) 27 sophists.
Solution.
1) If a sophist were a liar, he would have to tell the truth, and then he would say that there are $25+1=26$ liars. If he were a knight, there would be 25 liars, and he would lie.
2) Let C be the total number of sophists on the island. If ... | 26 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. A sequence of 2016 numbers is written. Each one, except the first and the last, is equal to the sum of its neighbors. Find the sum of all 2016 numbers. | Answer: the sum of all numbers is zero.
Solution. Let the first number be $a$, and the second $b$. Denote the third as $x$. Then $a+x=b$ and, therefore, $x=b-a$. Now, express the fourth in terms of $a$ and $b$: $b+y=b-a$ and, therefore, $y=-a$. Continuing, we get: $a, b, b-a, -a, -b, a-b, a, b, \ldots$ This means the ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. On a $5 \times 5$ board, a cross consisting of five cells (a cell and all its neighbors) is located. What is the minimum number of detectors needed to place on the board cells to accurately determine the position of the cross? (A detector indicates whether a cell belongs to the cross or not, and the detectors trigge... | Answer: 4.
Solution. There are 9 possible positions for the cross on the board, which is the same as the number of positions for the central cell of the cross. A detector has two states, so the total number of possible states for three detectors is $2^{3}=8$, and thus, they cannot distinguish between 9 positions of th... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3.6. Petya can draw only 4 things: the sun, a ball, a tomato, and a banana. But he does it extremely realistically! Today he drew several things, among which there were exactly 15 yellow, 18 round, and 13 edible. What is the maximum number of balls he could have drawn?
Petya believes that all tomatoes are roun... | Answer: 18.
Solution. Since there are a total of 18 round objects, and all balls are round, there were no more than 18 balls. Note that there could have been exactly 18, if 2 yellow balls, 16 green balls, and 13 bananas were drawn. | 18 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 3.7. Katya passes the time while her parents are at work. On a piece of paper, she absentmindedly drew Cheburashkas in two rows (at least one Cheburashka was drawn in each row).
Then, after some thought, between every two adjacent Cheburashkas in a row, she drew a Gena the Crocodile. And then to the left of ea... | Answer: 11.
Solution. The Krokodilofes are drawn exactly in the gaps between the other characters. In each of the two rows of gaps, there is 1 less gap than there are characters, so
There are 2 fewer Krokodilofes than all the other characters. Therefore, Cheburashkas, Crocodile Gens, and Old Ladies Shapoklyak total $... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 3.8. A small raft was swaying by the riverbank. Five mice weighing 70 g each, three moles weighing 90 g each, and four hamsters weighing 120 g each approached the shore. What is the minimum number of grams the raft must be able to support so that all the animals can cross to the other side, possibly in several ... | Answer: 140.
Solution. If the raft can carry less than two animals, then they will not all be able to cross. Indeed, in this case, one animal would have to row back and forth on the raft, without the possibility of staying on the other side of the river (since the raft cannot return without a rower).
Therefore, the r... | 140 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 4.1. Assign each letter a digit $1,2,3,4,5$ so that all inequalities are satisfied.
$$
3>\mathrm{A}>\mathrm{M}<\mathrm{E}<\mathrm{H}<\mathrm{A}
$$
Different letters must correspond to different digits. Write the number ZAMENA as your answer. | Answer: 541234.
Solution. From the condition, it follows that $\mathrm{M}<\mathrm{E}<\mathrm{H}<\mathrm{A}<3$. The numbers from 1 to 5 are uniquely ordered, so $\mathrm{M}=1, \mathrm{E}=2, \mathrm{H}=3, \mathrm{~A}=4,3=5$. Then ZAMENA $=541234$. | 541234 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.3. How many rectangles exist on this picture with sides running along the grid lines? (A square is also a rectangle.)
 | Answer: 24.
Solution. In a horizontal strip $1 \times 5$, there are 1 five-cell, 2 four-cell, 3 three-cell, 4 two-cell, and 5 one-cell rectangles. In total, $1+2+3+4+5=15$ rectangles.
In a vertical strip $1 \times 4$, there are 1 four-cell, 2 three-cell, 3 two-cell, and 4 one-cell rectangles. In total, $1+2+3+4=10$ r... | 24 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Task 4.5. Write the largest eight-digit number in which all even digits are present. (Even digits: $0,2,4,6,8$.) | Answer: 99986420.
Solution. The number 99986420 meets the condition of the problem. Suppose there exists a larger eight-digit number. It is clear that its first three digits must be nines. Since there are only 5 even digits, the last five digits are $0,2,4,6,8$ in some order. But then such an eight-digit number cannot... | 99986420 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.6. Some digits in the rectangle are already placed. Place the remaining digits so that:
- the sum of the digits in each column is the same;
- the sum of the digits in each row is the same;
- the sum of the digits in the red cells is equal to the sum of the digits in any row. Enter the three-digit number $A B... | Answer: 652.
Solution. If we add 3 to the sum of the two upper red numbers, we get the same number as if we added the sum of the two lower red numbers. Therefore, the unknown lower red number is 2. Then, in each row, the sum is $1+2+6=$ 9. This means that the unknown upper red number complements $4+3$ to 9, i.e., it i... | 652 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 4.8. If in the number 79777 the digit 9 is crossed out, the number 7777 is obtained. How many different five-digit numbers exist from which 7777 can be obtained by crossing out one digit? | Answer: 45.
Solution. Note that 77777 is one of such numbers. Next, we will consider five-digit numbers where to get 7777, we need to strike out a digit that is not 7.
If the first digit in the number is struck out, there are 8 options for it: $1,2,3,4,5,6,8,9$. If, however, the second, third, fourth, or fifth digit ... | 45 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.1. Sasha wrote down on the board all two-digit numbers divisible by 6, and then erased those that do not end in 4. What is the largest number that ended up on the board? | Answer: 84.
Solution. Let's take the largest two-digit number divisible by 6, which is 96, and sequentially subtract 6 from it. This way, we will go through all numbers divisible by 6 in descending order. We will stop at the first one that ends in 4, which is what we need.
$$
96 \rightarrow 90 \rightarrow 84
$$ | 84 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.3. A cuckoo clock is hanging on the wall. When a new hour begins, the cuckoo says "cuckoo" a number of times equal to the number the hour hand points to (for example, at 19:00, "cuckoo" sounds 7 times). One morning, Maxim approached the clock when it was 9:05. He started turning the minute hand until he advan... | Answer: 43.
Solution. The cuckoo will say "cuckoo" from 9:05 to 16:05. At 10:00 it will say "cuckoo" 10 times, at 11:00 - 11 times, at 12:00 - 12 times. At 13:00 (when the hand points to the number 1) "cuckoo" will sound 1 time. Similarly, at 14:00 - 2 times, at 15:00 - 3 times, at 16:00 - 4 times. In total
$$
10+11+... | 43 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.4. At the end-of-the-year school dance, there were twice as many boys as girls. Masha counted that there were 8 fewer girls, besides herself, than boys. How many boys came to the dance? | Answer: 14.
Solution. Let there be $d$ girls at the disco, then there are twice as many boys, i.e., $d+d$. The number of girls, excluding Masha, is $d-1$ (all except herself).
Since there are 8 more boys than the other girls, excluding Masha, then $d+d=(d-1)+8$. From this, it follows that $d=7$, so the total number o... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 5.5. From a $6 \times 6$ grid square, gray triangles have been cut out. What is the area of the remaining figure? The side length of each cell is 1 cm. Give your answer in square centimeters.
. Moreover, after three years, each city is the end of exactly three flights.
In the "mountain" cities, flights have a total of $30 \cdot 3=90$ ends. Among these, 21 flights connect two "mountain" cities. Each such flight between "mountain" citi... | 81 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.1. Masha placed the numbers from 1 to 16 in the cells of a $4 \times 4$ table so that any two numbers differing by one were in adjacent cells.
And Sasha erased all the numbers except for 1, 4, 9, and 16. What number was in the cell with the question mark?
| | 1 | | $?$ |
| :---: | :---: | :---: | :---: |
... | Answer: 13.
Solution. Since any two numbers that differ by one are in adjacent cells, the number in the upper left corner must be either 2 (neighbor of 1) or a number that has only one neighbor - 1 or 16. The second option does not work for us, so we can write the numbers 2 and 3 in the table.
| 2 | 1 | | $?$ |
| :-... | 13 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.2. For preparing one portion of salad, 2 cucumbers, 2 tomatoes, 75 grams of feta cheese, and 1 pepper are required. The restaurant's warehouse has 60 peppers, 4.2 kg of feta cheese, 116 tomatoes, and 117 cucumbers. How many portions can be made? | Answer: 56.
Solution. The pepper will last for $60: 1=60$ servings.
The feta cheese will last for $4200: 75=56$ servings (kilograms converted to grams).
The tomatoes will last for $116: 2=58$ servings.
The cucumbers will last for $117: 2=58.5$, i.e., 58 whole servings.
Since all ingredients must be included in eac... | 56 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.3. Vitya and his mother left home at the same time and walked in opposite directions at the same speed: Vitya - to school, and his mother - to work. After 10 minutes, Vitya realized he didn't have the keys to the house, and he would return from school earlier than his mother, so he started to catch up with he... | Answer: 5 minutes.
Solution. Let Vitya and his mother initially walk at a speed of $s$ meters per minute. After 10 minutes, when Vitya realized he had forgotten his keys, the distance between him and his mother became $10 s + 10 s = 20 s$. When Vitya started to catch up with his mother, he walked at a speed of $5 s$ m... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.5. In the park, paths are laid out as shown in the figure. Two workers started to asphalt them, starting simultaneously from point $A$. They lay asphalt at constant speeds: the first on the section $A-B-C$, the second on the section $A-D-E-F-C$. In the end, they finished the work simultaneously, spending 9 ho... | Answer: 45.
Solution. Let the line $A D$ intersect the line $C F$ at point $G$, as shown in the figure below. Since $A B C G$ and $D E F G$ are rectangles, we have $A B=C G, B C=A G, E F=D G$ and $D E=F G$.
 were picked from a tree. When they were weighed, it turned out that the weight of any three oranges taken together is less than $5 \%$ of the total weight of the remaining oranges. What is the smallest number of oranges that could have been picked? | Answer: 64.
Solution. 64 oranges could have been, for example, if all of them had the same mass $m$, since $\frac{3 m}{61 m}<0.05$.
Suppose there were no more than 63 oranges. Consider the three heaviest oranges with a total mass of $M$. All the other oranges (no more than 60) can be divided into no more than 20 grou... | 64 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Problem 6.8. There are exactly 120 ways to color five cells in a $5 \times 5$ table so that each column and each row contains exactly one colored cell.
There are exactly 96 ways to color five cells in a $5 \times 5$ table without a corner cell so that each column and each row contains exactly one colored cell.
How ma... | Answer: 78.
Solution. Consider the colorings of a $5 \times 5$ table described in the problem (i.e., such that in each column and each row exactly one cell is colored).
For convenience, let's introduce some notations. The top-left corner cell of the $5 \times 5$ table will be called $A$, and the bottom-right corner c... | 78 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.2. Three pirates were dividing a treasure. The first one got a third of the initial number of coins and one more coin, the second one got a quarter of the initial number of coins and five more coins, the third one got a fifth of the initial number of coins and twenty more coins (thus all the coins were distri... | Answer: 120.
Solution. Let $m$ be the initial total number of coins in the treasure. The pirates collectively took $\left(\frac{m}{3}+1\right)+\left(\frac{m}{4}+5\right)+\left(\frac{m}{5}+20\right)$ coins. Since they divided the entire treasure, this sum equals $m$. Expanding the brackets in the equation
$$
\left(\fr... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 7.5. If a platoon of soldiers is divided into brigades of 7 people, then 2 people will not fit into any brigade. If the platoon is divided into brigades of 12 people, then again 2 people will not fit into any brigade. What is the minimum number of soldiers that need to be added to the platoon so that it can be ... | Answer: 82.
Solution. A platoon can be entirely divided into brigades of 7 and brigades of 12 if and only if the number of people in it is divisible by $7 \cdot 12=84$.
Remove two people from the platoon. Then the remaining people can be divided into brigades both ways, meaning their number is divisible by 84. Return... | 82 | 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 |
Problem 8.1. In a $5 \times 5$ square, some cells have been painted black as shown in the figure. Consider all possible squares whose sides lie along the grid lines. In how many of them is the number of black and white cells the same?
. There are only two non-fitting $2 \times 2$ squares (both of which contain th... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.2. The arithmetic mean of three two-digit natural numbers $x, y, z$ is 60. What is the maximum value that the expression $\frac{x+y}{z}$ can take? | Answer: 17.
Solution. From the condition, it follows that the sum of the numbers $x, y, z$ is $60 \cdot 3=180$. Then
$$
\frac{x+y}{z}=\frac{180-z}{z}=\frac{180}{z}-1 \leqslant \frac{180}{10}-1=17
$$
since $z \geqslant 10$. Note also that when $x=90, y=80, z=10$, the value 17 is achieved. | 17 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.3. In triangle $ABC$, the sides $AC=14$ and $AB=6$ are known. A circle with center $O$, constructed on side $AC$ as the diameter, intersects side $BC$ at point $K$. It turns out that $\angle BAK = \angle ACB$. Find the area of triangle $BOC$.
Then
$$
\angle BAC = \angle BAK + \angle CAK = \angle BCA ... | 21 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.4. Find any solution to the puzzle
$$
\overline{A B C A}=182 \cdot \overline{C D}
$$
where $A, B, C, D$ are four different non-zero digits (the notation $\overline{X Y \ldots Z}$ means the decimal representation of the number).
As an answer, write the four-digit number $\overline{A B C D}$. | Answer: 2916.
Solution. $\overline{A B C A}=1001 A+10 \cdot \overline{B C}$. Note that 1001 and 182 are divisible by 91, therefore, $10 \cdot \overline{B C}$ is also divisible by 91, i.e., $\overline{B C}=91$.
Substituting $B=9, C=1$ and dividing by 91, we get the equation $11 A=10+2 D$. On the left side of this equa... | 2916 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.5. In a race, several people participated, including Andrey, Dima, and Lёnya. No two participants of this race arrived at the same time.
- The number of people who arrived before Andrey is 2 times less than the number of people who arrived after him.
- The number of people who arrived before Dima is 3 times ... | Answer: 61.
Solution. Let $x$ be the number of people who ran before Andrey, then $2x$ people ran after him. We get that there were $3x+1$ participants in the race.
Let $y$ be the number of people who ran before Dima, then there were $4y+1$ participants in total.
Finally, let $z$ be the number of people who ran befo... | 61 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.6. A natural number is called interesting if all its digits are different, and the sum of any two adjacent digits is a square of a natural number. Find the largest interesting number. | Answer: 6310972.
Solution. Mark 10 points on the plane, representing the digits from 0 to 9, and connect those that sum to the square of a natural number.
On such a diagram, we need to fin... | 6310972 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.7. Given an isosceles triangle $A B C$, where $A B=A C$ and $\angle A B C=53^{\circ}$. Point $K$ is such that $C$ is the midpoint of segment $A K$. Point $M$ is chosen such that:
- $B$ and $M$ are on the same side of line $A C$;
- $K M=A B$
- angle $M A K$ is the maximum possible.
How many degrees does angl... | Answer: 44.
Solution. Let the length of segment $AB$ be $R$. Draw a circle with center $K$ and radius $R$ (on which point $M$ lies), as well as the tangent $AP$ to it such that the point of tangency $P$ lies on the same side of $AC$ as $B$. Since $M$ lies inside the angle $PAK$ or on its boundary, the angle $MAK$ does... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Problem 8.8. A computer can apply three operations to a number: "increase by 2", "increase by 3", "multiply by 2". The computer was given the number 1 and was made to try all possible combinations of 6 operations (each of these combinations is applied to the initial number 1). After how many of these combinations will ... | Answer: 486.
Solution. If the number was odd before the last operation, then one of the two operations "increase by 3" or "multiply by 2" will make it even, but the third will not. Similarly, if the number was even before the last operation, then one of the two operations "increase by 2" or "multiply by 2" will make i... | 486 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.1. In a $3 \times 3$ table, the digits from 1 to 9 are arranged (each digit is written in exactly one cell). The digits 1, 2, 3, 4 are placed as shown in the figure. It is also known that the sum of the digits in the cells adjacent to the digit 5 is 9. Find the sum of the digits in the cells adjacent to the d... | Answer: 29.
Solution. Note that the smallest possible sum of the digits around the digit 5 is exactly $1+2+6=9$, and this is only possible when it is below the digit 1 and above the digit 2, with the digit 6 in the center.
Indeed, if 5 is in the center, then the sum of the surrounding digits is $6+7+8+$ $9>9$. And if... | 29 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.2. In the first hour of the shift, the master made 35 parts. Then he realized that, maintaining the current speed, he would have to stay an extra hour to complete the shift's plan. By increasing his speed by 15 parts per hour, he completed the plan half an hour before the end of the shift. How many parts shou... | Answer: 210.
Solution. Let $N$ be the number of parts he still needs to manufacture. If he continued to make 35 parts per hour, he would finish in $\frac{N}{35}$ hours. When he sped up and started producing 50 parts per hour, he completed the work in $\frac{N}{50}$ hours.
From the condition, it follows that $\frac{N}... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.3. The farmer said: "I have $N$ rabbits. Long ears are exactly on 13 of them. And exactly 17 of them can jump far."
The traveler rightly noted: "Therefore, among your rabbits, there are at least 3 rabbits that simultaneously have long ears and can jump far."
What is the largest value that the number $N$ can... | Answer: 27.
Solution. If $N \geqslant 28$, then the traveler is wrong: it is possible that the farmer had only 2 long-eared rabbits that can jump far. In addition to them, there were 11 long-eared rabbits that cannot jump far, and 15 rabbits without long ears that can jump far. The remaining $N-28$ rabbits neither hav... | 27 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.4. At a round table, 40 knights and 10 samurais are sitting. Exactly 7 knights have a samurai as their right neighbor. What is the maximum number of knights that could be sitting next to two knights? | # Answer: 32.
Solution. Let's divide the people sitting at the table into alternating groups of consecutive knights and consecutive samurais. From the condition, it follows that there are exactly 14 such groups (7 groups of knights and 7 groups of samurais), since the 7 knights mentioned in the condition are precisely... | 32 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 9.6. The teacher wrote a number on the board. Sasha decided to divide it by 102 with a remainder, and Masha - by 103. It turned out that the quotient obtained by Sasha and the remainder obtained by Masha add up to 20. What remainder did Sasha get? List all possible options. | Answer: 20.
Solution. Let $n$ be the number on the board, divide it by 102 and by 103 with remainders:
$$
n=102 a+b ; \quad n=103 c+(20-a)
$$
Subtract the second equation from the first:
$$
0=103(a-c)+(b-20)
$$
From this, it follows that $b-20$ is divisible by 103. Since $0 \leqslant b \leqslant 101$, we get that ... | 20 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.1. An equilateral triangle with a side of 10 is divided into 100 small equilateral triangles with a side of 1. Find the number of rhombi consisting of 8 small triangles (such rhombi can be rotated).
It is clear that the number of rhombuses of each orientation will be the same, so let's consider ... | 84 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.2. How many real numbers $x$ exist such that the value of the expression $\sqrt{123-\sqrt{x}}$ is an integer? | Answer: 12.
Solution. From the condition, it follows that the value $s=123-\sqrt{x} \leqslant 123$ is a square of an integer. Since $11^{2}<123<12^{2}$, this value can take one of 12 values $0^{2}$, $1^{2}, 2^{2}, \ldots, 11^{2}$. And for each of these 12 values of $s$, there is a unique value of $x=(123-s)^{2}$ (obvi... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.3. At exactly noon, a truck left the village and headed for the city, at the same time, a car left the city and headed for the village. If the truck had left 45 minutes earlier, they would have met 18 kilometers closer to the city. And if the car had left 20 minutes earlier, they would have met $k$ kilometer... | Answer: 8.
Solution. We will express all distances in kilometers, time in hours, and speed in kilometers per hour. Let the distance between the village and the city be $S$, the speed of the truck be $x$, and the speed of the car be $y$. The time until the first meeting is $\frac{S}{x+y}$, so the distance from the vill... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.4. Consider the sequence
$$
a_{n}=\cos (\underbrace{100 \ldots 0^{\circ}}_{n-1})
$$
For example, $a_{1}=\cos 1^{\circ}, a_{6}=\cos 100000^{\circ}$.
How many of the numbers $a_{1}, a_{2}, \ldots, a_{100}$ are positive? | Answer: 99.
Solution. Note that for an integer $x$ divisible by 40, the cosines of the angles $x^{\circ}$ and $10 x^{\circ}$ coincide, since the difference between these angles $9 x^{\circ}$ is divisible by $360^{\circ}$. Therefore, $\cos \left(\left(10^{k}\right)^{\circ}\right)=$ $\cos \left(\left(10^{k+1}\right)^{\c... | 99 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.5. A circle $\omega$ is inscribed in trapezoid $A B C D$, and $L$ is the point of tangency of $\omega$ and side $C D$. It is known that $C L: L D=1: 4$. Find the area of trapezoid $A B C D$, if $B C=9$, $C D=30$.
 than X's (crosses), then in the entire table there are no fewer O's than X's. Similarly, considering the rows, it follows that in the entire table there are no fewer X's than O's. Therefore, there are an equal number of X's and O's in the entir... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Problem 10.7. Olya drew $N$ different lines on the plane, any two of which intersect. It turned out that among any 15 lines, there are always two lines such that the angle between them is $60^{\circ}$. What is the largest $N$ for which this is possible? | Answer: 42.
Solution. First, we will show that 42 such lines exist. Draw three lines so that the angle between any two of them is $60^{\circ}$, and take 14 such triplets at different angles:
... | 42 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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